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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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)?
2
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
3 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
4 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:
5
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
6 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
7 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
8 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)
9 = (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,
10
.
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
11 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
12
.
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
13 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
7 L
Plasma volume
4 L
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
14
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
15 pressure is the determinant of glomerular filtration rate most subject to physiological control.