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Flashcards in quiz 3 Deck (28):
1

The figure below represents the double reciprocal plot of an enzyme-catalyzed reaction carried out at different substrate concentrations. What is the calculated Vmax of the reaction?

A.0.1 M/min

B.0.25 M/min

C.1 M/min

D.2.0 M/min

E.2.5 M/min

Q image thumb

In the above double-reciprocal plot, the Y-axis intercept represents 1 / V max, which has the value 0.4 µM/min. Since 1 / V max = 0.4 V max = 2.5 µM/min Therefore the correct answer is “e”

2

The human androgen dihydroxytestosterone directly interacts with androgen receptors in the cytoplasm and relocalizes them to the nucleus, where binding to hormone receptor elements on the chromosomal DNA allows transcription of several genes that are responsible for maintenance of the secondary sexual characters. If the cellular concentration of the hormone reaches 4 times the value of its dissociation constant, what fraction of the receptors remain hormone-bound?
 

 A.0.2
 B.0.4
 C.0.6
 D.0.8
 E.1.0
 

D

B = B max [H] / (K d + [H]), where B is bound hormone, [H] is hormone     concentration, and K d  is the dissociation constant for hormone-receptor binding.               For determining fractional saturation, the equation can be rearranged to
B/ B max =[H] / (K d + [H]).
We know that [H] = 4K d
Therefore, B/ B max = 4K d / (K d + 4K d) = 4K d / 5K d = 4/5 = 0.8
Therefore the correct answer is “d”
 

3

If an enzyme-catalyzed reaction (E + S ES E + P) has a starting substrate concentration of 5 mM and initial velocity (V0) of 25% Vmax, the calculated Km for the substrate is

 A.2.5 mM.
 B.5 mM.
C.15 mM.
 D.20 mM.
 E.25 mM.

Answer Key: C

Feedback:
The reaction scheme above shows Michelis-Menten kinetics:
                V 0 = V max [S] / (K m + [S])
This expression can be rearranged to
                V 0 / V max = [S] / (K m + [S]) = fractional Vmax = %Vmax/100 = 25/100 = 1/4
               
Plugging into the formula and solving for Km 
                1 / 4   = 5 / (K m + 5)
                 (K m + 5) = 4 x 5 = 20 – 5 = 15
                Therefore the correct answer is “c”

4

The use of alcohol-based hand sanitizers by hospital care givers has been blamed in part for an increased incidence of hospital-acquired Clostridium difficle diarrhea. The lack of efficacy of these sanitizers in controlling the spread of this disease is MOST LIKELY attributable to their poor killing of bacteria that are

 A.anaerobic.
 B.encapsulated.
 C.Gram-positive.
 D.multi-drug resistant.
 E.spore formers.

Answer Key: E

Feedback: Explanation: Many forms of sanitization and disinfection are active against vegetative cells, but only the most rigorous forms of disinfection, such as autoclaving, will kill the inert, thick-walled spore. While alcohol does kill many types of organisms, it is not effective at penetrating the spore or persisting long on surfaces do to evaporation. Answers A and C are true of C. difficile, it is anaerobic and Gram-positive, but those attributes do not explain resistance to killing. Answers B and D. do not pertain to C. difficle, nor do they contribute to resistance to killing with disinfectants.

5

In a study of patients admitted to Walter Reed after blast injuries in combat, researchers found that length of hospitalization (measured in days) followed a skewed distribution with most patients discharged within 60 days but a few extremely long hospital stays (outliers).  Based on this information, what is the best way to summarize length of hospitalization?

 A.mean (standard deviation)
 B.median (interquartile range)
 C.mode (range)
 D.percent (number of patients)

a. incorrect: the mean will be highly influenced by the outliers in the data and will not reflect a typical hospitalization

b. correct: the median is an appropriate summary for quantitative data, and is not likely to be influenced by a few outliers

c. incorrect: the mode is the most common value, and is generally useful for summarizing data that can only take on a small number of values.

d. incorrect: percents are useful for summarizing categorical data, but length of hospitalization is quantitative.

6

Valves that control the flow of fluid in vessels are found in:

 A.only lymphatic vessels
 B.only capillaries
 C.only veins.
 D.only arteries.
 E.both arteries and veins.
  F.both veins and lymphatic vessels.

F

7

In primary hemostasis, which of the following is central to forming the primary hemostatic plug?

 A.Reflex vasoconstriction secondary to injury.
 B.Platelets binding to the site of injury, platelet activation and recruitment.
 C.Activation of the coagulation cascade via tissue factor.
 D.Activation of the coagulation cascade via the Hageman factor.
 E.Production of insoluble fibrin to firmly bind the platelets.
 F.Activation of protein C and S, which will inactivate factors Va and VIIIa.

Answer Key: B

Feedback:

Answer:
Platelets are the central process in primary hemostasis. Von Willebrands factor binding to the extracellular matrix allows for platelets to bind to bind to the site of injury. Transient vasoconstriction is the first step in the process. After the platelets bind (primary hemostasis), the coagulation cascade (c, d and e above) allows for the formation of the secondary hemostatic plug (secondary hemostasis). Once hemostasis is initiated, it needs to be tightly controlled, and part of this process includes activation of protein C and S.
 

8

A clinician orders a Gram stain and culture on a throat swab taken from a 3 year old with pharyngitis to rule out strep throat (caused by the Gram positive coccus,Streptococcus pyogenes) . You are the lab director and suggest to the physician that a Gram stain will not help her because

 A.bacteria are too small to be seen by light microscopic examination of a 
Gram strain.
 B.bacteria in the throat cannot be Gram stained.
 C.many normal flora organism in the throat look like the agent of strep throat.
 D.most of the pharynx is sterile, and, therefore it will be hard to find 
anything on a Gram stain.
 E.saliva in the throat interferes with the Gram stain reaction.

Answer Key: C

Feedback:

ANS: C
Explanation: A is incorrect because most bacteria can be seen on Gram stain (they are not too small). B is incorrect because normal flora bacteria in the throat can be seen on Gram stain. D is wrong because the pharynx is not sterile. E is incorrect because saliva does not affect the Gram stain. C is correct.

9

An otherwise healthy 92-year-old female became bedridden for three months due to a pelvic fracture and developed respiratory distress. Functional lung studies disclosed localized changes consistent with decreased perfusion. Which of the following statements is most likely?

 A.Turbulence is the most important factor in the pathogenesis of this process.
 B.Stasis is a major contributor in the development of this process.
 C.Endothelial injury is the most important pathogenetic factor.
 D.Hypercoagulability is the most likely contributor factor since the process is systemic (affects pelvis and lungs).
 E.Deficiencies of anticoagulants such as antithrombin III, protein C or protein S should be ruled out as soon as possible.

Answer Key: B

Feedback: a. Incorrect. Turbulence contributes to arterial and cardiac thrombosis by causing endothelial injury or dysfunction. 
b. Correct. Stasis is a major contributor in the development of venous thrombi. 
c. Incorrect. Endothelial injury is important for thrombus formation in the heart or arterial circulation where high flow rates may impede clotting by preventing platelet adhesion and washing activated coagulation factors. 
d. Incorrect. Hypercoagulability is a less frequent contributor to thrombosis than endothelial injury or alterations of normal blood flow.
e. Incorrect. Deficiencies of anticoagulants such as antithrombin III, protein C or protein S typically present with venous thrombosis and recurrent thromboembolism starting in adolescence or early adulthood. 

10

You are preparing your patient with leukemia to receive a bone marrow transplant. The donor marrow sample must be treated to isolate the stem cells. The enrichment process should result in cells that are strongly positive for which of the following proteins?

 A.HLA DR
 B.CD34
 C.fibronectin
 D.epinephrine receptor
 E.fibrinogen

Answer Key: B

Feedback: Correct Ans B. A is incorrect because HLA DR expression increases as the cell differentiates, so cells with high expression are more differentiated. C is incorrect because fibronectin is an extracellular matrix protein. D. Epinephrine receptor is not directly related to differentiation stage. E is incorrect because fibrinogen is asoluble protein in the plasma.

11

Which imaging modality has the highest population-based radiation exposure risk attributed to it based on current practice?

 A.MRI
 B.CT
 C.Radiography
 D.SPECT

a. Incorrect: MRI does not deliver a radiation dose to the patient.

b. Correct: CT exposes patients to relatively large amounts of radiation, compared with other imaging modalities.  This is especially true for some types of CT protocols (eg, whole body CT).  Given the radiation exposure and the frequency with which CT is ordered, it has been identified as a significant cause of iatrogenic harm, mostly due to radiation-induced excess cancer.

c. Incorrect: Although radiography does expose the patient to some radiation it is small compared to the potential dose delivered by CT. 

d. Incorrect: SPECT exposes patients to about half-as much radiation than CT, and it is used much less often in practice.  Thus the excess cancer risk attributed to SPECT is much lower than for CT.

12

Sickle cell anemia results from a point mutation in the oxygen-carrying protein hemoglobin, converting the native structure to a form that can no longer bind oxygen efficiently. Oxygen binding to hemoglobin is mediated through tightly coordinated heme groups. Which one of the following structural motifs is specially designed for heme binding?

 A.Rossman Fold
 B.Leucine Zipper
 C.Globin Folds
 D.ß Barrels
 E.Zinc Fingers

Answer Key: C

Feedback: Globin folds are present in hemoglobin and myoglobin and bind heme tightly for oxygen transport. The Rossman Fold is responsible for nucleotide binding; the leucine zipper allows protein-protein interaction through heptad repeats; the beta-barrel structures are specialized for enzyme active sites; the zinc finger is a specialized structure for DNA binding. Therefore the correct answer is ???c???

13

Assume that diastolic blood pressure of 35-44 year old males is normally distributed with mean = 80 mmHg and standard deviation 10 mm Hg. If we define a “borderline hypertensive” as a person whose diastolic pressure is between 90 and 100 mmHg inclusive, what percent of persons from this population of 35-44 year old men are borderline hypertensive?

 A.2.5%
 B.13.5%
 C.16%
 D.27%
 E.32%

Answer Key: B

Feedback:

a. incorrect: 2.5% of these men have diastolic pressure above 100 mmHg (more than 2 standard deviations above the mean)

b. correct:  68% have diastolic pressure within 1 standard deviation of the mean (between 70 and 90 mmHg), so (100% - 68%)/2 = 16% have diastolic pressure above 90 mmHg.  Of these, (100%-95%)/2 = 2.5% have diastolic pressure above 100 mmHg.  By subtraction, 13.5% have diastolic pressure between 90 and 100 mmHg.

c. incorrect:  68% have diastolic pressure within 1 standard deviation of the mean (between 70 and 90 mmHg), so (100% - 68%)/2 = 16% have diastolic pressure above 90 mmHg.  This 16% includes some men with diastolic pressure above 100 mmHg.

d. incorrect:  68% have diastolic pressure within 1 standard deviation of the mean (between 70 and 90 mmHg), and 95% have diastolic pressure within 2 standard deviations of the mean (between 60 and 100 mmHg).  27% represents the difference between these two groups, i.e. both men between 90 and 100 mmHg and men between 60 and 70 mmHg.

e. incorrect:   68% have diastolic pressure within 1 standard deviation of the mean (between 70 and 90 mmHg), so (100%-68% =) 32% have diastolic pressure more than 1 standard deviation above or below the mean (above 90 OR below 70 mmHg).

14

It is 35 degrees Fahrenheit outside as you stand guard duty outside of the commander's tent at Ft Indiantown Gap. Your fingers are cold and pale. This is a result of:

 A.swelling of pericytes surrounding capillaries in the connective tissue 
of the digits as a result of the cold.
 B.elastic recoil of the tunica media of the muscular arteries of the hand
 C.constriction of smooth muscle of arterioles feeding into the 
capillaries in the digits
 D.impedance of venous flow out of the hand by venous valves.
 E.necrosis of fibroblasts in the connective tissue underlying the epidermis.

Answer Key: C

Feedback:

Correct Ans is C. Smooth muscle layer of arteriole is critical in controlling the flow of blood to organs under different conditions. A. pericytes don't swell in response to the cold, B elastic recoil of arteries smoothes out the pulsatile nature of the circulation but is not a regulatory factor at the level of the fingers. D. there should not be any venous impedance to blood flow here . Veins generally do not regulate blood flow. E. fibroblasts should not be dying under these circumstances and they do not contribute thermoregulation.

15

When a new rapid test for dengue was administered to 292 Vietnamese patients with febrile illness, 151 patients tested positive using the rapid test, and all of these patients were subsequently found to have dengue using the “gold standard” test. 141 patients tested negative using the rapid test, and 94 of these patients were subsequently found to have dengue using the “gold standard” test (BMC Infectious Diseases 2010, 10:142). What is the sensitivity of the rapid test?

 A.33.3%
 B.38.4%
 C.61.6%
 D.100%

Answer Key: C

Feedback: a. incorrect: this is the negative predictive value. From the question, TP = 151, (TP + FP) = 151, FN = 94, (FN + TN) = 141, so FP = 0 and TN = 47. Sensitivity is TP / (TP + FN) 

b. incorrect: this is the false negative rate, FN / (TP + FN) or 1-sensitivity. 

c. correct: From the question, TP = 151, FN = 94. Sensitivity is TP / (TP + FN) = 151 / (151+94) = 61.6%. 

d. incorrect: this is the specificity (also the positive predictive value). From the question, TP = 151, (TP + FP) = 151, FN = 94, (FN + TN) = 141, so FP = 0 and TN = 47. Sensitivity is TP / (TP + FN) 

16

Which of the following correctly identifies Kurt Lewin’s original leadership styles?

  A.Authoritarian, Laissez Faire, Democratic
 B.Dictator, Democratic, Deliberate
 C.Military, Civilian, Political
 D.Despot, Executive, Manager
 E.Taciturn, Angry, Accepting

A

17

A 36-year-old man who has just returned from a safari vacation in Kenya is hospitalized with an acute-onset, debilitating febrile illness that is accompanied by hemorrhagic lesions on this trunk, legs, and arms. The patient dies. Post-mortem blood samples from the patient are transferred to 3 monkeys, each of which succumbs to an illness similar to that of the human. An infectious agent is isolated from the blood that can be seen with an electron microscope but not a light microscope. The agent is passaged in tissue culture, purified , and, on characterization, is found to contain protein and RNA but no DNA. The causative agent of the man's illness is MOST LIKELY a

 A.bacterium
 B.fungus
 C.parasite
 D.prion
 E.virus

E

18

A pregnant 22-year-old female in her first trimester presents to the emergency room with right lower quadrant abdominal pain.  What is your first imaging modality of choice to further evaluate her right lower quadrant pain?

 A.CT
 B.Ultrasound
 C.Fluoroscopy
 D.PET

Answer Key: B

Feedback:

 

a. Incorrect: In a female patient with a first trimester pregnancy, CT would not be your first imaging modality of choice due to radiation concerns.

b. Correct: Ultrasound is the correct modality to pursue in a first trimester pregnancy as it has no radiation exposure to the fetus and has increased accuracy in evaluation of the pelvic structures as well.

c. Incorrect: In a female patient with a first trimester pregnancy, fluoroscopy would not be your first imaging modality of choice due to radiation concerns.  Additionally, it would not provide enough information regarding the right lower quadrant of the abdomen compared to other available modalities.

d. Incorrect: In a female patient with a first trimester pregnancy, PET would not be your first imaging modality of choice due to radiation concerns.

19

A 35yo female USN JAG officer (lawyer) was in her usual state of good health until 2 weeks ago when she developed low grade lower abdominal discomfort that has persisted. There are no other associated symptoms but she has been under a lot of stress recently. She is single, sexually active, and has had multiple partners in the past year. She tends to avoid doctors except for mandated military health requirements, but felt she needed to be evaluated and scheduled an appointment with you. Her opening statement of the interview is, “Lets get this over with!”.

Which of the following approaches is most likely to build rapport and facilitate communication:

 A.Reassure the patient
 B.Probe for more information about her strong statement
 C.Proceed with a standard history and physical
 D.Defer addressing her strong statement until later in the interview
 E.Excuse yourself and review her medical record before proceeding

Answer Key: B

Feedback: Answer: b. “Probe for more information…” When patients express strong emotions it is important to acknowledge such emotion and probe for more background information about their feelings. Visiting the doctor is a highly stressful experience for patients, especially if they have an undifferentiated symptom and are worried about a serious condition. Attending to emotionally charged signs or expressions is associated with enhanced rapport and higher quality interactions which in turn leads to better information transfer, enhanced patient-physician relationships, patient satisfaction, and higher quality of care. All the other choices are dismissive of the patient’s strong expression and lead to an implicit message that her anxiety and fear are unimportant.

20

A 58-year-old female dies 26 hours following a myocardial infarction. The heart has a localized and well demarcated area of softening. As you review the myocardial sections you notice a preserved tissue architecture, increased eosinophilia, karyolysis and karyorrhexis involving the full thickness of the sampled tissue, neutrophils are noted at the periphery. Which of the following statements is correct.

 A.An infarct is an area of fibrinoid necrosis resulting from occlusion of arterial supply or venous drainage.
 B.Nearly all infarcts result from embolic venous occlusions.
 C.Red infarcts occur with venous occlusions, in tissues with dual circulations, or when blood flow is re-established to a site of previous arterial occlusion and necrosis.
 D.White infarcts occur when leukocytes infiltrate previously necrotic tissues.
 E.Coagulative necrosis describes this process best.

Answer Key: E

Feedback: a. Incorrect. An infarct is an area of ischemic necrosis resulting from occlusion of arterial supply or venous drainage. 
b. Incorrect. Nearly all infarcts result from thrombotic or embolic arterial occlusions. 
c. Incorrect. Red infarcts occur with venous occlusions, in tissues with dual circulations, or when blood flow is re-established to a site of previous arterial occlusion and necrosis. 
d. Incorrect. White infarcts occur in organs with end-arterial circulation. 
e. Correct. The dominant histologic characteristic of infarction is ischemic coagulative necrosis. 

21

In a study of the association between prediagnostic IgE levels and risk of glioma, (J Natl Cancer Inst (2012) doi: 10.1093/jnci/djs315), 77 of 594 case subjects (patients with glioma) had elevated IgE in a prediagnostic serum sample. 198 of 1,177 control subjects (patients without glioma) had elevated IgE. The odds ratio for the association between glioma and elevated IgE is:

 A.0.16
 B.0.34
 C.0.74
 D.0.77
 E.0.81

Answer Key: C

Feedback: Answers: 
a. incorrect: this is the risk of exposure among all subjects 
b. incorrect: this is the proportion of cases among all subjects 
c. correct: the odds a case was exposed is 77 / (594 ??? 77) = .149. The odds a control subject was exposed is 198 / (1,177 ??? 198) = 0.202. The odds ratio is .149 / .202 = .74. 
d. incorrect: this is the proportion of cases exposed, divided by the proportion of controls exposed. The odds ratio is the odds a case was exposed divided by the odds a control was exposed. 
e. incorrect: this is the proportion of cases among the exposed divided by the proportion of cases among the unexposed. In a case-control study, the proportion of cases is set by the experimenter and is not a true measure of incidence or prevalence, so this measure is not generally used for case-control studies. 

22

A patient who suffered a below the knee amputation develops fever and gas gangrene at the operative site. The laboratory identifies a bacterial species from the stump wound as the strict anaerobe Clostridium perfringens. Which of the statements is MOST LIKELY to describe Clostridium perfringens.

 A.It produces the enzyme superoxide dismutase.
 B.It produces peroxidase.
 C.Metabolism via the tricarboxylic acid cycle occurs when it is exposed to O2.
 D.Minute amounts of oxygen are beneficial for replication.
 E.Minute amounts of oxygen are lethal to it.

Answer Key: E

Feedback: Explanation: The definition of a strict anaerobe is an organism that is killed by even the slightest exposure to oxygen. This is because such organisms have no way to defend themselves from toxic oxygen species. That is, they cannot inactivate oxygen radicals because they do not produce superoxide dismutase, and they cannot degrade peroxide because they lack peroxidase or catalase enzymes. Some organisms are benefited by small amounts of oxygen and those organisms are termed microaerophilic. 

23

For a leader, which of the following is important to consider?

 A.Self-awareness of biases, beliefs, and attitudes
 B.Verbal and non-verbal communication
 C.Communication style
 D.Interpersonal interactions
 E.All of the above

E

24

A 56-year-old female diabetic came to a clinic with an abscess on her left foot. She was treated with antibiotics and sent home. The next day she came to the hospital emergency room with chills and a spiking fever. She was immediately admitted into the intensive care unit where she developed severe hypotension and rapidly progressed into multi-organ failure. The patient died within 48 hours despite supportive therapy. A Gram-negative bacillus was isolated from her blood and the foot abscess. Which ONE of the following bacterial components was directly responsible for her death?

 A.Capsule
 B.Lipid A
 C.Peptidoglycan
 D.Pili
 E.Teichoic acid

Answer Key: B

Feedback: Explanation: Although she was treated with an appropriate antibiotic, the patient died due to the release of endotoxin from the dying bacteria. The component of the Gram-negative cell wall that mediates shock like symptoms and death is ???endotoxin??? or the lipopolysaccharide (LPS). The active component of LPS is lipid A. Teichoic acid can mediate a shock-like response in patients as well, though less frequently and it is associated with Gram-positive cell walls. Capsules can mediate virulence by blocking phagocytosis, but they do not cause death directly. Peptidoglycan is a non-inflammatory molecule of the cell wall, and pili are adherence factors which are not responsible for severe inflammatory reactions.

25

True totipotent embryonic stems cells are obtained from:

 A.the inner cell mass of a blastocyst.
 B.embryonic testicles.
 C.carcinomas of the ovaries.
 D.endothelium of the liver.
 E.sheep fibroblasts.

Answer Key: A

Feedback: Correct Ans is A. Other cell types are not true embryonic stem cells.

26

Which of the following best describes a biofilm?

 A.a polysaccharide capsule
 B.an adherent community of bacteria
 C.the bridge made by sex pili during conjugation
 D.the outer membrane of a Gram-negative bacterium
 E.the outer surface of a bacterial spore.

Answer Key: B

Feedback: Explanation: A biofilm is an aggregation of bacteria that becomes coated with polysaccharides and metabolic byproducts of bacterial growth such that they grow within a mass that adheres readily to moist surfaces. Within the biofilm, some bacteria may cease to produce flagella which are not necessary within the community. Also, biofilm formation enables the community of bacteria to resist penetration of antimicrobials. 

27


Which of the following substances can readily diffuse across the cell membrane with not transporter being involved?

 A. Acetate 
 B.Water
 C.Cholesterol
 D.Aspartate
 E.Glucose

Answer Key: C

Feedback:
A is a wrong answer. Acetate is a charged molecule and cannot readily diffuse through hydrophobic membrane interior
 
B is a wrong answer. Water is a polar molecule and cannot readily diffuse through nonpolar (hydrophobic) membrane interior
 
C is the right answer. Cholesterol is largely hydrophobic and can diffuse through the hydrophobic core of the plasma membrane utilizing hydrophobic interaction. 
 
D is a wrong answer. Aspartate is charged molecule and cannot readily diffuse through hydrophobic membrane interior
 
E is a wrong answer Glucose is a polar molecule and will not interact well with the hydrophobic core the of the plasma membrane

28

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