quiz 3 Flashcards
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
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”
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”
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”
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.
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.
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
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.
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.
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.
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.
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.
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???
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).
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.
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)
