1
RENAL
A 45-year-old man with uncontrolled diabetes mellitus is admitted to the hospital for observation. His breath smells
like nail polish remover. The following laboratory results were obtained from this patient.
Increased plasma levels of which of the following is the most likely cause of this patient's acid-base disorder?
A. Formic acid
B. Glycolic acid
C. Ketoacids
D. Lactic acid
E. Oxalic acid
Explanation:
The correct answer is C.This patient has metabolic acidosis (low pH, low HCO3-). The decrease in arterial CO2 is a
compensatory response to the acidosis. The smell of acetone (which is the primary solvent of nail polish) on the breath of a
diabetic suggests ketoacidosis. Overproduction of ketoacids and high levels of plasma acetone commonly occur
with uncontrolled diabetes mellitus. Because acetone is highly volatile, it is excreted mainly by the lungs. Other
common causes of ketoacidosis include starvation and the acute drinking-vomiting binge of the alcoholic.
Plasma levels of formic acid (choice A) increase when methanol is ingested. Plasma levels of glycolic acid (choice B) and
oxalic acid (choice E) increase when ethylene glycol is ingested.
Because lactic acid (choice D) is a product of anaerobic metabolism, its rate of production is increased by
decreases in tissue blood flow and oxygen delivery, as well as when oxygen utilization is impaired. Lactic acidosis
commonly occurs in circulatory failure. Although diabetes mellitus can cause lactic acidosis, it is far less common
than ketoacidosis, and the smell of acetone on the breath should remove any doubt as to the cause of the
metabolic acidosis.
A 23-year-old man has an intracellular fluid volume of 28 L, an extracellular fluid volume of 14 L, a plasma
volume of 3 L, and an extracellular fluid osmolarity of 285 mOsm/L. The man drinks 3 L of water and consumes
10 mEq of sodium (in the form of potato chips). What is his approximate extracellular osmolarity (assuming
osmotic equilibrium)?
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A. 266 mOsm/L
B. 285 mOsm/L
C. 291 mOsm/L
D. 300 mOsm/L
E. Cannot be determined
Explanation:
The correct answer is A. This problem appears to be complex at first, but the astute student will note that
consuming 3 L of water and essentially no sodium (10 mEq) will cause the extracellular osmolarity to decrease.
With this knowledge, the problem becomes very simple because there is only one value of plasma osmolarity
below the starting value of 285 mOsm/L, ie, 266 mOsm/L. (Recall that each liter of plasma normally contains
about 140 mEq sodium. Therefore, ingestion of 10 mEq of sodium has very little effect on the osmolarity of the
extracellular fluid when 3 L of water is consumed).
The answer to this problem can also be more rigorously calculated as the total number of milliosmoles in the
body fluid (11,980 mOsm) divided by the total body water (45 L) = 11,980/45 = 266 mOsm/L. The total number
of milliosmoles is calculated as follows: 42 L (initial total body water) x 285 mOsm/L (initial extracellular
osmolarity) = 11,970 mOsm + 10 mOsm (10 mEq sodium means 10 mOsm) = 11,980 mOsm. The total body
water is calculated as follows: 28 L (intracellular volume) + 14 L (extracellular volume) + 3 L (water consumed)
= 45 L.
As part of an experimental study, a volunteer agrees to have 10 grams of mannitol injected intravenously. After
sufficient time for equilibration, blood is drawn and the concentration of mannitol in the plasma is found to be 65
mg/100 mL. Urinalysis reveals that 10% of the mannitol had been excreted into the urine during this time period.
What is the approximate extracellular fluid volume of this volunteer?
A. 10 L
B. 14 L
C. 22 L
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D. 30 L
E. 42 L
Explanation:
The correct answer is B. Volume = amount/concentration.
The amount of mannitol in the volunteer is equal to the amount injected minus the amount excreted: 10 g −
1 g = 9 g = 9000 mg. Therefore,
A 24-year-old woman with a large appetite for salt consumes about 25 g of salt each day. What is the
approximate amount of salt (in grams) that is excreted each day by her kidneys?
A. 4
B. 12
C. 23
D. 50
E. 250
F. Cannot be determined
Explanation:
The correct answer is C. About 95% of the salt (sodium chloride) that is consumed by a person is excreted by
the kidneys; the remaining 5% is excreted in the sweat and feces. The total intake of salt (amount of salt
consumed each day) must equal the total output of salt (amount of salt excreted each day) under normal
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steady conditions, ie, salt intake = salt output. Therefore, it is clear that 25 g of salt must be excreted by the
kidneys each day when 25 g of salt is consumed each day. 95% of 25 is around 23 g.
The following data was collected from a normal patient before and after an intervention. Assume that plasma
osmolarity and glomerular filtration rate remain constant.
Before
After
Urine osmolarity (mOsm/L
900
250
Urine flow rate (mL/min)
0.65
2.3
Fractional clearance of sodium
1%
1%
Osmolar clearance (mL/min)
2.0
2.0
The intervention that would best account for the observed changes is
A. administration of furosemide
B. administration of hydrochlorothiazide
C. administration of lithium
D. a high dietary intake of potassium
E. a transfusion of 2 liters of isotonic saline
Explanation:
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The correct answer is C. Lithium inhibits the action of ADH (vasopressin) on the V2 receptors in the collecting
duct that regulate the permeability to water. Therefore, lithium administration will decrease water permeability in
the collecting duct, which will increase urine flow rate and decrease urine osmolarity. Since ADH has minimal
effects on sodium reabsorption in humans, the fractional clearance of sodium and the osmolar clearance are
unaffected (osmolar clearance refers to the clearance of all particles, including sodium and anions, from the
plasma per minute).
Furosemide (choice A) inhibits the active reabsorption of sodium, potassium, and chloride in the thick ascending
limb. This will deplete the medullary gradient, which could result in a slightly hypotonic urine, but furosemide will
significantly increase the fractional clearance of sodium, and hence the osmolar clearance.
Hydrochlorothiazide (choice B) inhibits the active reabsorption of sodium chloride from the distal convoluted
tubule. Since the distal tubule is in the renal cortex, it will not inhibit the ability of the kidney to concentrate urine,
and so would not decrease the urine osmolarity so dramatically. Additionally, it will also increase the fractional
clearance of sodium, and hence the osmolar clearance, but not as much as with furosemide.
High dietary intake of potassium (choice D) will increase plasma potassium levels, which will increase
aldosterone secretion by direct action on the adrenal cortical cells. The aldosterone would decrease fractional
clearance of sodium and would not increase urine flow rate.
Increased isotonic plasma volume (choice E) will increase atrial natriuretic factor (ANF), which will inhibit sodium
reabsorption in the nephron, and thus will increase the fractional clearance of sodium as well as osmolar
clearance.
The clearance of several substances was measured at a constant glomerular filtration rate and constant urine
flow rate, but at increasing plasma concentrations of the substance. Under these conditions, clearance will
increase at high plasma concentrations for which of the following substances?
A. Creatinine
B. Mannitol
C. Penicillin
D. Phosphate
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E. Urea
Explanation:
The correct answer is D. Clearance of a substance will change with increasing plasma concentration if that
substance is secreted or reabsorbed by a facilitated mechanism. As the concentration of the substance
increases, the transporter becomes saturated, and its contribution to excretion changes, changing the
clearance. If the substance is reabsorbed by a facilitated mechanism, clearance will eventually increase with
increasing plasma concentrations. Approximately 80% of filtered phosphate is reabsorbed in the proximal
tubule by a sodium-phosphate cotransporter, which is a facilitated mechanism.
In a patient with a urine flow rate of 1 mL/minute, the tubular fluid with the lowest osmolarity would be found in the
A. beginning of the proximal tubule
B. end of the cortical collecting tubule or duct
C. end of the papillary collecting tubule or duct
D. macula densa
E. tip of the loop of Henle
Explanation:
The correct answer is D. Tubular fluid first becomes hypotonic toward the end of the thick ascending limb of the
loop of Henle and will therefore be hypotonic by the macula densa (which is the border between the thick
ascending limb and the distal convoluted tubule).
Tubular fluid is isotonic at the beginning of the proximal tubule (choice A).
Tubular fluid is isotonic by the end of the cortical collecting duct (choice B) in the presence of antidiuretic
hormone (ADH), since water is reabsorbed until the tubular fluid osmolarity is the same as the peritubular fluid
in the cortex (which has the same osmolarity as plasma). A person with a urine flow rate of 1 mL/minute is
typically making hypertonic urine, and so has a significant amount of ADH present. The urine is assumed to be
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hypertonic since osmolar clearance (Cosm) is usually 2 mL/minute, and urine osmolarity must be greater than
plasma osmolarity if Cosm > urine flow rate.
Tubular fluid at the end of the papillary collecting duct (choice C) will be hypertonic in the presence of ADH.
(See explanation of choice B for why ADH is present.)
Tubular fluid at the tip of the loop of Henle is always hypertonic; essentially no water or solute is reabsorbed
along the thin descending limb (choice E).
Which of the following can be determined by calculating the clearance of para-aminohippuric acid (PAH)?
A. ECF volume
B. Effective renal plasma flow (ERPF)
C. Glomerular filtration rate (GFR)
D. Plasma volume
E. Total body water (TBW)
Explanation:
The correct answer is B. At less than saturating concentrations, PAH is completely secreted into the proximal
tubule and excreted into the urine. Therefore, the volume of plasma cleared of PAH is approximately equal to
the volume of plasma flowing through the peritubular capillaries, also called the effective renal plasma flow or
ERPF.
ERPF= Upah X V / P (pah)
The ECF volume (choice A) can be calculated by measuring the volume of distribution of solutes that move
freely across capillary walls but cannot permeate cell membranes (e.g., inulin and mannitol).
The GFR (choice C) is best calculated using a substance that is freely filtered at the glomerulus, not
reabsorbed, and only minimally secreted into the urine. Creatinine fits the bill and is used clinically to measure
the GFR (inulin also works and is used experimentally). While the creatinine excretion exceeds filtration by
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10-20% (because of the secretion), creatinine clearance is still a good approximation for GFR because the
error due to secretion is balanced by an overestimation of plasma creatinine inherent in the measurement
technique.
GFR = U (creatinine) X V / P (creatinine)
The plasma volume (choice D) can be measured by measuring the volume of distribution of radioactively
labeled serum albumin or of Evans blue dye (binds to albumin).
Total body water (choice E) can be measured by measuring the volume of distribution of tritium, deuterium, or
antipyrine.
A substance that is filtered, but not secreted or reabsorbed (substance X), is infused into a volunteer until a
steady state plasma level of 0.1 mg/mL is achieved. The subject then empties his bladder and waits one hour, at
which time he urinates again. The volume of urine in the second specimen is 60 mL and the concentration of
substance X is 10 mg/mL. What is the glomerular filtration rate (GFR) in this individual?
A. 30 mL/min
B. 60 mL/min
C. 100 mL/min
D. 300 mL/min
E. 600 mL/min
Explanation:
The correct answer is C. Because substance X is filtered, but not secreted or reabsorbed (like inulin), the
clearance of substance X can be used to approximate GFR.
GFR = [U]x× V / [P]x, therefore,
GFR = (10 mg/mL) ×(60 mL/hour) / (0.1 mg/mL)
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= (10 mg/mL) × (1 mL/min) / (0.1 mg/mL)
= 100 mL/min.
Note that you need to convert 60 mL/hour to 1 mL/min to get the correct answer in the correct units. Checking
to make sure the units are correct will help make sure you are using the formula properly.
A 30-year-old woman is given 0.1 g inulin intravenously. One hour later the plasma inulin concentration is 1 mg/100 mL.
Which of the following is the extracellular fluid volume (in liters) of this woman?
A. 8
B. 10
C. 12
D. 14
E. 16
Explanation:
The correct answer is B. The volume of a fluid compartment can be measured by placing a substance into the
compartment, allowing it to disperse evenly throughout the compartment, and then measuring the extent to which the
indicator is diluted in the fluid. The volume of a compartment can be determined using the following formula:
.
The extracellular fluid volume can be measured using inulin as the indicator: 0.1 g inulin was administered
intravenously and the concentration of inulin in the compartment was 1 mg/100 mL an hour later (when the inulin had
dispersed evenly in the extracellular fluid compartment). Therefore,
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.
A 66-year-old male has a cardiopulmonary arrest, and is transported to the hospital by paramedics. The data
shown below are derived from an arterial blood sample obtained upon admission.
Plasma pH
7.09
Plasma Bicarbonate
15 mEq/L
Arterial Carbon Dioxide
50 mm Hg
What type of acid-base abnormality is present in this man?
A. Metabolic acidosis
B. Metabolic alkalosis
C. Mixed acidosis
D. Mixed alkalosis
E. Respiratory acidosis
F. Respiratory alkalosis
Explanation:
The correct answer is C. A mixed acidosis commonly occurs with cardiopulmonary arrest. Cardiac arrest victims
experience some degree of lactic acidosis (metabolic acidosis) as a result of poorly perfused tissues. A
simultaneous respiratory acidosis due to ventilatory standstill also occurs. This combination of metabolic acidosis
and respiratory acidosis is referred to as a "mixed acidosis." A metabolic acidosis (choice A) is present when
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plasma pH and HCO3- concentration are low, and a
respiratory acidosis (choice E) is present when plasma pH is
low and arterial CO2 is high.
The table below shows changes in plasma pH, plasma HCO3-, and arterial CO2 for the various acid-base
disturbances.
Acid Base Status
Plasma pH
Plasma Bicarbonate
Arterial Carbon Dioxide
Metabolic acidosis (choice A)
low
low
low
Metabolic alkalosis (choice B)
high
high
high
Mixed acidosis (choice C)
low
low
high
Mixed alkalosis (choice D)
high
high
low
Respiratory acidosis (choice E)
low
high
high
Respiratory alkalosis (choice F)
high
low
low
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.
A healthy 36-year-old woman lost at sea is deprived of water for several days, but continues to excrete a small
volume of highly concentrated urine each day. Her plasma level of antidiuretic hormone (ADH) is 5 times greater
than normal. Which part of the tubule shown above would have the lowest tubular fluid osmolarity?
A. Site A
B. Site B
C. Site C
D. Site D
E. Site E
Explanation:
The correct answer is C. The osmolarity of fluid in the early distal tubule is usually less than 150 mOsm/L, even
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during water deprivation when ADH levels are high. The
early distal tubule is called the "diluting segment"
because the fluid is always hypotonic. In the proximal tubule (choice A), the tubular membrane is so highly
permeable to water that transport of solutes from the tubule causes a proportionate osmosis of water from the
tubule. Thus, the osmolarity of the tubular fluid is essentially the same as that of glomerular filtrate (which is the
same as that of plasma). The osmolarity of tubular fluid in the cortical collecting duct (choice D) depends entirely
on the presence or absence of ADH, reaching hypertonic levels (i.e., greater than 300 mOsm/L) when ADH levels
are high. The osmolarity of tubular fluid in the tip of the thin loop of Henle (choice B) and in the medullary
collecting duct (choice E) can be as great as 1200-1400 mOsm/L when ADH levels are high.
The data shown in the table below were collected from a 21-year-old college football player involved in a clinical
study.
Inulin space
20 L
Blood volume
7L
Plasma volume
4L
Plasma osmolarity
285 mOsm/L
Body weight
100 kg
What is his approximate interstitial fluid volume?
A. 9 L
B. 13 L
C. 16 L
D. 40 L
E. Cannot be determined
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Explanation:
The correct answer is C. The interstitial fluid volume cannot be measured directly because it occupies the
spaces between the cells and is part of the extracellular fluid volume along with the plasma volume. Interstitial
fluid volume is calculated by subtracting the plasma volume from the extracellular fluid volume. Extracellular fluid
volume was estimated in the subject using inulin as the indicator. Therefore, interstitial fluid volume = 20 L
(inulin space) - 4 L (plasma volume) = 16 L.
Inulin is a reasonable indicator (or marker) for the extracellular space because it disperses relatively evenly
throughout the extracellular fluid, but does not enter the cells to a significant extent. Because the various
substances used to estimate extracellular fluid volume (e.g., inulin, chloride, sodium, and sucrose) provide
different values, especially when these substances enter the cells (e.g., sodium and chloride), one often speaks
of the inulin space, the sodium space, the chloride space, or the sucrose space instead of the true extracellular
fluid volume.
A 23-year-old man with diabetes mellitus has a glomerular filtration rate (GFR) significantly greater that normal,
especially when he consumes excessive amounts of sweets. A decrease in which of the following parameters
would tend to increase the glomerular capillary hydrostatic pressure?
A. Afferent arteriolar resistance
B. Bowman's capsular hydrostatic pressure
C. Capillary filtration coefficient
D. Efferent arteriolar resistance
E. Plasma colloid osmotic pressure
Explanation:
The correct answer is A. A decrease in the resistance of the afferent arteriole (i.e., arteriolar dilation) directly
increases glomerular capillary hydrostatic pressure by lessening the drop in blood pressure that normally
occurs along the vasculature proximal to the glomerulus. [Recall that the afferent arteriole is upstream from the
glomerulus; the efferent arteriole is downstream from the glomerulus.] The glomerular capillary hydrostatic
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pressure is the determinant of glomerular filtration rate
most subject to physiological control.
Bowman's capsular hydrostatic pressure (choice B), capillary filtration coefficient (choice C), and plasma colloid
osmotic pressure (choice E) are important determinants of GFR but they do not have any direct effect to
increase or decrease the glomerular capillary hydrostatic pressure.
A decrease in efferent arteriolar resistance (choice D) would tend to decrease the glomerular capillary
hydrostatic pressure because the efferent arteriole is downstream from the glomerular capillaries.
A healthy 20-year-old man deprived of water for several days has an arterial pressure of 118/78 mm Hg and a
plasma concentration of antidiuretic hormone (ADH) 5 times above normal. Which of the following is the most
likely explanation for the increase in ADH concentration?
A. Decreased plasma aldosterone
B. Decreased plasma renin activity
C. Increased extracellular fluid volume
D. Increased left atrial pressure
E. Increased plasma osmolality
Explanation:
The correct answer is E. An obligatory loss of water from the body continues to occur even when a person is
deprived of water. This loss of water from the body tends to concentrate the extracellular fluid, causing it to
become hypertonic. Both the decrease in extracellular fluid (compare with choice C) and the increase in
osmolarity act as stimuli for increased thirst and increased secretion of ADH. The decrease in extracellular fluid
volume also tends to decrease arterial pressure, which in turn increases plasma renin activity (compare with
choice B) as well as aldosterone levels in the plasma (compare with choice A).
Water deprivation tends to decrease left atrial pressure (compare with choice D).
Q 16
Diagrams A-E show the relative osmolarity (Y-axis) and volume (X-axis) of the intracellular and extracellular fluid
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compartments during normal conditions (solid line) and
after various disturbances in the body fluids (shaded area,
dashed line). Which of the following diagrams depicts a man who has cholera without fluid replacement?
A.
B.
C.
D.
E.
Explanation:
The correct answer is A. Diagram A shows a disturbance in body fluid balance commonly referred to as "isotonic
contraction," which is caused by loss of isotonic fluid from the body. Cholera toxin causes a high rate of fluid
secretion in the small intestine, which leads to loss of large amounts of diarrhea fluid from the bowels. A
decrease in the extracellular fluid volume without a significant effect on the intracellular fluid volume (as shown in
diagram A) occurs with diarrhea because diarrhea fluid is isotonic to extracellular fluid (ie, loss of isotonic fluid
does not create an osmotic gradient for water loss from cells). The decrease in extracellular fluid volume shown
in diagram A is minimal, considering that as much as 10-12 L of fluid can be lost each day in cholera patients.
Death can occur within a short time when fluid replacement is not available.
Choice B (hypertonic expansion) can be caused by excessive intake of sodium chloride without drinking water.
Choice C (hypotonic contraction) is characteristic of sodium chloride loss from the body (eg, secondary to lack
of aldosterone).
Choice D (hypertonic contraction) can be caused by sweating (without water replacement) and other
perturbations in which a hypotonic fluid is lost from the body.
Choice E (hypotonic expansion) can be caused by retention of water by the kidneys, eg, inappropriate secretion
of antidiuretic hormone.
A healthy 38-year-old woman is found unconscious and severely dehydrated. Her plasma levels of antidiuretic
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hormone (ADH) are increased about 5-fold above
normal. In which portion of her kidney tubule is most of the
water being reabsorbed?
A. Cortical collecting duct
B. Distal tubules
C. Loops of Henle
D. Medullary collecting duct
E. Proximal tubule
Explanation:
The correct answer is E. About 65% of the water filtered by the glomeruli (180 L/day) is reabsorbed in the
proximal tubule, 15% in the loops of Henle (choice C), 10% in the distal tubules (choice B), and less than 10%
in the collecting ducts (choices A and D); about 1 L of water is normally excreted as urine each day. The
amount of water reabsorbed in the proximal tubule and loop of Henle is not affected by ADH, because ADH
does not affect tubular permeability in these segments of the nephron. However, ADH increases the
permeability of the distal tubules and collecting duct, which increases reabsorption of water. When ADH levels
are high, the urine output can decrease to less than 0.5 L/day; when ADH levels are low, the output of urine
can increase to more than 30 L/day. Even at these extremes, however, most of the water in the glomerular
filtrate is still reabsorbed in the proximal tubule.
Glomerular hydrostatic pressure = 44 mm Hg,
Bowman's capsule hydrostatic pressure = 9 mm Hg,
Osmotic pressure of plasma = 28 mm Hg,
Osmotic pressure of tubular fluid = 0.
Given this data, what is the net filtration pressure at the glomerulus?
A. – 5 mm Hg
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B. 7 mm Hg
C. 25 mm Hg
D. 63 mm Hg
E. 81 mm Hg
Explanation:
The correct answer is B. There is more than one way to think about this question. One way is to determine
which of each of the descriptions corresponds to Pc, Pi, πc, and πi and then to use the Starling equation
for net filtration pressure: (Pc− Pi) − (πc−πi). Perhaps faster and more intuitive is to
just envision that the filtration pressure will be the difference between the forces pushing fluid out and the
forces pulling fluid back into the glomerulus. The pushing forces are the hydrostatic pressure of the glomerulus
(44 mm Hg) and the osmotic pressure of the tubular fluid (0). So the total pressure forcing fluid from the
glomerulus into the tubular fluid is 44 mm Hg. The forces pulling the fluid back are the hydrostatic pressure of
the Bowman's capsule (9 mm Hg) and the osmotic pressure of the plasma (28 mm Hg). So the total pressure
pushing the fluid back into the glomerulus is 9 + 28 = 37 mm Hg. The difference between the forces favoring
filtration and those opposing it are therefore 44-37 = 7 mm Hg.
Q 19
Diagrams A-E show the relative osmolarity (Y-axis) and volume (X-axis) of the intracellular and extracellular fluid
compartments during normal conditions (solid line) and following various disturbances in the body fluids (shaded
area, dashed line). Which of the following diagrams depicts a woman who runs a marathon on a hot summer day
and replaces all volume lost in sweat by drinking water during the race?
A.
B.
C.
D.
E.
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Explanation:
The correct answer is C. Diagram C shows a disturbance in body fluid balance commonly referred to as
"hypotonic contraction" which is characteristic of sodium chloride loss from the body. Because the fluid lost as
sweat was replaced entirely with water and because sweat contains sodium chloride (as well as other
electrolytes), the body fluid osmolarity has decreased and the total volume of water in the body has remained at
a normal level, as shown in diagram C. The extracellular fluid volume has decreased (because the electrolytes
were lost from the extracellular fluid compartment) and the intracellular fluid volume has increased due to
movement of water into the cells. (The astute student will note that this question can be answered very quickly
because the total body water volume is unchanged only in diagram C.)
Choice A (isotonic contraction) can be caused by diarrhea.
Choice B (hypertonic expansion) can be caused by excessive intake of sodium chloride without water
supplementation.
Choice D (hypertonic contraction) can be caused by sweating (without water replacement) and other
perturbations in which a hypotonic fluid is lost from the body.
Choice E (hypotonic expansion) can be caused by retention of water by the kidneys, e.g., inappropriate secretion
of antidiuretic hormone.
The ratio of urinary concentration to plasma concentration of inulin {(U/P) inulin} decreases. Which of the
following is true if the glomerular filtration rate remains constant?
A. Aldosterone levels have increased
B. Inulin clearance has decreased
C. Positive free water clearance has decreased
D. Reabsorption of inulin has increased
E. Urine flow rate has increased
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Explanation:
The correct answer is E. Inulin is freely filtered, but is neither reabsorbed nor secreted. Since all inulin filtered in
the glomerulus will appear in the urine, the amount of water in the urine will determine the concentration of the
inulin. Therefore, (U/P) inulin will decrease if the urine flow rate increases.
In the presence of adequate amounts of antidiuretic hormone (ADH or vasopressin), increased aldosterone
(choice A) increases reabsorption of sodium in the collecting duct. Water will follow the sodium chloride, which
will increase (U/P) inulin.
Inulin clearance = glomerular filtration rate, which has not changed (choice B).
Reabsorbing less water in the collecting duct represents either a decrease in negative free water clearance (if
concentrated urine was being made), or an increase in positive free water clearance (choice C).
Inulin is neither reabsorbed nor secreted (choice D).
A 64-year-old man has severe polyuria and polydipsia, drinking 3 to 4 glasses of water and producing over 0.5
liters of urine each hour. A new internal medicine resident places the patient on overnight water restriction for
further analysis. The test results shown below were obtained the following morning.
Plasma sodium concentration: 155 mEq/L
Urine osmolarity: 90 mOsmol/L
Urine glucose concentration: 0 mg/dL
Which of the following is the most likely diagnosis?
A. Addison's disease
B. Diabetes insipidus
C. Diabetes mellitus
D. Fanconi syndrome
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Explanation:
The correct answer is B. Diabetes insipidus is characterized by the excretion of abnormally large volumes of
dilute urine (polyuria) with a commensurate increase in fluid intake (polydipsia). The most common type is due
to inadequate secretion of antidiuretic hormone (also called vasopressin) and is usually referred to as
"neurogenic" diabetes insipidus. This condition rarely causes severe problems as long as the person has
plenty of water to drink. Placing the patient on overnight water restriction caused severe dehydration and a
greatly elevated plasma sodium concentration. The possibility of diabetes mellitus (choice C), which can also
be associated with polyuria and polydipsia, is easily excluded by the lack of glucosuria.
Addison's disease (choice A) results from failure of the adrenal cortices to produce adrenocortical hormones.
The lack of aldosterone leads to decreases in sodium reabsorption allowing large amounts of sodium to be lost
into the urine. Polyuria and polydipsia are not characteristic of Addison's disease.
Fanconi's syndrome (choice D) is associated with multiple transport defects in the proximal tubule. Large
amounts of glucose (as well as other substances normally reabsorbed in the proximal tubule) are usually
present in the urine.
A healthy adult participating in a clinical research study increases his daily sodium intake greatly, but his plasma
sodium remains at a constant level. Which of the following substances is most responsible for this constancy in
plasma sodium concentration when large amounts of sodium are ingested?
A. Aldosterone
B. Angiotensin II
C. Antidiuretic hormone (ADH)
D. Atrial natriuretic factor (ANF)
E. Epinephrine
Explanation:
The correct answer is C. A 5-fold increase in sodium intake causes the plasma sodium concentration to
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increase by less than 1%, indicating the existence of a
powerful mechanism for maintaining extracellular sodium
concentration at a constant level. However, when the ADH-thirst mechanism is blocked, a 5-fold increase in
sodium intake causes the plasma sodium concentration to increase by more than 10%. Therefore, the major
mechanism for controlling extracellular sodium concentration (as well as extracellular osmolarity) is the
ADH-thirst mechanism. You should recall that ADH increases the permeability of the late distal tubule and
collecting duct to water, which allows water to be retained by the body and a concentrated urine to be excreted.
Aldosterone (choice A) and angiotensin II (choice B) are powerful salt-retaining hormones. They regulate the
total amount of sodium in the body, but have relatively little effect on plasma sodium concentration under
normal conditions for the following reasons: (1) they increase reabsorption of sodium and water to an equal
extent, and (2) any tendency for sodium concentration to change is immediately compensated for by changes
in ADH levels, which return sodium concentration to a normal value.
Atrial natriuretic factor (choice D) is released from the atria when blood volume increases. It acts on the kidneys
to increase the excretion of sodium and water. However, ANF does not have an important role in regulating
plasma sodium concentration because any tendency for sodium concentration (as well as osmolarity) to change
is immediately compensated for by changes in ADH levels, as discussed above.
Epinephrine (choice E) does not have an important role in regulating extracellular sodium concentration.
A research physiologist decides to use a marker to measure the volume of total body water in a volunteer
medical student. Which of the following substances would he most likely use?
A. Antipyrine
B. Cresyl violet
C. Evans blue
D. I131-albumin
E. Inulin
Explanation:
The correct answer is A. Antipyrine and tritium are both markers for total body water.
23
Cresyl violet (choice B) is a histological dye used to stain Nissl substance in neurons. It stains cell bodies.
Evans blue (choice C) is used to measure the plasma compartment.
I131-albumin (choice D) is used to measure the plasma compartment.
Inulin (choice E) is used to measure the extracellular fluid compartment.
A 28-year-old man decides to donate a kidney to his brother, who is in chronic renal failure, after HLA typing
suggests that he would be a suitable donor. He is admitted to the hospital, and his right kidney is removed and
transplanted into his brother. Which of the following indices would be expected to be decreased in the donor
after full recovery from the operation?
A. Creatinine clearance
B. Creatinine production
C. Daily excretion of sodium
D. Plasma creatinine concentration
E. Renal excretion of creatinine
Explanation:
The correct answer is A. Because creatinine is freely filtered by the glomerulus, but not secreted or reabsorbed
to a significant extent, the renal clearance of creatinine is approximately equal to the glomerular filtration rate.
In fact, creatinine clearance is commonly used to assess renal function in the clinical setting. When a kidney is
removed, the total glomerular filtration rate decreases because 50% of the nephrons have been removed,
which causes the creatinine clearance to decrease. In turn, the plasma creatinine concentration (choice D)
increases until the rate of creatinine excretion by the kidneys (choice E) is equal to the rate of creatinine
production by the body. Recall that creatinine excretion = GFR x plasma creatinine concentration. Therefore,
creatinine excretion is normal when GFR is decreased following removal of a kidney because the plasma
concentration of creatinine is elevated.
24
Creatinine is a waste product of metabolism. Creatinine
production (choice B) is directly related to the muscle
mass of an individual, but is independent of renal function.
The daily excretion of sodium (choice C) is unaffected by the removal of a kidney. The amount of sodium
excreted each day by the remaining kidney exactly matches the amount of sodium entering the body in the diet.
In a normal patient, if renal vascular resistance is decreased to 50% of its initial value, with no change in renal
artery or renal vein pressure, which of the following combinations of changes will occur?
Renal blood flow
Renal artery [oxygen]
Renal oxygen use
A. double
increase
no change
B. double
no change
increase
C. increase 50%
decrease
increase
D. increase 50%
increase
no change
E. decrease 50%
no change
decrease
Explanation:
25
The correct answer is B. Renal blood flow = (renal artery
pressure - renal vein pressure)/renal vascular
resistance (RVR). Therefore if RVR is decreased to half its original value, with no pressure changes, renal blood
flow must double (not increase 50%). Increased blood flow to the kidney will increase renal oxygen use by
increasing glomerular filtration rate, which increases the filtered load of sodium and other solutes. Since active
sodium reabsorption is load-dependent, increased tubular fluid sodium will increase all active sodium
reabsorption, which requires more ATP hydrolysis and synthesis (and hence more oxygen use). Renal artery
oxygen concentration will not change, since it is dependent on normal lung function, not oxygen extraction by the
kidney.
A 45-year-old woman is hospitalized after an automobile accident. The physician collects 1.44 L of urine from the
patient in a 24-hour period. The clinical lab returns the following results:
Plasma creatinine concentration
2.0 mg/mL
Plasma urea concentration
15 μmol/mL
Urine creatinine concentration
100 mg/mL
Urine urea concentration
160 μmol/mL
What is the approximate glomerular filtration rate (GFR; in mL/min) of this patient?
A. 10
B. 25
C. 50
D. 75
E. 100
Explanation:
The correct answer is C. Creatinine is formed from muscle creatine and released into the plasma at a fairly constant
rate. Creatinine passes freely through the glomerular membrane but is not reabsorbed to a significant extent and is
26
secreted only in small amounts by the kidney tubules.
Consequently, the glomerular filtrate has about the same
concentration of creatinine as the plasma. As the tubular fluid moves along the tubule, all the filtered creatinine
continues on into the urine. Therefore, the excretion rate of creatinine is approximately equal to the filtration rate of
creatinine. The rate of creatinine excretion and filtration can be determined from the creatinine concentration in the
urine (Ucreatinine) and plasma (Pcreatinine) and the rate of urine flow (V) as follows:
Excretion rate = Ucreatinine × V
Filtration rate = Pcreatinine × GFR
Because excretion rate and filtration rate of creatinine are approximately equal, as discussed above, GFR can be
determined as follows:
GFR X P craetinine
A healthy 22-year-old female medical student with normal kidneys decreases her sodium intake by 50% for a
period of 2 months. Which of the following parameters is expected to increase in response to the reduction in
sodium intake?
A. Arterial pressure
B. Atrial natriuretic peptide release
C. Extracellular fluid volume
D. Renin release
E. Sodium excretion
27
Explanation:
The correct answer is D. A reduction in sodium intake leads to a decrease in extracellular fluid volume (choice
C) and therefore a decrease in arterial pressure (choice A). The decrease in arterial pressure stimulates renin
release, which in turn leads to an increase in the formation of angiotensin II. The angiotensin II increases the
renal retention of salt and water (ie, decreases sodium excretion, choice E), which returns the extracellular fluid
volume nearly back to normal.
Atrial natriuretic peptide (choice B) is released from the two atria of the heart as a result of an increase in the
extracellular fluid volume. Therefore, a decrease in sodium intake would tend to decrease the release of atrial
natriuretic peptide.
PICTURE NEPHRON
Under normal conditions virtually 100% of the filtered load of glucose is reabsorbed by the kidney tubules. Which
part of the tubule shown above is expected to have the highest concentration of glucose under normal conditions?
A. A
B. B
C. C
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D. D
E. E
Explanation:
The correct answer is A. Glucose is freely filtered by the glomerular capillary membrane and totally reabsorbed in
the proximal tubule under normal conditions. Therefore, the concentration of glucose is highest in the proximal
portion of the proximal tubule. The concentration of glucose is essentially zero in the thin descending limb of loop
of Henle (choice B), distal convoluted tubule (choice C), cortical collecting tubule (choice D), and medullary
collecting tubule (choice E).
A 34-year-old man with severe diarrhea is admitted to the hospital through the emergency department. The
laboratory results shown in the table indicate that he has developed metabolic acidosis. What is the anion gap in
this patient?
A. 5
B. 10
C. 15
D. 20
E. 25
Explanation:
The correct answer is B. The anion gap is useful clinically to determine the basic cause of metabolic acidosis.
The anion gap is defined as the difference between the plasma Na+ concentration and the sum of the plasma Cland HCO3- concentrations: anion gap = [Na+] - ([Cl-] + [HCO3-]) = 140 - (116 + 14) = 10. The anion gap
normally ranges from 8 to 16 (no units are used). Because the plasma (like all solutions) must contain an equal
29
number of positive and negative charges in accordance
with the law of electroneutrality, the anion gap estimates
the unmeasured anions in the plasma. Included among these unmeasured anions are organic acids such as
lactate, acetoacetate, and others. Therefore, overproduction of endogenous acid or decreased ability to excrete
acid increases the anion gap (and is referred to as "normochloremic metabolic acidosis" because plasma
chloride is normal). Common causes of anion gap (normochloremic) metabolic acidosis (anion gap > 16) include
diabetic ketoacidosis, lactic acidosis, renal failure, salicylate poisoning, and methanol ingestion.
Non-anion gap (hyperchloremic) metabolic acidosis (anion gap between 8 and 16) is usually caused by a primary
loss of HCO3-. Common causes of hyperchloremic metabolic acidosis are diarrhea and renal tubular acidosis.
Hyperchloremic metabolic acidosis can also be caused by ingestion of HCl.
In a person weighing 75 kg, the volumes of total body water, intracellular fluid, and extracellular fluid are,
respectively
A. 40L, 30L, 10L
B. 45L, 30L, 15L
C. 45L, 35L, 10L
D. 50L, 25L, 25L
E. 50L, 35L, 15L
Explanation:
The correct answer is B. This is a straightforward fact question that will win you a quick point on the exam if you
get it right.
Total body water (TBW) in liters equals approximately 60% of body weight in kilograms and therefore equals 45
liters in a 75-kilogram person. Intracellular volume = 2/3 of TBW and is therefore 30 liters in this case.
Extracellular volume = 1/3 of TBW and is therefore 15 liters in this case.
Choices A, C, D, and E do not satisfy these conditions.
30
A certain substance, which is both freely filtered and secreted, is being maximally secreted. As the plasma
concentration of the substance increases, the renal clearance
A. decreases and approaches that of inulin
B. increases and approaches that of inulin
C. increases to the renal plasma flow
D. will decrease to zero
E. will remain the same
Explanation:
The correct answer is A. The clearance of any substance = urinary excretion/plasma concentration. For a
secreted substance, urinary excretion is the sum of the excretion of the filtered substance and the secreted
substance. As the plasma concentration initially rises, both the filtered and secreted components rise in
proportion to the plasma concentration, so clearance remains constant. Once the secretion mechanism is
saturated, its contribution to urinary concentration can no longer increase and remains constant. Although the
filtered contribution rises, urinary excretion is no longer rising as quickly in relationship to the plasma
concentration, so clearance falls. As the plasma concentration rises further, the contribution of the filtered
substance to its excretion becomes more and more dominant, so its clearance comes closer to that of inulin (a
substance that is filtered but not secreted or reabsorbed). This is the pattern seen with para-aminohippuric acid
(PAH); the clearance of PAH equals effective renal plasma flow unless the secreting mechanism is saturated.
The clearance of this substance will decrease (compare with choices B, C, and E) and approach (but never
equal) the clearance of inulin.
Since the clearance of inulin equals glomerular filtration rate, it will not be zero (choice D).
The obligatory urine volume is the minimal volume of urine in which the excreted solute can be dissolved. What is
the obligatory urine volume in a patient who has a maximum urine osmolarity of 1000 mOsmol/L and has 500
mOsmol of solute that must be excreted each day to maintain electrolyte balance?
31
A. 0.5 L/day
B. 1.0 L/day
C. 1.5 L/day
D. 2.0 L/day
E. Cannot be determined
Explanation:
The correct answer is A. If the maximum concentration of solute in the urine is 1000 mOsmol/L and if 500
mOsmol of solute must be excreted each day, the patient must excrete at least 0.5 L of urine each day (i.e.,
500 mOsmol/1000 mOsmol/L = 0.5 L).
A dialysis patient is admitted to the hospital with peripheral edema that has accumulated over a long holiday
weekend. Which of the following pairs of substances is best suited to determine the interstitial fluid volume of this
patient?
A. 51Cr-red cells and 125I-albumin
B. Heavy water and 125I-albumin
C. Inulin and 125I-albumin
D. Inulin and 22Na
E. Inulin and heavy water
Explanation:
The correct answer is C. The interstitial fluid volume cannot be measured directly, because interstitial fluid
occupies the spaces between the cells and along with the plasma volume, makes up the extracellular fluid
32
volume. Therefore, it is calculated by subtracting the
plasma volume from the extracellular fluid volume. Plasma
volume can be measured by the indicator dilution method using 125I-albumin as the indicator; extracellular fluid
volume can be measured using inulin as the indicator. Therefore, inulin and 125I-albumin are the substances
best suited to calculate interstitial fluid volume.
51Cr-red cells and 125I-albumin (choice A) are used to determine blood volume and plasma volume,
respectively (these volumes can be calculated from each other when hematocrit is known).
Heavy water and 125I-albumin (choice B) are used to determine total body water and plasma volume,
respectively.
Inulin and 22Na (choice D) are both used to determine extracellular fluid volume.
Inulin and heavy water (choice E) are used to determine extracellular fluid volume and total body water,
respectively.
A hypertensive patient is found to have a partial obstruction of the renal artery due to an atherosclerotic plaque.
The resultant decrease in blood flow causes the increased release of an enzyme from which of the following
structures?
A. Afferent arterioles
B. Arcuate arteries
C. Juxtaglomerular cells
D. Kupffer cells
E. Proximal convoluted tubule
Explanation:
The correct answer is C. The juxtaglomerular cells are in the wall of the afferent arteriole, close to the
glomerulus. In response to decreased blood pressure, they secrete renin, an enzyme that converts
angiotensinogen to angiotensin I. Angiotensin converting enzyme, found in the lungs, converts angiotensin I to
angiotensin II. Angiotensin II increases peripheral vascular resistance directly and stimulates aldosterone
33
secretion, resulting in increased reabsorption of sodium
and water in the distal convoluted tubules.
The afferent arteriole (choice A) carries blood from the interlobular arteries to the glomerulus. Filtration of blood
occurs in the glomerulus, with the filtrate entering Bowman's capsule.
The arcuate arteries (choice B) are branches of the interlobar arteries of the kidney. The arcuate arteries lie in
the corticomedullary junction of the kidney and give rise to interlobular arteries, which enter the cortex of the
kidney and supply the glomeruli.
Kupffer cells (choice D) are found in the liver, along the sinusoids. They are phagocytic cells that are part of the
reticuloendothelial system.
The proximal convoluted tubule (choice E) is directly continuous with Bowman's capsule. Most of the resorption
of the glomerular filtrate occurs in this part of the nephron.
A 56-year-old woman with a 25-year history of alcoholism and liver disease visits her physician complaining of
abdominal swelling and back pain. The physician notes that she has severe ascites and administers a loop diuretic.
The woman loses 5 L of fluid in a relatively short time. The data shown below were collected from the patient before
receiving the diuretic and after the diuretic had caused the loss of fluid.
.
Which of the following is the primary cause of the acid-base abnormality indicated in the above table?
A. Decreased HCO3- excretion
B. Decreased plasma aldosterone
C. Hyperkalemia
D. Hyperventilation
34
E. Loss of extracellular fluid
Explanation:
The correct answer is E. Diuretic therapy has resulted in the type of metabolic alkalosis called "contraction
alkalosis." Loop diuretics increase salt and water excretion by inhibiting tubular reabsorption of Na+ and Cl- in the
kidney. The salt and water that are lost from the body contain very little HCO3- so that virtually all of the HCO3contained in the ECF is retained in the body. In effect, the HCO3- present in the ECF (which includes the edema
fluid) becomes more and more concentrated as urine containing relatively little HCO3- is excreted from the body.
For example, if the ECF volume before diuretic therapy were 16 L, the total amount of HCO3- in the ECF would be
384 mEq (16 L x 24 mEq/L = 384 mEq). The excretion of 5 L of HCO3--free urine would cause the HCO3contained in the ECF to be concentrated into 11 L, thereby increasing the HCO3- concentration in the ECF to 35
mEq/L (384 mEq/11 L = 35 mEq/L). Therefore, it is the loss of extracellular fluid from the body that increases the
concentration of HCO3- in the plasma.
Because only small amounts of HCO3- are normally excreted in the urine, decreasing HCO3- excretion (choice A)
would have a small effect on the concentration of HCO3- in the plasma.
Diuretics lead to increased levels of aldosterone (choice B) and potassium (ie, hypokalemia, not hyperkalemia,
choice C) in the plasma by a variety of mechanisms.
Hyperventilation (choice D) decreases arterial PCO2; note that diuretic therapy has increased arterial PCO2.
A 35-year-old man is referred to the renal clinic for evaluation of proteinuria. He has no complaints other than
foamy urine. The following data are obtained from the patient:
Inulin clearance
100 mL/min
Plasma osmolarity
286 mOsm/L
Plasma sodium concentration
140 mEq/L
Urine flow
1.44 L / 24 hour
Urine osmolarity
205 mOsm/L
Urine sodium concentration
100 mEq/L
35
How much sodium does this patient reabsorb each day?
A. 14 mEq
B. 144 mEq
C. 244 mEq
D. 20,016 mEq
E. 20,160 mEq
Explanation:
The correct answer is D. The amount of sodium reabsorbed is equal to the amount filtered minus the amount
excreted in the urine. The amount of sodium filtered can be calculated as the product of the plasma sodium
concentration and the glomerular filtration rate (which is equal to the inulin clearance): sodium filtration = 140
mEq/L x 100 mL/min = 20,160 mEq/day. This is a normal tubular load of sodium equal to about 463 g sodium
filtered each day. The amount of sodium excreted by the kidneys can be calculated as the product of the urine
sodium concentration and urine flow rate: sodium excretion = 100 mEq/L x 1.44 L/day =144 mEq/day. This is a
normal amount of sodium excretion amounting to about 3.3 g/day. The amount of sodium reabsorbed is equal to
the amount filtered minus the amount excreted: sodium reabsorption = 20,160 mEq/day - 144 mEq/day = 20,016
mEq/day (this is about 460 g). Note that more than 99% of the filtered load of sodium is reabsorbed, and less
than 1% is excreted. Therefore, the renal handling of sodium in this patient appears to be normal.