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.
