Block 6 Renal Physiology Part 7
Questions and Answers
- 1.
RBF = 1 L/min Uin = 18 mg/ml Hematocrit = 40% Pin = 0.2 mg/ml V = 2 ml/min Filtration fraction is
- A.
A. 0.15.
- B.
B. 0.20.
- C.
C. 0.25.
- D.
D. 0.30.
- E.
E. Cannot be determined from the available data.
Correct Answer
D. D. 0.30.Explanation
The filtration fraction can be calculated by dividing the glomerular filtration rate (GFR) by the renal plasma flow (RPF). In this case, the GFR is equal to the product of the ultrafiltration coefficient (Kf), the net filtration pressure (NFP), and the filtration surface area (A). The RPF is equal to the product of the renal blood flow (RBF) and (1 - hematocrit). Since the values for Kf, NFP, and A are not given, we cannot calculate the GFR. Therefore, the filtration fraction cannot be determined from the available data.Rate this question:
-
- 2.
RBF = 1 L/min Uin = 18 mg/ml Hematocrit = 40% Pin = 0.2 mg/ml V = 2 ml/min If the interstitial osmotic gradient varies from 300 mOsm/L at the cortico-medullary boundary to 700 mOsm/L at the papillary tips, which of the following statements is true?
- A.
A. Less than 50% of the filtered urea is reabsorbed by the proximal tubule.
- B.
B. Glucose reabsorption is reduced in the proximal tubule
- C.
C. ADH levels are maximal
- D.
D. The patient is extremely dehydrated.
- E.
E. The osmolality of the patient’s urine is 700 mOsm/L or less.
Correct Answer
E. E. The osmolality of the patient’s urine is 700 mOsm/L or less.Explanation
The given information states that the interstitial osmotic gradient varies from 300 mOsm/L at the cortico-medullary boundary to 700 mOsm/L at the papillary tips. This gradient drives the reabsorption of water in the renal tubules. Since the osmolality of the patient's urine is determined by the amount of water reabsorbed, it can be inferred that if the osmotic gradient is higher, more water will be reabsorbed and the urine osmolality will be lower. Therefore, the statement "The osmolality of the patient’s urine is 700 mOsm/L or less" is true.Rate this question:
-
- 3.
Solute X is freely filtered at the glomerulus. Fifty percent (50%) of its filtered load is reabsorbed by the proximal tubule. Assuming normal glomerulotubular balance, what will the TF/P ratio be for solute X at the end of the proximal tubule?
- A.
A. 0
- B.
B. 0.33
- C.
C. 0.67
- D.
D. 1.0
- E.
E. 1.5
Correct Answer
E. E. 1.5Explanation
The TF/P ratio represents the amount of solute that is filtered by the glomerulus and reaches the end of the proximal tubule. In this case, since 50% of the filtered load of solute X is reabsorbed by the proximal tubule, only 50% of the initial amount of solute X remains at the end of the proximal tubule. Therefore, the TF/P ratio will be 1.5, indicating that 1.5 times the amount of solute X that was filtered reaches the end of the proximal tubule.Rate this question:
-
- 4.
The most potent stimulus for the secretion of ADH from the posterior pituitary is
- A.
A. 20% loss of blood volume
- B.
B. Posm = 290 mOsm/L, if the set point is 292 mOsm/L.
- C.
C. Hyponatremia
- D.
D. Ingestion of 2L of pure water
- E.
E. Hypokalemia
Correct Answer
A. A. 20% loss of blood volumeExplanation
The correct answer is a. 20% loss of blood volume. When there is a significant loss of blood volume, the body activates the release of antidiuretic hormone (ADH) from the posterior pituitary gland. ADH helps to conserve water by reducing urine production and increasing water reabsorption in the kidneys. This response helps to maintain blood pressure and prevent further fluid loss. The other options do not directly stimulate the secretion of ADH.Rate this question:
-
- 5.
A patient with uncontrolled diabetes mellitus comes into the emergency room in a coma. His laboratory values are listed below. Which value must represent a laboratory error?
- A.
A. PK = 3.5 mEq/L
- B.
B. PHCO3- = 9 mEq/L
- C.
C. PCO2 = 24 mm Hg
- D.
D. Arterial pH = 7.20
- E.
E. Pglucose = 10 mg/ml
Correct Answer
A. A. PK = 3.5 mEq/LExplanation
The laboratory value that must represent a laboratory error is a. PK = 3.5 mEq/L. This is because the normal range for potassium (K) levels in the blood is typically between 3.5-5.0 mEq/L. A value of 3.5 mEq/L is within the normal range and would not be indicative of a laboratory error. Therefore, the correct answer is a laboratory error.Rate this question:
-
- 6.
In a patient with nephritic syndrome, the normal permeability barrier of the glomerular capillaries is greatly reduced allowing the filtration of plasma proteins. The reduction in plasma proteins in the circulation will lead to
- A.
A. Increased reabsorption of fluid by peritubular capillaries.
- B.
B. Decreased ISF.
- C.
C. Increased filtration of fluid by systemic capillaries
- D.
D. Increased plasma oncotic pressure.
- E.
E. No change in TBW.
Correct Answer
C. C. Increased filtration of fluid by systemic capillariesExplanation
In nephritic syndrome, the permeability barrier of the glomerular capillaries is greatly reduced, leading to increased filtration of plasma proteins. This reduction in plasma proteins in the circulation will result in increased filtration of fluid by systemic capillaries. This is because the decreased protein levels in the blood will decrease the plasma oncotic pressure, which normally opposes filtration. As a result, more fluid will be filtered out of the capillaries into the interstitial space and ultimately into the systemic circulation.Rate this question:
-
- 7.
Which clearance is largest in a patient with central diabetes insipidus?
- A.
A. Free water
- B.
B. Osmolar
- C.
C. Glucose
- D.
D. Glycine
- E.
E. Serum albumin
Correct Answer
A. A. Free waterExplanation
In a patient with central diabetes insipidus, there is a deficiency of antidiuretic hormone (ADH), leading to excessive production of dilute urine. Free water clearance refers to the ability of the kidneys to excrete water without affecting the excretion of solutes. Since central diabetes insipidus is characterized by the inability to concentrate urine, the clearance of free water would be the largest in these patients. This means that the kidneys are unable to reabsorb water effectively, resulting in a high clearance of free water.Rate this question:
-
- 8.
A healthy young woman weighs 132 lb. (60 kg). What is the volume of her ICW?
- A.
A. 2.5 L
- B.
B. 7.5 L
- C.
C. 10 L
- D.
D. 20 L
- E.
E. 30 L
Correct Answer
D. D. 20 LExplanation
The volume of ICW (Intracellular Water) in a healthy young woman weighing 132 lb (60 kg) is estimated to be around 20 L.Rate this question:
-
- 9.
In a patient suffering from metabolic acidosis, which compensation does not occur?
- A.
A. Buffering of H+ by intracellular proteins.
- B.
B. Movement of K+ from the interstitium into muscle cells.
- C.
C. Hyperventilation
- D.
D. Increased reabsorption of K+ by intercalated cells.
- E.
E. Increased formation of ammonia and ammonium ion.
Correct Answer
B. B. Movement of K+ from the interstitium into muscle cells.Explanation
In a patient suffering from metabolic acidosis, compensation occurs through various mechanisms. Hyperventilation (option c) is a compensatory response where the patient breathes rapidly to decrease the carbon dioxide levels and increase the pH. Buffering of H+ by intracellular proteins (option a) helps maintain the pH balance by binding excess hydrogen ions. Increased formation of ammonia and ammonium ion (option e) also helps in buffering the excess acid. Increased reabsorption of K+ by intercalated cells (option d) occurs to maintain the acid-base balance. However, movement of K+ from the interstitium into muscle cells (option b) is not a compensation mechanism for metabolic acidosis.Rate this question:
-
- 10.
In hemorrhage, which normal kidney function will not occur?
- A.
A. Urea recycling.
- B.
B. Glomerulotubular feedback.
- C.
C. Countercurrent multiplication.
- D.
D. Ultrafiltration
- E.
E. Concentration of the urine.
Correct Answer
B. B. Glomerulotubular feedback.Explanation
Glomerulotubular feedback refers to the mechanism by which the macula densa cells in the distal convoluted tubule sense the sodium chloride concentration in the filtrate and regulate the glomerular filtration rate (GFR) accordingly. In hemorrhage, there is a decrease in blood volume, leading to a decrease in GFR. However, the macula densa cells are unable to sense this decrease in sodium chloride concentration due to reduced blood flow, resulting in a lack of appropriate feedback to regulate GFR. Therefore, glomerulotubular feedback does not occur in hemorrhage.Rate this question:
-
- 11.
Since we must excrete approximately 600 mEq of solutes per day, what is the minimal volume of urine per day that must be excreted?
- A.
A. 0 ml.
- B.
B. 1 ml
- C.
C. 100 ml.
- D.
D. 500 ml.
- E.
E. 1 L.
Correct Answer
D. D. 500 ml.Explanation
maximum concentration is 1200mOSM/L, = 1200mOSM/1000ml therefore;
600meq/ [1200mOSM/1000ml] =500mlRate this question:
-
- 12.
V = 3 ml/min Pin = 0.3 mg/ml Uin = 20 mg/ml Posm = 300 mOsm/L Uosm = 900 mOsm/L Pglu = 5 mg/ml Uglu = 25 mg/ml PaPAH = 1.1 mg/ml UPAH = 220 mg/ml PvPAH = 0.1 mg/ml In this patient, CH2O equals
- A.
A. 9 ml/min.
- B.
B. 6 ml/min
- C.
C. 0.
- D.
D. -6 ml/min.
- E.
E. -9 ml/min.
Correct Answer
D. D. -6 ml/min.Explanation
Free water clearance:
V = Cosm + Cwater
Cwater = V – Cosm
Cosm = (Uosm x V)/Posm
900x3/300 = 9
3-9=-6Rate this question:
-
- 13.
V = 3 ml/min Pin = 0.3 mg/ml Uin = 20 mg/ml Posm = 300 mOsm/L Uosm = 900 mOsm/L Pglu = 5 mg/ml Uglu = 25 mg/ml PaPAH = 1.1 mg/ml UPAH = 220 mg/ml PvPAH = 0.1 mg/ml GFR equals
- A.
90 ml/min.
- B.
180ml/min
- C.
60
- D.
200 ml/min.
- E.
120 ml/min.
Correct Answer
D. 200 ml/min.Explanation
20x3/.3Rate this question:
-
- 14.
V = 3 ml/min Pin = 0.3 mg/ml Uin = 20 mg/ml Posm = 300 mOsm/L Uosm = 900 mOsm/L Pglu = 5 mg/ml Uglu = 25 mg/ml PaPAH = 1.1 mg/ml UPAH = 220 mg/ml PvPAH = 0.1 mg/ml RPF equals
- A.
900 ml/min.
- B.
600 ml/min
- C.
120 ml/min
- D.
60 ml/min.
- E.
90 ml/min.
Correct Answer
B. 600 ml/minExplanation
RPF = clearance of PAH, = (220/1.1) x 3 = 600Rate this question:
-
- 15.
V = 3 ml/min Pin = 0.3 mg/ml Uin = 20 mg/ml Posm = 300 mOsm/L Uosm = 900 mOsm/L Pglu = 5 mg/ml Uglu = 25 mg/ml PaPAH = 1.1 mg/ml UPAH = 220 mg/ml PvPAH = 0.1 mg/ml Tubular fluid flow in this patient is
- A.
A. Normal
- B.
B. Increased because glucose is working as an osmotic diuretic.
- C.
C. Increased because the patient is hypervolemic.
- D.
D. Decreased because glucose vasoconstricts afferent arterioles.
- E.
E. Decreased because Kf is reduced in uncontrolled diabetes mellitus.
Correct Answer
B. B. Increased because glucose is working as an osmotic diuretic.Explanation
The patient has a high concentration of glucose in their urine (Uglu = 25 mg/ml) compared to their urine concentration of sodium (Uosm = 900 mOsm/L). This indicates that glucose is being excreted in the urine, which is not normal. Glucose acts as an osmotic diuretic, meaning it increases the osmotic pressure in the tubular fluid, leading to increased urine flow. Therefore, the tubular fluid flow in this patient is increased due to glucose acting as an osmotic diuretic.Rate this question:
-
- 16.
A patient is severely dehydrated following prolonged diarrhea that lasted several days but was not associated with vomiting. However, malaise and nausea prevented the patient form taking much food and drink during this period. After 3 days of this diarrhea the blood values are pHa 7.34, [HCO3-] = 12 mM, PaCO2 = 20 mm Hg, [Na+] = 138 mM, [K+] = 4.8 mM, [Cl-] = 114 mM. The above patient has
- A.
A. uncompensated metabolic acidosis.
- B.
B. partially compensated metabolic acidosis
- C.
C. uncompensated respiratory acidosis
- D.
D. partially compensated respiratory acidosis
- E.
E. a significantly increased anion gap
Correct Answer
B. B. partially compensated metabolic acidosisExplanation
The patient is experiencing metabolic acidosis, as indicated by the low pH (7.34) and low bicarbonate (HCO3-) level (12 mM). The low bicarbonate level suggests that there is an excess of acid in the body. The low PaCO2 (20 mm Hg) indicates that the respiratory system is trying to compensate by increasing ventilation to eliminate carbon dioxide. However, this compensation is only partial, as the pH is still below the normal range. This is consistent with the diagnosis of partially compensated metabolic acidosis. There is no information provided to suggest an increased anion gap, so option e can be ruled out.Rate this question:
-
- 17.
A patient is severely dehydrated following prolonged diarrhea that lasted several days but was not associated with vomiting. However, malaise and nausea prevented the patient form taking much food and drink during this period. After 3 days of this diarrhea the blood values are pHa 7.34, [HCO3-] = 12 mM, PaCO2 = 20 mm Hg, [Na+] = 138 mM, [K+] = 4.8 mM, [Cl-] = 114 mM. In the above patient, which one of the following would you expect to be decreased?
- A.
A. Plasma [aldosterone].
- B.
B. Plasma [atrial natriuretic peptide].
- C.
C. Plasma [angiotensin II].
- D.
D. Plasma [ADH].
- E.
E. Hematocrit.
Correct Answer
B. B. Plasma [atrial natriuretic peptide].Explanation
Prolonged diarrhea can lead to dehydration and loss of fluids and electrolytes, including sodium and chloride. This can result in a decrease in blood volume and blood pressure. In response to low blood volume and pressure, the body releases atrial natriuretic peptide (ANP), a hormone that promotes the excretion of sodium and water by the kidneys, leading to increased urine output. Therefore, in this patient, it would be expected that plasma ANP levels would be increased, not decreased. The correct answer is b. Plasma [atrial natriuretic peptide].Rate this question:
-
- 18.
A patient is severely dehydrated following prolonged diarrhea that lasted several days but was not associated with vomiting. However, malaise and nausea prevented the patient form taking much food and drink during this period. After 3 days of this diarrhea the blood values are pHa 7.34, [HCO3-] = 12 mM, PaCO2 = 20 mm Hg, [Na+] = 138 mM, [K+] = 4.8 mM, [Cl-] = 114 mM. In the above patient, comparing the relative changes in the volumes of various compartments, which statement is correct?
- A.
A. The ICF volume decreases more than the ECF volume
- B.
B. The packed RBC volume decreases more than the plasma volume
- C.
C. The ISF volume decreases more than the plasma volume
- D.
D. The volumes of all compartment decrease to an equal degree.
- E.
E. The plasma volume decreases more than any other volume decreases.
Correct Answer
C. C. The ISF volume decreases more than the plasma volumeExplanation
In this scenario, the patient is experiencing severe dehydration due to prolonged diarrhea. The malaise and nausea have prevented the patient from consuming adequate food and drink. The blood values indicate metabolic acidosis with a low bicarbonate concentration and low carbon dioxide levels. The sodium and chloride levels are within normal range.
Based on this information, option c is correct. The interstitial fluid (ISF) volume decreases more than the plasma volume. This is because the patient is losing fluid through diarrhea, which primarily affects the ISF volume. The plasma volume may also decrease, but to a lesser extent compared to the ISF volume.Rate this question:
-
- 19.
Aldosterone secretion in the adrenal cortex is
- A.
A. from cells of the zona fasciculata.
- B.
B. directly stimulated by angiotensin II.
- C.
C. directly stimulated by an increase in ECF [Na+].
- D.
D. directly stimulated by a a decrease in ECF [K+].
- E.
E. indirectly stimulated by increasing the GFR.
Correct Answer
B. B. directly stimulated by angiotensin II.Explanation
Aldosterone secretion in the adrenal cortex is directly stimulated by angiotensin II. Angiotensin II is a hormone that is produced in response to low blood pressure or low blood volume. It acts on the adrenal cortex to stimulate the release of aldosterone, which helps to increase sodium reabsorption and potassium excretion in the kidneys. This ultimately leads to an increase in blood volume and blood pressure.Rate this question:
-
- 20.
Renin
- A.
A. is released from macula densa cells.
- B.
B. secretion is stimulated by angiotensin II
- C.
C. secretion is inhibited by renal sympathetic stimulation
- D.
D. is a steroid hormone.
- E.
E. converts angiotensinogen to angiotensin I.
Correct Answer
E. E. converts angiotensinogen to angiotensin I.Explanation
Renin is an enzyme that is responsible for converting angiotensinogen, a protein produced by the liver, into angiotensin I. This conversion is the first step in the renin-angiotensin-aldosterone system, a hormonal cascade that regulates blood pressure and fluid balance in the body. Angiotensin I is then further converted into angiotensin II, a potent vasoconstrictor that increases blood pressure. Therefore, the correct answer is that renin converts angiotensinogen to angiotensin I.Rate this question:
-
- 21.
The major determinant of the plasma potassium concentration is
- A.
A. the dietary potassium intake
- B.
B. aldosterone
- C.
C. ADH
- D.
D. ANP
- E.
E. angiotensin II.
Correct Answer
B. B. aldosteroneExplanation
Aldosterone is a hormone produced by the adrenal glands that plays a crucial role in regulating the body's potassium levels. It acts on the kidneys to increase the reabsorption of sodium and the excretion of potassium, leading to an increase in plasma potassium concentration. Therefore, the major determinant of the plasma potassium concentration is aldosterone. Dietary potassium intake can also influence potassium levels, but it is not the primary determinant. ADH, ANP, and angiotensin II do not directly affect potassium levels.Rate this question:
-
- 22.
Angiotensin II
- A.
A. inhibits ADH secretion.
- B.
B. inhibits Na+-H+ exchange.
- C.
C. increases the RPF.
- D.
D. stimulates thirst.
- E.
E. causes generalized vasodilation.
Correct Answer
D. D. stimulates thirst.Explanation
Angiotensin II is a hormone that is released in response to low blood pressure or low blood volume. One of its main functions is to stimulate thirst. When angiotensin II is released, it acts on the brain to increase the sensation of thirst, leading to increased water intake. This helps to restore blood volume and maintain blood pressure within normal range. Therefore, the correct answer is d. stimulates thirst.Rate this question:
-
Back to top