OptoPrep Flashcards
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. 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.
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. 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.
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. Tylenol (acetaminophen) B. Omega III fish oil capsules C. Tums (calcium carbonate) D. Accutane (isotretinoin)
D. Explanation: 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.
An increased rate of molecular movement down its respective concentration gradient via help from carrier proteins refers to which type of transportation?
A. Facilitated diffusion
B. Active transport
C. Passive diffusion
D. Group translocation
A. Explanation: Facilitated diffusion is described as the net movement of molecules down its concentration gradient whose rate of diffusion is increased via the use of carrier proteins.
Passive diffusion refers to the movement of molecules through a plasma membrane from an area of high concentration to an area of low concentration without the use of carrier molecules. Active transport implies the movement of material against its respective concentration gradient. This type of transport requires energy and enlists the use of specific carrier proteins. Lastly, group translocation is defined as the chemical modification of a molecule while it is being transported into a cell; for example, sugars are often phosphorylated during transportation.
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. Visual Hallucinations B. Headache C. Gastrointestinal discomfort D. Tachycardia E. Depression
B. Explanation: 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.
A 63-year old female is seen at your office with a chief concern of blurry vision in the morning that takes about an hour to resolve before she can see clearly again. Biomicroscopy reveals endothelial guttata. You correctly diagnose her with moderate Fuch’s dystrophy. Which ophthalmic drop would be of MOST benefit to her?
A. Muro-128 (5% sodium chloride)
B. Vigamox (moxifloxacin)
C. Tobrex (tobramycin)
D. 1% Pred-Forte (prednisolone acetate)
A. Explanation: Sodium chloride is a topical hyperosmotic agent used to relieve stromal edema caused by endothelial decompensation. Topical steroids work well to decrease swelling caused by inflammation. In the above case, the corneal edema is not mitigated by an inflammatory response. Tobramycin and Vigamox would be of no benefit since there is no active infection, and prescribing either of these would only lead to corneal toxicity or increased pathogen resistance over time.
A deficiency of which vitamin leads to prolonged dark adaptation? A. Vitamin A B. Vitamin K C. Vitamin C D. Vitamin B E. Vitamin E
A. Explanation: A deficiency of vitamin A causes prolonged dark adaptation. Vitamin A is classified as a retinoid, and its active form is 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.
+1.50-1.50 x 090 is required to neutralize a reflex in retinoscopy with a working distance of 50 cm. What is the resulting NET retinoscopy finding?
A. -0.50-0.50 x 090
B. -0.50-1.50x 090
C. +1.50-1.50 x 090
D. Pl-0.50 x 090
B. Explanation: A working distance of 50 cm creates a divergent wave of 2.00 D that is neutralized by retinoscopy in addition to the patient’s refractive error. Therefore, + 2.00 D must be subtracted from the spherical portion of the findings. To determine how much to subtract from the gross findings, one must first calculate the reciprocal of the working distance in meters. In our case, 1/0.5 = 2. Therefore +1.50 (the spherical gross findings) -2 = -0.50-1.50 x 090. Remember NET is the final result, this is found after the working distance has been accounted for by subtracting the working distance from the spherical portion of the findings.
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. The a-wave will remain while the b-wave will disappear
B. Both the a-wave and the b-wave will remain
C. Both the a-wave and the b-wave will disappear
D. The a-wave will disappear while the b-wave will remain
A. Explanation: 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.
Free radicals can cause severe damage to tissue. Which of the following electrolytes can function as an antioxidant in the aqueous?
A. IgG
B. Albumin
C. 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.
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. Injection of epinephrine (EpiPen)
B. Olopatadine (Patanol)
Administration of Benadryl (oral)
C. Prednisone (oral)
a. 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.
Which of the following situations will result in the creation of a real, magnified image?
A. Placing an object between the center of the curvature of a concave mirror and its corresponding focal point
B. Placing an object at the center of curvature of a concave mirror
C. Placing an object beyond the center of curvature of a concave mirror
D. Placing an object between a concave mirror and its corresponding focal point
A. Explanation - Concave mirrors, also known as converging mirrors, act like a plus lens and converge light. When an object is located between the focal point of the mirror and the center of curvature, the resulting image will be real and will appear inverted and magnified. An object that is located beyond the center of curvature (and the focal length of the mirror) will result in the formation of an image that is real, inverted and minified. An object that is located between the mirror and its focal point will result in an image that is virtual and appears upright and magnified. An object that is located at the center of curvature of a convex mirror will form an image that is real, the same size as the object, and inverted.
A ray of light is deviated 5.50 cm by a prism made of crown glass located 7.0m away. Which of the following equations will CORRECTLY determine the total prism power?
A. P=(100)(5.50/700) B. P=(100)(700/0.55) C. P=(100)(7.0/5.50) D. P=(100)(700/5.50) E. P=(100)(0.55/700) F. P=(100)(5.50/7.0)
A. Explanation - Prism power is found by using the equation P=(100)(x/d), where P= power of the prism (in prism diopters, pd), x=the total distance that a ray of light is deviated, and d=the total distance from the prism to where the deviation is measured. For the above question, P= (100)(5.50 cm/700 cm)=0.786 pd. Key: both the distances must be in either meters or centimeters. If the units of the distances are not the same then 100 must be dropped from the formula. For example, P=5.50 cm/7.0 m=0.786 pd.
A concave mirror, located in water, has a radius of curvature of 13.0 cm. What is the power of the mirror (rounded to the nearest 0.25 D)?
A. 20.50 D B. -15.50 D C. -7.75 D D. -20.50 D E. 15.50 D
A. Explanation - A concave mirror converges light and therefore acts like a convex lens; hence, concave mirrors possess positive 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)/-0.13= 20.46 D, or 20.50 D rounded to the nearest diopter. Remember, a concave mirror will have a negative radius of curvature; a convex mirror will have a positive radius of curvature. The index of refraction of water is 1.33.
Which of the following ophthalmological imaging instruments is an example of a system utilizing confocal laser coherence tomography (CSLT)?
A. Optos 174 Retinal Camera
B. GDx VCC (Variable Corneal Compensator)
C. Correct answer HRT (Heidelberg Retina Tomograph)
D. Cirrus OCT (Optical Coherence Tomographer)
C. Explanation - The HRT (Heidelberg Retina Tomograph) utilizes a type of laser technology known as confocal scanning laser coherence tomography (CSLT).
The GDx VCC (Variable Corneal Compensator) is an example of a system that uses scanning laser polarimetry (SLP) technology.
The Cirrus OCT (along with all other brands of OCTs) utilizes optical coherence tomography technology to image the retina and optic nerve.
The Optos 174 retinal camera is a scanning laser ophthalmoscope that gives only a 2-dimensional photograph of the fundus, while the other devices allow for 3-dimensional retinal imaging.
In order to determine the most appropriate reading add for your presbyopic patient you decide to balance relative accommodation. With a +1.00 D tentative reading add in place, the NRA measures +1.50 and the PRA measures -0.50. Given the above results what should the final add be?
A. +1.25 D
B. +1.75 D
C. +2.00 D
D. +1.50 D
D. Explanation - In order to determine the reading add by balancing the relative accommodation, one must first determine what number needs to be added to the PRA and subtracted from the NRA to equalize the two. In the above example that number is 0.50 D. Taking away 0.50 D from the NRA yields +1.00D and adding it to the PRA gives -1.00 D. Therefore in order to balance both the NRA and the PRA one must increase the reading add by 0.50 D giving a final add of +1.50 D.
While practicing ray-tracing, you find which of the following to be TRUE in regards to object and image relationships?
A. For a real object, incident (object) rays are convergent
B. An object does not need to be located in object space; it only must be associated with incident light
C. For a virtual image, emergent (image) rays are convergent
D. The set of all points or rays associated with the light incident on an optical system is called image space
B. Explanation - The following is true of object and image relationships (assuming that object space is to the right of the optical system and image space is to the left for simplicity):
- Real Object: Incident (object) rays are divergent; it exists to the left of the optical surface or system; there are physically real objects and optically real objects.
- Virtual Object: Incident (object) rays are convergent on the surface; it exists to the right of the surface (where the incident rays would come to a point).
- Real Image: Emergent (image) rays are convergent; it exists to the right of a clear optical surface or system (for a mirror, it is to the left). There are physically and optically real images as well.
- Virtual Image: Emergent (image) rays are divergent; virtual images would have originated on the left for a clear optical surface (right of the reflecting surface for a mirror image).
- Object Space: Set of all points or rays associated with the light incident on an optical system; object does not need to be located in object space.
- Image space: Set of all points or rays associated with light leaving the optical system; image does not need to be located in image space.
A window in water is located 2 meters to the left of a light source (the index of refraction for water is 1.33). What is the reduced vergence for the pencil of rays falling at the surface of the window (the index of refraction for the window is 1.54)?
A. 0.77 D
B. -0.77 D
C. 0.67 D
D. -0.67 D
C. Explanation - The formula for reduced vergence (L) is L=n/x where n= the index of refraction and x is equal to the distance of the surface from the source (in meters). Remember measurements are to be taken from the surface to the point of convergence. Surfaces located to the right of the convergence point result in negative vergences. Surfaces located to the left yield positive vergences. For the above example, the index of refraction of the window does not matter, since the question addresses the light falling at its surface and not the light penetrating the window. The answer is determined as follows: 1.33/2=0.665 D or 0.67 D.
Which of the following spectacle prescriptions can be classified as compound hyperopic astigmatism?
A. +3.25 -3.25 x 070 B. +1.00 -1.75 x 090 C. -0.50 +2.00 x 150 D. +2.00 -0.50 x 120 E. Plano +3.00 x 010
Explanation - Refractive astigmatism is classified by the relationship of the two focal lines that are produced by the refractive components of the eye with respect to the retina (when accommodation is relaxed). The different clinical types of astigmatism are described below.
Simple astigmatism: When accommodation is relaxed, one focal line is located at the retina; the other is located either in front of or behind the retina
- If the second focal line is in front of the retina, this is simple myopic astigmatism (one principle meridian is plano, the other is myopic)
- If the second focal line is located behind the retina, this is simple hyperopic astigmatism (one principle meridian is plano, the other is hyperopic)
Compound astigmatism: When accommodation is relaxed, both focal lines are located either in front of or behind the retina
- If both are located in front of the retina, this is considered compound myopic astigmatism (both principle meridians are myopic)
- If both are located behind the retina, this is classified as compound hyperopic astigmatism (both principle meridians are hyperopic)
Mixed astigmatism: With relaxed accommodation, one focal line will be located in front of the retina and one line will be located behind the retina (one principle meridian is myopic, the other is hyperopic).
In order to determine where the focal points are located for a given spectacle correction, place the values on an optical cross. Plus values indicate that the focal line is behind the retina, minus values indicate focal points in front of the retina, and a value of plano indicates a focal line that is located on the retina.
In the above question, the spectacle prescription of +2.00 -0.50 x 120 will have hyperopic principle meridians (+2.00 and +1.50) when placed on an optical cross. This corresponds to the definition of compound hyperopic astigmatism.
A lens with a single refracting surface has negative power and a virtual object that lies between the vertex of the lens and its first focal point. What are the characteristics of the image that the system creates?
A. Virtual, upright, and smaller than the object
B. There is not enough information to determine these characteristics of the image
C. Real, upright, and smaller than the object
D. Virtual, upright, and larger than the object
E. Real, upright, and larger than the object
E. Explanation - Keep these rules in mind when ray-tracing:
- F is that unique object point that produces a final parallel pencil
- F’ is that unique image point produced by an incident parallel pencil
- V is the intersection of the optical axis with a surface, which is also called the vertex of the surface
- C is the center of curvature of a lens, which is the location where any ray perpendicular to the surface will pass through undeviated
- The surface vertex (V) and the center of curvature (C) are symmetrically positioned between the focal points. (FC = VF’ or FV=CF’)
In retinoscopy, which of the following reflexes is indicative of a high myopic prescription?
A. A narrow, slow against motion reflex B. A broad, fast against motion reflex C. A narrow, fast with motion reflex D. A narrow, fast against motion reflex E. A narrow, slow with motion reflex F. A broad, fast with motion reflex
A. Explanation - The location of the far point of the eye will change the intensity, speed and width of the reflex. As one nears neutrality, the reflex becomes wider/broader, brighter and faster-moving. Therefore, an uncorrected high myope will have a narrower, slower against motion reflex than a mild uncorrected myope.
A circle of light is projected onto a patient’s cornea from a distance of 9.0 cm. The resulting Purkinje image I is elliptical, with its long axis located horizontally. Which type of astigmatism is this patient MOST likely to possess?
A. This is the expected result, and the patient does not have astigmatism
B. Against-the-rule
C. Oblique
D. With-the-rule
D. Explanation - The patient’s cornea is steeper vertically and therefore minimizes the image in this meridian, creating an ellipse with a horizontal axis.
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
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.
You are looking through a microscope and focus on a mark on the lower surface of a glass plate that is 2.5mm thick. To focus on a mark on the top surface of the glass, you have to move the microscope up a distance of 1.5mm. What is the index of refraction of the glass?
A. 1.52 B. 1.67 C. 1.79 D. 1.48 F. 1.74
B. Explanation - This question is asking you to calculate the index of refraction using the equation for reduced thickness. If you are already looking through the microscope with the glass (or other medium) in place, you are viewing the image and the question requires that you use the reduced thickness equation.
reduced thickness = x’ = t /n
x’ is the distance from the object to the image which is 1.5mm
t= thickness of the medium which is 2.5mm
1.5 = 2.5/n
n= 1.67
A ray of light is deviated by a crown glass prism with a power of 4.0 pd that is located 4.0 m away from a screen. Which of the following will CORRECTLY determine the amount of deviation experienced by the ray of light projected at the screen?
A. 4.0 pd= (100)(400.0 cm/x)
B. 400.0 cm= (4.0 pd)(x/100)
C. 4.0 pd= (400cm)(100/x)
D. 4.0 pd= (100)(x/400.0 cm)
D. Explanation - Using the equation P= (100)(x/d), where P= power of the prism (in prism diopters, pd), x=the total distance that a ray of light is deviated and d=the total distance from the prism to the location where the deviation is measured, we can determine the total deviation of a ray of light. For the above question, 4.0 pd= (100)(x cm/400.0 cm), solving for x yields 16.0 cm.
Which of the following is NOT true regarding the conventional outflow route of the aqueous humor?
A. It declines with increasing age
B. Aqueous exits via the iris root, uveal meshwork and the anterior face of ciliary muscle
C. It is also called the trabecular route
D. It is the major component of outflow facility of the aqueous humor
B. Explanation - The non-conventional outflow route of the aqueous humor passes through the iris root, uveal meshwork, and the anterior face of the ciliary muscle. The conventional outflow is also called the trabecular route and, although there is a debate as to how much the trabecular route contributes to the outflow facility, it is agreed that this route is the major component of outflow facility of aqueous humor. Both the conventional and unconventional outflow facilities decline with increasing age.
Which perimetric target/technique given below tests the magno cellular pathway of the retina?
A. A white target on a white background such as that used in standard automated perimetry
B. A blue target on a yellow background utilized in short wavelength automated perimetry (SWAP)
C. Counterphase flickering targets found in frequency doubling technology
C. Explanation - When a stimulus is presented to the retina, it stimulates the photoreceptors, and the information is transmitted to the brain by either magno, parvo, or konio cellular pathway. A white stimulus on a white background will not isolate any particular pathway, and all mechanisms will be stimulated. A blue stimulus on a yellow background suppresses the mid-range and long-range cones and selectively stimulates the blue/yellow opponent pathway or konio cellular pathway. A flickering target like that of frequency doubling technology readily isolates the magno cells, as they are the most sensitive to flicker.
Which of the following types of visual field defects occurs EARLIEST in patients exhibiting glaucomatous optic nerve damage? A. Paracentral scotoma B. Arcuate scotoma C. Nasal step D. Ring scotoma E. Blind spot enlargement
A. 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:
- Glaucoma hemifield test that is “outside normal limits” on at least 2 consecutive occasions.
- 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
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. 30-2 B. 24-2 C. 40 point screener D. 10-2
D. 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
Which of the following ocular conditions is MOST commonly associated with the presence of an optic disc hemorrhage? A. Ocular hypertension B. Neovascular glaucoma C. Angle recession glaucoma D. Normal tension glaucoma E. Acute angle closure
D. Explanation - Optic disc hemorrhages occurring at the margin of the optic nerve are most commonly associated with normal tension glaucoma. These hemorrhages appear at the edge of the optic disc and typically extend onto the nerve fiber layer (called Drance hemorrhages). Additionally, they are most often found at the infero-temporal quadrant of the optic nerve head. It is thought that patients who exhibit progressive optic neuropathy with intraocular pressures (IOPs) consistently within the normal range are likely to have some form of vascular insufficiency or low systemic blood pressure which can lead to a decrease in the ocular perfusion pressure. It is for this reason that optic disc hemorrhages are more commonly observed in individuals exhibiting normal IOPs. When an optic disc hemorrhage is present, it is common to eventually manifest a visual field defect in the region corresponding to the location of the previous hemorrhage (often several months later).
Which of the following represents the most POSTERIOR structure of the anterior chamber angle? A. Ciliary body B. Posterior trabecular meshwork C. Schwalbe's line D. Scleral spur E. Anterior trabecular meshwork
A. 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.
What does angle recession of the eye refer to?
A. A tear between the longitudinal and circular muscle layers of the ciliary body
B. A detachment between the iris root and the ciliary body
C. A detachment between the ciliary body and the scleral spur
D. A tear between the posterior aspect of the zonules and the lenticular equator
A. Explanation - Iridodialysis involves a detachment between the iris root and the ciliary body. Cyclodialysis cleft involves a detachment between the ciliary body and the scleral spur.
A patient with a cataract undergoes a visual field test. Which of the following areas on the visual field print-out is likely to reveal generalized depression? A. The total deviation plot B. The pattern deviation plot C. The glaucoma hemifield test (GHT) D. The visual field index (VFI)
A. Explanation - Ptosis and cataracts will most likely present on this area of the printout (see figure 1A).
The pattern deviation plot gives a more accurate rendering of whether or not a true defect exists by removing excess “noise” such as that seen with a cataract. By removing the effects of generalized depression, it is easier to determine the size, shape and depth of a visual field defect (see figure 1B).
The VFI relates information pertaining to the overall amount of field loss. The VFI ranges from 100% (normal) to 0% (a blind eye). This index is purported to be less affected by cataracts than the other indices. The VFI can be used to monitor progression when compared to previous visual field results.
The glaucoma hemifield test compares test points above the horizontal midline to those below the midline. Any asymmetry is considered suspect and appropriate action should be taken (see figure 1C).
Cataracts, if they are advanced enough, may cause a generalized depression of the visual field because they will act like a filter and will also scatter light.
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. Atrophy in the alpha zone occurs more frequently and is typically larger in patients with glaucoma
B. Atrophy in the beta zone occurs more frequently and is typically larger in patients with glaucoma
C. The alpha zone borders the disc margin and is surrounded by the beta zone
D. The beta zone borders the disc margin and is surrounded by the alpha zone
B and D. 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.
Which of the following visual functions is present and continues to develop in a newborn infant? A. Vertical saccadic eye movements B. Stereopsis C. Pursuits D. Horizontal saccadic eye movements
D. Explanation - Horizontal saccadic eye movements first emerge at birth and continue to develop until roughly 24 months. Vertical saccades require a longer time to mature and emerge at around 2 months of age and continue to develop until about 24 months. Pursuits may be seen as early as 2 months but are more commonly evident at 4 months of age. Stereopsis is absent at birth and does not become apparent until 3-5 months of age (some sources quote 4-6 months), reaching adult levels at 5-7 years of age.
Pepsinogen and fibrinogen are examples of which type of protein? A. Isozymes B. Zymogens C. Allosteric enzymes D. Sphingolipids
B. Explanation - Pepsinogen and fibrinogen are categorized as zymogens. Zymogens, also called proenzymes, are synthesized as inactive proteins that later become activated via some type of transformation, for example, cleavage of a portion of the protein resulting in a functional enzyme. The enzymes cannot be reverted back to their inactive forms. Blood-clotting factors and digestive enzymes are common examples of zymogens. The zymogen pepsinogen is secreted from chief cells in the stomach and is converted to its active form, pepsin, via exposure to hydrochloric acid. The enzyme pepsin aids in the digestion of proteins.
Allosteric enzymes are regulatory proteins that are controlled via modulators. The modulators activate or inhibit the enzyme function by binding to a location on the enzyme that is not the active site.
Isoenzymes are enzymes that are molecularly different but catalyze the same reaction. Isozymes have different amino acid sequences than other enzymes, but both used for the same reaction.
Sphingolipids are important for central nervous system (CNS) function. They are generally incorporated into CNS membranes.
In order to fully reproduce a sine wave grating, four attributes must be specified. What are these four attributes?
A. Grating, Fourier analysis, reflectance, and transmittance
B. Frequency, phase, orientation, and contrast
C. Spacing, color, luminance, and brightness
D. LogMAR, illuminance, sensitivity, and position
Explanation - Sine wave gratings are specified by frequency, phase, orientation and contrast. Frequency denotes the number of cycles of alternating dark and light bands per degree. Contrast describes the luminance profile of the grating; this is basically the difference in luminance between the light and the dark bands. Mathematically, this can be calculated as contrast= the difference in luminance between the peak and trough/the total luminance of the peak and the trough. The phase of the sine wave grating is relative to another grating. Two gratings are in phase if their positions are the same (i.e., their peaks and troughs align) or 180 degrees out of phase (i.e., the trough of one grating is aligned with the peak of the other). The orientation refers to the angle of the grating in reference to horizontal or vertical.
Which of the following is NOT an indication of wet age-related macular degeneration (ARMD)?
A. Submacular hemorrhage
B. Drusenoid pigment epithelial detachment
C. Occult choroidal neovascular membrane (CNVM)
D. Classic choroidal neovascular membrane (CNVM)
E. Exudates
B. 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.
Which hormone is secreted by the heart and helps to regulate blood pressure? A. Atrial natriuretic hormone B. Erythropoietin C. Calcitonin D. Melatonin
A. 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.
Your rigid gas-permeable contact lens patient presents with dimple veiling of the cornea. What modification can be made to the contact lens in order to decrease the occurrence of this finding? A. Increase the optic zone diameter B. Decrease the optic zone diameter C. Increase the overall diameter D. Steepen the base curve E. Flatten the peripheral curves
B. Explanation - Removing the contact lens for several hours will typically allow for resolution of dimple veiling and any associated symptoms. Long-term treatment, however, will require modification of the contact lens parameters. Dimple veiling is a result of poor tear exchange under the gas permeable contact lens, which leads to the entrapment of bubbles of carbon dioxide under the central curvature of the lens. Decreasing the overall diameter of the contact lens and flattening the base curve of the lens may help prevent the formation of dimple veiling, but the treatment of choice is to decrease the diameter of the optic zone or in some manner reduce the vault of the lens. This allows for better tear flow that may have been impeded by a tight mid-peripheral junction of the optic zone and intermediate and peripheral curvatures of the contact lens
You correctly diagnose your 31 year-old female patient with acute unilateral anterior uveitis. When should laboratory testing typically be initiated in otherwise healthy patients presenting with this condition?
A. After the third episode
B. Laboratory testing is not indicated in acute cases of uveitis
C. After 1 year of recurrent episodes
D. After the second episode
E. After the first episode
D. Explanation - When the patient presents with an anterior uveitis that has an initial presentation, occurs unilaterally, and the patient is otherwise healthy with no other systemic symptoms, no further laboratory testing is usually indicated. On the other hand, patients with recurrent, chronic, or bilateral presentations, or patients that have other symptoms possibly indicating the presence of an associated systemic condition, should undergo a thorough work-up to determine the underlying etiology. Typically, in conjunction with the patient’s primary care provider, laboratory testing should be initiated in a patient where anterior uveitis recurs after cessation of treatment (2nd episode). In these cases, there is a much higher risk of being associated with a systemic disease, as compared to a patient with a single episode of intraocular inflammation.
When indicated, minimal lab testing should include the following:
- Complete blood cell count (CBC)
- Urinalysis
- Lyme titers
- HLA-B27 test (spondylarthropathies)
- Antinuclear antibody (ANA) test (juvenile arthritis)
- Angiotensin converting enzyme (ACE) test (sarcoidosis)
- Venereal disease research laboratory (VDRL) test (syphilis)
- Fluorescent treponemal antibody absorption (FTA-ABS) test (syphilis)
