SSN SBPM Workshop~ Exam 4
Multiple Choice Questions
**Last year’s handouts are available at
http://cpmcnet.columbia.edu/student/ssn/sbpmd/block4.htm
For Questions 1-4: Match the changes in volume of the intracellular and extracellular
compartments with the associated change. Assume the starting osmolarity of both
compartments is 290 mOsm/L.
1.
2.
3.
4.
Diarrhea
Profuse sweating
Saline infusion
High Salt Diet
A) Decreased extracellular volume, no change in intracellular volume, and no
change in osmolarity.
B) Increased extracellular volume, decreased intracellular volume, and
increased osmolarity.
C) Increased extracellular volume, no change in intracellular volume, and no
change in osmolarity.
D) Decreased extracellular volume, decreased intracellular volume, and
increased osmolarity.
E) Decreased extracellular volume, increased intracellular volume, and
decreased osmolarity.
5.
A rat is fed a salt diet of 7 g/day for a period of two weeks. During this
period his urinary salt excretion is measured. The rat is then switched to a
low salt diet (1gm/day). What was the rat’s excretion of sodium likely to be
on day 14 just prior to switching to the low salt diet, and then on day 16 two
days after switching?
A) On day 14 his urinary excretion is 5 g/day (positive salt balance), and on
day 16 it will be approximately 1 g/day (salt equilibrium).
B) On day 14 it was 7.0 g/day (salt equilibrium) and on day 16 it will be
1g/day (salt equilibrium).
C) On day 14 it was 7.0 g/day (salt equilibrium) and on day 16 it will be
reduced but greater than 1.0g/day (negative salt balance).
D) On day 14 it was 7.0 g/day, but it is impossible to say where his salt
excretion will be on day 16.
E) None of the above.
6.
Switching to the low salt diet caused a series of neurohormonal changes in
the rat that were induced by the decrease in the extracellular fluid
osmolarity. What answer choice best describes those changes?
I. Increased sympathetic activity
II. Increased ANP secretion
III. Activation of the renin-angiotensin-aldosterone system
IV. Change in starling forces to increase Na+ reabsorption in the
proximal tubule
A)
B)
C)
D)
7.
The effects of ADH on the Kidney include which of the following:
I. Insertion of aquaporins in the outer medullary collecting ducts.
II. Increased activity of the Na+/K+/2Cl- transporter in the thick
ascending limb.
III. Increased reabsorption of sodium in exchange for hydrogen in
the proximal tubule.
IV. Increased urea recycling.
A)
B)
C)
D)
E)
8.
I only
I, III, IV only
All of the above
None of the above
I only
I and II only
I, II and III only
I, II and IV only
All of the above
Which one of the following will DECREASE in a person with V/Q
abnormalities?
A) Arterial pH
B) Arterial carbon dioxide tension
C) Alveolar-arterial (A-a) gradient for oxygen
D) Alveolar ventilation
9.
Which one of the following is higher at the apex of the lung than at the base
when a person is standing?
A) V/Q ratio
B) Blood flow
C) Ventilation
D) PaCO2
E) Lung compliance
10.
An increase in pulmonary blood flow during exercise
A) Causes alveolar oxygen tension to decrease
B) Causes the V/Q ratio at the top of the lung to increase
C) Causes pulmonary arterial resistance to decrease
D) Causes diffusion capacity to decrease
E) Causes arterial oxygen saturation to decrease
11.
Emphysema is associated with loss of elastic fibers in the lungs, leading to
increased lung compliance. As a result, the FRC…
A) Increases
B) Decreases
C) Stays the same
For questions 12-16: You are given the following data for a patient breathing room air
(FIO2 = 21%). Assume Patm = 760 mm Hg, PH2O = 47 mm Hg, and R = 0.8.
Tidal volume
Dead space volume
Breathing frequency
PaCO2
12.
The patient’s alveolar oxygen tension PaO2 is approximately
A)
B)
C)
D)
E)
13.
70 mm Hg
80 mm Hg
90 mm Hg
100 mm Hg
110 mm Hg
The patient’s minute ventilation is
A)
B)
C)
D)
E)
14.
= 400 mL
= 100 mL
= 10 breaths/min
= 50 mm Hg
3 L/min
4 L/min
5 L/min
6 L/min
7 L/min
The patient’s alveolar ventilation is
A)
B)
C)
D)
E)
3 L/min
4 L/min
5 L/min
6 L/min
7 L/min
15. If the patient doubles her alveolar ventilation without changing her CO2
production, then her PaCO2 will be approximately
A)
B)
C)
D)
E)
15 mm Hg
20 mm Hg
25 mm Hg
35 mm Hg
40 mm Hg
16. If the patient doubles her tidal volume without changing her CO2 production,
then her PaCO2 will be approximately
A)
B)
C)
D)
E)
15 mm Hg
20 mm Hg
25 mm Hg
35 mm Hg
40 mm Hg
17. A women with a history of alcoholism presents to the ER complaining of blurred
vision, dizziness and vomiting after having drank a bottle of what she assumed
was EtOH. She is very tachypneic (breathing quickly)and has an increased
serum anion gap (this means that she has an increased concentration of some
anion besides chloride and bicarbonate in her blood). Which of the following
pCO2/pH combinations would be most likely in her case?
A) pCO2 30, pH 7.50
B) pCO2 30, pH 7.32
C) pCO2 50, pH 7.50
D) pCO2 50, pH 7.32
18. Which of the following can be measured by spirometry?
A) FRC
B) RV
C) FVC
D) TLC
E) All of the above
F) None of the above
19. Which of the following is/are true?
I. A stroke in the dorsal medulla might require ventilatory support.
II. An infarction of the apneustic center might require ventilatory support.
III. Central chemoreceptors are most sensitive to changes in blood pH.
IV. It is physiologically possible to have a negative alveolar-arterial gradient (eg
higher pO2 in the blood as compared to the alveolus).
A) I, II, III, IV
B) I, III
C) I only
D) I, III, IV
E) I, IV
20. A patient has an arterial blood gas shown below. Normal values in parentheses.
pH = 7.29
(7.40)
pCO2 = 70 mmHG
(40 mmHg)
[HCO3-] = 33 meq / L
(24 meq / L)
The patient’s acid-base status can best be described as:
A)
B)
C)
D)
Respiratory Acidosis.
Respiratory Alkalosis.
Metabolic Acidosis.
Metabolic Alkalosis
21. A patient takes an unusually large amount of aspirin (acetylsalicylic acid.)
Which of the following statements about the patient is false?
A) The patient’s acid-base status is best described as metabolic acidosis.
B) The patient is excreting H+ bound to HPO42-, which is coupled to the new
reabsorbtion of HCO3-.
C) The patient reabsorbs filtered HCO3- in the proximal tubule, which is coupled to
H+ excretion.
E) The patient is excreting NH4 in the urine in a process involving the proximal
tubules, thick ascending loop of henle, and the collecting duct.
22. A freely filterable substance that is neither secreted nor reabsorbed has a renal
artery concentration of 15 mg/ml and a renal vein concentration of 5 mg/ml.
What is the filtration fraction?
A)
B)
C)
D)
0.25
0.33
0.5
0.67
23. Both GFR and renal blood flow will increase if
A) The efferent and afferent arterioles both dilate to the same degree
B) The afferent constricts and the efferent dilates to the same degree
C) The efferent and afferent arterioles both constrict to the same degree
D) Only the efferent arteriole constricts
SSN SBPM Workshop~ Exam 4
Answers to Multiple Choice Questions
1. A
Diarrhea can be considered to be loss of isosmotic fluid from the extracellular space. As
a result, there will be a decrease in extracellular volume, but because there is no change
in osmolarity there will be no compensatory volume change in the intracellular
compartment to offset the loss.
2. D
Sweating can be considered loss of a hyposmotic fluid from the extracellular space. As a
result, the osmolarity of the ECF will become transiently hyperosmotic (more
concentrated) to the intracellular fluid. This results in water flow from the ICF to the
ECF. The net change is to reduce the volume of both spaces, and to increase the
osmolarity.
3. C
Isosmotic infusion of fluid into the ECF. No change in osmolarity or fluid shifts result.
4. B
Increased sodium in the ECF results in water flows into the ECF from the ICF. The net
result is increased ECF volume, decreased ICF volume, and an increase in the overall
osmolarity.
5. C
The rat is at equilibrium after two weeks on the high salt diet. Switching to the low salt
diet requires several days for the adjustment in neurohormonal activation to take place
and for his urinary excretion to decrease to 1.0 g/day. During this period of adjustment,
the rat is said to be in negative salt balance (excreting more than he can taken in).
6. B
ANP is secreted in response to an increase in ECFV. Switching to a low salt diet would
tend to cause a decrease in the ECF and thus decrease ANP secretion.
7. D
Reabsorption of Na+ in the proximal tubule depends on angiotensin not ADH. ADH has
all of the other effects listed.
8. B
V/Q mismatches will cause arterial oxygen levels (Pao2) to decrease. Decreased Pao2
will stimulate the chemoreceptors which, in turn, will increase alveolar ventilation and
decrease PaCO2. The decreased PaCO2 will cause respiratory alkalosis (increasing pH).
The fall in PaO2 causes the A-a gradient to rise.
9. A
The alveoli at the apex of the lung are larger than those at the base so their compliance is
less. Because the compliance is reduced, less inspired gas goes to the apex than to the
base. Also, because the apex is above the heart, less blood flows through the apex than
through the base. However, the reduction in airflow is less than the reduction in blood
flow, so that the V/Q ratio at the top of the lung is greater than it is at the bottom. The
increased V/Q ratio at the apex makes PaCO2 lower and PaO2 higher at the apex than
they are at the base.
10. C
The increase in blood flow through the pulmonary circulation during exercise increases
the diameter of the pulmonary vessels and therefore decreases their resistance. The
increased blood flow to the lungs evens out the V/Q ratio, decreasing it at the top of the
lung and increasing it at the bottom. Making the V/Q ratio more similar throughout the
lung may decrease some of the shunting effect that normally occurs at the base where the
V/Q ratio is quite low and may actually increase the O2 saturation of arterial blood. The
increased blood flow will also increase the surface area for respiratory exchange and
therefore increase the diffusing capacity.
11. A
As a result of the increase in compliance, the collapsing (elastic recoil) force on the lungs
is decreased at a given volume. At the original value for FRC, the tendency of the lungs
to collapse is less than the tendency of the chest wall to expand, and these opposing
forces will no longer be balanced. In order for the opposing forces to be balanced,
volume must be added to the lungs to increase their collapsing force. Thus, the combined
lung and chest-wall system seeks a new higher FRC, where the two opposing forces can
be balanced; the new intersection point, where airway pressure is zero, is increased.
12. C
Use the alveolar gas equation:
PAO2 = FIO2 (Patm – PH2O) – (PACO2/R)
= .21 (760 mm Hg – 47 mm Hg) – (50 mm Hg/0.8)
= 149.75 mm Hg – 62.5 mm Hg
= 87.5 mm Hg ~ 90 mm Hg
13. B
Minute ventilation is equal to tidal volume VT times breathing frequency f, so
Minute ventilation = VT × f = 0.4 L × 10 breaths/min = 4 L/min
14. A
Alveolar ventilation is minute ventilation corrected for the physiologic dead space VD, so
Alveolar ventilation = (VT – VD) × f = (0.4 L – 0.1 L) × 10 breaths/min = 3 L/min
15. C
PaCO2 is inversely proportional to alveolar ventilation, so doubling the latter would halve
the former, which would decrease from 50 mm Hg to 25 mm Hg.
16. B
Again, PaCO2 is inversely proportional to alveolar ventilation, so:
PaCO2 × alveolar ventilation = constant
If the patient doubles her tidal volume to 800 mL (0.8 L), her alveolar ventilation will
increase from 3 L/min to:
Alveolar ventilation = (0.8 L – 0.1 L) × 10 breaths/min = 7 L/min
Thus, because PaCO2 × alveolar ventilation = constant, we get:
50 mm Hg × 3 L / min = PaCO2 × 7 L /min
PaCO2 = 21 mm Hg ~ 20 mm Hg
17. B
This woman has methanol poisoning, which leads to severe metabolic acidosis. As pH
decreases, the compensatory respiratory response is to blow off CO2 by hyperventilating
(remember that alveolar ventilation is inversely proportional to PaCO2; also, peripheral
chemoreceptors are going to be involved here because the problem is NOT increased
PaCO2). For those who are interested, the treatment for methanol poisoning is to
administer ethanol, which competes with MeOH for access to alcohol dehydrogenase.
Choice A would be a situation of respiratory alkalosis, choice C would be metabolic
alkalosis and choice D would be respiratory acidosis (try to think of clinical scenarios
that would lead to each of these conditions).
18. C
This is mostly just to familiarize you with the multiple abbreviations having to do with
capacities and volumes and to reiterate that you CANNOT measure residual volume (a
component of FRC and TLC) by spirometry. You can measure the forced vital capacity.
19. C
The dorsal respiratory area in the medulla is needed for spontaneous inspiration without
cortical control (central hypoventilatory syndrome). The apneustic center in the pons is
not necessary for respiration. The central chemoreceptors do not directly see decreased
blood pH; they are sensitive to changes in CSF pH. The last one is kind of tricky, but in
general, if you see a negative A-a gradient, it probably means that some room air
contaminated the blood sample to increase the PaO2.
20. A.
The pH is low, therefore the patient is acidotic. The high PCO2 suggests a respiratory
cause. A low pCO2 would suggest a metabolic cause. The high HCO3- suggests that the
condition is chronic.
21. C
It is false because HCO3-reabsorbtion isn’t associated with H+ excretion.
[Note: H+ excretion is associated with new HCO3- reabsorbtion.]
22. D (FF = GFR/RBF = 10/15)
23. A
SSN SBPM Workshop~ Exam 4
Short Answer Questions
Note: These questions may not be similar to the on-line short
answer questions for your exam. The purpose of these questions is
to explain topics in more detail than is allowed by the Multiplechoice format. Also, these questions offer an opportunity to go over
some topics that the SSN teachers do not have time to teach during
the SSN session.
1. How does the kidney produce concentrated urine? Dilute urine?
The production of concentrated urine depends on the presence of ADH, and of a
concentration gradient within the medulla (maintained by the countercurrent
multiplier and urea recycling).
1. Osmolarity of glomerular filtrate equals that of the blood (300 mOsm/L).
2. There is no change in osmolarity in the proximal tubule as water and salt
are reabsorbed in proportion.
3. In the thick ascending limb of the loop of Henle the Na+/K+/2Cl- cotransporter reabsorbs NaCl from the lumen. These cells are impermeable
to water, however, and the resulting fluid is diluted significantly
(100mOsm/L). It is this co-transporter that is responsible for establishing
the concentration gradient in the medulla (countercurrent multiplier).
4. In the early distal tubule the process continues with the reabsorption of
salt by the Na/Cl co-transporter. This leads to further dilution of the fluid
as the cells are still impermeable to water.
5. In the late distal tubule, the cells are now permeable to water in the
presence of ADH (aquaporins). As a result, water flows down its gradient
until it is isosmotic with the surrounding fluid (300mOsm/L again).
6. In the collecting ducts, the cells are again permeable to water in the
presence of ADH. Fluid therefore flows out of the lumen and into the
interstitium as it moves down the tubule through the medulla in the
collecting duct (interstitial osmolarity is increasing in the inner medulla).
As a result, it will equilibrate until it reaches a concentration equal to that
of the surrounding environment (as much as 1200 mOsm/L).
The production of dilute urine is much simpler. Without ADH present, water
cannot be reabsorbed in the distal tubule or collecting duct. As a result, there is
no opportunity to concentrate the urine as it flows down through the medulla.
The result is a dilute Urine that may be as low as 75mOsm/L (even lower than the
100 mOsm/L in the early distal tubule due to further reabsorption of sodium in the
late distal tubule and collecting duct).
2. The diagram below represents a spirometry tracing illustrating the changes in
lung volume that occurred when a subject inhaled maximally and then rapidly
exhaled as much gas as possible.
If the patient’s total lung capacity is 6 L, calculate the following:
A) FRC
B) FEV1
C) Inspiratory capacity
D) Inspiratory reserve volume
E) Expiratory reserve volume
a) FRC = 3 L. Functional residual capacity (FRC) is the volume of gas in the lung at
the end of a normal expiration, when the muscles of inspiration and expiration
are relaxed. At FRC, the elastic forces acting on the lungs are in equilibrium; that
is, the tendency of the lungs to collapse is balanced by the tendency of the chest to
expand. When the volume is at FRC, the spirometry trace reads 1 L; note from the
trace that FRC is 3 liters less than total lung capacity. Therefore, FRC is 6 L – 3
L = 3 L.
b) FEV1 = 3.5 L. The FEV1 is the amount of gas that can be expelled from the lungs
in 1 s during a forced expiration from total lung capacity. Based on the tracing,
FEV1 = 3.5 L (go from 4 L to 0.5 L in the first second of the forced expiration
maneuver).
c) Inspiratory capacity = 3 L. The inspiratory capacity is the maximum amount of
gas than can be inhaled when the subject starts from FRC. In this subject, the
inspiratory capacity is 3 L.
d) Inspiratory reserve volume = 2.5 L. The inspiratory reserve volume is the
maximum amount of gas that can be inhaled when the subject starts at the end of
a normal inspiration.
e) Expiratory reserve volume = 1 L. The expiratory reserve volume is the maximum
amount of gas that can be exhaled when the subject starts from FRC.
3. A man comes to the ER complaining of difficulty breathing. His vitals are as follows:
Pulse 90/min, Resp. rate 24 and BP 140/90. A blood gas is drawn and the following
values are seen: pCO2 34 mmHg, pO2 54 mmHg. He is given 100% oxygen by nasal
cannula at 6 L/min. When his blood gases are taken one hour later, they are as follows:
pCO2 45 mmHg, pO2 85 mmHg. Describe a plausible mechanism to explain the increase
in pCO2 (hint: think about breathing control mechanisms).
One likely explanation is that originally, his rapid respiratory rate (nb: 24
breaths/min is rapid) was being driven by his low pO2 (hypoxemic peripheral drive).
His pCO2 is low, becauseof his hyperventilation, so his central receptors are not
driving respiration. After oxygenating him, his pO2 rises and, since his pCO2 is still
low, his hypoxemic hyperventilation would resolve. This would cause his pCO2 to
rise.
4. Fill in the chart below with “up” or “down” reflecting the appropriate changes.
pH
pCO2
HCO3-Respitory Acidosis
Respitory Alkalosis
Metabolic Acidosis
Metabolic Alkalosis
Respraitory Acidosis
Respiratory Alkalosis
Metabolic Acidosis
Metabolic Alkalosis
pH
down
up
down
up
pCO2
up
down
down
up
HCO3up (more chronic than acute)
down more chronic than acute)
down
up
5. The pKa of H2CO3 is 6.1 and the pKa of H2PO4- is 6.8. Based solely upon their
pKa, which is the better buffer? Why? In the human body, which is the better
buffer? Why?
Based on pKa alone, H2PO4- is the better buffer because it’s pKa is closest to
physiologic pH (7.4). That allows the steepest part of it’s titration curve to fall closer
to physiologic pH. However, in the human body H2CO3 is the better buffer because it
is present in a much higher concentration; also, because of carbonic anhydrase, the
acid form of CO2 can be blown off by respiration.
6. How does the rate of glomerular plasma flow relate to the rate of glomerular
filtration?
The greater the glomerular blood flow, the greater the rate of glomerular filtration.
The faster the plasma flows along the glomerular capillary, the less the oncotic
pressure of the plasma in the glomerular capillary increases. Thus the pressure of
ultrafiltration is greater, and the conditions for filtration are more favorable.
7. How does Angiotensin II affect the GFR?
Angiotensin II constricts the efferent arteriole. This will increase the hydrostatic
pressure in the glomerular capillary, which increases the GFR.