timed test #1 Flashcards

1
Q

Which organism represents the MOST common etiology for acute bacterial conjunctivitis in adults?

 Staphylococcus aureus   
 Haemophilus influenza  
 Chlamydia trachomatis  
 Streptococcus pneumoniae  
 Moraxella catarrhalis
A

Staphylococcus aureus

Explanation
Acute bacterial conjunctivitis in adults is primarily due to Staphylococcus aureus, Streptococcus pneumoniae, and Haemophilus influenza, with Staphylococcus aureus representing the most common etiology.
In children, the most common causes of acute bacterial conjunctivitis are Haemophilus influenza, Streptococcus pneumoniae, and Moraxella catarrhalis (with Haemophilus representing the most common etiology).
Hyperacute conjunctivitis is primarily due to Neisseria gonorrhoeae, and chronic conjunctivitis is primarily caused by Chlamydia trachomatis.

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

The measure of how tightly a drug binds to its receptor is known as which of the following?

Potency
Affinity
Efficacy
Reliability

A

Affinity

Explanation
The affinity of a drug for a receptor is a measure of how tightly and how able the drug in question binds to the receptor. Affinity is typically determined by the chemical structure of the drug.

The efficacy of a drug is a measure of the drug’s ability to produce a biological effect or initiate a biological change once it is bound to the receptor. Efficacy is used to characterize the level of maximal response by a drug.

It is important to note that drug affinity and efficacy do not go hand-in-hand. A drug may have a high affinity to a receptor but have minimal efficacy and vice versa.

The potency of a drug refers to the amount of a drug that is required to achieve a desired biological effect. A drug that requires a lower dose is considered more potent than a drug that requires a higher dosage to have the same effect.

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

You wish to perform a visual field on a patient with advanced glaucoma using a Humphrey Field Analyzer. Which of the following visual field programs will BEST allow for monitoring of changes to or progression of visual field loss?

40 point screener
30-2
24-2
10-2

A

10-2 Correct Answer

Explanation
A patient with advanced glaucoma should be monitored using a threshold 10-2 test which tests 68 test points, 2 degrees apart in a central 10 degree area. Patients with advanced glaucoma possess a central island of vision, and therefore it is important to monitor for change within this region. A 40 point screener, 30-2 or 24-2 threshold visual field presents too many test points that lay outside of the patient’s area of useable vision and therefore do not provide useful information. A 30-2 field tests 76 points, 6 degrees apart within a central 30 degree radius. A 24-2 field presents a total of 54 points within a central 24 degree region (except nasally where it extends out to 30 degrees). When treating patients with advanced glaucoma, the objective is to save the central acuity. If the glaucoma is not adequately controlled, progression will cause a splitting of fixation due to inward advancement of the superior nasal field defect followed by inferior field loss due to encroachment of the nasal defect.

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

How will over-correction of myopia change the observed near phoria?

Increase exophoria and increase the esophoria measured

Decrease esophoria or increase the exophoria measured

The resultant phoria will not change

Increase esophoria or reduce the exophoria measured

A

Increase esophoria or reduce the exophoria measured

Explanation
Over-correction of myopia stimulates accommodation. As the viewing distance decreases, the eyes will converge reflexively (accommodative convergence). In addition to the expected accommodative convergence, there will be increased convergence as the eye accommodates to bring the target in focus compensating for the distance over-correction. This over-convergence will either cause increased esophoria or reduce the amount of exophoria present.

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

The circulatory system of the retina is capable of autoregulation. This property is attributable to which cell?

Glial cell
Pericyte
Astrocyte
Schwann cell

A

Pericyte

Explanation
Pericytes are smooth muscle cells that are generally adhered to small blood vessels. They are found at the junction between arterioles and capillaries. Their muscle action alters the flow of blood through the retina’s microcirculation. Pericytes are not found in the choroid.

In poorly controlled diabetics, the pericytes become damaged causing leakage through the microvasculature as seen in diabetic retinopathy.

Glial cells are non-neuronal cells that serve several functions, depending on the type of glial cell. On the whole, glial cells are responsible for structural and nutritional support as well as insulating and protecting neurons.

Astrocytes are a type of glial cell that serve many functions including the potential to regulate blood flow, metabolic support, structural support of the neural cells and formation of the blood-brain barrier.

Schwann cells are also a type of glial cell. These cells are primarily responsible for the production of the myelin sheath in the peripheral nervous system.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

A colored filter is placed in front of an illuminated Tungsten light bulb, resulting in emergent light that appears red. Which of the following wavelengths is LEAST likely to be absorbed by the filter?

 592 nm   
 648 nm   
 401 nm  
 518 nm  
 471 nm
A

648 nm

Explanation
Colored filters are used to block (absorb) some wavelengths while transmitting others, resulting in the perception of transmitted light that is a different color than the light that was incident upon the filter. A red filter that is placed in front of a white light source will strongly absorb the majority of the wavelengths, except for those in the red region (roughly 620 to 700 nm).

The following are wavelengths and their associated color:
Red= 650nm
Orange= 600nm
Yellow= 570nm
Green= 520nm
Blue= 470nm
Violet= 400nm
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

Diurnal refractive fluctuations are MOST likely to be experienced by a patient who has undergone which of the following procedures?

 Photorefractive keratectomy (PRK)  
 Radial keratotomy (RK)   
 Laser-assisted in situ keratomileusis (LASIK)  
 Conductive keratoplasty (CK)
A

Radial keratotomy (RK)

Explanation
Radial keratotomy is performed by creating corneal incisions in a radial fashion, which serves to flatten the cornea, thereby decreasing the amount of myopia. However, many post-RK patients experience irregular astigmatism, dry eyes, and diurnal refractive error fluctuation. Studies have suggested that post RK patients suffer from unstable refractive errors secondary to normal variations in intraocular pressure (IOP) that typically occur throughout the day. Because the structure of the cornea has been surgically weakened, it is more susceptible to refractive instability due to an inability to compensate for IOP fluctuations. Treatment options include hard and soft contact lenses, hybrid lenses, several pairs of glasses with different prescriptions depending on refractive error and time of day, and suturing of RK incisions and lasso suture (circular suture that parallels the limbus to help stabilize the cornea). Other treatment options include the pharmaceutical use of brimonidine (off-label), which causes pupil miosis and helps to relieve glare and haloes, corneal cross-linking to help stabilize the cornea, and controlling IOP fluctuations via the use of topical prostaglandins (off-label).

Patients who have undergone LASIK, PRK and CK are much less likely to experience refractive variability throughout the day.

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

While attempting to determine the amplitude of accommodation via the push up method on your patient, she reports that the target is blurry at a distance of 14 cm for the right eye and at 13 cm for the left eye. What is her amplitude of accommodation in diopters for the right eye and the left eye respectively (to the nearest 0.25 D)?

  1. 14 D and 7.69 D
  2. 25 D and 7.75 D
  3. 725 D and 0.775 D
  4. 714 D and 0.769 D
A

7.25 D and 7.75 D

Explanation
In order to determine the amplitude of accommodation, one must take the reciprocal of the blur point in meters. For the above example, 1/0.14 = 7.14 and 1/0.13 = 7.69. Do not forget to convert to meters!! Also, the question asked that the answer be rounded to the nearest 0.25 D. Be sure to read questions carefully to avoid careless mistakes. Clinically, one always rounds to the nearest 0.25 D.

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

Which 2 of the following alterations to the cornea commonly occur after several years of contact lens wear? (Select 2)

The cornea becomes less sensitive to touch
The cornea becomes more sensitive to touch
The cornea epithelium thins
The corneal epithelium thickens

A

The cornea becomes less sensitive to touch
The cornea epithelium thins

Explanation
Long-term extended wear of contact lenses has been shown to lead to thinning of the corneal epithelium. Additionally, the corneal nerves exhibit a decrease in sensitivity over time.

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

Which of the following properly describes the movement of sodium and potassium ions in one cycle of the Na-K ATPase pump?

3 sodium out, 2 potassium in
2 sodium out, 3 potassium in
2 sodium in, 3 potassium out
3 sodium in, 2 potassium out

A

3 sodium out, 2 potassium in Correct Answer

Explanation
The sodium potassium ATPase pump, an integral membrane protein, is central to the functioning of most transporting epithelia. It is normally open to the cytoplasmic side with three Na binding sites. When those three are occupied, ATP phosphorylates the protein, prompting a change in configuration that now opens the protein’s binding sites to the outside, where the Na ions exit. The protein remains in its configuration until its two K binding sites are occupied. This causes dephosphorylation of the protein and reversal of the protein configuration so that it is again open to the inside of the cell. The 2 K ions enter the cytoplasm, completing one pump cycle.

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

What is the correct mechanism of action employed by fluoroquinolones that enables them to be strong antimicrobial agents?

Inhibition of cell wall synthesis
Inhibition of bacterial DNA gyrase and topoisomerase IV
Blocks elongation of peptidoglycan
Interruption of protein synthesis

A

Inhibition of bacterial DNA gyrase and topoisomerase IV

Explanation
Fluoroquinolones, such as Vigamox® and Zymar®, are effective antibacterial agents because they inhibit DNA gyrase and topoisomerase, and therefore ultimately inhibit bacterial DNA synthesis. Penicillin-based drugs inhibit cell wall synthesis.

Aminoglycosides, such as tobramycin, interfere with protein synthesis by binding to bacterial ribosomes. Vancomycin disrupts cell wall synthesis by binding to pentapeptides, thereby inhibiting elongation of peptidoglycan, which is necessary for cell wall synthesis.

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

Which 3 of the following represent common symptoms associated with hyperthyroidism? (Select 3)

 Heart palpitations   
 Hair loss  
 Weight loss   
 Dry skin  
 Heat intolerance   
 Elevated cholesterol
A

Heart palpitations
Weight loss
Heat intolerance

Explanation
These are the most frequently-observed symptoms of hyperthyroidism: 
-Heart palpitations 
-Fatigue 
-Tremors 
-Anxiety 
-Disturbed sleep 
-Weight loss 
-Heat intolerance 
-Sweating 
-Polydipsia
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

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

 1st trimester   
 Post-delivery  
 3rd trimester  
 2nd trimester  
 Conception
A

1st trimester

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

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

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

Which of the following 3 activities is associated with a decrease in intraocular pressure? (Select 3)

 Coughing  
 Blinking  
 Alcohol consumption   
 Lying down  
 General anesthesia   
 Exercise
A

Alcohol consumption
General anesthesia
Exercise

Explanation
Exercise, general anesthesia, alcohol consumption, and marijuana use cause a temporary decrease in intraocular pressure (IOP). Transitioning from sitting to a lying down position, blinking, coughing, blepharospasm and tobacco usage may cause an increase in IOP.

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

An 8-year old Hispanic male comes in with the following habitual prescription: +0.75-0.75 x 180 OU. His entering visual acuity with correction was 20/40 OD, OS, OU at both distance and near. After cycloplegic examination, you find the following on wet retinoscopy:
OD: +4.00-2.75 x 180 20/25 (Distance)
OS: +4.50-2.75 x 180 20/25 (Distance)
Which of the following would be the most appropriate treatment for this patient?

OD: +3.50-2.75x180
OS: +3.50-2.75x180
Return for follow-up appointment in 4-6 weeks

Continue with current habitual prescription; alternate patching therapy 3hrs/day 5 days/week. Return for follow-up appointment in 4-6 weeks.

OD: +0.75-2.75 x 180
OS: +0.75-2.75 x 180
Return for follow-up appointment in 1 year

Spectacle prescription
OD: +4.00-1.00 x 180
OS: +4.00-1.00 x 180
Return for follow up appointment in 4-6 weeks

Spectacle prescription:
OD: +3.25-2.75 x 180
OS: +3.75-2.75 x 180
Return for a follow-up appointment in 4-6 weeks

A

Spectacle prescription:
OD: +3.25-2.75 x 180
OS: +3.75-2.75 x 180
Return for a follow-up appointment in 4-6 weeks

Explanation
In this case, poor entering visual acuities were attributed to refractive amblyopia in both eyes. The patient was not wearing his full potential prescription. When deciding on the spectacle prescription, one should keep in mind that the prescription should have full cylinder and no less than half of the hyperopic prescription. For full visual acuity potential, the retinal images should be equal and balanced, hence keeping the +0.50 DS difference between the two eyes. It is important to follow up on these types of prescription changes in case the patient presents with more hyperopia after adapting to his/her new spectacle prescription. Part-time patching is not warranted at this initial visit as his visual acuity may improve drastically by correcting his refractive error alone, and his refractive error is relatively equal in both eyes.

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

During which phase of mitosis do chromosomes line up at the cellular equator?

 Metaphase   
 Telophase  
 Prophase  
 Anaphase   
 Interphase
A

Metaphase

Explanation
The first stage of mitosis is prophase. During early prophase, DNA begins to condense inside of a nuclear envelope. During late prophase, the process involves further condensation of the chromosomes, the nuclear envelope, and the movement of one of a pair of centrioles to the other pole of the cell. During the transition into metaphase (sometimes called prometaphase), the centrioles send out microtubules, forming a spindle apparatus which attaches to sister chromatids of each chromosome. The microtubules attach at a site called a kinetochore on the centromere. At this time, the nuclear envelop has completely disintegrated. Metaphase is marked by the chromosomes aligning along the cellular equator, guided by the microtubules. Anaphase follows. During this phase the sister chromatids of each chromosome are pulled apart and are directed to opposite poles towards the centrioles by the microtubules. This is now telophase. During this stage, the chromosomes are no longer attached to the microtubules and begin to become less condensed. Nuclear envelopes also form around the de-condensing chromosomes, separating the DNA from the cytoplasm. At this point, cytoplasmic division begins and is generally complete before the end of telophase. Both new cells are diploid and contain the same number of chromosomes as the parent cell. The cells then transition into interphase, in which the DNA is duplicated and the cytoplasm and its components increase in number. Interphase is the longest portion of the cell cycle and is not a part of mitosis. After interphase, the cell will then either move on to mitosis, or it will no longer divide.

17
Q

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

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

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

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

Serum creatinine

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

Blood urea nitrogen (BUN)

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

Glomerular filtration rate (GFR)

  • Measurement of how well the kidneys are removing waste and excess fluid from the blood
  • It may be calculated from serum creatinine level using patient age, weight, gender, and body size
  • Normal GFR values are 90 or above; a GFR below 60 indicates kidney disease; GFR below 15 indicates that treatment for kidney failure is necessary
18
Q

Neuroglial cells are important for neuronal protection and support. Which of the following glial cells is responsible for wrapping cellular axons in the CENTRAL nervous system with myelin sheath?

 Microglia  
 Oligodendrocytes   
 Ependymal cells  
 Schwann cells  
 Astrocytes
A

Oligodendrocytes

Explanation
Oligodendrocytes are non-neuronal cells of the central nervous system and serve to coat axons in myelin sheath, which greatly increases the speed of conduction. Oligodendrocytes wrap numerous axons simultaneously, whereas Schwann cells insulate on a one-to-one ratio the cellular axons of the peripheral nervous system.

Astrocytes are the most numerous and offer support and structure to the brain. Astrocytes also play a large role in the formation of the blood-brain barrier, which serves to inhibit toxic substances of the blood from entering the brain. The cells also play an important role in removing neurotransmitters from synaptic zones as well as removing extracellular potassium excess.

Microglial cells are a type of macrophage that are called into action in the event of injury, disease or infection and are capable of phagocytosis.

Ependymal cells are found lining cavities of the central nervous system as well as the walls of the ventricles of the brain. Collectively, these cells form the epithelium of the choroid plexus, which secretes cerebrospinal fluid.

19
Q

When analyzing a rigid gas-permeable (RGP) contact lens, you measure base curves of 7.71 (43.75) and 8.44 (40.00) with a radiuscope, and -4.25 and +2.50 with lensometry. What type of toric gas-permeable contact lens design do you have?

 SPE Bitoric  
 Front Surface (F1) Toric  
 CPE Bitoric   
 Spherical   
 Toric Base Curve
A

CPE Bitoric

Explanation
The first measurement that should be made when analyzing RGP contact lenses is lensometry. Lenses should be placed concave side down on the aperture stop and the power and axis wheels should be rotated until mires are clear and lined up. If cylinder is present, powers should be recorded in both major meridians (record gross amounts). After lensometry, lenses should be analyzed using the radiuscope. Again, once the sharpest focus of mires is found, both base curves should be measured and recorded. Base curves should be converted from millimeters to diopters. The steep meridian is recorded over the flat meridian, and the most plus meridian goes with the flat meridian.
For the above patient, the measurements are as follows:
43.75 -4.25
40.00 +2.50
The next step is to determine the values of the difference in base curves and the difference in contact lens power. For this patient, the difference in base curves is 3.75D and the difference in contact lens power is 6.75D.
- If the difference in base curve is multiplied by 3/2 and that value is equivalent to the contact lens power then you have a Toric Base Curve lens (with spherical front surface)
o For this patient (3/2) x (3.75) = 5.62 and difference in CL power = 6.75
o 5.62 is not equivalent to 6.75 (This lens is not a toric base curve design)
o Note that “equivalent” is defined as less than or equal to 0.50D
- If the difference in base curve is equal to the difference in contact lens power then you have a spherical power effect (SPE) bitoric lens design
o For the above patient difference in base curve is 3.75D and difference in CL power is 6.75D
o 3.75 is not equivalent to 6.75 (This lens is not an SPE bitoric design)
o Again, note that “equivalent” is defined as less than or equal to 0.50D
- If the difference in base curve is not equivalent to the difference in CL power, and 3/2 x the difference in base curve is also not equal to the difference in CL power then you have a cylinder power effect (CPE) bitoric lens
o This lens follows these rules; therefore, it can be considered a CPE bitoric lens design

20
Q

Which of the following tumors is considered the MOST common human malignancy?

 Basal cell carcinoma   
 Kaposi sarcoma  
 Squamous cell carcinoma  
 Sebaceous gland carcinoma  
 Malignant melanoma
A

Basal cell carcinoma

Explanation
Basal cell carcinoma is that most common human malignancy. About 90% of all cases present on the head and neck, with 10% of those involving the eyelid. Basal cell carcinoma is also the most prevalent malignant eyelid tumor, accounting for close to 90% of all cases. Most occur in the lower eyelid, followed by the medial canthus, upper eyelid, and lateral canthus.

Squamous cell carcinoma is much less common than basal cell carcinoma, but it is potentially more aggressive due to a higher risk of regional lymph node metastasis.

Sebaceous gland carcinoma is a very rare tumor that typically arises from the meibomian glands. Malignant melanomas rarely occur on the eyelids but are potentially lethal; Kaposi sarcoma is a vascular tumor that most commonly affects patients infected with AIDS.