NBEO Part 1

What is the power of a concave mirror (in diopters) located in air with a radius of curvature of 20 cm? -5.00 D -10.00 D +10.00 D +5.00 D

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

A convex mirror in water (n=1.33) has a radius of curvature of 12 cm. What is the dioptric power of the mirror? -22.17 D +22.17 D +8.33 D -11.08 D

A concave mirror converges light and therefore acts like a convex lens, hence concave mirrors have positive dioptric values whereas convex mirrors diverge light and possess negative dioptric powers. The equation used to determine the power of a mirror is P=-2n/r, where P= the power of the mirror in diopters, n= the index of refraction of the surrounding medium, and r= the radius of curvature of the mirror in meters. P=-2(1.33)/0.12=-22.17 D. Remember, a convex mirror will have a positive radius of curvature and a concave mirror will have a negative radius of curvature.

A convex mirror in water (n=1.33) has a radius of curvature of 12 cm. What is the dioptric power of the mirror? -22.17 D -11.08 D +22.17 D +8.33 D

A concave mirror converges light and therefore acts like a convex lens, hence concave mirrors have positive dioptric values whereas convex mirrors diverge light and possess negative dioptric powers. The equation used to determine the power of a mirror is P=-2n/r, where P= the power of the mirror in diopters, n= the index of refraction of the surrounding medium, and r= the radius of curvature of the mirror in meters. P=2(1.33)/-0.12=-22.17 D. Remember, a convex mirror will have a positive radius of curvature and a concave mirror will have a negative radius of curvature.

A cranial nerve VI palsy will cause what type of deviation? Exodeviation worse with near viewing Exodeviation worse with distance viewing Esodeviation worse with distance viewing Esodeviation worse with near viewing

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

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

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

Which of the following actions does not require input from the brain and therefore is referred to as a spinal reflex? Maintaining balance while riding a bicycle Pulling your hand away from a hot stove Dodging out of the way when an object is thrown in your direction Painting a picture Catching a ball

A spinal reflex does not require input from the brain to achieve the goal at hand. The simplest form of a spinal reflex is a reflex arc that is monosynaptic. This type of pathway involves only one synapse. In its most basic form, a sensory neuron synapses directly onto a motor neuron in the spinal cord, allowing for an almost instant reaction to sensory input. These types of reflexes can be inhibited. For instance, when grasping a hot plate your initial reaction is to drop that plate; however, your brain can override this reaction because cognitively there are negative consequences to dropping the plate (i.e., it may break or the food may spill). Catching a ball, painting and dodging out of the way require the integration of visual input from the brain in order to correctly accomplish the task. Riding a bicycle successfully can be done without visual input; however feedback from the cerebellum is crucial in order to maintain balance.

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

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

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

A unilateral lesion to a lower motor neuron of the facial nerve will result in weakness of upper and lower portions of the face on the ipsilateral side of the lesion. Therefore, in this case, because the upper and lower portions of the right side of the face are affected, the involved motor neuron would be a right lower motor neuron. The branch of the facial nerve that serves the upper portion of the face receives innervation from both the right and left corticobulbar tracts, while the branch of the facial nerve that serves the lower portion of the face only receives innervation from the contralateral side of the brain. If a lower motor neuron lesion occurs, both of these branches are affected, and all muscles of facial expressions beyond that point on the ipsilateral side of the face will experience weakness. Motor Speech Disorders: Diagnosis and Treatment (Second edition). Freed, D.B. 2012. Delmar Cengage Learning. Clifton Park, NY.

Hutchinson's sign associated with herpes zoster is an indication of reactivation of which of the following cranial nerves? Nasociliary branch of the trigeminal nerve Frontal branch of the trigeminal nerve Lacrimal branch of the trigeminal nerve Facial nerve

A vesicular eruption on the tip of the nose in patients with herpes zoster indicates the reactivation of herpes involving the nasocilary branch of the ophthalmic division of the trigeminal nerve. This sign may provide an early predictive factor of impending eye involvement (herpes zoster ophthalmicus).

+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 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.

Which of the following components of the AREDS I ocular vitamin formula used for dry age-related macular degeneration is contraindicated in smokers? Zinc Vitamin C Copper Beta-carotene Vitamin E

Based upon research, there may be a link between increased risk of lung cancer and beta-carotene supplementation when used with smokers.

Which of the following infections must be reported to the CDC (Centers for Disease Control and Prevention)? Syphilis Herpes Simplex Epidemic keratoconjunctivitis (EKC) Acanthamoeba

According to the CDC, all health care providers and laboratories must report new cases of syphilis to local and state health departments. Gonorrhea and Chlamydia trachomatis must also be reported along with many other serious types of infections that are easily spread and can produce long-term systemic effects.

Many skin anomalies may mimic malignant lesions. Which of the following skin conditions has the HIGHEST risk of becoming malignant? Seborrhoeic keratosis Actinic keratosis Cutaneous horn Papilloma

Actinic keratosis is a precursor to squamous cell carcinoma and appears as scaly, dry skin that does not heal. People with skin that is of lighter pigmentation along with excessive exposure to ultraviolet light tend to be most at risk for development of this condition. Papillomas may take on various forms and may be viral or non-viral in origin. They can commonly be found on the eyelids or surrounding orbital skin. Viral warts tend to grow at an accelerated rate while non-viral papillomas are fairly slow to grow. Papillomas can mimic neoplastic growths so be sure to rule this out while watching carefully for color change, ulceration, lash loss, bleeding, and vascularization. Cutaneous horns or tags are also benign and are likely a form of papilloma but appear to involve more keratin. Treatment is similar to that of a papilloma. Seborrhoeic keratosis is more commonly seen in middle-aged and elderly persons. This benign, epidermal growth is quite superficial and does not extend into the dermis. It appears like a brown plaque that has been stuck onto someone's skin. The borders are very distinct and there may be some elevation. The lesions may be removed if the patient is concerned about cosmesis.

Which of the following walls of the orbit is MOST susceptible to a blowout fracture secondary to blunt ocular trauma? Lateral wall Medial wall Floor Roof

All of the walls of the orbit are relatively thin, making them susceptible to injury and fracture when blunt force trauma to the eye is incurred. Increased pressure in the orbit often results in one or more areas of the orbit to break and "blow out" into the sinuses. The inferior and medial walls tend to be the most susceptible to this type of injury, although an orbital floor fracture is the most common. In certain patients, a blowout fracture of the floor will require surgical repair. These circumstances include cases in which there is rectus muscle entrapment, enophthalmus greater than 2mm, and large fractures that include over 50% of the involved wall (this is typically decided after a 2-week waiting period).

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

Allergic conjunctivitis is an ocular inflammatory response that is triggered by IgE-induced mast cell and basophil degranulation in response to an antigen. These cells, once degranulated, release histamines, cytokines, leukotrienes, prostaglandins, interleukins, chemokines and other mediators (early phase), which produce the classical symptoms of allergic conjunctivitis. In the later phase of the allergic response, infiltration of inflammatory cells such as eosinophils, neutrophils, and lymphocytes also occur, with eosinophils predominating. Therefore, if the conjunctival secretions were to be stained, one would likely observe numerous eosinophils present. Eosinophils are not typically present in the conjunctival scrapings of normal patients, so their presence is consistent with a diagnosis of allergic conjunctivitis. Patterson, R., Grammer, L.C., Greenberger, P.A. (2009). Patterson's Allergic Diseases (7th ed.). Baltimore, MD: Lippincott, Williams & Wilkins. Volcheck, G.W. (2009). Clinical Allergy: Diagnosis and Management. Rochester, MN: Mayo Foundation.

Alpha helices and beta sheets are considered which level of protein structure? Tertiary Primary Secondary Quaternary

Alpha helices and beta sheets are categorized as secondary structures. Primary structures are the initial sequences of amino acids that will eventually code for the final structure of the protein. Primary structures are held together via peptide bonds. Secondary structures are formed by folding chains of amino acids into a helix or sheet. The helices are coiled so that the R side chains face away from the center of the helix. The beta sheets can be folded in a parallel pattern, where all of the N terminals are located at the same end of the sheet, or in an anti-parallel pattern, where the N and C terminals alternate at the ends of the sheet. Secondary structures are stabilized by hydrogen bonds between the peptide bond groups. Tertiary structure refers to the final overall three-dimensional shape of the protein. Bonding between the side chains causes folding and twisting of the protein. There are four types of bonding that occur at this level; hydrogen bonds, hydrophobic interactions between non-polar side groups (weak), disulfide bonds (strong, covalent) and salt bridges (also called electrostatic attraction). Electrostatic forces deals with the overall charge of the side groups. A positively charged group will be attracted to a negatively charged group. Disulfide bonds form when the sulfhydryl groups of cysteine become oxidized. Quaternary proteins can be classified as either globular or fibrous. Quaternary proteins are oligomeric, meaning that they are created through the bonding of several peptide units. Globular proteins (for example hemoglobin) are generally compact, water-soluble and ball-shaped. Most globular proteins are enzymes. Fibrous proteins tend to be elongated, strong, and not water-soluble. Fibrous proteins (for example collagen) tend to serve a structural role.

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

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

An amphipathic molecule will react in what manner when exposed to water? It will form micelles It will combust It will completely dissolve It will form a gas

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

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

An iris coloboma is an inferonasal, keyhole-shaped defect. The remainder of the iris is normal. Atypical colobomas may develop at sites other than the inferonasal area. Ref: Trachimowicz, R. Review of Embryology and its Relation to Ocular Disease in the Pediatric Population. Optometry and Vision Science. 71:3 pp 154-163

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?

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.

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? Administration of Benadryl (oral) Olopatadine (Patanol) Injection of epinephrine (EpiPen) Prednisone (oral)

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.

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

As the red lens is placed over the right eye, your patient reports that his right eye sees a red light and the left eye sees a white light. In this case, the red light is perceived to the left of the white light demonstrating crossed diplopia. Crossed diplopia is characteristic of an exo deviation. The easiest way to remember this is that a cross makes an 'x', and 'exo' contains an 'x' so the 'x's always go together.

A back surface toric (spherical front surface) gas-permeable (GP) contact lens is ordered with base curve radii of 7.85 mm (43.00 D) and 8.44 mm (40.00 D). When verifying this lens with a lensometer you would expect to find approximately how many diopters of "induced cylinder"? 4.50 D 1.50 D 3.00 D 2.00 D

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

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

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

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? Edge thickness Ultraviolet (UV) inhibitor Contact lens material Contact lens solution

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.

Berger's space is created by an interval between which two structures? The equator of the lens and the ciliary body The anterior face of the lens and the posterior surface of the iris The posterior surface of the cornea and the anterior face of the iris The posterior face of the lens and the anterior vitreous

Berger's space is created by the separation between the posterior face of the lens and the anterior face of the vitreous. The space between the equator of the lens and the ciliary body is known as the circumlental space.

Berger's space is created by an interval between which two structures? The posterior surface of the cornea and the anterior face of the iris The anterior face of the lens and the posterior surface of the iris The equator of the lens and the ciliary body The posterior face of the lens and the anterior vitreous

Berger's space is created by the separation between the posterior face of the lens and the anterior face of the vitreous. The space between the equator of the lens and the ciliary body is known as the circumlental space.

Which area of the extrastriate cortex is involved in the perception of motion? The inferotemporal cortex (IT) Visual area 4 (V4) Visual area 1 (V1) Visual area 5 (V5) Visual area 2 (V2)

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

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

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

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

Blepharitis is a condition caused by pathogens, usually of Staphylococcus origin, that colonize along the eyelid margins. The bacteria produce exotoxins which take the form of flakes and are generally seen along the base of the eyelashes. Unfortunately, this condition is chronic but will wax and wane in its presentation. Long-term complications include madarosis (missing lashes), trichiasis, neovascularization of the eyelid margin, keratitis, erythema, phlyctenule formation and infiltrates. Patients may complain of dry, irritated eyes, stinging, pain, itching, frequent eye infections, foreign body sensation, and decreased acuity (if there is corneal involvement). Treatment includes eye lid scrubs, antibiotic ointments and sometimes transient topical steroid use to decrease lid inflammation (usually used in conjunction with a topical antibiotic). Occasionally oral antibiotics are prescribed, especially in the event of poor compliance. Distichiasis is a rare congenital phenomenon marked by an absence of meibomian glands. In the place of the meibomian glands is an extra row of eyelashes. Hypertelorism is a term used to describe the incidence in which the orbits are located quite far apart. This generally occurs along with other congenital cranium anomalies. Tristichiasis is a very rare occurrence in which a person possesses three rows of eyelashes.

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

CCTVs work on the principle of relative size magnification (or projection). It operates by enlarging the text without lenses in front of the patient or the patient moving closer to the device. When the print is enlarged electronically in this matter, the image of the print subtends a larger area on the retina and thus a larger size. An example of relative distance magnification would be if you were holding a newspaper at 40 cm and you moved it closer to 20 cm. The print now appears 2 times as large relative to the 40 cm distance. Rated magnification is often used by manufacturers of some hand magnifiers and stand magnifiers using a 25 cm reference distance.

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

CCTVs work on the principle of relative size magnification (or projection). It operates by enlarging the text without lenses in front of the patient or the patient moving closer to the device. When the print is enlarged electronically in this matter, the image of the print subtends a larger area on the retina and thus a larger size. An example of relative distance magnification would be if you were holding a newspaper at 40 cm and you moved it closer to 20 cm. The print now appears 2 times as large relative to the 40 cm distance. Rated magnification is often used by manufacturers of some hand magnifiers and stand magnifiers using a 25 cm reference distance.

Central serous retinopathy (CSR) is associated with an acute decrease in vision along with central distortion. The condition usually occurs unilaterally. Which gender and age group tends to have the highest incidence of CSR? Males; ages 50-70 Females; ages 20-40 Females; ages 10-20 Males; ages 30-50

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

A positive catalase test indicates that a bacteria is capable of breaking down which of the following? Carbon dioxide Pyruvate Hydrogen peroxide Glucose

Catalase is an enzyme commonly found in organisms that are exposed to oxygen. Catalase breaks down hydrogen peroxide into oxygen and water. The catalase test is performed by applying a drop of hydrogen peroxide to a microscope slide. A colony of bacteria is then exposed to the hydrogen peroxide via an applicator stick. The presence of bubbles or froth yields a positive catalase test. Staphylococci and Micrococci are catalase-positive organisms. Campylobacter and Escherichia coli are catalase-negative organisms. Ref: Brock, Madigan et al. Biology of Microorganisms 9th edition (2000) pg 160-161

What substance is secreted by the alveolar epithelium to reduce the surface tension between alveoli of the lungs? Intrapleural fluid Carbon dioxide Angiotensin Surfactant

Cells of the alveolar epithelium secrete pulmonary surfactant to reduce the surface tension of the alveoli; this helps to prevent their collapse. The pleural membranes are separated by an intrapleural space filled with intrapleural fluid that serves to prevent the two membranes from rubbing against one another. Some illnesses like pneumonia can cause inflammation of the pleurae, leading to chafing and painful ventilation called pleurisy. Angiotensin causes constriction of blood vessels, increasing blood pressure. Angiotensin I is converted to Angiotensin II by the angiotensin-converting enzyme, which is found in the greatest density in the lungs.

Which of the following is the MOST common early pattern of a glaucomatous visual field defect? Para-central scotoma Inferior nasal step Inferior arcuate Superior nasal step Enlarged blind spot Superior arcuate

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), which is summarized below: 1. Glaucoma hemifield test that is "outside normal limits" on at least 2 consecutive occasions. 2. A cluster of 3 or more non-edge points in a location typical for glaucoma, all of which are depressed on pattern standard deviation (PSD) at a P < 5% level and one of which is depressed at a P <1% level, on 2 consecutive occasions. 3. Corrected pattern standard deviation (CPSD) that occurs in less than 5% of normal individuals on two consecutive fields.

Which of the following ocular signs is virtually pathognomonic for trachoma caused by chlamydia? Lymphadenopathy Superior tarsal follicles Inferior tarsal papillae Tranta's dots

Chlamydia causes two forms of conjunctivitis, trachoma and inclusion. Trachoma is more common in lesser-developed countries and can cause blindness if not treated appropriately. Trachoma presents in several stages, initially starting with mucopurulent discharge, lymphadenopathy, red eye, small superior tarsal follicles, and mild superior pannus. As the condition evolves, the formation of limbal follicles occurs and will eventually scar causing Hebert's pits, which are characteristic of this infection. This condition, if left untreated, ultimately progresses to horrible scarring of the eyelid (Arlt's line) and cornea, causing extremely poor visual acuity. Diagnosis is made with the observation of two or more of the following: follicles on the upper tarsus, pannus (particularly superiorly), limbal follicles or Herbert's pits and typical conjunctival scarring of the upper lid. Treatment includes oral doxycycline, tetracycline, or erythromycin along with topical tetracycline or erythromycin ointment. Azithromycin is also a good choice because it is given as 1000 mg PO which delivers exceptional compliance; however, this is not to be prescribed to those with liver disease or to young adults under the age of 16. Inclusion conjunctivitis is linked to venereal disease and can present either unilaterally or bilaterally (which is more common) as follicles on the upper and lower tarsal plates (lower follicles will be larger and more prominent), lymphadenopathy, possible mucopurulent discharge, lid edema, micropannus, superior corneal sub-epithelial infiltrates, superficial punctate keratitis, and scarring of the upper eyelid (sometimes called Arlt's line or "basketweave" because of its appearance). This type of conjunctivitis is less severe than trachoma. Treatment is similar to that of trachoma. Follicles are related to cellular immunity which serves to protect against viruses. Many types of viral infections can cause inferior palpebral follicles, such as EKC, Herpes simplex and molluscum contagiosum. Superior tarsal follicles are highly suggestive of a chlamydial infection. A superior papillary response is generally associated with an allergic response. Inferior tarsal papillae are frequently seen in bacterial infections and allergic responses as papillae act as the release sites for both eosinophils (associated with allergies) and polymorphonuclear leukocytes which destroy bacteria. References: Clinical Ophthalmology, A Systematic Approach 5th edition. Kanski, J. Butterworth-Heinemann, Elsevier Science, 2003, pages 70-73. Vaughan & Ashbury's General Ophthalmology 16th edition, Riordan-Eva, P., Vaughan, D. & Asbury T., McGraw-Hill, 2004, pages 105-108.

Which organism can be contracted in a newborn via an infected mother and was previously treated prophylactically with silver nitrate? Neisseria meningiditis Staphylococcus aureus Corynebacterium spp Neisseria gonorrhoeae

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

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

Coenzyme Q, all steroids, cholesterol, and vitamins A, D, E, and K are derived from isoprenes or terpenes. These agents contain or at some point originated from precursors that were comprised of isoprene units. Isoprene units have the chemical formula C5H8. Sphingolipids are important in cell membranes, especially those located in the central nervous system such as myelin sheath. Shingolipids contain shingosine as a backbone and are then further classified depending upon which molecules are attached to that backbone, such are ceremides, gangliosides, sphingomyelin, etc. Phospholipids contain a polar and non-polar end, thus making them amphoteric. This property allows for the formation of bilayers (polar ends aligned together and pointed outwards) resulting in the lipid bilayer commonly seen in cell membranes. Phospholipids are generally comprised of a phosphate group, a choline group (polar), and two fatty acid chains (non-polar) attached to glycerol, which serves as the backbone. Triglycerides are comprised of three fatty acid chains attached to a glycerol backbone. Triglycerides are important in long-term energy storage for use by cells.

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

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

Antibiotic resistance that is rapidly spread within a population of bacteria is due to what mechanism? Transformation Budding Binary fission Conjugation

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

The lens changes significantly as we age. One of these changes is the formation of vacuoles. What causes this? Decreased function of the sodium/potassium pump causing decreased protein synthesis Glutathione loss with age causing cross linking between proteins The separation of water from proteins causing agglutination of proteins and pooling of water Loss of lenticular capsule elasticity causing decreased accommodation

Crystalline proteins are normally water-soluble. With age, they begin to clump together altering the amount of water that they are capable of binding to. The water begins to pool together creating vacuoles. The pooling of water causes light scattering because the index of refraction differs between that of the lens proteins and the water aggregations. Glutathione loss and decreased function of the ion pump do indeed contribute to cataract formation, but the direct cause of vacuoles is pooling of water. Loss of capsular elasticity leads to presbyopia.

If a thin lens with an index of 1.5 has a dioptric power of +4.00 in air, what is its power if placed in water? +11.76 diopters +1.36 diopters +1.18 diopters +4.00 diopters

D (air) / D (water) = n (lens) - n (air) / n (lens) - n (water) +4.00 / D (water) = 1.5 - 1.0 / 1.5 - 1.33 D (water) = +1.36 diopters

Dacryoadenitis refers to an inflammation or infection of which of the following ocular structures? The lacrimal sac The lacrimal gland The nasolacrimal sac The puncta

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

Dacryoadenitis refers to an inflammation or infection of which of the following ocular structures? The nasolacrimal sac The lacrimal sac The lacrimal gland The puncta

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

A lens system in air consists of a +4.00 diopter and +6.00 diopter lens separated by 10 cm. What is the equivalent power of this optical system? +7.6 diopters +10.0 diopters +12.4 diopters +10.24 diopters +9.76 diopters

De = D1 + D2 - (t/n) x D1D2 De = equivalent power, D1 = front surface power, D2 = back surface power t = thickness of lens system, n = index between the 2 surfaces In the above question, t = 10 cm (0.1 m), n=1.0, D1 = +4.00 and D2 = +6.00 De = 4 + 6 - ((0.1/1.0) x (4) x (6)) De = 10 - (0.1 x 4 x 6) De = 10 - (2.4) De = +7.6 diopters

A patient is using a stand magnifier of +16D with a +2.00 add. If the distance separating the two lenses is 25 cm what is the equivalent power of this combination? 22D 10D 18D 26D

De= D1+D2 -tD1D2 where De=equivalent power;D1=power of magnifier;D2=power add;t=separation in meters between the lenses De = (16+2) - 0.25(16)(2) De= 18-8 = 10D 18D- incorrect answer -would come up with this if added the stand magnifier power to the power of the add 22D -incorrect answer - would come up with this if added 16D for stand mag 2D for add and 4D for equivalent of 25cm. 26D - if added the 18 +8 in the De equation instead of subtracting

A patient is using a stand magnifier of +16D with a +2.00 add. If the distance separating the two lenses is 25 cm what is the equivalent power of this combination? 10D 18D 26D 22D

De= D1+D2 -tD1D2 where De=equivalent power;D1=power of magnifier;D2=power add;t=separation in meters between the lenses De = (16+2) - 0.25(16)(2) De= 18-8 = 10D 18D- incorrect answer -would come up with this if added the stand magnifier power to the power of the add 22D -incorrect answer - would come up with this if added 16D for stand mag 2D for add and 4D for equivalent of 25cm. 26D - if added the 18 +8 in the De equation instead of subtracting

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

Diffuse SPK likely signals either a viral origin or a toxic reaction to solution or topical ophthalmic drops. A bacterial etiology will cause staining of the inferior third of the cornea. Interpalpebral corneal staining is usually caused by lagophthalmos or environmental desiccation. A foreign body will either leave small punctate staining or tracks (especially if it is trapped under a contact lens).

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

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.

Drusen typically deposit between which layers of the retina? The inner and outer plexiform layers The ganglion cell layer and the nerve fiber layer The retinal pigment epithelium and Bruch's membrane The inner and outer nuclear layers

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

How is the temporal modulation transfer function expected to change in a person with early glaucoma not yet manifesting any defects on visual field testing? Decreased sensitivity to low and moderate frequencies Decreased sensitivity to moderate and high temporal frequencies Decreased sensitivity to low temporal frequencies only Decreased sensitivity to moderate frequencies only

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

Which of the following antibiotics is classified as a macrolide? Amoxicillin Tobramycin Erythromycin Tetracycline

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, Augmentin®, cloxacillin, dicloxacillin are common anti-bacterials belonging to a class of drugs termed penicillins. Penicillins disrupt cell wall synthesis, making them valuable and commonly 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 they cause the creation of openings in bacterial cell membranes, allowing for increased antibiotic uptake.

Which of the following classifications refers to an organism that can survive in an environment with or without oxygen? Facultative anaerobe Microaerophile Strict anaerobe Obligate aerobe

Facultative anaerobes are capable of using oxygen as a life source but can also thrive using anaerobic respiration. When culturing this type of bacteria in a test tube of liquid media, one would observe the bacteria dispersed throughout the entire tube but with greatest concentration at the top, as aerobic respiration yields a more favorable energy result. Obligate aerobes require oxygen to survive. Test tube culturing would display bacteria gathering at the top of the tube where the oxygen levels are greatest. Microaerophiles are capable of utilizing oxygen, but only at lower concentrations. A test tube culture of microaerophillic bacteria would result in the greatest concentration of bacteria towards the upper portion of the test tube, but not at the very top. Strict anaerobes cannot tolerate oxygen and rely on anaerobic respiration to survive. A test tube culture of strict anaerobes would display the greatest concentration of bacteria towards the bottom of the tube where the least amount of oxygen is present. Aerotolerant organisms are capable of growing in the presence of oxygen but do not actually utilize it for respiration. One would observe these bacteria evenly distributed throughout a test tube liquid culture.

Dietary triglycerides are metabolized primarily by which organ of the body? Stomach Kidney Liver and gall bladder Intestine

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

Fatty acid synthesis is activated during which of the following situations? Increased levels of citrate and glucagon, decreased levels of insulin Increased levels of citrate, decreased levels of glucagon and insulin Decreased levels citrate and insulin, increased levels of glucagon Increased levels of citrate and insulin, decreased levels of glucagon Decreased levels of citrate, increased levels of glucagon and insulin Decreased levels of citrate and glucagon, increased levels of insulin

Fatty acid synthesis is stimulated by insulin and inhibited by glucagon and epinephrine. The formation of fatty acids is catalyzed via the enzyme acetyl CoA reductase, which is activated by citrate. Acetyl CoA reductase is allosterically inhibited by palmitoyl-CoA. Fatty acid synthesis occurs in the cytosol and requires the use of ATP and NADPH, as this is a complex anabolic process. A brief overview and important points of the pathway are as follows: Essentially, acetyl CoA combines with oxaloacetate (OAA) in the mitochondria to form citrate, which is then shuttled out into the cytosol by the citrate shuttle. Once in the cytosol, citrate is broken down again into its constituents, OAA and acetyl CoA. OAA is converted back into pyruvate to re-gain entry into the mitochondria. Acetyl CoA is converted to malonyl-CoA via acetyl CoA carboxylase (rate-limiting step and uses biotin as a cofactor). Malonyl CoA and the acyl carrier protein undergo several reactions to create fatty acid.

What is the main mechanism of action of fluticasone and salmeterol, which are components found in Advair®, respectively? Anti-histamine; anti-inflammatory Anti-inflammatory; anti-histamine Bronchodilator; anti-inflammatory Leukotriene inhibitor; anti-inflammatory Anti-inflammatory; bronchodilator Anti-inflammatory; leukotriene inhibitor

Fluticasone is a synthetic corticosteroid that has a very potent anti-inflammatory action. Inflammation plays an important role in the pathogenesis of asthma; therefore, this anti-inflammatory is very effective in helping to control asthmatic complications. As is the case with all other steroids, patients using Advair® should be examined periodically to monitor intraocular pressure and possible development of cataracts. Salmeterol is a long-acting beta2-adrenergic agonist. Beta2-adrenoceptors are the predominant adrenergic receptors involved in bronchial smooth muscle, with little association to the heart. Salmeterol acts to increase the levels of circulating cyclic AMP in the bloodstream; this causes relaxation of bronchial smooth muscle and inhibition of the release of mediators of immediate hypersensitivity from cells (especially mast cells).

A patient walks into your office with a mild corneal abrasion, what is the correct healing sequence? Mitosis of cells surrounding wound area-> attachment of cells via fibronectin and laminin-> hemidesmosome formation between basal cells-> basal cells at the wound margin flatten and spread Basal cells at the wound margin flatten and spread-> mitosis of cells surrounding wound area-> attachment of cells via fibronectin and laminin-> hemidesmosome formation between basal cells Hemidesmosome formation between basal cells-> basal cells at the wound margin flatten and spread-> mitosis of cells surrounding wound area-> attachment of cells via fibronectin and laminin Basal cells at the wound margin flatten and spread-> attachment of cells via fibronectin and laminin-> mitosis of cells surrounding wound area-> hemidesmosome formation between basal cells

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

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

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

A patient brings you an old pair of glasses and asks you how much prism is in the lenses. With the lensometer, you measure 7 prism diopters base up and 4 base out in the right eye and 4 prism diopters base up and 5 base out in the left eye. What is the total amount of prism in the glasses? 11 prism diopters base up total and 9 base out 3 prism diopters base up in the right eye and 1 base out 3 prism diopters base up in the right eye and 9 base out 11 prism diopters base up total and 1 base out

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

A patient brings you an old pair of glasses and asks you how much prism is in the lenses. With the lensometer, you measure 7 prism diopters base up and 4 base out in the right eye and 4 prism diopters base up and 5 base out in the left eye. What is the total amount of prism in the glasses? 3 prism diopters base up in the right eye and 9 base out 11 prism diopters base up total and 1 base out 11 prism diopters base up total and 9 base out 3 prism diopters base up in the right eye and 1 base out

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

Glycolysis occurs at which location within a eukaryotic cell? Cellular membrane Mitochondrion Cytosol Endoplasmic reticulum Nucleus

Glycolysis is an important metabolic pathway that breaks down glucose into pyruvate. Then, pyruvate can either be converted anaerobically into lactate or undergo oxidative phosphorylation yielding 36 moles of ATP. Glycolysis occurs in the cytosol of a cell. To help remember the above, think of this little rhyme: glycol occurs in the cytosol. Ref: Guyton and Hall, Textbook of Medical Physiology 10th ed. (2000) page 775-778.

Chrysiasis of the cornea occurs secondary to administration of medication used to treat which of the following conditions? Cardiac arrhythmias Diabetes Cancer Rheumatoid arthritis Hypertension

Gold salts are used to manage rheumatoid arthritis, primarily when other treatment options have failed. Chrysiasis occurs secondary to the deposition of gold in the skin, lens, and cornea, causing a gray discoloration of the skin and brown/gold deposits in the deep stroma of the cornea. Amiodarone is an anti-arrhythmic medication. Use of this drug commonly causes yellow/brown or white powdery corneal epithelial deposits located inferocentrally that appear to swirl outward while sparing the limbus. These deposits affect visual acuity minimally, if at all. The aforementioned deposits can also be observed in Fabry's disease and in patients taking tamoxifen, chlorpromazine, chloroquine, or indomethacin.

According to the Keith-Wagener-Barker method of classification, hypertensive retinopathy is categorized as stage four when which ocular sign is present? Hard exudates in a star configuration Swelling of the optic disc Flame hemorrhages Retinal edema

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

Which of the following microorganisms is associated with peptic ulcer formation? Vibrio cholerae Clostridium tetani Campylobacter spp Helicobacter pylori

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

Hypoxia associated with hydrophilic (soft) contact lens wear can result in which of the following? Corneal swelling Corneal decompensation 3/9 staining, scarring and pseudoptyergium Blepharitis

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.

How does uncorrected myopia affect a near phoria? It results in greater hypophoria It results in greater esophoria or less exophoria It will not change the degree of phoria It results in greater hyperphoria It results in greater exophoria or less esophoria

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

Which of the following antibodies is the first to be secreted during an immune response? IgG antibodies IgE antibodies IgD antibodies IgA antibodies IgM antibodies

IgE antibodies are involved in allergic reactions as well as inflammatory responses. IgM is the antibody responsible for activation of the complement pathway and is the first antibody produced in response to an infection. IgA antibodies are known as the secretory antibodies and are located in mucous membranes. They are primarily responsible for protection of the lungs and gastrointestinal tract from infection. IgA antibodies can also be detected in some mammalian milk transferred for passive immunity. IgD antibodies are typically bound to the surfaces of B-lymphocytes and are only a minor blood component. IgG antibodies are responsible for long-term protection (i.e., from viruses) and is the predominant antibody type found in blood. IgG is the only type of immunoglobulin that can cross the placenta to help protect the fetus. IgG is also found in milk produced early after the delivery of a child.

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

IgG is the predominant type of antibody found in the bloodstream and is the only class of immunoglobulin that crosses the blood-placenta barrier. IgA is the antibody that is primarily found in secretions. It is the second most abundant type of antibody in the body. IgM is usually the first antibody made in the immune response and is the largest antibody of the five. IgM can form pentamers with other IgM molecules in the secreted form. IgE is implicated for an important role in hypersensitivity and allergic reactions and interacts strongly with mast cells. The primary role of IgD is unknown.

How much image jump will be created by a +2.00 D flat top 25 mm segment add with a carrier lens of +2.25 DS? 2.5 prism diopters 1.125 prism diopters 5.31 prism diopters 1 prism diopter 2.125 prism diopters

Image jump is created by the vertical prismatic effect when looking through the reading addition of a bifocal lens. When the patient looks from the distance portion of their lenses into the reading area, the viewed image will appear to move or jump. The greater the distance between the reading add optical center and the bifocal line, the greater the image jump experienced by the patient. The total amount of image jump depends on the reading add and the distance between the optical center of the reading add and the segment line. The optical center of a flat top reading segment is located 5 mm below the segment line. Next apply the Prentice rule to solve this problem: prism diopters(pd) =d*F where d= the distance from the optical center in centimeters and F= the power of the add. Pd= 0.5(2.00) = 1 prism diopter.

Which of the bifocal lens designs below will create the largest amount of image jump? FT 35 bifocal Round 28 bifocal Executive bifocal FT 28 bifocal

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

A 6-foot tall man wishes to buy a plane mirror in which he can visualize his whole length at the same time. How tall must the mirror measure in order for the above to occur? 5.2 feet tall 2.3 feet tall 3 feet tall 4.5 feet tall 6 feet tall

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

While performing the astigmatic clock dial, your patient reports that the 1-7 and 2-8 lines are equally blacker and clearer than all of the other lines. What would be the corresponding axis of astigmatism? 30 degrees 60 degrees 45 degrees 90 degrees

In order to determine the corresponding axis of astigmatism utilizing the clock dial, one must multiply the smallest number of the clearest clock position by 30 degrees. In this instance, because the two lines are equally clear and black, the axis would lie in between them; therefore 1 x 30= 30 degrees. Each clock hour on the clock dial is separated by 30 degrees. Midway between the 1 o'clock and 2 o'clock position is 30 degrees divided by 2 thus, 30/2= 15 degrees. We must now add 15 degrees to our previous number of 30 degrees because the astigmatism axis lies in between the 1 and 2 o'clock positions. This would give us a total of 45 degrees.

When analyzing a gas-permeable lens, you measure base curves of 7.58 and 7.84 with a radiuscope, and -1.00 and -2.50 on lensometry. What type of toric gas-permeable contact lens design do you have? Thin-flex Front surface (F1) toric Cylinder power effect (CPE) bitoric Spherical power effect (SPE) bitoric Back surface (base curve) toric

In order to determine which design of toric gas-permeable contact lens you have once the lens has been analyzed, the difference in base curve (BC) values and contact lens power (CLP) readings must be calculated first. BC1 = 7.58 = 337.5/7.58 = 44.50 BC2 = 7.84 = 337.5/7.84 = 43.00 Change in BC = 1.50 D CLP1 = -1.00 CLP2 = -2.50 Change in CLP = 1.50 D The differences in base curves and contact lens powers for the above gas-permeable contact lens are equal (both 1.50 D), indicating that the design of the lens is a spherical power effect (SPE) bitoric type. If the change in BC does not equal the change in CLP, the lens type may either be a base curve toric or cylinder power effect (CPE) bitoric. The way to tell these two apart is that if 3/2 change in BC = change in CLP, then it may be considered a base curve toric.

Which of the following correctly lists the layers of the retina beginning with the retinal pigment epithelium and moving anteriorly? Retinal pigment epithelium, photoreceptor layer, outer nuclear layer, external limiting membrane, outer plexiform layer, inner plexiform layer, inner nuclear layer, ganglion cell layer, nerve fiber layer, internal limiting membrane Retinal pigment epithelium, photoreceptor layer, external limiting membrane, outer nuclear layer, outer plexiform layer, inner nuclear layer, inner plexiform layer, ganglion cell layer, nerve fiber layer, internal limiting membrane Retinal pigment epithelium, external limiting membrane, outer nuclear layer, photoreceptor layer, outer plexiform layer, inner nuclear layer, inner plexiform layer, nerve fiber layer, ganglion cell layer, internal limiting membrane Retinal pigment epithelium, outer nuclear layer, external limiting membrane, photoreceptor layer, outer plexiform layer, inner plexiform layer, inner nuclear layer, nerve fiber layer, ganglion cell layer, internal limiting membrane

In order: 1) The retinal pigment epithelium is a single layer of pigmented cells that functions to form part of the blood-retinal barrier, phagocytose fragments from shedding of photoreceptor discs, and metabolize and store vitamin A which is used in forming photopigment 2) The photoreceptor cell layer contains the outer and inner segments of rods and cones 3) The external limiting membrane consists of intercellular junctions of photoreceptor cells 4) The outer nuclear layer contains the cell bodies and nuclei of the rods and cones 5) The outer plexiform (synaptic) layer is made up of the fibers of rods and cones and synapses between photoreceptor cells and cells from the inner nuclear layer 6) The inner nuclear layer contains cell bodies of several types of neurons and Muller cells. 7) The inner plexiform (synaptic) layer contains the synapses of bipolar cell axons and gangion cell dendrites 8) The ganglion cell layer contains amacrine cells, Muller cell bodies, and astroglial cells 9) The nerve fiber layer contains the ganglion cell axons 10) The internal limiting membrane forms the innermost boundary and is composed of footplates of Muller cells Ref: Remington, LA. Clinical Anatomy of the Visual System, 1998 p 59-65

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

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

Illumination is one of the most important considerations to discuss in the case disposition for a visually impaired patient. A patient with chronic open angle glaucoma moves a 60 watt bulb on a flexible mounted arm from three feet to one foot from the page. The illumination on the page will appear to have been increased by how much? Should be the same brightness Decreased by 1/3 of the original brightness Increased by 3 times the original brightness Increased by 9 times the original brightness Decreased by 1/9 of the original brightness

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

Illumination is one of the most important considerations to discuss in the case disposition for a visually impaired patient. A patient with chronic open angle glaucoma moves a 60 watt bulb on a flexible mounted arm from three feet to one foot from the page. The illumination on the page will appear to have been increased by how much? Should be the same brightness Increased by 3 times the original brightness Decreased by 1/3 of the original brightness Decreased by 1/9 of the original brightness Increased by 9 times the original brightness

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

A common cause of epiphora in infants is caused by a small membrane that covers over which of the following structures? The canaliculus The valve of Hasner The puncta The lacrimal gland

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

A 42-year old construction worker presents to you with a history of getting plaster in his right eye. He complains of pain, foreign body sensation, and photophobia. His acuity is reduced in that eye to 20/50 with a normal pupillary response. What type of chemical trauma did the worker experience, and what would be your first therapeutic intervention? Alkali burn and lavage eye with balanced saline solution for 30 minutes Ultraviolet (UV) burn and lavage eye with balanced saline solution for 30 minutes Thermal burn and lavage eye with balanced saline solution for 30 minutes Acid burn and lavage eye with balanced saline solution for 30 minutes

Lime, particularly in the form of plaster, is the most commonly encountered alkali injury. Fortunately, it tends to cause a less severe burn than other types of alkali burns. The rapidity with which pH abnormalities of the ocular surface are neutralized has a significant impact on the subsequent clinical course. Irrigation for a minimum of 30 minutes and checking pH of tears for evidence of neutrality is recommended. The explanations for the alternative answers are as follows: Acid burn and Lavage eye with balanced saline solution for 30 minutes; plaster contains lime, which is alkali and not acid. Thermal burn and lavage eye with balanced saline solution for 30 minutes; there is no mention of heat (cigarette or matches) during the injury to the eye. Ultraviolet (UV) burn and lavage eye with balanced saline solution for 30 minutes; UV burns usually are associated with delayed response after exposure to UV light. References: 1. McCulley JP, Chemical Injuries. Smolin G, Thoft RA eds. The Cornea: scientific foundation and clinical practice, 2nd ed. 1987: 527-542 2. Herr RD, White GL, Clinical comparison of ocular irrigation fluids following chemical injury, Am J of Emerg Med 1991: 9:228-231

A 2.5x Galilean telescope has a -25D ocular lens. When focused for infinity, what is the length of the telescope? 14 cm 5 cm 6 cm 4 cm 10 cm

M = -Doc/Dobj where Doc=power ocular; Dobj= power objective; t= separation of lenses 2.5= - (-)25/Dobj Dobj= 10D t= f'obj + f'oc f'obj=1/10D = 0.10 m f'oc = 1/-25D = 0.04 m t=0.10 + -0.04 = 0.06 m or 6 cm 10 cm - incorrect- would come up with this answer if only took in account the focal length of the objective lens. 4 cm - incorrect- would come up with this answer if only took in account the focal length of the ocular lens. 14 cm - incorrect - would come up with this answer if thought equation was t= f'obj + f'oc

A 2.5x Galilean telescope has a -25D ocular lens. When focused for infinity, what is the length of the telescope? 5 cm 4 cm 6 cm 10 cm 14 cm

M = -Doc/Dobj where Doc=power ocular; Dobj= power objective; t= separation of lenses 2.5= - (-)25/Dobj Dobj= 10D t= f'obj + f'oc f'obj=1/10D = 0.10 m f'oc = 1/-25D = 0.04 m t=0.10 + -0.04 = 0.06 m or 6 cm 10 cm - incorrect- would come up with this answer if only took in account the focal length of the objective lens. 4 cm - incorrect- would come up with this answer if only took in account the focal length of the ocular lens. 14 cm - incorrect - would come up with this answer if thought equation was t= f'obj + f'oc

Midget ganglion cells receive information pre-synaptically from which cells? Horizontal bipolar cells Rod bipolar cells Midget bipolar cells Flat bipolar cells

Midget bipolar cells synapse onto midget ganglion cells. These are very selective and exclusive channels as one cone cell synapses with one midget bipolar cell which then in turn relays the information to a midget ganglion cell. There is no additional input from other cells or synapses. These types of monosynaptic cells are most common in the central retina thus explaining the ability to visually discern fine details. Flat bipolar cells receive information from many cone cells and in turn synapse with many ganglion cells. Rod bipolar cells, as their name suggests, convey information from many rod cells to several ganglion cells. Rods relay information only to rod bipolar cells.

When examining a patient, a pinpoint spot of the posterior surface of the lens known as Mittendorf's dot is seen. What is this a remnant of? Pupillary membrane Hyaloid artery Glial tissue of the optic nerve Vitreous

Mittendorf's dot is a remnant of the hyaloid artery and appears as a black dot on the posterior surface of the lens. Pupillary membrane remnants would be present in front of the lens and are a complex of fibers. Glial tissue of the optic nerve head is a remnant that is known as Bergmeister's papilla. Ref: Remington, LA. Clinical Anatomy of the Visual System, 1998 pp 110

A patient is seen at your office reporting constant diplopia. The patient notes that the diplopia is still present when you cover her right eye. Based upon this information, what is the MOST likely etiology of her diplopia? Uncorrected refractive error Superior oblique palsy Aneurysm Lateral rectus palsy

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

A patient is seen at your office reporting constant diplopia. The patient notes that the diplopia is still present when you cover her right eye. Based upon this information, what is the MOST likely etiology of her diplopia? Uncorrected refractive error Superior oblique palsy Lateral rectus palsy Aneurysm

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

A 24-year old female wears soft contact lenses with a Dk/t of 175 and admits to sleeping in her lenses. She is very satisfied with both the comfort and the vision of her lenses. Biomicroscopy reveals mucin balls under her lenses bilaterally that leave impressions in her central corneas upon removal of her lenses. Which of the following actions would BEST help to eliminate the formation of mucin balls? Altering the power of the contact lens but maintaining the same lens material Changing her multi-purpose solution Instructing the patient to increase her blinking frequency Maintaining the same lens material but changing to a steeper base curve

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

Myasthenia gravis is an autoimmune disease that affects which of the following types of receptors in the body? Muscarinic acetylcholine receptors Nicotinic acetylcholine receptors Alpha-adrenergic Beta-adrenergic receptors Adenosine receptors

Myasthenia gravis is autoimmune neuromuscular disease in which circulating antibodies mediate damage and destruction of nicotinic acetylcholine receptors in striated muscle. The subsequent impairment of function at the neuromuscular junction results in weakness and fatigue of the skeletal musculature (with no affect on cardiac or involuntary muscles). This disease most commonly occurs in women and typically presents in the third decade of life. The earliest presenting signs of this condition are usually the development of ptosis or diplopia.

Which of the following genus of organisms is responsible for tuberculosis and leprosy? Salmonella Klebsiella Mycobacterium Borrelia

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

An elderly patient presents in your office with decreased visual acuity. He remarks that he can read better without his glasses and his refraction denotes a large myopic shift. Dilated fundus exam is unremarkable. Which of the following slit lamp findings would MOST likely explain the above findings? Bilateral limbal girdle of Vogt Bilateral corneal arcus Bilateral crocodile shagreen Bilateral 3+ nuclear sclerosis of the lens

Nuclear sclerosis is caused by changes to the optical clarity of the lens. As we age, proteins precipitate out of the lens matrix, causing the lens to become cloudy and altering its density. As time passes, the lens will also begin to change color from clear to a yellow/brown in a process called lens brunescence. Cataracts also generally cause a myopic shift with an increase in against-the-rule astigmatism, leading to decreased distance vision but improved near vision. Corneal arcus is caused by lipid deposition in the peripheral cornea. There remains a characteristic clear zone between the lipid and the limbus. Arcus does not generally interfere with vision. Crocodile shagreen and limbal girdle of Vogt are also benign corneal findings commonly seen in the elderly. Crocodile shagreen appears in the peripheral cornea as polygonal white opacities. Limbal girdle of Vogt is noted at the 3 o'clock and 9 o'clock interpalpebral positions as white crescent-shaped opacities.

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

Oculoglandular syndrome can be caused by a myriad of organisms and presents as a unilateral follicular conjunctivitis along with lymphadenopathy on the same side as the affected eye. Other signs and symptomology vary depending on the causative organism. Causes include but are not limited to: cat-scratch disease, tularemia, syphilis, tuberculosis, sprotrichosis, mononucleosis, coccidioidomycosis, sarcoidosis, Hansen's disease, mumps, actinomycosis, Listeria and Herpes simplex. Based solely upon the age of the child, one would first assume cat-scratch disease, which is the most common cause of oculoglandular syndrome. This assumption would be verified by asking if the child had recently been scratched by a cat and by performing the Hanger-Rose skin test for confirmation. Coccidioidomycosis can cause oculoglandular syndrome, but it is contracted via exposure to a fungus that is found in soil and vegetables through an opening in the skin and is frequently encountered in gardeners, farm workers, or botanists. Children do not generally contract oculoglandular syndrome in this fashion. Diabetes, measles, and toxoplasmosis do not display a correlation with the development of oculoglandular syndrome. Reference: Holland, E. Cornea 2nd edition, volume 1, Fundamentals, Diagnosis and Management, 2005 pp 653-659.

A newborn presenting with symptoms of ophthalmia neonatorum 3 days after birth is MOST likely infected with which of the following organisms? Neisseria gonorrhoeae Chlamydia trachomatis Streptococcus pneumonia Herpes simplex virus Haemophilus influenza Staphylococcus aureus

Ophthalmia neonatorum is a conjunctivitis that typically develops within the first 3 weeks after birth as a result of transmission of infection from mother to child during delivery. This condition is particularly serious due to the lack of immunity in infants as well as the immaturity of the ocular surface (poor tear film and undeveloped lymphoid tissue). Ophthalmia neonatorum secondary to N. gonorrhoeae typically develops within 2-5 days postpartum as hyperacute conjunctivitis. Most cases present bilaterally with periorbital edema, conjunctival chemosis, and excessive amounts of purulent discharge. It is extremely important to quickly and aggressively treat this infection due to the ability of N. gonorrhoeae to penetrate an intact corneal epithelium. When C. trachomatis is the organism responsible for ophthalmia neonatorum, mild to moderate symptoms of unilateral or bilateral conjunctivitis commonly occur between 5 to 14 days after birth. C. trachomatis is the most common cause of ophthalmia neonatorum. These patients present with lid edema, conjunctival chemosis, punctate corneal opacities, and occasionally micropannus formation. Other etiologies of ophthalmia neonatorum can include S. aureus, Haemophilus species, S. pneumoniae, E. coli, and P. aeruginosa. These pathogens are part of the normal bacterial flora of the female genital tract and are likely acquired as the newborn travels through the birth canal.

In order to maximize drug penetration through the cornea an ophthalmic drug or its vehicle should possess which property? Highly polar component High pH Lipid solubility High alcohol content

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

Oral acyclovir is most effective for patients presenting with eyelid findings associated with herpes zoster if administered within which of the following periods following the onset of the disease? 10-12 hours 72 hours 4-5 days 24 hours 7-10 days

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

Oral acyclovir is most effective for patients presenting with eyelid findings associated with herpes zoster if administered within which of the following periods following the onset of the disease? 72 hours 24 hours 4-5 days 7-10 days 10-12 hours

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

A 12 year-old patient is seen at your office complaining of distance blur. Cover test reveals 4 prism diopters of exophoria at near. Subjective refraction reveals -2.25 DS OU. The patient returns 4 weeks later and reports that her vision is clear at distance with the glasses, but when she reads her eyes become fatigued and she reports frontal headaches, all of which she did not experience prior to getting her glasses. What is the MOST likely etiology of her headaches while reading? Her glasses are too tight and are putting pressure on her temples The pantoscopic tilt is incorrect and is inducing unwanted astigmatism The optical center of her glasses was measured too low Resultant esophoria at near induced by her glasses

Patients with myopia who are newly corrected may display esophoria at near when wearing their spectacle correction. Reading through minus-powered lenses causes an increase in the accommodative response as well as the vergence system, resulting in less exophoria or more esophoria at near. If enough esophoria is induced, symptoms of asthenopia can occur. For patients with mild amounts of myopia, the removal of their glasses while performing near, visually oriented tasks will typically eliminate ocular discomfort. Other options include prescribing a bifocal or a progressive addition lens (PAL). Research has demonstrated that myopic patients who possess esophoria at near may benefit from a bifocal or PAL, which may serve to slow down the rate of progression of myopia. Primary Care Optometry 5th edition. Grosvenor, T. Butterworth-Heinemann, 2007, page 252.

A 22-year old male presents in your office for his annual exam. Biomicroscopy reveals bilateral Krukenberg spindles and iris transillumination defects. Given the above findings, what is the correct diagnosis for this patient? Pseudoexfoliation Pigmentary dispersion syndrome Salzmann's nodular degeneration Posner-Schlossman syndrome

Pigmentary dispersion syndrome generally occurs in young, myopic males with deep anterior chambers. This condition is caused by a disruption of the posterior iris pigment epithelium, causing this layer to rub against the ciliary zonules and release pigment into the anterior chamber and its associated structures. The aqueous in the anterior chamber displays convection currents. Warming inferiorly, aqueous rises and migrates forward towards the posterior surface of the endothelium, carrying pigment granules with it, then falls as it cools, causing the pigment to deposit vertically in a linear-fashion on the endothelium. Monitor these patients for blockage of the trabecular meshwork causing a rise in intraocular pressure (IOP) and potential glaucomatous damage to the optic nerve. Physical exertion as well as mydriasis may exacerbate pigment release; therefore, when dilating (especially with phenylephrine), you must be sure to measure post-dilated IOPs. Salzmann's nodular degeneration appears as blue/white hyaline plaque deposits between the epithelium and Bowman's membrane, generally around the pupillary area of the cornea. This condition stems from other pathologies, primarily old phlyctenula. Treatment is generally not required unless vision is affected. Posner-Schlossman syndrome causes an acute IOP spike, usually unilaterally and lasting for hours to weeks with recurrent episodes. Patients are often young and report decreased vision due to corneal edema, mild pain, and ocular redness. This syndrome is also known as glaucomatocyclitic crisis. Biomicroscopy will often reveal ciliary flush, a sluggish or dilated pupil, a mild anterior chamber reaction, potentially with keratic precipitates, corneal edema, open angles and normal optic nerves. IOP readings will normally range from 40-60 mmHg. The etiology is still uncertain; some have postulated that it may be of viral origin. Treatment includes (unless contraindicated) topical steroid drops, beta-blockers, and carbonic anhydrase inhibitors. Pseudoexfoliation appears as white, flaky material that deposits along the pupillary margin, the anterior surface of the lens and other structures of the anterior chamber. This condition is usually unilateral and is seen in the elderly with a concurrent cataract. Transillumination defects, if present, are limited to the iris sphincter region. The pseudoexfoliative material can accumulate in the trabecular meshwork, causing an increase in IOP leading to glaucoma.

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

Pred Forte® is an ocular medication that is in suspension form; therefore, it is important to shake this medication well before use. Other forms of prednisolone that are suspensions include Pred Mild®, Econopred®, and Econopred Plus®. On the other hand, there are prednisolone ocular medications that are solutions, making shaking of the bottle unnecessary. These include AK-Pred®, Inflamase Mild®, and Inflamase Forte®. Zymaxid® is a 4th generation fluoroquinolone antibiotic that is bottled in solution form, as well as Acular®, which is an ocular non-steroidal anti-inflammatory drug (NSAID). FreshKote® is a prescription artificial tear that also does not need to be shaken before use.

Ptosis can be caused by dysfunction or damage to which of the following muscles? Pars ciliaris (Riolan's muscle) Inferior rectus Muscle of Horner Superior tarsal muscle (muscle of Muller)

Ptosis is a condition in which the upper eyelid sags. It can be caused by dysfunction of either the superior palpebral levator or the superior tarsal muscle (muscle of Muller). Because the levator is the major muscle responsible for raising the upper eyelid, ptosis from levator damage is often more severe then ptosis from dysfunction of the muscle of Muller. The muscle of Horner (also known as the pars lacrimalis) is part of the palpebral portion of the orbicularis oculi. The fibers for the muscle of Horner come from the lacrimal crest and encircle the lacrimal canaliculi. This assists the flow of tears into the nasolacrimal drainage system when the orbicularis oculi contracts to close the eye. The muscle of Riolan (also known as the pars ciliaris) is another section of the palpebral portion of the orbicularis oculi; it lies near the lid margin to maintain the margins next to the globe. The orbicularis oculi is the major muscle responsible for closing the eyelids.

Which of the following wavelengths of visible light has an increased association with the development of macular degeneration? 415nm-455nm 520nm-555nm 570nm-620nm 485nm-510nm

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

What type of agar is commonly used to culture fungi? Cetrimide agar Hay infusion agar Thayer-Martin agar Sabouraud's agar Blood agar plate

Sabouraud's agar is useful for culturing fungi. Culturing and sensitivity testing are important diagnostic tools used to determine the offending pathogen and help to select the appropriate medication for treatment. Sabouraud's agar is unique in that it possesses a low pH that causes the inhibition of most bacterial growth, allowing for better isolation of the fungus. Hay infusion agar is frequently utilized to culture slime moulds. Cetrimide agar is designed to isolate Gram-negative bacteria. Blood agar plates are useful for the detection of pathogenic organisms via the presence of hemolytic activity (the ability of organisms to lyse or destroy red blood cells). Blood agar plates are not a selective medium, as many different types of organisms are capable of growth on this type of agar. Thayer-Martin agar is a type of chocolate agar used to isolate Neisseria gonorrhoeae.

Which of the following muscle pairings and actions follows Sherrington's law of reciprocal innervation? Looking to the right causes contraction of the right medial rectus and contraction of the left medial rectus Looking to the right causes contraction of the right medial rectus and inhibition of the right lateral rectus Looking to the right causes contraction of the right lateral rectus and inhibition of the left medial rectus Looking to the right causes contraction of the right lateral rectus and inhibition of the right medial rectus

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

A ray of light traveling in water (n=1.33) strikes a flat, transparent surface (n= 1.59) at an angle of 32 degrees from the normal. What is the angle of refraction?

Snell's law of refraction states that when light travels through a material that possesses an index of refraction greater than 1.0, the light rays change direction and become bent (or refracted). Snell's law is depicted as the following: n sin i= n' sin i' where n= the index of refraction of the first medium, i= the angle of incidence, n'= the index of the second medium, and i' = the angle of the refracted ray. All angles are measured with respect to the normal, which lies perpendicular to the interface between the different media. For the above example, 1.33(sin 32)=1.59 sin i', solving for i'= 26.31 degrees. It is important to commit the index of refraction of water to memory; it is 1.33.

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?

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.

Numerous reports have suggested that increased tear film osmolarity is a key consequence in dry eye. Although osmolarity is not easily measured in the clinical setting, tear osmolarity increases in most dry eye sub-types due to which of the following processes? Loss of tear stability induces an increased evaporation rate, leading to increased osmolarity Decreased capillary exchange leads to ionic bonding In aqueous tear deficiency, the lacrimal gland produces more ionic species Patients with dry eye tend to blink less than normals, leading to increased evaporation Reactive oxygen species are increased in the tears of most dry eye sub-types; this increases osmolarity The lipid layer is altered in most dry eye states, leading to ion pairing

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

Tear volume in a normal, healthy, young adult measures approximately between which of the following values? 2.0-5.0 microliters 17.0-20.0 microliters 9.0-12.0 13.0-16.0 microliters 6.0-8.0 microliters

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

Which of the following medications should be taken with an empty stomach? Doxycycline Amoxicillin with clavulanate (Augmentin®) Cephalexin (Keflex®) Tetracycline

Tetracycline has a half-life of 8 hours and is classified as short-acting. It is most effective while taken on an empty stomach due to the fact that calcium, iron, and magnesium can decrease effectiveness or inactivate this medication. Remember, tetracycline should NEVER be prescribed to children because it will deposit in developing teeth and bones. Doxycycline has a half-life of 18 hours and is therefore categorized as a long-acting tetracycline. Doxycycline can be taken with or without food. Although it is more costly, doxycycline is generally favored over tetracycline because it brings about less gastrointestinal upset, therapeutic benefits can be achieved with smaller doses, less phototoxicity is experienced by the patient, and there is a decreased risk of kidney and liver toxicity. The effectiveness of Augmentin® and Keflex® remain unaltered if food is ingested.

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

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

What is the relationship between the Abbe number and chromatic aberration? No relationship Inversely proportional Equal Proportional

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

What is the name for the phenomenon in which a flickering light that is 10 Hz is seen as brighter than a steady light (one that does not flicker) that possesses the same average luminance? The Granit-Harper law The Brucke-Bartley effect The Troxler effect The Purkinje tree

The Brucke-Bartley effect describes the fact that a flickering stimulus that is 10 Hz will appear brighter than a non-flickering light with the same average luminance. This fact also holds true for stimulus presentation duration. A light that is presented for 50 milliseconds will appear brighter than stimuli presented for longer or shorter durations. The Troxler effect occurs when the eye is fixated (although the eye is truly always moving) on a point in space and the surrounding background begins to blend together. There must be several factors that come into play in order for the Troxler effect to transpire. The best example of this phenomenon is the figure in which there are two squares that are superimposed. The smaller square is centered in the larger square and is slightly lighter than the larger surrounding square. The border between the squares is blurred, resulting in a distinction of the two squares based upon brightness alone. When fixating upon an X placed in the center of the smaller square, the border completely disappears as does the smaller square, resulting in the perception of one uniformly-colored large square. Some patients experience the Troxler effect while performing the FDT visual field and they report, especially during testing of the second eye, that the entire field appears to go gray. When this occurs make sure to inform the patient to blink. The peripheral retina is most sensitive to flicker because it is a part of the magno system which is also known as the "where" system. The magno system is noted for its poor spatial resolution but good temporal resolution. The fact that the peripheral retina is more sensitive to flicker is in accordance with the Granit-Harper law, which states that as the log of the area of the stimulus is increased, the critical flicker fusion frequency also increases accordingly. This helps to explain why the chances of perceiving flicker are greater for a larger stimulus for a given modulation. Remember that the receptors of the peripheral retina display increased summation; therefore, a larger stimulus will take up more area of the retina, increasing the chances of detection. Due to the fact that the eye does not respond well to low temporal frequencies, stabilized retinal images are unable to be detected. Blood vessels that lie on top of the photoreceptors are considered stable relative to the retina. However, if you shine a penlight against your eyelid and move it rapidly, you will be able to see the shadows cast by the blood vessels on your retina. This is called a Purkinje tree. The penlight must be moving to ensure that the temporal frequency is great enough to be visualized.

The visual acuity of a 77 year-old female patient with age-related macular degeneration is 2/16 in the right eye on the ETDRS chart. Why is this chart useful in monitoring the response to treatment with anti-vascular endothelial growth factor (VEGF)? The Snellen construction of the chart enables the examiner to quickly note that a two-line increment represents a factor of a two time increase in the size of the letters Each line has 5 Sloan letters throughout the chart with equal spacing and is 1.26 times larger than the line below it; each line is .1 log units larger than the line below it when moving up the chart A three-line decrease represents a factor of a two time decrease in the size of the letters Each line is 1.0 log units larger than the previous line

The ETDRS chart is a logarithmic eye chart modeled after the Bailey-Lovie chart. It is the primary standardized eye chart used in evaluating the visual acuity of low vision patients. The ETDRS charts are logMAR (log of the minimum angle of resolution) in design and are constructed with 10 Sloan sans serif letters. Each line is 1.26 times larger than the line below, and the construction of each line is such that the difficulty is theoretically equivalent on every line. The construction of the ETDRS chart is made to eliminate the inherent errors in the measurement of visual acuity found in the traditional non-standardized Snellen test charts. The Snellen test charts have variations in legibility of different letters as well as differences in the spacing between the lines of letters and between adjacent letters on single lines. The ETDRS logarithmic chart is constructed in such a way that each line of letters is 0.1 log units (about 1.26 times) larger than the previous line. This is a geometric progression.

A concerned father reports that one of his 12-month-old infant's eyes does not appear straight. You decide to perform the Hirschberg test to evaluate for strabismus. The corneal reflex of the right eye is centered, while the left reflex is displaced 0.5 mm superiorly relative to the center of the pupil. Angle Kappa (Lambda) is zero for each eye. What is the correct deviation and magnitude of the observed strabismus? Left hypertropia of 22 prism diopters Left hypotropia of 22 prism diopters Left hypotropia of 11 prism diopters Left hypertropia of 11 prism diopters

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

According to the Van Herick technique of angle estimation, which angle grade is considered MOST narrow and potentially capable of closure? Grade 3 Grade 1 Grade 4 Grade 2

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

According to the Van Herick technique of angle estimation, which angle grade is considered MOST narrow and potentially capable of closure? Grade 3 Grade 1 Grade 4 Grade 2

The Van Herick method of angle estimation, although not infallible, is a good way of assessing the probability of angle closure with dilation. The temporal angle of a patient is evaluated by having the patient look straight ahead, placing the oculars of the slit lamp in the straight ahead position and shifting the lighthouse temporally so that an angle of 60 degrees is created. The beam is then narrowed to an optic section and placed at the temporal limbus. The width of the anterior chamber space at the angle is then compared to the width of the corneal optic section. The procedure is repeated and the illumination source is placed nasally to assess the grade of the nasal angle. If the angle space is equal to 1/2 or greater than the corneal optic section, the angle is not considered occludable and is given a grade of 4. A grade three displays an angle space that is less than 1/2 but greater than 1/4 of the width of the corneal optic section, and is considered safe to dilate. A grade two has an angle space that is roughly equal to 1/4 the width of the corneal optic section, and may be dilated but with caution. A grade 1 has an angle space of less than 1/4 the width of the corneal optic section, and requires gonioscopy to ensure that the patient is safe to dilate. Reference: Casser, L., Fingeret, M., Woodcome, T. Atlas of Primary Eye Care Procedures, 2nd edition 1997, pp. 28-29.

A father brings in his two-year old son for evaluation at your office. The father remarks that his son was born with what appears as a small red birthmark on his forehead. He wishes to know if it requires treatment or warrants removal. Applying pressure over the area of interest causes blanching of the lesion, and the father reports that the birthmark appears darker when his son cries. What is your prognosis? Malignant and requires immediate biopsy A pre-cursor to a malignant condition and will continue to increase in size with time Benign and will likely regress by the time the child is 5 years of age Benign but is likely to be permanent and will darken will time

The above findings describe a capillary hemangioma (also called a strawberry mark), which is a patch of blood vessels or a vascular birthmark. Most hemangiomas occur at birth or shortly thereafter. There are several different types of hemangiomas, but the most common is the capillary type. The capillary hemangioma will blanch with applied pressure and tends to change color with Valsalva maneuvers, crying, or straining due to changes in blood flow. The majority of these types of vascular lesions will regress by the time the child is 5 years old. If the lesion is still present past puberty, then the lesion is likely permanent and may be removed if the patient is unhappy with the cosmesis.

What wavelength of light is most readily absorbed by the photopigment rhodopsin? 480 nm 507 nm 460 nm 555 nm

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

What is the total cumulative dosage of chloroquine that is MOST commonly associated with retinal toxicity? 1,000 grams 100 grams 460 grams 1500 grams 4,000 grams

The amount of chloroquine that is usually required to produce retinal toxicity is a total cumulative dosage of 460 grams. Symptoms of chloroquine toxicity typically involve decreased vision, abnormal color vision, and difficulty dark-adapting. Retinal evaluation commonly reveals a bull's-eye appearance of the macula in which there is a ring of depigmentation surrounded by a ring of increased pigmentation. Foveal reflex is also diminished in patients presenting with this condition. It is also possible to develop retinal toxicity as a result of hydroxychloroquine use, but this is much less common than with chloroquine. Typically, more than 400 mg/day taken over a period of months to years, or a cumulative dose of 1,000 grams, is required to produce clinical signs and symptoms.

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

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.

Free radicals can cause severe damage to tissue. Which of the following electrolytes can function as an antioxidant in the aqueous? Sodium ions Albumin Ascorbate IgG Chloride ions

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.

All else being equal, cells found in which layer of the cornea consume the GREATEST amount of oxygen? Stroma Bowman's membrane Epithelium Endothelium

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

The blood-aqueous barrier is formed by tight junctions between which cells of the ciliary body? The pigmented epithelial cells The stromal cells The non-pigmented epithelial cells The circular cells Basal laminar cells

The cells of the non-pigmented epithelium have many different attachment sites to each other to ensure that proteins and other macromolecules do not enter. The side walls of the cells are connected via desmosomes. The superior portion of the cell's lateral wall towards the apex are firmly joined by a connection termed zona occludens. The non-pigmented epithelial cells and the epithelial cells are attached to each other by desmosomes and puncta adherens to ensure that the layers are not easily separated.

A chiasmal lesion or mass, such as a pituitary tumor, generally causes what type of visual field defect? Binasal heminopsia Left homonymous hemianopsia Right homonymous hemianopsia Bitemporal hemianopsia

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

A chiasmal lesion or mass, such as a pituitary tumor, generally causes what type of visual field defect? Left homonymous hemianopsia Right homonymous hemianopsia Binasal heminopsia Bitemporal hemianopsia

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

Dissection of the eye reveals that the ciliary body is actually triangular in shape. The apex of this triangle points in which direction and is continuous with which structure? Points posteriorly ; continuous with scleral spur Points anteriorly: continuous with ora serrata Points posteriorly; continuous with choroid Points anteriorly; continuous with iris

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

Congenital cataracts can be caused by a viral infection of the mother with rubella virus (German measles) during development of the primary lens fibers. At which time period in embryonic development can infection cause congenital cataracts? 2nd trimester 1st trimester 3rd trimester Conception Post-delivery

The developing lens is susceptible to rubella virus when the lens fibers are forming, which occurs around weeks 4-7 of gestation. Earlier infection will occur prior to lens fiber development, and the lens is resistant to later infection because the virus is unable to penetrate the lens capsule. The fetus is most susceptible to lenticular damage during the first trimester. Contraction of the rubella virus will cause the greatest amount of damage during this time period. Congenital cataracts are usually detectable at birth but may be seen later because the virus can persist in the lens. Ref: Remington, LA. Clinical Anatomy of the Visual System, 1998 pp 109

What is the net overall moles of ATP produced by the electron transport chain (i.e. not including glycolysis)? 2 moles of ATP 30 moles of ATP 34 moles of ATP 6 moles of ATP 38 moles of ATP

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

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

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

A convex crown glass lens in air has a radius of curvature of 4 cm. What is the dioptric power of the lens? -13 diopters +4 diopters +8.25 diopters +13.0 diopters -4 diopters -8.25 diopters

The equation for the power of a curved surface is: D= (n'-n/r) D= power of the lens (may also be notated as F or P) n'= 2nd index (image), n= 1st index (object) r= radius of curvature (in meters) D= (1.52-1.0)/ 0.04 D= (+)13.0 diopters Remember that converging (or convex) surfaces will have positive power, and diverging (or concave) surfaces will have negative power. If this question was changed to a concave glass lens, the radius of curvature would be negative; therefore, the power of the lens would be (-) 13 diopters.

What is the vergence demand using a Variable Tranaglyph when the separation is measured as 4 cm at a distance of 80 cm when training divergence? 20 prism diopters base-in 20 prism diopters base-out 5 prism diopters base-in 5 prism diopters base-out

The equation to calculate vergence demand is: demand = target separation in centimeters/training distance in meters demand = 4 cm/0.80 m demand = 5 prism diopters Because divergence is being trained in the above patient, the demand will be base-in. If convergence was being trained, the demand would be base-out.

A 42-year old patient reports that her right eye has been watery and she has mild pain, redness, and swelling in the lower medial canthal region. You suspect dacryocystitis as the cause of her symptoms. Which of the following procedures is NOT appropriate when further evaluating this possible diagnosis? Digital palpation of the medial canthal area Dilation and irrigation of the lacrimal system Extraocular muscle motility Gram stain and blood agar cultures of discharge Exophthalmometry

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

A 42-year old patient reports that her right eye has been watery and she has mild pain, redness, and swelling in the lower medial canthal region. You suspect dacryocystitis as the cause of her symptoms. Which of the following procedures is NOT appropriate when further evaluating this possible diagnosis? Dilation and irrigation of the lacrimal system Digital palpation of the medial canthal area Extraocular muscle motility Exophthalmometry Gram stain and blood agar cultures of discharge

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

Any optical system in air has first and second nodal points that coincide with which of the following corresponding points? First and second principal points Front and back surfaces of the optical system The geometrical center of the optical system First and second focal points

The first and second nodal points (N and N') of an optical system are unique conjugate points such that an incident ray directed at N yields a final ray emerging from N' that is undeviated and parallel to the initial ray. For any optical system in air, the first and second nodal points (N and N') correspond to the first and second principal points (P and P').

The presence of foam at the canthus is thought to be pathognomonic for blepharitis. What is the direct etiology of the foam? Mucin balls form from increased ocular surface shear forces exerted by the eyelids in dry eye Increased lysozymes react with free radicals to produce a froth-like material Tear film debris such as sloughed epithelial cellular material that is increased in dry eye Bacterial lipases resulting from low-grade infection within the meibomian glands A detergent effect from altered meibomian gland lipids

The foam is considered to be caused by an alteration in the chemical composition of the meibomian lipids through a process termed saponification. Upon contact of these abnormal lipids with the calcium located in the tear film, a frothy tear film develops. Foam at the canthi is mentioned as a key diagnostic sign of MGD in most literature on the topic.

In addition to the meibomian glands which other accessory glands secrete oil? Zeiss and Moll Moll and Krause Zeiss and Wolfring Wolfring and Krause

The glands of Zeiss and Moll are accessory oil glands located on the lid margins adjacent to the base of the lash follicles. The lipid layer of the tear film is superficial and as such it is exposed to the environment protecting the aqueous layer from evaporation. The glands of Wolfring and Krause are located deep in the fornix of the eyelids and serve to secrete a portion of the aqueous layer of the tear film.

In addition to the meibomian glands which other accessory glands secrete oil? Zeiss and Wolfring Zeiss and Moll Wolfring and Krause Moll and Krause

The glands of Zeiss and Moll are accessory oil glands located on the lid margins adjacent to the base of the lash follicles. The lipid layer of the tear film is superficial and as such it is exposed to the environment protecting the aqueous layer from evaporation. The glands of Wolfring and Krause are located deep in the fornix of the eyelids and serve to secrete a portion of the aqueous layer of the tear film.

What is the mean horizontal and vertical diameter of the human cornea, respectively (when viewed ANTERIORLY)? 10.2mm, 11.5mm 11.5mm, 10.2mm 10.6mm, 11.7mm 11.7mm, 10.6mm

The human cornea has an elliptical configuration in which the mean horizontal diameter is 11.7mm and the mean vertical diameter is 10.6mm, when viewed anteriorly. Contrastingly, when viewed posteriorly, the cornea is actually circular, with mean horizontal and vertical diameters of 11.7mm. This discrepancy is due to the anterior extension of the opaque sclera superiorly and inferiorly.

Which two layers of the iris are derived from mesoderm? The anterior limiting layer and the stroma The epithelium and the posterior pigmented epithelium The stroma and the anterior epithelium The posterior pigmented epithelium and the anterior limiting layer The stroma and the posterior pigmented epithelium

The iris can be classified into four layers. The most external is the anterior limiting layer, followed by the stroma, the anterior epithelium and the posterior pigmented epithelium which is the most internal layer. The first two layers are derived from mesoderm. The anterior limiting layer contains numerous melanocytes and fibroblasts. Interestingly, this layer is not present over Fuch's crypts. The anterior limiting layer is not uniform and varies in density across the entirety of the iris. There is some debate as to whether or not this layer exists and therefore can be simply classified as stroma. The stroma is the thickest layer of the iris and is comprised mostly of collagen. The stroma houses many structures including; blood vessels, pigmented cells, nerves and the sphincter muscle. The epithelial layers are derived from neuroectoderm.

The lymphatic system serves many important roles in the human body. The lateral portion of the eyelid lymphatics drain into which of the following structures? The pre-auricular lymph node The puncta The conjunctiva The submandibular lymph node

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

The lymphatic system serves many important roles in the human body. The lateral portion of the eyelid lymphatics drain into which of the following structures? The puncta The conjunctiva The submandibular lymph node The pre-auricular lymph node

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

Which extraocular rectus muscle has its insertion site CLOSEST to the limbus? The superior rectus The inferior rectus The lateral rectus The medial rectus

The medial rectus inserts into the sclera roughly 5.3 mm from the limbus, followed by the inferior rectus, which inserts 6.8 mm from the limbus. The lateral rectus inserts 6.9 mm from the limbus, and the superior rectus has the furthest insertion point at 7.9 mm from the limbus. Remember MILS (Medial rectus, Inferior rectus, Lateral rectus, Superior rectus). If one draws a line connecting the insertion points of the muscles, an imaginary spiral is created called the spiral of Tillaux. Reference: Chen, W. Oculoplastic Surgery: the Essentials (2001) page 330.

What is the MOST common side effect associated with intravenous administration of sodium fluorescein in patients requiring a fluorescein angiography? Elevated temperature Anaphylaxis Localized tissue necrosis Nausea and vomiting Bronchospasm

The most common reaction observed in patients undergoing intravenous injection of sodium fluorescein is transient nausea and occasional vomiting, which typically occurs within 30-60 seconds after administration, in fewer than 5% of patients. Moderate adverse side effects can occur in up to 1% of patients, which include thrombophlebitis, localized tissue necrosis, nerve palsies, and elevated temperature. Severe and potentially life-threatening complications can result in laryngeal edema, bronchospasm, circulatory shock, myocardial infarction, and anaphylaxis; however, these adverse reactions occur in an extremely low percentage of patients.

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

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

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

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

Which of the following structures serves as the strongest attachment point of the vitreous? The macula Blood vessels The ora serrata The optic nerve head

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.

Long-term use of corticosteroids can lead to the formation of which of the following types of cataract? Cortical Posterior subcapsular Nuclear sclerotic Anterior subcapsular

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

Long-term use of corticosteroids can lead to the formation of which of the following types of cataract? Nuclear sclerotic Posterior subcapsular Cortical Anterior subcapsular

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

What is the Interval of Sturm for a spherocylindrical lens with a power of +6.00 -2.00 x 090? 20 cm 16.7 cm 41.7 cm 8.3 cm 25 cm

The powers in each meridian of the lens are +6.00 and +4.00. The Interval of Sturm is simply the distance between the focal point of each power. 1/+6.00 = 16.7cm 1/+4.00 = 25.0cm 25cm - 16.7cm = 8.3cm

At what time period of the day would the aqueous humor production be least? Midnight to 6 AM Afternoon 8 AM to noon

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

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

The primary Jones dye test can be utilized to determine the patency of the nasolacrimal system. 1-2 drops of fluorescein are instilled into the inferior fornix of the eyes while the patient is in an upright position and blinking her eyes normally. After a period of 5 to 10 minutes, a cotton-tipped applicator is used to swab the undersurface of the inferior turbinate on each side of the nasal passage. When the primary Jones dye test is positive (dye is recovered from the inferior turbinate of the nose), practitioners may conclude that the system is patent and that no significant blockage of the nasolacrimal drainage structure is likely. However, minor stenosis or physiologic dysfunctions cannot be completely ruled out. Patients who have a positive result on the Jones I test are more likely to experience symptoms of epiphora that are secondary to primary oversecretion of tears, rather than a dysfunction in lacrimal drainage (as in the above question). When the primary Jones dye test is negative, the probability of an obstruction or dysfunction in lacrimal drainage is much greater; however, this test alone is not sufficient to document this conclusion. The secondary Jones dye test is then necessary to determine the severity and location of the obstruction

Which of the following is the correct order of structures through which the pupillary fiber pathway passes? Optic nerve -> optic chiasm -> brachium of the superior colliculus -> pretectal region of the midbrain -> Edinger-Westphal nucleus Optic nerve -> optic chiasm -> optic tract -> pretectal region of the midbrain -> Lateral geniculate nucleus in the thalamus Optic nerve -> optic chiasm -> optic tract -> Lateral geniculate nucleus in the thalamus Optic nerve-> optic chiasm -> Lateral geniculate nucleus in the thalamus-> Edinger-Westphal nucleus

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

A 47-year old man sustained orbital trauma and now presents with complaints of retro-orbital pain, impaired ability to move the eye, a droopy eyelid, and diplopia. These signs are most consistent with damage to which of the following structures? Internal auditory meatus Superior orbital fissure Stylomastoid foramen Superficial temporal artery

The superior orbital fissure is a cleft between the lesser and greater wings of the sphenoid. Structures traveling through the superior orbital fissure include the oculomotor nerve (CN III), trochlear nerve (CN IV), abducens nerve (CN VI), lacrimal nerve, frontal nerve, nasociliary nerve, and the ophthalmic vein (superior and inferior divisions). These structures can be damaged when there is orbital trauma causing fractures through the floor of the orbit into the maxillary sinus. This leads to superficial orbital fissure syndrome (also known as Rochon-Duvigneaud's syndrome). Signs include paralysis of extraocular muscles, diplopia, ptosis, exophthalmia and decreased sensation of the upper eyelid and forehead. Vision loss or blindness implies a more serious injury involving the orbital apex (orbital apex syndrome). Tolusa-Hunt syndrome (THS) is an inflammatory condition within the cavernous sinus or superior orbital fissure causing damage to the structures in those regions. Signs are usually acute and unilateral at onset in adults and the most common presenting signs are pain and ophthalmoparesis. The internal auditory (or acoustic) meatus is a canal in the petrous portion of the temporal bone through which the facial (CN VII) and vestibulocochlear nerves (CN VIII) and the labyrinthine artery travel. Damage to these structures can result in deafness and facial muscle paralysis. Acoustic neuromas will commonly expand the internal auditory meatus and damage these structures. Other signs may include tinnitus or vertigo. The stylomastoid foramen is the termination of the facial canal between the styloid and mastoid processes of the temporal bone. The facial nerve and stylomastoid artery travel through this area. Damage to this area can result in facial drooping and paralysis. Bell's palsy (idiopathic facial nerve paralysis) is an inflammatory condition that may lead to swelling of the facial nerve in this region. The superficial temporal artery is a major artery arising from the bifurcation of the external carotid artery. The artery begins within the parotid salivary gland and passes over the zygomatic process of the temporal bone. It is often affected in cases of giant cell arteritis (which is also known as temporal arteritis for this reason). This condition is a vasculitis of the medium and large arteries of the head and is not necessarily restricted to the temporal artery. Temporal arteritis is seen predominantly in older patients and is characterized by fever, headache, sensitivity on the scalp, jaw pain, reduced visual acuity or vision loss, diplopia, and acute tinnitus. Due to potentially rapid progressive vision loss, this disease is a medical emergency. Treatment usually consists of high-dose corticosteroids.

A +3.00 D hyperope is corrected with a +1.50 D contact lens bilaterally. If he views a near object located at 22.0 cm, what degree of accommodation is required to achieve a clear retinal image (rounded to the nearest 0.25 D)? +4.50 D +7.50 D +1.50 D +6.00 D

The target requires +4.50 D of accommodation to be viewed clearly, which is determined by the taking the reciprocal of the target distance (in meters) (1/0.22= 4.55 D or +4.50 D). However, in addition to the target demand, he must also accommodate 1.50 D more as he is undercorrected by this amount at distance in order to ensure a clear retinal image.

Which of the primary germ layers of embryonic development take part in the development of ocular structures? Ectoderm only Endoderm only Mesoderm and ectoderm Mesoderm and endoderm Ectoderm and endoderm

The three primary germ layers are the endoderm, mesoderm, and ectoderm. Of these layers, only the ectoderm and mesoderm are involved in ocular development. The endoderm is involved in development of the gastrointestinal tract, respiratory tract, endocrine organs, urinary bladder, and auditory epithelium. Ref: Remington, LA. Clinical Anatomy of the Visual System, 1998 pp 103

The ligaments that suspend the lens (zonules) are embryonically derived from what structure? The primary vitreous The lens capsule The tertiary vitreous The lens epithelium

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

Which of the following systemic conditions can cause a falsely low measurement of a patient's hemoglobin A1c level? Pregnancy Alcoholism Iron deficient anemia Hyperbilirubinemia Chronic opioid use

There are a few conditions that may cause a reading of hemoglobin A1c (HbA1c) to measure either falsely high or falsely low, therefore altering the reliability of the lab test. Conditions that falsely elevate A1c levels: -Iron deficiency anemia -Any process that slows erythropoiesis increases A1c by maintaining an older erythrocyte cohort in the blood plasma (e.g. aplastic anemia) -Alcoholism -Hyperbilirubinemia -Certain medications (high doses of salicylates, chronic opioid use) Conditions that falsely lower A1c levels: -Any process that shortens the lifespan of erythrocytes (e.g. hemolytic anemia, chronic kidney or liver disease) -Vitamins C and E (by inhibiting glycosylation of glucose to hemoglobin) -Pregnancy -Splenomegaly -Rheumatoid arthritis -Certain medications (antiretrovirals, ribavirin, and dapsone) -Hypertriglyceridemia In these cases, HbA1C levels may still be used to monitor blood sugar levels in patients with diabetes by comparison to previous readings in the same patient. However, goal values must be altered with respect to the underlying condition that is the cause of the unreliable result.

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

There are four Purkinje images. The first image is caused by reflection from the anterior corneal surface and is the brightest of the images. The first image is roughly the same size as the object. The second Purkinje image is formed by the posterior surface of the cornea and almost coincides with the first Purkinje image. The third Purkinje image is the largest and is caused by reflection off of the anterior plane of the crystalline lens. The fourth Purkinje image is the smallest and is inverted, formed by reflection off of the posterior surface of the lens. During the process of accommodation the anterior surface of the lens moves forward. The image that is reflected off of this surface is Purkinje III. Purkinje image III will be seen to move forward during accommodation.

Instillation of too much fluorescein when attempting Goldmann applanation tonometry will result in what type of erroneous result? Mires that are misaligned superiorly Mires that are misaligned inferiorly A measured intraocular pressure that is lower than the true intraocular pressure A measured intraocular pressure that is higher than the true intraocular pressure A corneal abrasion

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

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

This patient is suffering from a recurrent corneal erosion. These types of corneal defects frequently occur in response to a corneal abrasion incurred by something organic (like a fingernail or a tree branch). The initial abrasion heals, but a short time afterwards the patient will experience another episode without any incidence of trauma. The second occurrence tends to transpire first thing in the morning as the eyelids stick to that unstable flap of tissue overnight and rip it off like a band-aid when the eyes open. The best way to treat a recurrent corneal erosion is through the use of a topical antibiotic (unpreserved is best) to ensure sterility (as the cornea is exposed) as well as a bandage contact lens to speed up the healing process if the area of erosion is large. Hyperosmotic drops or artificial tears (preservative-free of course) should be prescribed for roughly 6-8 weeks (sometimes longer) to ensure healing and to allow for proper formation of hemidesmosomes that will help to alleviate future episodes. Other treatments include stromal micropuncture, debridement, phototherapeutic keratectomy (PTK), or oral tetracycline, which inhibits matrix metalloproteinases and allows for proper corneal healing. A corneal abrasion occurs secondary to some type of trauma or injury, and this was not the case in the above example. Recurrent corneal erosions are a common occurrence with epithelial basement membrane dystrophies, but that is not the resultant diagnosis of the current problem experienced by the patient. An ulcer is ruled out on the basis that there is no active infection and is unlikely, as these more commonly occur in patients who wear contact lenses.

In which of the following conditions are bandage contact lenses NOT typically utilized? Bullous keratopathy Recurrent corneal erosion Filamentary keratitis Post-LASIK surgery Eyelid entropion

There are several indications for the use of therapeutic bandage contact lenses; however, when deciding to place a contact lens on an already compromised cornea, the risks and benefits should be carefully considered. Bandage contact lenses may be used to promote corneal epithelial healing in cases where an epithelial defect persists (abrasion or recurrent corneal erosion), as the lens acts to protect the epithelium from the rubbing action of the eyelids, allowing the hemidesmosomes to create a strong attachment to the basement membrane. In addition to promoting corneal healing, bandage contact lenses are also commonly utilized for pain relief in certain conditions such as bullous keratopathy, Thygeson's superficial keratitis, filamentary keratitis, and trichiasis. The bandage contact lens relieves pain by protecting exposed corneal nerves from the shearing forces of the eyelids during blinking or via mechanical protection of the cornea from inwardly turned eyelashes. Bandage contact lenses are also used post-surgically in all cases of photorefractive keratoplasty (PRK) to allow healing of the corneal epithelium that has undergone the surgical procedure. Contact lenses are not typically applied after LASIK surgery unless there has been a serious complication.

As the interpupillary distance increases what happens to the amount of convergence (in prism diopters) needed to maintain fusion for a near target ? It decreases In decreases, but only for targets closer than 4 cm It increases It increases, but only for targets closer than 4 cm It remains the same

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

Which of the following organisms can penetrate an INTACT cornea? Salmonella enterica Haemophilus influenza Staphylococcus aureus Staphylococcus epidermis

There is currently much debate in the literature regarding Pseudomonas because it was previously believed that Pseudomonas aeruginosa was capable of penetrating an intact cornea by secreting proteolytic enzymes that could break down the corneal barrier. However, current research suggests that this line of thinking no longer holds true. At present, it is believed that N. gonorrhoeae, Corynebacterium diphtheriae, Listeria species, and Haemophilus species are capable of infecting an intact cornea. Treatment commonly involves topical antibiotic ophthalmic drops. Staphylococcus aureus, S. Epidermis, and Salmonella cannot penetrate an intact epithelium. References: Clinical Applications of the Limulus Amoebocyte Lysate Test. Prior, R. CRC Press, 1990 page 122. Clinical Infectious Disease.Schlossberg, D. Cambridge Medicine, 2008, page 87 Cornea and External Eye Disease. Reinhard, T. et al. Springer, 2008, page 22. Haemophilus: Advances in Research and Treatment: 2011 Edition: Scholarly Paper. http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2715680/

A constant ringing of the ears is known as which of the following terms? Malleus Sinusitis Otitis media Tinnitus

Tinnitus is caused by damage to hair cells in the inner ear from exposure to excessive noise, medications (like aspirin), aging, and some diseases. Sound waves cause the hair cells to bend, releasing a neurotransmitter and causing action potentials of the auditory nerve. Sometimes the hair cells break or are left in the "on" position, causing the perception of ringing. Otitis media is an infection or inflammation of the middle ear. Sinusitis is an inflammation of the sinuses. The malleus, also known as the hammer, is one of the tiny bones in the ear that help to transmit and amplify sound to the auditory nerve.

A Galilean telescope has an ocular lens with a power of -32.00 D and an objective lens with a power of +8.00 D. What is the magnification provided by the telescope? 8x 256x 4x 0.25x

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

What is the front surface power of a lens in air with a refractive index of 1.50 and radius of curvature of 50 cm? 2.00 D 1.00 D 3.00 D 1.50 D

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

A patient returns to your office reporting that her eyes feel strange when she reads 6 mm below the optical centers of her new glasses. The prescription in her right eye is -4.00 DS and -7.00 DS in her left eye. How much vertical prism is induced when she reads? 1.8 prism diopters base down 4.2 prism diopters base down 6.6 prism diopters base down 2.4 prism diopters base down

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

A patient returns to your office reporting that her eyes feel strange when she reads 6 mm below the optical centers of her new glasses. The prescription in her right eye is -4.00 DS and -7.00 DS in her left eye. How much vertical prism is induced when she reads? 4.2 prism diopters base down 1.8 prism diopters base down 2.4 prism diopters base down 6.6 prism diopters base down

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

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

VKC is a condition of the young and presents with an increased frequency in males. This type of allergy typically develops before age 14 and lasts for 4-10 years before the child outgrows it; VKC occurs predominantly in the spring and summer. The condition progressively lessens in severity, with the first episode being the worst. Usually VKC is seen in patients who are prone to atopy; therefore they suffer from eczema, asthma, or hay fever. Patients typically suffer from itchy eyes and photophobia. The condition basically presents as a very severe type of allergic conjunctivitis. Signs include cobblestone papillae of the upper lid, lid swelling, and ropy discharge that is worse in the morning. Corneal defects (usually superiorly) known as keratitis of Togby may also be present. Occasionally, patients will develop a shield ulcer and Trantas' dots (calcified eosinophils seen circumlimbally that appear as chalky concretions), which may lead to the feeling of an associated foreign body sensation. Treatment includes mast cell stabilizers that should be started several weeks prior to re-occurring episodes, pulse steroid therapy, cool compresses, and sunglasses to help alleviate ensuing photophobia. EKC and bacterial conjunctivitis typically do not cause extreme itching and should not present with cobblestone papillae. The number one symptom associated with iritis is photophobia, and the patient should be neither complaining of itching nor have cobblestone papillae present on biomicroscopy.

A young strabismic child presents at your office. Using visuoscopy you ask the patient to fixate the center of the target with their right eye (the left eye is occluded). The foveal reflex is positioned three hash marks to the LEFT of the center circle of the target. This finding suggests which type of fixation? (assume each hash mark is equal to 1 prism diopter) (Picture of visuoscopy target) 4 prism diopters temporal eccentric fixation 4 prism diopters inferior eccentric fixation 4 prism diopters nasal eccentric fixation The patient does not possess any eccentric fixation 4 prism diopters superior eccentric fixation

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

A young strabismic child presents at your office. Using visuoscopy you ask the patient to fixate the center of the target with their right eye (the left eye is occluded). The foveal reflex is positioned three hash marks to the LEFT of the center circle of the target. This finding suggests which type of fixation? (assume each hash mark is equal to 1 prism diopter) 4 prism diopters temporal eccentric fixation 4 prism diopters inferior eccentric fixation 4 prism diopters nasal eccentric fixation The patient does not possess any eccentric fixation 4 prism diopters superior eccentric fixation

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

A person can just barely detect the difference between two weights; one weighs 12 pounds and the other weighs 10 pounds. What is the just noticeable difference for a weight of 70 pounds? 2 pounds 20 pounds 8 pounds 56 pounds 14 pounds

Weber's Law deals with the just noticeable difference and can be expressed mathematically as: K= delta I/I, where K= Weber's constant, delta I= that difference threshold, and I= the original stimulus intensity (weight etc.) For the above example, we must first solve for Weber's constant. 12-10/10=0.2. Using this Weber's constant we can then solve for the just noticeable difference or the increment threshold for 70 pounds. X-70/70=0.2, X= 14 pounds. Therefore, the above person will just be able to discern the difference between a 70-pound weight and an 84-pound weight.

What portion of the progressive lens should be used to verify the distance prescription when the lenses arrive from the lab? The center of the fitting cross The prism reference point (PRP) The major reference point (MRP) The center of the distance arc

When a progressive lens prescription is returned from the lab, it typically contains the removable markings that can be used for prescription verification and fitting purposes. Distance reference point (DRP) -Located at the center of the distance arc -Indicates the recommended position of the lens through which the distance prescription should be measured with a lensometer Fitting cross -Used to verify the fitting height -Should be centered on the patient's pupil -2 horizontal dashes to the right and left of the fitting cross help to determine whether the lens is level or tilted PRP (prism reference point) -Used to verify the prism power -This is the same as the MRP (major reference point) Near reference point (NRP) -Located at the center of the circle in the lower part of the lens -Used to verify the near add power

Which of the following BEST describes the reasoning for the need to taper topical ocular corticosteroids? Minimize the risk of developing steroid-induced elevation of IOP Minimize risk of adrenal insufficiency due to decreased production of natural cortisol Avoid signs and symptoms of rebound ocular inflammation Decrease the risk of posterior subcapsular cataract formation Prevent possible secondary ocular infections

When prescribing a topical ocular corticosteroid, the tapering schedule of the medication is just as important as the initial dose. It is imperative not to prematurely discontinue steroid medications and to closely monitor the patient for signs of rebound inflammation during and after the tapering process. The mechanism for developing rebound inflammation stems from the fact that prolonged use of corticosteroids causes a reduction in the amount of mature circulating leukocytic elements. If the steroid is stopped abruptly, the immature cells can proliferate, producing large quantities of antibodies in response to the residual antigen remaining in the tissue. This physiologic action is what causes rebound inflammation. The exaggerated inflammatory response is more commonly associated with systemic steroid use, but prolonged topical ocular therapy (usually longer than 1 month) can also induce this response. A typical steroid taper should begin once the inflammation is completely controlled. There are many different tapering schedules that may be implemented depending on the patient's condition, but classically, the dose of the steroid should be halved for each given time interval. For example, if you prescribed a steroid q.i.d. x 5 days, you could taper to b.i.d. x 5 days, then q.day x 5 days. The longer the treatment period or more frequent the dosage, the longer the taper will be. With long-term oral corticosteroid use, there is a risk of developing adrenal insufficiency if the steroid is not tapered properly. This is due to the fact that the adrenal glands decrease their normal production of natural cortisol when an individual is taking a dose of steroids. Tapering the steroid is important such that the adrenal glands can return to producing habitual levels of cortisol before complete cessation of the oral steroids.

Cyclosporine is commonly used topically to treat certain eye conditions including keratoconjunctivitis sicca (KCS). Which of the following is a known mechanism of action for cyclosporine? Stabilization of mast cells Inhibition of T-cell activation Antagonism of vitamin K Inhibition of cyclooxygenase (COX)

While the exact mechanism of action of cyclosporine remains controversial, it is generally accepted that cyclosporine inhibits calcineurin in CD4+ T helper cells which, under normal circumstances, are responsible for production of interleukin-2 (IL-2). IL-2 normally stimulates activation and proliferation of cytotoxic T cells and other helper T cells. Cyclosporine prevents this activation and acts as an anti-inflammatory. One of the major causes of KCS is autoimmune destruction of lacrimal cells by T cells. Therefore, topical cyclosporine can be protective. In all likelihood, there are likely additional mechanisms of action for cyclosporine in the treatment of KCS, but this is one of the most accepted mechanisms and is the best option from the choices given here.

A patient is seen at your office complaining of distance blur with his glasses. With his current prescription of -1.25 D in place, you determine that his far point is 50 cm from the spectacle plane for his left eye. Given this information, which of the following is the MOST appropriate spectacle prescription to obtain a clear retinal image when an object is viewed at optical infinity (rounded to the nearest 0.25 D)? -1.25 D -2.00 D +0.75 D -3.25 D -0.75 D

With the current prescription in place, the patient's far point is 50 cm. The far point vergence at the spectacle plane necessary to obtain a clear image is the reciprocal of the far point, in meters. 1/0.50= 2.00 D. Therefore to achieve a clear retinal image for an object focused at optical infinity requires -3.25 D at the spectacle plane. Because the far point of 50 cm was determined with his current prescription in place, his prescription needs to be strengthened by the amount of the far point vergence required at the spectacle plane, which was -2.00 D in this case, and this value should be added to his prescription of -1.25 D.

A cornea that displays toricity is said to have with-the-rule astigmatism if it possesses which of the following keratometry readings? The horizontal meridian has a shorter radius of curvature The horizontal meridian is steeper than the other principal meridian The vertical meridian is steeper than the other principal meridian The vertical meridian is flatter than the other principal meridian

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

What is the name of the pigmented line that represents the leading edge of a pterygium? Fleischer's ring Krukenberg's line Stocker's line Hudson-Stahli line Coat's white ring Ferry's line

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

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 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.

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? The a-wave will remain while the b-wave will disappear Both the a-wave and the b-wave will remain Both the a-wave and the b-wave will disappear The a-wave will disappear while the b-wave will remain

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.

A cranial nerve VI palsy will cause what type of deviation? Exodeviation worse with distance viewing Esodeviation worse with distance viewing Esodeviation worse with near viewing Exodeviation worse with near viewing

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

A deficiency of which vitamin leads to prolonged dark adaptation?

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.

Which of the following types of refractive error would have the greatest tendency to lead to amblyopia? A four-year old boy with an uncorrected refractive error of OD: +6.00 DS and OS: +1.50 DS A five-year old girl with an uncorrected refractive error of OD: -3.25 DS and OS: -0.75 DS A four-year old girl with an uncorrected refractive error of OD: +1.00-1.50 x 180 and OS: +1.50-1.25 x 180 A three-year old boy with an uncorrected refractive error of OD: +1.50 DS and OS: -2.00 DS

A prescription in which there is a big refractive difference between the eyes, especially if both eyes are hyperopic, is most likely to cause amblyopia. Consider the prescription of OD: +6.00 DS and OS: +1.50 DS. The left eye will be able to accommodate 1.50 diopters to obtain a clear distance image and, because accommodation is bilateral and equal, the right eye will still be 4.50 diopters out of focus. This defocus will cause the left eye to dominate the cortical neurons, causing a decreased amount of binocular neurons and leading to poor stereopsis and amblyopia of the right eye. A prescription of OD: -3.25 DS and OS:-0.75 DS will not lead to amblyopia because even though the right eye is blurry in the distance, at 30 cm its image will be clear and in focus. This also applies to the prescription of OD: +1.50 DS and OS:-2.00 DS. Although both patients will have good monocular corrected acuities, they will most likely possess poor stereopsis due to a decreased amount of binocular neurons because the eyes are never in focus at the same distance. It is important to note that in order for amblyopia to occur, ametropia must be present during the critical period.

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

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

Many skin anomalies may mimic malignant lesions. Which of the following skin conditions has the HIGHEST risk of becoming malignant? Cutaneous horn Papilloma Actinic keratosis Seborrhoeic keratosis

Actinic keratosis is a precursor to squamous cell carcinoma and appears as scaly, dry skin that does not heal. People with skin that is of lighter pigmentation along with excessive exposure to ultraviolet light tend to be most at risk for development of this condition. Papillomas may take on various forms and may be viral or non-viral in origin. They can commonly be found on the eyelids or surrounding orbital skin. Viral warts tend to grow at an accelerated rate while non-viral papillomas are fairly slow to grow. Papillomas can mimic neoplastic growths so be sure to rule this out while watching carefully for color change, ulceration, lash loss, bleeding, and vascularization. Cutaneous horns or tags are also benign and are likely a form of papilloma but appear to involve more keratin. Treatment is similar to that of a papilloma. Seborrhoeic keratosis is more commonly seen in middle-aged and elderly persons. This benign, epidermal growth is quite superficial and does not extend into the dermis. It appears like a brown plaque that has been stuck onto someone's skin. The borders are very distinct and there may be some elevation. The lesions may be removed if the patient is concerned about cosmesis.

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? Prednisone (oral) Administration of Benadryl (oral) Olopatadine (Patanol) Injection of epinephrine (EpiPen)

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 methods can be used to test for the presence of eccentric fixation? Binocular versions The Hirschberg test The Bruckner test Visuoscopy

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

You are fitting a toric soft contact lens to your patient's right eye. The patient's manifest refraction is -2.00 -1.50 X 095. You apply a -1.75 -1.25 X 085 diagnostic toric soft contact lens. It fits well, and the prism base down marking consistently locates halfway between the 6 o'clock and 7 o'clock hours. What axis should you order? 70 degrees 80 degrees 110 degrees 95 degrees 100 degrees

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

Which of the following will occur if you increase the water content of a soft hydrogel contact lens? The patient will report an increase in dry eye symptoms The tendency of lens deposits will decrease The oxygen permeability will decrease The lens durability will increase

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

Which of the following BEST describes the definition of irregular astigmatism? The principal meridians of the cornea are located 90 degrees apart The principal meridians of the cornea are not perpendicular to each other The axis of astigmatism is located along an oblique axis The axis of astigmatism is located along the 90 degree meridian

Astigmatism can be classified as either being regular or irregular. Regular astigmatism occurs in individuals in which the principal meridians of the cornea are located 90 degrees apart. That is, the area of the cornea with the flattest curvature (the axis) is oriented perpendicular to the meridian of the steepest curvature. In certain ocular conditions such as keratoconus, corneal scarring, or post-surgical corneas, the steep and flat meridians may not be oriented 90 degrees apart. This type of corneal curvature can be considered irregular astigmatism. In these cases, the refractive error is typically not well corrected with spectacles, in comparison to correction with gas-permeable contact lenses.

A chin fissure is a dominant trait. If a father who is homozygous-dominant for this trait and a mother who is homozygous-recessive for this trait mate, what are the chances that their first child will have a chin fissure? 50% 0% 25% 100% 75%

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

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

Blepharitis is a condition caused by pathogens, usually of Staphylococcus origin, that colonize along the eyelid margins. The bacteria produce exotoxins which take the form of flakes and are generally seen along the base of the eyelashes. Unfortunately, this condition is chronic but will wax and wane in its presentation. Long-term complications include madarosis (missing lashes), trichiasis, neovascularization of the eyelid margin, keratitis, erythema, phlyctenule formation and infiltrates. Patients may complain of dry, irritated eyes, stinging, pain, itching, frequent eye infections, foreign body sensation, and decreased acuity (if there is corneal involvement). Treatment includes eye lid scrubs, antibiotic ointments and sometimes transient topical steroid use to decrease lid inflammation (usually used in conjunction with a topical antibiotic). Occasionally oral antibiotics are prescribed, especially in the event of poor compliance. Distichiasis is a rare congenital phenomenon marked by an absence of meibomian glands. In the place of the meibomian glands is an extra row of eyelashes. Hypertelorism is a term used to describe the incidence in which the orbits are located quite far apart. This generally occurs along with other congenital cranium anomalies. Tristichiasis is a very rare occurrence in which a person possesses three rows of eyelashes.

What is the MOST common type of oculomotor deviation? Exophoria Hyperphoria Hypophoria Esophoria

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

A patient is concerned with an acute reduction of the acuity in the right eye. You correctly diagnose central serous retinopathy, and confirm your diagnosis with an optical coherence tomography (OCT). What is the standard treatment protocol? Monitor monthly for resolution Refer for intravitreal steroid injection Treat the patient with prism as they are likely to develop diplopia Refer for laser treatment of the retina Refer for cryotherapy of the retina

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

A positive catalase test indicates that a bacteria is capable of breaking down which of the following? Pyruvate Glucose Carbon dioxide Hydrogen peroxide

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

Which of the following types of congenital cataracts are characteristic of galactosemia? Christmas tree cataracts Blue dot (Cerulean) opacities Oil droplet opacities Sunflower cataracts

Central oil droplet opacities are a type of congenital cataract that is associated with galactosemia, a genetic metabolic disorder that affects the body's ability to metabolize galactose properly. Blue dot (Cerulean) opacities are congenital cataracts and are not usually associated with systemic disease but are thought to be due to autosomal dominant mutations in several genes. Christmas tree cataracts are not considered a congenital type of cataract; they are a rare variant of senile cataracts that have a strong association with myotonic dystrophy. Sunflower cataracts are also not considered congenital cataracts and are due to the abnormal deposition of copper in patients suffering from Wilson's disease.

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

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

Antibiotic resistance that is rapidly spread within a population of bacteria is due to what mechanism? Transformation Budding Conjugation Binary fission

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

Which one of the following bitoric GP contact lenses would NOT induce cylinder if rotated to a misaligned position on the eye? 7.54 mm / +1.50 D --------------------- 7.99 mm / +2.75 D 7.63 mm / -1.50 D --------------------- 8.11 mm / +1.12 D 7.46 mm / -4.25 D --------------------- 8.13 mm / -1.75 D All of the options listed would induce cylinder if rotated off axis 7.58 mm / -5.37 D --------------------- 8.18 mm / -0.50 D

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

An increased rate of molecular movement down its respective concentration gradient via help from carrier proteins refers to which type of transportation?

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.

What separation distance will make the combination of a +3.00 and a +10.00 thin lens afocal? 23 cm 0.43 cm 2.3 cm 43 cm 1.7 cm 17 cm

For this question, the equation for equivalent power of a thick lens system should be used, solving for thickness (t). De = D1 + D2 - (t/n) x D1D2 De = equivalent power, D1 = front surface power, D2 = back surface power t = thickness of lens system, n = index between the 2 surfaces An afocal system has its focal points (F and F') located at infinity. Therefore, an incident parallel pencil of light rays will emerge into image space as a parallel pencil as well. Another way to characterize an afocal system is that the equivalent power (De) is 0. In the above question, De = 0, D1 = +3.00, D2 = +10.00, n= 1 0 = 3 + 10 - ((t/1) x (3) x (10)) 0 = 13 - (t x 3 x 10) 0 = 13 -30t 30t = 13 t = 0.43 m (or 43 cm) If the two lenses are separated by 43 cm, the lens system can be considered afocal. This type of combination of two plus lenses is also an example of a simple astronomical (Keplerian) telescope. Keep in mind that the image in this type of optical system is inverted.

Which of the following conditions would be categorized as causing amblyopia due to deprivation? A child born with a large congenital cataract in one eye only A child born with a monocular 2 mm ptosis A five-year old with an uncorrected prescription of OD: +7.00 D 20/400 OS: +0.50 20/20 A three-year with a constant right 30 prism diopter esotropia

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

Which of the following drugs decrease intraocular pressure by increasing uveoscleral outflow? Pilocarpine Brinzolamide Timolol Brimonidine Dorzolamide

Glaucoma medications lower intraocular pressure by either decreasing aqueous production or by increasing aqueous outflow. There are three classes of drugs for which the mechanism of action is increasing aqueous outflow: cholinergic agonists, prostaglandin analogs, and alpha-2 agonists. Cholinergic agonists, such as pilocarpine, work by increasing trabecular outflow, whereas prostaglandin analogs and alpha-2 agonists work by increasing uveoscleral outflow. The other classes of glaucoma medications, such as beta-blockers and carbonic anhydrase inhibitors, work by decreasing aqueous production. It is important to note that alpha-2 agonists, such as Brimonidine (Alphagan®) and Apraclonidine (Iopidine®), have dual mechanisms of action. This class of medication decreases intraocular pressure by both increasing uveoscleral outflow and decreasing aqueous production.

What is the name of the surgical procedure in which thermal laser burns are placed in the mid-periphery of the cornea in an attempt to steepen the corneal curvature? Limbal relaxation incisions Laser-assisted in-situ keratomileusis Radial keratotomy Photorefractive keratectomy Conductive keratoplasty

In cases where the corneal curvature must be steepened in order to correct for refractive error (hyperopia or presbyopia), conductive keratoplasty (CK) is a viable surgical option. Although this surgical procedure was used more often in earlier years, it is not currently as widely used as laser-assisted in-situ keratomileusis (LASIK) and photorefractive keratectomy (PRK). In comparison to CK, LASIK and PRK tend to be safe, have long-standing results, and more predictable outcomes. The CK technique involves using a radiofrequency probe to create burns in either one or two concentric rings in the mid-peripheral region of the cornea. These thermal laser burns cause subsequent stromal shrinkage, which results in an increase in the curvature of the cornea. This change in curvature typically decays over time, but the procedure can be repeated. Radial keratotomy is also an older surgical procedure in which a diamond blade is used to create several radial corneal incisions (the number and depth of the incisions depends on the refractive error) in order to flatten the corneal curvature in patients with myopic refractive errors. Limbal relaxation incisions are similar in that arcuate incisions are made on opposite sides of the corneal periphery in the meridian of the "plus" cylinder axis in order to create flattening of the steep corneal curvature (with some smaller steepening of the flat meridian) in an attempt to reduce the amount of corneal astigmatism. Photorefractive keratotomy (PRK) and laser-assisted in-situ keratomileusis (LASIK) are refractive surgery techniques that use an excimer laser to ablate corneal tissue to a certain depth in either the central cornea (to correct myopia) or peripherally (to correct hyperopia).

A 6-foot tall man wishes to buy a plane mirror in which he can visualize his whole length at the same time. How tall must the mirror measure in order for the above to occur? 3 feet tall 6 feet tall 5.2 feet tall 4.5 feet tall 2.3 feet tall

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

A patient who has a high spatial frequency cut-off of 40 cycles per degree will have what predicted Snellen acuity? 20/30 20/40 20/20 20/15

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

While performing the astigmatic clock dial, your patient reports that the clearest/blackest line is the 2-8 line while the 5-11 line is the least clear. What would be the corresponding axis of astigmatism? 60 degrees 30 degrees 180 degrees 150 degrees

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

A contracting muscle that develops tension but does not shorten displays which type of muscle tension? Isometric Isovelocity Isotonic Isovolume

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

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? Tums® (calcium carbonate) Accutane® (isotretinoin) Tylenol® (acetaminophen) Omega III fish oil capsules

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.

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?

Isotretinoin, birth control pills, and diuretics, among many other drugs, can cause myopia in some patients. Myopia mostly 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.

A common cause of epiphora in infants is caused by a small membrane that covers over which of the following structures? The lacrimal gland The canaliculus The valve of Hasner The puncta

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

Which of the following skin conditions is considered to be benign and has the LOWEST risk of malignancy? Squamous cell carcinoma Keratoacanthoma Actinic keratosis Basal cell carcinoma

Keratoacanthoma appears very much like squamous cell carcinoma (SCC) in that it tends to progress rapidly and appears to ulcerate. This condition typically occurs in middle-aged and elderly patients of Caucasian descent on areas of the skin that are exposed. The lesion appears elevated, and eventually the center will produce a scab-like plug of keratin. The margins surrounding the plug will be rolled. At some point the keratin plug will fall out, resulting in the formation of a pit, and the lesion will regress. Most patients and clinicians do not like to wait this condition out due to its similarities to SCC. Actinic keratosis is a pre-cursor to squamous cell carcinoma and appears as scaly, dry skin that does not heal. People with skin that is of lighter pigmentation along with excessive exposure to ultraviolet light tend to be most at risk for development of this condition. Squamous cell carcinoma (SSC) is thankfully one of the rarest malignancies but due to its ability to metastasize can be quite dangerous. This malignancy has the ability to progress rapidly and has a high affinity for people who spend a lot of time in the sun, especially those who are light-skinned. The only way to definitively diagnose SCC is to refer for a biopsy and ensuring the use of Mohs technique. This strategy takes more time but ensures that the lesion is removed. Essentially, Mohs procedure calls for removal of tissue and biopsy of the surrounding borders. If the borders prove to be malignant then more tissue is removed and biopsied. This continues until the borders prove to be free of any carcinoma. Basal cell carcinoma (BCC) is the most common malignant lid lesion and mercifully tends to be very slow-growing. BCC generally appears as a waxy, translucent nodule. Eventually the nodule will ulcerate. Patients may bring these to your attention and tell you that they have "had it for years and it just does not seem to heal". Whenever you hear this it is best to send out for biopsy via Mohs technique. BCC very rarely metastasizes.

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

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

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

Many bacteria possess flagella, or thread-like appendages, which allow for movement. Certain chemicals attract bacteria (chemoattractants), while others repel them (chemorepellents). Chemotaxis refers to the response of the bacteria to either chemoattractants or chemorepellents. In the absence of either of the aforementioned chemicals, bacteria will move in random patterns. Some bacteria possess genes and proteins which allow for the sensing of concentration gradients in their environment. In the presence of a chemoattractant, bacteria will have longer runs in the appropriate direction. Apoptosis is defined as programmed cell death. Phagocytosis refers to the engulfment of a particle (for example, bacteria) by a phagocyte (for example, a macrophage). Transposition refers to the rare phenomenon in which genes move from one place on the genome to another position.

An elderly patient presents in your office with decreased visual acuity. He remarks that he can read better without his glasses and his refraction denotes a large myopic shift. Dilated fundus exam is unremarkable. Which of the following slit lamp findings would MOST likely explain the above findings? Bilateral corneal arcus Bilateral limbal girdle of Vogt Bilateral 3+ nuclear sclerosis of the lens Bilateral crocodile shagreen

Nuclear sclerosis is caused by changes to the optical clarity of the lens. As we age, proteins precipitate out of the lens matrix, causing the lens to become cloudy and altering its density. As time passes, the lens will also begin to change color from clear to a yellow/brown in a process called lens brunescence. Cataracts also generally cause a myopic shift with an increase in against-the-rule astigmatism, leading to decreased distance vision but improved near vision. Corneal arcus is caused by lipid deposition in the peripheral cornea. There remains a characteristic clear zone between the lipid and the limbus. Arcus does not generally interfere with vision. Crocodile shagreen and limbal girdle of Vogt are also benign corneal findings commonly seen in the elderly. Crocodile shagreen appears in the peripheral cornea as polygonal white opacities. Limbal girdle of Vogt is noted at the 3 o'clock and 9 o'clock interpalpebral positions as white crescent-shaped opacities.

Which type of anterior scleritis is associated with the highest risk of perforation? Diffuse Necrotizing Nodular Scleromalacia perforans

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. References: Schwartz, G. Around the eye in 365 days (2009) page 314. Pavan-Langston, D. Manual of Ocular Diagnosis and Therapy, 6th edition (2008) pp 133-136.

Which of the following is a precursor to steroid hormones such as testosterone? Phospholipids Triglycerides Sphingolipids Cholesterol

Progesterone, aldosterone, testosterone, estradiol and cortisol are all derived from cholesterol. Cholesterol has a unique configuration comprised of four joined cycloalkane rings. Because these hormones are fat-soluble, they readily pass through cell membranes. They diffuse into the blood and are generally bound to carrier proteins, which transport the hormones to their designated target site where they may further undergo processing or transformation. Sphingolipids are important in cell membranes, especially those located in the central nervous system, such as myelin sheath. Sphingolipids contain sphingosine as a backbone and are then further classified depending on which molecules are attached to that backbone, such as ceremides, gangliosides, sphingomyelin, etc. Phospholipids contain a polar and non-polar end, thus making them amphoteric. This property allows for the formation of bilayers (polar ends aligned together and pointed outwards) resulting in the lipid bilayer commonly seen in cell membranes. Phospholipids are generally comprised of a phosphate group, a choline group (polar), and two fatty acid chains (non-polar) attached to glycerol, which serves as the backbone. Triglycerides are comprised of three fatty acid chains attached to a glycerol backbone. Triglycerides are important in long-term energy storage for use by cells.

Ptosis can be caused by dysfunction or damage to which of the following muscles? Inferior rectus Muscle of Horner Superior tarsal muscle (muscle of Muller) Pars ciliaris (Riolan's muscle)

Ptosis is a condition in which the upper eyelid sags. It can be caused by dysfunction of either the superior palpebral levator or the superior tarsal muscle (muscle of Muller). Because the levator is the major muscle responsible for raising the upper eyelid, ptosis from levator damage is often more severe then ptosis from dysfunction of the muscle of Muller. The muscle of Horner (also known as the pars lacrimalis) is part of the palpebral portion of the orbicularis oculi. The fibers for the muscle of Horner come from the lacrimal crest and encircle the lacrimal canaliculi. This assists the flow of tears into the nasolacrimal drainage system when the orbicularis oculi contracts to close the eye. The muscle of Riolan (also known as the pars ciliaris) is another section of the palpebral portion of the orbicularis oculi; it lies near the lid margin to maintain the margins next to the globe. The orbicularis oculi is the major muscle responsible for closing the eyelids.

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

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

Which of the following types of scleritis presents without ocular inflammation, has a low risk for perforation, and does not typically result in pain or decreased visual acuity? Granulomatous necrotizing scleritis Scleromalacia perforans Posterior scleritis Vaso-occlusive necrotizing scleritis Nodular scleritis Anterior non-necrotizing diffuse scleritis

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

Which of the following correctly describes the autonomic innervation of the iris muscles? The iris sphincter is innervated sympathetically and the iris dilator is innervated parasympathetically The iris sphincter and iris dilator are both innervated sympathetically The iris sphincter and iris dilator are both innervated parasympathetically The iris sphincter is innervated parasympathetically and the iris dilator is innervated sympathetically

Stimulation of the sympathetic nervous system results in pupil dilation and the parasympathetic nervous system pupil constriction. Accordingly, the sphincter muscle (which constricts the pupil) is innervated by the parasympathetic nervous system and the dilator muscle (which dilates the pupil) is innervated by the sympathetic nervous system. Ref: Remington, LA. Clinical Anatomy of the Visual System, 1998 p 44

Numerous reports have suggested that increased tear film osmolarity is a key consequence in dry eye. Although osmolarity is not easily measured in the clinical setting, tear osmolarity increases in most dry eye sub-types due to which of the following processes? The lipid layer is altered in most dry eye states, leading to ion pairing Decreased capillary exchange leads to ionic bonding Patients with dry eye tend to blink less than normals, leading to increased evaporation Reactive oxygen species are increased in the tears of most dry eye sub-types; this increases osmolarity In aqueous tear deficiency, the lacrimal gland produces more ionic species Loss of tear stability induces an increased evaporation rate, leading to increased osmolarity

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

Tear volume in a normal, healthy, young adult measures approximately between which of the following values? 13.0-16.0 microliters 6.0-8.0 microliters 17.0-20.0 microliters 2.0-5.0 microliters 9.0-12.0

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

The visual acuity of a 77 year-old female patient with age-related macular degeneration is 2/16 in the right eye on the ETDRS chart. Why is this chart useful in monitoring the response to treatment with anti-vascular endothelial growth factor (VEGF)? A three-line decrease represents a factor of a two time decrease in the size of the letters Each line is 1.0 log units larger than the previous line The Snellen construction of the chart enables the examiner to quickly note that a two-line increment represents a factor of a two time increase in the size of the letters Each line has 5 Sloan letters throughout the chart with equal spacing and is 1.26 times larger than the line below it; each line is .1 log units larger than the line below it when moving up the chart

The ETDRS chart is a logarithmic eye chart modeled after the Bailey-Lovie chart. It is the primary standardized eye chart used in evaluating the visual acuity of low vision patients. The ETDRS charts are logMAR (log of the minimum angle of resolution) in design and are constructed with 10 Sloan sans serif letters. Each line is 1.26 times larger than the line below, and the construction of each line is such that the difficulty is theoretically equivalent on every line. The construction of the ETDRS chart is made to eliminate the inherent errors in the measurement of visual acuity found in the traditional non-standardized Snellen test charts. The Snellen test charts have variations in legibility of different letters as well as differences in the spacing between the lines of letters and between adjacent letters on single lines. The ETDRS logarithmic chart is constructed in such a way that each line of letters is 0.1 log units (about 1.26 times) larger than the previous line. This is a geometric progression.

Which layer of the cornea, if penetrated, will leave a scar? The tear film The epithelium The wing cell layer The stroma

The corneal epithelium is comprised of 3 major layers. The outermost layer is composed of superficial cells (2-3 layers) followed by wing cells (2-3 layers) and, lastly, basal cells (1 layer). Damage to the epithelium will heal without keloid formation. The epithelial basement membrane is made up of collagen types IV, VII and XII. The stroma makes up the bulk of the cornea and is comprised of keratocytes, nerves, type I collagen fibers and mucopolysaccharides. If injured, the stroma will heal but a scar will remain at the site of trauma. The tear film lies anterior to the cornea and is not composed of tissue and as such cannot scar, nor is it considered a part of the cornea.

Congenital cataracts can be caused by a viral infection of the mother with rubella virus (German measles) during development of the primary lens fibers. At which time period in embryonic development can infection cause congenital cataracts? Conception 1st trimester 3rd trimester 2nd trimester Post-delivery

The developing lens is susceptible to rubella virus when the lens fibers are forming, which occurs around weeks 4-7 of gestation. Earlier infection will occur prior to lens fiber development, and the lens is resistant to later infection because the virus is unable to penetrate the lens capsule. The fetus is most susceptible to lenticular damage during the first trimester. Contraction of the rubella virus will cause the greatest amount of damage during this time period. Congenital cataracts are usually detectable at birth but may be seen later because the virus can persist in the lens.

What is the minimum thickness necessary for an antireflective coating (n=1.9) to be useful against incident light of 530 nm wavelength? 58.3 nm 69.7 nm 139.5 nm 132.5 nm 278.9 nm

The equation for finding the minimum antireflective coating thickness is: thickness = wavelength/(4 x index of coating) thickness = 530 / (4 x 1.9) thickness = 530 / 7.6 thickness = 69.73 nm

Which of the following is the correct pathway for the drainage of tears through the nasolacrimal drainage system? Nasolacrimal duct, lacrimal sac, valve of Hasner, lacrimal canaliculus, ampulla, lacrimal punctum Lacrimal punctum, lacrimal canaliculus, ampulla, lacrimal sac, nasolacrimal duct, valve of Hasner Lacrimal sac, lacrimal punctum, lacrimal canaliculus, ampulla, nasolacrimal duct, valve of Hasner Lacrimal punctum, lacrimal canaliculus, ampulla, valve of Hasner, nasolacrimal duct, lacrimal sac

The lacrimal punctum is a small aperture located in the lacrimal papilla, the slight elevation at the junction of the lacrimal and ciliary portions of the eyelid margin. Initially, the tear film drains through this aperture. The lacrimal punctum leads into the lacrimal canaliculus, the tube connecting the punctum to the lacrimal sac. The ampulla is a slight dilation in the initial portion of the lacrimal canaliculus. The canaliculi from the upper and lower lids run horizontally along the lid margin, connecting into a common canaliculus that then enters the lateral aspect of the lacrimal sac located in the anterior portion of the medial orbital wall. The lacrimal sac empties into the nasolacrimal duct in the maxillary bone. The valve of Hasner is located at the terminus of the nasolacrimal duct in the inferior nasal meatus. The Valve of Hasner is a fold of mucosal tissue that ensures that fluid flows anterograde out of the duct. Ref: Remington, LA. Clinical Anatomy of the Visual System, 1998 pp 153-154

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

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

You are measuring the palpebral fissure height in a patient reporting drooping of his upper eyelid. Which of the following BEST describes the normal positioning of the upper and lower eyelids in comparison to the limbus? The upper lid normally rests about 2mm lower than the upper limbus, and the lower lid rests about 1mm lower than the lower limbus The upper lid normally rests about 2mm lower than the upper limbus, and the lower lid rests about 1mm above the lower limbus The upper lid normally rests about 1mm lower than the upper limbus, and the lower lid rests about 2mm above the lower limbus The upper lid normally rests about 1mm lower than the upper limbus, and the lower lid rests about 2mm lower than the lower limbus

The palpebral fissure height is a measurement of the distance between the upper and lower eyelid margins when the patient is looking in primary gaze. This particular measurement is typically less in males (7-10mm) as compared to females (8-12mm). The normal positioning of the upper and lower eyelids are as follows: the upper eyelid usually rests about 2mm below the superior limbus, while the lower eyelid position is typically 1mm above the lower limbus. A unilateral ptosis can be quantified by comparing these measurements to the contralateral eye. A ptosis up to 2mm may be graded as mild; a 3mm ptosis is considered moderate; a ptosis of 4mm or more is deemed severe. Another important measurement in evaluating a ptosis is the marginal-reflex distance (MRD). The MRD can be defined as the distance between the upper eyelid margin and the resultant corneal reflection caused by directing a patient's gaze at a penlight held by the examiner. This measurement is normally 4-4.5mm.

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

The primary Jones dye test can be utilized to determine the patency of the nasolacrimal system. 1-2 drops of fluorescein are instilled into the inferior fornix of the eyes while the patient is in an upright position and blinking her eyes normally. After a period of 5 to 10 minutes, a cotton-tipped applicator is used to swab the undersurface of the inferior turbinate on each side of the nasal passage. When the primary Jones dye test is positive (dye is recovered from the inferior turbinate of the nose), practitioners may conclude that the system is patent and that no significant blockage of the nasolacrimal drainage structure is likely. However, minor stenosis or physiologic dysfunctions cannot be completely ruled out. Patients who have a positive result on the Jones I test are more likely to experience symptoms of epiphora that are secondary to primary oversecretion of tears, rather than a dysfunction in lacrimal drainage (as in the above question). When the primary Jones dye test is negative, the probability of an obstruction or dysfunction in lacrimal drainage is much greater; however, this test alone is not sufficient to document this conclusion. The secondary Jones dye test is then necessary to determine the severity and location of the obstruction.

During gestation, when does the secondary vitreous begin to develop? The 20th week of gestation The 30th week of gestation The 9th week of gestation The 1st week of gestation

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

A cortical hypercolumn is comprised of which of the following? An ocular dominance column for one eye only and orientation columns for one specific orientation only An ocular dominance column for one eye only and a complete set of orientation columns Ocular dominance columns for both eyes and orientation columns for one specific orientation only Ocular dominance columns for both eyes and a complete set of orientation columns A complete set of ocular dominance columns for one eye only and a complete set of orientation columns

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

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

The trochlear nerve (CN IV) innervates the superior oblique muscle. The abducens nerve (CN VI) innervates the lateral rectus, which is not involved in the motion described in this question. The oculomotor nerve has two divisions; the inferior division innervates the inferior rectus, inferior oblique and medial rectus, while the superior division innervates the superior rectus and levator palpebrae superioris. The functions and anatomy of the extraocular muscles are as follows: Superior rectus - turns eye up, adducts, and medially rotates (intorsion) Inferior rectus - turns down, adducts, and laterally rotates (extorsion) Lateral rectus - abducts eye (laterally) Medial rectus - adducts eye (medially) Superior oblique - medially rotates (intorsion), abducts and turns eye down Inferior oblique - laterally rotates (extorsion), abducts and turns eye up In the case described here, when the patient adducts the eye medially with the medial rectus as well as the superior and inferior rectus, only the superior oblique and inferior oblique can move the eye down or up respectively because the superior and inferior rectus muscles are already contracted. The same is true if a patient abducts the eye with the obliques and lateral rectus: only the superior and inferior rectus can move the eye up or down respectively.

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

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

The ligaments that suspend the lens (zonules) are embryonically derived from what structure? The lens capsule The tertiary vitreous The lens epithelium The primary vitreous

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

Which of the following alterations will help to loosen a tightly-fitting gas-permeable lens? Steepen the base curve of the lens Increase the overall diameter Steepen the peripheral curve system Reduce the width of the peripheral curve system Reduce the size of the optic zone

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

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

There are four Purkinje images. The first image is caused by reflection from the anterior corneal surface and is the brightest of the images. The first image is roughly the same size as the object. The second Purkinje image is formed by the posterior surface of the cornea and almost coincides with the first Purkinje image. The third Purkinje image is the largest and is caused by reflection off of the anterior plane of the crystalline lens. The fourth Purkinje image is the smallest and is inverted, formed by reflection off of the posterior surface of the lens. During the process of accommodation the anterior surface of the lens moves forward. The image that is reflected off of this surface is Purkinje III. Purkinje image III will be seen to move forward during accommodation.

A person who is missing the photopigment chlorolabe is categorized as which of the following? A protanope A deuteranomalous trichromat A tritanope A protanomalous trichromat A deuteranope

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

A Galilean telescope has an ocular lens with a power of -32.00 D and an objective lens with a power of +8.00 D. What is the magnification provided by the telescope? 256x 8x 0.25x 4x

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

What is the equivalent of a Reduced Snellen 20/50 optotype in metric notation (assuming a working distance of 40 cm)? 2M 1M 0.5M 0.67M

To convert from Reduced Snellen to metric notation one must divide the denominator by 50. In the above example 50/50 = 1M. To convert from Reduced Snellen to Printer's point, divide the denominator by 6. To convert from Printer's point to metric, divide by 8. To convert from Metric notation to Reduced Snellen, multiply by 50; this will give you the denominator of Reduced Snellen. A good rule of thumb is 1M = RS 20/50 = 8 point. https://www.optoprep.com/simboards/qod/dailydose-question.jsp?userId=PmTcfyyfQCtjxP202754&questionId=FvgPZqTQFlksXwWy2726&answerOrder=1023&historyId=9260171&answerId=tBmYxVAFllSUAiKi2727&

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?

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 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? Gastrointestinal discomfort Headache Depression Tachycardia Visual Hallucinations

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.