Sample Questions Best site for GRE, LSAT, SAT, GMAT, TOEFL, CCNA, CCSA and interview sample questions  

MCAT Sample Questions ›› Biological Sciences

MCAT Sample Questions : Biological Sciences

Passage IX

Plasma clearance refers to the capacity of the kidney to remove a substance from the plasma.It is determined by comparing the concentrations of the substance in the plasma and the urine, and then calculating the rate at which the substance appears in the urine.

Plasma clearance is affected by the tubular transport maximum (Tm) of a substance.The Tm is the maximum rate of transport (mg/min) at which a substance can be reabsorbed by the kidney.That is, if the filtration rate of a substance exceeds its Tm, the substance will begin to appear in the urine.The Tm for glucose averages 320 mg/min in an adult human.

Two experiments were done to study the plasma clearance of Substance A, a hexose that readily enters the glomerular filtrate.

Experiment 1

First, researchers measured the concentrations of Substance A in the plasma and urine of a volunteer.Later, they determined the concentrations of glucose in the volunteer's plasma and urine.In each trial, the test substance was injected intravenously and continuously during measurement.Results are shown in Figure 1.

Figure 1 Plasma clearance of Substance A and glucose

Experiment 2

The effect of antidiuretic hormone (ADH) on the plasma clearance of Substance A was tested.The volunteer was given two dosages of ADH during two separate trials.The first dosage was within the normal physiological range of concentrations and affected the renal reabsorption of water.The second dosage, which was much higher than the normal physiological concentration, affected urine output by increasing blood pressure.

Following are some sample questions on this passage:

  1. According to the passage, the Tm represents the rate of plasma filtration that just exceeds the:

    1. rate of concentration of the substance in the glomerular filtrate.
    2. rate of concentration of the substance in the urine.
    3. capacity of the kidney tubules to reabsorb the substance.
    4. capacity of the bladder to store and excrete the substance.

    Answer: C

    Explanation: The glomerulus is a tuft of capillaries that bulges into the capsular space (also referred to in some texts by its eponym, Bowman's space), which is a potential space lined with simple squamous epithelium.Fluid is expressed from the blood across the capillary endothelium.It enters first the capsular space, then the kidney tubule.The part of the tubule closest to the glomerulus is the proximal tubule, the next part is the U-shaped loop of Henle and the last part of the tubule is the distal tubule.Suppose one gradually increased the rate at which fluid was expressed from the bloodstream (the glomerular filtrate rate) and measured the concentration of some substance reabsorbed by the kidney as it left the distal tubule as urine.The concentration might not rise at first, because the cells lining the tubule might be completely reabsorbing the substance and putting it back into the blood stream.Eventually, however, the rate of flow would reach a point at which it exceeds the rate at which the tubule cells could reabsorb the substance.The fluid would be flowing too rapidly through the tubule for the cells to reabsorb all the substance.The rate of flow through the tubule at which the substance begins to be observed in the urine is Tm.At that point the rate of flow of fluid through the tubule begins to exceed the capacity of the kidney tubule cells to reabsorb the substance.This description of Tm is given only in answer choice C, the correct answer.

  2. Under normal conditions, the tubular load of glucose (the amount/min that filters into the kidney tubules) is approximately 125 mg/min.The amount of glucose in the urine under these conditions is approximately:

    1. 0 mg/min.
    2. 125 mg/min.
    3. 195 mg/min.
    4. 515 mg/min.

    Answer: A

    Explanation: The amount of glucose in the fluid in the capsular space as it enters the tubule system would be the same as the amount of glucose in the plasma, but by the time the fluid reaches the distal tubule much of the glucose will have been reabsorbed.If the amount being filtered per minute is less than Tm, all of it will have been reabsorbed.If it is more than Tm, then the amount will be linearly related to the amount in the plasma.In the case proposed in this question, the amount being filtered per minute (125mg/min) is less than the Tm for glucose given in the passage (320 mg/min), so all of the glucose should be reabsorbed.The correct answer, therefore, is 0 mg/min, answer choice A.

  3. A lower-than-normal blood pressure will cause which of the following effects on the rate of plasma clearance of Substance A?

    1. An increase, because the concentration of Substance A in the urine will increase
    2. An increase, because the ADH levels will be very low
    3. A decrease, because the decreased rate of urine output will allow more reabsorption by the kidney
    4. A decrease, because ADH levels will be very high

    Answer: C

    Explanation: Low blood pressure could have an effect on ADH levels (as suggested in answer choices B and D), which could, in turn, affect the amount of substance A in the urine, but there is not enough information in the passage to decide if such an effect exists and, if it does, whether its effect would overcome the certain affect of lowering the blood pressure.The best answer is answer choice C: that low blood pressure decreases the glomerular filtration rate, allowing more time for reabsorption and decreasing the amount of substance A in the urine.Blood pressure is the source of the energy that forces fluid into the capsular space.If the heart stopped and the blood in the glomerular capillaries had no hydrostatic pressure, fluid in the space around the glomerulus would flow back into the capillary bloodstream.This would occur because the protein-rich blood would be hypertonic with respect to the protein-poor fluid in the capsular space so that the fluid would flow down the osmotic gradient into the blood.

  4. Equal concentrations of 8 mg/mL of Substance A and glucose are found in a volunteer's plasma.Based on Figure 1, which substance will the kidney clear from the plasma more rapidly?

    1. Substance A, because the slope of the clearance line for Substance A is higher than that for glucose
    2. Substance A, because Substance A reaches its Tm at a lower plasma concentration than does glucose
    3. Glucose, because glucose reaches its Tm at a higher plasma concentration than does Substance A
    4. Glucose, because the slope of the clearance line for glucose is lower than that for Substance A

    Answer: B

    Explanation: The Tm is a characteristic of the individual substances in the tubule system and a measure of how efficiently each substance can be reabsorbed.A high Tm indicates a high capacity for reabsorption of substances in the kidney tubules.In figure 1, the Tm for each substance can be read as the concentration in plasma when the concentration in the urine is zero.In this case it looks like the Tm for Substance A is a little over 6 mg/mL and that of glucose is 10 mg/mL.So substance A has a lower Tm.This means the tubules will not reabsorb it very efficiently.Much of it will be spilling into the urine, thus being eliminated from the body.The question asks which will clear from the blood more rapidly at a concentration of 8 mg/mL and why.The answer is that all glucose will be reabsorbed at that concentration, none will appear in the urine and none will be cleared from the plasma.Glucose has such a high Tm, that all of the glucose will be reabsorbed into the bloodstream, perhaps to reenter the kidney tubule again.A rate of 8mg/mL is above the Tm of substance A, so there will be some substance A in the urine at this plasma concentration.The answer to the question depends on the value of Tm, not the slope of the clearance line, so answer choices A and D can be eliminated.Substance A will clear more rapidly than glucose, therefore, answer choice B is correct.

  5. According to Figure 1, at approximately what plasma concentration of glucose is the Tm (320 mg/min) reached?

    1. 6.5 mg/mL
    2. 10.0 mg/mL
    3. 11.5 mg/mL
    4. 12.5 mg/mL

    Answer: B

    Explanation: Above 10 mg/mL, glucose begins to be found in the urine.The Tm for glucose is therefore 10 mg/mL, answer choice B.This can be read from the graph by looking at the concentration in the plasma where the concentration in the urine is zero.In other words, where the clearance line for glucose crosses the axis.

  6. In Experiment 2, the increased blood pressure resulting from the higher-than-normal concentration of ADH most likely affected the urinary output of Substance A by increasing the:

    1. glomerular filtration rate.
    2. Tm of solutes.
    3. water reabsorption from the tubules.
    4. concentrating ability of the loop of Henle.

    Answer: A

    Explanation: The best answer is that increased blood pressure will affect the glomerular filtration rate, answer choice A.Tm is a characteristic that depends on the characteristics of the cells lining the renal tubules and independent of blood pressure, so answer choice B is not correct.Water resorption and concentrating ability are the same, so answer choices C and D are essentially the same.Increasing blood pressure should increase flow of fluid through the kidney system and decrease, rather than increase, water reabsorption, so these answer choices are incorrect.

  7. Where in the nephron is most reabsorption of glucose likely to occur?

    1. Glomerulus
    2. Proximal tubule
    3. Distal tubule
    4. Loop of Henle

    Answer: B

    Explanation: The glomerulus functions as a filter, removing fluid from the capillary blood and leaving behind molecules and formed elements too large to pass through.This fluid is called glomerular filtrate.The capsular space is lined with simple squamous cells unsuited for reabsorption, so answer choice A is incorrect.The first reabsorptive segment of the tubule is the proximal tubule.Most of the glucose should be reabsorbed in the proximal tubule since it is the first segment to process the filtrate.In fact, normally, all glucose reabsorption takes place in the proximal tubule (which has been calculated to remove half a pound of glucose a day from the glomerular filtrate).The proximal tubule is lined with simple columnar cells with a brush border on the luminal side consisting of microvilli.The microvilli give the cell an extensive surface area in contact with the filtrate to facilitate reabsorption.Therefore, the correct answer is answer choice B.

« Previous || Next »

MCAT Sample Question Number : 1-7| 8-11| 12-16| 17-22| 23-29| 30-34| 35-38| 39-43| 44-50| 51-55
Sample Test Questions
GRE Sample Questions
CAT Sample Questions
GMAT Sample Questions
TOEFL Sample Questions
ACT Sample Questions
SAT Sample Questions
LSAT Sample Questions
PSAT Sample Questions
MCAT Sample Questions
PMP Sample Questions
GED Sample Questions
ECDL Sample Questions
DMV Sample Questions
CCNA Sample Questions
MCSE Sample Questions
Network+ Sample Questions
A+ Sample Questions
Technical Sample Questions
WASL Sample Questions
CISA Sample Questions

Other Sample Questions
Sample Interview Questions
Sample Teacher Interview Questions
Sample Citizenship Questions
Accuplacer Sample Questions
Science Bowl sample Questions
Driving Test Sample Questions
Sample Survey Questions Sample Essay Questions
Sample Behavioral Interview Questions

Copyright © 2004-2013, Best BSQ. All Rights Reserved.