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.

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.

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.

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).

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 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 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 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.

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 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.

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 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.

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 correct answer is C. A 5-fold increase in sodium intake causes the plasma sodium concentration to 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.

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.

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.

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 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

 

 

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 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.

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).

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 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).