Optoprep Flashcards

1
Q

Which area of the extrastriate cortex is involved in the perception of motion?

A. The inferotemporal cortex (IT)
B. Visual area 5 (V5)
C. Visual area 2 (V2)
D. Visual area 4 (V4)
E. Visual area 1 (V1)
A

B. Visual area 5 (V5)

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2
Q

What is the name for the phenomenon in which a flickering light that is 10 Hz is seen as brighter than a steady light (one that does not flicker) that possesses the same average luminance?

A. The Troxler effect
B. The Purkinje tree
C. The Granit-Harper law
D. The Brucke-Bartley effect

A

D. The Brucke-Bartley effect

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3
Q

Which of the following is the correct order of structures through which the pupillary fiber pathway passes?

A. Optic nerve -> optic chiasm -> brachium of the superior colliculus -> pretectal region of the midbrain -> Edinger-Westphal nucleus

B. Optic nerve-> optic chiasm -> Lateral geniculate nucleus in the thalamus-> Edinger-Westphal nucleus

C. Optic nerve -> optic chiasm -> optic tract -> Lateral geniculate nucleus in the thalamus

D. Optic nerve -> optic chiasm -> optic tract -> pretectal region of the midbrain -> Lateral geniculate nucleus in the thalamus

A

A. Optic nerve -> optic chiasm -> brachium of the superior colliculus -> pretectal region of the midbrain -> Edinger-Westphal nucleus

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4
Q

Which of the bifocal lens designs below will create the largest amount of image jump?

A. Round 28 bifocal
B. Executive bifocal
C. FT 35 bifocal
D. FT 28 bifocal

A

A. Round 28 bifocal

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5
Q

What is the relationship between the Abbe number and chromatic aberration?

A. No relationship
B. Equal
C. Inversely proportional
D. Proportional

A

C. Inversely proportional

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6
Q

What is the power of a concave mirror (in diopters) located in air with a radius of curvature of 20 cm?

A. +10.00 D
B. -5.00 D
C. -10.00 D
D. +5.00 D

A

A. +10.00 D

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7
Q

What is the front surface power of a lens in air with a refractive index of 1.50 and radius of curvature of 50 cm?

A. 1.50 D
B. 3.00 D
C. 2.00 D
D. 1.00 D

A

D. 1.00 D

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8
Q

Which organism can be contracted in a newborn via an infected mother and was previously treated prophylactically with silver nitrate?

A. Neisseria gonorrhoeae
B. Staphylococcus aureus
C. Neisseria meningiditis
D. Corynebacterium spp

A

A. Neisseria gonorrhoeae

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9
Q

Which of the following microorganisms is associated with peptic ulcer formation?

A. Vibrio cholerae
B. Clostridium tetani
C. Helicobacter pylori
D. Campylobacter spp

A

C. Helicobacter pylori

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10
Q

Which of the following genus of organisms is responsible for tuberculosis and leprosy?

A. Mycobacterium
B. Klebsiella
C. Salmonella
D. Borrelia

A

A. Mycobacterium

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11
Q

Which of the following classifications refers to an organism that can survive in an environment with or without oxygen?

A. Facultative anaerobe
B. Strict anaerobe
C. Obligate aerobe
D. Microaerophile

A

A. Facultative anaerobe

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12
Q

What type of agar is commonly used to culture fungi?

A. Cetrimide agar
B. Sabouraud's agar
C. Hay infusion agar
D. Blood agar plate
E. Thayer-Martin agar
A

B. Sabouraud’s agar

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13
Q

What term describes the phenomenon in which a bacterium directs its movement TOWARD a chemical in its environment?

A. Phagocytosis
B. Apoptosis
C. Chemotaxis
D. Transposition

A

C. Chemotaxis

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14
Q

Which patient would be considered legally blind?

A. A retinitis pigmentosa patient who has 20/20 central vision in each eye and a 30 degree in diameter visual field

B. A wet macular degeneration patient with best central acuities of 10/120 OD and 10/200 OS

C. A patient with a total retinal detachment of the right eye, no light perception and a best corrected central acuity of 8/60 due to wet macular degeneration

D. A patient with Best’s disease with best corrected central acuities measure OD 10/80 and OS 10/100

E. A myopic patient with acuities of 20/400 OD and OS uncorrected

A

B. A wet macular degeneration patient with best central acuities of 10/120 OD and 10/200 OS

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15
Q

Which of the following skin conditions is considered to be benign and has the LOWEST risk of malignancy?

A. Keratoacanthoma
B. Actinic keratosis
C. Squamous cell carcinoma
D. Basal cell carcinoma

A

A. Keratoacanthoma

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16
Q

Which of the following is the correct pathway for the drainage of tears through the nasolacrimal drainage system?

A. Nasolacrimal duct, lacrimal sac, valve of Hasner, lacrimal canaliculus, ampulla, lacrimal punctum

B. Lacrimal punctum, lacrimal canaliculus, ampulla, valve of Hasner, nasolacrimal duct, lacrimal sac

C. Lacrimal punctum, lacrimal canaliculus, ampulla, lacrimal sac, nasolacrimal duct, valve of Hasner

D. Lacrimal sac, lacrimal punctum, lacrimal canaliculus, ampulla, nasolacrimal duct, valve of Hasner

A

C. Lacrimal punctum, lacrimal canaliculus, ampulla, lacrimal sac, nasolacrimal duct, valve of Hasner

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17
Q

You are measuring the palpebral fissure height in a patient reporting drooping of his upper eyelid. Which of the following BEST describes the normal positioning of the upper and lower eyelids in comparison to the limbus?

A. The upper lid normally rests about 2mm lower than the upper limbus, and the lower lid rests about 1mm above the lower limbus

B. The upper lid normally rests about 1mm lower than the upper limbus, and the lower lid rests about 2mm lower than the lower limbus

C. The upper lid normally rests about 1mm lower than the upper limbus, and the lower lid rests about 2mm above the lower limbus

D. The upper lid normally rests about 2mm lower than the upper limbus, and the lower lid rests about 1mm lower than the lower limbus

A

A. The upper lid normally rests about 2mm lower than the upper limbus, and the lower lid rests about 1mm above the lower limbus

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18
Q

Which of the following types of congenital cataracts are characteristic of galactosemia?

A. Blue dot (Cerulean) opacities
B. Oil droplet opacities
C. Christmas tree cataracts
D. Sunflower cataracts

A

B. Oil droplet opacities

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19
Q

When examining a patient, a pinpoint spot of the posterior surface of the lens known as Mittendorf’s dot is seen. What is this a remnant of?

A. Pupillary membrane
B. Hyaloid artery
C. Vitreous
D. Glial tissue of the optic nerve

A

B. Hyaloid artery

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20
Q

Which of the following drugs decrease intraocular pressure by increasing uveoscleral outflow?

A- Dorzolamide
B- Timolol
C- Brinzolamide
D- Pilocarpine
E- Brimonidine
A

E. Brimonidine

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21
Q

Which type of anterior scleritis is associated with the highest risk of perforation?

A. Necrotizing
B. Scleromalacia perforans
C. Nodular
D. Diffuse

A

A. Necrotizing

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22
Q

Which of the following types of scleritis presents without ocular inflammation, has a low risk for perforation, and does not typically result in pain or decreased visual acuity?

A. Granulomatous necrotizing scleritis
B. Vaso-occlusive necrotizing scleritis
C. Anterior non-necrotizing diffuse scleritis
D. Posterior scleritis
E. Nodular scleritis
F. Scleromalacia perforans
A

F. Scleromalacia perforans

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23
Q

Which of the following correctly describes the autonomic innervation of the iris muscles?

A. The iris sphincter and iris dilator are both innervated parasympathetically

B. The iris sphincter is innervated parasympathetically and the iris dilator is innervated sympathetically

C. The iris sphincter is innervated sympathetically and the iris dilator is innervated parasympathetically

D. The iris sphincter and iris dilator are both innervated sympathetically

A

B. The iris sphincter is innervated parasympathetically and the iris dilator is innervated sympathetically

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24
Q

You are fitting a toric soft contact lens to your patient’s right eye. The patient’s manifest refraction is -2.00 -1.50 X 095. You apply a -1.75 -1.25 X 085 diagnostic toric soft contact lens. It fits well, and the prism base down marking consistently locates halfway between the 6 o’clock and 7 o’clock hours. What axis should you order?

A. 70 degrees
B. 95 degrees
C. 100 degrees
D. 80 degrees
E. 110 degrees
A

B. 95 degrees

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25
Q

Which one of the following bitoric GP contact lenses would NOT induce cylinder if rotated to a misaligned position on the eye?

A. 7.58 mm / -5.37 D ——————— 8.18 mm / -0.50 D
B. All of the options listed would induce cylinder if rotated off axis
C. 7.63 mm / -1.50 D ——————— 8.11 mm / +1.12 D
D. 7.54 mm / +1.50 D ——————— 7.99 mm / +2.75 D
E. 7.46 mm / -4.25 D ——————— 8.13 mm / -1.75 D

A

C. 7.63 mm / -1.50 D ——————— 8.11 mm / +1.12 D

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26
Q

Which of the following will occur if you increase the water content of a soft contact lens?

A. The tendency of lens deposits will decrease
B. The patient will report an increase in dry eye symptoms
C. The oxygen permeability will decrease
D. The lens durability will increase

A

B. The patient will report an increase in dry eye symptoms

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27
Q

Which of the following alterations will help to loosen a tightly-fitting gas-permeable lens?

A. Steepen the peripheral curve system
B. Reduce the width of the peripheral curve system
C. Reduce the size of the optic zone
D. Increase the overall diameter
E. Steepen the base curve of the lens
A

C. Reduce the size of the optic zone

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28
Q

Which of the following ocular signs is virtually pathognomonic for trachoma caused by chlamydia?

A. Superior tarsal follicles
B. Inferior tarsal papillae
C. Tranta’s dots
D. Lymphadenopathy

A

A. Superior tarsal follicles

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29
Q

Which layer of the cornea, if penetrated, will leave a scar?

A. The tear film
B. The epithelium
C. The stroma
D. The wing cell layer

A

C. The stroma

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30
Q

What is the name of the surgical procedure in which thermal laser burns are placed in the mid-periphery of the cornea in an attempt to steepen the corneal curvature?

A. Laser-assisted in-situ keratomileusis
B. Radial keratotomy
C. Conductive keratoplasty
D. Photorefractive keratectomy
E. Limbal relaxation incisions
A

C. Conductive keratoplasty

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31
Q

What is the name of the pigmented line that represents the leading edge of a pterygium?

A. Stocker's line
B. Fleischer's ring
C. Krukenberg's line
D. Ferry's line
E. Hudson-Stahli line
F. Coat's white ring
A

A. Stocker’s line

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32
Q

Which of the following is a precursor to steroid hormones such as testosterone?

A. Sphingolipids
B. Triglycerides
C. Phospholipids
E. Cholesterol

A

E. Cholesterol

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33
Q

What is the net overall moles of ATP produced by the electron transport chain (i.e. not including glycolysis)?

A. 30 moles of ATP
B. 6 moles of ATP
C. 2 moles of ATP
D. 34 moles of ATP
E. 38 moles of ATP
A

D. 34 moles of ATP

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34
Q

While performing the astigmatic clock dial, your patient reports that the clearest/blackest line is the 2-8 line while the 5-11 line is the least clear. What would be the corresponding axis of astigmatism?

A. 180 degrees
B. 150 degrees
C. 60 degrees
D. 30 degrees

A

C. 60 degrees

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35
Q

Which type of light scattering is responsible for the reddish-orange colors that are often observed during sunsets?

A. Raman scattering
B. Rayleigh scattering
C. Brillouin scattering
D. Mie scattering
E. Tyndall scattering
A

B. Rayleigh scattering

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36
Q

Which of the following BEST describes the definition of irregular astigmatism?

A. The principal meridians of the cornea are not perpendicular to each other
B. The axis of astigmatism is located along the 90 degree meridian
C. The principal meridians of the cornea are located 90 degrees apart
D. The axis of astigmatism is located along an oblique axis

A

A. The principal meridians of the cornea are not perpendicular to each other

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37
Q

What separation distance will make the combination of a +3.00 and a +10.00 thin lens afocal?

A. 2.3 cm
B. 17 cm
C. 43 cm
D. 0.43 cm
E. 1.7 cm
F. 23 cm
A

C. 43 cm

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38
Q

What is the minimum thickness necessary for an antireflective coating (n=1.9) to be useful against incident light of 530 nm wavelength?

A. 139.5 nm
B. 58.3 nm
C. 132.5 nm
D. 278.9 nm
E. 69.7 nm
A

E. 69.7 nm

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39
Q

What is the Interval of Sturm for a spherocylindrical lens with a power of +6.00 -2.00 x 090?

A. 41.7 cm
B. 16.7 cm
C. 25 cm
D. 8.3 cm
E. 20 cm
A

D. 8.3 cm

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40
Q

What is the equivalent of a Reduced Snellen 20/50 optotype in metric notation (assuming a working distance of 40 cm)?

A. 2M
B. 0.5M
C. 0.67M
D. 1M

A

D. 1M

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41
Q

Which of the following types of refractive error would have the greatest tendency to lead to amblyopia?

A. A four-year old boy with an uncorrected refractive error of OD: +6.00 DS and OS: +1.50 DS

B. A five-year old girl with an uncorrected refractive error of OD: -3.25 DS and OS: -0.75 DS

C. A three-year old boy with an uncorrected refractive error of OD: +1.50 DS and OS: -2.00 DS

D. A four-year old girl with an uncorrected refractive error of OD: +1.00-1.50 x 180 and OS: +1.50-1.25 x 180

A

A. A four-year old boy with an uncorrected refractive error of OD: +6.00 DS and OS: +1.50 DS

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42
Q

A chin fissure is a dominant trait. If a father who is homozygous-dominant for this trait and a mother who is homozygous-recessive for this trait mate, what are the chances that their first child will have a chin fissure?

A. 100%
B. 0%
C. 75%
D. 25%
E. 50%
A

A. 100%

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43
Q

A 31-year old male patient presents to your office for a photorefractive keratectomy (PRK) pre-operative examination. As you review his required ocular medication schedule, which of the following prescribed drops must you remember to tell him to “shake well” before instillation?

A. FreshKote®
B. Acular®
C. Pred Forte®
D. Zymaxid®

A

C. Pred Forte®

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44
Q

A 24-year old female patient presents at your office complaining of side effects that began when she started using Patanol® to treat her ocular allergies. She reports complete compliance with her eye drop administration. Which of the following symptoms is MOST likely associated with olopatadine (Patanol®) use?

A. Headache
B. Visual Hallucinations
C. Depression
D. Gastrointestinal discomfort
E. Tachycardia
A

A. Headache

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45
Q

A person who is missing the photopigment chlorolabe is categorized as which of the following?

A. A deuteranomalous trichromat
B. A protanomalous trichromat
C. A tritanope
D. A protanope
E. A deuteranope
A

E. A deuteranope

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46
Q

What is the MOST common side effect associated with intravenous administration of sodium fluorescein in patients requiring a fluorescein angiography?

A. Localized tissue necrosis
B. Bronchospasm
C. Nausea and vomiting
D. Anaphylaxis
E. Elevated temper
A

C. Nausea and vomiting

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47
Q

What is the total cumulative dosage of chloroquine that is MOST commonly associated with retinal toxicity?

A. 1,000 grams
B. 1500 grams
C. 4,000 grams
D. 300 grams
E. 100 grams
A

D. 300 grams

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48
Q

Which of the following antibiotics is classified as a macrolide?

A. Tetracycline
B. Amoxicillin
C. Erythromycin
D. Tobramycin

A

C. Erythromycin

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49
Q

Which of the following BEST describes the reasoning for the need to taper topical ocular corticosteroids?

A. Minimize the risk of developing steroid-induced elevation of IOP

B. Decrease the risk of posterior subcapsular cataract formation

C. Minimize risk of adrenal insufficiency due to decreased production of natural cortisol

D. Avoid signs and symptoms of rebound ocular inflammation

E. Prevent possible secondary ocular infections

A

D. Avoid signs and symptoms of rebound ocular inflammation

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50
Q

Which of the following medications should be taken with an empty stomach?

A. Tetracycline

B. Cephalexin (Keflex®)

C. Doxycycline

D. Amoxicillin with clavulanate (Augmentin®)

A

Tetracycline

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51
Q

What substance is secreted by the alveolar epithelium to reduce the surface tension between alveoli of the lungs?

A. Intrapleural fluid
B. Surfactant
C. Angiotensin
D. Carbon dioxide

A

B. Surfactant

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52
Q

Which class of antibody crosses the blood-placenta barrier and protects newborns for the first months of life until they produce their own antibodies?

A. IgD
B. IgA
C. IgM
D. IgE
E. IgG
A

E. IgG

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53
Q

Which of the following actions does not require input from the brain and therefore is referred to as a spinal reflex?

A. Catching a ball
B. Painting a picture
C. Maintaining balance while riding a bicycle
D. Pulling your hand away from a hot stove
E. Dodging out of the way when an object is thrown in your direction

A

D. Pulling your hand away from a hot stove

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54
Q

Which of the following antibodies is the first to be secreted during an immune response?

A. IgD antibodies
B. IgM antibodies
C. IgE antibodies
D. IgA antibodies
E. IgG antibodies
A

B. IgM antibodies

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55
Q

Which of the following walls of the orbit is MOST susceptible to a blowout fracture secondary to blunt ocular trauma?

A. Medial wall
B. Floor
C. Lateral wall
D. Roof

A

B. Floor

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56
Q

Which of the primary germ layers of embryonic development take part in the development of ocular structures?

A. Ectoderm only
B. Endoderm only
C. Mesoderm and ectoderm
D. Mesoderm and endoderm
E. Ectoderm and endoderm
A

C. Mesoderm and ectoderm

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57
Q

Which two layers of the iris are derived from mesoderm?

A. The anterior limiting layer and the stroma

B. The stroma and the anterior epithelium

C. The stroma and the posterior pigmented epithelium

D. The epithelium and the posterior pigmented epithelium

E. The posterior pigmented epithelium and the anterior limiting layer

A

A. The anterior limiting layer and the stroma

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58
Q

Which of the following components of the AREDS I ocular vitamin formula used for dry age-related macular degeneration is contraindicated in smokers?

A. Copper
B. Beta-carotene
C. Zinc
E. Vitamin C
D. Vitamin E
A

B. Beta-carotene

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59
Q

Which of the following correctly lists the layers of the retina beginning with the retinal pigment epithelium and moving anteriorly?

A. Retinal pigment epithelium, outer nuclear layer, external limiting membrane, photoreceptor layer, outer plexiform layer, inner plexiform layer, inner nuclear layer, nerve fiber layer, ganglion cell layer, internal limiting membrane

B. Retinal pigment epithelium, photoreceptor layer, outer nuclear layer, external limiting membrane, outer plexiform layer, inner plexiform layer, inner nuclear layer, ganglion cell layer, nerve fiber layer, internal limiting membrane

C. Retinal pigment epithelium, photoreceptor layer, external limiting membrane, outer nuclear layer, outer plexiform layer, inner nuclear layer, inner plexiform layer, ganglion cell layer, nerve fiber layer, internal limiting membrane

D. Retinal pigment epithelium, external limiting membrane, outer nuclear layer, photoreceptor layer, outer plexiform layer, inner nuclear layer, inner plexiform layer, nerve fiber layer, ganglion cell layer, internal limiting membrane

A

C. Retinal pigment epithelium, photoreceptor layer, external limiting membrane, outer nuclear layer, outer plexiform layer, inner nuclear layer, inner plexiform layer, ganglion cell layer, nerve fiber layer, internal limiting membrane

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60
Q

Which of the following structures serves as the strongest attachment point of the vitreous?

A. The ora serrata
B. Blood vessels
C. The optic nerve head
D. The macula

A

A. The ora serrata

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61
Q

Which of the following organisms can penetrate an INTACT cornea?

A. Streptococcus aureus
B. Haemophilus influenza
C. Salmonella enterica
D. Staphylococcus epidermis

A

B. Haemophilus influenza

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62
Q

What is the vergence demand using a Variable Tranaglyph when the separation is measured as 4 cm at a distance of 80 cm when training divergence?

A. 20 prism diopters base-out
B. 5 prism diopters base-in
C. 20 prism diopters base-in
D. 5 prism diopters base-out

A

B. 5 prism diopters base-in

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63
Q

When analyzing a gas-permeable lens, you measure base curves of 7.58 and 7.84 with a radiuscope, and -1.00 and -2.50 on lensometry. What type of toric gas-permeable contact lens design do you have?

A. Cylinder power effect (CPE) bitoric
B. Spherical power effect (SPE) bitoric
C. Thin-flex
D. Back surface (base curve) toric
E. Front surface (F1) toric
A

B. Spherical power effect (SPE) bitoric

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64
Q

Which of the following is the MOST common early pattern of a glaucomatous visual field defect?

A. Para-central scotoma
B. Inferior arcuate
C. Enlarged blind spot
D. Superior nasal step
E. Inferior nasal step
F. Superior arcuate
A

A. Para-central scotoma

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65
Q

What portion of the progressive lens should be used to verify the distance prescription when the lenses arrive from the lab?

A. The major reference point (MRP)
B. The prism reference point (PRP)
C. The center of the fitting cross
D. The center of the distance arc

A

D. The center of the distance arc

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66
Q

Which of the following acquired color vision deficiencies would you MOST expect to see in optic nerve disease and macular disease, respectively?

A. Red-green, blue-yellow
B. Red-green, rod monochromacy
C. Blue-yellow, rod monochromacy
D. Blue-yellow, red-green

A

A. Red-green, blue-yellow

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67
Q

What is the main mechanism of action of fluticasone and salmeterol, which are components found in Advair®, respectively?

A. Bronchodilator; anti-inflammatory
B. Anti-histamine; anti-inflammatory
C. Anti-inflammatory; anti-histamine
D. Anti-inflammatory; bronchodilator
E. Anti-inflammatory; leukotriene inhibitor
F. Leukotriene inhibitor; anti-inflammatory

A

D. Anti-inflammatory; bronchodilator

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68
Q

In which of the following conditions are bandage contact lenses NOT typically utilized?

A. Eyelid entropion
B. Filamentary keratitis
C. Post-LASIK surgery
D. Recurrent corneal erosion
E. Bullous keratopathy
A

C. Post-LASIK surgery

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69
Q

Which of the following infections must be reported to the CDC (Centers for Disease Control and Prevention)?

A. Acanthamoeba
B. Herpes Simplex
C. Syphilis
D. Epidemic keratoconjunctivitis (EKC)

A

C. Syphilis

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70
Q

Which of the following systemic conditions can cause a falsely low measurement of a patient’s hemoglobin A1c level?

A. Chronic opioid use
B. Alcoholism
C. Hyperbilirubinemia
D. Pregnancy
E. Iron deficient anemia
A

D. Pregnancy

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71
Q

A father brings in his two-year old son for evaluation at your office. The father remarks that his son was born with what appears as a small red birthmark on his forehead. He wishes to know if it requires treatment or warrants removal. Applying pressure over the area of interest causes blanching of the lesion, and the father reports that the birthmark appears darker when his son cries. What is your prognosis?

A. Benign but is likely to be permanent and will darken will time
B. Malignant and requires immediate biopsy
C. Benign and will likely regress by the time the child is 5 years of age
D. A pre-cursor to a malignant condition and will continue to increase in size with time

A

C. Benign and will likely regress by the time the child is 5 years of age

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72
Q

A +3.00 D hyperope is corrected with a +1.50 D contact lens bilaterally. If he views a near object located at 22.0 cm, what degree of accommodation is required to achieve a clear retinal image (rounded to the nearest 0.25 D)?

A. +4.50 D
B. +1.50 D
C. +6.00 D
D. +7.50 D

A

C. +6.00 D

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73
Q

A 12 year-old patient is seen at your office complaining of distance blur. Cover test reveals 4 prism diopters of exophoria at near. Subjective refraction reveals -2.25 DS OU. The patient returns 4 weeks later and reports that her vision is clear at distance with the glasses, but when she reads her eyes become fatigued and she reports frontal headaches, all of which she did not experience prior to getting her glasses. What is the MOST likely etiology of her headaches while reading?

A. Resultant esophoria at near induced by her glasses
B. The pantoscopic tilt is incorrect and is inducing unwanted astigmatism
C. The optical center of her glasses was measured too low
D. Her glasses are too tight and are putting pressure on her temples

A

A. Resultant esophoria at near induced by her glasses

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74
Q

You ask your patient to place a red lens in front of their right eye and proceed to perform the red lens test. Your patient reports seeing two images with the red image being perceived to the left of the white light. What type of deviation corresponds with the above findings?

A. Esophoria
B. Exophoria or exotropia
C. Hyperphoria
D. Hypophoria

A

B. Exophoria or exotropia

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75
Q

While performing the astigmatic clock dial, your patient reports that the 1-7 and 2-8 lines are equally blacker and clearer than all of the other lines. What would be the corresponding axis of astigmatism?

A. 90 degrees
B. 45 degrees
C. 30 degrees
D. 60 degrees

A

B. 45 degrees

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76
Q

Myasthenia gravis is an autoimmune disease that affects which of the following types of receptors in the body?

A. Beta-adrenergic receptors
B. Muscarinic acetylcholine receptors
C. Nicotinic acetylcholine receptors
D. Alpha-adrenergic
E. Adenosine receptors
A

C. Nicotinic acetylcholine receptors

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77
Q

You suspect allergic conjunctivitis as the cause of your 22 year-old patient’s symptoms of red, watery, itchy eyes. If you were to perform a conjunctival scraping, the presence of which of the following types of cells would confirm your diagnosis?

A. Monocytes
B. Basophils
C. Neutrophils
D. Eosinophils
E. Lymphocytes
A

D. Eosinophils

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78
Q
What is the mean horizontal and vertical diameter of the human cornea, respectively (when viewed ANTERIORLY)?
A. 11.7mm, 10.6mm
B. 10.6mm, 11.7mm
C. 10.2mm, 11.5mm
D. 11.5mm, 10.2mm
A

A. 11.7mm, 10.6mm

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79
Q

You suspect allergic conjunctivitis as the cause of your 22 year-old patient’s symptoms of red, watery, itchy eyes. If you were to perform a conjunctival scraping, the presence of which of the following types of cells would confirm your diagnosis?

A. Monocytes
B. Basophils
C. Neutrophils
D. Eosinophils
E. Lymphocytes
A

D. Eosinophils

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80
Q

Your last patient of the day insists on being fit with a soft toric contact lens. Her subjective refraction is -3.00 -2.50 x 178. Which of the following lenses would be the BEST choice to use as a trial lens for her fitting evaluation?

A. -3.00 -0.75 x 180
B. -2.75 -2.25 x 140
C. -2.75 -1.75 x 090
D. -1.00 -2.25 x 180
E. +2.00 -2.25 x 180
A

A. -3.00 -0.75 x 180

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81
Q

The blood-aqueous barrier is formed by tight junctions between which cells of the ciliary body?

A. The non-pigmented epithelial cells
B. The pigmented epithelial cells
C. The stromal cells
D. Basal laminar cells
E. The circular cells
A

A. The non-pigmented epithelial cells

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82
Q

The lens changes significantly as we age. One of these changes is the formation of vacuoles. What causes this?

A. Loss of lenticular capsule elasticity causing decreased accommodation
B. Glutathione loss with age causing cross linking between proteins
C. Decreased function of the sodium/potassium pump causing decreased protein synthesis
D. The separation of water from proteins causing agglutination of proteins and pooling of water

A

D. The separation of water from proteins causing agglutination of proteins and pooling of water

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83
Q

The presence of foam at the canthus is thought to be pathognomonic for blepharitis. What is the direct etiology of the foam?

A. A detergent effect from altered meibomian gland lipids
B. Tear film debris such as sloughed epithelial cellular material that is increased in dry eye
C. Increased lysozymes react with free radicals to produce a froth-like material
D. Mucin balls form from increased ocular surface shear forces exerted by the eyelids in dry eye
E. Bacterial lipases resulting from low-grade infection within the meibomian glands

A

A. A detergent effect from altered meibomian gland lipids

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84
Q

A patient is seen at your office complaining of distance blur with his glasses. With his current prescription of -1.25 D in place, you determine that his far point is 50 cm from the spectacle plane for his left eye. Given this information, which of the following is the MOST appropriate spectacle prescription to obtain a clear retinal image when an object is viewed at optical infinity (rounded to the nearest 0.25 D)?

A. +0.75 D
B. -2.00 D
C. -0.75 D
D. -1.25 D
E. -3.25 D
A

E. -3.25 D

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85
Q

How much image jump will be created by a +2.00 D flat top 25 mm segment add with a carrier lens of +2.25 DS?

A. 1.125 prism diopters
B. 1 prism diopter
C. 2.125 prism diopters
D. 5.31 prism diopters
E. 2.5 prism diopters
A

B. 1 prism diopter

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86
Q

Hutchinson’s sign associated with herpes zoster is an indication of reactivation of which of the following cranial nerves?

A. Nasociliary branch of the trigeminal nerve
B. Lacrimal branch of the trigeminal nerve
C. Facial nerve
D. Frontal branch of the trigeminal nerve

A

A. Nasociliary branch of the trigeminal nerve

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87
Q

Cyclosporine is commonly used topically to treat certain eye conditions including keratoconjunctivitis sicca (KCS). Which of the following is a known mechanism of action for cyclosporine?

A. Inhibition of T-cell activation
B. Stabilization of mast cells
C. Inhibition of cyclooxygenase (COX)
D. Antagonism of vitamin K

A

A. Inhibition of T-cell activation

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88
Q

In order to maximize drug penetration through the cornea an ophthalmic drug or its vehicle should possess which property?

A. High alcohol content
B. Highly polar component
C. High pH
D. Lipid solubility

A

D. Lipid solubility

Explanation - Ophthalmic drops that are comprised of both lipid (non-polar) and water-soluble (polar) components result in the most effective preparation. The tight junctions of the corneal epithelium keep hydrophilic drugs out but allow good penetration for lipid-soluble drugs. In contrast, the corneal stroma demonstrates good penetration for water-soluble agents. However, a HIGHLY polar agent will not cross the corneal epithelium; therefore, a mildly polar substance is advantageous over a highly polar agent. One should NEVER place alcohol on the eye as it will instantly debride the corneal epithelium; ALWAYS rinse all instruments used for ocular procedures with saline solution after sterilization with any agent that could be toxic to the cornea. Solutions with high pH (basic) are more damaging to the cornea than solutions with a lower pH (acidic). It is important to use solutions that are close to a neutral pH (7.0) to eliminate possible damage to the cornea.

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89
Q

A 66 year-old male presents with a facial nerve palsy resulting in weakness of both the upper and lower portions of the right side of his face. Which of the following BEST describes the characteristics of the involved motor neuron?

A. Right lower motor neuron
B. Left lower motor neuron
C. Left upper motor neuron
D. Right upper motor neuron

A

A. Right lower motor neuron

Explanation - A unilateral lesion to a lower motor neuron of the facial nerve will result in weakness of upper and lower portions of the face on the ipsilateral side of the lesion. Therefore, in this case, because the upper and lower portions of the right side of the face are affected, the involved motor neuron would be a right lower motor neuron.

The branch of the facial nerve that serves the upper portion of the face receives innervation from both the right and left corticobulbar tracts, while the branch of the facial nerve that serves the lower portion of the face only receives innervation from the contralateral side of the brain. If a lower motor neuron lesion occurs, both of these branches are affected, and all muscles of facial expressions beyond that point on the ipsilateral side of the face will experience weakness.

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90
Q

Central serous retinopathy (CSR) is associated with an acute decrease in vision along with central distortion. The condition usually occurs unilaterally. Which gender and age group tends to have the highest incidence of CSR?

A. Males; ages 50-70
B. Males; ages 30-50
C. Females; ages 20-40
D. Females; ages 10-20

A

B. Males; ages 30-50

This condition causes fluid to leak from the choriocapillaries into the subretinal area, causing a serous detachment of the neurosensory retina. There is an associated loss of the foveal reflex, a hyperopic shift, a potential relative scotoma, and metamorphopsia. Flourescein angiography will reveal hyperfluorescence that appears like a smoke-stack. Evaluation of the posterior pole will typically display a blister-like elevation of the neurosensory retina. The patient is monitored monthly and intervention is rarely required, as most cases of CSR will resolve within roughly 6 months.

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91
Q

Which of the following wavelengths of visible light has an increased association with the development of macular degeneration?

A. 415nm-455nm
B. 520nm-555nm
C. 485nm-510nm
D. 570nm-620nm

A

A. 415nm-455nm

Explanation - Recent studies have demonstrated a correlation between blue-violet light that lies within the range of 415nm-455nm and the development of macular degeneration. Excessive exposure to light that falls within this bandwidth is associated with death of the retinal pigment epithelial cells. However, blue-turquoise light (465-495nm) does not appear to possess detrimental effects to ocular health. Blue-turquoise light is important in activation of the pupillary reflex as well as management of the circadian sleep/wake cycle. There is increasing evidence that compact fluorescent lamps, LED lights as well as sunlight all transmit blue-violet light, which over time may be linked with retinal damage

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92
Q

A 24-year old female wears soft contact lenses with a Dk/t of 175 and admits to sleeping in her lenses. She is very satisfied with both the comfort and the vision of her lenses. Biomicroscopy reveals mucin balls under her lenses bilaterally that leave impressions in her central corneas upon removal of her lenses. Which of the following actions would BEST help to eliminate the formation of mucin balls?

A. Maintaining the same lens material but changing to a steeper base curve
B. Changing her multi-purpose solution
C. Altering the power of the contact lens but maintaining the same lens material
D. Instructing the patient to increase her blinking frequency

A

A. Maintaining the same lens material but changing to a steeper base curve

Explanation - Mucin balls appear as small, white, pearl-like debris that occur behind the posterior surface of contact lenses. They generally occur with silicone hydrogel lenses that are fit too flat and are used for extended wear purposes. Mucin balls do not actually pose a threat to vision and do not generally compromise the integrity of the cornea. However, if they are severe enough, there are several options available to clinicians to combat their formation. An easy way to decrease generation of mucin balls is to steepen the base curve of the lens. Alternatively, one can decrease the amount of extended wear or add re-wetting drops to the patient’s contact lens regimen. Upon removal, mucin balls will cause pooling of sodium fluorescein but will not cause staining of the cornea.

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93
Q

A 23-year old female is seen at your office with concerns of eye fatigue, diplopia and headaches after 30 minutes of computer use. Her subjective refraction is +0.25 D OU. Her near point of convergence is 15 centimeters. What is the most likely diagnosis based solely upon this information?

A. Brain tumor
B. Convergence excess
C. Latent hyperopia
D. Convergence insufficiency

A

D. Convergence insufficiency

Explanation - The most notable of the exam findings is the near point of convergence (NPC) -generally one expects an NPC of less than 6cm (closer is better with NPC). Her NPC is receded to 15 cm. Normally, if patients display a receded NPC but they are asymptomatic, no treatment is necessary. In the above case, the patient is experiencing diplopia and asthenopia with prolonged near work which infers some type of intervention may be required (likely vision therapy). Brain tumors typically present as headaches that are present in the morning and worsen over a time period of weeks to months. Latent hyperopia normally presents as blurry vision and headaches with prolonged near work, but diplopia is generally absent.

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94
Q

A 47-year old man sustained orbital trauma and now presents with complaints of retro-orbital pain, impaired ability to move the eye, a droopy eyelid, and diplopia. These signs are most consistent with damage to which of the following structures?

A. Internal auditory meatus
B. Superficial temporal artery
C. Superior orbital fissure
D. Stylomastoid foramen

A

C. Superior orbital fissure

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95
Q

A 6-year old white male presents with a mild left head turn. Wet retinoscopy reveals OD: +0.25 OS: +0.50 with best corrected acuities of 20/20 in each eye. Extraocular movements show limited adduction of the left eye in right gaze. It is also noted that the left eye retracts with a narrowing of the eyelid fissure. What is the most appropriate diagnosis for this patient?

A. Brown syndrome OS
B. Duane Syndrome Type II OD
C. Bilateral Brown Syndrome
D. Duane Syndrome Type II OS
E. Duane Syndrome Type III OS
F. Duane Syndrome Type I OS
A

D. Duane Syndrome Type II OS

Explanation - Limited adduction to an affected eye usually points to a lateral rectus problem. In this case, Duane Syndrome is suspected due to the additional sign of eye retraction and narrowing of the eyelid fissure. Duane Syndrome Type III is the most appropriate diagnosis due to the limited ADDuction of the affected eye on right gaze, along with the left head tilt, which also implies limited ABDuction as well.
Duane Syndrome Type I describes limited ABDuction of the affected eye (the most common) as well as a possible compensatory head tilt toward the involved side.
Duane Syndrome Type II describes limited ADDuction of the affected eye, as well as a possible compensatory head tilt toward the uninvolved side.
Duane Syndrome Type III describes limited ABDuction AND limited ADDuction of the affected eye. It also usually presents with a compensatory head tilt toward the involved side.
A good way to remember the difference between the three types is that type I results in an aBDuction deficit (aBDuction has one D therefore it is type I).
Type II causes an aDDuction deficit (aDDuction has two Ds therefore it is type II). Type III has three Ds, aBDuction and aDDuction-the number of the types matches the number of Ds in the deficit.

Brown syndrome describes a limitation of elevation in adduction. It is a limitation of the superior oblique tendon.

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96
Q

A patient with a high AC/A ratio (8/1) displays esophoria at a 6 m distance. Based on the AC/A ratio, how would you expect the phoria to change as the target is brought closer to the patient?

A. Increase in eso deviation
B. Increase in exo deviation
C. Decrease in eso deviation
D. Remain unchanged

A

A. Increase in eso deviation

Explanation - The AC/A ratio denotes the amount of change to convergence resulting from a change in accommodation. If a patient possesses a high AC/A ratio (6/1 is considered the normal range), a one-diopter increase in accommodation will theoretically cause a greater increase in convergence. Regardless of the initial phoria, with decreasing viewing distance the phoria will become more eso (or less exo). The opposite holds true for a low AC/A ratio (less than 6/1); as the target gets closer, the resultant phoria becomes more exo or less eso.

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97
Q

As the interpupillary distance increases what happens to the amount of convergence (in prism diopters) needed to maintain fusion for a near target ?

A. It increases, but only for targets closer than 4 cm
B. It increases
C. It remains the same
D. In decreases, but only for targets closer than 4 cm
E. It decreases

A

B. It increases

Explanation - There is a direct correlation between the amount of convergence necessary to maintain fusion on a near target and interpupillary distance (IPD). As the IPD increases, so does the amount of convergence required to maintain fusion. Logically, this makes sense: if the eyes are further apart then they must rotate to a greater degree around the horizontal axis to maintain fusion if the target distance remains the same.

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98
Q

A concerned father reports that one of his 12-month-old infant’s eyes does not appear straight. You decide to perform the Hirschberg test to evaluate for strabismus. The corneal reflex of the right eye is centered, while the left reflex is displaced 0.5 mm superiorly relative to the center of the pupil. Angle Kappa (Lambda) is zero for each eye. What is the correct deviation and magnitude of the observed strabismus?

A. Left hypertropia of 11 prism diopters
B. Left hypertropia of 22 prism diopters
C. Left hypotropia of 11 prism diopters
D. Left hypotropia of 22 prism diopters

A

C. Left hypotropia of 11 prism diopters

Explanation - The Hirshberg test is performed at a distance of 50 cm. A penlight or transilluminator is held just below the doctor’s preferred eye and the doctor then sits in front of the patient and directs the beam towards the patient’s nose while the patient is instructed to fixate on the light. The position of the corneal reflexes relative to the center of the pupil is assessed in each eye. Superior displacement of the corneal reflex suggests hypotropia, while inferior displacement infers hypertropia. Each millimeter of displacement of the reflex from the center of the pupil equates to roughly 22 prism diopters of deviation.

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99
Q

In order to determine if environmental elements aid in the development of ametropia, researchers reared an infant monkey with a clear lens placed over one eye. These studies determined that putting a minus lens over one eye induced which type of refractive error?

A. Large amounts of astigmatism
B. Myopia
C. Presbyopia
D. Hyperopia

A

B. Myopia

Explanation - Although there has been much debate in the past over whether or not the etiology of refractive errors was environmental versus inherited, it is now believed that both factors contribute to the development of ametropias. Hung et al, 1995 demonstrated that by placing a prescription lens over one eye of an infant monkey and removing it after the eye had reached maturity, the mature eye developed the refractive error that the lens is normally meant to neutralize. A minus lens induced myopia and a plus lens induced hyperopia. Regardless of whether the etiology of the refractive error is inherited or environmental, it is absolutely essential that a clear retinal image be present in order for emmetropization to occur.

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100
Q

A young strabismic child presents at your office. Using visuoscopy you ask the patient to fixate the center of the target with their right eye (the left eye is occluded). The foveal reflex is positioned three hash marks to the LEFT of the center of the target. This finding suggests which type of fixation? (assume each hash mark is equal to 1 prism diopter)

A. 3 prism diopters superior eccentric fixation
B. 3 prism diopters inferior eccentric fixation
C. 3 prism diopters nasal eccentric fixation
D. The patient does not possess any eccentric fixation
E. 3 prism diopters temporal eccentric fixation

A

C. 3 prism diopters nasal eccentric fixation

Explanation - Visuoscopy is an excellent technique to evaluate for eccentric fixation. This is performed by using the cross-hair target of your direct ophthalmoscope and projecting it onto the macula of the unoccluded eye. The patient is asked to fixate on the center of the target. No eccentric fixation is present if the foveal reflex aligns with the center of the cross-hairs. If the foveal reflex is to the left of the center (for the right eye), then the patient has nasal eccentric fixation. If the foveal reflex is located to the right of the target center, the patient possesses temporal eccentric fixation. The opposite holds true for the left eye (if the foveal reflex is to the right of the target center, the patient has nasal eccentric fixation). If the foveal reflex is located above the target, then the patient has inferior eccentric fixation, whereas a foveal reflex below the target is classified as superior eccentric fixation.
In order to calculate the amount of eccentric fixation, you will have to know that from the center of the circle on the visuoscopy target to the edge of the circle is one prism diopter, and then each hash mark away from the center circle is an additional prism diopter. Therefore, the above patient has a total of 4 prism diopters of eccentric fixation (1 to the edge of the circle, and 3 for each additional hash mark).

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101
Q

A convex crown glass lens in air has a radius of curvature of 4 cm. What is the dioptric power of the lens?

A. -4 diopters
B. -13 diopters
C. +13.0 diopters
D. -8.25 diopters
E. +8.25 diopters
F. +4 diopters
A

C. +13.0 diopters

Explanation - The equation for the power of a curved surface is:
D= (n’-n/r)
D= power of the lens (may also be notated as F or P)
n’= 2nd index (image), n= 1st index (object)
r= radius of curvature (in meters)

D= (1.52-1.0)/ 0.04 
D= (+)13.0 diopters 

Remember that converging (or convex) surfaces will have positive power, and diverging (or concave) surfaces will have negative power. If this question was changed to a concave glass lens, the radius of curvature would be negative; therefore, the power of the lens would be (-) 13 diopters.

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102
Q

A cornea that displays toricity is said to have with-the-rule astigmatism if it possesses which of the following keratometry readings?

A. The vertical meridian is flatter than the other principal meridian
B. The horizontal meridian has a shorter radius of curvature
C. The horizontal meridian is steeper than the other principal meridian
D. The vertical meridian is steeper than the other principal meridian

A

D. The vertical meridian is steeper than the other principal meridian

Explanation - With-the-rule astigmatism occurs when the vertical meridian is steeper than the other principal meridian; that is, the horizontal meridian is flatter and corresponds with the axis of the astigmatism. In this case, the vertical meridian would have a shorter radius of curvature indicating that it possesses a greater dioptric power than the horizontal meridian. If the steeper meridian lies between 60 and 120 degrees, the cornea is said to have with-the-rule astigmatism. If the steeper meridian lies between 150 and 30 degrees, the cornea displays against-the-rule astigmatism. Anything outside of these meridians is considered oblique astigmatism.

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103
Q

A lens system in air consists of a +4.00 diopter and +6.00 diopter lens separated by 10 cm. What is the equivalent power of this optical system?

A. +12.4 diopters
B. +9.76 diopters
C. +7.6 diopters
D. +10.0 diopters
E. +10.24 dioptersC
A

C. +7.6 diopters

Explanation - De = D1 + D2 - (t/n) x D1D2
De = equivalent power, D1 = front surface power, D2 = back surface power
t = thickness of lens system, n = index between the 2 surfaces

In the above question, t = 10 cm (0.1 m), n=1.0, D1 = +4.00 and D2 = +6.00

De = 4 + 6 - ((0.1/1.0) x (4) x (6)) 
De = 10 - (0.1 x 4 x 6) 
De = 10 - (2.4) 
De = +7.6 diopters
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104
Q

A patient views a bichrome visual acuity chart with no lenses before his eyes. With the patient’s left eye occluded, he reports that the letters on the red side of the chart appear blacker and darker. With his right eye occluded, the letters on the green side appear blacker and darker. Given these observations and assuming that accommodation is at rest, what would likely be the refractive condition of the right eye and left eye respectively?

A. Hyperopia and myopia
B. Myopia and myopia
C. Myopia and hyperopia
D. Hyperopia and emmetropia
F. Myopia and emmetropia
E. Emmetropia and hyperopia
A

C. Myopia and hyperopia

Explanation - For an emmetropic eye that is not accommodating, the chromatic interval within the eye would be positioned so that the anterior (green) and posterior (red) ends of the interval are equidistance from the retina (i.e. the midpoint of the interval would be at the retina) and thus the letters on the red and green sides of the chart would appear equally black and dark. For uncorrected myopia, the interval would shift forward and thus the red end of the interval moves closer to the retina. In this case, the letters on the red side would appear blacker and darker. For uncorrected hyperopia, the interval would shift backward and thus the green end of the interval moves closer to the retina. In this case, the letters on the green side would appear blacker and darker.

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105
Q

Any optical system in air has first and second nodal points that coincide with which of the following corresponding points?

A. Front and back surfaces of the optical system
B. The geometrical center of the optical system
C. First and second focal points
D. First and second principal points

A

D. First and second principal points

Explanation - The first and second nodal points (N and N’) of an optical system are unique conjugate points such that an incident ray directed at N yields a final ray emerging from N’ that is undeviated and parallel to the initial ray.

For any optical system in air, the first and second nodal points (N and N’) correspond to the first and second principal points (P and P’).

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106
Q

How does uncorrected myopia affect a near phoria?

A. It results in greater hypophoria
B. It results in greater hyperphoria
C. It results in greater esophoria or less exophoria
D. It will not change the degree of phoria
E. It results in greater exophoria or less esophoria

A

E. It results in greater exophoria or less esophoria

Explanation - If myopia is not corrected the stimulus for accommodation is decreased resulting in less “accommodative” convergence; this will likely lead to less esophoria, or more exophoria. The opposite is true for an uncorrected hyperope. Uncorrected hyperopia will cause an increased accommodative convergence response leading to greater convergence and therefore less exophoria or more esophoria

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107
Q

If a thin lens with an index of 1.5 has a dioptric power of +4.00 in air, what is its power if placed in water?

A. +11.76 diopters
B. +4.00 diopters
C. +1.36 diopters
D. +1.18 diopters

A

C. +1.36 diopters

Explanation - D (air) / D (water) = n (lens) - n (air) / n (lens) - n (water)

+4.00 / D (water) = 1.5 - 1.0 / 1.5 - 1.33

D (water) = +1.36 diopters

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108
Q

Alpha helices and beta sheets are considered which level of protein structure?

A. Quaternary
B. Primary
C. Tertiary
D. Secondary

A

D. Secondary

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109
Q

An amphipathic molecule will react in what manner when exposed to water?

A. It will completely dissolve
B. It will form micelles
C. It will form a gas
D. It will combust

A

B. It will form micelles

Explanation - Amphipathic molecules, such as phosphatidylcholine, contain both polar and non-polar components. Because water is polar, it will dissolve other polar compounds (like dissolves like). Upon exposure to water, the hydrophilic polar elements of the amphipathic molecule will begin to align and point outwards towards the water, eventually forming small spheres called micelles. The inner portion of the micelle is comprised of the non-polar elements which are hydrophobic. Micelle formation increases the entropy of water molecules; therefore, the water molecules are less ordered; this is a more favorable outcome. If the solvent is non-polar, then an inverse micelle will form, with the hydrophobic portions pointing outwards. Entropy is a measure of disorder; higher entropy entails less order. Lipid bilayers are formed based on the principals of micelle formation.

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110
Q

Coenzyme Q, vitamins A, D, E, and K, and cholesterol are all derived from which of the following lipids?

A. Isoprenes and terpenes
B. Sphingolipids
C. Phospholipids
D.Triglycerides

A

A. Isoprenes and terpenes

Explanation
Coenzyme Q, all steroids, cholesterol, and vitamins A, D, E, and K are derived from isoprenes or terpenes. These agents contain or at some point originated from precursors that were comprised of isoprene units. Isoprene units have the chemical formula C5H8.

Sphingolipids are important in cell membranes, especially those located in the central nervous system such as myelin sheath. Shingolipids contain shingosine as a backbone and are then further classified depending upon which molecules are attached to that backbone, such are ceremides, gangliosides, sphingomyelin, etc.

Phospholipids contain a polar and non-polar end, thus making them amphoteric. This property allows for the formation of bilayers (polar ends aligned together and pointed outwards) resulting in the lipid bilayer commonly seen in cell membranes. Phospholipids are generally comprised of a phosphate group, a choline group (polar), and two fatty acid chains (non-polar) attached to glycerol, which serves as the backbone.

Triglycerides are comprised of three fatty acid chains attached to a glycerol backbone. Triglycerides are important in long-term energy storage for use by cells.

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111
Q

Dietary triglycerides are metabolized primarily by which organ of the body?

A. Stomach
B. Liver and gall bladder
C. Intestine
D. Kidney

A

C. Intestine

Explanation
Fat is digested primary in the intestine. It enters the intestine from the stomach, where it becomes emulsified by bile salts and hydrolyzed by lipases released from the pancreas. These end products are then absorbed by enterocytes that line the walls of the intestine. Once in the enterocytes, triglycerides are then rebuilt and packaged along with cholesterol and protein to form chylomicrons. The chylomicrons are able to enter the lymph system, where they are absorbed into the blood stream and transported to the liver or adipose tissue. Triglycerides cannot directly diffuse through the cell membranes of the liver or adipose tissue; they must first be broken down into fatty acids and glycerol. This process is made possible by lipoprotein lipases located on the walls of blood vessels. The fatty acids and glycerol are then absorbed by the liver for energy or taken up by adipose tissue and re-synthesized into triglycerides for storage purposes.

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112
Q

Fatty acid synthesis is activated during which of the following situations?
A. Increased levels of citrate and insulin, decreased levels of glucagon
B. Increased levels of citrate, decreased levels of glucagon and insulin
C. Decreased levels of citrate, increased levels of glucagon and insulin
D. Increased levels of citrate and glucagon, decreased levels of insulin
E. Decreased levels citrate and insulin, increased levels of glucagon
F. Decreased levels of citrate and glucagon, increased levels of insulin

A

A. Increased levels of citrate and insulin, decreased levels of glucagon

Explanation
Fatty acid synthesis is stimulated by insulin and inhibited by glucagon and epinephrine. The formation of fatty acids is catalyzed via the enzyme acetyl CoA reductase, which is activated by citrate. Acetyl CoA reductase is allosterically inhibited by palmitoyl-CoA. Fatty acid synthesis occurs in the cytosol and requires the use of ATP and NADPH, as this is a complex anabolic process.

A brief overview and important points of the pathway are as follows: Essentially, acetyl CoA combines with oxaloacetate (OAA) in the mitochondria to form citrate, which is then shuttled out into the cytosol by the citrate shuttle. Once in the cytosol, citrate is broken down again into its constituents, OAA and acetyl CoA. OAA is converted back into pyruvate to re-gain entry into the mitochondria. Acetyl CoA is converted to malonyl-CoA via acetyl CoA carboxylase (rate-limiting step and uses biotin as a cofactor). Malonyl CoA and the acyl carrier protein undergo several reactions to create fatty acid.

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113
Q

Glycolysis occurs at which location within a eukaryotic cell?

A. Mitochondrion
B. Cellular membrane
C. Endoplasmic reticulum
D. Cytosol
E. Nucleus
A

D. Cytosol

Explanation
Glycolysis is an important metabolic pathway that breaks down glucose into pyruvate. Then, pyruvate can either be converted anaerobically into lactate or undergo oxidative phosphorylation yielding 36 moles of ATP. Glycolysis occurs in the cytosol of a cell.

To help remember the above, think of this little rhyme: glycol occurs in the cytosol.

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114
Q

A 10-year old child presents in your office with a unilateral follicular conjunctivitis along with ipsilateral adenopathy. You correctly diagnose oculoglandular syndrome. Because it is the most common etiology, which of the following causes are you MOST likely to suspect?

A. Cat-scratch disease
B. Toxoplasmosis
C. Coccidioidomycosis
D. Measles
E. Diabetes
A

A. Cat-scratch disease

Explanation
Oculoglandular syndrome can be caused by a myriad of organisms and presents as a unilateral follicular conjunctivitis along with lymphadenopathy on the same side as the affected eye. Other signs and symptomology vary depending on the causative organism. Causes include but are not limited to: cat-scratch disease, tularemia, syphilis, tuberculosis, sprotrichosis, mononucleosis, coccidioidomycosis, sarcoidosis, Hansen’s disease, mumps, actinomycosis, Listeria and Herpes simplex.

Based solely upon the age of the child, one would first assume cat-scratch disease, which is the most common cause of oculoglandular syndrome. This assumption would be verified by asking if the child had recently been scratched by a cat and by performing the Hanger-Rose skin test for confirmation.

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115
Q

A 10-year old male is seen at your office complaining of itchy eyes and severe photophobia. He has a history of eczema and hay fever. Biomicroscopy reveals bilateral cobblestone papillae of the superior eyelids, ropy discharge, and mild superior corneal disruption that stains with sodium fluorescein. Given the above findings, what is your diagnosis?

A. Vernal keratoconjunctivitis (VKC)
B. Epidemic keratoconjunctivitis (EKC)
C. Bacterial conjunctivitis
D. Iritis

A

A. Vernal keratoconjunctivitis (VKC)

Explanation
VKC is a condition of the young and presents with an increased frequency in males. This type of allergy typically develops before age 14 and lasts for 4-10 years before the child outgrows it; VKC occurs predominantly in the spring and summer. The condition progressively lessens in severity, with the first episode being the worst. Usually VKC is seen in patients who are prone to atopy; therefore they suffer from eczema, asthma, or hay fever. Patients typically suffer from itchy eyes and photophobia. The condition basically presents as a very severe type of allergic conjunctivitis. Signs include cobblestone papillae of the upper lid, lid swelling, and ropy discharge that is worse in the morning. Corneal defects (usually superiorly) known as keratitis of Togby may also be present. Occasionally, patients will develop a shield ulcer and Trantas’ dots (calcified eosinophils seen circumlimbally that appear as chalky concretions), which may lead to the feeling of an associated foreign body sensation. Treatment includes mast cell stabilizers that should be started several weeks prior to re-occurring episodes, pulse steroid therapy, cool compresses, and sunglasses to help alleviate ensuing photophobia.

EKC and bacterial conjunctivitis typically do not cause extreme itching and should not present with cobblestone papillae.

The number one symptom associated with iritis is photophobia, and the patient should be neither complaining of itching nor have cobblestone papillae present on biomicroscopy.

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116
Q

A 24-year old female patient is seen at your office and reports that her eyes have been red for a few days. Biomicroscopy reveals bilateral diffuse superficial punctate keratitis (SPK) that stains with sodium fluorescein with no mucopurulent discharge. Based ONLY upon the corneal staining pattern, what is the MOST likely origin of her condition?

A. Bacterial
B. Viral
C. Dry eyes
D. Foreign body

A

B. Viral

Explanation
Diffuse SPK likely signals either a viral origin or a toxic reaction to solution or topical ophthalmic drops.

A bacterial etiology will cause staining of the inferior third of the cornea.

Interpalpebral corneal staining is usually caused by lagophthalmos or environmental desiccation.

A foreign body will either leave small punctate staining or tracks (especially if it is trapped under a contact lens).

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117
Q

A 32-year old male is seen at your office and is in a fair amount of pain. He can barely open his right eye and reports that the pain began this morning when he first opened his eyes. His medical history is unremarkable, and he does not wear contact lenses. His ocular history is remarkable for a mild corneal abrasion of the right eye from a tree branch that occurred over a month ago but had since healed. Biomicroscopy (after instillation of a topical anesthetic) reveals an epithelial defect 1.5 mm wide and 1.0 mm long that stains with sodium fluorescein. There is no anterior chamber reaction and no visible discharge. What is the MOST appropriate diagnosis?

A. Corneal abrasion
B. Corneal ulcer (microbial keratitis)
C. Recurrent corneal erosion
D. Epithelial basement membrane dystrophy

A

C. Recurrent corneal erosion

Explanation
This patient is suffering from a recurrent corneal erosion. These types of corneal defects frequently occur in response to a corneal abrasion incurred by something organic (like a fingernail or a tree branch). The initial abrasion heals, but a short time afterwards the patient will experience another episode without any incidence of trauma. The second occurrence tends to transpire first thing in the morning as the eyelids stick to that unstable flap of tissue overnight and rip it off like a band-aid when the eyes open. The best way to treat a recurrent corneal erosion is through the use of a topical antibiotic (unpreserved is best) to ensure sterility (as the cornea is exposed) as well as a bandage contact lens to speed up the healing process if the area of erosion is large. Hyperosmotic drops or artificial tears (preservative-free of course) should be prescribed for roughly 6-8 weeks (sometimes longer) to ensure healing and to allow for proper formation of hemidesmosomes that will help to alleviate future episodes. Other treatments include stromal micropuncture, debridement, phototherapeutic keratectomy (PTK), or oral tetracycline, which inhibits matrix metalloproteinases and allows for proper corneal healing.

A corneal abrasion occurs secondary to some type of trauma or injury, and this was not the case in the above example. Recurrent corneal erosions are a common occurrence with epithelial basement membrane dystrophies, but that is not the resultant diagnosis of the current problem experienced by the patient. An ulcer is ruled out on the basis that there is no active infection and is unlikely, as these more commonly occur in patients who wear contact lenses.

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118
Q

A 42-year old construction worker presents to you with a history of getting plaster in his right eye. He complains of pain, foreign body sensation, and photophobia. His acuity is reduced in that eye to 20/50 with a normal pupillary response. What type of chemical trauma did the worker experience, and what would be your first therapeutic intervention?

A. Acid burn and lavage eye with balanced saline solution for 30 minutes
B. Alkali burn and lavage eye with balanced saline solution for 30 minutes
C. Ultraviolet (UV) burn and lavage eye with balanced saline solution for 30 minutes
D. Thermal burn and lavage eye with balanced saline solution for 30 m

A

B. Alkali burn and lavage eye with balanced saline solution for 30 minutes

Explanation
Lime, particularly in the form of plaster, is the most commonly encountered alkali injury. Fortunately, it tends to cause a less severe burn than other types of alkali burns.

The rapidity with which pH abnormalities of the ocular surface are neutralized has a significant impact on the subsequent clinical course. Irrigation for a minimum of 30 minutes and checking pH of tears for evidence of neutrality is recommended.

The explanations for the alternative answers are as follows:

Acid burn and Lavage eye with balanced saline solution for 30 minutes;
plaster contains lime, which is alkali and not acid.

Thermal burn and lavage eye with balanced saline solution for 30 minutes;
there is no mention of heat (cigarette or matches) during the injury to the eye.

Ultraviolet (UV) burn and lavage eye with balanced saline solution for 30 minutes; UV burns usually are associated with delayed response after exposure to UV light.

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119
Q

A newborn presenting with symptoms of ophthalmia neonatorum 3 days after birth is MOST likely infected with which of the following organisms?

A. Chlamydia trachomatis
B. Staphylococcus aureus
C. Haemophilus influenza
D. Streptococcus pneumonia
E. Neisseria gonorrhoeae
F. Herpes simplex virus
A

E. Neisseria gonorrhoeae

Explanation
Ophthalmia neonatorum is a conjunctivitis that typically develops within the first 3 weeks after birth as a result of transmission of infection from mother to child during delivery. This condition is particularly serious due to the lack of immunity in infants as well as the immaturity of the ocular surface (poor tear film and undeveloped lymphoid tissue).

Ophthalmia neonatorum secondary to N. gonorrhoeae typically develops within 2-5 days postpartum as hyperacute conjunctivitis. Most cases present bilaterally with periorbital edema, conjunctival chemosis, and excessive amounts of purulent discharge. It is extremely important to quickly and aggressively treat this infection due to the ability of N. gonorrhoeae to penetrate an intact corneal epithelium.

When C. trachomatis is the organism responsible for ophthalmia neonatorum, mild to moderate symptoms of unilateral or bilateral conjunctivitis commonly occur between 5 to 14 days after birth. C. trachomatis is the most common cause of ophthalmia neonatorum. These patients present with lid edema, conjunctival chemosis, punctate corneal opacities, and occasionally micropannus formation.

Other etiologies of ophthalmia neonatorum can include S. aureus, Haemophilus species, S. pneumoniae, E. coli, and P. aeruginosa. These pathogens are part of the normal bacterial flora of the female genital tract and are likely acquired as the newborn travels through the birth canal.

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120
Q

A patient walks into your office with a mild corneal abrasion, what is the correct healing sequence?

A. Basal cells at the wound margin flatten and spread-> attachment of cells via fibronectin and laminin-> mitosis of cells surrounding wound area-> hemidesmosome formation between basal cells

B. Basal cells at the wound margin flatten and spread-> mitosis of cells surrounding wound area-> attachment of cells via fibronectin and laminin-> hemidesmosome formation between basal cells

C. Mitosis of cells surrounding wound area-> attachment of cells via fibronectin and laminin-> hemidesmosome formation between basal cells-> basal cells at the wound margin flatten and spread

D. Hemidesmosome formation between basal cells-> basal cells at the wound margin flatten and spread-> mitosis of cells surrounding wound area-> attachment of cells via fibronectin and laminin

A

B. Basal cells at the wound margin flatten and spread-> mitosis of cells surrounding wound area-> attachment of cells via fibronectin and laminin-> hemidesmosome formation between basal cells

Explanation
Following minor corneal injuries, the main concern is for the cornea to seal off the wound and form tight junctions as quickly as possible to prevent edema and infection. The first phase following an abrasion is the sloughing off of superficial cells at the wound margin. This is followed by a flattening of basal cells surrounding the wound which then move into the damaged area via filpodia (extensions that allow for amoeboid-type motion). Mitosis rates increase surrounding the injury site to replace those cells that have migrated. The basal cells then secrete fibronectin and laminin to achieve transitory cellular attachment until the formation of hemidesmosomes. Once a basal cell comes in contact with another basal cell, a hemidesmosome forms signaling for a decrease in mitosis.

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121
Q

All else being equal, cells found in which layer of the cornea consume the GREATEST amount of oxygen?

A. Endothelium
B. Epithelium
C. Bowman’s membrane
D. Stroma

A

A. Endothelium

Explanation
The cells of the endothelium require the greatest amount of oxygen. This is due to the fact that endothelial cells maintain a steady state of corneal clarity and hydration. They actively pump out ions into the anterior chamber, which sets up an osmotic gradient that causes water to flow down its concentration gradient, thus preventing corneal swelling and opacification. However, because the endothelium is only one layer thick, in total this layer consumes 21% of the oxygen provided to the cornea. The stroma utilizes 39% of oxygen made available to the cornea, which is a low consumption rate considering that it makes up the bulk of the cornea. The epithelium is responsible for 40% of the oxygen consumed by the cornea. However, all else being equal, the endothelial cells consume the greatest amount of oxygen (140 X 10-5 ml of oxygen per sec), vs. stromal cells (2.85 X 10-5 ml of O2/sec) and epithelial cells (26.5 X 10-5 ml of O2/sec).

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122
Q

Axenfeld loops are smooth, dome-shaped, greyish-appearing nodules located under the bulbar conjunctiva that are a result of intrascleral looping of which of the following nerves?

A. Long anterior ciliary nerves
B. Long posterior ciliary nerves
C. Short posterior ciliary nerves
D. Short anterior ciliary nerves

A

B. Long posterior ciliary nerves

Explanation
Axenfeld loops are common findings on slit lamp examination that present as small, smooth, dome-shaped nodules most commonly located in the superior sclera. They often appear greyish in color, with occasional brown pigment surrounding the loop (observed particularly in patients with a darker iris). Axenfeld loops represent an anastomosis of a long ciliary nerve that turns to enter the sclera before looping back again to continue to its insertion at the ciliary body.

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123
Q

Chrysiasis of the cornea occurs secondary to administration of medication used to treat which of the following conditions?

A. Hypertension
B. Cancer
C. Diabetes
D. Rheumatoid arthritis
E. Cardiac arrhythmias
A

D. Rheumatoid arthritis

Explanation
Gold salts are used to manage rheumatoid arthritis, primarily when other treatment options have failed. Chrysiasis occurs secondary to the deposition of gold in the skin, lens, and cornea, causing a gray discoloration of the skin and brown/gold deposits in the deep stroma of the cornea.

Amiodarone is an anti-arrhythmic medication. Use of this drug commonly causes yellow/brown or white powdery corneal epithelial deposits located inferocentrally that appear to swirl outward while sparing the limbus. These deposits affect visual acuity minimally, if at all. The aforementioned deposits can also be observed in Fabry’s disease and in patients taking tamoxifen, chlorpromazine, chloroquine, or indomethacin.

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124
Q

During periods of severe hypoxia, the cornea will revert to anaerobic metabolism and break down glycogen. What layer of the cornea is capable of storing glycogen for use during times of hypoxia?

A. Endothelial layer
B. Epithelial layer
C. Descemet’s membrane
D. Stroma

A

B. Epithelial layer

Explanation
The epithelial cells store glycogen, which is used as an energy source when oxygen is not available. The stores can last for about 2 hours before being depleted. Once glycogen is no longer accessible, the cornea will not produce enough ATP and the epithelial cells will begin to die.

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125
Q

A back surface toric (spherical front surface) gas-permeable (GP) contact lens is ordered with base curve radii of 7.85 mm (43.00 D) and 8.44 mm (40.00 D). When verifying this lens with a lensometer you would expect to find approximately how many diopters of “induced cylinder”?

A. 1.50 D
B. 4.50 D
C. 3.00 D
D. 2.00 D

A

B. 4.50 D

Explanation
Assuming the 1-2-3 rule is correct, a base curve toric GP lens with a spherical front surface when analyzed will exhibit a difference in lensometry readings that are 3/2 the amount of the base curve difference measured in diopters. In the above case, the difference in the two measured base curve meridians is 3 diopters; therefore, if there is no toricity on the front surface (that is, this is not a bitoric GP); optically, the difference in the two raw powers measured by lensometry will be 4.50 diopters. For example, the powers could be measured to be 1.00 D in one meridian and -5.50 D in the meridian 90 degrees away, or +2.00 D and -2.50 D. Keep in mind that the 1-2-3 rule is based on the index of refraction (n) of the lens material. Most of today’s GP lenses have ‘n’ values that are in the 1.40 to 1.48 range. This range of ‘n’ will result in less difference in the measured lensometry powers. For our example of a 3.00 D base curve toric GP lens, the difference in lensometry powers might be 4.00 D for a GP lens fabricated in a material with a lower ‘n’.

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126
Q

A patient has a spectacle correction of +12.00 D. The vertex distance is 13 mm. What is the power at the corneal plane?

A. +12.00 D
B. +14.25 D
C. +11.25 D
D. +10.37 D

A

B. +14.25 D

Explanation
When a plus lens is moved from the spectacle plane to the cornea, you need to adjust the power by adding plus. So the only option in this question that is more plus than +12.00 D is +14.25 D. You can also use the following formula to arrive at the correct answer:

Fc = Fs/1-dFs Fc = Power at the corneal plane (diopters) 
Fs = Power at the spectacle plane (diopters) 
d = vertex distance (meters) 

For our question Fc = +12.00/1-(0.013)(+12.00) = +14.22 D

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127
Q

An aphakic patient is seen at your office and wishes to be fit with contact lenses. What is an important contact lens parameter that MUST be considered in this patient’s care?

A. Ultraviolet (UV) inhibitor
B. Contact lens solution
C. Edge thickness
D. Contact lens material

A

A. Ultraviolet (UV) inhibitor

Explanation
Because this patient is aphakic, their retinas no longer receive the UV protection that is naturally provided by the crystalline lens. Although all of the above options should be included when deciding which type of lens to order, it is essential that you provide a UV inhibitor on the contact lens as well as sunglasses for this patient. When the contact lens power will be a high plus prescription, one should order a lenticular lens design to reduce lens thickness, help enhance centration, increase comfort as well as increase the Dk/t of the contact lens.

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128
Q

Hypoxia associated with hydrophilic (soft) contact lens wear can result in which of the following?

A. Corneal swelling
B. Corneal decompensation
C. 3/9 staining, scarring and pseudoptyergium
D. Blepharitis

A

A. Corneal swelling

Explanation
Hypoxia can cause corneal swelling (edema) acutely and corneal thinning chronically (by mobilization of glycosaminoglycans), can lead to secondary cornea neovascularization, both superficial pannus and occasionally deep stromal vessels, and endothelial changes including polymegathism and decreased cell numbers. Contact lens hypoxia, however, does not lead to corneal decompensation, blepharitis or peripheral 3/9 lesions which are more related to chronic rigid lens-induced exposure keratitis.

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129
Q

Dissection of the eye reveals that the ciliary body is actually triangular in shape. The apex of this triangle points in which direction and is continuous with which structure?

A. Points anteriorly; continuous with iris
B. Points posteriorly ; continuous with scleral spur
C. Points posteriorly; continuous with choroid
D. Points anteriorly: continuous with ora serrata

A

C. Points posteriorly; continuous with choroid

Explanation
The ciliary body is a band roughly 6mm wide and runs circumferentially internal to the sclera and posterior to the limbus. This structure is triangular in shape whose apex points posteriorly and is continuous with the choroid.

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130
Q

Free radicals can cause severe damage to tissue. Which of the following electrolytes can function as an antioxidant in the aqueous?

A. Sodium ions
B. Ascorbate
C. Albumin
D. IgG
E. Chloride ions
A

B. Ascorbate

Explanation
The aqueous humor contains many electrolytes including Na+, K+ , Cl-, HCO3-, glucose, lactate, amino acids, and ascorbate. Ascorbate is found in high concentrations in the aqueous (20x greater when compared to the concentration found in plasma). Ascorbate can serve as an antioxidant to eradicate free radicals reducing potential damage from ultraviolet light. Interesting note: the aqueous humor and tears of uncontrolled diabetics display higher levels of glucose than those of non-diabetics.

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131
Q

Iris colobomas form due to incomplete closure of the choroidal fissure. This usually results in a keyhole-shaped defect in which region of the iris?

A. Inferonasal
B. Inferotemporal
C. Superotemporal
D. Superonasal

A

A. Inferonasal

Explanation
An iris coloboma is an inferonasal, keyhole-shaped defect. The remainder of the iris is normal. Atypical colobomas may develop at sites other than the inferonasal area.

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132
Q

A 22-year old male presents in your office for his annual exam. Biomicroscopy reveals bilateral Krukenberg spindles and iris transillumination defects. Given the above findings, what is the correct diagnosis for this patient?

A. Posner-Schlossman syndrome
B. Pigmentary dispersion syndrome
C. Salzmann’s nodular degeneration
D. Pseudoexfoliation

A

B. Pigmentary dispersion syndrome

Explanation
Pigmentary dispersion syndrome generally occurs in young, myopic males with deep anterior chambers. This condition is caused by a disruption of the posterior iris pigment epithelium, causing this layer to rub against the ciliary zonules and release pigment into the anterior chamber and its associated structures. The aqueous in the anterior chamber displays convection currents. Warming inferiorly, aqueous rises and migrates forward towards the posterior surface of the endothelium, carrying pigment granules with it, then falls as it cools, causing the pigment to deposit vertically in a linear-fashion on the endothelium. Monitor these patients for blockage of the trabecular meshwork causing a rise in intraocular pressure (IOP) and potential glaucomatous damage to the optic nerve. Physical exertion as well as mydriasis may exacerbate pigment release; therefore, when dilating (especially with phenylephrine), you must be sure to measure post-dilated IOPs.

Salzmann’s nodular degeneration appears as blue/white hyaline plaque deposits between the epithelium and Bowman’s membrane, generally around the pupillary area of the cornea. This condition stems from other pathologies, primarily old phlyctenula. Treatment is generally not required unless vision is affected.

Posner-Schlossman syndrome causes an acute IOP spike, usually unilaterally and lasting for hours to weeks with recurrent episodes. Patients are often young and report decreased vision due to corneal edema, mild pain, and ocular redness. This syndrome is also known as glaucomatocyclitic crisis. Biomicroscopy will often reveal ciliary flush, a sluggish or dilated pupil, a mild anterior chamber reaction, potentially with keratic precipitates, corneal edema, open angles and normal optic nerves. IOP readings will normally range from 40-60 mmHg. The etiology is still uncertain; some have postulated that it may be of viral origin. Treatment includes (unless contraindicated) topical steroid drops, beta-blockers, and carbonic anhydrase inhibitors.

Pseudoexfoliation appears as white, flaky material that deposits along the pupillary margin, the anterior surface of the lens and other structures of the anterior chamber. This condition is usually unilateral and is seen in the elderly with a concurrent cataract. Transillumination defects, if present, are limited to the iris sphincter region. The pseudoexfoliative material can accumulate in the trabecular meshwork, causing an increase in IOP leading to glaucoma.

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133
Q

According to the Van Herick technique of angle estimation, which angle grade is considered MOST narrow and potentially capable of closure?

A. Grade 1
B. Grade 3
C. Grade 2
D. Grade 4

A

A. Grade 1

Explanation
The Van Herick method of angle estimation, although not infallible, is a good way of assessing the probability of angle closure with dilation. The temporal angle of a patient is evaluated by having the patient look straight ahead, placing the oculars of the slit lamp in the straight ahead position and shifting the lighthouse temporally so that an angle of 60 degrees is created. The beam is then narrowed to an optic section and placed at the temporal limbus. The width of the anterior chamber space at the angle is then compared to the width of the corneal optic section. The procedure is repeated and the illumination source is placed nasally to assess the grade of the nasal angle. If the angle space is equal to 1/2 or greater than the corneal optic section, the angle is not considered occludable and is given a grade of 4. A grade three displays an angle space that is less than 1/2 but greater than 1/4 of the width of the corneal optic section, and is considered safe to dilate. A grade two has an angle space that is roughly equal to 1/4 the width of the corneal optic section, and may be dilated but with caution. A grade 1 has an angle space of less than 1/4 the width of the corneal optic section, and requires gonioscopy to ensure that the patient is safe to dilate.

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134
Q

At what time period of the day would the aqueous humor production be least?

A. 8 AM to noon
B. Midnight to 6 AM
C. Afternoon

A

B. Midnight to 6 AM

Explanation
The precise role and receptors specificity of adrenergic mechanisms in regulating the rate of aqueous humor formation are unclear. Studies using fluorophotometry have shown that beta adrenergic antagonists unequivocally decrease aqueous humor formation, particularly aqueous humor production during sleeping hours, during which production is decreased by up to 50%. This decrease is due to the B-arrestin/cAMP cascade regulation from the beta-adrenergic receptors. This does not mean that the intraocular pressure necessarily decreases at night; in fact, nocturnal intraocular pressure spikes have recently been reported; these may have a role in causing progressive glaucomatous damage.

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135
Q

Goldmann applanation tonometry is calibrated for a central corneal thickness of roughly 520 micrometers. How will a cornea measuring 600 micrometers thick affect the intraocular pressure (IOP) measurements, and how should the IOP reading be compensated?

A. IOP will be falsely low; the measured IOP should be adjusted up
B. IOP will be falsely elevated; the measured IOP should be adjusted up
C. IOP will be falsely elevated; the measured IOP should be adjusted down
D. IOP will be falsely low; the measured IOP should be adjusted down

A

C. IOP will be falsely elevated; the measured IOP should be adjusted down

Explanation
A thicker than average cornea will cause a falsely elevated IOP measurement due to greater than normal corneal rigidity requiring more pressure to achieve proper indentation with the tonometer. Therefore, the resultant IOP should be adjusted down. The magnitude of the adjustment depends on how much the central corneal thickness deviates from the calibrated thickness of 520 micrometers.

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136
Q

How is the temporal modulation transfer function expected to change in a person with early glaucoma not yet manifesting any defects on visual field testing?

A. Decreased sensitivity to low and moderate frequencies
B. Decreased sensitivity to moderate frequencies only
C. Decreased sensitivity to low temporal frequencies only
D. Decreased sensitivity to moderate and high temporal frequencies

A

D. Decreased sensitivity to moderate and high temporal frequencies

Explanation
Early glaucomatous damage can be difficult to detect because intraocular pressure and visual fields can be normal. Recent studies demonstrate that the magno cells may be damaged early on in glaucoma. The magno cells are a part of the “where” pathway and therefore display excellent temporal resolution. Due to this factor, there is evidence that clearly reveals a correlation between an altered temporal modulation transfer function and early glaucoma, with a marked decrease noted for the moderate and high temporal frequencies, even though the respective visual field is free of defects.

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137
Q

Instillation of too much fluorescein when attempting Goldmann applanation tonometry will result in what type of erroneous result?

A. Mires that are misaligned superiorly

B. A corneal abrasion

C. A measured intraocular pressure that is higher than the true intraocular pressure

D. Mires that are misaligned inferiorly

E. A measured intraocular pressure that is lower than the true intraocular pressure

A

C. A measured intraocular pressure that is higher than the true intraocular pressure

Explanation
There are many potential sources of error when performing Goldmann applanation tonometry. If the mires created by the prisms in the tonometer are not aligned properly, the resulting IOP measurement will not be accurate. This also applies if too little or too much fluorescein is instilled. Too little fluorescein causes thin mires and an underestimation of the IOP reading. Too much fluorescein results in large mires and a falsely elevated measurement. If the mires are misaligned inferiorly (that is, the majority of the mires are located below the prism line) then the slit lamp needs to be lowered to center the mire images. If the mires are misaligned superiorly then the slit lamp needs to be raised to center the mires.

A corneal abrasion may occur if the slit lamp is moved excessively or if too much force is placed by the tonometer head against the cornea.

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138
Q

An elderly patient presents in your office with decreased visual acuity. He remarks that he can read better without his glasses and his refraction denotes a large myopic shift. Dilated fundus exam is unremarkable. Which of the following slit lamp findings would MOST likely explain the above findings?

A. Bilateral limbal girdle of Vogt
B. Bilateral corneal arcus
C. Bilateral 3+ nuclear sclerosis of the lens
D. Bilateral crocodile shagreen

A

C. Bilateral 3+ nuclear sclerosis of the lens

Explanation
Nuclear sclerosis is caused by changes to the optical clarity of the lens. As we age, proteins precipitate out of the lens matrix, causing the lens to become cloudy and altering its density. As time passes, the lens will also begin to change color from clear to a yellow/brown in a process called lens brunescence. Cataracts also generally cause a myopic shift with an increase in against-the-rule astigmatism, leading to decreased distance vision but improved near vision.

Corneal arcus is caused by lipid deposition in the peripheral cornea. There remains a characteristic clear zone between the lipid and the limbus. Arcus does not generally interfere with vision.

Crocodile shagreen and limbal girdle of Vogt are also benign corneal findings commonly seen in the elderly. Crocodile shagreen appears in the peripheral cornea as polygonal white opacities. Limbal girdle of Vogt is noted at the 3 o’clock and 9 o’clock interpalpebral positions as white crescent-shaped opacities.

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139
Q

Long-term use of corticosteroids can lead to the formation of which of the following types of cataract?

A. Cortical
B. Posterior subcapsular
C. Anterior subcapsular
D. Nuclear sclerotic

A

B. Posterior subcapsular

Explanation
The possible formation of posterior subcapsular cataracts (PSC) is a common concern in patients undergoing long-term treatment with corticosteroid therapy. PSCs have been associated with the use of systemic, topical, ophthalmic, topical dermatologic, nasal aerosol, and inhalation type steroids. This relationship is likely dose-dependent, and the usual time from beginning steroid treatment to the onset of lens changes is 1 year (with a dosage of 10 mg/day of prednisone) but has been observed in as short as 2 months with as little as 5 mg/day. Patients with PSC formation may complain of an increase in light sensitivity, photophobia, glare, or difficulty reading. If visual acuity is notably decreased, surgical removal of the lens may be warranted.

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140
Q

Berger’s space is created by an interval between which two structures?

A. The anterior face of the lens and the posterior surface of the iris
B. The equator of the lens and the ciliary body
C. The posterior surface of the cornea and the anterior face of the iris
D. The posterior face of the lens and the anterior vitreous

A

D. The posterior face of the lens and the anterior vitreous

Explanation
Berger’s space is created by the separation between the posterior face of the lens and the anterior face of the vitreous.

The space between the equator of the lens and the ciliary body is known as the circumlental space

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141
Q

Patients with a history of homocystinuria are MOST likely to experience crystalline lens subluxation in which of the following directions?

A. Up and inward
B. Down and inward
C. Down and outward
D. Up and outward

A

B. Down and inward

Explanation
Common ocular sequelae that have been associated with a diagnosis of homocystinuria include ectopia lentis (bilateral crystalline lens subluxation), retinal detachment, and secondary glaucoma. In most cases of ectopia lentis, the lens is more likely to be displaced downward and inward in homocystinuria (as compared to upward and outward in Marfan’s syndrome). Additionally, in homocystinuria, the lens zonules are markedly abnormal, the lens does not accommodate, and up to 1/3 of the cases of lens subluxation eventually completely dislocate into the vitreous or anterior chamber. Due to the severity of systemic and cardiovascular complications associated with homocystinuria (thrombosis and occlusion), patients presenting with ectopia lentis should be screened for this disease using the sodium nitroprusside test to measure homocysteine in the urine.

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142
Q

Purkinje images are caused by reflections of objects on the cornea and lens. Which of the four images moves forward with accommodation?

A. I
B. III
C. IV
D. II

A

B. III

Explanation
There are four Purkinje images. The first image is caused by reflection from the anterior corneal surface and is the brightest of the images. The first image is roughly the same size as the object. The second Purkinje image is formed by the posterior surface of the cornea and almost coincides with the first Purkinje image. The third Purkinje image is the largest and is caused by reflection off of the anterior plane of the crystalline lens. The fourth Purkinje image is the smallest and is inverted, formed by reflection off of the posterior surface of the lens.

During the process of accommodation the anterior surface of the lens moves forward. The image that is reflected off of this surface is Purkinje III. Purkinje image III will be seen to move forward during accommodation.

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143
Q

The ligaments that suspend the lens (zonules) are embryonically derived from what structure?

A. The primary vitreous
B. The tertiary vitreous
C. The lens capsule
D. The lens epithelium

A

B. The tertiary vitreous

Explanation
The zonules are attached to the posterior and anterior surfaces of the lens and connect to the pars plana of the ciliary body. The primary vitreous develops from weeks 3 through 9. The secondary vitreous then begins to form and condenses the primary vitreous forming Cloquet’s canal. Developmentally, the tertiary vitreous is secreted last; the zonules are comprised of condensed tertiary vitreous.

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144
Q

A 42-year old patient reports that her right eye has been watery and she has mild pain, redness, and swelling in the lower medial canthal region. You suspect dacryocystitis as the cause of her symptoms. Which of the following procedures is NOT appropriate when further evaluating this possible diagnosis?

A. Exophthalmometry
B. Dilation and irrigation of the lacrimal system
C. Extraocular muscle motility
D. Digital palpation of the medial canthal area
E. Gram stain and blood agar cultures of discharge

A

B. Dilation and irrigation of the lacrimal system

Explanation
The evaluation of a patient suspected of dacryocystitis should involve a detailed case history including a discussion of any previous episodes with similar symptoms, or the presence of any concomitant ear, nose, or throat irritation/infection. External examination of the patient should include the application of gentle pressure to the lacrimal sac region in order to attempt to express any discharge from the punctum; this should be done bilaterally. If any discharge can be recovered, a Gram stain or blood agar culture is helpful in determining the type of bacteria present. In addition to these tests, extraocular motility and evaluation for the presence of proptosis should be completed to rule out orbital cellulitis. In atypical, severe, or non-responding cases, a computed tomography scan (CT) should be considered. It is important to remember that probing, dilation, and/or irrigation of the lacrimal system should not be attempted during an acute infection of the lacrimal gland. This may cause the infection to spread to other areas such as the throat

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145
Q

A common cause of epiphora in infants is caused by a small membrane that covers over which of the following structures?

A. The puncta
B. The valve of Hasner
C. The lacrimal gland
D. The canaliculus

A

B. The valve of Hasner

It is common for mothers of young infants to note that one eye (or both eyes) of her infant constantly tears in conjunction with the presence of mucopurulent discharge. This epiphora results from a blockage of the nasolacrimal passageway caused by a membrane covering the valve of Hasner. The majority of blockages will self-resolve without intervention (80-90% of infants) within the first 12 months of life. Treatment may include massage of the nasolacrimal sac several times a day in an effort to rupture the membrane.

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146
Q

An 81-year old female reports that her eye has been watering more frequently over the past month; you decide to administer the primary Jones dye test (Jones I). After 5 minutes, the application of a cotton-tipped applicator to the inferior turbinate reveals the presence of dye in the area. Taking this into consideration, what is the MOST likely cause of the patient’s epiphora complaint?

A. Dysfunction of the valve of Hasner
B. Punctal stenosis
C. Hypersecretion of tears
D. Partial nasolacrimal duct obstruction
E. Complete nasolacrimal duct obstruction
A

C. Hypersecretion of tears

The primary Jones dye test can be utilized to determine the patency of the nasolacrimal system. 1-2 drops of fluorescein are instilled into the inferior fornix of the eyes while the patient is in an upright position and blinking her eyes normally. After a period of 5 to 10 minutes, a cotton-tipped applicator is used to swab the undersurface of the inferior turbinate on each side of the nasal passage.

When the primary Jones dye test is positive (dye is recovered from the inferior turbinate of the nose), practitioners may conclude that the system is patent and that no significant blockage of the nasolacrimal drainage structure is likely. However, minor stenosis or physiologic dysfunctions cannot be completely ruled out. Patients who have a positive result on the Jones I test are more likely to experience symptoms of epiphora that are secondary to primary oversecretion of tears, rather than a dysfunction in lacrimal drainage (as in the above question).

When the primary Jones dye test is negative, the probability of an obstruction or dysfunction in lacrimal drainage is much greater; however, this test alone is not sufficient to document this conclusion. The secondary Jones dye test is then necessary to determine the severity and location of the obstruction.

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147
Q

Chronic blepharitis, if left untreated, can cause which of the following structural changes to the anterior ocular segment?

A. Tristichiasis
B. Madarosis
C. Distichiasis
D. Hypertelorism

A

B. Madarosis

Blepharitis is a condition caused by pathogens, usually of Staphylococcus origin, that colonize along the eyelid margins. The bacteria produce exotoxins which take the form of flakes and are generally seen along the base of the eyelashes. Unfortunately, this condition is chronic but will wax and wane in its presentation. Long-term complications include madarosis (missing lashes), trichiasis, neovascularization of the eyelid margin, keratitis, erythema, phlyctenule formation and infiltrates. Patients may complain of dry, irritated eyes, stinging, pain, itching, frequent eye infections, foreign body sensation, and decreased acuity (if there is corneal involvement). Treatment includes eye lid scrubs, antibiotic ointments and sometimes transient topical steroid use to decrease lid inflammation (usually used in conjunction with a topical antibiotic). Occasionally oral antibiotics are prescribed, especially in the event of poor compliance.

Distichiasis is a rare congenital phenomenon marked by an absence of meibomian glands. In the place of the meibomian glands is an extra row of eyelashes.

Hypertelorism is a term used to describe the incidence in which the orbits are located quite far apart. This generally occurs along with other congenital cranium anomalies.

Tristichiasis is a very rare occurrence in which a person possesses three rows of eyelashes.

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148
Q

Dacryoadenitis refers to an inflammation or infection of which of the following ocular structures?

A. The puncta
B. The lacrimal sac
C. The lacrimal gland
D. The nasolacrimal sac

A

C. The lacrimal gland

Dacryoadenitis describes inflammation of the lacrimal gland, generally due to infection. The swelling is categorized as either chronic or acute. Acute presentations appear more commonly as a unilateral swelling of the upper eyelid, along with pain, excessive lacrimation, probable ipsilateral lymphadenopathy, and potential proptosis. If the condition is bilateral it is likely due to a systemic infection. Chronic dacryoadenitis is generally bilateral and presents with hard masses that are palpable at the location of the lacrimal gland. This form is often painless and caused by inflammatory diseases such as Grave’s, Sjogren’s, or sarcoidosis. The chronic type warrants further investigation in order to rule out a lacrimal gland tumor.

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149
Q

In addition to the meibomian glands which other accessory glands secrete oil?

A. Wolfring and Krause
B. Zeiss and Moll
C. Zeiss and Wolfring
D. Moll and Krause

A

B. Zeiss and Moll

The glands of Zeiss and Moll are accessory oil glands located on the lid margins adjacent to the base of the lash follicles. The lipid layer of the tear film is superficial and as such it is exposed to the environment protecting the aqueous layer from evaporation.

The glands of Wolfring and Krause are located deep in the fornix of the eyelids and serve to secrete a portion of the aqueous layer of the tear film.

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150
Q

Many skin anomalies may mimic malignant lesions. Which of the following skin conditions has the HIGHEST risk of becoming malignant?

A. Papilloma
B. Seborrhoeic keratosis
C. Actinic keratosis
D. Cutaneous horn

A

C. Actinic keratosis

Actinic keratosis is a precursor to squamous cell carcinoma and appears as scaly, dry skin that does not heal. People with skin that is of lighter pigmentation along with excessive exposure to ultraviolet light tend to be most at risk for development of this condition.

Papillomas may take on various forms and may be viral or non-viral in origin. They can commonly be found on the eyelids or surrounding orbital skin. Viral warts tend to grow at an accelerated rate while non-viral papillomas are fairly slow to grow. Papillomas can mimic neoplastic growths so be sure to rule this out while watching carefully for color change, ulceration, lash loss, bleeding, and vascularization.

Cutaneous horns or tags are also benign and are likely a form of papilloma but appear to involve more keratin. Treatment is similar to that of a papilloma.

Seborrhoeic keratosis is more commonly seen in middle-aged and elderly persons. This benign, epidermal growth is quite superficial and does not extend into the dermis. It appears like a brown plaque that has been stuck onto someone’s skin. The borders are very distinct and there may be some elevation. The lesions may be removed if the patient is concerned about cosmesis

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151
Q

Numerous reports have suggested that increased tear film osmolarity is a key consequence in dry eye. Although osmolarity is not easily measured in the clinical setting, tear osmolarity increases in most dry eye sub-types due to which of the following processes?

A. Reactive oxygen species are increased in the tears of most dry eye sub-types; this increases osmolarity

B. The lipid layer is altered in most dry eye states, leading to ion pairing

C. Decreased capillary exchange leads to ionic bonding
Patients with dry eye tend to blink less than normals, leading to increased evaporation

D. Loss of tear stability induces an increased evaporation rate, leading to increased osmolarity

E. In aqueous tear deficiency, the lacrimal gland produces more ionic species

A

D. Loss of tear stability induces an increased evaporation rate, leading to increased osmolarity

Tear instability leads to greater evaporation and higher osmolarity through a mechanism of concentration of the remaining tears, since only the aqueous tear portion evaporates rather than the ionic species. Several studies have indicated that normal tear osmolarity is less than or equal to 300 Osm/L, with values exceeding 308 Osm/L indicating increased osmolarity. As a single measure, tear osmolarity has recently been found to correlate the best (r squared 0.55) to dry eye severity of several clinical tests in a large, multi-center study (Sullivan et al., IOVS 51:6125-6130, 2010).

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152
Q

Oral acyclovir is most effective for patients presenting with eyelid findings associated with herpes zoster if administered within which of the following periods following the onset of the disease?

A. 4-5 days
B. 72 hours
C. 10-12 hours
D. 24 hours
E. 7-10 days
A

B. 72 hours

Oral acyclovir is the mainstay of therapy for patients diagnosed with herpes zoster ophthalmicus. This systemic treatment is maximally beneficial of it is initiated within 72 hours from the onset of the disease (usually the appearance of eyelid lesions). The use of oral acyclovir typically results in quick resolution of skin vesicles, decreases the amount of pain the patient experiences, and reduces the duration of viral shedding and appearance of new lesions. Acyclovir has also been shown to significantly reduce the incidence of ocular findings such as episcleritis, keratitis, and iritis. The recommended dosage is 800mg orally 5 times per day for 7-10 days.

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153
Q

Ptosis can be caused by dysfunction or damage to which of the following muscles?

A. Superior tarsal muscle (muscle of Muller)
B. Muscle of Horner
C. Inferior rectus
D. Pars ciliaris (Riolan’s muscle)

A

A. Superior tarsal muscle (muscle of Muller)

Ptosis is a condition in which the upper eyelid sags. It can be caused by dysfunction of either the superior palpebral levator or the superior tarsal muscle (muscle of Muller). Because the levator is the major muscle responsible for raising the upper eyelid, ptosis from levator damage is often more severe then ptosis from dysfunction of the muscle of Muller.

The muscle of Horner (also known as the pars lacrimalis) is part of the palpebral portion of the orbicularis oculi. The fibers for the muscle of Horner come from the lacrimal crest and encircle the lacrimal canaliculi. This assists the flow of tears into the nasolacrimal drainage system when the orbicularis oculi contracts to close the eye. The muscle of Riolan (also known as the pars ciliaris) is another section of the palpebral portion of the orbicularis oculi; it lies near the lid margin to maintain the margins next to the globe. The orbicularis oculi is the major muscle responsible for closing the eyelids.

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154
Q

Tear volume in a normal, healthy, young adult measures approximately between which of the following values?

A. 17.0-20.0 microliters
B. 2.0-5.0 microliters
C. 13.0-16.0 microliters
D. 9.0-12.0
E. 6.0-8.0 microliters
A

E. 6.0-8.0 microliters

Tear volume has been measured by several methods to be approximately 6-7 microliters in normal individuals, with lesser values occurring in conditions of aqueous tear deficiency. This has implications for drug delivery, since the normal ophthalmic drop volume varies between 25 and 50 microliters, effectively overwhelming the native tear value upon instillation.

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155
Q

The lymphatic system serves many important roles in the human body. The lateral portion of the eyelid lymphatics drain into which of the following structures?

A. The puncta
B. The submandibular lymph node
C. The conjunctiva
D. The pre-auricular lymph node

A

D. The pre-auricular lymph node

The lateral 2/3 of the upper lid and the lateral 1/3 of the lower lid lymphatics drain into the pre-auricular lymph node located directly in front of the ear. The medial 1/3 of the upper eye lid and the medial 2/3 of the lower lid lymphatics drain into the submandibular node located just under the jaw-line. Therefore, it is very important to evaluate these two nodes separately, especially when a condition of viral etiology is suspected

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156
Q

A 2.5x Galilean telescope has a -25D ocular lens. When focused for infinity, what is the length of the telescope?

A. 5 cm
B. 6 cm
C. 14 cm
D. 4 cm
E. 10 cm
A

B. 6 cm

Explanation
M = -Doc/Dobj where Doc=power ocular; Dobj= power objective; t= separation of lenses 
2.5= - (-)25/Dobj 
Dobj= 10D 
t= f'obj + f'oc 
f'obj=1/10D = 0.10 m 
f'oc = 1/-25D = 0.04 m 

t=0.10 + -0.04 = 0.06 m or 6 cm

10 cm - incorrect- would come up with this answer if only took in account the focal length of the objective lens.
4 cm - incorrect- would come up with this answer if only took in account the focal length of the ocular lens.
14 cm - incorrect - would come up with this answer if thought equation was t= f’obj + f’oc

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157
Q

A patient is using a stand magnifier of +16D with a +2.00 add. If the distance separating the two lenses is 25 cm what is the equivalent power of this combination?

A. 22D
B. 26D
C. 10D
D. 18D

A

C. 10D

Explanation
De= D1+D2 -tD1D2 where De=equivalent power;D1=power of magnifier;D2=power add;t=separation in meters between the lenses
De = (16+2) - 0.25(16)(2) De= 18-8 = 10D

18D- incorrect answer -would come up with this if added the stand magnifier power to the power of the add
22D -incorrect answer - would come up with this if added 16D for stand mag 2D for add and 4D for equivalent of 25cm.
26D - if added the 18 +8 in the De equation instead of subtracting

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158
Q

Illumination is one of the most important considerations to discuss in the case disposition for a visually impaired patient. A patient with chronic open angle glaucoma moves a 60 watt bulb on a flexible mounted arm from three feet to one foot from the page. The illumination on the page will appear to have been increased by how much?

A. Decreased by 1/9 of the original brightness
B. Should be the same brightness
C. Increased by 9 times the original brightness
D. Decreased by 1/3 of the original brightness
E. Increased by 3 times the original brightness

A

C. Increased by 9 times the original brightness

It has been said that prescribed optical devices without consideration of the appropriate lighting will often doom the patient to failure. Unfortunately, there are no good tests to determine the exact type of lighting. Generally, different light levels are tried during the examination (as well as during the training session) with the patient using an adjustable light. The distance from the page is very important because of the inverse-square law of illumination: the intensity varies inversely as the square of the distance from the page. If the light is moved from 1 foot to 3 feet from the page, a bulb will be needed that is approximately nine times as bright to keep the same illumination on the page. (It should be noted that technically, this relationship is only true for a point source of light.) Clinically, however, it gives a good approximation of the change in brightness (illumination) seen on the page when the distance of the light is changed. The illumination in the above example would therefore increase by 9X when the bulb is moved towards the page.

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159
Q

One of your tech-savvy low vision patients wishes to use a CCTV for reading. The CCTV operates on what principle of magnification?

A. Relative distance magnification
B. Rated magnification
C. Equivalent magnification
D. Relative size magnification

A

D. Relative size magnification

CCTVs work on the principle of relative size magnification (or projection). It operates by enlarging the text without lenses in front of the patient or the patient moving closer to the device. When the print is enlarged electronically in this matter, the image of the print subtends a larger area on the retina and thus a larger size.

An example of relative distance magnification would be if you were holding a newspaper at 40 cm and you moved it closer to 20 cm. The print now appears 2 times as large relative to the 40 cm distance.

Rated magnification is often used by manufacturers of some hand magnifiers and stand magnifiers using a 25 cm reference distance.

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160
Q

The visual acuity of a 77 year-old female patient with age-related macular degeneration is 2/16 in the right eye on the ETDRS chart. Why is this chart useful in monitoring the response to treatment with anti-vascular endothelial growth factor (VEGF)?

A. Each line has 5 Sloan letters throughout the chart with equal spacing and is 1.26 times larger than the line below it; each line is .1 log units larger than the line below it when moving up the chart
B. The Snellen construction of the chart enables the examiner to quickly note that a two-line increment represents a factor of a two time increase in the size of the letters
C. A three-line decrease represents a factor of a two time decrease in the size of the letters
D. Each line is 1.0 log units larger than the previous line

A

A. Each line has 5 Sloan letters throughout the chart with equal spacing and is 1.26 times larger than the line below it; each line is .1 log units larger than the line below it when moving up the chart

Explanation
The ETDRS chart is a logarithmic eye chart modeled after the Bailey-Lovie chart. It is the primary standardized eye chart used in evaluating the visual acuity of low vision patients. The ETDRS charts are logMAR (log of the minimum angle of resolution) in design and are constructed with 10 Sloan sans serif letters. Each line is 1.26 times larger than the line below, and the construction of each line is such that the difficulty is theoretically equivalent on every line.

The construction of the ETDRS chart is made to eliminate the inherent errors in the measurement of visual acuity found in the traditional non-standardized Snellen test charts. The Snellen test charts have variations in legibility of different letters as well as differences in the spacing between the lines of letters and between adjacent letters on single lines. The ETDRS logarithmic chart is constructed in such a way that each line of letters is 0.1 log units (about 1.26 times) larger than the previous line. This is a geometric progression.

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161
Q

A positive catalase test indicates that a bacteria is capable of breaking down which of the following?

A. Glucose
B. Hydrogen peroxide
C. Pyruvate
D. Carbon dioxide

A

B. Hydrogen peroxide

Catalase is an enzyme commonly found in organisms that are exposed to oxygen. Catalase breaks down hydrogen peroxide into oxygen and water. The catalase test is performed by applying a drop of hydrogen peroxide to a microscope slide. A colony of bacteria is then exposed to the hydrogen peroxide via an applicator stick. The presence of bubbles or froth yields a positive catalase test. Staphylococci and Micrococci are catalase-positive organisms. Campylobacter and Escherichia coli are catalase-negative organisms.

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162
Q

Antibiotic resistance that is rapidly spread within a population of bacteria is due to what mechanism?

A. Binary fission
B. Transformation
C. Budding
D. Conjugation

A

D. Conjugation

Conjugation occurs between a donor (possesses a conjugative plasmid) and recipient bacteria. The donor bacterium initiates contact with the recipient via a sex pilus, allowing for cell-to-cell contact and transfer of DNA. The plasmids often contain genes that encode for toxin production, virulence factors, and antibiotic resistance. Genetic transformation is achieved by very few strains of bacteria and may only occur during certain phases of growth; therefore, rapid antibiotic resistance is not feasible. Budding and binary fission are means of reproduction but are not directly responsible for antibacterial resistance. Genes must have been transferred that code for resistance prior to budding and binary fission in order for the progeny to contain genes that allow for drug resistance.

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163
Q

Congenital cataracts can be caused by a viral infection of the mother with rubella virus (German measles) during development of the primary lens fibers. At which time period in embryonic development can infection cause congenital cataracts?

A. 2nd trimester
B. Post-delivery
C. 1st trimester
D. Conception
E. 3rd trimester
A

C. 1st trimester

Explanation
The developing lens is susceptible to rubella virus when the lens fibers are forming, which occurs around weeks 4-7 of gestation. Earlier infection will occur prior to lens fiber development, and the lens is resistant to later infection because the virus is unable to penetrate the lens capsule.

The fetus is most susceptible to lenticular damage during the first trimester. Contraction of the rubella virus will cause the greatest amount of damage during this time period. Congenital cataracts are usually detectable at birth but may be seen later because the virus can persist in the lens.

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164
Q

A 12-year old male is sitting in your waiting room while his mother undergoes her annual eye exam. While waiting, he eats a candy bar containing peanuts, and, as luck would have it, he is deathly allergic to nuts. To counter anaphylactic shock, what would be the BEST course of action?

A.Prednisone (oral)
B. Injection of epinephrine (EpiPen)
C. Olopatadine (Patanol)
D. Administration of Benadryl (oral)

A

B. Injection of epinephrine (EpiPen)

Explanation
Anaphylactic shock is defined as a severe, multi-system, type I hypersensitive, acute allergic reaction that may be life-threatening. Signs of an allergic reaction include tingling, itching, hives, swelling of lips and tongue, constriction of the airway, vasodilation, myocardial depression, and a decrease in blood pressure. The EpiPen is injected intramuscularly to the upper lateral thigh to ensure rapid delivery. Epinephrine (Adrenaline) activates both alpha and beta adrenergic receptors causing an increase in peripheral vascular resistance and allowing for an increase in blood pressure and coronary artery perfusion. Adrenaline also serves to reverse vasodilation and decrease urticaria and angioedema. For severe, life-threatening reactions, Benadryl (diphenhydramine) will not work quickly enough. Topical antihistamines have little if any systemic absorption and therefore will not be effective in counteracting the anaphylaxis. While oral steroids may be useful in the post-management of anaphylactic shock, they will not yield the desired immediate response.

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165
Q

A Galilean telescope has an ocular lens with a power of -32.00 D and an objective lens with a power of +8.00 D. What is the magnification provided by the telescope?

A. 8x
B. 0.25x
C. 4x
D. 256x

A

C. 4x Correct Answer

Explanation
To calculate the magnification (M) of a telescope divide the power of the ocular lens (Doc) by the power of the objective lens (Dobj): M=-Doc/Dobj. In the example above, M=-(-32 D)/8 D= 4x. The magnification of a Galilean telescope is positive due to the fact that its ocular has a minus powered lens. The magnification of an astronomical telescope is negative and therefore its image will be upside down.

166
Q

A 6-foot tall man wishes to buy a plane mirror in which he can visualize his whole length at the same time. How tall must the mirror measure in order for the above to occur?

A. 4.5 feet tall
B. 5.2 feet tall
C. 3 feet tall
D. 2.3 feet tall
E. 6 feet tall
A

C. 3 feet tall

In order for a person to see their entire reflection, a plane mirror must be half as tall as the person. This holds true regardless of the position of the person. For the above example, 6/2= 3 feet.

167
Q

A convex mirror in water (n=1.33) has a radius of curvature of 12 cm. What is the dioptric power of the mirror?

A. +22.17 D
B. +8.33 D
C. -22.17 D
D. -11.08 D

A

C. -22.17 D

Explanation
A concave mirror converges light and therefore acts like a convex lens, hence concave mirrors have positive dioptric values whereas convex mirrors diverge light and possess negative dioptric powers.

The equation used to determine the power of a mirror is P=-2n/r, where P= the power of the mirror in diopters, n= the index of refraction of the surrounding medium, and r= the radius of curvature of the mirror in meters. P=2(1.33)/-0.12=-22.17 D. Remember, a convex mirror will have a positive radius of curvature and a concave mirror will have a negative radius of curvature.

168
Q

A patient is seen at your office complaining that her right eye is physiologically higher than her left eye. She would like to know if glasses would help improve the cosmesis of her predicament. You know that prism will shift the image of an object. How would you orient a prism to help her appearance?

A. Prescribe base down prism over the left eye
B. Prescribe base in prism over the left eye
C. Prescribe base up prism over the right eye
D. Prescribe base out prism over the left eye

A

C. Prescribe base up prism over the right eye

Explanation
A prism will bend light towards its base, but the image will be shifted towards the apex of the prism. Therefore, by prescribing base up prism over her right eye, its image will be shifted down towards the apex of the prism. Another way of remembering this is to think of the prism as an arrow that will point in the direction of the deviation (i.e., exotropia is neutralized with base in prism, the eye points outwards, the apex of the prism also points out). Prescribing prism for cosmetic purposes may not always be an option as significant vertical prism may induce diplopia or visual discomfort.

169
Q

A patient brings you an old pair of glasses and asks you how much prism is in the lenses. With the lensometer, you measure 7 prism diopters base up and 4 base out in the right eye and 4 prism diopters base up and 5 base out in the left eye. What is the total amount of prism in the glasses?

A. 3 prism diopters base up in the right eye and 9 base out
B. 3 prism diopters base up in the right eye and 1 base out
C. 11 prism diopters base up total and 9 base out
D. 11 prism diopters base up total and 1 base out

A

A. 3 prism diopters base up in the right eye and 9 base out

For vertical prism, if the bases are oriented in the same direction, they will cancel each other out. When the bases are oriented in opposite directions in the vertical meridian (i.e., base up and base down), the powers will add together. The opposite holds true for prisms with their bases oriented horizontally. If the prism bases are both base out or base in, the powers are additive, while if they are opposite (that is, one is base in and one is base out), the powers will cancel.

170
Q

A patient returns to your office reporting that her eyes feel strange when she reads 6 mm below the optical centers of her new glasses. The prescription in her right eye is -4.00 DS and -7.00 DS in her left eye. How much vertical prism is induced when she reads?

A. 1.8 prism diopters base down
B. 6.6 prism diopters base down
C. 4.2 prism diopters base down
D. 2.4 prism diopters base down

A

A. 1.8 prism diopters base down

Use the Prentice rule to solve this problem: prism diopters(pd) =d*F, where d= the distance from the optical center in centimeters and F= the power of the lens in the desired meridian in diopters. In this instance, the patient is looking through base down prism in both eyes, which will cancel some of the prismatic effect as the bases are aligned. Solving for the amount of prism on the right eye, pd=0.6(-4.00)= 2.4 base down prism. Solve for the left eye: pd=0.6(-7.00)=4.2 base down prism. Subtract the two to determine the total prismatic effect experienced by the patient: 4.2-2.4=1.8 base down prism. Alternatively, you can omit one of the steps by initially determining the total power difference in the vertical meridian between the two lenses, which is 3.00 D (7-4=3). Then you can multiply this power difference by the distance between the patient’s line of sight and the optical center, which is 6 mm in this question. Pd=0.6(3)= 1.8 prism diopters base down over the left eye. Generally, vertical imbalances of smaller magnitudes do not pose too much of a problem for single vision lenses as the patient can tilt her head to re-align the optical centers with her line of sight, thus eliminating any possible diplopia.

171
Q

A chiasmal lesion or mass, such as a pituitary tumor, generally causes what type of visual field defect?

A. Left homonymous hemianopsia
B. Bitemporal hemianopsia
C. Binasal heminopsia
D. Right homonymous hemianopsia

A

B. Bitemporal hemianopsia

The center of the chiasm contains axons of decussating ganglion cells that originate from the nasal retinas, which process temporal visual field information. The lateral portion of the chiasm is comprised of the axons of the temporal aspect of the retinas, which do not cross over. Lesions most commonly occur in the central portion of the chiasm and not the lateral aspects. Any central chiasmal mass or lesion will cause a bitemporal visual field defect that respects the vertical midline.

172
Q

A cranial nerve VI palsy will cause what type of deviation?

A. Esodeviation worse with near viewing
B. Exodeviation worse with distance viewing
C. Esodeviation worse with distance viewing
D. Exodeviation worse with near viewing

A

C. Esodeviation worse with distance viewing

A cranial nerve (CN) VI palsy, or a palsy of the abducens nerve, will cause an esodiviation on the affected side and will result in horizontal diplopia, which worsens with distance viewing since this nerve innervates the lateral rectus muscle. The patient may present with a head turn towards the same side as the affected eye. For instance, if the patient has a right lateral rectus palsy, he or she may present with a head turn to the right to help eliminate diplopia. It helps to think in terms of function. The lateral recti serve to abduct the eyes, and distance viewing requires divergence, or a turning out of both eyes simultaneously. A CN VI palsy will therefore be more evident when the patient looks far away, because the eye cannot abduct.

173
Q

A healthy retinal nerve fiber layer is thickest at which portion of the optic nerve head?

A. Superiorly
B. Nasally
C. Temporally
D. Inferiorly

A

D. Inferiorly

The nerve fiber layer is thickest at the inferior and superior regions of the nerve, respectively. The inferior and superior arcades are composed of large diameter axons with little overlap of the receptive fields, thus explaining why a field defect occurs in these regions first for early cases of glaucoma. Inferior or superior notching of the nerve is highly suspect for glaucomatous damage, and must undergo further testing in order to rule out glaucoma. The next thickest area of nerve fiber layer tissue is nasally, which is comprised of the nasal radial fibers. These axons are affected in the later stages of glaucoma, thus explaining why a temporal island of the visual field is often left remaining in advanced cases of glaucoma. Lastly, the temporal rim area is the thinnest. Temporal rim tissue is comprised of the papillomacular bundle. The fibers in this area are very small and compact, with a high degree of receptive field overlap, therefore because of the receptive field redundancy, a visual field defect correlating to this region will occur only after significant fiber loss has occurred. Due to the fact that these fibers are so small in diameter, even though they are numerous, the fibers do not occupy a lot of space in the optic nerve. The thickness of the nerve fiber layer rim tissue is best remembered as ISNT, with inferior being the thickest and temporal rim tissue being the thinnest.

174
Q

A patient is seen at your office reporting constant diplopia. The patient notes that the diplopia is still present when you cover her right eye. Based upon this information, what is the MOST likely etiology of her diplopia?

A. Aneurysm
B. Lateral rectus palsy
C. Superior oblique palsy
D. Uncorrected refractive error

A

D. Uncorrected refractive error

Monocular diplopia is never caused by any type of cranial nerve dysfunction. The most common cause of monocular diplopia is an uncorrected refractive error. Other causes of monocular diplopia include corneal irregularities, lens irregularities, lens subluxation (very rare), or an improper glasses prescription. Whenever you are confronted with a recent onset of diplopia, the first thing you must determine is whether the diplopia is present monocularly or binocularly.

175
Q

A cortical hypercolumn is comprised of which of the following?

A. Ocular dominance columns for both eyes and a complete set of orientation columns

B. Ocular dominance columns for both eyes and orientation columns for one specific orientation only

C. A complete set of ocular dominance columns for one eye only and a complete set of orientation columns

D. An ocular dominance column for one eye only and a complete set of orientation columns

E. An ocular dominance column for one eye only and orientation columns for one specific orientation only

A

A. Ocular dominance columns for both eyes and a complete set of orientation columns

The striate cortex is organized into discrete rows and columns that help to code for specifics of the stimulus. A hypercolumn consists of both a right and left eye ocular dominance column as well as orientation columns for every orientation. An electrode that penetrates the cortex perpendicularly will encounter cells with the same ocular dominance, and they all respond to stimuli of the same orientation. However, an electrode that penetrates the cortex parallel to its surface will encounter neurons that all possess the same ocular dominance but respond preferentially to stimuli of different orientations. In order for ocular dominance columns to form properly, it is essential that normal vision is present in both eyes during the early years of life.

176
Q

A patient who has a high spatial frequency cut-off of 40 cycles per degree will have what predicted Snellen acuity?

A. 20/15
B. 20/20
C. 20/40
D. 20/30

A

A. 20/15

In order to convert from cycles per degree to Snellen acuity, simply divide 600 by the cycles per degree; this will solve for the denominator of the Snellen acuity. For the above example 600/40 = 15. Therefore, the predicted Snellen acuity would be 20/15.

177
Q

A person can just barely detect the difference between two weights; one weighs 12 pounds and the other weighs 10 pounds. What is the just noticeable difference for a weight of 70 pounds?

A. 20 pounds
B. 14 pounds
C. 2 pounds
D. 56 pounds
E. 8 pounds
A

B. 14 pounds

Weber’s Law deals with the just noticeable difference and can be expressed mathematically as:
K= delta I/I, where K= Weber’s constant, delta I= that difference threshold, and I= the original stimulus intensity (weight etc.)

For the above example, we must first solve for Weber’s constant. 12-10/10=0.2. Using this Weber’s constant we can then solve for the just noticeable difference or the increment threshold for 70 pounds.
X-70/70=0.2, X= 14 pounds. Therefore, the above person will just be able to discern the difference between a 70-pound weight and an 84-pound weight.

178
Q

A person who is missing the photopigment chlorolabe is categorized as which of the following?

A. A protanomalous trichromat
B. A deuteranomalous trichromat
C. A protanope
D. A tritanope
E. A deuteranope
A

E. A deuteranope

There are several classifications of color-vision defects; hereditary defects are the most common. The two broad categories are dichromacy and anomalous trichromacy. In dichromacy, one of the photopigments is missing; the type of dichromacy is categorized based on which photopigment is lacking. A deuteranope is missing chlorolabe, a tritanope is missing cyanolabe, and a protanope is missing erythrolabe. It is theorized that the missing photopigment is replaced by the photopigments that are present; otherwise, the person would likely suffer a deficit in visual acuity. Anomalous trichromats are in possession of all three photopigments but the absorption spectrum of one of the pigments has been shifted. For a protanomalous trichromat, the spectrum for erythrolabe is shifted towards the shorter wavelengths. A deuteranomalous trichromat displays a shift of the maximum sensitivity of chlorolabe towards the longer wavelengths. Protans and deutans are said to be red-green colorblind while tritans tend to mix up blues and yellows and are said to possess a blue-yellow defect; this is usually acquired rather than hereditary.

179
Q

A 24-year old female patient presents at your office complaining of side effects that began when she started using Patanol® to treat her ocular allergies. She reports complete compliance with her eye drop administration. Which of the following symptoms is MOST likely associated with olopatadine (Patanol®) use?

A. Tachycardia
B. Headache
C. Visual Hallucinations
D. Gastrointestinal discomfort
E. Depression
A

B. Headache

Topical antihistamines and mast cell stabilizers such as Patanol® (olopatadine) are commonly prescribed to relieve the symptoms associated with ocular allergies. They are a very effective class of medication due to their dual action mechanisms. Topical antihistamines that possess this dual action are olopatadine (Patanol®), ketotifen fumarate (Zaditor®), azelastine (Optivar®), and epinastine (Elestat®). The aforementioned drops serve to alleviate itching and redness by blocking H1 receptors as well as inhibiting mast cell and basophil degranulation. Side effects of topical antihistamine/mast cell stabilizers include stinging upon instillation, headaches, and adverse taste (don’t forget to inform your patients about punctual occlusion!). Tachycardia, depression, gastrointestinal discomfort, and visual hallucinations have not been reported with Patanol® use.

180
Q

A 31-year old male patient presents to your office for a photorefractive keratectomy (PRK) pre-operative examination. As you review his required ocular medication schedule, which of the following prescribed drops must you remember to tell him to “shake well” before instillation?

A. Acular®
B. Zymaxid®
C. FreshKote®
D. Pred Forte®

A

D. Pred Forte®

Pred Forte® is an ocular medication that is in suspension form; therefore, it is important to shake this medication well before use. Other forms of prednisolone that are suspensions include Pred Mild®, Econopred®, and Econopred Plus®.

On the other hand, there are prednisolone ocular medications that are solutions, making shaking of the bottle unnecessary. These include AK-Pred®, Inflamase Mild®, and Inflamase Forte®.

Zymaxid® is a 4th generation fluoroquinolone antibiotic that is bottled in solution form, as well as Acular®, which is an ocular non-steroidal anti-inflammatory drug (NSAID). FreshKote® is a prescription artificial tear that also does not need to be shaken before use.

181
Q

A 32-year old female is seen at your office complaining of a recent onset of blurred vision, only at a distance. A thorough case history reveals that she recently began taking a new medication which you correctly assume has induced myopia. Which of the following medications is MOST likely to be the culprit?

A. Omega III fish oil capsules
B. Accutane® (isotretinoin)
C. Tums® (calcium carbonate)
D. Tylenol® (acetaminophen)

A

B. Accutane® (isotretinoin)

Isotretinoin, birth control pills, and diuretics, among many other drugs, can cause myopia in some patients. Myopia most likely results from corneal swelling, which steepens the curvature of the cornea. Drugs that cause swelling of the lens, accommodative spasm, or edema of the ciliary body will also result in myopia. A reduction in the dose of the medication or cessation of the offending drug will usually result in reversal of nearsightedness. Fish oil, Tylenol®, and Tums® have not been shown to have a correlation with transient myopia development.

182
Q

A chin fissure is a dominant trait. If a father who is homozygous-dominant for this trait and a mother who is homozygous-recessive for this trait mate, what are the chances that their first child will have a chin fissure?

A. 75%
B. 100%
C. 0%
D. 50%
E. 25%
A

B. 100%

Because the father is homozygous-dominant, it is indicated that he possesses a dominant gene pair for the chin fissure trait (FF). On the other hand, the mother is homozygous-recessive for this trait; therefore, phenotypically she would have a “normal chin” because she has an identical gene pair that does not code for a chin fissure (ff). Each child would receive an allele from each parent, but the pair of genes would not be identical (this is termed heterozygous (Ff)).

However, because they would inherit a dominant form of the allele, this is the form of the gene that would influence the phenotype, resulting in the appearance of a chin fissure.

183
Q

A constant ringing of the ears is known as which of the following terms?

A. Otitis media
B. Tinnitus
C. Malleus
D. Sinusitis

A

B. Tinnitus

Tinnitus is caused by damage to hair cells in the inner ear from exposure to excessive noise, medications (like aspirin), aging, and some diseases. Sound waves cause the hair cells to bend, releasing a neurotransmitter and causing action potentials of the auditory nerve. Sometimes the hair cells break or are left in the “on” position, causing the perception of ringing.

Otitis media is an infection or inflammation of the middle ear.

Sinusitis is an inflammation of the sinuses.

184
Q

A contracting muscle that develops tension but does not shorten displays which type of muscle tension?

A. Isovelocity
B. Isotonic
C. Isovolume
D. Isometric

A

D. Isometric

Isometric contraction occurs when a muscle is contracting but is not shortening. This type of muscle tension is used for load-bearing situations such as holding a plate of food in front of you. Muscles that shorten but maintain the same amount of tension are said to display isotonic contraction. An isovelocity contraction follows when the force of the contraction varies while the velocity remains constant.

185
Q

During gestation, when does the secondary vitreous begin to develop?

A. The 1st week of gestation
B. The 20th week of gestation
C. The 9th week of gestation
D. The 30th week of gestation

A

C. The 9th week of gestation

The primary vitreous develops at around the third week of gestation. It is formed by mesoderm. The secondary vitreous begins to develop during the ninth embryonic week and later becomes the mature vitreous. The secondary vitreous stems from primary vitreal cells and retinal glial cells and therefore originates from neuroectoderm. The secondary vitreous expands to fill the globe while compacting the primary vitreous in the center of the globe.

186
Q

A central retinal artery occlusion (CRAO) causes tremendous damage to the retina. How will the electroretinogram (ERG) of a person who has suffered a CRAO be affected?

A. Both the a-wave and the b-wave will disappear
B. Both the a-wave and the b-wave will remain
C. The a-wave will remain while the b-wave will disappear
D. The a-wave will disappear while the b-wave will remain

A

C. The a-wave will remain while the b-wave will disappear

A central retinal artery occlusion will cause a loss of the b-wave which is formed by responses from the bipolar and Muller cells, both of which are nourished by the central retinal artery. The a-wave results from excitation of the photoreceptors. The a-wave will not be lost in the event of a CRAO due to the fact that photoreceptors receive their oxygen supply via the choroid.

187
Q

A patient is concerned with an acute reduction of the acuity in the right eye. You correctly diagnose central serous retinopathy, and confirm your diagnosis with an optical coherence tomography (OCT). What is the standard treatment protocol?

A. Monitor monthly for resolution
B. Refer for laser treatment of the retina
C. Refer for cryotherapy of the retina
D. Treat the patient with prism as they are likely to develop diplopia
E. Refer for intravitreal steroid injection

A

A. Monitor monthly for resolution

CSR is more commonly seen in middle-aged males under high-stress, who are very anxious, or with type A personalities. This condition causes fluid to leak from the choriocapillaries into the subretinal area, causing a serous detachment of the neurosensory retina. There is an associated loss of the foveal reflex, a hyperopic shift, a potential relative scotoma, and metamorphopsia. Flourescein angiography will reveal hyperfluorescence that appears like a smoke-stack. Evaluation of the posterior pole will typically display a blister-like elevation of the neurosensory retina. The patient is monitored monthly and intervention is rarely required, as most cases of CSR will resolve within roughly 6 months.

188
Q

According to the Keith-Wagener-Barker method of classification, hypertensive retinopathy is categorized as stage four when which ocular sign is present?

A. Hard exudates in a star configuration
B. Flame hemorrhages
C. Retinal edema
D. Swelling of the optic disc

A

D. Swelling of the optic disc

Grading of hypertensive retinopathy according to the Keith-Wagener-Barker system is as follows:
Stage 1- narrowing of the retinal arteries
Stage 2- stage 1, plus focal constriction of the retinal vasculature (arteriovenous nicking)
Stage 3- stage 2, plus retinal hemorrhages, hard exudates (likely in a star configuration), cotton wool spots, and retinal edema
Stage 4- stage 3, plus swelling of the optic disc. This patient must be hospitalized immediately.

189
Q

Bipolar cells receive information from photoreceptors. Which type of neurotransmitter do bipolar cells respond to?

A. Dopamine
B. Serotonin
C. Glutamate
D. Glycine

A

C. Glutamate

Bipolar cells respond to glutamate released by photoreceptor cells. Glutamate release in the dark causes on-center bipolar cells to hyperpolarize (inhibition) and off-center bipolar cells to depolarize (excitation).

190
Q

Drusen typically deposit between which layers of the retina?

A. The retinal pigment epithelium and Bruch’s membrane
B. The inner and outer plexiform layers
C. The inner and outer nuclear layers
D. The ganglion cell layer and the nerve fiber layer

A

A. The retinal pigment epithelium and Bruch’s membrane

Drusen deposits collect between the retinal pigment epithelium (RPE) and Bruch’s membrane. The retinal pigment epithelium plays a very important role in phagocytosis of shed outer segments of photoreceptors. If the RPE fails to rid the retina of this debris, it will begin to accumulate, which can have a significant impact on vision and may lead to macular degeneration.

191
Q

Midget ganglion cells receive information pre-synaptically from which cells?

A. Rod bipolar cells
B. Horizontal bipolar cells
C. Midget bipolar cells
D. Flat bipolar cells

A

C. Midget bipolar cells

Midget bipolar cells synapse onto midget ganglion cells. These are very selective and exclusive channels as one cone cell synapses with one midget bipolar cell which then in turn relays the information to a midget ganglion cell. There is no additional input from other cells or synapses. These types of monosynaptic cells are most common in the central retina thus explaining the ability to visually discern fine details.

192
Q

What is the MOST common type of oculomotor deviation?

A. Exophoria
B. Hyperphoria
C. Hypophoria
D. Esophoria

A

A. Exophoria

By far the most common oculomotor deviations are exo in nature ( about 95%), however most do not pose a problem. The least common type of deviation is vertical.

193
Q

While examining a patient with diplopia, you ask him to look downward and toward his nose. He is able to move the eye toward his nose (medially) but not down. Based on the isolated agonist model of eye movement by extraocular muscles, which nerve and muscle are not functioning appropriately?

A. Oculomotor nerve, superior rectus
B. Trochlear nerve, superior oblique
C. Abducens nerve, inferior oblique
D. Abducens nerve, lateral rectus
E. Trochlear nerve, inferior oblique
A

B. Trochlear nerve, superior oblique

The trochlear nerve (CN IV) innervates the superior oblique muscle. The abducens nerve (CN VI) innervates the lateral rectus, which is not involved in the motion described in this question. The oculomotor nerve has two divisions; the inferior division innervates the inferior rectus, inferior oblique and medial rectus, while the superior division innervates the superior rectus and levator palpebrae superioris.

The functions and anatomy of the extraocular muscles are as follows:
Superior rectus - turns eye up, adducts, and medially rotates (intorsion)
Inferior rectus - turns down, adducts, and laterally rotates (extorsion)
Lateral rectus - abducts eye (laterally)
Medial rectus - adducts eye (medially)
Superior oblique - medially rotates (intorsion), abducts and turns eye down
Inferior oblique - laterally rotates (extorsion), abducts and turns eye up

In the case described here, when the patient adducts the eye medially with the medial rectus as well as the superior and inferior rectus, only the superior oblique and inferior oblique can move the eye down or up respectively because the superior and inferior rectus muscles are already contracted. The same is true if a patient abducts the eye with the obliques and lateral rectus: only the superior and inferior rectus can move the eye up or down respectively.

194
Q

Which of the following muscle pairings and actions follows Sherrington’s law of reciprocal innervation?

A. Looking to the right causes contraction of the right medial rectus and contraction of the left medial rectus

B. Looking to the right causes contraction of the right lateral rectus and inhibition of the right medial rectus

C. Looking to the right causes contraction of the right medial rectus and inhibition of the right lateral rectus

D. Looking to the right causes contraction of the right lateral rectus and inhibition of the left medial rectus

A

B. Looking to the right causes contraction of the right lateral rectus and inhibition of the right medial rectus

Sherrington’s law of reciprocal innervation states that when a muscle is stimulated to contract, its antagonist is inhibited. Based upon this law, looking to the right causes contraction of the right lateral rectus and inhibition of the right medial rectus.

195
Q

Which extraocular rectus muscle has its insertion site CLOSEST to the limbus?

A. The lateral rectus
B. The medial rectus
C. The inferior rectus
D. The superior rectus

A

B. The medial rectus

The medial rectus inserts into the sclera roughly 5.3 mm from the limbus, followed by the inferior rectus, which inserts 6.8 mm from the limbus. The lateral rectus inserts 6.9 mm from the limbus, and the superior rectus has the furthest insertion point at 7.9 mm from the limbus. Remember MILS (Medial rectus, Inferior rectus, Lateral rectus, Superior rectus). If one draws a line connecting the insertion points of the muscles, an imaginary spiral is created called the spiral of Tillaux.

196
Q

When performing a unilateral cover test on your patient, you note the following: upon covering the right eye, the left eye moves in 1/10 times. Upon covering the left eye, the right eye moves in 4/10 times. The alternating cover test measures 25 prism diopters base-in. What is your diagnosis based on these findings?

A. 25 prism diopter intermittent alternating exotropia; right eye preferred
B. 25 prism diopter intermittent alternating esotropia; right eye preferred
C. 25 prism diopter intermittent alternating esotropia; left eye preferred
D. 25 prism diopter intermittent alternating exotropia; left eye preferred

A

D. 25 prism diopter intermittent alternating exotropia; left eye preferred

The unilateral cover test will tell you the eye (or eyes) affected, the direction, and the frequency of the ocular deviation. In this case, the eyes lose fixation 5/10 times, which shows that the frequency is intermittent (it would be constant if at least 1 eye moved 10/10 times). Since each eye moves at least once when the other is covered, the deviation is considered to be alternating. The uncovered eye is noted to move “in” on unilateral cover test, meaning that the deviation is an exotropia (if the eye moves “out” it is an esotropia). Since the right eye loses fixation more than the left eye (4/10 vs. 1/10), the left eye is considered to be the preferred eye. Also, the alternating cover test will tell you the full amount of the deviation, which is 25 prism diopters in this case.

197
Q

Which of the following conditions would be categorized as causing amblyopia due to deprivation?

A. A child born with a large congenital cataract in one eye only
B. A child born with a monocular 2 mm ptosis
C. A three-year with a constant right 30 prism diopter esotropia
D. A five-year old with an uncorrected prescription of OD: +7.00 D 20/400 OS: +0.50 20/20

A

A. A child born with a large congenital cataract in one eye only

Form deprivation amblyopia results when a clear and focused retinal image is blocked to one eye during the critical period. This can occur by a complete congenital cataract in one eye, a large ptosis that covers most or all of the pupil or by some other element that occludes the eye. The lack of visual information to the retina causes the other eye (non-occluded eye) to become dominant and thusly have stronger and a greater number of synaptic connections to the brain. Amblyopia causes a disproportionate amount of cortical neurons to respond preferentially to the non-deprived eye. The occlusion must occur during the critical period, and the earlier the occlusion is detected and removed, the better the prognosis. A small ptosis (i.e. 2 mm) would not be expected to cause amblyopia because the pupil would not be occluded. An unequal prescription such as the one in the above question would cause anisometropic amblyopia in which one eye would receive a clear image while the other would receive a blurry image. The brain would favor the clear retinal image, resulting in a strong dominance of cortical neurons for the least ametropic eye. Strabismus results in the perception of two images that are not fusible by the brain, causing diplopia. In order to eliminate double vision, the eye will suppress an eye (usually the deviated eye). This suppression leads to amblyopia.

198
Q

Which of the following methods can be used to test for the presence of eccentric fixation?

A. Visuoscopy
B. Binocular versions
C. The Hirschberg test
D. The Bruckner test

A

A. Visuoscopy

Angle Kappa (Lambda), visuoscopy, Haidinger’s Brush, and the Brock-Givner afterimage transfer tests are all methods of investigating for the presence of monocular fixation. The Hirschberg test allows for the determination of the direction, magnitude, and frequency of the ocular deviation. The Bruckner test may be used to detect small angle deviations, media opacities, anisometropia, and tumors. Binocular versions allows for the determination of the comitancy of the deviation.

199
Q

Which of the following types of refractive error would have the greatest tendency to lead to amblyopia?

A. A four-year old boy with an uncorrected refractive error of OD: +6.00 DS and OS: +1.50 DS
B. A five-year old girl with an uncorrected refractive error of OD: -3.25 DS and OS: -0.75 DS
C. A three-year old boy with an uncorrected refractive error of OD: +1.50 DS and OS: -2.00 DS
D. A four-year old girl with an uncorrected refractive error of OD: +1.00-1.50 x 180 and OS: +1.50-1.25 x 180

A

A. A four-year old boy with an uncorrected refractive error of OD: +6.00 DS and OS: +1.50 DS

A prescription in which there is a big refractive difference between the eyes, especially if both eyes are hyperopic, is most likely to cause amblyopia. Consider the prescription of OD: +6.00 DS and OS: +1.50 DS. The left eye will be able to accommodate 1.50 diopters to obtain a clear distance image and, because accommodation is bilateral and equal, the right eye will still be 4.50 diopters out of focus. This defocus will cause the left eye to dominate the cortical neurons, causing a decreased amount of binocular neurons and leading to poor stereopsis and amblyopia of the right eye.

A prescription of OD: -3.25 DS and OS:-0.75 DS will not lead to amblyopia because even though the right eye is blurry in the distance, at 30 cm its image will be clear and in focus. This also applies to the prescription of OD: +1.50 DS and OS:-2.00 DS. Although both patients will have good monocular corrected acuities, they will most likely possess poor stereopsis due to a decreased amount of binocular neurons because the eyes are never in focus at the same distance.

It is important to note that in order for amblyopia to occur, ametropia must be present during the critical period.

200
Q

What is the equivalent of a Reduced Snellen 20/50 optotype in metric notation (assuming a working distance of 40 cm)?

A. 2M
B. 0.5M
C. 0.67M
D. 1M

A

D. 1M

To convert from Reduced Snellen to metric notation one must divide the denominator by 50. In the above example 50/50 = 1M.

To convert from Reduced Snellen to Printer’s point, divide the denominator by 6. To convert from Printer’s point to metric, divide by 8. To convert from Metric notation to Reduced Snellen, multiply by 50; this will give you the denominator of Reduced Snellen.

A good rule of thumb is 1M = RS 20/50 = 8 point.

201
Q

What is the Interval of Sturm for a spherocylindrical lens with a power of +6.00 -2.00 x 090?

A. 16.7 cm
B. 20 cm
C. 8.3 cm
D. 41.7 cm
E. 25 cm
A

C. 8.3 cm

The powers in each meridian of the lens are +6.00 and +4.00.
The Interval of Sturm is simply the distance between the focal point of each power.

1/+6.00 = 16.7cm 
1/+4.00 = 25.0cm 

25cm - 16.7cm = 8.3cm

202
Q

What is the minimum thickness necessary for an antireflective coating (n=1.9) to be useful against incident light of 530 nm wavelength?

A. 69.7 nm
B. 278.9 nm
C. 58.3 nm
D. 139.5 nm
E.  132.5 nm
A

A. 69.7 nm

The equation for finding the minimum antireflective coating thickness is:

thickness = wavelength/(4 x index of coating) 
thickness = 530 / (4 x 1.9) 
thickness = 530 / 7.6 
thickness = 69.73 nm
203
Q

What separation distance will make the combination of a +3.00 and a +10.00 thin lens afocal?

A. 2.3 cm
B. 0.43 cm
C. 23 cm
D. 43 cm
E. 17 cm
F. 1.7 cm
A

D. 43 cm

For this question, the equation for equivalent power of a thick lens system should be used, solving for thickness (t).

De = D1 + D2 - (t/n) x D1D2 
De = equivalent power, D1 = front surface power, D2 = back surface power 
t = thickness of lens system, n = index between the 2 surfaces 

An afocal system has its focal points (F and F’) located at infinity. Therefore, an incident parallel pencil of light rays will emerge into image space as a parallel pencil as well. Another way to characterize an afocal system is that the equivalent power (De) is 0.

In the above question, De = 0, D1 = +3.00, D2 = +10.00, n= 1 
0 = 3 + 10 - ((t/1) x (3) x (10)) 
0 = 13 - (t x 3 x 10) 
0 = 13 -30t 
30t = 13 
t = 0.43 m (or 43 cm) 

If the two lenses are separated by 43 cm, the lens system can be considered afocal. This type of combination of two plus lenses is also an example of a simple astronomical (Keplerian) telescope. Keep in mind that the image in this type of optical system is inverted.

204
Q

Which of the following BEST describes the definition of irregular astigmatism?

A. The principal meridians of the cornea are not perpendicular to each other
B. The principal meridians of the cornea are located 90 degrees apart
C. The axis of astigmatism is located along an oblique axis
D. The axis of astigmatism is located along the 90 degree meridian

A

A. The principal meridians of the cornea are not perpendicular to each other

Astigmatism can be classified as either being regular or irregular. Regular astigmatism occurs in individuals in which the principal meridians of the cornea are located 90 degrees apart. That is, the area of the cornea with the flattest curvature (the axis) is oriented perpendicular to the meridian of the steepest curvature.

In certain ocular conditions such as keratoconus, corneal scarring, or post-surgical corneas, the steep and flat meridians may not be oriented 90 degrees apart. This type of corneal curvature can be considered irregular astigmatism. In these cases, the refractive error is typically not well corrected with spectacles, in comparison to correction with gas-permeable contact lenses.

205
Q

Which type of light scattering is responsible for the reddish-orange colors that are often observed during sunsets?

A. Brillouin scattering
B. Tyndall scattering
C. Mie scattering
D. Raman scattering
E. Rayleigh scattering
A

E. Rayleigh scattering

Scattering of light occurs when the medium through which light or other electromagnetic radiation travels is not homogenous. In the case of Rayleigh scattering, the particles that scatter the light are smaller than the wavelength of the light passing through. The particles may be individual atoms or molecules of a solid, liquid, or most commonly, a gas. The appearance of the blue sky during the day and the reddish hue of the sunset are due to Rayleigh scattering of light.

206
Q

While performing the astigmatic clock dial, your patient reports that the clearest/blackest line is the 2-8 line while the 5-11 line is the least clear. What would be the corresponding axis of astigmatism?

A. 180 degrees
B. 150 degrees
C. 60 degrees
D. 30 degrees

A

C. 60 degrees

In order to determine the corresponding axis of astigmatism utilizing the clock dial, one must multiply the smallest number of the clearest clock position by 30 degrees. In our case 2 x 30= 60 degrees. In general the line perpendicular to the clearest line is generally the least clear as this corresponds to the second principal meridian of the eye. Remember the principal meridians of the eye are 90 degrees apart.

207
Q

What is the net overall moles of ATP produced by the electron transport chain (i.e. not including glycolysis)?

A. 34 moles of ATP
B. 30 moles of ATP
C. 2 moles of ATP
D. 38 moles of ATP
E. 6 moles of ATP
A

A. 34 moles of ATP

The electron transport chain yields a total of 34 moles of ATP. Glycolysis produces a total of 2 moles of ATP. The overall net of cellular respiration is 36 moles of ATP.

208
Q

Which of the following is a precursor to steroid hormones such as testosterone?

A. Cholesterol
B. Sphingolipids
C. Phospholipids
D. Triglycerides

A

A. Cholesterol

Progesterone, aldosterone, testosterone, estradiol and cortisol are all derived from cholesterol. Cholesterol has a unique configuration comprised of four joined cycloalkane rings. Because these hormones are fat-soluble, they readily pass through cell membranes. They diffuse into the blood and are generally bound to carrier proteins, which transport the hormones to their designated target site where they may further undergo processing or transformation.

Sphingolipids are important in cell membranes, especially those located in the central nervous system, such as myelin sheath. Sphingolipids contain sphingosine as a backbone and are then further classified depending on which molecules are attached to that backbone, such as ceremides, gangliosides, sphingomyelin, etc.

Phospholipids contain a polar and non-polar end, thus making them amphoteric. This property allows for the formation of bilayers (polar ends aligned together and pointed outwards) resulting in the lipid bilayer commonly seen in cell membranes. Phospholipids are generally comprised of a phosphate group, a choline group (polar), and two fatty acid chains (non-polar) attached to glycerol, which serves as the backbone.

Triglycerides are comprised of three fatty acid chains attached to a glycerol backbone. Triglycerides are important in long-term energy storage for use by cells.

209
Q

What is the name of the pigmented line that represents the leading edge of a pterygium?

A. Stocker's line
B. Ferry's line
C. Hudson-Stahli line
D. Fleischer's ring
E. Krukenberg's line
F. Coat's white ring
A

A. Stocker’s line

  • Stocker’s line is a deposition of iron in the corneal epithelium that is located at the leading edge of a pterygium
  • A Hudson-Stahli line is an iron line that is commonly observed at the junction of the middle and lower third of the cornea (where lid closure occurs upon blinking)
  • Ferry’s line is found in front of a filtering bleb
  • Coat’s white ring is a small, white, oval ring at the level of Bowman’s membrane that is associated with a previous corneal foreign body
  • A Fleischer ring is an iron pigment that encircles the base of a cone in keratoconus
  • A Krukenberg spindle is a deposition of pigment on the corneal endothelium that is associated with pigment dispersion syndrome
210
Q

What is the name of the surgical procedure in which thermal laser burns are placed in the mid-periphery of the cornea in an attempt to steepen the corneal curvature?

A.Laser-assisted in-situ keratomileusis
B. Conductive keratoplasty
C. Limbal relaxation incisions
D. Radial keratotomy
E. Photorefractive keratectomy
A

B. Conductive keratoplasty

In cases where the corneal curvature must be steepened in order to correct for refractive error (hyperopia or presbyopia), conductive keratoplasty (CK) is a viable surgical option. Although this surgical procedure was used more often in earlier years, it is not currently as widely used as laser-assisted in-situ keratomileusis (LASIK) and photorefractive keratectomy (PRK). In comparison to CK, LASIK and PRK tend to be safe, have long-standing results, and more predictable outcomes. The CK technique involves using a radiofrequency probe to create burns in either one or two concentric rings in the mid-peripheral region of the cornea. These thermal laser burns cause subsequent stromal shrinkage, which results in an increase in the curvature of the cornea. This change in curvature typically decays over time, but the procedure can be repeated.

Radial keratotomy is also an older surgical procedure in which a diamond blade is used to create several radial corneal incisions (the number and depth of the incisions depends on the refractive error) in order to flatten the corneal curvature in patients with myopic refractive errors. Limbal relaxation incisions are similar in that arcuate incisions are made on opposite sides of the corneal periphery in the meridian of the “plus” cylinder axis in order to create flattening of the steep corneal curvature (with some smaller steepening of the flat meridian) in an attempt to reduce the amount of corneal astigmatism.

Photorefractive keratotomy (PRK) and laser-assisted in-situ keratomileusis (LASIK) are refractive surgery techniques that use an excimer laser to ablate corneal tissue to a certain depth in either the central cornea (to correct myopia) or peripherally (to correct hyperopia).

211
Q

Which layer of the cornea, if penetrated, will leave a scar?

A. The stroma
B. The tear film
C. The epithelium
D. The wing cell layer

A

A. The stroma

The corneal epithelium is comprised of 3 major layers. The outermost layer is composed of superficial cells (2-3 layers) followed by wing cells (2-3 layers) and, lastly, basal cells (1 layer). Damage to the epithelium will heal without keloid formation. The epithelial basement membrane is made up of collagen types IV, VII and XII.

The stroma makes up the bulk of the cornea and is comprised of keratocytes, nerves, type I collagen fibers and mucopolysaccharides. If injured, the stroma will heal but a scar will remain at the site of trauma.

The tear film lies anterior to the cornea and is not composed of tissue and as such cannot scar, nor is it considered a part of the cornea.

212
Q

Which of the following ocular signs is virtually pathognomonic for trachoma caused by chlamydia?

A. Inferior tarsal papillae
B. Lymphadenopathy
C. Tranta’s dots
D. Superior tarsal follicles

A

D. Superior tarsal follicles

Chlamydia causes two forms of conjunctivitis, trachoma and inclusion. Trachoma is more common in lesser-developed countries and can cause blindness if not treated appropriately. Trachoma presents in several stages, initially starting with mucopurulent discharge, lymphadenopathy, red eye, small superior tarsal follicles, and mild superior pannus. As the condition evolves, the formation of limbal follicles occurs and will eventually scar causing Hebert’s pits, which are characteristic of this infection. This condition, if left untreated, ultimately progresses to horrible scarring of the eyelid (Arlt’s line) and cornea, causing extremely poor visual acuity. Diagnosis is made with the observation of two or more of the following: follicles on the upper tarsus, pannus (particularly superiorly), limbal follicles or Herbert’s pits and typical conjunctival scarring of the upper lid. Treatment includes oral doxycycline, tetracycline, or erythromycin along with topical tetracycline or erythromycin ointment. Azithromycin is also a good choice because it is given as 1000 mg PO which delivers exceptional compliance; however, this is not to be prescribed to those with liver disease or to young adults under the age of 16.

Inclusion conjunctivitis is linked to venereal disease and can present either unilaterally or bilaterally (which is more common) as follicles on the upper and lower tarsal plates (lower follicles will be larger and more prominent), lymphadenopathy, possible mucopurulent discharge, lid edema, micropannus, superior corneal sub-epithelial infiltrates, superficial punctate keratitis, and scarring of the upper eyelid (sometimes called Arlt’s line or “basketweave” because of its appearance). This type of conjunctivitis is less severe than trachoma. Treatment is similar to that of trachoma.

Follicles are related to cellular immunity which serves to protect against viruses. Many types of viral infections can cause inferior palpebral follicles, such as EKC, Herpes simplex and molluscum contagiosum. Superior tarsal follicles are highly suggestive of a chlamydial infection. A superior papillary response is generally associated with an allergic response. Inferior tarsal papillae are frequently seen in bacterial infections and allergic responses as papillae act as the release sites for both eosinophils (associated with allergies) and polymorphonuclear leukocytes which destroy bacteria.

213
Q

Which of the following alterations will help to loosen a tightly-fitting gas-permeable lens?

A. Steepen the peripheral curve system
B. Increase the overall diameter
C. Reduce the width of the peripheral curve system
D. Reduce the size of the optic zone
E. Steepen the base curve of the lens
A

D. Reduce the size of the optic zone

There are a multitude of alterations that can be made when a lens is fitting too tightly, many of which can be done in-office if a modification unit is available. If a gas-permeable lens is fit too tightly, the most commonly altered parameter is flattening of the base curve. One can also decrease the optic zone, decrease the overall diameter (OAD), widen the peripheral curve system, or flatten the peripheral curve system. In order to modify a lens that is fitting too loosely, simply reverse all of the above: steepen the base curve, increase the OAD, increase the optic zone, steepen the peripheral curve system, and narrow the width of the peripheral curves.

214
Q

Which of the following will occur if you increase the water content of a soft hydrogel contact lens?

A. The tendency of lens deposits will decrease
B. The lens durability will increase
C. The patient will report an increase in dry eye symptoms
D. The oxygen permeability will decrease

A

C. The patient will report an increase in dry eye symptoms

As the water content of a soft hydrogel contact lens increases, generally the durability of the lens will decrease, the permeability of the lens will increase as will deposit formation and dry eye symptoms.
This is also mostly try for silicone-hydrogel lenses, except for the fact that with these lenses, as water content increases, the permeability of the lens tends to decrease.

215
Q

Which one of the following bitoric GP contact lenses would NOT induce cylinder if rotated to a misaligned position on the eye?

A. 7.63 mm / -1.50 D ——————— 8.11 mm / +1.12 D
B. 7.46 mm / -4.25 D ——————— 8.13 mm / -1.75 D
C. All of the options listed would induce cylinder if rotated off axis
D. 7.54 mm / +1.50 D ——————— 7.99 mm / +2.75 D
E. 7.58 mm / -5.37 D ——————— 8.18 mm / -0.50 D

A

A. 7.63 mm / -1.50 D ——————— 8.11 mm / +1.12 D

Cylinder power effect (CPE) bitoric and base curve toric (with a spherical front-surface) gas-permeable (GP) lenses will induce unwanted cylinder if the lens rotates off axis. The resulting cylinder is due to cross-cylinder effects. However, a spherical power effect (bitoric) will not induce unwanted cylinder regardless of lens rotation. To determine whether a GP lens is a spherical power effect (SPE) or cylinder power effect (CPE) bitoric, measure the two base curves using a radiuscope and the two raw contact lens powers using a lensometer. If the difference between the two base curve meridians in diopters is the same as the difference between the two raw powers, the lens is an SPE bitoric. This is the case for only one of the above answers. Converting mm of base curve radius to diopters results in 7.63 mm = 44.25 D and 8.11 mm = 41.62; a difference of 2.62 D. The difference between the two raw powers of +1.12 D and -1.50 D is also 2.62 D. Therefore, this lens is a spherical power effect (SPE) bitoric GP contact lens.

216
Q

You are fitting a toric soft contact lens to your patient’s right eye. The patient’s manifest refraction is -2.00 -1.50 X 095. You apply a -1.75 -1.25 X 085 diagnostic toric soft contact lens. It fits well, and the prism base down marking consistently locates halfway between the 6 o’clock and 7 o’clock hours. What axis should you order?

A. 95 degrees
B. 70 degrees
C. 80 degrees
D. 110 degrees
E. 100 degrees
A

D. 110 degrees

Applying LARS to compensate for lens rotation, since the lens is rotated to the Left, you would Add the amount of left rotation to the manifest refraction axis. Every hour on the clock dial would translate to 30 degrees rotation. In the above example, the lens is rotated to the doctor’s left by 15 degrees (between the 6 and 7 o’clock hours). Add the amount of rotation (15 degrees) to the cylinder axis of the manifest refraction (95 degrees). This results in a cylinder axis order of 110 degrees.

217
Q

Which of the following correctly describes the autonomic innervation of the iris muscles?

A. The iris sphincter and iris dilator are both innervated parasympathetically
B. The iris sphincter is innervated parasympathetically and the iris dilator is innervated sympathetically
C. The iris sphincter is innervated sympathetically and the iris dilator is innervated parasympathetically
D. The iris sphincter and iris dilator are both innervated sympathetically

A

B. The iris sphincter is innervated parasympathetically and the iris dilator is innervated sympathetically

Stimulation of the sympathetic nervous system results in pupil dilation and the parasympathetic nervous system pupil constriction. Accordingly, the sphincter muscle (which constricts the pupil) is innervated by the parasympathetic nervous system and the dilator muscle (which dilates the pupil) is innervated by the sympathetic nervous system.

218
Q

Which of the following types of scleritis presents without ocular inflammation, has a low risk for perforation, and does not typically result in pain or decreased visual acuity?

A. Vaso-occlusive necrotizing scleritis
B. Anterior non-necrotizing diffuse scleritis
C. Granulomatous necrotizing scleritis
D. Posterior scleritis
E. Scleromalacia perforans
F. Nodular scleritis
A

E. Scleromalacia perforans

Scleromalacia perforans is a type of necrotizing scleritis that typically presents without vascular congestion or pain. Clinical observations commonly include yellow-colored necrotic plaques that occur near the limbus without inflammation and very slow progression of scleral thinning that eventually exposes the underlying uveal tissue. Patients commonly complain of a mild non-specific irritation but no pain. Visual acuity is also not usually affected in these patients. Scleromalacia perforans typically affects elderly women with a long-standing history of rheumatoid arthritis. By the time patients are correctly diagnosed with this condition, treatment is usually not needed or is ineffective. Even though the name contains the word “perforans,” the risk of perforation is extremely rare, as the integrity of the globe is usually well maintained.

219
Q

Which type of anterior scleritis is associated with the highest risk of perforation?

A. Necrotizing
B. Diffuse
C. Nodular
D. Scleromalacia perforans

A

A. Necrotizing

Scleritis is an inflammation of the sclera that generally occurs secondarily to a systemic condition, usually of collagen vascular origin. Diffuse scleritis has a gradual onset and presents as a boring pain which may radiate to other structures such as the jaw and forehead. Patients will present with distension of the scleral vascular pattern, causing a deep pinkish hue of the sclera. Nodular scleritis appears similar to diffuse scleritis, but the areas of inflammation are localized to painful, raised nodules. Scleromalacia perforans is the least common form and is almost always seen in association with rheumatoid arthritis. Patients with scleromalacia perforans generally do not experience pain or inflammation. Necrotizing scleritis is the most severe form and has a higher mortality rate than the other types due to the fact that it usually stems from autoimmune diseases.

220
Q

Which of the following drugs decrease intraocular pressure by increasing uveoscleral outflow?

A. Dorzolamide
B. Brimonidine
C. Brinzolamide
D. Timolol
E. Pilocarpine
A

B. Brimonidine

Glaucoma medications lower intraocular pressure by either decreasing aqueous production or by increasing aqueous outflow. There are three classes of drugs for which the mechanism of action is increasing aqueous outflow: cholinergic agonists, prostaglandin analogs, and alpha-2 agonists. Cholinergic agonists, such as pilocarpine, work by increasing trabecular outflow, whereas prostaglandin analogs and alpha-2 agonists work by increasing uveoscleral outflow. The other classes of glaucoma medications, such as beta-blockers and carbonic anhydrase inhibitors, work by decreasing aqueous production.

It is important to note that alpha-2 agonists, such as Brimonidine (Alphagan®) and Apraclonidine (Iopidine®), have dual mechanisms of action. This class of medication decreases intraocular pressure by both increasing uveoscleral outflow and decreasing aqueous production.

221
Q

When examining a patient, a pinpoint spot of the posterior surface of the lens known as Mittendorf’s dot is seen. What is this a remnant of?

A. Hyaloid artery
B. Glial tissue of the optic nerve
C. Vitreous
D. Pupillary membrane

A

A. Hyaloid artery

Mittendorf’s dot is a remnant of the hyaloid artery and appears as a black dot on the posterior surface of the lens. Pupillary membrane remnants would be present in front of the lens and are a complex of fibers. Glial tissue of the optic nerve head is a remnant that is known as Bergmeister’s papilla.

222
Q

Which of the following types of congenital cataracts are characteristic of galactosemia?

A. Blue dot (Cerulean) opacities
B. Sunflower cataracts
C. Christmas tree cataracts
D. Oil droplet opacities

A

D. Oil droplet opacities

Central oil droplet opacities are a type of congenital cataract that is associated with galactosemia, a genetic metabolic disorder that affects the body’s ability to metabolize galactose properly.

Blue dot (Cerulean) opacities are congenital cataracts and are not usually associated with systemic disease but are thought to be due to autosomal dominant mutations in several genes.

Christmas tree cataracts are not considered a congenital type of cataract; they are a rare variant of senile cataracts that have a strong association with myotonic dystrophy.

Sunflower cataracts are also not considered congenital cataracts and are due to the abnormal deposition of copper in patients suffering from Wilson’s disease.

223
Q

You are measuring the palpebral fissure height in a patient reporting drooping of his upper eyelid. Which of the following BEST describes the normal positioning of the upper and lower eyelids in comparison to the limbus?

A. The upper lid normally rests about 1mm lower than the upper limbus, and the lower lid rests about 2mm above the lower limbus
B. The upper lid normally rests about 2mm lower than the upper limbus, and the lower lid rests about 1mm above the lower limbus
C. The upper lid normally rests about 2mm lower than the upper limbus, and the lower lid rests about 1mm lower than the lower limbus
D. The upper lid normally rests about 1mm lower than the upper limbus, and the lower lid rests about 2mm lower than the lower limbus

A

B. The upper lid normally rests about 2mm lower than the upper limbus, and the lower lid rests about 1mm above the lower limbus

The palpebral fissure height is a measurement of the distance between the upper and lower eyelid margins when the patient is looking in primary gaze. This particular measurement is typically less in males (7-10mm) as compared to females (8-12mm). The normal positioning of the upper and lower eyelids are as follows: the upper eyelid usually rests about 2mm below the superior limbus, while the lower eyelid position is typically 1mm above the lower limbus. A unilateral ptosis can be quantified by comparing these measurements to the contralateral eye. A ptosis up to 2mm may be graded as mild; a 3mm ptosis is considered moderate; a ptosis of 4mm or more is deemed severe.

Another important measurement in evaluating a ptosis is the marginal-reflex distance (MRD). The MRD can be defined as the distance between the upper eyelid margin and the resultant corneal reflection caused by directing a patient’s gaze at a penlight held by the examiner. This measurement is normally 4-4.5mm.

224
Q

Which of the following is the correct pathway for the drainage of tears through the nasolacrimal drainage system?

A. Lacrimal punctum, lacrimal canaliculus, ampulla, valve of Hasner, nasolacrimal duct, lacrimal sac
B. Nasolacrimal duct, lacrimal sac, valve of Hasner, lacrimal canaliculus, ampulla, lacrimal punctum
C. Lacrimal punctum, lacrimal canaliculus, ampulla, lacrimal sac, nasolacrimal duct, valve of Hasner
D. Lacrimal sac, lacrimal punctum, lacrimal canaliculus, ampulla, nasolacrimal duct, valve of Hasner

A

C. Lacrimal punctum, lacrimal canaliculus, ampulla, lacrimal sac, nasolacrimal duct, valve of Hasner

The lacrimal punctum is a small aperture located in the lacrimal papilla, the slight elevation at the junction of the lacrimal and ciliary portions of the eyelid margin. Initially, the tear film drains through this aperture. The lacrimal punctum leads into the lacrimal canaliculus, the tube connecting the punctum to the lacrimal sac. The ampulla is a slight dilation in the initial portion of the lacrimal canaliculus. The canaliculi from the upper and lower lids run horizontally along the lid margin, connecting into a common canaliculus that then enters the lateral aspect of the lacrimal sac located in the anterior portion of the medial orbital wall. The lacrimal sac empties into the nasolacrimal duct in the maxillary bone. The valve of Hasner is located at the terminus of the nasolacrimal duct in the inferior nasal meatus. The Valve of Hasner is a fold of mucosal tissue that ensures that fluid flows anterograde out of the duct.

225
Q

Which of the following skin conditions is considered to be benign and has the LOWEST risk of malignancy?

A. Squamous cell carcinoma
B. Actinic keratosis
C. Basal cell carcinoma
D. Keratoacanthoma

A

D. Keratoacanthoma

Keratoacanthoma appears very much like squamous cell carcinoma (SCC) in that it tends to progress rapidly and appears to ulcerate. This condition typically occurs in middle-aged and elderly patients of Caucasian descent on areas of the skin that are exposed. The lesion appears elevated, and eventually the center will produce a scab-like plug of keratin. The margins surrounding the plug will be rolled. At some point the keratin plug will fall out, resulting in the formation of a pit, and the lesion will regress. Most patients and clinicians do not like to wait this condition out due to its similarities to SCC.

Actinic keratosis is a pre-cursor to squamous cell carcinoma and appears as scaly, dry skin that does not heal. People with skin that is of lighter pigmentation along with excessive exposure to ultraviolet light tend to be most at risk for development of this condition.

Squamous cell carcinoma (SSC) is thankfully one of the rarest malignancies but due to its ability to metastasize can be quite dangerous. This malignancy has the ability to progress rapidly and has a high affinity for people who spend a lot of time in the sun, especially those who are light-skinned. The only way to definitively diagnose SCC is to refer for a biopsy and ensuring the use of Mohs technique. This strategy takes more time but ensures that the lesion is removed. Essentially, Mohs procedure calls for removal of tissue and biopsy of the surrounding borders. If the borders prove to be malignant then more tissue is removed and biopsied. This continues until the borders prove to be free of any carcinoma.

Basal cell carcinoma (BCC) is the most common malignant lid lesion and mercifully tends to be very slow-growing. BCC generally appears as a waxy, translucent nodule. Eventually the nodule will ulcerate. Patients may bring these to your attention and tell you that they have “had it for years and it just does not seem to heal”. Whenever you hear this it is best to send out for biopsy via Mohs technique. BCC very rarely metastasizes.

226
Q

Which patient would be considered legally blind?

A. A wet macular degeneration patient with best central acuities of 10/120 OD and 10/200 OS
B. A patient with Best’s disease with best corrected central acuities measure OD 10/80 and OS 10/100
C. A patient with a total retinal detachment of the right eye, no light perception and a best corrected central acuity of 8/60 due to wet macular degeneration
D. A myopic patient with acuities of 20/400 OD and OS uncorrected
E. A retinitis pigmentosa patient who has 20/20 central vision in each eye and a 30 degree in diameter visual field

A

A. A wet macular degeneration patient with best central acuities of 10/120 OD and 10/200 OS

Legal blindness must take into account central best corrected visual acuity, with the better eye 20/200 or worse. If the central acuity is normal such as in the RP patient then the field would be the restricting qualification, which, at 30 degrees, does not qualify.

227
Q

What term describes the phenomenon in which a bacterium directs its movement TOWARD a chemical in its environment?

A. Chemotaxis
B. Apoptosis
C. Phagocytosis
D. Transposition

A

A. Chemotaxis

Many bacteria possess flagella, or thread-like appendages, which allow for movement. Certain chemicals attract bacteria (chemoattractants), while others repel them (chemorepellents). Chemotaxis refers to the response of the bacteria to either chemoattractants or chemorepellents. In the absence of either of the aforementioned chemicals, bacteria will move in random patterns. Some bacteria possess genes and proteins which allow for the sensing of concentration gradients in their environment. In the presence of a chemoattractant, bacteria will have longer runs in the appropriate direction.

Apoptosis is defined as programmed cell death.

Phagocytosis refers to the engulfment of a particle (for example, bacteria) by a phagocyte (for example, a macrophage).

Transposition refers to the rare phenomenon in which genes move from one place on the genome to another position.

228
Q

What type of agar is commonly used to culture fungi?

A. Cetrimide agar
B. Blood agar plate
C. Sabouraud's agar
D. Hay infusion agar
E. Thayer-Martin agar
A

C. Sabouraud’s agar

Sabouraud’s agar is useful for culturing fungi. Culturing and sensitivity testing are important diagnostic tools used to determine the offending pathogen and help to select the appropriate medication for treatment. Sabouraud’s agar is unique in that it possesses a low pH that causes the inhibition of most bacterial growth, allowing for better isolation of the fungus. Hay infusion agar is frequently utilized to culture slime moulds.

Cetrimide agar is designed to isolate Gram-negative bacteria.

Blood agar plates are useful for the detection of pathogenic organisms via the presence of hemolytic activity (the ability of organisms to lyse or destroy red blood cells). Blood agar plates are not a selective medium, as many different types of organisms are capable of growth on this type of agar.

Thayer-Martin agar is a type of chocolate agar used to isolate Neisseria gonorrhoeae

229
Q

Which of the following classifications refers to an organism that can survive in an environment with or without oxygen?

A. Strict anaerobe
B. Obligate aerobe
C. Facultative anaerobe
D. Microaerophile

A

C. Facultative anaerobe

Facultative anaerobes are capable of using oxygen as a life source but can also thrive using anaerobic respiration. When culturing this type of bacteria in a test tube of liquid media, one would observe the bacteria dispersed throughout the entire tube but with greatest concentration at the top, as aerobic respiration yields a more favorable energy result.

Obligate aerobes require oxygen to survive. Test tube culturing would display bacteria gathering at the top of the tube where the oxygen levels are greatest.

Microaerophiles are capable of utilizing oxygen, but only at lower concentrations. A test tube culture of microaerophillic bacteria would result in the greatest concentration of bacteria towards the upper portion of the test tube, but not at the very top.

Strict anaerobes cannot tolerate oxygen and rely on anaerobic respiration to survive. A test tube culture of strict anaerobes would display the greatest concentration of bacteria towards the bottom of the tube where the least amount of oxygen is present.

Aerotolerant organisms are capable of growing in the presence of oxygen but do not actually utilize it for respiration. One would observe these bacteria evenly distributed throughout a test tube liquid culture.

230
Q

Which of the following genus of organisms is responsible for tuberculosis and leprosy?

A. Klebsiella
B. Salmonella
C. Borrelia
D. Mycobacterium

A

D. Mycobacterium

Mycobacterium is an aerobic, Gram-positive (although this is somewhat debatable) genus that includes several pathogenic species. Mycobacteria are extremely difficult to treat due to the nature of their cell walls which are truly neither Gram-negative nor Gram-positive (although they are classified as Gram-positive because they are acid-fast). M. tuberculosis is responsible for causing tuberculosis. M. leprae is the culprit that causes Hansen’s disease (also termed Leprosy). A species of Borrelia can cause Lyme disease. Salmonella has been known to cause food poisoning. Contraction of Klebsiella can lead to the development of pneumonia.

231
Q

Which of the following microorganisms is associated with peptic ulcer formation?

A. Vibrio cholerae
B. Clostridium tetani
C. Helicobacter pylori
D. Campylobacter spp

A

C. Helicobacter pylori

H. pylori is a helix-shaped gram-negative microaerophillic organism that colonizes in cells that secrete the mucosal lining of the stomach. These bacteria protect themselves from stomach acid by secreting urease, which, through a series of reactions, helps to neutralize the lower pH levels. H. pylori causes chronic gastritis, which in turn can lead to ulcer formation if pepsin and stomach acid levels increase to the point that the protective mucosal lining is broken down. C. tetani is associated with the contraction of tetanus. V. cholerae causes cholera, resulting in abdominal cramps, diarrhea, fever, and vomiting.

232
Q

Which organism can be contracted in a newborn via an infected mother and was previously treated prophylactically with silver nitrate?

A. Staphylococcus aureus
B. Neisseria gonorrhoeae
C. Corynebacterium spp
D. Neisseria meningiditis

A

B. Neisseria gonorrhoeae

Chlamydia trachomatis and N. gonorrhoeae are both organisms that can lead to neonatal eye infections. Passage of a newborn through an infected birth canal results in contraction of the organism. Most countries currently treat the newborn with eye ointments or drops containing erythromycin or some type or antibacterial. Silver nitrate is now rarely used as it can lead to toxic conjunctivitis.

233
Q

What is the front surface power of a lens in air with a refractive index of 1.50 and radius of curvature of 50 cm?

A. 1.00 D
B. 2.00 D
C. 3.00 D
D. 1.50 D

A

A. 1.00 D

To solve this problem, input the values into the equation for a single surface power, F = n’-n/r where F= the power of the lens, n’= the index of the medium that light is entering (the lens), n= the index of the medium in which light is exiting (medium surrounding the lens; in this case, air), and r = the radius of curvature (in meters) of the lens. Solve for F = 1.50-1.0/0.5 = 1.00 D.

234
Q

What is the power of a concave mirror (in diopters) located in air with a radius of curvature of 20 cm?

A. +10.00 D
B. -5.00 D
C. -10.00 D
D. +5.00 D

A

A. +10.00 D

A concave mirror converges light and therefore acts like a convex lens hence concave mirrors have positive dioptric values whereas convex mirrors diverge light and possess negative dioptric powers. The equation used to determine the power of a mirror is P=-2n/r where P= the power of the mirror in diopters, n= the index of refraction of the surrounding medium, and r= the radius of curvature of the mirror in meters. Unless otherwise stated, always assume that the index of refraction of air is 1.0. Input the values from the above question: P= -2(1)/-0.2=+10.00 D.

235
Q

What is the relationship between the Abbe number and chromatic aberration?

A. Proportional
B. Inversely proportional
C. Equal
D. No relationship

A

B. Inversely proportional

The Abbe number is the reciprocal of the dispersion of a spectacle lens. As the dispersion increases, a patient might experience seeing a rainbow of color as they view in the lens periphery. This is known as chromatic aberration. Crown glass has a high Abbe number of 59 and thus a low chromatic aberration.

236
Q

Which of the bifocal lens designs below will create the largest amount of image jump?

A. FT 28 bifocal
B. Executive bifocal
C. FT 35 bifocal
D. Round 28 bifocal

A

D. Round 28 bifocal

Image jump is the phenomenon where the patient’s eyes are passing from the distance prescription and are now viewing just below the top of the bifocal segment. The eyes are not quite at the near optical center (OC), therefore prism is induced. The farther away the near OC is from the top of the segment, the larger the amount of image jump. The round 28 bifocal has its optical center 14 mm below the top of the segment (the optical center for a round segment is located at its center, which is half of its diameter). The flat top 28 and flat top 35 bifocal optical centers are located 5 mm below the segment line. The Executive OC is on the segment line, so no image jump is created.

237
Q

Which of the following is the correct order of structures through which the pupillary fiber pathway passes?

A. Optic nerve -> optic chiasm -> optic tract -> Lateral geniculate nucleus in the thalamus
B. Optic nerve -> optic chiasm -> optic tract -> pretectal region of the midbrain -> Lateral geniculate nucleus in the thalamus
C. Optic nerve -> optic chiasm -> brachium of the superior colliculus -> pretectal region of the midbrain -> Edinger-Westphal nucleus
D. Optic nerve-> optic chiasm -> Lateral geniculate nucleus in the thalamus-> Edinger-Westphal nucleus

A

C. Optic nerve -> optic chiasm -> brachium of the superior colliculus -> pretectal region of the midbrain -> Edinger-Westphal nucleus

The pupillary fibers exit the eye through the optic nerve and pass through the optic chiasm, where they then exit the optic tract and enter into the brachium of the superior colliculus and synapse onto cells in the pretectal area of the midbrain. The pathway then continues and stimulates intercalated neurons, which in turn stimulate cells in the Edinger-Westphal nucleus. The axons of the pupillary pathway never enter the lateral geniculate nucleus.

238
Q

What is the name for the phenomenon in which a flickering light that is 10 Hz is seen as brighter than a steady light (one that does not flicker) that possesses the same average luminance?

A. The Granit-Harper law
B. The Brucke-Bartley effect
C. The Troxler effect
D. The Purkinje tree

A

B. The Brucke-Bartley effect

The Brucke-Bartley effect describes the fact that a flickering stimulus that is 10 Hz will appear brighter than a non-flickering light with the same average luminance. This fact also holds true for stimulus presentation duration. A light that is presented for 50 milliseconds will appear brighter than stimuli presented for longer or shorter durations.

The Troxler effect occurs when the eye is fixated (although the eye is truly always moving) on a point in space and the surrounding background begins to blend together. There must be several factors that come into play in order for the Troxler effect to transpire. The best example of this phenomenon is the figure in which there are two squares that are superimposed. The smaller square is centered in the larger square and is slightly lighter than the larger surrounding square. The border between the squares is blurred, resulting in a distinction of the two squares based upon brightness alone. When fixating upon an X placed in the center of the smaller square, the border completely disappears as does the smaller square, resulting in the perception of one uniformly-colored large square. Some patients experience the Troxler effect while performing the FDT visual field and they report, especially during testing of the second eye, that the entire field appears to go gray. When this occurs make sure to inform the patient to blink.

The peripheral retina is most sensitive to flicker because it is a part of the magno system which is also known as the “where” system. The magno system is noted for its poor spatial resolution but good temporal resolution. The fact that the peripheral retina is more sensitive to flicker is in accordance with the Granit-Harper law, which states that as the log of the area of the stimulus is increased, the critical flicker fusion frequency also increases accordingly. This helps to explain why the chances of perceiving flicker are greater for a larger stimulus for a given modulation. Remember that the receptors of the peripheral retina display increased summation; therefore, a larger stimulus will take up more area of the retina, increasing the chances of detection.

Due to the fact that the eye does not respond well to low temporal frequencies, stabilized retinal images are unable to be detected. Blood vessels that lie on top of the photoreceptors are considered stable relative to the retina. However, if you shine a penlight against your eyelid and move it rapidly, you will be able to see the shadows cast by the blood vessels on your retina. This is called a Purkinje tree. The penlight must be moving to ensure that the temporal frequency is great enough to be visualized.

239
Q

What wavelength of light is most readily absorbed by the photopigment rhodopsin?

A. 480 nm
B. 555 nm
C. 460 nm
D. 507 nm

A

D. 507 nm

The absorption spectrum curve for rhodopsin shows peak absorption for wavelengths that are roughly 507 nm. This is determined by taking a sample of rhodopsin and projecting a fixed quantity of monochromatic light onto it and measuring the amount of light that is transmitted. Light that is not transmitted is absorbed; therefore the absorption curve and the transmitted curve are reciprocals of each other. The procedure is repeated with many different wavelengths. The wavelength that results in the greatest amount of absorption (or the least amount of transmitted light) is obviously the one that has the highest probability of absorption by rhodopsin. 555 nm is the peak sensitivity for photopic conditions.

240
Q

Which area of the extrastriate cortex is involved in the perception of motion?

A. Visual area 5 (V5)
B. The inferotemporal cortex (IT)
C. Visual area 4 (V4)
D. Visual area 1 (V1)
E. Visual area 2 (V2)
A

A. Visual area 5 (V5)

Beyond the striate cortex, visual information is thought to split into two separate streams: the ventral (also known as the “what” or temporal stream), and the dorsal (also known as the “where” or parietal stream). V5, also known as the middle temporal cortex (or MT), is considered to be a part of the dorsal stream and is believed to code for motion. V4 and IT are a part of the ventral stream. V4 serves to play a role in processing color while IT is important in classifying complex shapes and form recognition such as faces. V1 is the primary visual cortex and is not considered a part of the extrastriate cortex.

241
Q

You decide to prescribe a beta-blocker for your patient diagnosed with primary open angle glaucoma. Taking into consideration the mechanism of action of this medication, what time of day is a beta-blocker MOST effective in lowering intraocular pressure?

A. In the evening
B. In the middle of the night
C. Right before the patient goes to bed
D. In the morning

A

D. In the morning

Explanation - Beta-blockers serve to decrease aqueous production. Aqueous production is measurably reduced at night when a patient is laying down and therefore beta-blockers are least effective at this time. Research has demonstrated that beta-blockers prove to be most effective in the morning.

242
Q

Upon examination of your 63 year-old female glaucoma suspect, you notice what appears to be an area of peripapillary atrophy surrounding the optic nerve of her left eye. Which 2 of the following statements are TRUE in regards to chorioretinal atrophy in this region? (Select 2)

A. The beta zone borders the disc margin and is surrounded by the alpha zone

B. The alpha zone borders the disc margin and is surrounded by the beta zone

C. Atrophy in the beta zone occurs more frequently and is typically larger in patients with glaucoma

D. Atrophy in the alpha zone occurs more frequently and is typically larger in patients with glaucoma

A

A &C

A. The beta zone borders the disc margin and is surrounded by the alpha zone
C. Atrophy in the beta zone occurs more frequently and is typically larger in patients with glaucoma

Explanation - Chorioretinal atrophy surrounding the optic nerve (also known as peripapillary atrophy) can be divided into two separate zones. The beta zone directly borders the margin of the optic disc, while the alpha zone is the outer region that concentrically surrounds the beta zone.

Chorioretinal atrophy that occurs in the beta zone will present with increased visibility of the sclera and large choroidal blood vessels, in contrast to the appearance of alpha zone atrophy, which displays an irregular variation of both hyper and hypopigmentation of the retinal pigmented epithelium.

Typically, when comparing peripapillary atrophy characteristics in normal and glaucomatous patients, the alpha zone is usually larger in patients with glaucoma; however, its frequency is similar in both groups. Conversely, both the size and frequency of beta zone atrophy is greater in patients with glaucoma, and the changes appear to be more pronounced in more severely affected eyes.

243
Q

At what time period of the day would the aqueous humor production be least?

A. Afternoon
B. Midnight to 6 AM
C. 8 AM to noon

A

B. Midnight to 6 AM

Explanation - The precise role and receptors specificity of adrenergic mechanisms in regulating the rate of aqueous humor formation are unclear. Studies using fluorophotometry have shown that beta adrenergic antagonists unequivocally decrease aqueous humor formation, particularly aqueous humor production during sleeping hours, during which production is decreased by up to 50%. This decrease is due to the B-arrestin/cAMP cascade regulation from the beta-adrenergic receptors. This does not mean that the intraocular pressure necessarily decreases at night; in fact, nocturnal intraocular pressure spikes have recently been reported; these may have a role in causing progressive glaucomatous damage.
Next Question
End Practice

244
Q

Which of the following side effects is MOST likely to be FIRST observed after the initiation of topical prostaglandin therapy used in the management of glaucoma?

A. Atrophy of periorbital fat
B. Darkening of the irides
C. Darkening of the periorbital skin
D. Ptosis
E. Conjunctival hyperemia
A

E. Conjunctival hyperemia

Explanation - Potential side effects associated with prostaglandin use include increased lash length, increased iris and periorbital tissue pigmentation, and periorbital fat cell atrophy. Signs of prostaglandin-associated periorbitopathy include increased inferior scleral show, ptosis of the upper lid, dermatochalasis involution, slight enophthalmos, atrophy of periorbital fat cells, and a deepening of the superior lid sulcus. However, the most common initial side effect is conjunctival hyperemia. This side effect can be bothersome to many patients, and therefore it is important for the primary eye care provider to determine which prostaglandin analog is best suited for the patient. Studies have demonstrated that latanoprost may be associated with the least amount of conjunctival hyperemia. Conjunctival hyperemia is purportedly mediated by nitric-oxide causing vasodilation of the conjunctival vasculature. Typically, the level of hyperemia decreases with increased time and continued use of the medication.

245
Q

Which of the following sentences regarding the aqueous humor is NOT true?
A. The aqueous humor is a source of antioxidants for the corneal endothelium and the lens
B. The aqueous humor is a carefully controlled filtrate of the blood
C. The aqueous humor carries oxygen via hemoglobin to the ocular structures
D. The aqueous humor helps to maintain the shape of the globe and also serves as a shock absorber

A

C. The aqueous humor carries oxygen via hemoglobin to the ocular structures

Explanation - Aqueous humor is a carefully controlled filtrate of the blood that supplies nourishment to both the cornea and crystalline lens. It has a high turnover rate, as it is produced and removed continuously from the eye. If either the production of aqueous humor stops or there is a sudden pathological leaking of fluid, the globe becomes soft and loses its shape. Aqueous humor functions as a shock absorber and provides nutrition including glucose and oxygen to both the lens and cornea.

246
Q

When assessing the anterior chamber angle, what is an advantage to using a Goldmann 3 mirror as opposed to a four mirror gonioscopy lens?

A. The 3 mirror lens does not need to be rotated on the eye
B. The 3 mirror provides greater optics
C. The 3 mirror allows for greater patient comfort
D. The 3 mirror does not require the use of a cushioning agent

A

B. The 3 mirror provides greater optics

Explanation - A four mirror is a very convenient tool used to assess the anterior chamber angle. This lens affords a faster exam, does not require a cushioning agent, and allows for greater patient comfort, as is it smaller and does not require rotation on the eye. The 3 mirror, however, offers greater and clearer optics

247
Q

Frequency doubling technology (FDT) has been thought to be an advantageous screening tool for early detection of visual field loss associated with glaucomatous damage. Which of the following statements BEST describes the basis for this concept?

A. FDT targets the large bistratified retinal ganglion cells, which are thought to be damaged in early glaucoma
B. FDT targets the magnocellular pathway, which is thought to be preferentially damaged in the early glaucoma process
C. FDT targets the parvocellular pathway, which is thought to be preferentially damaged in the early glaucoma process
D. FDT targets the small bistratified retinal ganglion cells, which are thought to be damaged in early glaucoma

A

B. FDT targets the magnocellular pathway, which is thought to be preferentially damaged in the early glaucoma process

Explanation - Frequency doubling technology (FDT) perimetry utilizes a low spatial frequency sinusoidal grating combined with a high temporal frequency counterphase flicker to produce a stimulus target. The FDT has been shown to specifically activate the magnocellular ganglion cells in order to see this type of stimulus. This is of great importance due to the fact that the magnocellular pathway is thought to be preferentially damaged in the early glaucoma process.

Recent studies have shown that the FDT demonstrates 85% sensitivity and 90% specificity for detecting early glaucomatous visual field loss, and 95% sensitivity and 95% specificity for detecting visual field loss in patients with moderate glaucomatous damage. Furthermore, this degree of effectiveness in early detection has allowed the FDT to reveal the presence of abnormalities in the visual field before they could be detected by standard automated perimetry.

It is also believed that short-wave automated perimetry (SWAP) is able to detect early glaucomatous functional loss by isolating the small bistratified ganglion cells that have also been thought to be damaged early in the glaucoma process.

248
Q

You wish to perform a visual field on a patient with advanced glaucoma using a Humphrey Field Analyzer. Which of the following visual field programs will BEST allow for monitoring of changes to or progression of visual field loss?

A. 40 point screener
B. 24-2
C. 10-2
D. 30-2

A

C. 10-2

Explanation - A patient with advanced glaucoma should be monitored using a threshold 10-2 test which tests 68 test points, 2 degrees apart in a central 10 degree area. Patients with advanced glaucoma possess a central island of vision, and therefore it is important to monitor for change within this region. A 40 point screener, 30-2 or 24-2 threshold visual field presents too many test points that lay outside of the patient’s area of useable vision and therefore do not provide useful information. A 30-2 field tests 76 points, 6 degrees apart within a central 30 degree radius. A 24-2 field presents a total of 54 points within a central 24 degree region (except nasally where it extends out to 30 degrees). When treating patients with advanced glaucoma, the objective is to save the central acuity. If the glaucoma is not adequately controlled, progression will cause a splitting of fixation due to inward advancement of the superior nasal field defect followed by inferior field loss due to encroachment of the nasal defect.

249
Q

Haab’s striae are clinical signs typically associated with congenital glaucoma that indicate ruptures of which of the following layers of the cornea?

A. Descemet's membrane
B. Bowman's layer
C. Endothelium
D. Epithelium
E. Stroma
A

A. Descemet’s membrane

Explanation - Haab’s striae are caused by horizontal breaks or ruptures in Descemet’s membrane. These striae may be single or multiple and are usually associated with congenital glaucoma and buphthalmos. It is important to distinguish these from the vertically oriented breaks in Descemet’s membrane that are associated with trauma from forceps delivery during birth.

250
Q

Which of the following topical nonsteroidal anti-inflammatory drugs is also approved for seasonal allergic conjunctivitis?

A. Diclofenac (Voltaren®)

B. Nepafenac (Nevanac®)

C. Flurbiprofen (Ocufen®)

D. Ketorolac (Acular®)

A

D. Ketorolac (Acular®)

Explanation - Acular® is used to decrease conjunctival and post-cataract inflammation, prevent and treat cystoid macular edema, and control discomfort associated with seasonal allergic conjunctivitis

251
Q

Which of the following types of aberrations is primarily responsible for results found using the Red-Green test when performing a subjective refraction?

A. Radial astigmatism

B. Chromatic aberration

C. Spherical aberration

D. Coma

E. Curvature of field

A

B. Chromatic aberration

Explanation - Chromatic aberration is a type of distortion in which a lens is unable to focus all wavelengths of light to the same convergence point. It occurs due to the fact that the refractive index of any medium (other than a vacuum) varies with wavelength. Therefore, shorter wavelengths (blues) will bend more as they pass through a lens, while longer wavelengths (reds) bend less. This leads to a variation in the image location (longitudinal chromatic aberration) and image size (lateral chromatic aberration).

In a subjective refraction, it is ideal to have the retina centered between the image point for the shorter and longer wavelengths. The use of the Red-Green test allows this to occur. If a patient reports that the letters on the red side are clearer, this indicates that the center point is too far in front of the retina, and a minus lens will improve this balance (Red Add Minus-RAM). If the patient reports that the green side is sharper, the center point is too far behind the retina, and a plus lens will shift the balance (Green Add Plus-GAP). This entire test is dependent on the principles of chromatic aberrations.

252
Q

Which hormone is secreted by the heart and helps to regulate blood pressure?

A. Erythropoietin

B. Calcitonin

C. Atrial natriuretic hormone

D. Melatonin

A

C. Atrial natriuretic hormone

Explanation - The atrial natriuretic hormone (ANH) is secreted by the atria in response to increased fluid volume signaled by atrial stretch. ANH causes the kidney to excrete greater quantities of sodium as well vasodilation of blood vessels, leading to a decrease in blood pressure.

Calcitonin is released by the thyroid; it inhibits osteoclast activity in bones and absorption of calcium by the intestines, leading to diminished blood calcium levels.

Melatonin is produced by the pineal gland in response to diminished light levels and causes the hypothalamus to ready the body for sleep.

Erythropoietin is released by the kidneys and stimulates the production of red blood cells by the bone marrow.

253
Q

All else being equal, what pupil diameter is associated with the least amount of spherical aberration?

A. 6.0 mm

B. 4.0 mm

C. 3.0 mm

D. 5.0 mm

A

C. 3.0 mm

Explanation - Spherical aberration increases with increased pupil size. Multiple sources state that the ideal pupil size is between 2.0-3.0 mm in size. Pupils greater than this size experience a larger degree of spherical aberration, while a pupil smaller than this size is optically limited by diffraction.

254
Q

During fetal development, which quadrant of the peripheral retina is the last to become fully vascularized?

A. Superior

B. Temporal

C. Inferior

D. Nasal

A

B. Temporal

Explanation - The retina is a unique tissue due to the fact that it is avascular until about the fourth month of gestation. At this time, arteries and veins begin to emanate from the hyaloid vessels at the optic disc and grow towards the peripheral retina. These blood vessels ultimately first reach the nasal periphery after 8 months gestation. The last area to become vascularized is the temporal retina, which occurs about 1 month after birth. The incompletely developed retinal vascular system at birth is what causes a premature infant to be particularly susceptible to oxygen damage, which can lead to the development of retinopathy of prematurit

255
Q

Which of the following skin conditions is considered to be benign and has the LOWEST risk of malignancy?

A. Keratoacanthoma

B. Actinic keratosis

C. Basal cell carcinoma

D. Squamous cell carcinoma

A

A. Keratoacanthoma

Explanation - Keratoacanthoma appears very much like squamous cell carcinoma (SCC) in that it tends to progress rapidly and appears to ulcerate. This condition typically occurs in middle-aged and elderly patients of Caucasian descent on areas of the skin that are exposed. The lesion appears elevated, and eventually the center will produce a scab-like plug of keratin. The margins surrounding the plug will be rolled. At some point the keratin plug will fall out, resulting in the formation of a pit, and the lesion will regress. Most patients and clinicians do not like to wait this condition out due to its similarities to SCC.

Actinic keratosis is a pre-cursor to squamous cell carcinoma and appears as scaly, dry skin that does not heal. People with skin that is of lighter pigmentation along with excessive exposure to ultraviolet light tend to be most at risk for development of this condition.

Squamous cell carcinoma (SSC) is thankfully one of the rarest malignancies but due to its ability to metastasize can be quite dangerous. This malignancy has the ability to progress rapidly and has a high affinity for people who spend a lot of time in the sun, especially those who are light-skinned. The only way to definitively diagnose SCC is to refer for a biopsy and ensuring the use of Mohs technique. This strategy takes more time but ensures that the lesion is removed. Essentially, Mohs procedure calls for removal of tissue and biopsy of the surrounding borders. If the borders prove to be malignant then more tissue is removed and biopsied. This continues until the borders prove to be free of any carcinoma.

Basal cell carcinoma (BCC) is the most common malignant lid lesion and mercifully tends to be very slow-growing. BCC generally appears as a waxy, translucent nodule. Eventually the nodule will ulcerate. Patients may bring these to your attention and tell you that they have “had it for years and it just does not seem to heal”. Whenever you hear this it is best to send out for biopsy via Mohs technique. BCC very rarely metastasizes.

256
Q

Which of the following vessels provides the majority of the blood supply to the optic nerve head?

A. Anterior ciliary arteries

B. Short posterior ciliary arteries

C. Long posterior ciliary arteries

D. Central retinal artery

A

B. Short posterior ciliary arteries

Explanation - The ophthalmic artery provides the majority of the supply of blood to both the inner retina and the optic nerve.

  • The central retinal artery (CRA) branch of the ophthalmic artery enters the optic nerve approximately 12mm behind the globe
  • In the retina, the retinal ganglion cell bodies and the nerve fiber layer are primarily supplied by capillary branches of the central retinal artery that emerge from the optic nerve head
  • As the CRA courses over the optic disc, it provides partial perfusion to some of the superficial optic disc; however, it provides minimal perfusion to the optic nerve itself, through which it courses
  • Branches of the medial and lateral short posterior ciliary arteries (SPCAs) also originate from the ophthalmic artery
  • These provide the majority of the blood supply to the optic nerve head as well as the choroid
  • Most notably, anastomoses of the SPCAs create the Circle of Zinn-Haller, which provides significant perfusion to the optic nerve head
257
Q

According to the American Diabetes Association, which of the following statements BEST describes the criteria for diagnosis of diabetes based on hemoglobin A1c (HbA1c) level?

A. HbA1c of 6.5% or greater is diagnostic for diabetes

B. HbA1c of 6.8% or greater is diagnostic for diabetes

C. HbA1c of 5.0% or greater is diagnostic for diabetes

D. HbA1c of 6.0% or greater is diagnostic for diabetes

E. HbA1c of 5.7% or greater is diagnostic for diabetes

A

A. HbA1c of 6.5% or greater is diagnostic for diabetes

Explanation - Current criteria for the diagnosis of diabetes include the following, explained in detail below:

  • HbA1c of 6.5%. The HbA1c test is a blood test that reflects the average blood glucose levels over a 3-month period and does not show daily fluctuations. This test does not require fasting and may be performed at any time during the day.
  • Fasting Plasma Glucose (FPG) of 126 mg/dL (7.0 mmol/L). Fasting is defined as no caloric intake for at least 8 hours. This test has been the most commonly used for diagnosing diabetes because it is convenient and cost-effective.
  • 2-hour plasma glucose of 200 mg/dL (11.1 mmol/L) during an oral glucose tolerance test (OGTT). This measures blood glucose after a person fasts for at least 8 hours then drinks a liquid containing 75 grams of glucose dissolved in water; the patient is then tested 2 hours after ingestion.

In a patient with classic symptoms of hyperglycemia or hyperglycemic crisis, a random plasma glucose of 200 mg/dL (11.1 mmol/L) may be observed.

In the absence of unequivocal hyperglycemia, the result should be confirmed by repeat testing.

258
Q

Which of the following astigmatic dials is fixed?

A. Rotary-T dial

B. Clock dial

C. Arrowhead dial

D. Raubitschek (paraboline) dial

A

B. Clock dial

Explanation - The clock dial is an example of a fixed astigmatism dial. The target is not mobile. Whereas the rotary-T dial, the arrowhead dial and the Raubitschek (paraboline) dial are all rotary. The rotary dials contain fewer lines than the clock dial. These targets are rotated to locate the principal meridians. During rotation of the target, when the patient reports that one of the lines is the clearest and blackest and that the line perpendicular to it is least clear, the endpoint has been reached. At this point cylinder power is added to neutralize the astigmatism

259
Q

What is the name of the funnel-shaped region anterior to the optic disc that represents the posterior termination of Cloquet’s canal?

A. Berger’s space

B. Erggelet’s space

C. Hyaloid space

D. Area of Martegiani

F. Patellar fossa

A

D. Area of Martegiani

Explanation - The area of Martegiani signifies the funnel-shaped dilation surrounding the optic disc representing the posterior termination of Cloquet’s canal (also known as the hyaloid canal).

Berger’s space is an area between that anterior face of the vitreous body and the posterior lens capsule. It represents the anterior termination of Cloquet’s canal. It is also known as Erggelet’s space.

The patellar fossa characterizes the anterior depression of the vitreous body in which the crystalline lens resides.

260
Q

A goiter is caused by a deficiency of which of the following substances?

A. Riboflavin

B. Calcium

C. Iron

D. Vitamin D

E. Iodine

A

E. Iodine

Explanation - The thyroid secretes thyroxine (T4), triiodothyronine (T3) and calcitonin. Proper feedback is required to ensure that adequate levels are synthesized and secreted. A lack of iodine causes circulating levels of thyroid hormones to decrease, causing the anterior pituitary to secrete the thyroid-stimulating hormone thyrotropin. This cycle causes overstimulation of the thyroid gland, resulting in glandular enlargement and the formation of a goiter.

261
Q

Traveler’s diarrhea, also known as Montezuma’s revenge, is most commonly caused by which organism?

A. Escherichia

B. Streptococcus

C. Salmonella

D. Staphylococcus

A

A. Escherichia

Explanation - Traveler’s diarrhea (TD) is usually linked to ingestion of contaminated water or food which results in loose stools, abdominal cramping, and nausea. The most frequent cause of TD is enterotoxigenic E. coli. Treatment is not usually required, as most cases are self-limiting; however, it is important to maintain hydration and prevent loss of electrolytes. The best prevention is to avoid questionable sources of food and water.

262
Q

You decide to perform fluorescein angiography on a patient with decreased vision and central metamorphopsia in his left eye. Results show a “smoke-stack” appearance in which fluorescein enters the subretinal space, ascends vertically, then spreads laterally. Which of the following retinal conditions is your patient MOST likely suffering from?

A. Choroidal neovascular membrane

B. Central serous retinopathy

C. Pigment epithelial detachment

D. Cystoid macular edema

A

B. Central serous retinopathy

Explanation - Fluorescein angiography studies of patients with central serous retinopathy typically reveal a classic “smoke-stack” presentation. This appearance is a result of leakage of fluorescein dye through the retinal pigment epithelium, which causes a small hyperfluorescent spot in the early stage of angiography. During the later venous stage, dye will continue to pass into the subretinal space, where it ascends vertically to the upper limit of the detachment before extending laterally until the entire subretinal space is filled with fluorescein.

Patients with cystoid macular edema will reveal a “flower-petal” pattern of hyperfluorescence on fluorescein angiography studies. This is caused by the accumulation of the fluorescein dye inside the microcystic spaces that have developed within the retina.

Classic choroidal neovascular membranes will show a well-defined membrane that fills with fluorescein in the early phase, creating a lacy appearance. The dye will then eventually leak after 1-2 minutes into the subretinal space surrounding the neovascular membrane.

263
Q

A soft contact lens with a power of -7.00 D states on the package that it has a Dk/t equal to 35. However, given your knowledge of Dk/t, you know that the lens actually possesses which of the following parameters?

A. Because the lens is minus in power, it is thickest in the periphery; thus the periphery of the lens has a Dk/t lower than that stated on the package

B. Because the lens is minus in power, it is thickest in the center; thus the center of the lens has a Dk/t higher than that stated on the package

C. Because the lens is minus in power, it is thickest in the periphery; thus the periphery of the lens has a Dk/t higher than that stated on the package

D. Because the lens is minus in power, it is thickest in the center; thus the center of the lens has a Dk/t lower than that stated in the package

A

A. Because the lens is minus in power, it is thickest in the periphery; thus the periphery of the lens has a Dk/t lower than that stated on the package

Explanation - The Dk/t of a lens is a measurement of the oxygen permeability for a given thickness. Generally, this measurement is given for a lens with a power of -3.00 D. Therefore, a lens that possesses a power higher than -3.00 D will actually display a lower Dk/t, especially towards the edges, because a minus lens is thickest in the periphery. A high plus lens will display a lower Dk/t, especially towards the center, because the lens is thickest in this area.

264
Q

Which of the following systemic diseases are MOST commonly associated with the presence of angioid streaks observed in the retina?

A. Ehlers-Danlos syndrome

B. Paget’s disease

C. Sickle cell disease

D. Pseudoxanthoma elasticum

E. Marfan syndrome

A

D. Pseudoxanthoma elasticum

Explanation - Approximately 50% of patients who present with angioid streaks have an associated systemic disease; the other 50% of cases are considered idiopathic. Pseudoxanthoma elasticum (PXE) is by far the most commonly associated systemic disease in these patients. In general, PXE is a rather uncommon, inherited, generalized connective tissue disorder whereby tissues in the body containing elastin are significantly affected. Up to 85% of patients with PXE will develop ocular complications, usually after the second decade of life. The combination of PXE and ocular involvement with angioid streaks is referred to as “Gronblad-Strandberg syndrome.” Patients with PXE typically have characteristic signs of very loose skin folds and yellow skin papules that are commonly observed in the neck region, axillae, and on flexor aspects of joints. These patients also frequently suffer from cardiovascular disease caused by accelerated atherosclerosis and have an increased risk of developing gastrointestinal bleeds, which can be life-threatening.

Ehlers-Danlos syndrome is another systemic condition that is occasionally associated with the presence of angioid streaks. It is a rare, usually dominantly inherited disorder of collagen in the body that is caused by a deficiency of hydroxylysine. Systemic features include thin, hyperelastic skin, hyperextensible joints, cardiovascular disease, and other systemic lesions. Besides angioid streaks, patients with Ehlers-Danlos syndrome can also develop other ocular conditions such as lens subluxation, blue sclera, high myopia, keratoconus and retinal detachments.

Angioid streaks also occur in about 2-10% of patients diagnosed with Paget disease. Paget disease is a chronic, progressive (inherited in some cases) disease that is characterized by an enlarged skull, bone pain, frequent bone fractures, hearing loss, and cardiovascular complications. The disease may be localized to a few bones or may be generalized. In some cases, patients are even asymptomatic; however, in late stages, significant vision loss can ensue due to optic nerve compression from enlarging bone. Lab testing in these patients will show an increased serum alkaline phosphatase and urine calcium level.

Less common systemic disorders that may cause the formation of angioid streaks include sickle-cell disease, acromegaly, senile elastosis, lead poisoning, and Marfan syndrome.

265
Q

Neovascularization associated with contact lens wear is generally located in which layer of the cornea?

A. Posterior stromal layers

B. Endothelium

C. Epithelium

D. Posterior to the endothelium

A

C. Epithelium

Explanation - Neovascularization associated with soft contact lens wear is generally located superficially and presents as an extension of vessels from the superficial marginal arcade beyond the limbus into the cornea. Most commonly, neovascularization is the result of over-wear from a contact lens that possesses a low Dk/t.

Stromal neovascularization can occur with contact lens wear but this is not the norm. Most stromal neovascularization is typically the result of infections such as chronic blepharoconjunctivitis, keratitis, phlyctenulosis, trachoma, or graft rejection.

266
Q

Clinical measurement of Snellen acuity is an example of which type of visual acuity?

A. Geometrical

B. Detection

C. Vernier

D. Resolution

A

D. Resolution

Explanation - Snellen acuity is an example of both recognition and resolution acuity. Recognition is the ability to recognize a target such as a letter or number. However in order for recognition to occur one must first be able to resolve the target. Resolution acuity is defined as the capacity to discern two or more targets that are spatially separated. The human threshold for resolution acuity is approximately 30 seconds of arc.

Detection acuity is described as the ability to determine whether a target is present or absent in an empty field. The minimum threshold for humans is one second of arc.

Vernier acuity is defined as the capacity to detect lateral misalignment of two objects that are separated by space. The minimum threshold for humans is 3 seconds of arc.

267
Q

An accumulation of which of the following metabolic wastes is associated with gout?

A. Ammonia	
B. Nitric acid	
C. Carbon dioxide
D. Urea
E. Uric acid
A

E. Uric acid

Explanation - Uric acid is a waste product created when nucleic acids are broken down. If uric acid is not eliminated in the urine, it may accumulate and crystallize in joints, causing gout. Initially, patients who experience gout will often complain of pain in their big toe, knee, or ankle; these are the most common joints to be affected.

Ammonia is formed during the process of deamination, wherein the amino groups are removed from amino acids that contain nitrogen.

Protein metabolism results in the formation of urea in the liver; this is caused by the joining of two ammonia molecules with carbon dioxide. Some urea that is filtered by the kidneys is reabsorbed; the rest of it is eliminated in the urine.

Kidney function is measured by evaluation of blood urea nitrogen. If these levels are found to be elevated, then the culprit is either depressed kidney function, body dehydration, heart failure, or a poor diet.

The body rids itself of the waste product carbon dioxide via exhalation through the lungs.

268
Q

While performing the bichrome test with the left eye occluded, the patient initially reports that the letters on the red side appear bolder, blacker and sharper. -0.25 D is then placed before the right eye and the patient reports that the letters on the green side now appear clearer. Which of the following is the MOST appropriate next step?

A. Leave the findings in the phoropter, occlude the right eye and proceed to the left eye
B. Add an additional -0.25 D over the right eye
C. Add +0.25 D over the right eye
D. Re-explain the goal of the test, as the patient is likely unclear as to the test instructions

A

C. Add +0.25 D over the right eye

Explanation - The bichrome test is performed by occluding one eye and projecting letters on a red and green background. The patient is initially fogged by roughly + 0.50 D. Fogging over-plusses the patient and shifts the chromatic interval forwards (causing objects projected on a background with longer wavelengths (red) to be bolder) such that the patient should report that the letters are blacker and bolder on the red side of the chart. The addition of minus lenses should equate the two sides such that the letters appear equally black and bold on the red and green sides. If the patient initially reports that the letters on the green side are bolder, then either plus lenses need to be added or minus powered lenses need to be reduced to shift the chromatic interval forwards. If the patient does not report equal boldness/blackness/sharpness of the letters on the green and red sides of the chart but instead instantly reports reversal upon switching lenses, the majority of clinicians will leave the power of the lens in the phoropter at the setting where the patient last reported that the letters on the red side of the chart appeared bolder/blacker to minimize the chances of over-minussing the patient.

269
Q

A common cause of epiphora in infants is caused by a small membrane that covers over which of the following structures?

A. The lacrimal gland
B. The puncta
C. The valve of Hasner
F. The canaliculus

A

C. The valve of Hasner
Explanation - It is common for mothers of young infants to note that one eye (or both eyes) of her infant constantly tears in conjunction with the presence of mucopurulent discharge. This epiphora results from a blockage of the nasolacrimal passageway caused by a membrane covering the valve of Hasner. The majority of blockages will self-resolve without intervention (80-90% of infants) within the first 12 months of life. Treatment may include massage of the nasolacrimal sac several times a day in an effort to rupture the membrane.

270
Q

Your 50 year-old patient currently wears soft contact lenses as monovision with the right eye distance and left eye near. Manifest refraction is OD: -2.00 -0.50 x 180 and OS: -2.25 -0.50 x 180. If the patient requires a +1.75D add, what would you expect the over-refraction to be if spherical soft lenses were placed on the eyes to allow for this type of vision?

A. OD: +2.00 -0.50 x 180 OS: +0.25 -0.50 x 180
B. OD: -1.50 -0.50 x 180 OS: +0.25 -0.50 x 180
C. OD: +0.25 -0.50 x 180 OS: -1.50 -0.50 x 180
D. OD: +0.25 -0.50 x 180 OS: +2.00 -0.50 x 180

A

C. OD: +0.25 -0.50 x 180 OS: -1.50 -0.50 x 180

Explanation - In the above case, the patient wishes to wear monovision contact lenses, with right eye distance and left eye near with a +1.75 effective add.
Because the right eye is to be used for distance, her over-refraction should be plano equivalent sphere (EDS) for the right eye. She does show -0.50 cylinder on refraction, so it will also likely present in the over-refraction as well. Therefore, in order to give the best possible distance vision in this eye, we would expect the over-refraction to be +0.25 -0.50 x 180 (plano EDS).
The left eye is to be used for near and the patient requires a +1.75 add power to achieve her best reading vision. Therefore, in order to achieve this, we would expect the over-refraction to be -1.75 equivalent sphere (EDS). However, she also has -0.50D of cylinder in her left eye that will need to be taken into consideration; this will then produce an expected over-refraction of -1.50 -0.50 x 180 (-1.75 EDS).

271
Q

Patients diagnosed with vernal keratoconjunctivitis have an increased incidence of which of the following corneal diseases?

A. Thygeson’s superficial punctate keratitis

B. Terrien’s marginal degeneration

C. Superior limbic keratoconjunctivitis

D. Keratoconus

A

D. Keratoconus

Explanation - Patients with vernal keratoconjunctivitis are more likely than the general population to develop keratoconus. This is due to the association of atopic disease in both conditions. Additionally, patients with VKC and keratoconus tend to have a more severe form of keratoconus that is commonly complicated by corneal hydrops and a greater tendency for corneal neovascularization.

272
Q

Which of the following rigid gas-permeable (RGP) lens designs is MOST appropriate for a patient presenting with corneal toricity of 0.50D at axis 090, and 1.75D of residual cylinder axis 090 on over-refraction (when a spherical RGP diagnostic lens is placed on the eye)?

A. Front Surface (F1) Toric
B. Toric Base Curve
C. SPE Bitoric
D. Spherical
E. CPE Bitoric
A

A. Front Surface (F1) Toric

xplanation - The following chart represents guidelines for lens selection when prescribing rigid gas-permeable (RGP) contact lenses. This will aid in determining which type of RGP lens is best for your patient, depending on the patient’s corneal toricity and residual over-refraction cylinder values (when a spherical GP diagnostic lens is placed on the eye).

The above patient presents with a minimal amount of corneal toricity, so a lens with a toric back surface is not required. A spherical contact lens; however, will not provide adequate vision, as there is significant residual cylinder (lenticular) on over-refraction. In these cases, a Front Surface Toric (F1 toric) lens is most appropriate as the base curve will be spherical, and the front surface will be toric to account for the residual cylinder.

273
Q

What common ocular finding is seen in patients with Osteogenesis Imperfecta (OI)?

A. Cataracts

B. Iris colobomas

C. Blue Sclera

D. Glaucoma

A

C. Blue Sclera

Explanation - Osteogenesis Imperfecta (OI) is a congenital, autosomal dominant condition that is characterized by brittle bones due to the production of abnormal type I collagen. There are four subtypes of OI, each with different severities and prognoses. Common symptoms include multiple bone fractures (due to brittle bones), potential hearing loss at an early age, and a blue hue to the sclera. Currently a cure does not exist for this condition; however, some drugs such as bisphosphonates and calcium supplements are delivering promising results. Therapies include caution with and minimization of dangerous or high-impact activities and practicing exercises geared towards building strong muscles and bones. Occasionally leg bones may need to be surgically reinforced with metal rods or braces. Some sources state that patients with OI are more likely to suffer from a retinal detachment due to poor scleral rigidity as well as the presence of subdural hematomas and retinal hemorrhages.

274
Q

In what order do the lens structures develop embryologically?

A. Lens pit, lens vesicle, lens plate (placode), lens capsule, embryonic lens nucleus
B. Lens plate (placode), lens capsule, lens vesicle, embryonic lens nucleus, lens pit
C. Lens vesicle, lens pit, lens plate (placode), embryonic lens nucleus, lens capsule
D. Lens plate (placode), lens pit, lens vesicle, embryonic lens nucleus, lens capsule
E. Lens plate (placode), lens vesicle, lens pit, lens capsule, embryonic lens nucleus

A

D. Lens plate (placode), lens pit, lens vesicle, embryonic lens nucleus, lens capsule

Explanation - The lens plate (placode) forms from surface ectoderm adjacent to the optic vesicle, beginning around day 27 of development. The outer surface of the lens plate invaginates to form a lens pit which continues until it forms the lens vesicle. The lens vesicle which separates from the surface ectoderm around day 33; it is a single layer of cells forming a hollow sphere. After the lens vesicle is formed, posterior epithelial cells become the primary lens fibers and form the embryonic lens nucleus at the center of the developing lens. The anterior cells stay in place at this time, and the equatorial cells divide and elongate to form secondary lens fibers around the embryonic nucleus. The lens capsule is apparent at about 5 weeks; it originates from the basement membrane of the surface ectoderm and from secretions of the lens epithelium.

275
Q

An elderly patient is seen at your office for a follow-up visit. She was diagnosed with iritis 1 week ago and is currently being managed with Pred Forte® topical eye drops every 2 hours. She reports no change in her symptomology and is still very photophobic, and her eye still bothers her. Slit lamp evaluation reveals no change in her clinical signs. Which of the following conditions may be the underlying cause of a non-responsive iritis?

A. An intraocular tumor
B. Behcet’s syndrome
C. Tuberculosis
D. Rheumatoid arthritis

A

A. An intraocular tumor

Explanation - Patients, especially the elderly, may exhibit a non-responsive iritis, which is caused by an intraocular tumor, leukemia, metastatic cancer, or non-Hodgkin’s lymphoma. Therefore, if a patient is being treated for iritis and the condition is not responding to conventional topical therapeutic treatment, further evaluation is warranted.

Patients with iritis, also known as anterior uveitis, typically will report photophobia, excessive lacrimation, pain, and diminished visual acuity. This condition is caused by either inflammation of the iris or of both the iris and the anterior portion of the ciliary body (iridocyclitis). Clinical signs include: keratic precipitates (which are variable in size and distribution, depending upon the etiology of the iritis) deposited on the corneal endothelium, cells and flare (protein that has leaked from iris vessels into the anterior chamber), sluggish and slightly constricted pupils caused by swelling of the uveal tract, irregular pupil margins (if posterior synechiae are present), and in the event of granulomatous inflammation, iris nodules. An acute episode of anterior uveitis or chronic uveitis may cause the formation of posterior synechiae. Cells may also be present in the vitreal chamber; however, the number of cells in the anterior chamber should exceed the number seen in the vitreous cavity. Treatment of iritis includes a cycloplegic agent to help prevent formation or promote breakdown of synechiae. A cycloplegic also aids in pain management by controlling pupil size and avoiding unnecessary movement of the iris muscles, which can be very painful with this inflammatory condition. UV protection (tinted lenses) is also helpful in management of the associated photophobia. A topical steroid is prescribed to reduce the inflammatory response. It is very important to slowly taper the use of the steroid; otherwise there is a risk of rebound inflammation.

Although rheumatoid arthritis, Behcet’s syndrome, and tuberculosis can lead to the development of an iritis, this associated iritis can typically be managed via traditional methods such as topical steroid and cycloplegic drops.

276
Q

Of the four major variants of leukemia, which type typically presents in children and has the BEST prognosis following proper treatment?

A. Acute lymphocytic leukemia
B. Acute myelocytic leukemia
C. Chronic myelocytic leukemia
D. Chronic lymphocytic leukemia

A

A. Acute lymphocytic leukemia

Explanation - Acute lymphocytic (lymphoblastic) leukemia (ALL): This type of leukemia predominantly affects children but can affect adults as well (typically over the age of 65). The standard treatment for ALL involves chemotherapy and radiotherapy. Survival rates for children are very high, as about 85-90% of cases respond to treatment and close to 70% of patients become cured. Prognosis is more guarded in older populations, with an observed survival rate of about 50%.

An easy way to remember the answer to this question is that “ALL kids survive” (in ALL cases, the prognosis for children is very good).

Acute myelocytic (myeloblastic) leukemia (AML): AML is most frequently found in older populations and rarely affects children. The treatment of choice in these cases is chemotherapy, and remission rates are about 30% (for those under the age of 60).

Chronic lymphocytic leukemia (CLL): CLL typically affects adults over the age of 55. It may rarely affect younger adults, but almost never occurs in children. There is no cure for this type of leukemia, but there are many effective treatments. CLL has a very chronic course and is not typically fatal.

Chronic myelocytic leukemia (CML): CML more frequently affects older populations more frequently than younger adults or children. It has a progressive, chronic course with a less favorable prognosis. The mainstay of treatment for these patients is imatinib (a tyrosine-kinase inhibitor).

277
Q

An 81-year old female reports that her eye has been watering more frequently over the past month; you decide to administer the primary Jones dye test (Jones I). After 5 minutes, the application of a cotton-tipped applicator to the inferior turbinate reveals the presence of dye in the area. Taking this into consideration, what is the MOST likely cause of the patient’s epiphora complaint?
A. Hypersecretion of tears
B. Dysfunction of the valve of Hasner
C. Partial nasolacrimal duct obstruction
D. Punctal stenosis
E. Complete nasolacrimal duct obstruction

A

A. Hypersecretion of tears

Explanation - The primary Jones dye test can be utilized to determine the patency of the nasolacrimal system. 1-2 drops of fluorescein are instilled into the inferior fornix of the eyes while the patient is in an upright position and blinking her eyes normally. After a period of 5 to 10 minutes, a cotton-tipped applicator is used to swab the undersurface of the inferior turbinate on each side of the nasal passage.

When the primary Jones dye test is positive (dye is recovered from the inferior turbinate of the nose), practitioners may conclude that the system is patent and that no significant blockage of the nasolacrimal drainage structure is likely. However, minor stenosis or physiologic dysfunctions cannot be completely ruled out. Patients who have a positive result on the Jones I test are more likely to experience symptoms of epiphora that are secondary to primary oversecretion of tears, rather than a dysfunction in lacrimal drainage (as in the above question).

When the primary Jones dye test is negative, the probability of an obstruction or dysfunction in lacrimal drainage is much greater; however, this test alone is not sufficient to document this conclusion. The secondary Jones dye test is then necessary to determine the severity and location of the obstruction.

278
Q

Which of the following BEST describes the properties of the entrance pupil of an optical system?

A. The image of the field stop through all preceding lenses
B. The image of the field stop through all following lenses
C. The image of the aperture stop through all following lenses
D. The image of the aperture stop through all preceding lenses

A

D. The image of the aperture stop through all preceding lenses

Explanation - The definition of the entrance pupil of an optical system is the image of the aperture stop through all preceding lenses. In other words, it is the aperture stop placed into object space. In situations where no lenses precede the aperture stop, the aperture stop is also the entrance pupil.

The exit pupil of an optical system can be defined as the image of the aperture stop through all following lenses. Again, if no lenses follow the aperture stop, the aperture stop is also the exit pupil.

279
Q

Which of the following types of melanocytic nevi are formed by nevus cells located solely at the border of the epidermis and dermis?

A. Intradermal nevi

B. Compound nevi

C. Blue nevi

D. Spitz nevi

E. Junctional nevi

A

E. Junctional nevi

Explanation - Junctional nevi are composed of nevus cells that are located at the junction between the epidermis and dermis. They are usually flat and brown or black in color.

Compound nevi are comprised of nevus cells that extend from the epidermis into the dermis layers of the skin (mixture of junctional and intradermal proliferation). They are usually slightly raised and the shade of pigment varies from tan to dark brown.

Intradermal nevi are the most common form of an acquired nevus. The nevus cells are completely confined to the dermis. This type of nevi is typically raised, and most are flesh-colored (not pigmented).

Blue nevi are composed of spindle-shaped nevus cells that are located deep within the layers of the dermis. The covering epidermis is normal, leading to a bluish-colored lesion.

Spitz nevi are a variant of an intradermal nevus that commonly presents as a raised, reddish lesion.

280
Q

Antihyperlipidemic drugs such as Lipitor® (atorvastatin) serve to decrease the risk of cardiovascular disease primarily by which mechanism?

A. Increasing HMG-CoA (hydroxy-3-methyl-glutaryl-Coenzyme A) reductase

B. Increasing triglyceride synthesis

C. Decreasing high-density lipoproteins (HDL)

D. Decreasing low-density lipoproteins (LDL)

A

D. Decreasing low-density lipoproteins (LDL)

Explanation - Elevated levels of low-density lipoproteins and triglycerides are associated with increased risk of heart disease and stroke because they can lead to plaque formation and build-up on the inside of arteries, which narrows the lumen of the blood vessels. HDLs aid in decreasing the risk of cardiovascular disease; therefore, elevated levels of this type of cholesterol are desirable. HMG-CoA reductase is the enzyme found in the mevalonate pathway, which is responsible for the synthesis of cholesterol. Drugs that inhibit HMG-CoA reductase decrease the concentration of cholesterol in hepatic cells, which results in an increase in the synthesis of hepatic receptors for LDL. This in turn causes decreased levels of plasma LDL, thereby resulting in an increase in plasma HDL levels.

281
Q

At which location of the retina is the peak density of rod photoreceptors located?

A. 6.0mm from the foveal pit

B. 2.5mm from the foveal pit

C. 4.5mm from the foveal pit

D. 1.0mm from the foveal pit

E. 0.5mm from the foveal pit

A

C. 4.5mm from the foveal pit

Explanation - In the human fovea there are no rod photoreceptors present; cones are the only photoreceptors in this area, and they are perfectly arranged in a hexagonal mosaic pattern. Outside of the foveal region, the rod photoreceptors are introduced, breaking up this tight hexagonal cone packing. However, this architecture is still very organized, as cones are rather evenly spaced around the rods. Cone density rapidly falls outside of the fovea and remains at a steady density in the peripheral retina. As the cone density rapidly declines, the rod photoreceptor density quickly increases to a peak density in a ring around the fovea (also known as the “rod ring”). This is located about 4.5mm from the center of the fovea (or 18 degrees from the foveal pit).

282
Q

Which of the following best describes a monocyte?

A. A type of leukocyte responsible for mast cell release

B. A leukocyte that triggers the formation of membrane attack complexes

C. An immature macrophage located in the blood

D. A type of granulocytic white blood cell

A

C. An immature macrophage located in the blood

Explanation - There are two major classes of leukocytes: granulocytes and agranulocytes. Neutrophils, basophils and eosinophils all fall under the category of granulocytes due to the fact that they contain a lobed nucleus and possess cytoplasmic granules visible upon staining.

The second class of leukocytes is agranulocytes, aptly named because they do not possess cytoplasmic granules. This class contains two types of cells, monocytes and lymphocytes. Monocytes are immature macrophages that move into an infection site and divide and differentiate into macrophages and dendritic cells that destroy microbes and debris via phagocytosis. Lymphocytes, the T cells and the B cells, are involved in hosting specific immune responses.

Basophils release histamine, along with other substances, that prolong the inflammatory response.

283
Q

Which of the following types of visual field defects occurs EARLIEST in patients exhibiting glaucomatous optic nerve damage?

A. Arcuate scotoma

B. Ring scotoma

C. Paracentral scotoma

D. Blind spot enlargement

E. Nasal step

A

C. Paracentral scotoma

Explanation - Characteristic defects in glaucoma consist of damage to the optic nerve head, resulting in a retinal nerve fiber bundle defect. The configuration of nerve fibers served by the damaged bundle will correspond to a specific defect in the visual field. The earliest visual field changes that may suggest glaucomatous damage commonly consist of an increased variability of responses in an area that will eventually develop a defect.

When a glaucomatous visual field defect does occur, it tends to initially present as a paracentral scotoma. Paracentral scotomas are typically small and relatively steep depressions that are most commonly observed just supero-nasal to the fovea. Approximately 70% of all early glaucomatous field defects can be characterized as a paracentral scotoma. This type of defect is due to damage of the papillomacular bundle, which will respect the horizontal midline.

It is important to remember that a single visual field test cannot definitively prove that a visual field defect exists. For this reason, interpretation of visual fields should not be performed in isolation but rather in conjunction with other clinical findings (IOP, appearance of optic nerve, RNFL). According to Kanski’s Clinical Ophthalmology, there is a set of minimal criteria for determining glaucomatous damage (also known as Anderson’s criteria), summarized below:

  1. Glaucoma hemifield test that is “outside normal limits” on at least 2 consecutive occasions.
  2. A cluster of 3 or more non-edge points in a location typical for glaucoma, all of which are depressed on pattern standard deviation (PSD) at a P < 5% level and one of which is depressed at a P <1% level, on 2 consecutive occasions.
  3. Corrected pattern standard deviation (CPSD) that occurs in less than 5% of normal individuals on two consecutive fields.
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    End Practice
284
Q

Base-in prism can be induced by which of the following means?

A. Decentering the optical center of a plus powered lens inwards (towards the patient’s nose)
B. Decentering the optical center of a minus powered lens inwards (towards to patient’s nose)
C. Decentering the optical center of a minus powered lens downwards
D. Decentering the optical center of a plus powered lens upwards

A

A. Decentering the optical center of a plus powered lens inwards (towards the patient’s nose)

Explanation - Rather than specifically ordering prism, low amounts can sometimes be induced by decentering the optical center of the ophthalmic lens from the patient’s pupillary distance (PD), causing the patient to look through a different area of the lens other than the optical center. If a minus powered lens is decentered inwards (nasally), the induced prism will be will be base out; if the same lens is decentered outwards (temporally), the induced prism will be base in. A plus powered lens that is decentered nasally induces base in prism, while temporal decentration induces base out prism.

285
Q

The choriocapillaris is separated from the retina by a thin membrane. What is the name of this membrane?

A. Sattler’s membrane

B. Bowman’s membrane

C. Bruch’s membrane

D. Descemet’s membrane

A

C. Bruch’s membrane

Explanation - Bruch’s membrane lies between the choriocapillaris of the choroid and the retinal pigment epithelium of the retina. Although this membrane is very thin (about 2 microns thick) it is very complex. The membrane consists of five facets. The outermost component is the basement membrane of the choriocapillaris followed by the outer collagenous zone, the elastic layer, the inner collagenous zone and most internally the basement membrane of the retinal pigment epithelium.

Sattler’s membrane is actually a layer of vessels in the choroid located externally to choriocapillaris.

Bowman’s and Descemet’s membranes are found in the cornea.

286
Q

The unilateral cover test reveals the following results: cover right eye, left eye turns in: cover left eye, no movement of right eye: cover right eye, no movement of left eye. How would you classify these results?

A. Intermittent left exotropia

B. Constant left exotropia

C. Constant left exophoria

D. Intermittent left exophoria

A

A. Intermittent left exotropia

Explanation - Movement of the uncovered eye on the unilateral cover test indicates that eye was not directed at the target of regard. In this instance, movement of the left eye was noted on the first attempt at covering the right eye but the left eye remained stationary when the test was repeated. Any movement seen on the unilateral cover test is diagnosed as a tropia of the eye that moved. Since the left eye did not move each time the right eye is covered, it is described as an intermittent movement. Since the left eye rotated outward, this is described as a left exophoria. The presence of a phoria is tested for by administration of the alternating cover test. Phorias cannot be subdivided as intermittent or constant.

287
Q

If a real object located at optical infinity results in the formation of an image that is focused on the retinal surface of an emmetropic, reduced eye (n=1.33, radius of curvature= 5.55 mm), which of the following can be calculated?

A. The height of the object
B. The thickness of the retinal surface
C. The size of the exit pupil
D. The axial length of the reduced eye

A

D. The axial length of the reduced eye

Explanation - Initially, one can determine the power of the refracting surface of the reduced eye by using the equation P=(n’-n)/r where P= the power of the refracting surface, n’= the refractive index of the surface, n= the refractive index of the surrounding medium, and r= the radius of curvature of the surface in meters. Using the known values given in the problem, one can calculate the value of P, P=(1.33-1.0)/0.00555= +59.46 D; rounding up to the nearest whole diopter yields +60.00 D. The value of P is then inserted into the equation L’=L+P where L’=the image vergence (n’/l’ in meters), L=the object vergence (n/l in meters), and P=the power of the refracting surface. The value of the object vergence (L) is 0 because the object rays are parallel. Therefore, solving for L’=60.00 +0, (n’/l’)=60.00 D, 1.33/l’=60, l’=0.02216 or 22.2 mm.

288
Q

Which of the following is NOT TRUE in regards to the characteristics of wave motion?

A. The frequency of a wave slows when light travels from air into a denser medium

B. For a wave of a given velocity (e.g. light), frequency and wavelength are inversely related

C. The speed of light (or c) is a universal constant that is independent of the observer or the source

D The amount of energy of light is proportional to frequency, and inversely related to wavelength

A

A. The frequency of a wave slows when light travels from air into a denser medium

Explanation - The frequency of light is invariant across media; therefore, frequency remains constant when light travels from one medium to another. On the other hand, wavelength and velocity are dependent on the medium.

289
Q

Which of the following female reproductive organs secretes estrogen and progesterone?

A. The ovaries

B. The cervix

C. The oviducts

D. The uterus

A

A. The ovaries

Explanation - The ovaries produce oocytes as well as the female sex hormones estrogen and progesterone. The oviducts serve as a passageway for eggs from the ovaries to the uterus. Also, the oviducts are the primary site of fertilization. The uterus is a chamber capable of enlargement in order to accommodate a developing baby. The cervix secretes mucus to help sperm motility and also serves to help protect the developing fetus from bacterial infections (post-fertilization).

290
Q

Which one of the following bitoric GP contact lenses would NOT induce cylinder if rotated to a misaligned position on the eye?

A. All of the options listed would induce cylinder if rotated off axis

B. 7.46 mm / -4.25 D

- -------------------- 
8. 13 mm / -1.75 D

C. 7.63 mm / -1.50 D

  - -------------------- 
 8. 11 mm / +1.12 D

D. 7.58 mm / -5.37 D

 - -------------------- 
  8. 18 mm / -0.50 D

E. 7.54 mm / +1.50 D

  - -------------------- 
  7. 99 mm / +2.75 D
A

C. 7.63 mm / -1.50 D

  - -------------------- 
 8. 11 mm / +1.12 D

Explanation - Cylinder power effect (CPE) bitoric and base curve toric (with a spherical front-surface) gas-permeable (GP) lenses will induce unwanted cylinder if the lens rotates off axis. The resulting cylinder is due to cross-cylinder effects. However, a spherical power effect (bitoric) will not induce unwanted cylinder regardless of lens rotation. To determine whether a GP lens is a spherical power effect (SPE) or cylinder power effect (CPE) bitoric, measure the two base curves using a radiuscope and the two raw contact lens powers using a lensometer. If the difference between the two base curve meridians in diopters is the same as the difference between the two raw powers, the lens is an SPE bitoric. This is the case for only one of the above answers. Converting mm of base curve radius to diopters results in 7.63 mm = 44.25 D and 8.11 mm = 41.62; a difference of 2.62 D. The difference between the two raw powers of +1.12 D and -1.50 D is also 2.62 D. Therefore, this lens is a spherical power effect (SPE) bitoric GP contact lens.

291
Q

Why is ultraviolet light (UV) considered mutagenic?

A. It causes incorporation of acridine dyes, leading to frameshift mutations

B. It leads to the creation of base analogs that resemble bases, causing faulty base pairing

C. It leads to back mutations and reversions

D. It causes the formation of thymine dimers

A

D. It causes the formation of thymine dimers

Explanation - The formation of thymine dimers disrupts proper DNA replication. UV light causes covalent bonds to form between adjacent thymine bases which then inhibit polymerases and DNA replication. The offending nucleotides can be excised by the body but, if they are left, dimers can causes melanomas in humans.

Acridine used to be a main component in hair dye. This chemical would become incorporated via intercalation into DNA and lead to frameshift mutations secondary to insertion of nucleotides. A base analog is a chemical that resembles a nucleotide and may become incorporated into DNA, which leads to faulty base pairing. Back mutation or reversion describes a scenario in which a point mutation leads to the restoration of a wild-type phenotype or the original nucleotide sequence.

292
Q

What is the power of a concave mirror (in diopters) located in air with a radius of curvature of 20 cm?

A. +5.00 D

B. -5.00 D

C. +10.00 D

D. -10.00 D

A

C. +10.00 D

Explanation - A concave mirror converges light and therefore acts like a convex lens hence concave mirrors have positive dioptric values whereas convex mirrors diverge light and possess negative dioptric powers. The equation used to determine the power of a mirror is P=-2n/r where P= the power of the mirror in diopters, n= the index of refraction of the surrounding medium, and r= the radius of curvature of the mirror in meters. Unless otherwise stated, always assume that the index of refraction of air is 1.0. Input the values from the above question: P= -2(1)/-0.2=+10.00 D.

293
Q

From which of the following germ cell layers is the crystalline lens derived from?

A. Endoderm

B. Neural ectoderm

C. Mesoderm

D. Surface ectoderm

A

D. Surface ectoderm

Explanation - Unlike the rest of the human eye, which is mostly derived from the neural ectoderm, the crystalline lens is formed from surface ectoderm.

The first stage of lens differentiation begins at the 4mm embryonic stage, in which the lens plate (placode) forms from surface ectoderm adjacent to the optic vesicle (formed from neural ectoderm), beginning around day 27 of development. At this stage, the lens placode is a single monolayer of columnar cells. As development progresses, the outer surface of the lens plate deepens and invaginates to form a lens pit, which continues to invaginate until it forms the lens vesicle. By the 10mm stage, the lens vesicle completely separates from the surface ectoderm (around day 33) and forms a hollow sphere consisting of a single layer of cells. After the lens vesicle is formed, posterior epithelial cells become the primary lens fibers and form the embryonic lens nucleus at the center of the developing lens. The anterior cells stay in place at this time, and the equatorial cells divide and elongate to form secondary lens fibers around the embryonic nucleus. The lens capsule is apparent at about 5 weeks; it originates from the basement membrane of the surface ectoderm and from secretions of the lens epithelium.

294
Q

Which of the following represents the angle subtended by the visual and anatomical axes of the human eye?

A. Kappa

B. Lambda

C. Omega

D. Alpha

A

A. Kappa

Explanation - Angle kappa represents the angle subtended by the visual and anatomical axes of the eye. The visual axis is a line that passes from the center of the fovea, through the nodal point of the eye, and to the point of fixation. The anatomical axis of the eye is a line that passes from the center of the posterior pole through the center of the cornea. Because of the fact that the fovea is typically slightly temporal to the anatomical center of the eye, these 2 axes do not usually line up perfectly. The difference between these is angle kappa. Angle kappa is commonly about 5 degrees and is positive when the fovea is aligned temporally to the center of the posterior pole (resulting in a nasal displacement of the corneal reflex).

295
Q

A typical sperm contains how many chromosomes?

A. 46

B. 32

C. 23

D. 21

A

C. 23

Explanation - A typical sperm and a normal egg each contain 23 chromosomes. During reproduction the two join, eventually resulting in the creation of a baby with 46 chromosomes.

296
Q

As applied to color vision, what is the proper term for two stimuli that appear the same but are physically different?

A. Metamers

B. Dichromats

C. Scalar properties

D. Grassmann’s law

A

A. Metamers

Explanation - Two stimuli that are physically different but appear to be the same are called metamers. A dichromat is a person who, if given three wavelengths of light that are divided into two separate patches, is able to adjust the intensities of the wavelengths so that the two patches appear the same. For example, on one side of the patch is wavelength A and on the other side are wavelengths B and C. A dichromat would be able to adjust the intensities of wavelengths B and C so that they appear to be identical; however, a trichromat would not. Grassmann’s laws of metamers consists of the additivity property, the scalar property, and the associative property. The additive property states that if you add an equal amount of a new wavelength to each metamer, they will remain metamers. The scalar property of metamers states that if the intensities of the metamers are increased or decreased equally, they will remain metamers. Lastly, the law of associative property of metamers explains that if a metamer is substituted for another, the two metamers will still match. For example, wavelength A+B= wavelength C+D. However, wavelength A= wavelength K+J; therefore, wavelength K+J+B= wavelength C+D.

297
Q

You have a patient who you correctly diagnose with idiopathic demyelinating optic neuritis. The presence of which of the following characteristics on magnetic resonance imaging (MRI) testing represents the greatest risk for the future development of multiple sclerosis?

A. MRI shows one ovoid grey matter lesion greater than 3mm in size

B. Normal MRI

C. MRI shows six or more white matter lesions

D. MRI shows one ovoid white matter lesion greater than 3mm in size

E. MRI shows six or more grey matter lesions

A

C. MRI shows six or more white matter lesions

Explanation - Patients diagnosed with idiopathic optic neuritis are at a very high risk of subsequent development of multiple sclerosis; the risk can be as high as 74% at 15 years. The results of initial magnetic resonance imaging (MRI) testing are highly correlated with this risk and can help predict the chances of future development of multiple sclerosis (MS).

  • The overall risk for development of MS after optic neuritis is 50% (at 15 years)
  • When an MRI of the brain is normal, there is still a 23% chance of MS at 15 years
  • If a brain MRI shows one T-2 weighted ovoid lesion of the white matter (greater than 3mm in diameter), the 15-year risk of MS increases to 56%
  • The highest risk factor for MS is an MRI that shows at least six T2-weighted white matter lesions. These patients have a 74% chance of developing MS at 15 years.

Therefore, patients with normal MRIs may develop MS, and patients with abnormal MRIs may never develop MS. It is important to closely follow these patients with regular eye examinations and to coordinate care with their primary care physician.

298
Q

Which of the following BEST describes the design function of a fused crescent bifocal rigid gas-permeable (RGP) contact lens?

A. Alternating

B. Annular

C. Concentric

D. Aspheric

E. Simultaneous

F. Diffractive

A

A. Alternating

Explanation - Fused crescent bifocal rigid gas-permeable (RGP) contact lenses are fit as an alternating design. The appearance of the contact lens is similar to that of a bifocal spectacle lens in which there is a segment positioned in the inferior portion of the lens that posses a higher index of refraction than that of the main portion of the contact lens. The effectiveness of this type of lens is based on the idea that as the patient views in primary gaze, the pupil will be in the area of the distance Rx; further, as the patient looks down, the lens catches on the lower lid and stays in position while the pupil moves down into the add portion. In order for this to work properly, these lenses are typically prism ballasted so that to segment does not rotate, and truncated so as to catch on the lower lid. With this idea, the patient is “alternating” viewing through the distance and near portion of the lens as desired.

299
Q

During ocular development, the neural tube forms and the ectoderm can be divided into the neural ectoderm (which lines the neural tube) and the surface ectoderm (which is around the tube). The neural crest develops from a fold of ectoderm between the neural ectoderm and surface ectoderm. Which of the 3 following structures are derived from neural crest cells? (Select 3)

A. Corneal stroma

B. Retinal pigment epithelium

C. Uveal pigment cells

D. Ciliary body epithelium

E. Descemet’s membrane

F. Meibomian glands

A

A. Corneal stroma
C. Uveal pigment cells
E. Descemet’s membrane

Explanation - The neural crest gives rise to the corneal stroma, corneal endothelium including Descemet’s membrane, most of the sclera, trabecular structures and uveal pigment cells. The surface ectoderm gives rise to the lens, corneal epithelium, conjunctival epithelium, lacrimal gland, epithelium of the eyelids, meibomian glands and epithelial lining of the nasolacrimal system. The neural ectoderm gives rise to the retinal pigment epithelium, neural retinal fibers of the optic nerve, neuroglia, ciliary body epithelium, and the iris epithelium (including the iris sphincter and dilator).

300
Q

Which of the following statements BEST describes the changes in corneal diameter and curvature from birth to approximately age five?

A. The cornea steepens and overall diameter increases

B. The cornea steepens and overall diameter decreases

C. The cornea flattens and overall diameter increases

D. The cornea flattens and overall diameter decreases

A

C. The cornea flattens and overall diameter increases

Explanation - Corneal curvature from birth to age 6 months of age averages approximately 47.6D. As a child ages, the corneal curvature flattens to about 45.4D at 2 years of age, and continues to flatten to about 42.7 at 5 years old. Studies have shown that the corneal curvature from age 5 on is relatively stable, unless a person experiences a corneal injury, undergoes corneal surgery, or wears contact lenses. Most literature states that the average corneal curvature in adults is about 43D or 7.85mm.

301
Q

The lymphatic system serves many important roles in the human body. The lateral portion of the eyelid lymphatics drain into which of the following structures?

A. The conjunctiva
B. The puncta
C. The pre-auricular lymph node
D. The submandibular lymph node

A

C. The pre-auricular lymph node

The lateral 2/3 of the upper lid and the lateral 1/3 of the lower lid lymphatics drain into the pre-auricular lymph node located directly in front of the ear. The medial 1/3 of the upper eye lid and the medial 2/3 of the lower lid lymphatics drain into the submandibular node located just under the jaw-line. Therefore, it is very important to evaluate these two nodes separately, especially when a condition of viral etiology is suspected.

302
Q

Numerous reports have suggested that increased tear film osmolarity is a key consequence in dry eye. Although osmolarity is not easily measured in the clinical setting, tear osmolarity increases in most dry eye sub-types due to which of the following processes?

A. In aqueous tear deficiency, the lacrimal gland produces more ionic species

B. . The lipid layer is altered in most dry eye states, leading to ion pairing

C. Decreased capillary exchange leads to ionic bonding

D. Loss of tear stability induces an increased evaporation rate, leading to increased osmolarity

E. Reactive oxygen species are increased in the tears of most dry eye sub-types; this increases osmolarity

F. Patients with dry eye tend to blink less than normals, leading to increased evaporation

A

D. Loss of tear stability induces an increased evaporation rate, leading to increased osmolarity

Explanation
Tear instability leads to greater evaporation and higher osmolarity through a mechanism of concentration of the remaining tears, since only the aqueous tear portion evaporates rather than the ionic species. Several studies have indicated that normal tear osmolarity is less than or equal to 300 Osm/L, with values exceeding 308 Osm/L indicating increased osmolarity. As a single measure, tear osmolarity has recently been found to correlate the best (r squared 0.55) to dry eye severity of several clinical tests in a large, multi-center study (Sullivan et al., IOVS 51:6125-6130, 2010).

303
Q

Which of the following BEST describes the actions of the lacrimal system that occur when the eyes close during a blink?

A. Horizontal canaliculi are shortened, puncta moves medially, and lacrimal sac expands

B. Horizontal canaliculi are lengthened, puncta moves temporally, and lacrimal sac collapses

C. Horizontal canaliculi are lengthened, puncta moves temporally, lacrimal sac expands

D. Horizontal canaliculi are shortened, puncta moves medially, and lacrimal sac collapses

A

A. Horizontal canaliculi are shortened, puncta moves medially, and lacrimal sac expands

Explanation - As a person closes his eyes during a blink, the pre-tarsal orbicularis oculi compresses the vertical component of the canaliculi and shortens the horizontal canaliculi, which in turn causes the puncta to move medially. Simultaneously, the lacrimal portion of the orbicularis oculi also contracts, which results in expansion of the lacrimal sac. This action creates negative pressure, which draws the tears from the ocular surface through the canaliculi and into the sac.

304
Q

What is the name of the corneal surgical procedure in which the corneal epithelium and partial thickness of the stroma are transplanted, leaving the deep stroma and endothelium intact?

A. Descemet’s stripping endothelial keratoplasty

B. Lamellar keratoplasty

C. Penetrating keratoplasty

D. Deep lamellar keratoplasty

A

B. Lamellar keratoplasty

Explanation - A lamellar keratoplasty involves a partial thickness excision and transplantation of the corneal epithelium and stroma only, leaving the deep stroma and corneal endothelium intact. This type of procedure is typically indicated in patients with localized corneal thinning, marginal corneal thinning or infiltration, or opacification of the superficial 1/3 of the corneal stroma.

A deep anterior lamellar keratoplasty is a transplantation procedure in which all of the opaque corneal tissue is removed almost all the way to the level of Descemet’s membrane. It is indicated in patients who have a corneal disease involving 95% of the corneal thickness but who maintain a healthy corneal endothelium with an absence of breaks in Descemet’s membrane.

A penetrating keratoplasty is a surgical procedure in which the full thickness of the cornea is replaced by donor tissue. This procedure is used in cases where the endothelium is compromised, there is dense and deep corneal scarring, severely infected corneal tissue, or degenerative diseases that could potentially recur.

305
Q

How is the segment of a fused bifocal lens constructed compared to the segment of a one-piece bifocal lens, and what material is a fused bifocal made from?

A. Change of index of refraction, glass material

B. Change of curvature, plastic material

C. Change of index of refraction, plastic material

D. Change of curvature, glass material

A

A. Change of index of refraction, glass material

Explanation - A fused bifocal is constructed by joining a bifocal segment made of a higher index material (usually flint glass) with the carrier lens made of a lower index material (usually crown glass) together under high heat and pressure. A plastic lens could not withstand such heat and pressure without warping or melting; thus, a fused bifocal is typically made of glass. A one-piece bifocal is constructed out of the same material with the segment having a change of curvature. A one-piece bifocal is usually plastic such as CR-39. You can feel the change of curvature of a plastic bifocal by running your finger across the top of the bifocal segment.

306
Q

Which of the following procedures is NOT typically performed binocularly?

A. Randot Stereoacuity Test

B. Axis refinement with Jackson crossed-cylinder

C. Prism dissociation

D. Worth-4-dot

A

B. Axis refinement with Jackson crossed-cylinder

Explanation - During subjective refraction, cylindrical axis refinement is performed monocularly in order to achieve the most accurate results. The Worth-4-dot, prism dissociation, and Randot Stereoacuity Test are all completed binocularly.

307
Q

Which of the following vitamins is also known as thiamine?

A. Vitamin B2

B. Vitamin B1

C. Vitamin B6

D. Vitamin B3

A

B. Vitamin B1

Explanation - Vitamin B1 is also known as thiamine. Thiamine is important for carbohydrate metabolism and the maintenance of neuronal tissue. Although rarely seen in the United States due to the fact that most foods are enriched with vitamins, a B1 deficiency can result in beriberi. Symptoms of beriberi include difficulty walking, loss of sensation in hands and feet, paralysis (due to deterioration of myelin sheath), nystagmus, speech difficulty/mental confusion, ophthalmoplegia, and congestive heart failure. Alcoholism can lead to beriberi due to malnutrition. Vitamin B2, also known as riboflavin, aids in the metabolism of nutrients. Avitaminosis B2 can result in decreased visual acuity due to corneal neovascularization, keratoconjunctivitis sicca, or cataract formation. Vitamin B3 (a.k.a. niacin) is responsible for fat and carbohydrate metabolism. A deficiency of Vitamin B3 can cause pellagra, which results in diarrhea, dementia, and dermatitis. Niacin can be used to treat hyperlipoproteinemia. Vitamin B6, also called pyridoxine, serves an important role as a coenzyme for the formation of hemoglobin as well as amino acid and protein metabolism

308
Q

The vertical recti muscles are inserted in front of the equator, creating what angle with the visual axis?

A. 23 degrees

B. 51 degrees

C. 39 degrees

D. 67 degrees

E. 19 degrees

A

A. 23 degrees

Explanation - The medial and lateral walls of the orbit are positioned at an angle of 45 degrees from each other. The orbital axis then forms an angle of 22.5 degrees with the medial and lateral walls (for the sake of simplicity, this angle is usually regarded as 23 degrees). Therefore, in primary gaze, the visual axis also forms an angle of 23 degrees with the orbital axis. The actions of all of the extraocular muscles depend on the position of the eye as each undergoes contraction. The vertical recti muscles run along the same line as the orbital axis, thereby also creating an angle of 23 degrees with the visual axis at their attachment point anterior to the equator. When the globe is abducted 23 degrees, the vertical recti muscles act purely to elevate or depress the eye.

309
Q

What is the equivalent base curve in millimeters of a rigid gas permeable contact lens with a base curve noted to be 49.75D?

A. 6.85mm

B. 6.68mm

C. 6.92mm

D. 6.78mm

E. 6.98mm

A

D. 6.78mm

Explanation - The formula for converting a base curve represented in diopters to millimeters is: 
mm = 337.5 / D 
mm = 337.5 / 49.75 
mm = 6.783 
Round to the nearest 0.01 = 6.78mm
310
Q

You are performing a Haidinger’s brush test on the right eye of your young amblyopic patient and he reports to you that he visualizes the center of the brush pattern to be 5 degrees to the left of to the fixation target. Which of the following BEST describes the result of this test?

A. The patient has nasal eccentric fixation

B. The patient has nasal fixation disparity

C. The patient has anomalous retinal correspondence

D. The patient has temporal fixation disparity

E. The patient has temporal eccentric fixation

A

A. The patient has nasal eccentric fixation

Explanation - Eccentric fixation refers to an anomalous condition in which some retinal point other than the fovea is used for fixation. Eccentric fixation can occur under both binocular and monocular conditions, but it is best diagnosed with the normally fixating eye occluded. When present, eccentric fixation is observed in the strabismic or amblyopic eye. In a patient with esotropia, the abnormal fixation point is typically located in the nasal retina; with exotropia, the fixation point is located in the temporal retina. Additionally, there can also be a vertical component to eccentric fixation.

There are several different tests that can be utilized to aid in the diagnosis of eccentric fixation, including the Haidinger’s brush test. During this test, the normally fixating eye is occluded while the patient views a blank surface through a rotating polarizer. The patient will then see a small rotating figure eight or bow-tie pattern that represents where the fovea projects in object space. The patient is asked to fixate on a reference dot on the display pattern and report where the brush pattern is located with respect to this point. If the patient states that the brush is placed nasal to the fixation point (as in the above question), this indicates that he or she is using an eccentric fixation point that is nasal to the fovea; the opposite is true if the patient states that the brush is rotating temporal to the reference point.

Anomalous retinal correspondence is a binocular condition in which there is an abnormal sensory adaptation that results in the fovea of the normal eye being paired with a non-foveal retinal point of the deviating eye.

Fixation disparity is also a binocular condition in which there is a very small misalignment of the eyes that does not result in a breakdown of single binocular vision.

311
Q

Which of the following is considered to be a part of the primary visual cortex?

A. The middle temporal cortex (MT)

B. Visual area 2 (V2)

C. The inferotemporal cortex (IT)

D. Brodmann’s area 17

E. Visual area 4 (V4)

A

D. Brodmann’s area 17

Explanation - The striate cortex is also known as the primary visual cortex, visual area 1 (V1), and Brodmann’s area 17. The striate cortex sends its projections to the extrastriate cortex, which itself contains many areas including IT, MT, V2, V4, and V5. MT and V5 are different names for the same region.

312
Q

You see a patient with a subluxated lens; you suspect which of the following conditions?

A. Coloboma

B. Grave’s disease

C. Marfan’s syndrome

D. Posner-Schlossman

A

C. Marfan’s syndrome

Explanation - Marfan’s syndrome is a genetic condition demarcated by height, a high arch palate, and very long fingers. With Marfan’s syndrome there are collagen deficiencies, one of which has lead to the subluxation of the lens. Homocystinuria is another condition that may cause lens subluxation and is associated with very high levels of homocysteine and susceptibility to heart disease.

313
Q

What is a clinical term for symptoms including, eye strain, eye fatigue, headaches, pain in/around the eyes and photophobia?

A. Ocular discomfort

B. Malingering

C. Asthenopia

D. Ametropia

A

C. Asthenopia

Explanation - Asthenopia is the clinical term used to encompass the above symptoms. Ametropia refers to the condition of an uncorrected refractive error but not the symptoms. Ocular discomfort is a symptom. Malingering refers to a situation in which a patient pretends (for whatever reason) that they cannot see clearly when there is no evidence of organic pathology.

314
Q

Which of the following medications should be prescribed to a patient suffering from shingles to help decrease the chances of postherpetic neuralgia?

A. Prednisolone

B. Natacyn® (natamycin)

C. Viroptic® (trifluridine)

D. Zovirax® (acyclovir)

A

D. Zovirax® (acyclovir)

Explanation - Postherpetic neuralgia is a painful condition that can last for months to years after the resolution of lesions that occur with shingles. Shingles is caused by the virus Varicella zoster, which causes chicken pox. Shingles are generally experienced by patients who have already had chicken pox and are elderly or immunocompromised. After resolution of chicken pox, the virus lays dormant in the root of the nerves and, for reasons that remain unclear at this time, becomes reactivated later on in life, causing shingles. Shingles presents as lesions on only one side of the body or is limited to a specific dermatome. Initially, one may only first experience a headache followed by tingling, itching, or sensitivity in the affected area. A rash will then develop in this area, followed by lesions that eventually blister over. Lesions do not always appear. Shingles cannot be contracted from one person to another; however, if a person has never been infected with chicken pox, it is possible to contract chicken pox from a person suffering from shingles. Oral antiviral agents like acyclovir, famciclovir or valacyclovir should be initiated within 72 hours of the onset of skin lesions to help minimize the chances of postherpetic neuralgia. Viroptic® is a topical antiviral frequently utilized for Herpes simplex. Natacyn® is an antifungal. Prednisolone, although useful for treating pain associated with shingles, does not help in preventing postherpetic neuralgia.

315
Q

An equiconcave thin lens has a power of -10.00 D, an index of refraction of 1.49 and a radius of curvature of 9.8 cm. If the back surface of the lens is coated to create a reflective surface, what would be the resulting power of the lens-mirror combination?

A. -10.00 D

B. -40.40 D

C. -30.50 D

D. +20.40 D

A

B. -40.40 D

Explanation - The total power of a lens-mirror combination can be determined using the formula Dlm = 2 D1 + P(n’), where Dlm= the power of the lens-mirror combination, D1= the front surface power, and n’= the index of refraction of the lens. However, the power of the mirror must first be determined using the formula P= -2n/r, where P= the power of the mirror in diopters, n= the index of refraction and r= the radius of curvature of the mirror in meters. Solve for P = -2(1.49)/0.098 = -30.40 D. Solve for Dlm= 2(-5.00) + -30.40 D = -40.40 D. It is important to note that because the lens is equiconcave, the power of the front surface is equivalent to the back surface D1=D2. Because the total power of the lens is -10.00 D, we know that the front surface power is -5.00 D. Also, when determining the power of the mirror, you must be sure to watch your signs!! The radius of curvature is positive and the power of the mirror is negative because the mirror is convex and would diverge light.

316
Q

Yellowing of the sclera is MOST commonly caused by excessive plasma levels of which of the following substances?

A. Blood urea nitrogen

B. Creatinine

C. Alkaline phosphatase

D. Albumin

E. Bilirubin

A

E. Bilirubin

Explanation - Yellowish pigmentation of the conjunctiva, skin, and other mucous membranes is known as jaundice. Jaundice is caused by increased levels of bilirubin in the blood (hyperbilirubinemia), which subsequently leads to an elevated concentration of bilirubin in the extracellular fluid. Normal blood plasma levels of bilirubin should be below 1.2 mg/dL. Concentrations above about 3mg/dL typically lead to jaundice. Jaundice is most often seen in liver disease (such as hepatitis) or liver cancer. Therefore, if a patient presents with yellowish discoloration of the sclera, liver function tests should be ordered immediately.

317
Q

Which of the following represents the most POSTERIOR structure of the anterior chamber angle?

A. Anterior trabecular meshwork

B. Scleral spur

C. Ciliary body

D. Schwalbe’s line

E. Posterior trabecular meshwork

A

C. Ciliary body

Explanation - Starting with the iris and moving anteriorly, the anterior chamber angle structures are as follows:
Ciliary body –> scleral spur –> posterior trabecular meshwork –> anterior trabecular meshwork –> Schwalbe’s line

When evaluating the anterior chamber angle using gonioscopy, it is best to orient oneself by beginning with the iris structure and following it out to the ciliary body. The ciliary body will vary in color, depending on the pigmentation of the iris, but is typically darker than the iris itself. Moving anteriorly, the next observable structure is the scleral spur, which will be bright white in color because it is a projection of the white sclera. This is a great landmark when assessing the angle because it does not vary much from person to person. Adjacent to the scleral spur is the trabecular meshwork. This can be greyish or pinkish in color (not as bright as the scleral spur). If there is pigment in the trabecular meshwork, one will be able to differentiate between the posterior and anterior trabecular meshwork; the posterior will be more pigmented than the anterior trabecular meshwork. The most anterior identifiable structure in the angle is Schwalbe’s line, which is the termination of Descemet’s membrane. It is a white ridge (which may be difficult to observe in some patients) that may occasionally have pigment lying on surface of the ridge. The area anterior to Schwalbe’s line represents reflections from the surface of the cornea.

318
Q

The Gullstrand #1 eye is also known as the exact eye. How many refracting surfaces does this model possess?

A. 2

B. 3

C. 1

D. 6

A

D. 6

Explanation - The Gullstrand #1 eye possesses a total of 6 refracting surfaces. The surfaces are the anterior and posterior aspect of the cornea, the anterior and posterior aspect of the lens cortex, and the anterior and the posterior aspect of the lenticular nucleus. The Gullstrand #2 eye, which is also known as the simplified eye, has 3 refracting surfaces: the cornea and the anterior and posterior aspect of the lens. The reduced eye is in possession of 1 refracting surface, at which all refraction is assumed to take place.

319
Q

Which 3 of the following retinal cells develop from the inner neuroblastic layer of the developing neural retina? (Select 3)

A. Amacrine cells

B. Photoreceptor cells

C. Bipolar cells

D. Muller cells

E. Horizontal cells

F. Ganglion cells

A

A. Amacrine cells
D. Muller cells
E. Horizontal cells

Explanation - The inner neuroblastic layer becomes the ganglion cells, amacrine cells, and Muller cells. The outer neuroblastic layer becomes the bipolar cells, horizontal cells, and photoreceptor cells. Note that in the developed retina, the nerve fiber layer and ganglion cell layer are composed of only cells from the inner neuroblastic layer (ganglion cells). The inner plexiform layer, inner nuclear layer, and outer plexiform layer of the developed retina consist of cells from both developmental layers (amacrine cells, Muller cells, bipolar cells, and horizontal cells). The outer nuclear layer and photoreceptor layer are composed only of cells from the outer neuroblastic layer (photoreceptors).

320
Q

A college student presents with bilateral internuclear ophthalmoplegia (INO). This finding is pathognomonic for which one of the following?

A. Lyme disease
B. Pseudotumor cerebri
C. Multiple sclerosis
D. Bell’s palsy

A

C. Multiple sclerosis

Explanation
An internuclear ophthalmoplegia is caused by an interruption or dysfunction in the medial longitudinal fasciculus. This is a heavily myelinated tract, and the clinical disturbance is a failure of conjugate lateral gaze. When an attempt to look to the right is made, the affected eye (in this case, the left eye) cannot adduct (move in to maintain conjugate gaze) while the unaffected eye (in this case, the right eye) will usually manifest a nystagmus due to overcompensation during abduction. In young patients, either a one-sided or bilateral INO is highly suggestive of multiple sclerosis. In older patients, it may represent a vascular defect or stroke.

Lyme disease does not produce INOs but can present with facial palsy; it is said that bilateral facial palsies are pathognomonic for Lyme disease. Bell’s palsy is a paralysis of the seventh cranial nerve (CN VII). Pseudotumor cerebri is a defect in the reabsorption of cerebrospinal fluid leading to elevated intracranial pressure and the finding of papilledema. It is most commonly seen in young overweight females.

321
Q

While performing keratometry, your patient’s cornea is steeper than 52.00D and therefore extends past the range of the keratometer. Which of the following steps should be taken to properly measure your patient’s corneal toricity?

A. Extend the range using a +1.25 D trial lens and add 9.00 D to the drum reading

B. Extend the range using a -1.00 D trial lens and add 6.00 D to the drum reading

C. Extend the range using a +1.25 D trial lens and add 6.00 D to the drum reading

D. Extend the range using a -1.00 D trial lens and add 9.00 D to the drug reading

A

A. Extend the range using a +1.25 D trial lens and add 9.00 D to the drum reading

Explanation
In the event that your patient has a cornea steeper than 52.00 D, a +1.25 D trial lens must be placed in front of the Keratometer’s aperture and 9.00 D should be added to the drum reading. A good way to remember this is that the + signs go together. You need to use a +1.25 lens and you need to add + 9.00 D.

If your patient’s cornea is flatter than 36.00 D then you will need to place a -1.00 D trial lens in front of the Keratometer’s aperture and subtract 6.00 D from the drum reading. A good way to remember this is that both six and subtract start with the letter ‘S’. Therefore, use a minus lens and subtract six once (the once part will help you to remember that you need to use a -1.00).

322
Q

Which area of the retina naturally corresponds with a sensitivity of 0 decibels on visual field testing?

A. The optic nerve head
B. The inferior arcade
C. The macula
D. The superior arcade
E. The papillomacular bundle
A

A. The optic nerve head

The optic nerve head, also known as the blind spot, will demonstrate a sensitivity of 0 decibels, because this area of the eye does not contain any visual receptors, and therefore does not possess any viable vision. The area of the retina with the highest decibel level (hence the highest sensitivity) as seen with visual field testing is the fovea. The sensitivity to stimuli decreases with increasing distance from the fovea.

323
Q

During administration of the red lens test your patient initially reports a red light and a white light which then fuse and appear as a pink light after 5 seconds of viewing. This type of response is classified as what grade of fusion?

A. Grade C
B. Grade B
C. Grade A
D. Grade D

A

B. Grade B

Explanation
Grade B fusion describes the event in which fusion is preceeded by momentary diplopia. Grade A fusion is the normal response and is depicted by instant fusion. Grade C fusion is defined as constant diplopia with the absence of fusion. Grade D is observed when one eye is suppressed or the eyes alternately suppress resulting in the perception of only one light at a time.

324
Q

The ophthalmic portion (V1) of cranial nerve V (the trigeminal nerve) possesses three main sub-divisions. Which of the following is a subdivision of V1?

A. Mandibular division
B. Supratrochlear division
C. Maxillary division
D. Frontal division

A

D. Frontal division

Explanation
Cranial nerves and their respective divisions can be very confusing and are best understood and remembered with diagrams and mnemonics.

The trigeminal nerve has three main divisions: ophthalmic (V1), maxillary (V2), mandibular (V3).

The ophthalmic division (V1) has three sub-divisions (think NFL) which are the nasociliary, frontal and lacrimal.

The frontal division of the V1 branch possesses two sub-divisions: the supraorbital and the supratrochlear divisions. The frontal division is the most commonly affected in herpes zoster ophthalmicus.

The nasociliary division has four sub-divisions (think LINE): long ciliary, infratrochlear, nasal, and the posterior ethmoid.

325
Q

Which of the following tests for stereopsis does not require the use of polarizing glasses?

A. The Random dot E test
B. The Randot stereo test
C. The Lang stereo test
D. The Titmus stereo test

A

C. The Lang stereo test

Explanation
The Random dot E, the Titmus stereo test, and the Randot stereo test all require the use of polarizing glasses to achieve stereopsis. The Lang stereo test is administered without the use of polarizing glasses, which is helpful with anxious or non-cooperative children. The test plate uses a random dot stereogram pattern to test for bifixation. The card is held at 40 cm and the child is asked to find a star, a car, and a cat (for Lang I) which all represent different levels of stereopsis (600”, 550”, and 1200”, respectively).

326
Q

Using the Von Graefe method to determine the amount of phoria present at near on your patient he reports that the targets are lined up horizontally with 5 base up prism before the left eye. What conclusion can you deduce from the above results?

A. Right hyper deviation or left hypo deviation
B. Exophoria of 5 prism diopters
C. Esophoria of 5 prism diopters
D. Left hyper deviation or right hypo deviation

A

A. Right hyper deviation or left hypo deviation

Explanation
When performing the Von Graefe method, a 4 base up dissociating prism is placed before the left eye and an 18 base in prism (biasing prism) is placed before the right eye. A single letter or target is projected at far (or near). The patient should initially perceive two targets with one being up and to the right and the other down and to the left . The 4 base up prism is then slowly reduced (either via intermittent occlusion or no occlusion) until the patient reports that the two targets are directly beside one another lined up horizontally. When the targets are aligned horizontally, the amount of prism before the left eye is recorded. If there is base up prism in front of the left eye then the deviation is described as either right hyperphoria or left hypophoria in the amount equal to the neutralizing prism before the left eye.

327
Q

The phenomenon, sometimes called tobacco dust, in which pigment is visible in the vitreal chamber and is fairly pathognomonic for a retinal tear is known as which of the following?

A. Ascension phenomenon
B. Cloquet’s canal
C. Schafer’s sign
D. Photopsia

A

C. Schafer’s sign

Explanation
Schafer’s sign occurs when there is a release of pigment into the vitreal chamber. The excessive pigment may be noted by the patient as a sudden increase in floaters. This sign usually indicates the presence of a retinal tear because a break in the retina may release retinal pigment epithelium. Pigment in the vitreal chamber may also be associated with a posterior uveitis or trauma or can occur secondary to intraocular surgery. It is of vital importance when a patient enters your clinic complaining of a recent increase in floaters, especially monocularly, or if you visualize pigment in the vitreous, that you rule out a retinal tear. Remember, many patients who suffer from a retinal tear are asymptomatic.

The ascension phenomenon describes the ability of the practitioner to better visualize vitreal cells or other suspended granules by asking the patient to move their eyes rapidly and then upon stabilization, using the slit lamp, the doctor waits and watches for cells or other details to move across the viewing beam.

Photopsias are flashes of light that may be perceived by the patient during periods of retinal traction.

Cloquet’s canal is comprised of the primary vitreous, which develops from weeks 3 through 9. The secondary vitreous then begins to form and condenses the primary vitreous, producing Cloquet’s canal.

328
Q

The choriocapillaris is separated from the retina by a thin membrane. What is the name of this membrane?

A. Bruch’s membrane
B. Sattler’s membrane
C. Bowman’s membrane
D. Descemet’s membrane

A

A. Bruch’s membrane

Explanation
Bruch’s membrane lies between the choriocapillaris of the choroid and the retinal pigment epithelium of the retina. Although this membrane is very thin (about 2 microns thick) it is very complex. The membrane consists of five facets. The outermost component is the basement membrane of the choriocapillaris followed by the outer collagenous zone, the elastic layer, the inner collagenous zone and most internally the basement membrane of the retinal pigment epithelium.

Sattler’s membrane is actually a layer of vessels in the choroid located externally to choriocapillaris.

Bowman’s and Descemet’s membranes are found in the cornea.

329
Q

When evaluating a nevus of the eyelid, which of the following characteristics is indicative of stability (i.e., benign)?

A. Uniformity of color within the lesion
B. Irregular, indistinct borders
C. Ulceration
D. Greater than 6 mm in diameter
E. The presence of inactive hair follicles within the lesion
A

A. Uniformity of color within the lesion

Explanation
When evaluating a suspicious nevus of the skin surrounding the eye, it is important to remember your “ABCDEs”. A=asymmetry, B=borders, C=color, D=diameter, E=evolution. A benign nevus tends to be symmetrical such that one can almost fold it in half and the sides would be evenly matched. The borders of a benign nevus are typically regular and distinct. The uniformity of the color should be assessed, rather than the actual level of pigmentation; evenness of pigmentation indicates stability. In general, the diameter of a benign nevus does not exceed 6 mm; although this is obviously not true all of the time, some small nevi may be malignant, and some large nevi (i.e., greater than 6 mm) can be benign. Lastly, most benign nevi tend to be rather flat (as opposed to elevated) with no change or evolution over time. Generally, hair growing out of the lesion is a positive sign, as this indicates that the hair follicle is intact. A cancerous lesion tends to kill the follicle; in this case, hair will no longer sprout. Remember, nevi should be stable and not change in shape, color, or size. ANY change in the nevus warrants a biopsy. Photo documentation is key when following a nevus or with any suspicious lesion.

330
Q

Which 3 of the following characteristics are considered the MOST common risk factors for the development of a retinal detachment? (Select 3)

A. Myopia
B. Aphakia
C. Hyperopia
D. Glaucoma
E. Ocular trauma
A

A. Myopia
B. Aphakia
E. Ocular trauma

Explanation
The incidence of retinal detachments in patients who are aphakic and have no significant history of ocular trauma is about 1 in 10,000 persons per year. The incidence of retinal breaks in the general population is about 3.3% per year.

According to the American Optometric Association, the most common risk factors for the development of a retinal detachment are myopia (40-55%), aphakia (30-40%), and ocular trauma (10-20%).

Trauma significantly increases the risk of detachment because retinal tears or dialysis can occur, both of which can lead to the formation of a retinal detachment. Myopia greater than 8 diopters or patients with an axial length greater than 24mm have been shown to be at an increased risk for detachments as well. Studies have also shown a greater incidence of retinal detachments in patients who are aphakic; this typically occurs within the first year following surgery. It is presumed that this may be correlated to posterior vitreous detachments, which commonly occur during this period.

331
Q

A young man 28 years of age is seen at your office and wishes to undergo LASIK. His subjective refraction is OD: -9.00-5.00 x 178 and OS: -3.50-7.25 x 172. His corneas are clear and there are no apparent ocular or systemic contraindications to surgery. Is this patient a good candidate for LASIK?

A. Only the right eye could undergo LASIK
B. Yes, both eyes could safely undergo LASIK
C. Only the left eye could undergo LASIK
D. No, neither eye should undergo LASIK

A

D. No, neither eye should undergo LASIK

Explanation
Unfortunately, this patient is not a good candidate as his high prescription falls outside the specified parameters for LASIK. Reportedly, LASIK can be used to correct up to 12.00 D of myopia, 6.00 D of astigmatism, and 6.00 D of hyperopia. The right eye of our patient has 14.00 D of myopia in the vertical meridian (when in doubt, always place the prescription on an optical cross). His left eye possesses a high amount of astigmatism which also falls outside the range of current approval for LASIK.

332
Q

Myopia, hyperopia, and astigmatism refractive errors are known as which of the following types of aberrations?

A. First-order aberrations
B. Second-order aberrations
C. Zero-order aberrations
D. Third-order aberrations
E. Fourth-order aberrations
F. Higher-order aberrations
A

D. Third-order aberrations

Explanation
Lower order aberrations: can be corrected by glasses, contact lenses, or refractive surgery
- Zero order aberrations: a constant (characterized by axial symmetry and a flat wavefront
- First order aberrations: lineal aberrations (correspond to tilting around a horizontal or a vertical axis and describe a tilt or prismatic error)
- Second order aberrations: defocus (myopia and hyperopia) and astigmatism

Higher order aberrations: cannot be corrected for by spherocylindrical lenses

  • Third order aberrations: coma (horizontal and vertical) and trefoil (triangular astigmatism with base along the x or y axis)
  • Fourth order aberrations: spherical aberration, tetrafoil, and secondary astigmatism (oblique/irregular astigmatism)
  • Fifth through tenth order aberrations: only important with extremely dilated pupils
333
Q

When determining the near point of convergence, which of the following sentences is TRUE?

A. As the target is moved closer, diplopia is typically reported prior to blurring of the target
B. The maximal point of divergence is recorded in centimeters as the ‘break point’
C. As the object is moved closer, the accommodative system maintains the target clarity while the convergence system preserves fusion of the object
D. If diplopia is not reported by the patient, the test should be repeated until the patient reports blurring of the target

A

C. As the object is moved closer, the accommodative system maintains the target clarity while the convergence system preserves fusion of the object

Explanation
The near point of convergence (NPC) is the point in which the patient’s eyes are maximally converged. The NPC is determined by presenting the patient with an appropriate accommodative target at near and with the patient’s near correction in place, the patient is asked to keep the target clear and single. The target is then brought closer to the patient until either the patient reports that the image of the object appears double or the clinician notes that one of the patient’s eyes turns out. This point is measured and recorded from the spectacle plane in centimeters. While performing the NPC, the convergence system maintains fusion of the target and accommodation sustains clarity. Some patients may report blurring of the target (near point of accommodation) prior to experiencing diplopia, but rarely do patients report blurring of the target after experiencing diplopia. The majority of clinicians record ‘TTN’ or ‘to the nose’ for patients who do not report diplopia or do not exhibit a break in fusion.

334
Q

A patient returns to your office reporting that her eyes feel strange when she reads 6 mm below the optical centers of her new glasses. The prescription in her right eye is -4.00 DS and -7.00 DS in her left eye. How much vertical prism is induced when she reads?

A. 6.6 prism diopters base down
B. 2.4 prism diopters base down
C. 1.8 prism diopters base down
D. 4.2 prism diopters base down

A

C. 1.8 prism diopters base down

Explanation
Use the Prentice rule to solve this problem: prism diopters(pd) =d*F, where d= the distance from the optical center in centimeters and F= the power of the lens in the desired meridian in diopters. In this instance, the patient is looking through base down prism in both eyes, which will cancel some of the prismatic effect as the bases are aligned. Solving for the amount of prism on the right eye, pd=0.6(-4.00)= 2.4 base down prism. Solve for the left eye: pd=0.6(-7.00)=4.2 base down prism. Subtract the two to determine the total prismatic effect experienced by the patient: 4.2-2.4=1.8 base down prism. Alternatively, you can omit one of the steps by initially determining the total power difference in the vertical meridian between the two lenses, which is 3.00 D (7-4=3). Then you can multiply this power difference by the distance between the patient’s line of sight and the optical center, which is 6 mm in this question. Pd=0.6(3)= 1.8 prism diopters base down over the left eye. Generally, vertical imbalances of smaller magnitudes do not pose too much of a problem for single vision lenses as the patient can tilt her head to re-align the optical centers with her line of sight, thus eliminating any possible diplopia.

335
Q

ou are evaluating the function of the levator muscle in a patient with a suspected myogenic ptosis. A measurement less than what value would be considered abnormal?

A. 5mm
B. 9mm
C. 16mm
D. 12mm

A

D. 12mm

Explanation
Measuring levator function (upper eyelid excursion) is a very helpful test in aiding in the diagnosis of a ptosis and identifying the underlying etiology. The measurement is achieved by having the patient look down while placing a thumb firmly against the brow to negate the action of the frontalis muscle. The patient is then instructed to look up as far as possible, and the amount of excursion is measured with a ruler.
- Average levator function is about 15mm
- A value of 12mm or above is considered “good”
- “Fair” levator function is 5-11mm
- 4mm or less is considered “poor”
- Typically anything below 12mm is considered abnormal

336
Q

An equiconcave thin lens has a power of -10.00 D, an index of refraction of 1.49 and a radius of curvature of 9.8 cm. If the back surface of the lens is coated to create a reflective surface, what would be the resulting power of the lens-mirror combination?

A. -30.50 D
B. +20.40 D
C. -10.00 D
D. -40.40 D

A

D. -40.40 D

Explanation
The total power of a lens-mirror combination can be determined using the formula Dlm = 2 D1 + P(n’), where Dlm= the power of the lens-mirror combination, D1= the front surface power, and n’= the index of refraction of the lens. However, the power of the mirror must first be determined using the formula P= -2n/r, where P= the power of the mirror in diopters, n= the index of refraction and r= the radius of curvature of the mirror in meters. Solve for P = -2(1.49)/0.098 = -30.40 D. Solve for Dlm= 2(-5.00) + -30.40 D = -40.40 D. It is important to note that because the lens is equiconcave, the power of the front surface is equivalent to the back surface D1=D2. Because the total power of the lens is -10.00 D, we know that the front surface power is -5.00 D. Also, when determining the power of the mirror, you must be sure to watch your signs!! The radius of curvature is positive and the power of the mirror is negative because the mirror is convex and would diverge light.

337
Q

An 81 year-old female patient presents with a chief complaint of blurred vision and glare at night. Slit-lamp examination reveals fibrosis on the posterior surface of her intraocular lens. What is her diagnosis?

A. Posterior capsular opacification
B. Posterior subcapsular cataract
C. Anterior capsular contraction
D. Posterior polar cataract

A

A. Posterior capsular opacification

Explanation
The most common late complication (of uncomplicated cataract surgery) is visually significant posterior capsular opacification. This finding commonly causes complaints of decreased vision, impaired contrast sensitivity, glare, and in some cases, monocular diplopia. Elschnig pearls are a typical finding that appear as vacuolated clusters on the back surface of the lens implant. They are a result of migration and proliferation of residual equatorial lens epithelial cells along the posterior capsule. In addition to Elschnig pearls, capsular fibrosis is another common finding of posterior capsular opacification that is due to fibrous metaplasia of epithelial cells. It is less common than Elschnig pearls but usually appears earlier.

Posterior subcapsular cataracts and posterior polar cataracts only occur in the natural crystalline lens.

Anterior capsular contraction is a post-operative complication that typically occurs several weeks after cataract surgery. Contraction and fibrosis of the anterior capsule can continue to progress and affect visual acuity; this progression may require a laser capsulotomy.

338
Q

You are able to accurately screen for a heterophoria by performing a cover test on your 4 year-old patient. To properly complete this examination, which of the following tests does NOT need to be performed?

A. Monocular estimate method (MEM)  
B.  Determining the near point of convergence  
C. Color vision testing  
D. Stereoacuity  
E.  Hirschberg testing
A

E. Hirschberg testing

Explanation
Hirschberg testing and cover testing are both utilized to denote the presence of an ocular deviation; therefore, Hirschberg testing does not need to be performed on this patient.

339
Q

While performing the alternate cover test at near on a patient with esophoria, you ask the patient to note any target movement. Which of the following responses is this patient MOST likely to report?

A. The target does not appear to move
B. The target appears to move in the same direction of the occluder movement
C. The target appears to move upwards
D. The target appears to move in the opposite direction of the occluder movement
E. The target appears to move downwards

A

D. The target appears to move in the opposite direction of the occluder movement

Explanation
If a phoria is too small in magnitude to be observed while performing the alternate cover test, the clinician can ask the patient to fixate on the target and report any movement seen. Patients with esophoria will report that the target appears to become displaced in the opposite direction of the occluder movement. For example, when the occluder moves from the patient’s right eye to the left eye, the patient will note that the fixation target appeared to move from left to right or “against jump”. A patient with exophoria will report that the target is displaced in the same direction as the movement of the occluder or “with jump”.

340
Q

What is the equivalent logMAR acuity of 20/60?

A. 0.60
B. 0.50
C. 0.33
D. 0.48

A

D. 0.48

Explanation
The logMAR acuity is determined by taking the log of the reciprocal of the decimal acuity. For the above Snellen fraction 20/60, the decimal acuity is found by dividing the numerator by the denominator. 20/60 = 0.333. Now take the log of the reciprocal which gives us log (1/0.333) = 0.478 or 0.48.

341
Q

What is the MOST common accommodative dysfunction seen in the general population?

 A. Accommodative ill-sustained  
 B. Accommodative excess  
 C. Accommodative infacility   
 D. Accommodative insufficiency 
 E. Paralysis of accommodation
A

D. Accommodative insufficiency

Explanation
Accommodative insufficiency is the most common accommodative dysfunction that the clinician will encounter, with accommodative excess and paralysis of accommodation being the least frequent.

342
Q

A 68-year old male patient presents to your office with a chief complaint of a dark spot in the center of his vision in his left eye. Upon dilated retinal examination, you observe what appears to be a macular hole; however, you do not have access to an OCT to determine if it is a full thickness or lamellar macular hole. Which of the following tests could you use in-office to aid in determining the thickness of the retinal defect?

 A. Randot stereo test  
 B. Color vision test  
 C. 10-2 Humphrey visual field  
 D. Amsler grid  
 E. Watzke-Allen test
A

E. Watzke-Allen test

Explanation
The most effective and sensitive method of evaluating a suspected macular hole is by use of optical coherence tomography (OCT); however, many general optometric practices do not have this instrument available in the their offices. Another helpful clinical test that is utilized to assess for a suspected full thickness macular hole is the Watzke-Allen slit lamp test. A narrow vertical slit-beam of light is placed over the foveal region using the slit-lamp and condensing lens. The patient is then asked to describe the line of light. The test is considered positive if the patient reports a break or significant thinning of the bar of light. Most patients with a full-thickness macular hole will report a positive Watzke-Allen test.

Amsler grid testing is great for macular disease in that it is sensitive for detecting the presence of macular lesions; however, it is not specific to the type of lesion that may be present.

343
Q

Which of the following is considered an example of the highest level of fusion?

A. Visual acuity via a Snellen chart
B. The red lens test
C. The Worth-4-Dot test
D. Stereoacuity test

A

D. Stereoacuity test

Explanation
There are 3 degrees of fusion, with third-degree fusion being the highest level. Stereopsis is an example of third-degree fusion. The red lens test and the Worth-4-Dot are both examples of second-degree fusion.

Third-degree fusion is the ability to fuse two identical images that are separated by space, creating the perception of depth. The images must be placed at points that cause retinal disparity in order for the perception of depth to occur.

Second-degree fusion is the ability to superimpose like objects (not necessarily identical objects), with the end result being the perception of a single object that is a composition of the two separate images.

First-degree fusion is defined as the ability to superimpose two dissimilar objects such that the two objects are perceived to occupy the same space and appear as a combination of the two objects.

344
Q

A 43-year old woman is seen at your office concerned about the fact that she constantly seems to have styes on her eyelids and her eyes are continually red and irritated. Biomicroscopy reveals meibomitis and blepharitis along with lid telangiectasia and a TBUT of 4 seconds for each eye. She also reports that her cheeks, nose, forehead and chin are easily flushed, especially when exposed to heat or cool temperatures. Given the above findings, what is your tentative diagnosis?

A. Hordeolum
B. Dry eye syndrome
C. Rosacea
D. Systemic lupus erythematosus

A

C. Rosacea

Explanation
Rosacea is a condition that causes excessive blushing of the face, with or without ocular involvement. Rhinophyma (a large, bulbous, red nose) is common in rosacea. Patients will often complain of facial flushing that is exacerbated with extreme temperature exposure, exertion, or the ingestion of hot beverages. This condition is seen two times more frequently in women. Ocular implications commonly include blepharitis, meibomitis, telangiectasia, dry eye syndrome, and occasional corneal involvement such as superficial punctate keratitis, pannus, and neovascularization. Treatment of facial rosacea includes oral tetracycline, topical metronidazole and retinoid compounds. Ocular rosacea is best treated by managing concomitant lid diseases along with artificial tears. While a hordeola and dry eye syndrome do occur in rosacea, these should not be your primary diagnoses. Remember to look at the overall broad picture and not just the pair of eyeballs.

Systemic lupus erythematosus (SLE) is an autoimmune disorder that has the capability of affecting many areas of the body. A common finding of SLE is called the malar (or butterfly) rash that is seen in roughly half of the individuals affected by this disorder. This rash generally occurs on the cheeks and over the nose bridge and worsens with ultraviolet light exposure. Concurrent eye conditions with SLE are primarily secondary to complications from the medications used to manage the condition such as oral corticosteroids and anti-malarials.

345
Q

Where is the thickest portion of the iris?

A. The collarette
B. The central pupillary zone
C. The pupillary ruff
D. The peripheral ciliary zone

A

A. The collarette

Explanation
The anterior surface of the iris is divided into two main zones; the central pupillary and the peripheral ciliary zones. The central pupillary zone is located just next to the pupillary ruff. The collarette lays adjacent to the central pupillary zone and serves as a junction between the central pupillary zone and the peripheral ciliary zone. The collarette typically has a distinctive zigzag appearance to its edges. The collarette is roughly 0.6 mm thick. The iris ruff surrounds the pupil edge and appears as a dark, scalloped border. The pupillary ruff is the visible portion of the posterior pigmented epithelium.

346
Q

Which of the four Purkinje images is inverted?

A. II
B. III
C. I
D. IV

A

D. IV

Explanation
There are four Purkinje images. The first image is caused by reflection from the anterior corneal surface and is the brightest of the images. The first image is roughly the same size as the object. The second Purkinje image is formed by the posterior surface of the cornea and almost coincides with the first Purkinje image. The third Purkinje image is the largest and is caused by reflection off of the anterior plane of the crystalline lens. The fourth Purkinje image is the smallest and is inverted, formed by reflection off of the posterior surface of the lens.

347
Q

Which 3 of the following are considered types of mutations? (Select 3)

A. Point
B. Silent
C. Blank
D. Frame-shift
E. Master
F. Recombination
A

A. Point
B. Silent
D. Frame-shift

Explanation
Recombination refers to a process in which genetic elements from two different genomes are combined together into one unit. A point mutation results from the change of a single base. Point mutations can lead to missense mutations, silent mutations, and nonsense mutations. Missense mutations are caused by the alteration of a single nucleotide, resulting in a codon that now codes for a different amino acid. Silent mutations occur when a single base is changed but the actual codon still codes for the same amino acid. For instance, ACC is altered to CGG both codons code for arginine. Nonsense mutations are the result of a changed nucleotide that leads to a stop codon. For example, CAG is altered to UAG. CAG codes for the amino acid lysine, but UAG is a stop codon, causing premature completion of a protein. Deletion or insertion of a single or multiple bases usually results in frame-shift mutations and may lead to a complete loss of ability to produce necessary proteins.

348
Q

A ray of light traveling in air strikes a flat transparent surface (n= 1.53) at an angle of 47 degrees from the normal. Which of the following equations can be used to determine the angle of refraction?

A. 1.53 (sin 47)= 1.00 sin i’
B. 1.00 (sin 47)= 1.53 sin i’
C. 1.00 (sin i)= 1.53 sin 43
D. 1.00 (sin 43)= 1.53 sin i’

A

B. 1.00 (sin 47)= 1.53 sin i’

Explanation
Snell’s law of refraction states that when light travels through a material that possesses an index of refraction greater than 1.0, the light rays change direction and become bent (or refracted). Snell’s law is depicted as the following: n sin i= n’ sin i’ where n= the index of refraction of the first medium, i= the angle of incidence, n’= the index of the second medium, and i’= the angle of the refracted ray. All angles are measured with respect to the normal which lies perpendicular to the interface between the different media. For the above example 1.00 (sin 47)= 1.53 sin i’, solving for i’= 28.55 degrees.

349
Q

You are performing binocular indirect ophthalmoscopy on a patient and drop and crack your 20D condensing lens. You are able to find an old 30D condensing lens to complete the examination. How will the magnification and field of view change with the 30D lens as compared to the 20D lens?

A. The magnification will increase, and the field of view will decrease
B. Both the magnification and field of view will decrease
C. The magnification will decrease, and the field of view will increase
D, Both the magnification and field of view will increase

A

C. The magnification will decrease, and the field of view will increase

Explanation
As the power of the condensing lens increases, the field of view increases but the magnification decreases. Additionally, the working distance will decrease as well.

It may be easier to think about this concept in terms of fundus lenses because we commonly switch back and forth between powers depending on pupil size, while most practitioners only use one condensing lens (20D). The field of view is larger with a 90D lens, and the magnification is less with a 90D lens as compared to a 78D lens; this can be applied to condensing lenses used with binocular indirect ophthalmoscopy as well.

350
Q

Increased intake of which of the following may result in nystagmus?

A. Aspirin (acetylsalicylic acid)
B. Benadryl® (diphenhydramine)
C. Lipitor® (atorvastatin)
D. Alcohol

A

C. Lipitor® (atorvastatin)

Explanation
Excessive amounts of alcohol intake can result in nystagmus. Law enforcement officers take advantage of this knowledge to test for sobriety. Alcohol is a central nervous system depressant, and excessive amounts can lead to cognitive and motor impairment. Alcohol is metabolized by the liver, which is capable of processing roughly 0.5 ounce of ethanol per hour (depending on height, weight, food intake etc.). The effects of alcohol on the brain are not uniform. The first area of the brain to be affected is the outer cerebral hemispheres, followed by the limbic system, and lastly, the brainstem (which is comprised of the midbrain, pons, and medulla). Areas of the brain are affected at different rates; this explains why systems that regulate some functions (i.e. emotions) become deficient before others (heart rate and blood pressure). Alcohol causes impairment of smooth pursuits and saccades. Lithium can cause linear waveform jerk nystagmus. Benadryl is sometimes used to treat vertigo, which can also be associated with nystagmus.

351
Q

A measure of drug activity that is expressed in terms of the amount of a drug that is required in order to produce a desired response is referred to as which of the following?

A. The reliability of the drug
B. The potency of the drug
C. The efficacy of the drug
D. The affinity of the drug

A

B. The potency of the drug

Explanation
The potency of a drug refers to the amount of a drug that is required to achieve a desired biological effect. A drug that requires a lower dose is considered more potent than a drug that requires a higher dosage to achieve the same effect.

The affinity of a drug for a receptor is a measure of how tightly and how able the drug in question binds to the receptor. Affinity is typically determined by the chemical structure of the drug.

The efficacy of a drug is a measure of the drug’s ability to produce a biological effect or initiate a biological change once it is bound to the receptor. Efficacy is used to characterize the level of maximal response by a drug.

352
Q

Which 2 of the following sentences are TRUE regarding the crystalline lens? (Select 2)

A. The clarity of the crystalline lens is attributable to high molecular weight aggregates that are not water-soluble
B. The development of cataracts is attributable to the formation of disulfide bonds secondary to oxidative damage
C. Increased levels of glutathione have been linked with cataract formation
D. The crystalline lens is formed of amino acids joined by peptide bonds arranged in multiple anti-parallel beta sheets

A

B. The development of cataracts is attributable to the formation of disulfide bonds secondary to oxidative damage

D. The crystalline lens is formed of amino acids joined by peptide bonds arranged in multiple anti-parallel beta sheets

Explanation
The crystalline lens is composed of proteins arranged in anti-parallel beta sheets. Crystallin protein is highly water-soluble and is the primary protein located within the lens. Crystallin is important in lens clarity. Oxidative damage, such as that from UV light, causes the formation of sulfide bonds between methionine groups and cysteine side chains (both of these amino acids contain sulfur). The sulfide bonds cause high molecular weight aggregates that precipitate as they lose their water solubility. The precipitates are opaque, causing reduced lens clarity. Glutathione is a powerful reductant produced by the body. It is implicated in the protection against oxidative damage by free radicals. Glutathione breaks the disulfide bonds, restoring proper protein configuration.

353
Q

What is the vergence demand using a Variable Tranaglyph when the separation is measured as 4 cm at a distance of 80 cm when training divergence?

A. 5 prism diopters base-out
B. 20 prism diopters base-out
C. 20 prism diopters base-in
D. 5 prism diopters base-in

A

D. 5 prism diopters base-in

Explanation
The equation to calculate vergence demand is:
demand = target separation in centimeters/training distance in meters
demand = 4 cm/0.80 m
demand = 5 prism diopters

Because divergence is being trained in the above patient, the demand will be base-in. If convergence was being trained, the demand would be base-out.

354
Q

Which of the following types of tumors represents the MOST common primary intraocular malignancy of childhood?

 A. Choroidal melanoma  
 B. Intraocular lymphoma  
 C. Retinoblastoma  
 D. Astrocytoma  
 E. Choroidal hemangioma
A

C. Retinoblastoma

Explanation
Retinoblastoma is the most common primary intraocular malignancy in children. It represents about 3-4% of all childhood cancers, occurring in about 1 in 17,000 live births. About 80% of all children diagnosed with retinoblastoma present with the condition before the age of 3, and an initial diagnosis over the age of 6 is extremely rare. Melanoma is the most common intraocular malignancy in adults.

355
Q

The bulk of aqueous humor is formed by which mechanism?

A. Osmosis
B. Ultrafiltration
C. Active transport
D. Diffusion

A

C. Active transport

Explanation
The ciliary processes of the pars plicata produce the aqueous humor. It is created by three different methods which are as described below.

Active transport produces the majority of aqueous humor. It requires the use of sodium, chloride and bicarbonate ions that serve to create an osmotic gradient between the interdigitations of the non-pigmented epithelial cells of the ciliary body. This creates a higher concentration of solutes inside the cells near their tight junctions which diminishes towards the posterior chamber, pulling water into the posterior chamber.

Ultrafiltration accounts for a small amount of aqueous produced. Fluid is pushed out of the ciliary processes via capillary hydrostatic forces (blood pressure).

Diffusion is responsible for producing a very small amount of aqueous. Particles in fluid flow down their concentration gradient moving from an area of high concentration to an area of low particle concentration.

356
Q

The change in direction of a light ray that occurs when it travels from one media through another is known as which of the following?

A. Absorption
B. Reflection
C. Rarefaction
D. Refraction
E. Deflection
A

D. Refraction

Explanation
In general, upon entering a material of which the index is different than the material the light ray is exiting, the ray will change its direction. This is called refraction. Refraction occurs because of the speed change that light undergoes when it changes mediums.

The angle of refraction is the angle that a refracted light ray makes with the normal to an interface separating different media. Keep in mind that when a ray is directed along the normal, it will pass straight through the medium undeviated (it will slow down but will not bend). If the light travels from a lower to higher index, the ray will bend towards the normal, and away from the normal if traveling from higher to lower index.

The law of refraction states that the incident, normal, and refracted rays all lie in the same plane. The angles of refraction are related by Snell’s law (n sin i = n’ sin i’). Snell’s law does not distinguish between incident and refracted rays; therefore, the path is reversible (it doesn’t matter which direction light is coming from).

357
Q

Which of the following BEST describes the properties of the entrance pupil of an optical system?

A. The image of the field stop through all following lenses
B. The image of the field stop through all preceding lenses
C. The image of the aperture stop through all preceding lenses
D. The image of the aperture stop through all following lenses

A

C. The image of the aperture stop through all preceding lenses

Explanation
The definition of the entrance pupil of an optical system is the image of the aperture stop through all preceding lenses. In other words, it is the aperture stop placed into object space. In situations where no lenses precede the aperture stop, the aperture stop is also the entrance pupil.

The exit pupil of an optical system can be defined as the image of the aperture stop through all following lenses. Again, if no lenses follow the aperture stop, the aperture stop is also the exit pupil.

358
Q

Your patient requires 2 BO prism. You decide to decenter the left lens in his glasses to induce the desired prism. The prescription is OD: -4.00 DS, OS: -4.75-0.75x178. The patient’s pupillary distance is 68 mm. How much and in which direction do you have to decenter the left lens?

A. 4.2 mm, in
B. 4.2 mm, out
C. 3.5 mm, in
D. 3.5 mm, out

A

A. 4.2 mm, in

Explanation
In order to create base out prism, one must decenter the left lens inwards (towards the patient’s nose). To determine the degree of nasal decentration, use Prentice’s rule: prism diopters= d*F where d is equal to the distance from the optical center in centimeters and F= the power of the lens in the desired meridian in diopters. The power of the left lens in the horizontal meridian is equal to -4.75 D. Put the values into Prentice’s rule, yielding: 2BO= (d)(-4.75) solve for d: d=2/-4.75, d=0.42 cm or 4.2 mm.

359
Q

Shining a small spot of light onto the center of the receptive field of a retinal ganglion cell results in an increased frequency of action potentials; however, increasing the size of the light causes a decrease in the frequency of action potentials. Why does this occur?

A. Because activation of rods causes the inhibition of ganglion cells
B. The retinal dark current, if overly exposed, results in spatial antagonism by cone cells
C. Rhodopsin responds preferentially to small sources of light allowing for high spatial resolution
D. Due to the fact that ganglion cells have receptive fields that are center-surround

A

D. Due to the fact that ganglion cells have receptive fields that are center-surround

Explanation
The majority of ganglion cells possess center-surround receptive fields and therefore exhibit lateral inhibition. This property serves to increase spatial resolution. The receptive fields can be either center on or center off. The receptive field appears like a donut, with the center being excited by light and the surrounding annulus inhibited by light or vice versa. Increasing the size of the stimulus causes summation of both parts of the receptive fields, resulting in a greater reduction in the frequency of action potentials than if the center were to be stimulated alone.

360
Q

Why is the formation of ketone bodies detrimental to the body?

A. Ketone bodies disrupt the function of fatty acid synthesis
B. Ketone bodies become incorporated into nucleic acids
C. Ketone bodies are acidic and lower the pH of blood
D. Ketone bodies become favorably metabolized over glucose

A

C. Ketone bodies are acidic and lower the pH of blood

Explanation
Ketoacidosis can occur during periods of starvation, low carbohydrate levels, when there is excessive breakdown of fats, or in uncontrolled diabetes. Carbohydrate catabolism is always favored over the formation of ketone bodies. When there is an excessive amount of acetyl-CoA present, ketone bodies are formed (for instance, acetone). When the body is starving, the brain uses ketones as a fuel source to maintain regulatory functions. The ketone bodies are acidic and can lower the pH of blood. Warning signs include: vomiting, flushed skin, difficulty breathing, and confusion. A key sign to watch and/or smell for is fruity breath or breath that smells like nail polish remover. Ketoacidosis can lead to a diabetic coma or death. People with Type I diabetes are more susceptible to ketoacidosis than those suffering from Type II.

361
Q

As a stimulus for accommodation increases, which of the following statements is TRUE in regards to the measured accommodative response?

A. The measured accommodative response is typically the same as the magnitude of the stimulus
B. The measured accommodative response is typically greater than the magnitude of the stimulus
C. The measured accommodative response is typically less than the magnitude of the stimulus

A

C. The measured accommodative response is typically less than the magnitude of the stimulus

Explanation
As a stimulus for accommodation increases (an object is brought closer to the eyes), studies have shown that the objectively measured response to accommodation is actually less than the overall magnitude of the stimulus. This is known as a lag of accommodation. This type of study can be done by placing negative trial lenses with increasing power in front of the eyes (stimulating accommodation), while measuring the accommodative response with a refractometer. As the stimulus is increased and plotted, the linear response will have a slope of less than 1 due to the observed lag of accommodation. Eventually, there comes a point in which the maximum accommodative response amplitude is reached. It is also important to note that there is typically an initial small lead in accommodation (the response is greater than the stimulus) for low magnitudes of accommodative stimuli.

362
Q

Which of the following frame materials requires HIGH heat for lens insertion and frame adjustments?

 A. Polyamide  
 B. Cellulose propionate  
 C. Polycarbonate  
 D. Optyl  
 E. Carbon fiber  
 F. Cellulose acetate
A

D. Optyl

Explanation
The steps utilized in inserting a lens into a frame during fabrication and adjusting the frame are standard for most plastic frames; however, the major variant is whether or not heat is used during the process (and how much). Some frame materials do not use heat at all when inserting lenses (this is called “cold snapping”) and some materials require high amounts of heat so that the material easily bends with only mild pressure.
Optyl is a frame material that requires high amounts of heat. Heat is typically applied to the frame until it can bend under it’s own weight and then lenses can be snapped into the frame, or adjustments to the frame can be made as necessary. Optyl material can safely be heated to temperatures of up to 200 degrees Celsius.
Most other materials utilize only minimal amounts of heat if lenses cannot be “cold snapped” (cellulose acetate, cellulose proprionate, and carbon fiber). Polycarbonate and polyamide materials should not be heated at all for the lens insertion process as this may lead to lens warpage; therefore, edged lenses must be fabricated exactly to size.

363
Q

Which of the following is NOT an indication of wet age-related macular degeneration (ARMD)?

A. Drusenoid pigment epithelial detachment
B. Submacular hemorrhage Exudates
C. Classic choroidal neovascular membrane (CNVM)
D. Occult choroidal neovascular membrane (CNVM)

A

A. Drusenoid pigment epithelial detachment

Explanation
Drusenoid pigment epithelial detachments are a sign of dry ARMD, as there is no leakage involved. Instead, a detachment in the retinal pigment epithelium occurs due to the thickness of the soft confluent drusen. The drusen push and displace the retinal pigment epithelium.

364
Q

Which of the following regions of the sclera is the weakest and most likely to cup outward when there is increased intraocular pressure (i.e. glaucoma)?

A. Palisades of Vogt
B. Limbus
C. Scleral spur
D. Lamina cribrosa

A

D. Lamina cribrosa

Explanation
The lamina cribrosa is the posterior scleral foramen through which the optic nerve passes. It is a sieve-like structure with interwoven collagen fibrils forming canals allowing optic nerve fibers to pass. It is the weakest area of the outer connective tissue tunic which is why cupping out or ectasia may be seen as a sign of increased intraocular pressure. The scleral spur is a region of circularly oriented collagen bundles which serves as the origin of the longitudinal ciliary muscle fibers. The limbus is the band located at the junction of the cornea and sclera. Here, there is a transition from squamous corneal epithelium to columnar conjunctival epithelium, as well as a transition from the very regular corneal stroma to the irregular scleral stroma, and the termination of Bowman’s layer, Descemet’s membrane and the corneal endothelium. The palisades of Vogt are radial projections of the limbal epithelium that extend into the cornea; it is an area where corneal epithelial stem cells reside.

365
Q

Which of the following plays a critical role in directing B lymphocytes to initiate immunoglobulin class switching?

A. Basophils
B. Helper T lymphocytes
C. Natural killer cells
D. Complement

A

B. Helper T lymphocytes

Explanation
Helper T lymphocytes are required for a B2 lymphocyte to switch from producing IgM to either IgG, IgA, or IgE. The type of helper cell, TH1 or TH2, determines if a cell will produce IgG, IgA, or IgE. The cytokines released by TH1 cells that promote switching to IgG1, IgG2, IgG3, and IgA include TGF-Beta and IL-5. The cytokines released by TH2 cells that promote switching to IgG4 and IgE include IL-4 and IL-13.

366
Q

Inhibitory postsynaptic potentials have which type of effect on a neuron?

A. Hypopolarizing
B. Depolarizing
C. Hyperpolarizing
D. Brings the neuronal membrane closer to threshold

A

C. Hyperpolarizing

Explanation
Many neurons receive input from multiple other neurons. Some of the input is excitatory in nature, termed excitatory postsynaptic potentials (EPSPs), and others are inhibitory (IPSPs). Inhibitory signals push the membrane of the neuron away from its threshold point and therefore cause hyperpolarization because the inside of the membrane becomes even more negatively charged. When excitatory potentials cause the membrane of the neuron to become more positively charged relative to its resting state, they are hypopolarizing. All of the synaptic signals that reach the input zone at the same time are integrated together. If the resulting message causes the membrane to reach its threshold, an action potential ensues.

367
Q

A patient walks into your office and would like to be fit with soft contact lenses. Your subjective refraction reveals: OD: +5.50 DS, OS: +4.75 DS. If the phoropter was placed 13 mm from the corneal plane, which of the following contact lens prescriptions would be the BEST choice for the right eye of this patient?

A. +6.00 DS
B. +5.00 DS
C. +5.50 DS
D.+5.75 DS

A

A. +6.00 DS

Explanation
To determine the proper power, you must calculate the power at the cornea rather than the spectacle plane by utilizing the vertex power formula. Vertexing is important when the power in any meridian exceeds +/- 4.00 D. In order to properly vertex, one must use the following formula: Fc= Fs/1-dFs where Fc= the dioptric power at the corneal plane, Fs= the dioptric power at the spectacle plane, and d= the vertex distance in meters. In general, the vertex distance of the phoropter from the corneal plane is 13 mm unless otherwise specified. Solving for Fc, Fc=+5.50/1- (0.013)(+5.50), Fc = +5.50/1 -(0.0715), Fc= +5.50/0.9285, Fc=+5.93; rounding to the nearest 0.25, Fc is +6.00.

368
Q

While several tests such as tear stability and ocular surface staining are global tests diagnostic for the presence of dry eye, what is a specific test that can differentiate aqueous tear deficiency from meibomian gland dysfunction (MGD)?

A. Tear clearance, which is normal in MGD
B. A detailed dry eye questionnaire to allow for score-based symptomology
C. The Schirmer I test, which is considered abnormal if the strip wetting is less than 10 mm in 5 minutes without anesthesia
D. Meibomian gland atrophy/dropout as measured by meiboscopy
E. Conjunctival staining by lissamine green, which is worse in MGD

A

C. The Schirmer I test, which is considered abnormal if the strip wetting is less than 10 mm in 5 minutes without anesthesia

Explanation
Whether aqueous tear production is normal or not can be determined using the cotton thread tear test or the Schirmer I test without anesthesia. The Schirmer I without anesthesia essentially becomes a stress test since it is so irritating; if the patient cannot produce aqueous tears under irritating conditions, the main lacrimal gland is probably compromised. Conversely, meibomian gland dropout can be observed in both aqueous tear deficiency and MGD, possibly due to inflammatory processes on the ocular surface.

369
Q

Your -10.00 D patient wishes to have the thinnest edges possible for his glasses. Which of the following actions will result in a reduced edge thickness?

A. Increasing the center thickness of the lens
B. Decreasing the refractive index of the lens
C. Increasing the minimum blank size of the lens
D. Choosing a smaller eyesize

A

D. Choosing a smaller eyesize

Explanation
For a myope, the goal is to reduce the overall edge thickness. Decreased edge thickness can be achieved by decreasing the eyesize, increasing the refractive index of the lens, or by minimizing the center thickness

370
Q

Which of the following alterations will help to loosen a tightly-fitting gas-permeable lens?

A. Steepen the peripheral curve system
B. Reduce the width of the peripheral curve system
C. Reduce the size of the optic zone
D. Steepen the base curve of the lens
E. Increase the overall diameter
A

C. Reduce the size of the optic zone

Explanation
There are a multitude of alterations that can be made when a lens is fitting too tightly, many of which can be done in-office if a modification unit is available. If a gas-permeable lens is fit too tightly, the most commonly altered parameter is flattening of the base curve. One can also decrease the optic zone, decrease the overall diameter (OAD), widen the peripheral curve system, or flatten the peripheral curve system. In order to modify a lens that is fitting too loosely, simply reverse all of the above: steepen the base curve, increase the OAD, increase the optic zone, steepen the peripheral curve system, and narrow the width of the peripheral curves.

371
Q

Which of the following systemic conditions is MOST commonly associated with a diagnosis of superior limbic keratoconjunctivitis?

 A. Hypertension  
 B. Hypercholesterolemia  
 C. Systemic lupus erythematosus  
D.  Hyperthyroidism 
E.  Diabetes
A

D. Hyperthyroidism

Explanation
Superior limbic keratoconjunctivitis (SLK) is predominantly found in middle-aged women that tend to have associated abnormal thyroid function (usually hyperthyroidism). Studies have shown a 50% prevalence of thyroid disease in patients diagnosed with SLK. The exact etiology and pathophysiology of SLK remains unclear, but the relation with hyperthyroidism and the pattern of exacerbations and remissions suggests a possible autoimmune association. It is thought that SLK occurs as a result of blink-related mechanical trauma that leads to irritation of the superior limbal region as loose conjunctival tissue rubs against the limbus during the blinking movement of the upper eyelids. Damage typically occurs to both the superior tarsal and conjunctival surfaces.

372
Q

How does one clinically differentiate between a true Von Graefe’s sign and a pseudo Von Graefe’s sign?

A. Ptosis disappears in down gaze
B. Ptosis is not present in straight-ahead gaze
C. Ptosis is not evident in adducted gaze
D. Ptosis is less prominent on abduction

A

C. Ptosis is not evident in adducted gaze

Explanation
Pseudo Von Graefe’s sign occurs due to aberrant regeneration of cranial nerve III. After incurring an insult (paralysis), upon recovery, instead of innervating the levator palpebrae superioris, fibers now connect to the medial rectus. Therefore, when one adducts the eye on the same side of the palsy, the ptosis is not evident. In straight-ahead gaze a ptosis is present on the same side of the palsy.

Von Graefe’s sign occurs in Grave’s disease (thyroid dysfunction). This term refers to the inability of the eyelid to move down when the globe is directed in a down gaze. It is often described as a lagging of the upper eyelid-not because it sags but because it lags behind movement-wise. The greater the down gaze, the greater the bearing of the sclera. Von Graefe’s results from excessive innervation of the sympathetic system causing upper lid retraction via Muller’s muscle.

373
Q

Your low vision patient’s best corrected acuity is 10/80. What is the diopter of the add she would need to see reduced Snellen 40 at near?

A.5 D
B. 8 D
C. 10 D
D. 4 D

A

C. 10 D

Explanation
10/80 = 20/160
Reciprocal of Vision = 160/40= 4x De=(4x)(2.5 D)= 10 D add

One could arrive at 5 D as the answer if acuity had not been converted to 20 foot test distance.
One could arrive at 4 D as an answer if 4x magnification had been mistaken for 4 D.

374
Q

In the absence of injury, how frequently are the superficial epithelial cells of the cornea completely regenerated?

A. Every 12 days
B. Every 7 days
C. Every 2 days
D. Every month

A

B. Every 7 days ???? (answer may have been wrongly keyed in optoprep)

Explanation
The cornea is truly an amazing structure. The corneal epithelial cells transition from basal to columnar to wing cells and are ultimately sloughed off over a 7 day period. A small, superficial corneal abrasion (that is, one that does not penetrate Bowman’s layer) can heal completely within 24-48 hours without leaving a scar!

375
Q

On binocular indirect ophthalmoscopy of your 43 year-old patient, you see what appears to be a small U-shaped retinal tear in the far periphery at the 12 o’clock position of the left eye. You decide to use your 3-mirror Goldmann lens to obtain another view of the retina. Which 2 of the following modifications can be made in order to allow for the furthest possible peripheral view? (Select 2)

A. Tilt the Goldmann lens up
B. Have the patient look up
C. Tilt the Goldmann lens down
D. Have the patient look down

A

B. Have the patient look up
C. Tilt the Goldmann lens down

Explanation
There are some cases where a more peripheral view of the retina is desired than that which can be achieved with the patient in primary gaze and the Goldmann lens held in the upright position. In these situations, certain modifications can be made such as tilting the Goldmann lens or having the patient move their eyes. The lens should be tilted in the opposite direction of the location of the lesion, and the patient should be instructed to look in the actual direction of the area being viewed.

For example, in the above question, the suspected lesion is at the 12 o’clock position. To obtain a more peripheral view, the lens should be tilted down, and the patient should be asked to look up.

376
Q

You are performing a Haidinger’s brush test on the right eye of your young amblyopic patient and he reports to you that he visualizes the center of the brush pattern to be 5 degrees to the left of to the fixation target. Which of the following BEST describes the result of this test?

A. The patient has temporal eccentric fixation
B. The patient has temporal fixation disparity
C. The patient has nasal fixation disparity
D. The patient has anomalous retinal correspondence
E. The patient has nasal eccentric fixation

A

E. The patient has nasal eccentric fixation

Explanation
Eccentric fixation refers to an anomalous condition in which some retinal point other than the fovea is used for fixation. Eccentric fixation can occur under both binocular and monocular conditions, but it is best diagnosed with the normally fixating eye occluded. When present, eccentric fixation is observed in the strabismic or amblyopic eye. In a patient with esotropia, the abnormal fixation point is typically located in the nasal retina; with exotropia, the fixation point is located in the temporal retina. Additionally, there can also be a vertical component to eccentric fixation.

There are several different tests that can be utilized to aid in the diagnosis of eccentric fixation, including the Haidinger’s brush test. During this test, the normally fixating eye is occluded while the patient views a blank surface through a rotating polarizer. The patient will then see a small rotating figure eight or bow-tie pattern that represents where the fovea projects in object space. The patient is asked to fixate on a reference dot on the display pattern and report where the brush pattern is located with respect to this point. If the patient states that the brush is placed nasal to the fixation point (as in the above question), this indicates that he or she is using an eccentric fixation point that is nasal to the fovea; the opposite is true if the patient states that the brush is rotating temporal to the reference point.

Anomalous retinal correspondence is a binocular condition in which there is an abnormal sensory adaptation that results in the fovea of the normal eye being paired with a non-foveal retinal point of the deviating eye.

Fixation disparity is also a binocular condition in which there is a very small misalignment of the eyes that does not result in a breakdown of single binocular vision.

377
Q

A patient with an axial length that measures 25.3mm is MOST likely to require correction with which of the following spectacle prescriptions (assuming that the average axial length of the adult eye is 24mm in length)?

 A. +1.00  
 B. +2.00  
 C. -1.00  
D.  -7.00  
E.  -4.00
A

E. -4.00

Explanation
The average axial length of the adult human eye is 24mm. In the above question, the patient’s axial length is 1.3mm longer than average. Several studies have shown that a difference in axial length of 1mm corresponds to approximately 3.00 D of refractive error; therefore, 1.3mm would equate to roughly 4.00 D. Due to the axial length measuring longer than average, this would result in a myopic refractive error that would be corrected with a -4.00D lens.

378
Q

Because the lungs of a fetus are collapsed in the womb, the majority of blood bypasses the lungs through a hole in the heart known as which of the following terms?

A. The foramen rotundum
B. The foramen cecum
C. The foramen lacerum
D. The foramen ovale

A

C. The foramen lacerum

Explanation
All of the above foramina are located at various locations in the bones that make up the skull; however the foramen ovale is a term that is also used for the heart. This foramen essentially acts as a door between the right and left atria, allowing the passage of oxygenated blood from the mother’s placenta to the brain of the fetus. The majority of blood enters the atria and is shunted into the arterial duct, which bypasses the lungs. However, once the newborn takes its first few breaths, the left atrium fills with blood that has been returned from the lungs, effectively closing the hole.

379
Q

Which of the following values most closely corresponds to the average eccentricity of the human cornea?

 A. 0.50  
 B. 1.00  
 C. 1.50  
 D. 0.00  
 E. 0.10
A

A. 0.50

Explanation
Eccentricity is the measurement of the asphericity of a curved surface. In the case of the human cornea, it indicates the way in which the cornea changes from a more curved central portion to a flatter periphery (also known as a prolate shape). Normal corneal eccentricity values (e values) range between +0.50 and +0.60 in humans. A higher corneal eccentricity value indicates that the cornea flattens more rapidly in the periphery; a lower eccentricity measurement would occur in a patient whose cornea flattened more slowly in the periphery.

Another way to think of eccentricity is that the ‘e’ value designates that amount by which the cornea deviates from a perfect sphere (which has an eccentricity value of 0). A parabola has an eccentricity value of 1; a prolate ellipse (as in the normal cornea) will have an eccentricity between 0.1 and 0.9 depending on how fast the peripheral flattening occurs. Lastly, an oblate ellipsoid (as in some post myopic refractive surgical corneas) will have an eccentricity value of between -0.1 and -0.9, depending on how fast the curvature steepens in the periphery.

380
Q

In refractive astigmatism, the horizontal line focus is formed by which of the following?

A. Light rays traversing the circle of least confusion
B. Light rays traversing the interval of Sturm
C. Light rays traversing the vertical principal meridian
D. Light rays traversing the horizontal principal meridian

A

C. Light rays traversing the vertical principal meridian

Explanation
Light rays of the horizontal line focus are formed by light traversing the vertical principal meridian. Remember that with astigmatism the orientation of the formed image is 90 degrees away from the power meridian/principal meridian! The interval of Sturm is created based on the fact that an eye with astigmatism has two principal meridians which will come into focus at different points. The interval between the two focal points is known as the interval of Sturm. The location within this interval that allows for the clearest image is known as the circle of least confusion.

381
Q

A patient suffering from dichromacy lacks one of the three photopigments necessary for normal color vision but possesses 20/20 vision. What is the MOST likely reason that this patient is able to retain good visual acuity despite the fact that he is missing a photopigment?

A. The missing photopigment is replaced by the remaining photopigments that are present
B. Patients with dichromacy possess more cone receptors than patients with normal color vision
C. The color processing pathway allocates more neurons to color vision in people with dichromacy
D. The cortical area associated with color vision is more complex in these individuals

A

A. The missing photopigment is replaced by the remaining photopigments that are present

Explanation
In dichromacy, one of the photopigments is missing; the type of dichromacy is categorized based on which photopigment is lacking. A deuteranope is missing chlorolabe, a tritanope is missing cyanolabe, and a protanope is missing erythrolabe. It is theorized that the missing photopigment is replaced by the photopigments that are present; otherwise, the person would likely suffer a deficit in visual acuity. In the instance of a protanope, it is likely that erythrolabe is replaced with chlorolabe. Protanopes and deuteranopes confuse red and green. Tritanopes confuse blues and yellows. Deuteranopes possess a neutral point of 498 nm. Wavelengths above the neutral point are perceived as various saturated levels of yellow, and those below the neutral point are viewed as shades of blue that become more saturated as the wavelength decreases. Protanopes possess a neutral point of 492 nm. Wavelengths above the neutral point are perceived as various hues of yellow, and those below the neutral point are seen as blue. A tritanope has a neutral point of 570 nm. Wavelengths above 570 nm are seen as shades of red, and those below are observed as green. Wavelengths at the neutral point are seen as white.

382
Q

Your patient is able to discern an optotype that is 17.46 mm high at a distance of 20 feet. What is the visual acuity of this patient?

A. 20/30
B. 20/60
C. 20/40
D. 20/50

A

C. 20/40

Explanation

First one must determine what the acuity would be at a distance of 20 feet. The height of the Snellen 20/20 optotype that subtends an angle of 5 minutes of arc at twenty feet is 8.73 mm. All other target heights will be multiples of this number. Therefore solve for x. (8.73mm)(x)=17.46mm yields 17.46/8.73 =x, giving us x=2, therefore the optotype is 2 times bigger than the 20/20 optotype. To figure out the visual acuity simply multiply 2 by the Snellen fraction 20/20. Hence 20 * 2 =40. The visual acuity for this person is 20/40. An easy formula that incorporates the above is;

Snellen denominator= (height of optotype/8.73)x20. Plugging our above example into this formula gives
Snellen denominator= (17.46/8.73)x20 = 40. Therefore the acuity is 20/40.

383
Q

Your 29 year-old rigid gas-permeable contact lens patient requests a new prescription as she lost her current contacts. Her keratometry readings are 44.62 @ 180 / 43.12 @ 90 and refraction is +2.75 -1.00 x 095 for her right eye. Given this data, which of the following diagnostic rigid contact lenses would provide a predicted fluorescein pattern of apical alignment and predicted spherical over-refraction of plano, respectively? (Assume that overall diameter of the lens is 9.0mm, and optic zone is 7.4mm)

A. Base curve: 45.37 D / Contact lens power: +3.50 D
B. Base curve: 43.87 D / Contact lens power: +2.00 D
C. Base curve: 43.12 D / Contact lens power: +2.75 D
D. Base curve: 42.37 D / Contact lens power: +0.50 D
E. Base curve: 44.62 D / Contact lens power: +1.25 D

A

C. Base curve: 43.12 D / Contact lens power: +2.75 D

Explanation
A common fitting philosophy for gas-permeable contact lenses is to use the average keratometry value minus 0.75 diopters. This is based upon the idea that the cornea is an aspheric surface with an eccentricity of close to 0.50. Therefore, to achieve an apical alignment fit on a surface with this type of curvature, the base curve of the contact lens must be flatter than the central cornea curvature (if the cornea was spherical, an alignment base curve would be equal to the average keratometry value). Assuming a corneal eccentricity of 0.50 and a contact lens optic zone diameter of 7.4mm, the base curve of an alignment-fitting lens should be close to 0.75 diopters flatter than the average keratometry reading. Another common fitting philosophy is to fit the contact lens so that the base curve is equal to the flat keratometry value (this would provide the same answer in the above case).

Therefore, in the above patient, the base curve of a diagnostic contact lens that would provide a predicted fluorescein pattern of apical alignment would be 43.12 D.
Average K= (44.62 + 43.12) / 2 = 43.87
Average K -0.75D = 43.87 - 0.75 = 43.12

When determining the contact lens power that would provide a predicted spherical over-refraction of plano, the spherical portion of the subjective refraction, flat keratometry value, and base curve of the contact lens are utilized for calculations. First, one must determine the effect of the tear layer created between the anterior corneal surface and back surface of the contact lens using the flat keratometry and contact lens base curve values. If the contact lens is fit flatter than K, a minus tear power is created; therefore, a corresponding change in plus power is necessary. On the other hand, if the lens is fit steeper than K, a plus tear lens is created that must be compensated by adding minus to the contact lens power.

In the above case, the contact lens base curve for ideal apical alignment was determined to be 43.12, which also corresponds to the patient’s flat K reading. When the contact lens is fit “on K,” or equal to the flat keratometry reading, the predicted lens power will be equal to the spherical portion of the subjective refraction (this must be vertexed if above 4D). Therefore; in order to provide a predicted over-refraction of plano, a diagnostic contact lens with a base curve of 43.12 should have a power of +2.75D.

384
Q

Your patient has keratoconus and is pregnant. She would like to know if this condition is hereditary. What is the MOST appropriate response to this question?

A. Yes, keratoconus is hereditary 100% of the time
B. No, keratoconus has a high correlation with hypertension
C. No, there is no hereditary component associated with keratoconus
D. Yes, there is a hereditary component to keratoconus but it presents with incomplete penetrance

A

D. Yes, there is a hereditary component to keratoconus but it presents with incomplete penetrance

Explanation
Although there still remains a lot of debate regarding the genetic link to keratoconus, according to the collaborative longitudinal evaluation in keratoconus study (CLEK) it is believed that the condition is autosomal dominant with incomplete penetrance. Therefore, there is a slightly higher chance of her child manifesting keratoconus; however, the odds are still quite small. Most patients with diagnosed cases of keratoconus do not profess a family history of the condition. Keratoconus does appear to be linked to eye rubbing, although whether this is a cause or simply a symptom is unclear at this point. There also appears to be a very high correlation between keratoconus and atopy. People who suffer from eczema, allergies, and hay fever tend to display a higher incidence of keratoconus compared to the rest of the general population. Again, this subset of people has a high correlation of eye rubbing associated with ocular irritation, which may contribute to the formation of keratoconus.

385
Q

The internal limiting membrane of the retina is formed by a combination of glial cell processes and which other cell type?

A. Bipolar cells
B. Horizontal cells
C. Amacrine cells
D. Muller cells

A

D. Muller cells

Explanation
The internal limiting membrane is formed by radial feet processes of Muller cells along with other glial cell constituents. Muller cell processes extend throughout the length of the retina except for the retinal pigment epithelium. Their cell bodies are found in the inner nuclear layer. Muller cells, along with segments of the photoreceptors, create tight junctions and serve to form the external limiting membrane.

386
Q

A patient views a bichrome visual acuity chart with no lenses before his eyes. With the patient’s left eye occluded, he reports that the letters on the red side of the chart appear blacker and darker. With his right eye occluded, the letters on the green side appear blacker and darker. Given these observations and assuming that accommodation is at rest, what would likely be the refractive condition of the right eye and left eye respectively?

 A. Hyperopia and emmetropia  
 B. Emmetropia and hyperopia  
 C.  Myopia and hyperopia   
 D. Myopia and myopia  
 E. Myopia and emmetropia  
 F. Hyperopia and myopia
A

C. Myopia and hyperopia

Explanation
For an emmetropic eye that is not accommodating, the chromatic interval within the eye would be positioned so that the anterior (green) and posterior (red) ends of the interval are equidistance from the retina (i.e. the midpoint of the interval would be at the retina) and thus the letters on the red and green sides of the chart would appear equally black and dark. For uncorrected myopia, the interval would shift forward and thus the red end of the interval moves closer to the retina. In this case, the letters on the red side would appear blacker and darker. For uncorrected hyperopia, the interval would shift backward and thus the green end of the interval moves closer to the retina. In this case, the letters on the green side would appear blacker and darker.

387
Q

Which of the following terms describes the vertical measurement taken from the top of the reading segment on a bifocal to the deepest part of the frame or lens?

A. The segment depth
B. The segment fusion
C. The segment height
D. The segment width

A

C. The segment height

Explanation
The segment height is measured as the distance between the top of the segment to the deepest portion of the frame (this measurement is frame dependent). The segment depth is measured as the longest vertical distance from the top of the seg to the bottom of the seg. The seg width is determined by measuring the widest horizontal portion of the seg. The segment fusion is fictional.

388
Q

Your patient had LASIK over one year ago, and her prescription has regressed slightly by 1.25 D of myopia. She wishes to get an enhancement and asks you what is the MOST common complication associated with a LASIK enhancement?

A. Presbyopia
B. Flap dislocation
C. Post sub-capsular opacification
D. Epithelial ingrowth

A

D. Epithelial ingrowth

Explanation
The most common complication associated with an enhancement after previously undergoing LASIK is epithelial ingrowth. Although this condition can occur after the first LASIK procedure due to poor flap adhesion or from stray epithelial cells remaining under the flap, it is far more common with enhancements. Some surgeons are attempting to decrease the incidence of ingrowth occurrence with enhancements by re-cutting a new flap or via refractive keratotomy (RK). In general, ingrowth does not cause a problem if it is isolated to a small area and if there is little elevation or change with time. If the ingrowth is significant and vision is compromised, treatment requires that the flap be lifted and the offending cells removed.

Posterior capsular opacification only occurs after cataract surgery when residual lenticular epithelial cells proliferate, causing opacification of the posterior aspect of the space between the posterior lens implant and the posterior capsule. If vision is compromised, treatment requires that the cells be removed via a YAG laser.

389
Q

Motor neurons carry information to and from which of the following structures respectively?

A. Neurons and the skin
B. The brain and the muscles
C. Muscles and the brain
D. Skin and neurons

A

C. Muscles and the brain

Explanation
There are three different classes of neurons: motor neurons, interneurons, and sensory neurons. Sensory neurons respond to stimuli and relay the information to the brain for interpretation. Interneurons receive sensory messages and combine them with other information to alter the activity of other neurons. Motor neurons relay information away from the brain and spinal cord to the muscles or glands to carry out the required actions.

390
Q

Which space in the meninges of the optic nerve is continuous with the space between the brain meninges such that an increase in cerebrospinal fluid pressure may cause the appearance of papilledema?

A. The subdural space
B. The subpial space
C. Subarachnoid space
D. Dural sinuses

A

C. Subarachnoid space

Explanation
The subarachnoid space of the optic nerve is continuous with that of the brain. The subarachnoid space exists between the arachnoid and the pia mater and ends at the lamina cribrosa where it folds back on itself and terminates in a cul-de-sac. This space is filled with cerebrospinal fluid (CSF) and therefore should anything cause an increase in CSF pressure, this can translate down to the nerve and manifest as papilledema.

The dural sinuses are venous channels that carry blood from the brain to the heart, not cerebrospinal fluid.

The dura is generally firmly attached to the skull or bones and the arachnoid is connected to the dura. In the event of injury, trauma or sickness the dura and arachnoid may become separated resulting in a space called the subdural space.

In a healthy individual there should not exist a space underneath the pia mater. The pia mater is an extremely thin membrane that adheres very closely to the brain, spinal cord and optic nerve such that it follows the sulci and gyri of the brain’s surface.

391
Q

A stimulus (the mask) is first presented to a subject followed by another stimulus (the target). The situation in which the mask precedes the presentation of a target is known as which type of masking?

A. Backward masking
B. Forward masking
C. Metacontrast
D. Simultaneous masking

A

B. Forward masking

Explanation
Masking involves the presentation of two stimuli to the visual system, one being the mask and the other being the target, to help gather information regarding how the visual system processes spatial and temporal information. The mask is supposed to interfere with the visibility of the target. Forward masking describes the situation in which the mask is presented before the target. Backward masking is when the target is presented first followed by the mask (the mask being brighter). Simultaneous masking occurs when the mask and the target are presented at the same time. Metacontrast is considered to be a type of backward masking in which the target is presented first followed by the mask, but the two stimuli are spatially adjacent. For example, an x is presented followed quickly by an O that is larger and surrounds the x but does not occupy the same space. Remember, the name of the masking type describes the location of the mask. Therefore, in forward masking the mask is presented before the target, and in backward masking the target precedes the mask, etc.

392
Q

A healthy retinal nerve fiber layer is thickest at which portion of the optic nerve head?

A. Inferiorly
B. Superiorly
C. Nasally
D. Temporally

A

A. Inferiorly

Explanation
The nerve fiber layer is thickest at the inferior and superior regions of the nerve, respectively. The inferior and superior arcades are composed of large diameter axons with little overlap of the receptive fields, thus explaining why a field defect occurs in these regions first for early cases of glaucoma. Inferior or superior notching of the nerve is highly suspect for glaucomatous damage, and must undergo further testing in order to rule out glaucoma. The next thickest area of nerve fiber layer tissue is nasally, which is comprised of the nasal radial fibers. These axons are affected in the later stages of glaucoma, thus explaining why a temporal island of the visual field is often left remaining in advanced cases of glaucoma. Lastly, the temporal rim area is the thinnest. Temporal rim tissue is comprised of the papillomacular bundle. The fibers in this area are very small and compact, with a high degree of receptive field overlap, therefore because of the receptive field redundancy, a visual field defect correlating to this region will occur only after significant fiber loss has occurred. Due to the fact that these fibers are so small in diameter, even though they are numerous, the fibers do not occupy a lot of space in the optic nerve. The thickness of the nerve fiber layer rim tissue is best remembered as ISNT, with inferior being the thickest and temporal rim tissue being the thinnest.

393
Q

Which 4 of the following nerves are strictly MOTOR in function? (Select 4)

 A. Cranial nerve XI  
 B. Cranial nerve I  
 C. Cranial nerve VI  
 D. Cranial nerve IV   
 E. Cranial nerve II  
 F. Cranial nerve III
A

A. Cranial nerve XI
C. Cranial nerve VI
D. Cranial nerve IV
F. Cranial nerve III

Explanation
Cranial nerves I (olfactory), II (optic), and VIII (vestibulocochlear) are sensory in function. Cranial nerves III (oculomotor), IV (trochlear), VI (abducens), XI (spinal-accessory) and XII (hypoglossal) have exclusively motor functions. Cranial nerves V (trigeminal), VII (facial), IX (glossopharyngeal) and X (vagus) possess both motor and sensory functions. A good mnemonic is: Some Say Marry Money But My Brother Says Big Brains Marry Money (The words correspond to the cranial nerves in proper numerical order, with all ‘S’ words representing sensory function: ‘M’ words depict exclusive motor function, and ‘B’ words signify nerves with both motor and sensory function).

394
Q

A 13 year-old patient is seen at your office reporting headaches for the past 2 months. She reports that the headaches are dull and occur above her eyes. The headaches are typically not present upon wakening and worsen towards mid-day, especially with reading. She typically does not experience any headaches on the weekend unless she is using the computer or reading. Based upon the patient’s case history alone, what is the MOST likely etiology of her headaches?

A. Sinusitis
B. Eyestrain
C. Tension
D. Migraine

A

B. Eyestrain

Explanation
Ocular-related headaches or those due to eyestrain present with prolonged use of the eyes. The majority of patients will note an absence of headache in the morning (although some will report headaches in the morning due to excessive ocular use the night before). The headaches are typically dull in nature and occur around or above the eyes. Occasionally, the eyestrain can occur concurrently with a tension headache, and the patient may also report tightness of the neck or pain in the occipital region.

Tension headaches, also known as muscular contraction headaches, are most commonly caused by stress and may be also be attributed to poor posture, hunger and fatigue. This type of headache is associated with stiffness of the muscles in the neck region and pain around the occipital region or a vice-like pain in the frontal area. Patients typically report the presence of a throbbing pain.

Migraine headaches are thought to be neurological in nature and can be triggered by several mechanisms such as hormonal changes or neurotransmitter imbalances. Typically patients will note a unilateral throbbing/pulsating pain. Patients will generally experience associated symptoms of nausea, vomiting, photophobia, and phonophobia. Some patients will perceive an aura immediately preceding the headache.

Sinus headaches are more commonly ‘associated acute’ rather than ‘chronic nasal sinusitis’ and usually present with a deep and constant pain in the frontal region of the head. Pain in these cases is usually associated with other sinus symptoms such as rhinorrhea, feeling of fullness in the ears, fever, facial swelling, and lacrimation. Patients typically report a transient loss of the sense of smell. The majority of patients will note an increase in pain/pressure when bending down.

395
Q

Topical application of a drug in solution form causes loss of medication retention in the eye. When the drop is introduced into the conjunctival cul-de-sac, what 2 locations does the MAJORITY of drug loss initially occur? (Select 2)

A. Drained through the nasolacrimal sac
B. Absorbed into the conjunctival fornixes and then lost to trabecular outflow
C. Absorbed into conjunctival vessels and lost to systemic absorption
D. Absorbed into the cornea and then diluted in the aqueous humor of the anterior chamber
E. Mixed and diluted in reflexive tears caused by drop instillation
F. Spillover onto the cheek and face

A

A. Drained through the nasolacrimal sac
F. Spillover onto the cheek and face

Explanation
When an ophthalmic medication is first instilled into the conjunctival sac, particularly suspensions and solutions, much of the drug is lost to spillover and drainage through the nasolacrimal duct, both of which are exacerbated by an increase in blinking. Research also reveals that some of the medication is lost to absorption into the conjunctival, anterior uveal, and episcleral vessels. A small portion of the medication is eliminated via trabecular and uveoscleral outflow once it becomes absorbed into the anterior chamber. Lastly, a minor amount is lost to absorption through limbal vessels and the tear film.

396
Q

Numerous reports have suggested that increased tear film osmolarity is a key consequence in dry eye. Although osmolarity is not easily measured in the clinical setting, tear osmolarity increases in most dry eye sub-types due to which of the following processes?

A. In aqueous tear deficiency, the lacrimal gland produces more ionic species
B. Patients with dry eye tend to blink less than normals, leading to increased evaporation
C. Loss of tear stability induces an increased evaporation rate, leading to increased osmolarity
D. Decreased capillary exchange leads to ionic bonding
E. Reactive oxygen species are increased in the tears of most dry eye sub-types; this increases osmolarity
F. The lipid layer is altered in most dry eye states, leading to ion pairing

A

C. Loss of tear stability induces an increased evaporation rate, leading to increased osmolarity Your Answer

Explanation
Tear instability leads to greater evaporation and higher osmolarity through a mechanism of concentration of the remaining tears, since only the aqueous tear portion evaporates rather than the ionic species. Several studies have indicated that normal tear osmolarity is less than or equal to 300 Osm/L, with values exceeding 308 Osm/L indicating increased osmolarity. As a single measure, tear osmolarity has recently been found to correlate the best (r squared 0.55) to dry eye severity of several clinical tests in a large, multi-center study (Sullivan et al., IOVS 51:6125-6130, 2010).

397
Q

While performing retinoscopy, how does the resultant reflex change as your approach neutral motion?

A. The reflex becomes brighter and appears narrower
B. The reflex appears to move more slowly
C. The reflex becomes brighter and broader
D. The reflex becomes dimmer and broader

A

C. The reflex becomes brighter and broader

Explanation
When performing retinoscopy, as one nears neutrality, the reflex becomes wider/broader, brighter and appears to move more quickly.

398
Q

During retinoscopy why is the contralateral eye not occluded?

A. To ensure a bright reflex
B. To control for myopia
C. To control for accommodation
D. To ensure good lighting conditions

A

C. To control for accommodation

Explanation
In order to achieve good retinoscopy results, both eyes must remain open and fogging (adding lenses causing the projected letters to be perceived as blurry but still readable) of the contralateral eye helps to control accommodation. Controlling accommodation is especially important with hyperopes who possess very good accommodative skills.

399
Q

Which of the following steroids is LEAST likely to contribute to the formation of posterior subcapsular cataract (PSC) formation?

 A. Prednisolone  
 B. Loteprednol
 C. Dexamethasone  
 D. Rimexolone  
 E. Fluorometholone
A

B. Loteprednol

Explanation
Research has demonstrated that ester-based steroids possess the least likelihood of causing a PSC. To date, the only ester-based topical ophthalmic steroid is loteprednol. Loteprednol possesses an ester group in the carbon 20 position rather than a ketone group. Posterior capsular cataract formation occurs as a result of the interaction between the ketone group and lens proteins, causing the formation of a Schiff base intermediate and eventually leading to PSC development. Because loteprednol does not contain a ketone group, the probability of PSC formation is significantly diminished.

400
Q

The semicircles created by the prism of the Goldmann tonometer probe and sodium fluorescein are most successfully viewed with the help of which of the following filters?

 A. Cobalt blue  
 B. Neutral density filter  
 C. Red-free  
 D. Green-free  
 E. A filter will not enhance the resultant image
A

A. Cobalt blue

Explanation
A cobalt blue filter is used in conjunction with sodium fluorescein to enhance the resultant image of the semicircles created by the tonometer prism, in order to maximize intraocular pressure measurement. The blue light emitted by the cobalt blue filter excites the sodium fluorescein, which causes it to glow a bright green color.

401
Q

Which of the following BEST describes the rotation of the eyeball during Bell’s phenomenon?

A. Downward and outward
B. Upward and outward
C. Upward and inward
D. Downward and inward

A

B. Upward and outward

Explanation
Bell’s phenomenon is a reflex coordinated between the facial nerve and the oculomotor nuclei in which the eyeball is rotated upward and outward upon closure of the eyelids. The reflexive movement occurs as a protective mechanism that repositions the cornea up under the eyelid and, therefore, away from potential danger. The neurological pathway of this phenomenon is not completely understood. It is also not present in about 10% of otherwise healthy individuals; thus, its absence cannot be relied upon as a cause or cited as a sign of disease.

402
Q

Which class/isotype of immunoglobulins is responsible for the primary immune response?

 A. IgD  
 B. IgA  
 C. IgG  
 D. IgM
 E. IgE
A

D. IgM

Explanation
There are 5 classes of immunoglobulins that are produced by B lymphocytes. IgM is found as a monomer on naive B lymphocytes or as a pentamer in its secreted form. Its role is to recognize antigens and initiate the primary immune response. IgD is found as cell bound monomer whose function is not understood.

IgG, IgE and IgA are only produced after the primary immune response and require isotype switching for their production. IgG has 4 subclasses and carries out a wide range of effector functions, including antibody-dependent cell-mediated cytotoxicity, opsonization, and complement fixation. IgG is the most abundant isotype present in the serum and is the only isotype that crosses the placenta.

IgA has 2 subclasses and is found in secreted form as monomers and dimers. It is present in the serum, but its main effector function occurs in the respiratory and intestinal tracts. It is the most actively produced isotype and confers mucosal immunity.

IgE is a monomer that mediates allergic disease. It is the only isotype that binds to mast cells and has the shortest half-life of all isotypes.

403
Q

During embryonic development, what is the name of the artery that supplies nutrients to the lens and later dissolves to form the canal of Cloquet?

A. Cilioretinal artery
B. Central retinal artery
C. Hyaloid artery
D. Carotid artery

A

C. Hyaloid artery

Explanation
The hyaloid artery is a branch of the ophthalmic artery and is important for delivering nutrients to the lens and structures behind the lens during development. The artery disintegrates before birth leaving a canal in the primary vitreous called Cloquet’s canal. Remnants of this artery are often perceived as “floaters”.

404
Q

A person who has type A blood can receive a successful blood transfusion using which of the following blood types?

A. Type BO
B. Type O
C. Type AB
D. Type B

A

B. Type O

Explanation
A person who has type A blood has A markers on their red blood cells and therefore does not possess antibodies against A markers but does manufacture antibodies for B markers. This individual can receive a transfusion from type O blood as red blood cells of this type do not display any markers and therefore do not result in antibody production.

A person who has type B blood can receive type B and type O blood but not type A because they possess antibodies for the A markers.

A person who has type AB blood can receive blood from types A, B, AB, and O. Type AB is considered the universal recipient because they will not surmount an immune reaction to any type of blood.

A person with type O blood is considered the universal donor (an easy way to remember this is that dOnor has an O in it!). Individuals with type O blood can only receive type O blood because they carry antibodies against both the A and B markers.

If two incompatible blood types are mixed, agglutination or clumping of the blood will occur, which can block blood vessels and lead to damage of the tissues and/or death.

405
Q

How are the end products of triglyceride metabolism transported from the intestine to the liver?

A. Packaged into chylomicrons
B. Malate-aspartate shuttle
C. Via exocytosis into the lymph system
E. Excreted directly into the blood stream

A

A. Packaged into chylomicrons

Explanation
Fat is digested primarily in the intestine. It enters the intestine from the stomach, where it becomes emulsified by bile salts and hydrolyzed by lipases released from the pancreas. These end products are then absorbed by enterocytes that line the walls of the intestine. Once inside, the enterocytes, triglycerides are then rebuilt and packaged along with cholesterol and protein to form chylomicrons. The chylomicrons are able to enter the lymph system, where they are absorbed into the blood stream and transported to the liver or adipose tissue. Triglycerides cannot directly diffuse through the cell membranes of the liver or adipose tissue; they must first be broken down into fatty acids and glycerol. This process is made possible by the presence of lipoprotein lipases on the walls of blood vessels. The fatty acids and glycerol are then absorbed by the liver to be used for energy or taken up by adipose tissue, where they are re-synthesized into triglycerides for storage purposes.

406
Q

Which of the following charts will allow for the measurement of a patient’s contrast sensitivity?

A. Jaeger chart
B. Pelli-Robson chart
C. Feinbloom chart
D. Snellen chart

A

B. Pelli-Robson chart

Explanation
Patients may report blurring of vision even though acuity may be measured to be 20/20 or 20/25. In these instances, one should consider performing a visual acuity test with low contrast, such as the Bailey-Lovie low contrast acuity chart or the Pelli-Robson test, to assess for a deficit in contrast sensitivity. The Pelli-Robson chart consists of optotypes that are the same size but decrease in contrast sensitivity (see image 1).

The Feinbloom, Snellen, and Jaeger charts, although all different, possess optotypes that do not vary in contrast. The Jaeger chart is used for assessment of near visual acuity.

407
Q

When performing automated perimetry, when one fixates at a particular point, after about 20 seconds or so, a stimulus away from the fixation point will fade away and disappear. What is the name of this phenomenon?

A. Riddoch phenomenon
B. Stiles Crawford effect I
C. Troxler effect
D. Raynaud’s phenomenon

A

C. Troxler effect

Explanation
The Troxler effect is noted when individuals stare at a central fixation point and the peripheral targets disappear. This is observed regularly by patients when performing perimetry. This phenomenon is particularly evident when observing low contrast dull targets. Raynaud’s phenomenon refers to an abnormal response of peripheral vasculature when an individual moves from a hot to a cold environment. This condition is caused by restriction of blood flow to the extremities, such as the fingers and toes, causing numbness or an excessively cool sensation when exposed to decreased temperatures or stress. The Riddoch phenomenon refers to the fact that a stimulus is only observed when it is in motion, and cannot be detected by the observer when it is static. The Riddoch phenomenon may be reported by individuals who suffer from an occipital lobe lesion, optic nerve damage, or chiasmal damage. The Stiles Crawford effect I refers to the fact that photoreceptors are more sensitive to light rays from the center of the pupil than to light rays entering from the periphery.

408
Q

In order to maintain binocular fixation and fusion on a target that is moved away from the patient in a linear fashion, assuming the target moves in one plane only, which of the following ocular movements must occur?

A. Divergence
B. Infraversion
C. Dextroversion of the right eye and levoversion of the left eye
D. Supravergence

A

A. Divergence

Explanation
In order to maintain fixation on a target that is moved away from the patient, his eyes must diverge (rotate outwards away from the midline). A vergence is a binocular movement of the eyes such that the eyes move in opposite directions. A version is a binocular eye movement in which the eyes move in the same direction and of equal magnitude. Infraversion is a movement where both eyes look downwards. Although during divergence, the right eye technically moves slightly to the right (away from the midline) and the left eye moves slightly to the left (away from the midline) these movements cannot be termed dextroversion and levoversion respectively as a version infers movement of the eyes in the same direction. Supravergence is the movement of one eye upwards with respect to the other.

409
Q

The fact that you no longer feel the pressure of your clothes on your skin after several minutes of wearing them is an example of which of the following phenomena?

A. Increased perception
B. Sensory adaptation
C. Sensory sensation
D. Somatic secretion

A

B. Sensory adaptation

Explanation
Sensory adaptation occurs when a stimulus is continually applied to a receptor or several receptors. Rather than continually signaling the presence of the stimulus, some receptors will cease to fire action potentials or decrease the frequency of action potentials.

410
Q

Flame hemorrhages that are characteristic of hypertensive retinopathy occur in what layer of the retina?

A. The ganglion cell layer
B. The nerve fiber layer
C. The retinal pigment epithelium
D. The inner nuclear layer

A

B. The nerve fiber layer

Explanation
Flame hemorrhages occur at the level of the nerve fiber layer and stem from blood loss from the inner capillary network of the retinal vasculature. These types of hemorrhages are commonly seen in hypertensive patients, vein occlusions, as well as optic neuropathies such as papilledema. The characteristic flame shape occurs because the blood will take the path of least resistance and follow the structure of the nerve fiber layer

411
Q

All else being equal, cells found in which layer of the cornea consume the GREATEST amount of oxygen?

A. Epithelium
B. Stroma
C. Endothelium
D. Bowman’s membrane

A

C. Endothelium

Explanation
The cells of the endothelium require the greatest amount of oxygen. This is due to the fact that endothelial cells maintain a steady state of corneal clarity and hydration. They actively pump out ions into the anterior chamber, which sets up an osmotic gradient that causes water to flow down its concentration gradient, thus preventing corneal swelling and opacification. However, because the endothelium is only one layer thick, in total this layer consumes 21% of the oxygen provided to the cornea. The stroma utilizes 39% of oxygen made available to the cornea, which is a low consumption rate considering that it makes up the bulk of the cornea. The epithelium is responsible for 40% of the oxygen consumed by the cornea. However, all else being equal, the endothelial cells consume the greatest amount of oxygen (140 X 10-5 ml of oxygen per sec), vs. stromal cells (2.85 X 10-5 ml of O2/sec) and epithelial cells (26.5 X 10-5 ml of O2/sec).

412
Q

Which agent is frequently added to non-opioid analgesics to enhance their effects?

A. Antacids
B. Mucolytics
C. Antihistamines
D. Caffeine

A

D. Caffeine

Explanation
Caffeine is a mild stimulant that, by itself, has little to no analgesic properties. Caffeine can be added to acetaminophen or to aspirin to enhance their analgesic effects. Excedrin® is a combination of acetaminophen, aspirin, and caffeine. Anacin® is also a combination medication comprised of aspirin and caffeine. Antacids and antihistamines are also frequently added to non-opioid analgesics but with the purpose of treating other ailments (i.e. allergy).