timed test #2 Flashcards

1
Q

Which TWO of the following are TRUE in regards to frame adjustments on a patient’s face? (Select 2)

If the right lens is too close to the face, bring the right temple in
If the right lens is too low, bend the left temple down
If the right lens is too far from the face, bring the left temple out
If the right lens is too high, bend the right temple up

A

Which TWO of the following are TRUE in regards to frame adjustments on a patient’s face? (Select 2)

If the right lens is too close to the face, bring the right temple in
If the right lens is too high, bend the right temple up

Explanation
If the temple spread of a pair of spectacles is uneven, or one side of the patient’s head is somewhat wider than the other, it is possible that one lens will be closer to the face than the other. For example, if the right lens is too far from the patient’s face, this could be due to the fact that either the right temple is not spread far enough, making that side fit too tightly, or that the left temple is too loose. The opposite is true if the right lens were to be sitting too close to the face.
A simple rule of thumb for frame adjustment is:
-If the right lens is in -> move the right temple in (or left temple out)
-If the left lens is in -> move the left temple in (or right temple out)
-If the right lens is out -> move the right temple out (or left temple in)
-If the left lens is out -> move the left temple out (or right temple in)

If the frame is not straight on the face, it could be because of incomplete standard alignment, or because one ear of the patient is positioned slightly higher than the other. The solution is the same regardless of the cause of the misalignment:

  • If the right lens is up -> bend the right temple up (or left temple down)
  • If the left lens is up -> bend the left temple up (or right temple down)
  • If the right lens is down -> bend the right temple down (or left temple up)
  • If the left lens is down -> bend the left temple down (or right temple up)
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Which 3 of the following blood tests are used in the analysis of kidney function? (Select 3)

 Glomerular filtration rate (GFR)   
 Bilirubin levels   
 Alkaline phosphatase (ALP)  
 Alanine and aspartate transaminase (ALT & AST)  
 Blood urea nitrogen (BUN)   
 Serum creatinine
A

Which 3 of the following blood tests are used in the analysis of kidney function? (Select 3)

Glomerular filtration rate (GFR)
Blood urea nitrogen (BUN)
Serum creatinine

Explanation
Healthy kidneys are responsible for removing wastes and excess fluids from the body. Several blood and urine tests can be utilized in order to assess proper kidney function. Three of the major blood tests include analysis of serum creatinine levels, blood urea nitrogen (BUN) and a measure of glomerular filtration rate (GFR).

Serum creatinine

  • Creatinine is a waste product produced by muscles of the body
  • A creatinine level greater than 1.2 for women and 1.4 for men may be an early indicator that the kidneys are not working properly
  • As kidney disease progresses, levels of creatinine in the blood will rise

Blood urea nitrogen (BUN)

  • Urea nitrogen originates from the breakdown of protein in food
  • A normal BUN level is between 7 and 20
  • As kidney function decreases, BUN levels will rise

Glomerular filtration rate (GFR)

  • Measurement of how well the kidneys are removing waste and excess fluid from the blood
  • It may be calculated from serum creatinine level using patient age, weight, gender, and body size
  • Normal GFR values are 90 or above; a GFR below 60 indicates kidney disease; GFR below 15 indicates that treatment for kidney failure is necessary
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

What is the approximate cross reactivity between penicillins and cephalosporins in individuals who reported a penicillin allergy?

 50%   
 30%  
 1%   
 10%  
 20%
A

What is the approximate cross reactivity between penicillins and cephalosporins in individuals who reported a penicillin allergy?

1%

Explanation
From a biochemistry perspective, first-generation cephalosporins and penicillin-based medications appear similar in that their physical pharmacological make-ups both possess a beta-lactam ring. Previous studies purport that roughly 10% of patients who are allergic to penicillin may also potentially exhibit an allergic response to the first-generation cephalosporins; however, recent studies have proved this data to be incorrect. It is now thought that there is approximately 1% or less cross-reactivity. The cross-reactivity between penicillins and cephalosporins is almost non-existent with the third-generation cephalosporins. Although the chances of a patient possessing a cross-reactivity between penicillin and cephalosporin drugs is very rare, it is still best to be cautious when a known allergy to penicillins exists.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

What is the name of the congenital retinal vascular anomaly in which a vessel that supplies perfusion to the macula and the papillomacular bundle can be observed emerging from the optic disc, separately from the central retinal artery?

Optociliary shunt vessel
Cilioretinal artery
Situs inversus
Arteriovenous anastomosis

A

What is the name of the congenital retinal vascular anomaly in which a vessel that supplies perfusion to the macula and the papillomacular bundle can be observed emerging from the optic disc, separately from the central retinal artery?

Cilioretinal artery

Explanation
Cilioretinal arteries are most frequently a result of an anomalous direct or indirect branch of the posterior ciliary arteries. In some cases, they can also arise from the peripapillary choroid. Cilioretinal arteries are typically observed as emerging from the temporal aspect of the optic disc, separately from the central retinal artery. The incidence of occurrence varies from 6-25% in the relevant literature. A cilioretinal artery has been said to be a possible protective feature that allows for a secondary route of blood flow to the macula in the event of a central retinal artery occlusion.

Optociliary shunt vessels are venous communications between the retinal and choroidal circulations; these vessels occur on the disc surface when there is impaired venous outflow through the retinal vein.

Situs inversus is a congenital vascular anomaly found at the optic disc, where the retinal vessels have followed an abnormal course. They enter the disc temporally and proceed towards the nasal retina before turning back and coursing out to the temporal retina.

An arteriovenous anastomosis of the retina is an abnormal communication between an artery and vein that occurs through collateral channels. It serves to shunt blood away from an artery to a vein in order to bypass a capillary bed.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

Which of the following antimicrobial agents has a mechanism of action that results in interruption of protein synthesis via binding to the 30S ribosomal subunit of the infecting bacteria?

 Cephalosporins  
 Tetracyclines   
 Fluoroquinolones  
 Aminoglycosides   
 Macrolides  
 Penicillins  
 Sulfonamides
A

Which of the following antimicrobial agents has a mechanism of action that results in interruption of protein synthesis via binding to the 30S ribosomal subunit of the infecting bacteria?

Tetracyclines

Explanation
Erythromycin and azithromycin belong to a class of drugs called macrolides. Macrolides are effective antibiotics because they bind to the 50S subunit of bacterial ribosomes, thus interfering with bacterial protein synthesis. 
Tetracycline and doxycycline are classified as tetracyclines. This class of drugs also interferes with protein synthesis via binding to the 30S ribosomal subunit. 
Amoxicillin, cloxacillin, and dicloxacillin are common anti-bacterials belonging to a class of drugs termed penicillins. Penicillins disrupt cell wall synthesis, making them valuable and widely used antibiotics. 
Tobramycin, gentamicin and neomycin are categorized as aminoglycosides, which serve as antimicrobials via two mechanisms; aminoglycosides inhibit bacterial protein synthesis as well as create openings in bacterial cell membranes, allowing for increased antibiotic uptake.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

Which of the following subnuclei of the third cranial nerve innervates its corresponding contralateral muscle?

Inferior oblique subnuclei
Inferior rectus subnuclei
Superior rectus subnuclei
Medial rectus subnuclei

A

Which of the following subnuclei of the third cranial nerve innervates its corresponding contralateral muscle?

Superior rectus subnuclei

Explanation
The nuclear complex of the oculomotor nerve resides in the midbrain at the level of the superior colliculus. It is composed of paired and unpaired subnuclei that innervate their corresponding muscles on either the ipsilateral or contralateral sides.

The medial rectus subnuclei, inferior rectus, and inferior oblique subnuclei are paired subnuclei (meaning that there is one on each side) and they innervate their corresponding muscles on the ipsilateral side. Lesions involving this area will result in ipsilateral medial rectus, inferior rectus, and inferior oblique weakness.

The superior rectus subnuclei are also paired; however, they innervate their respective contralateral superior rectus muscle. Therefore, a lesion affecting the third nerve nuclear complex will spare the ipsilateral superior rectus and will involve the contralateral superior rectus.

The levator subnucleus is the only subnucleus that is unpaired. It lies in the caudal midline and innervates both levator muscles. Lesions in this area will result in bilateral ptosis.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

A polycarbonate lens requires the addition of an anti-reflective coating. Which of the following equations would be used to find the MOST appropriate index of refraction for the coating?

√1.52
√1.586
1.586²
1.52²

A

A polycarbonate lens requires the addition of an anti-reflective coating. Which of the following equations would be used to find the MOST appropriate index of refraction for the coating?

√1.586

Explanation
Antireflective coatings are applied to ophthalmic lenses to increase the amount of light that is transmitted through the lens thereby minimizing the overall amount of light that is reflected off the surface of the lens. The higher the index of refraction of the lens, the greater the amount of reflection experienced. Antireflective coatings minimize reflections by creating destructive interference. The coating causes the reflected light from the surface of the lens to be 180 degrees out of phase with the light that is reflected from the coating. To optimize the antireflective property of the coating, the coating’s index of refraction should be equal to the square root of the index of refraction of the lens. For the above question, the index of refraction of polycarbonate lenses is 1.586 and the square root of 1.586 is 1.259.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Your 33 year-old patient with a manifest refraction of OD: -0.50 -2.75 x 172 and OS: -0.75 -1.75 x 010 wishes to try soft contact lenses. After discussion, you decide to PureVision2 monthly lenses. Which of the following prescriptions would you try first for his right eye?

  • 0.50 -2.75 x 170
  • 0.50 -2.25 x 170
  • 0.50 -2.50 x 170
  • 0.75 -2.75 x 170
  • 0.75 -2.50 x 170
  • 0.75 -2.25 x 170
A

Your 33 year-old patient with a manifest refraction of OD: -0.50 -2.75 x 172 and OS: -0.75 -1.75 x 010 wishes to try soft contact lenses. After discussion, you decide to PureVision2 monthly lenses. Which of the following prescriptions would you try first for his right eye?

-0.75 -2.25 x 170

Explanation
When determining the initial toric contact lens prescription to try on a patient, it is first important to know the contact lens parameters that are available for the specific brand of lenses that you would like to use. Most soft toric contact lenses all have very similar parameters; 0.25D steps in the spherical component (0.50 steps with higher Rxs), cylinder component is in 0.50 steps (starting with 0.75, up to 2.25 or 2.75 depending on brand), and cylinder axis is typically in 10-degree steps.
For the above patient, the cylinder component is -2.75D, but this is not available in PureVision2 toric lenses, as these lenses only go up to a cylinder value of -2.25. Therefore, for this patient, we would use -2.25D cyl. Because the cylinder component is being adjusted by 0.50D, in order to keep the same spherical equivalent, it is appropriate to add an additional -0.25 to the spherical component, making this -0.75D. When it comes to axis determination, this patient is between 170 and 180 but is closer to 170 degrees, so we would begin with a 170-degree axis lens.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Calculate the amount of image jump for the bifocal Rx below:

+3.00 -2.00 x 180 OU, +2.00 add
Flat top 28 bifocal
Distance PD 64 mm, Near PD 60 mm

1 prism diopter base down
1.5 prism diopter base up
1.5 prism diopter base down
1 prism diopter base up

A

Calculate the amount of image jump for the bifocal Rx below:

+3.00 -2.00 x 180 OU, +2.00 add
Flat top 28 bifocal
Distance PD 64 mm, Near PD 60 mm

1 prism diopter base down

Explanation
Image jump is based on the add power and the near optical center location and does not involve the distance Rx. As the person looks down into the segment to read, their eyes pass from the distance Rx into the bifocal segment. At that point they experience “jump”, since the eyes are not looking through the near optical center. Vertical prism is thus induced, causing the “jump” effect. Using Prentice’s rule, Prism = F x d(cm) where the power (F) is the add power and the distance (d) is determined by how far away the near optical center is from the top of the segment; i.e., prism = 2.00 x 0.5 or 1 prism diopter. Remember, for a flat top 28 bifocal, the optical center is located 5 mm below the segment line.

The add is a plus lens. Draw a plus lens in the segment so that the bases of the plus lens are coincident with the near optical center. If the patient’s eyes are just passing through the top of the segment, base down prism is induced.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

A mother reports that her 2-year-old child’s right eye does not appear straight. You decide to perform the Hirschberg test to evaluate for strabismus. The corneal reflex of the right eye is displaced 0.5 mm temporally while the left reflex is centered. Angle Kappa (Lambda) is zero for each eye. What is the correct deviation and magnitude of the observed strabismus?

Right esotropia of 11 prism diopters
Right esotropia of 22 prism diopters
Right exotropia of 11 prism diopters
Right exotropia of 22 prism diopters

A

A mother reports that her 2-year-old child’s right eye does not appear straight. You decide to perform the Hirschberg test to evaluate for strabismus. The corneal reflex of the right eye is displaced 0.5 mm temporally while the left reflex is centered. Angle Kappa (Lambda) is zero for each eye. What is the correct deviation and magnitude of the observed strabismus?

Right esotropia 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. Temporal displacement of the corneal reflex suggests esotropia, while nasal displacement infers exotropia. Each millimeter of displacement of the reflex from the center of the pupil equates to roughly 22 prism diopters of deviation.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

The magnitude of normal breathing is controlled by which structure in the brain stem?

The medulla oblongata
The pineal gland
The midbrain
The pons

A

The magnitude of normal breathing is controlled by which structure in the brain stem?

The medulla oblongata

Explanation
The medulla is surrounded by cerebrospinal fluid. Should the levels of carbon dioxide in the body become too elevated, the partial pressure of increased carbon dioxide causes a decrease in the pH of the cerebrospinal fluid. This stimulates the respiratory center in the brain to increase the breathing rate and depth to diminish carbon dioxide levels.

The pons is involved in regulating information between the cerebellum and the integration centers of the forebrain.

The pineal gland serves to regulate circadian rhythms

The midbrain mainly serves to coordinate visual, auditory, and tactile input.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Two spots of light are flashed simultaneously and are detected by the scotopic system, which transmits the information and signals as if there were one spot of light. This observation can be explained, in part, by which of the following?

The photopic system possesses greater temporal summation than the scotopic system
The receptive field of ganglion cells that receive information from rod cells is on-center surrounded by on-surround
The information from the second flash of light is not transmitted because the available rhodopsin has been bleached and therefore cannot signal the presence of the second light
The information from many rods cells is summated indirectly onto one ganglion cell

A

Two spots of light are flashed simultaneously and are detected by the scotopic system, which transmits the information and signals as if there were one spot of light. This observation can be explained, in part, by which of the following?

The information from many rods cells is summated indirectly onto one ganglion cell

Explanation
Spatial summation is utilized by the scotopic system and allows for greater sensitivity at the cost of resolution. Information from several rods across space is funneled indirectly onto one ganglion cell (rods do not directly synapse onto ganglion cells). Because the information is received from several rods, this combines to increase the chances of detecting an object (i.e., increased sensitivity). However, because the area over which the information is gathered is large and the information is sent to one ganglion cell, the resolution is poor. This is in opposition to the photopic system, which displays a much smaller degree of spatial summation, leading to decreased sensitivity but greater resolution. For example, two lights that are flashed close together in both time and space will be coded as one flash by the scotopic system because the information is sent to one ganglion cell. However, the same stimuli will be perceived as two lights by the photopic system due to the stimulation of two adjacent ganglion cells, thereby enabling this system to code for two different stimuli, resulting in greater resolution.

The receptive field of a ganglion cell is antagonistic and is either on-center surrounded by off-surround or off-center surrounded by on-surround.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

You are verifying a spectacle lens prescription that has returned from your lab. You ordered an Rx for the right eye of +1.00 -2.50 x 100. Which of the following values should be used for determining the ANSI standard for the sphere power?

 \+3.50  
 -1.50   
 \+1.00  
 -3.50  
 -2.50
A

You are verifying a spectacle lens prescription that has returned from your lab. You ordered an Rx for the right eye of +1.00 -2.50 x 100. Which of the following values should be used for determining the ANSI standard for the sphere power?

-1.50

Explanation
When determining the error tolerance for the sphere power of spectacle lenses, one must first find the meridian of highest absolute power. The easiest way to find this number is to put the ordered prescription on an optical cross. In this case, the highest absolute power is -1.50.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Which of the following tissue cells are insulin-independent?

 Cells that line the walls of blood vessels   
 Skeletal muscle cells  
 Heart muscle cells  
 Red blood cells   
 Adipose cells
A

Which of the following tissue cells are insulin-independent?

Red blood cells

Explanation
Insulin is a hormone secreted by the beta cells of the pancreas in response to elevated blood glucose. The target sites for insulin are the liver, adipose tissue, and muscle cells where the hormone stimulates anabolic pathways. In muscle cells, there is an increase in the uptake of glucose.

In the liver, glucokinase and glycogen synthase are activated; this results in an increase in the uptake of glucose and an increase in the synthesis of glycogen respectively. The enzyme acetyl-CoA carboxylase is also activated, causing a rise in fatty acid synthesis. Conversely, glycogen phosphorylase is inhibited, producing a decrease in the breakdown of glycogen.

Lipoprotein lipase enzymes are activated in adipose tissue triggering an increase in the synthesis of triglycerides.

Red blood cells, brain cells, lens fiber cells, and kidney cells do not require insulin for proper functioning and therefore do not possess insulin receptors. Glucose entry into these cells is not regulated by insulin but rather the concentration difference of glucose levels on the inside versus the outside of the cell. Muscle cells and adipose cells require insulin receptors which mediate the transportation of glucose into the cells. Insulin-independent cells are therefore more susceptible to damage in diabetes cases as the glucose intake into these tissues is not as regulated as those that possess insulin receptors.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

Which of the following oral antihistamines will most likely cause mydriasis and dry mouth?

Loratadine (Claritin®)
Cetirizine (Zyrtec®)
Fexofenadine (Allegra®)
Diphenhydramine (Benadryl®)

A

Which of the following oral antihistamines will most likely cause mydriasis and dry mouth?

Diphenhydramine (Benadryl®)

Explanation
Diphenhydramine is a first-generation antihistamine that also possesses significant anti-cholinergic activity; as such, it can induce dry mouth, urinary retention, constipation, delirium, and mydriasis. Furthermore, first-generation antihistamines freely cross the blood-brain barrier and by antagonizing histamine receptors in the CNS also cause significant somnolence.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Which of the following lipids is involved in inflammation?

Isoprenes
Eicosanoids
Cholesterol
Triglycerides

A

Which of the following lipids is involved in inflammation?

Eicosanoids

Explanation
Eicosanoids are derivatives of arachidonic acid. Examples of eicosanoids include leukotrienes, prostaglandins, and thromboxanes.

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 that transport the hormones to their designated target site, where they may further undergo further processing or transformation.

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.

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.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

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

 Chloride ions  
 Sodium ions  
 Albumin  
 Ascorbate   
 IgG
A

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

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.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

Approximately how many axons comprise the normal adult human optic nerve?

 4,000,000  
 2,200,000  
 700,000  
 1,200,000   
 3,500,000
A

Approximately how many axons comprise the normal adult human optic nerve?

1,200,000

Explanation
The normal adult human optic nerve is comprised of approximately 1.2 million retinal ganglion cell axons (using manual measuring techniques). These numbers have been shown to vary from about 700,000 to 1,400,000, utilizing automated counting algorithms. Factors that may affect the number of axons within the optic nerve include inherited differences, damage to the fibers from disease processes (such as optic neuropathy), and gradual loss due to the normal aging process.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

Which of the following colors possesses the greatest wavelength?

 Violet  
 Green  
 Blue  
 Red   
 Yellow
A

Which of the following colors possesses the greatest wavelength?

Red

Explanation
As one moves to the right in the electromagnetic spectrum, the wavelengths progressively decrease while the frequency of the emitted radiation increases. At the right end of the spectrum are radio waves; moving to the left we find microwaves, and infrared light (IR). We then see visible light and further to the left is ultraviolet light, X-ray and gamma ray. Sandwiched between IR and ultraviolet light is the visible portion of the spectrum; it is truly a very small slice of the entire spectrum. X-rays and gamma rays are very high in frequency but possess a small wavelength. Radiation emitted by rays at the far left of the spectrum can be quite dangerous. Remember the mnemonic ROY G BIV. This will help you to remember color wavelengths in the visible range as red possesses the longest wavelength while violet has the smallest wavelength. Red is roughly 650 nm, orange 600 nm, yellow 570 nm, green 520 nm, blue 470 nm and violet 400 nm.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

The finding of subretinal epithelial deposits between the basement membrane of the retinal pigment epithelium (RPE) and Bruch’s membrane in the macular area of a 64-year old white female with a slight loss of visual acuity (20/30) and pseudophakia is the clinical hallmark of what disease?

Retinal pigment epithelium (RPE) dropout or window defect
Dry or age-related macular degeneration (AMD)
Macular pucker or epiretinal membrane (ERM)
Soft exudates or clinically significant macular edema (CSME)

A

The finding of subretinal epithelial deposits between the basement membrane of the retinal pigment epithelium (RPE) and Bruch’s membrane in the macular area of a 64-year old white female with a slight loss of visual acuity (20/30) and pseudophakia is the clinical hallmark of what disease?

Dry or age-related macular degeneration (AMD)

Explanation
Drusen are the hallmark of AMD. Multiple types of drusen have been described, including large (>64 microns) and small (<63 microns), calcified, and basal laminar. Drusen are required to make the diagnosis of AMD as they are the destructive force behind RPE loss. Drusen become more apparent on fluorescein angiography.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

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

Blue-yellow, red-green
Red-green, blue-yellow
Red-green, rod monochromacy
Blue-yellow, rod monochromacy

A

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

Red-green, blue-yellow

Explanation
Dr. Kollner, an ophthalmologist, reviewed an extensive amount of literature on the nature of color vision impairment in patients with acquired ocular diseases. He concluded that most patients with diseases of the optic nerve tended to have difficultly discriminating red from green hues, while most patients with retinal disease (primarily macular) possessed a greater loss of discrimination between blue and yellow hues. This general dichotomy of red/green defects in optic nerve disease, and blue/yellow defects in macular disease has since come to be known as Kollner’s rule. It is important to note that not all cases of acquired color vision deficiencies conform to this rule, however this theory holds true for the majority of patients.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

You prescribe a patient 4 base down prism in the right eye. The patient wishes to order bifocal lenses. How should the segment height be adjusted to compensate for the prism?

The segment height of the left eye should be raised
The segment height of the right eye should be lowered
The segment height does not need to be altered
The segment height of the right eye should be raised

A

You prescribe a patient 4 base down prism in the right eye. The patient wishes to order bifocal lenses. How should the segment height be adjusted to compensate for the prism?

The segment height of the right eye should be raised

Explanation
Because the prism will shift the image towards the apex, the patient’s eye will be looking slightly up at the image. In order to ensure that the pupils view through the segments at the same time, one must raise the bifocal height of the right eye. Make sure that you inform the patient that their seg heights will be unequal so there are no surprises when they pick up their glasses! A good rule of thumb to follow is that the segment height must be altered by 0.3 mm for every diopter of vertical prism added to one eye only (if the prism is distributed fairly equally between the eyes, then there is no need to adjust the bifocal height). Remember that for base down prism you must raise the seg height and for base up prism you must lower the seg height.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

Which of the following glaucoma medications is considered a cholinergic?

Dipevefrin (Propine®)
Dichlorphenamide (Daranide®)
Pilocarpine
Epinephrine (Glaucon®)

A

Which of the following glaucoma medications is considered a cholinergic?

Pilocarpine

Explanation
Cholinergics decrease intraocular pressure by stimulation of the ciliary muscles, causing an increase in aqueous outflow. Side effects of cholinergics include accommodative myopia, band keratopathy, cataract formation, retinal detachment, miosis of pupils, and decreased night vision, among other side effects. Cholinergics are generally not the first line of choice for the management of glaucoma.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

Which 2 of the following are the most common etiologies associated with non-granulomatous intraocular inflammation? (Select 2)

 Idiopathic   
 Syphilis  
 HLA B27 autoimmune disease   
 Tuberculosis  
 Drug-induced  
 Herpetic viral infections
A

Which 2 of the following are the most common etiologies associated with non-granulomatous intraocular inflammation? (Select 2)

Idiopathic
HLA B27 autoimmune disease

Explanation
Non-granulomatous intraocular inflammation most commonly involves the anterior segment of the eye, has an acute onset, and is accompanied by a cellular reaction that represents an accumulation of cells due to a breakdown of the blood-ocular barrier. These are primarily white blood cells and protein that present as cells and flare. If these cells form a cluster on the corneal endothelium they are called keratic precipitates. In granulomatous uveitis these precipitates are much larger and are referred to as mutton-fat keratic precipitates. These are thought to be an inflammatory or auto-immune response to an infectious organism. Non- granulomatous presentations are typically idiopathic in nature, or are due to HLA B27 involvement. Non-granulomatous inflammation also occurs more frequently than does granulomatous uveitis (especially in patients with anterior uveitis).

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
Q

Grave’s disease can cause spontaneous diplopia. Which of the following extraocular muscles is the most commonly affected in a patient afflicted with Grave’s disease?

 The inferior oblique  
 The superior oblique  
 The lateral rectus  
 The inferior rectus   
 The medial rectus
A

Grave’s disease can cause spontaneous diplopia. Which of the following extraocular muscles is the most commonly affected in a patient afflicted with Grave’s disease?

The inferior rectus

Explanation
Grave’s disease is an auto-immune condition caused by hyperthyroidism (overproduction of thyroid hormones). The eye muscles may become thicker and shorter due to an infiltration of lymphocytes, which limits the ability of the muscles to contract, eventually resulting in fibrosis of the muscle. Restriction of the extraocular muscle (EOM) causes a perceived palsy of the opposing EOM. The eye muscles most frequently affected are the inferior rectus, medial rectus, and the superior rectus, with the inferior rectus being by far the most commonly involved.

26
Q

You receive a pair of glasses back from your fabrication lab that measures -1.25 -0.25 x 030. The original order was -1.25 -0.25 x 018. Before contacting your patient to let them know their glasses are ready for pick-up you check ANSI standards to see if this is within tolerance. According to ANSI standards, what is the tolerance for error in cylinder axis for this prescription?

 Axis may deviate up to 18 degrees  
 Axis may deviate up to 7 degrees   
 Axis may deviate up to 5 degrees  
 Axis may deviate up to 14 degrees   
 Axis may deviate up to 3 degrees
A

You receive a pair of glasses back from your fabrication lab that measures -1.25 -0.25 x 030. The original order was -1.25 -0.25 x 018. Before contacting your patient to let them know their glasses are ready for pick-up you check ANSI standards to see if this is within tolerance. According to ANSI standards, what is the tolerance for error in cylinder axis for this prescription?

Axis may deviate up to 14 degrees

Explanation
The ANSI standards for cylinder axis are dependent upon the strength of the cylinder power
- For small cylinder powers of 0.25 D, the axis may deviate up to 14 degrees in either direction and still fall within tolerance
- A cylinder power of 0.50 D may allow an axis variation of up to 7 degrees
- 0.75 D of cylinder power may have a deviation of axis up to 5 degrees
- For cylinder powers of 1.00, 1.25, and 1.50 D, the axis tolerance is 3 degrees in any direction
- Cylinder powers of 1.75 or above are only allowed an axis deviation of 2 degrees

27
Q

The principal fibers of the lens insert from which ocular structure onto the lens?

From the cornea onto the lens capsule
From the ciliary body onto the lens capsule
From the anterior face of the vitreous onto the lens capsule
From the iris onto the lens capsule

A

The principal fibers of the lens insert from which ocular structure onto the lens?

From the ciliary body onto the lens capsule

Explanation
The zonules can be divided into two different groups based upon their function and attachment site to the lens, the principal fibers and the association fibers. The principal fibers control accommodation and support the lens. This group can be further separated into two groups: the orbiculocapsular group and the ciliocapsular (both of which contain further subdivisions). The orbiculocapsular group arise from the posterior ciliary body and insert onto the lens. There are two sub-categories within the orbiculocapsular group. The first is the orbicular-anterior-capsular group which runs from the posterior pars plana and inserts onto the lens anterior to the equator. The second subdivision, the orbicular-posterior-capsular group, originates at the ora serrata and courses over the ciliary body to insert onto the lens posterior to the equator.

The ciliocapsular group arises from the sides and valleys of the ciliary processes and inserts onto the lens. This group also is further divided into two subdivisions; the first is the cilio-posterior-capsular group, which runs from the sides and valleys of the ciliary processes and inserts onto the lens posterior to the equator. The second division, the cilio-equatorial-capsular group, runs from the anterior sides of the ciliary processes and inserts onto the lenticular equator.

The association fibers serve to brace and support the zonules. This group is further divided into three parts. The first group is the orbiculociliary division, which rises from the pars plana and connects to the ciliary processes. The second division, the interciliary fibers, run between the ciliary processes and connect the processes to each other. The last subdivision is the circular fibers which run between the zonules and serve to connect them in a circular pattern similar to a spider web.

28
Q

What is the approximate volume of the vitreous?

  1. 0 ml
  2. 0 ml
  3. 0 ml
  4. 0 ml
A

What is the approximate volume of the vitreous?

4.0 ml

Explanation
The vitreous weighs approximately 4.0 grams and has a volume of roughly 4.0 ml. The essentially clear, gel-like fluid occupies about 2/3 of the bulk of the eye.

29
Q

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

Necrotizing
Scleromalacia perforans
Diffuse
Nodular

A

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

Necrotizing

Explanation
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.

30
Q

A patient presents with a right pupil that is 1 mm larger than the left pupil both in bright and dim illumination. The pupils are equally reactive to light with no afferent pupil defect (APD) detected. Light-near dissociation is not present, and the eyelid position is normal for both eyes. There is no history of trauma or headaches. All other ocular findings are normal. Photos from four years ago confirm the presence of a larger pupil on the right side. What is the most likely diagnosis?

Physiological anisocoria
Horner’s syndrome
Cranial nerve III palsy
Optic nerve damage of the right eye

A

A patient presents with a right pupil that is 1 mm larger than the left pupil both in bright and dim illumination. The pupils are equally reactive to light with no afferent pupil defect (APD) detected. Light-near dissociation is not present, and the eyelid position is normal for both eyes. There is no history of trauma or headaches. All other ocular findings are normal. Photos from four years ago confirm the presence of a larger pupil on the right side. What is the most likely diagnosis?

Physiological anisocoria

Explanation
When a patient presents with physiological anisocoria, it is important to rule out all other possible causes of unequal pupil size before concluding this diagnosis. The pupil size should evaluated in both bright and dim illumination; physiological anisocoria should be the same under both lighting conditions (or somewhat greater in dim light). The lid positions should then be evaluated for possible ptosis. The pupillary response to direct light and the consensual light response should also be assessed in order to rule out pathology. The near response may also be evaluated, however if the direct response is normal, then there is no need to check the near response. Lastly, the patient should be evaluated for the presence of an APD. If all of the findings are normal, it is best to get confirmation of the stability of the anisocoria by observing previous photos of the patient. Some studies report that physiological anisocoria can be present in up to 20% of patients.

31
Q

Distortion of perceived images in which straight lines appear wavy is known as which of the following?

 Micropsia  
 Metamorphopsia   
 Photopsia  
 Scotoma  
 Macropsia
A

Distortion of perceived images in which straight lines appear wavy is known as which of the following?

Metamorphopsia

Explanation
Metamorphopsia is a type of visual distortion in which perceived linear images appear wavy or broken. This commonly occurs in macular diseases such as macular degeneration, clinically significant macular edema, central serous retinopathy, etc. Clinicians typically test for this visual symptom using an Amsler Grid.

Micropsia is a type of visual distortion in which images appear to be decreased in size; this is caused by the spreading apart of foveal cones. Contrastingly, macropsia is a perceived increase in image size as a result of a crowding together of foveal cones. Both of these visual symptoms are very uncommon.

Photopsia is a perceived flash of light that is most commonly associated with traction of the retina in cases of a retinal break, tear, or detachment, or most commonly, a posterior vitreous detachment.

Scotomas are areas of the visual field whereby vision is a partially or entirely degenerated.

32
Q

In order for the cornea to maintain its transparency, the organization and spacing of the collagen fibrils within the corneal stroma MUST adhere to which of the following rules?

Distance between collagen fibrils must be less than 1/3 the wavelength of visible light
Distance between collagen fibrils must be greater than 1/3 the wavelength of visible light
Distance between collagen fibrils must be greater than 1/4 the wavelength of visible light
Distance between collagen fibrils must be less than 1/2 the wavelength of visible light
Distance between collagen fibrils must be greater than 1/2 the wavelength of visible light
Distance between collagen fibrils must be less than 1/4 the wavelength of visible light

A

In order for the cornea to maintain its transparency, the organization and spacing of the collagen fibrils within the corneal stroma MUST adhere to which of the following rules?

Distance between collagen fibrils must be less than 1/2 the wavelength of visible light

Explanation
The transparency of the cornea is due to the precise arrangement of the sheets of collagen fibrils that makeup the corneal stroma. Because of this configuration, light that is scattered by individual collagen fibrils is cancelled by destructive interference with light scattered by adjacent collagen fibrils. Therefore, light ends up only being scattered in the forward direction; this allows the cornea to remain transparent. In order for this type of destructive interference to occur, a few factors must be maintained. The diameter of the collagen fibrils must be of equal diameter, and each fibril must be equidistant from adjacent fibrils. Additionally, it is required that the distance between each collagen fibril must be less than the wavelength of visible light. When this arrangement is compromised (i.e. in an edematous cornea), the corneal stroma will swell, leading to loss of clarity due to the increase in distance between the collagen fibrils. Proteoglycans within the corneal stroma are responsible for preserving the regular packing and spacing of the fibrils, maintaining the transparency of this tissue.

33
Q

If a patient scores low on both the horizontal and vertical portions of the Developmental Eye Movement test (DEM), but the ratio and error scores are normal, what may be concluded regarding the patient’s oculomotor function?

Normal oculomotor function, normal automaticity
Abnormal oculomotor function, poor automaticity
Normal oculomotor function, poor automaticity
Abnormal oculomotor function, normal automaticity

A

If a patient scores low on both the horizontal and vertical portions of the Developmental Eye Movement test (DEM), but the ratio and error scores are normal, what may be concluded regarding the patient’s oculomotor function?

Normal oculomotor function, poor automaticity

Explanation
The DEM test consists of 3 subtests that should be given in a specific order. Subtests A and B are the vertical components, and Subtest C is the horizontal component.
° The vertical time score is determined by adding the total time it takes to complete tests A and B
° This score determines the automaticity of number calling ability
° Does not require oculomotor control for vertical eye movements
° The horizontal time score is determined by compensating the time it takes to complete test C for the presence of omission and addition errors
° Horizontal time = Test C time x [80/(80 - o + a)] (wherein o = omission, a = addition)
° This score evaluates number calling ability in a spatial array
• Requires a sophisticated level of oculomotor control
° The total errors are calculated by adding the occurrence of all individual errors
° Total errors = (s + o + a + t errors)
° s = substitution, o = omission, a = addition, t = transposition
° The ratio score represents a convenient method for evaluating horizontal and vertical time simultaneously
° Ratio scores that are significantly greater than expected normal values suggest a much greater difficulty in number naming when horizontal eye movements are required

° If a patient scores low on the vertical subtest, this simply indicates a problem with naming numbers out loud (automaticity)
° If the horizontal subtests are also lower than expected, this could indicate that the patient has difficulty with automaticity and/or oculomotor dysfunction
° In these instances the ratio score helps differentiate the two
• If ratio is normal -> patient has equal difficulty with horizontal and vertical subtests and the problem is automaticity
• If the ratio is abnormal -> patient has more difficulty with the horizontal subtest as compared to the vertical subtest and the problem is both automaticity and oculomotor dysfunction

34
Q

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

The macula
The ora serrata
The optic nerve head
Blood vessels

A

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

The ora serrata

Explanation
The ora serrata, which also serves as the anterior vitreous base is the strongest point of attachment of the vitreous. The second strongest point of attachment is the optic nerve head. A detachment of the posterior hyaloid from the optic nerve head results in a posterior vitreal detachment (PVD) commonly reported in the elderly as the sudden appearance of a large floater that is oval or circular in shape (Weiss ring). The macula is the third strongest point of attachment, followed by the blood vessels.

35
Q

A patient walks into your office and wishes to be fit with a soft contact lens. The keratometry readings of the right eye are 45.00 D x 44.00 D. Which of the following base curves would be the BEST choice for this patient?

  1. 6 mm
  2. 6 mm
  3. 5 mm
  4. 0 mm
  5. 2 mm
A

A patient walks into your office and wishes to be fit with a soft contact lens. The keratometry readings of the right eye are 45.00 D x 44.00 D. Which of the following base curves would be the BEST choice for this patient?

8.6 mm

Explanation
In order to determine the proper base curve, one must first determine the average keratometry reading which would be 44.50 D. To convert this reading from diopters to millimeters, one must divide 337.5 by the average keratometry readings. 337.5/44.50=7.58 mm. Rounding to the nearest tenth gives 7.6 mm. Soft contact lenses are typically fabricated flatter than the actual corneal curvature. In order to compensate for this factor, many authors suggest that adding 0.8 mm to 1.0 mm to the flat keratometry reading (or average K reading if the readings are close in magnitude) will allow for ideal movement of the lens. Applying this rule to our K reading yields 7.6 + 0.8 mm= 8.4 mm. The above choices do not have this number as an option. When faced with the decision to either go too flat or too steep, one should always error on the side of using a flatter-fitting lens. When a contact lens is fit too steeply, one risks sealing off the cornea because the contact lens acts like a suction cup and does not move when the patient blinks. Seal-off can cause corneal edema and/or neovascularization. The base curves of soft contact lenses are generally quite a bit flatter than the true corneal measurements; it is for this reason that keratometry tends not to play as much of a role with soft contact lenses as it does with rigid gas-permeable lenses.

Ultimately, the definitive criteria for fitting a soft contact lens is how much lens movement is observed after several hours of wear on the eye. Lenses with the same base curve and diameter will often fit and move differently depending upon lens materials, center thickness and edge profile. The soft contact lens base curve/corneal curvature relationship is especially important for lenses fabricated of silicone hydrogel material due to their greater modulus values

36
Q

Which 2 of the following statements are TRUE regarding corneal epithelial microcysts from low-Dk soft contact lens extended wear? (Select 2)

They are about 15 to 50 micrometers in diameter and are irregularly shaped
Patients with microcysts report extreme photophobia due to inflammatory anterior chamber reactions
They show unreversed illumination when viewed under high magnification and marginal retroillumination
They require about two months of extended contact lens wear to appear
They represent permanent corneal damage due to hypoxia

A

Which 2 of the following statements are TRUE regarding corneal epithelial microcysts from low-Dk soft contact lens extended wear? (Select 2)

They are about 15 to 50 micrometers in diameter and are irregularly shaped

They require about two months of extended contact lens wear to appear

Explanation
Epithelial microcysts are a chronic reaction to insufficient oxygen diffusion, usually observed with the wear of low Dk soft contact lenses worn on an extended wear basis. They appear roughly 15-50 micrometers in diameter and exhibit reversed illumination when viewed with the slit lamp. Microcysts generally appear after 2 months of extended contact lens wear and are transient in nature. Patients with corneal microcysts are usually asymptomatic.

37
Q

Your patient has a vertexed manifest refraction of -2.00 -3.00 x 090 OS. You place a diagnostic soft toric contact lens with a power of -2.00 -3.00 x 090 on her left eye. The over-refraction results reveal +3.00 -1.50 at an oblique axis. Without assessing the fit with a slit lamp, from the over-refraction results alone you correctly estimate the lens to manifest what degree of rotation?

 30 degrees   
 5 degrees  
 25 degrees  
 15 degrees   
 10 degrees
A

Your patient has a vertexed manifest refraction of -2.00 -3.00 x 090 OS. You place a diagnostic soft toric contact lens with a power of -2.00 -3.00 x 090 on her left eye. The over-refraction results reveal +3.00 -1.50 at an oblique axis. Without assessing the fit with a slit lamp, from the over-refraction results alone you correctly estimate the lens to manifest what degree of rotation?

15 degrees

Explanation
According to Snyder (1989), a contact lens that is rotated 10 degrees away from its proper axis will result in the manifestation of astigmatism that is equal to roughly 1/3 of its original power in the over-refraction located at some oblique angle. A contact lens that has rotated 15 degrees away from its axis will display an over-refraction with astigmatism that is equal to 1/2 of the original power located at an oblique angle. A lens that has rotated 30 degrees away from its proper axis will reveal the full amount of cylinder power in the over-refraction at an oblique angle.

38
Q

Which of the following conditions appears as a black-purple, oval-shaped thinning of the scleral tissue and is a common, benign finding in aged adults?

Senile hyaline plaque
Limbal girdle of Vogt
Necrotizing scleritis
Crocodile shagreen

A

Which of the following conditions appears as a black-purple, oval-shaped thinning of the scleral tissue and is a common, benign finding in aged adults?

Senile hyaline plaque

Explanation
A senile hyaline plaque (also known a senile scleral translucency) is commonly seen on the temporal aspect of the sclera in front of the insertion of the lateral rectus or nasally in front of the insertion of the medial rectus in patients over the age of 50. The color of the plaque is attributable to scleral thinning, which allows for increased visibility of the uveal coat. This condition is benign and generally does not require treatment.

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.

Necrotizing scleritis is the most severe form of scleritis and is associated with scleral thinning, but this condition is not considered benign. Necrotizing scleritis has a higher mortality rate than the other types presented due to the fact that it usually stems from autoimmune diseases.

39
Q

An object is located 40 cm from an emmetropic eye (measured from the corneal plane). What degree of accommodation is required to achieve a clear retinal image?

-0.25 D
+0.25 D
+2.50 D
-2.50 D

A

An object is located 40 cm from an emmetropic eye (measured from the corneal plane). What degree of accommodation is required to achieve a clear retinal image?

+2.50 D

Explanation
In order a clear retinal image of an object located at 40 cm to be achieved, an emmetropic eye must accommodate +2.50 D, which is the accommodative demand at that target distance. This is calculated by taking the reciprocal of the target distance (in meters). For the above example, 1/0.40= +2.50 D. The answer will be positive, as accommodation causes the lens to increase its refractive power.

40
Q

The reversible replacement of one differentiated cell type with another mature differentiated cell type (as seen in patients with Barrett’s esophagus) is known as which of the following?

 Neoplasia  
 Dysplasia   
 Anaplasia  
 Metaplasia   
 Hyperplasia
A

The reversible replacement of one differentiated cell type with another mature differentiated cell type (as seen in patients with Barrett’s esophagus) is known as which of the following?

Metaplasia

Explanation
Metaplasia refers to the reversible substitution of one differentiated cell type for another fully differentiated cell type within a given tissue. Metaplasia most commonly occurs in epithelial tissues and can be part of the normal maturation process or can be a result of some abnormal stimulus. This occurs in cases of Barrett’s esophagus, thought to be the result of cellular damage from chronic stomach acid exposure. The original cells lining the lower esophagus are not robust enough to withstand this harsh environment and change into a cell type that is better suited for the environment.

41
Q

Your optician is trying to insert lenses into a frame made of polycarbonate and therefore cannot use heat. Which of the following techniques offers the easiest method of lens insertion?

Insert the temporal edge of the lens first, followed by the nasal aspect from the back of the frame
Insert the nasal edge first, followed by the temporal edge of the lens from the front of the frame
Insert the temporal edge of the lens first, followed by the nasal aspect from the front of the frame
Insert the nasal edge first, followed by the temporal edge of the lens from the back of the frame

A

Your optician is trying to insert lenses into a frame made of polycarbonate and therefore cannot use heat. Which of the following techniques offers the easiest method of lens insertion?

Insert the temporal edge of the lens first, followed by the nasal aspect from the front of the frame

Explanation
Although there are many ways of inserting ophthalmic lenses into a frame, the most commonly used method is to insert the temporal aspect of the lens first followed by snapping the nasal edge into place from the front of the frame. Most frame materials can be heated, which allows for ease of insertion, but this must be done with caution. Most opticians and clinicians prefer to use a hot air warmer rather than glass beads or sand, allowing for better control of heat distribution and minimizing frame and lens damage. If using sand or glass beads, it is best to use a lower heat setting. Some opticians add baby powder to glass beads to help keep the beads from sticking to the frame, which may cause small bubble-like indentations. When using heat, it is best to remove the temples, thereby decreasing the chances of dislodging or twisting the hinge. Remember that dark-colored frames tend to heat more quickly than light-colored ones. Frames fabricated from polyamide, copolyamide, carbon fiber, carbon fiber graphite or polycarbonate should not be heated.

42
Q

If a -2.50 D sphere lens is decentered by 5 mm in and 6 mm down from the prism reference point, what is the resultant prism power?

  1. 75 prism diopters
  2. 375 prism diopters
  3. 95 prism diopters
  4. 8 prism diopters
A

If a -2.50 D sphere lens is decentered by 5 mm in and 6 mm down from the prism reference point, what is the resultant prism power?

1.95 prism diopters

Explanation
In order to determine the amount of compounded prism, one must first determine the length of the resultant vector. Using the Pythagorean theorem, we can find the third component of our triangle. We know that one side measures 5 mm and the other measures 6 mm; now we need to determine the hypotenuse. The Pythagorean theorem states that: a2+b2= c2; therefore, solving for c yields: 25+36 = c2, c = 7.8 mm. Now, to determine the amount of prism, input this value into 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. X=(0.78)(-2.50), x=1.95 prism diopters.

43
Q

You decide to make your own pinhole camera to take a picture of your classmate. He is 6 feet tall and is standing 10 feet from your camera. How long does your “camera” need to be in order to make sure that he perfectly fits on your photo paper that is 4 inches tall?

0.55 inches
6.6 inches
15 inches
22 inches

A

You decide to make your own pinhole camera to take a picture of your classmate. He is 6 feet tall and is standing 10 feet from your camera. How long does your “camera” need to be in order to make sure that he perfectly fits on your photo paper that is 4 inches tall?

6.6 inches

Explanation
In a pinhole camera, an object lies on the left of the pinhole, and the image screen lies on the right of the pinhole. The pinhole allows only 1 object ray through at a time; therefore, we are able to use the equation of similar triangles:
x’/x =h’/h
x’ = image distance x = object distance, h’ = image height, h = object height,
In this example, x = 10 feet, h’ = 4 inches (or 0.33 feet), h = 6 feet, solve for x’
x’/10 feet = 0.33 feet/6 feet
x’ = (10 x 0.33)/6 = 0.55 feet (or 6.6 inches)

44
Q

A person who is completely dark-adapted and is shown scotopic stimuli will report that light of which wavelength will appear the brightest?

555 nm
650 nm
610 nm
507 nm

A

A person who is completely dark-adapted and is shown scotopic stimuli will report that light of which wavelength will appear the brightest?

507 nm

Explanation
The scotopic system is meditated by rods and contains the photopigment rhodopsin, whose peak absorbency is 507 nm. The scotopic system cannot discern colors and is very sensitive to dim illumination; the scotopic system also possesses poor spatial resolution, good spatial summation, good temporal summation, poor temporal resolution, and poor contrast sensitivity. The photopic system, on the other hand, possesses three photopigments - erythrolabe, cyanolabe, and chlorolabe. This system has phenomenal color discrimination as well as spatial and temporal resolution. Cones display poor temporal and spatial summation. The photopic system displays a peak spectral sensitivity to wavelengths that are 555 nm.

45
Q

The cornea absorbs which of the following amounts of incident UV-A, UV-B, and UV-C light, respectively?

40%, 75%, 95%
100%, 90%, 60%
60%, 90%, 100%
95%, 75%, 40%

A

The cornea absorbs which of the following amounts of incident UV-A, UV-B, and UV-C light, respectively?

60%, 90%, 100%

Explanation
The cornea is a major filter for UV light that is incident on the eye.
- 100% of incident UV-C light (100-280nm) is absorbed by the cornea
- About 90% of incident UV-B light (280-315nm) is absorbed by the cornea
- Approximately 60% of incident UV-A light (315-400nm) is absorbed by the cornea

46
Q

In retinoscopy, a myope with 2.00 D of ametropia will have principal meridians located at which axes?

A spherical ametropic eye has one principal meridian at 180 degrees
A spherical ametropic eye does not have principal meridians Your Answer
A spherical ametropic eye has one principal meridian at 90 degrees
A spherical ametropic eye has two principal meridians at 180 and at 90 degrees

A

In retinoscopy, a myope with 2.00 D of ametropia will have principal meridians located at which axes?

A spherical ametropic eye does not have principal meridians

Explanation
A spherical ametropic eye does not have any principal meridians. In clinical retinoscopy, an astigmatic eye will have two principal meridians, one denotes the axis of greatest power of the eye while the other indicates the meridian of least power.

47
Q

Which of the following dyes is useful for staining corneal nerves?

Methylene blue

Explanation
Methylene blue is used to stain devitalized cells, mucus, and corneal nerves. Methylene blue is also employed to stain the lacrimal sac prior to surgery for dacryocystorhinoplasty. It is important to note that methylene blue is bacteriostatic, therefore, one should do a cell culture (if necessary) prior to any application of methylene blue. Lissamine green stains dead or degenerated cells and therefore is useful for early detection of dry eye syndrome. Rose Bengal is similar to lissamine green in that it stains degenerated and dead cells as well as mucous stands. Rose Bengal can be used to evaluate dry eye, corneal and conjunctival lesions, and dendrites. Lissamine green and Rose Bengal both have mild anti-viral properties. Sodium fluorescein is actually not a stain but a dye. This dye is useful for detection of conjunctival and corneal lesions because it accumulates in the defect and fluoresces when used in conjunction with a cobalt blue filter. It is also useful in fitting gas-permeable lenses and for evaluating the tear break-up time. Fluorescein can be administered intravenously to help detect retinal neovascularization, macular lesions, central serous retinopathy, malignant melanomas, etc.

A

Which of the following dyes is useful for staining corneal nerves?

Methylene blue

Explanation
Methylene blue is used to stain devitalized cells, mucus, and corneal nerves. Methylene blue is also employed to stain the lacrimal sac prior to surgery for dacryocystorhinoplasty. It is important to note that methylene blue is bacteriostatic, therefore, one should do a cell culture (if necessary) prior to any application of methylene blue. Lissamine green stains dead or degenerated cells and therefore is useful for early detection of dry eye syndrome. Rose Bengal is similar to lissamine green in that it stains degenerated and dead cells as well as mucous stands. Rose Bengal can be used to evaluate dry eye, corneal and conjunctival lesions, and dendrites. Lissamine green and Rose Bengal both have mild anti-viral properties. Sodium fluorescein is actually not a stain but a dye. This dye is useful for detection of conjunctival and corneal lesions because it accumulates in the defect and fluoresces when used in conjunction with a cobalt blue filter. It is also useful in fitting gas-permeable lenses and for evaluating the tear break-up time. Fluorescein can be administered intravenously to help detect retinal neovascularization, macular lesions, central serous retinopathy, malignant melanomas, etc.

48
Q

Which of the following drugs is MOST EFFECTIVE against Methicillin-resistant Staphylococcus aureus (MRSA)?

Pred-Forte®
Ciloxan®
Polytrim®
Erythromycin

A

Which of the following drugs is MOST EFFECTIVE against Methicillin-resistant Staphylococcus aureus (MRSA)?

Polytrim®

Explanation
Polytrim® is most effective against MRSA. MRSA is a bacterial strain that can cause severe infections and is highly resistant to many antibiotics. MRSA is most commonly seen in people that have been in a health care setting such as a hospital or nursing home, and when contracted by this method, it is termed Health care-associated MRSA (HA-MRSA). HA-MRSA is typically associated with invasive procedures such as surgeries. MRSA is also seen in healthy communities (termed community-associated MRSA or CA-MRSA) such as child care workers or high school wrestlers and is transmitted by direct skin-to-skin contact. Unfortunately, MRSA is becoming more prominent; therefore, it is of utmost importance that when a patient presents with conjunctivitis, we as clinicians should always have MRSA in the back of our minds. MRSA is best treated with fortified vancomycin; however, as optometrists, we rarely if ever have this at our disposal. Many studies have shown that bacitracin and Polytrim® display excellent penetration to ocular infections caused by MRSA. Bacitracin and Polytrim® are often overlooked as antimicrobials because they are slightly archaic; however, given that MRSA-related ocular infections are increasing, this is an important finding. Both erythromycin and ciprofloxacin (Ciloxan®) show very poor sensitivities for MRSA. Pred-Forte®, especially by itself, should not be prescribed due to the fact that it suppresses the immune system and treats only inflammation, further enabling the infection. The fourth and fifth generation fluoroquinolones such as Zymar® and Besivance® demonstrate good efficacy against MRSA and are good alternatives; however, Polytrim® remains more effective and therefore should be the first choice when treating ocular infections associated with MRSA.

49
Q

Which of the following landmarks of an electrocardiogram represents repolarization of the atria?

 S wave  
 P wave  
 Q wave  
 Atrial repolarization is obscured by the QRS complex   
 PR interval  
 T wave
A

Which of the following landmarks of an electrocardiogram represents repolarization of the atria?

Atrial repolarization is obscured by the QRS complex

Explanation
An electrocardiogram machine processes electrical signals picked up through electrodes placed on the patient’s skin. It then produces a graphical representation of the electrical activity of the patient’s heart. The basic pattern of electrical activity is comprised of three waves: the P wave, QRS complex, and T wave. These landmarks are summarized below:

  • P wave: small upward deflection wave that represents atrial depolarization
  • Q wave: beginning of the QRS complex (depolarization of ventricles) that corresponds to the depolarization of the interventricular septum
  • S wave: the end of the QRS wave complex that represents the final depolarization of the ventricles at the base of the heart
  • T wave: signifies ventricular repolarization
  • PR interval: time between the first deflection of the P wave and first deflection of the QRS wave complex
  • There is no specific wave or landmark for the repolarization of the atria as it occurs during the same time at the main QRS complex and is obscured by this large wave
50
Q

The hip joint is an example of which type of joint?

Fibrous joint
Synovial joint
Cartilaginous joint
Hinge-like joint

A

The hip joint is an example of which type of joint?

Synovial joint

Explanation
Joints are points of contact or near contact between bones. Synovial joints are the most common type. This joint type separates bones via a cavity, and the involved ends of the bones are covered with a layer of cartilage that helps to cushion the joint. Cells that line the capsules also secrete fluid called synovial fluid into the cavity of the joint. Ball-and-socket joints allow for a full range of movement. Hinge-like joints, like the knee and elbow, allow for movement in one plane.

Cartilaginous joints are found between the vertebrae and some of the ribs. In this joint type, the space between the bones is filled with cartilage, which allows for little if any movement.

Fibrous joints do not exhibit cavities. This joint type serves to unite bones and is found on the skull bones of a newborn baby.

51
Q

Which of the following extraocular muscles is considered the antagonist pairing to the right lateral rectus?

 Left medial rectus  
 Left lateral rectus  
 Right superior oblique  
 Left superior oblique  
 Right medial rectus
A

Which of the following extraocular muscles is considered the antagonist pairing to the right lateral rectus?

 Left medial rectus  
 Left lateral rectus  
 Right superior oblique  
 Left superior oblique  
 Right medial rectus   

Explanation
Agonist-antagonist extraocular muscle pairs are those muscles of the same eye that move the eye in opposite directions. The agonist muscle is the primary muscle that contracts to move the eye in a given direction. The antagonist is the muscle of the ipsilateral eye that acts to move the eye in the opposite direction when contracted. Therefore, in this question, the agonist is the right lateral rectus, which moves the eye to the right, and the antagonist is the right medial rectus, which will move the eye to the left if contracted.

52
Q

Which of the following antibiotics have been proven to be MOST effective against methicillin-resistant Staphylococcus aureus (MRSA) infections?

 Tetracycline  
 Vancomycin   
 Amoxicillin  
 Clindamycin  
 Sulfacetamide  
 Cephalexin
A

Which of the following antibiotics have been proven to be MOST effective against methicillin-resistant Staphylococcus aureus (MRSA) infections?

Vancomycin

Explanation
Methicillin-resistant Staphylococcus aureus (MRSA) is a dangerous pathogen that is most commonly found in hospital settings and long-term care facilities. Studies have shown that approximately 35% of Staphylococcus aureus isolates in the U.S. are considered methicillin-resistant. MRSA is resistant to all of the beta-lactam antibiotics, which include penicillin derivatives and cephalosporins. Additionally, many MRSA strains are also resistant to all generations of fluoroquinolones, as well as multiple other classes of antibiotics. At the present time, vancomycin has emerged as the most reliable agent for the treatment of MRSA infections and is therefore the drug of choice under these circumstances.

53
Q

You are verifying a pair of spectacles that arrived from your fabrication lab to make sure that they fall within ANSI standards. If a prescription reads -5.25 -0.67 x 166 by lensometry, and -5.00 -0.75 x 170 was ordered, is this prescription acceptable according to ANSI?

No, this Rx fails ANSI standards due to the difference in sphere power
No, this Rx fails ANSI standards due to the difference in cylinder axis
No, this Rx fails ANSI standards due to the difference in cylinder power
Yes, this Rx falls within ANSI standards

A

You are verifying a pair of spectacles that arrived from your fabrication lab to make sure that they fall within ANSI standards. If a prescription reads -5.25 -0.67 x 166 by lensometry, and -5.00 -0.75 x 170 was ordered, is this prescription acceptable according to ANSI?

No, this Rx fails ANSI standards due to the difference in sphere power

Explanation
According to ANSI standards regarding error tolerances for spectacle lenses, the above prescription is considered to fall outside of standards due to the difference in the sphere power of the measured and ordered lens prescriptions.
When determining the error tolerances for the sphere power of spectacle lenses, one must first find the meridian of the highest absolute power. In this question (when placed on an optical cross), the meridian of highest absolute power is axis 080 in which the power should be -5.75 D. This should then be compared to the prescription that was sent from the lab, which is -5.92 (at axis 076). The difference between these two powers is 0.17.
ANSI standards read that for sphere powers equal to or less than +/- 6.50 D, the tolerance is +/- 0.13 D. Therefore, this prescription fails ANSI standards because the difference is -0.17 D.
For powers greater than -6.50, the standard is +/- 2% of the sphere power in the meridian of highest absolute power.
ANSI standards for error tolerance in cylinder power is +/- 0.13 D for cylinder powers of 2.00 D or less (this prescription just passes on cylinder power).
Additionally, the ANSI standard for cylinder axis with cylinder power of 0.75 D is a variation of up to 5 degrees (this prescription also passes on cylinder axis).

54
Q

A new patient is seen at your office and has worn gas-permeable lenses for years. He is very happy with the vision and the comfort of his current lenses. Biomicroscopy reveals nasal and temporal injection along with a slight loss of corneal transparency at the 3 and 9 o’clock positions on the cornea. How can you alter his current gas-permeable lens to prevent further damage to his corneas?

Change the power of the lens
Flatten the secondary curve
Increase the overall diameter of the lens
Increase the oxygen transmissibility of his gas-permeable lens

A

A new patient is seen at your office and has worn gas-permeable lenses for years. He is very happy with the vision and the comfort of his current lenses. Biomicroscopy reveals nasal and temporal injection along with a slight loss of corneal transparency at the 3 and 9 o’clock positions on the cornea. How can you alter his current gas-permeable lens to prevent further damage to his corneas?

Change the power of the lens
Flatten the secondary curve
Increase the overall diameter of the lens
Increase the oxygen transmissibility of his gas-permeable lens

Explanation
Peripheral corneal desiccation syndrome, also known as 3-9 staining, can occur in patients who wear rigid lenses. It can be caused by incomplete blinking, excessive edge lift or poor lens design, to name a few. Slit lamp exam will generally reveal hyperemia at the nasal and temporal aspects of the cornea, usually due to poor tear distribution and subsequent desiccation. If left alone these areas can lose transparency and may slough epithelium, causing an indentation on the corneal surface called a dellen. One can decrease the occurrence of 3-9 staining by increasing the overall diameter of the lenses, decreasing the edge thickness to allow a more even distribution of tears, steepening of the secondary curve, or refitting the patient into a soft contact lens. Some suggest that adding ocular lubricants helps with this issue, but this may be of little value since the problem is created by a poorly fitted lens.

55
Q

When performing Goldmann applanation tonometry, why is it important to control for patient accommodation?

Accommodation increases the size of the mires
Accommodation can falsely decrease intraocular pressure
Accommodation changes the contractile properties of the cornea, offsetting the mechanical forces of the tonometer tip
Accommodation neutralizes the effects of the tonometer prism

Explanation
Accommodation can cause a rapid decrease in intraocular pressure, and therefore it is important to direct your patient’s fixation towards a distant object in order to maximize the reliability of your results.

A

When performing Goldmann applanation tonometry, why is it important to control for patient accommodation?

Accommodation can falsely decrease intraocular pressure

Explanation
Accommodation can cause a rapid decrease in intraocular pressure, and therefore it is important to direct your patient’s fixation towards a distant object in order to maximize the reliability of your results.

56
Q

The cells of which layer of the retina serve as a storage site for vitamin A?

 The retinal pigment epithelium   
 The outer plexiform layer  
 The inner nuclear layer  
 The external limiting membrane  
 The nerve fiber layer
A

The cells of which layer of the retina serve as a storage site for vitamin A?

The retinal pigment epithelium

Explanation
Cells found in the retinal pigment epithelial layer serve as a storage site for vitamin A. Vitamin A is classified as a retinoid with its active form being retinol. Retinol is necessary for the formation of rhodopsin, a pigment used by rods. Rods are most active in situations with dim illumination. Less rhodopsin results in fewer rods being able to respond in low levels of light causing prolonged dark adaption.

The outer plexiform layer is the location at which rods and cones synapse onto horizontal and bipolar cells.

The nerve fiber layer is comprised of ganglion cell axons.

The inner nuclear layer contains the cell bodies of bipolar, horizontal and amacrine cells.

The external limiting membrane serves as a distinction between the inner segments of the photoreceptors and their respective nuclei.

57
Q

Which of the following conditions is MOST likely to lead to the observation of what is known as ‘scissor motion’ during retinoscopy?

Accommodative insufficiency
Primary open-angle glaucoma
Amblyopia
Keratoconus

A

Which of the following conditions is MOST likely to lead to the observation of what is known as ‘scissor motion’ during retinoscopy?

Accommodative insufficiency
Primary open-angle glaucoma
Amblyopia
Keratoconus

Explanation
Performing retinoscopy on a patient with an irregular cornea, such as in cases of keratoconus, can be difficult and usually results in a reflex called ‘scissor-motion’ in which the reflex appears to separate or converge together. Regardless of the etiology, when challenged with an aberrated reflex, the majority of clinicians recommend focusing on the reflex seen at the center of the pupil.

58
Q

Which of the following lenses has a nominal power of +4.00 D?

A meniscus lens with a front surface power of +5.00 D and an ocular surface power of -2.00 D
A converging lens with a front surface power of +6.00 D and an ocular surface power of -3.00 D
A bi-convex lens with a front surface power of +2.00 D and an ocular surface power of +2.00 D
A plano-convex lens with a front surface power of +5.00 D

A

Which of the following lenses has a nominal power of +4.00 D?

A bi-convex lens with a front surface power of +2.00 D and an ocular surface power of +2.00 D

Explanation
The nominal or approximate power of a lens can be determined by adding the powers of the front surface of the lens to the power of the back (ocular) surface. The only lens that fulfills the desired power of +4.00 D from the above combinations is a bi-convex lens with a front surface power of +2.00 D and a back surface power of +2.00 D.

59
Q

The colon is divided into four regions. What is the order that undigested matter will pass through it prior to expulsion from the anus?

Ascending colon, descending colon, transverse colon, sigmoid colon, rectum
Ascending colon, transverse colon, descending colon, sigmoid colon, rectum
Ascending colon, transverse colon, sigmoid colon, descending colon, rectum
Ascending colon, sigmoid colon, transverse colon, descending colon, rectum

A

The colon is divided into four regions. What is the order that undigested matter will pass through it prior to expulsion from the anus?

Ascending colon, transverse colon, descending colon, sigmoid colon, rectum

Explanation
The colon appears like an inverted U. Undigested matter will first pass through the ascending colon located on the right side of the abdominal cavity, pass through the transverse colon to the left side of the body, and travel down through the descending colon. Lastly the matter will pass through the s-shaped sigmoid colon into the rectum and out the anus.

60
Q

Your patient wishes to know how much magnification is created by her glasses. The back vertex power of her glasses measures -7.50 D, the back vertex distance is 13 mm from the entrance pupil of the eye, the center thickness of her lenses is 1.8 mm, the index of refraction is 1.61, and the power of the front surface of her lens is +1.00. What is the total magnification (in percentage) created by her lenses?

  • 10.01%
  • 8.8 %
  • 1.6%
  • 9.1%
A

Your patient wishes to know how much magnification is created by her glasses. The back vertex power of her glasses measures -7.50 D, the back vertex distance is 13 mm from the entrance pupil of the eye, the center thickness of her lenses is 1.8 mm, the index of refraction is 1.61, and the power of the front surface of her lens is +1.00. What is the total magnification (in percentage) created by her lenses?

-8.8 %

Explanation
In order to determine the total magnification (Mspec) experienced by this patient, we require two pieces of data: the shape magnification (Ms) and the power magnification (Mp). The formula for Ms=1/[1-F1(tc/n)] where F1= the power of the front surface of the lens, tc= the center thickness of the lens (in meters), and n= the index of the lens. Mp=1/[1-h(Fv)] where h= the vertex distance to the entrance pupil of the eye (in meters) and Fv= the back vertex power of the lens. Calculate Mp= 1/[1-0.013(-7.50), Mp= 0.911. Calculate Ms=1/[1-1(0.0018/1.61], Ms=1.001. To determine the total magnification, multiply Ms and Mp: Mspec=1.0011 x 0.911 = 0.912. Expressed as a percentage, Mspec%= (Mspec-1)x100.