Block 6 Renal Physiology Part 3
Questions and Answers
- 1.
Px= 5mg/ml Ux= 50mg/ml Pglu= 4mg/ml Uglu= 65 mg/ml Pin=2mg/ml Uin=120 mg/ml V= 2ml/min The filtered load of glucose is
- A.
A. 0ml/min
- B.
B. 32.5 ml/min
- C.
C. 120 ml/min
- D.
D. 480 mg/min
- E.
E. 600 mg/min
Correct Answer
D. D. 480 mg/minExplanation
FL = GFR x [Plasma con] , GFR determined by Inulin clearance. (Uin x V)/Pin=120
therefore FLglu = 120 x 4Rate this question:
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- 2.
Px= 5mg/ml Ux= 50mg/ml Pglu= 4mg/ml Uglu= 65 mg/ml Pin=2mg/ml Uin=120 mg/ml V= 2ml/min 500 mg/min is the rate at which Solute X is
- A.
A. Filtered
- B.
B. Reabsorbed
- C.
C. Secreted
- D.
D. Excreted
- E.
E. Cleared
Correct Answer
B. B. ReabsorbedExplanation
The rate at which Solute X is reabsorbed can be calculated by multiplying the concentration of Solute X in the urine (Ux) by the urine flow rate (V). In this case, Ux is given as 50 mg/ml and V is given as 2 ml/min. Therefore, the rate at which Solute X is reabsorbed is 50 mg/ml * 2 ml/min = 100 mg/min. Since the given answer is 500 mg/min, it does not match the calculated rate of reabsorption.Rate this question:
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- 3.
Px= 5mg/ml Ux= 50mg/ml Pglu= 4mg/ml Uglu= 65 mg/ml Pin=2mg/ml Uin=120 mg/ml V= 2ml/min The excretion rate for inulin is
- A.
A. The same as its filtration rate
- B.
B. 100 mg/min
- C.
C. 130 mg/min
- D.
D. 120 ml/min
- E.
E. 240 mg/min
Correct Answer
E. E. 240 mg/minExplanation
The excretion rate for inulin can be calculated by multiplying the concentration of inulin in urine (Uin) by the urine flow rate (V). In this case, Uin is given as 120 mg/ml and V is given as 2 ml/min. Multiplying these values gives us an excretion rate of 240 mg/min. Therefore, the correct answer is e. 240 mg/min.Rate this question:
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- 4.
The normal range for this value is 280-295 mOsm/L
- A.
The concentration of ADH in the plasma
- B.
Urine flow rate
- C.
Plasma osmolality
- D.
Free water clearance
- E.
The clearance of Na+
Correct Answer
C. Plasma osmolalityExplanation
Plasma osmolality refers to the concentration of solutes in the plasma, specifically the number of particles per unit of solvent. The normal range for plasma osmolality is 280-295 mOsm/L. This value is important for maintaining the balance of fluids and electrolytes in the body. Deviations from the normal range can indicate dehydration or overhydration. Therefore, monitoring plasma osmolality is crucial in assessing a person's hydration status and overall health.Rate this question:
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- 5.
A typical value for a well-nourished and well-hydrated person is 1 L/day
- A.
The concentration of ADH in the plasma
- B.
Urine flow rate
- C.
Plasma osmolality
- D.
Free water clearance
- E.
The clearance of Na+
Correct Answer
B. Urine flow rateExplanation
Urine flow rate refers to the volume of urine excreted by the kidneys per unit of time. A typical value for a well-nourished and well-hydrated person is 1 L/day. This means that a healthy individual would excrete approximately 1 liter of urine in a 24-hour period. The urine flow rate is influenced by various factors including fluid intake, hydration status, and kidney function. It is an important parameter to assess renal function and can provide insights into the body's fluid balance.Rate this question:
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- 6.
The value that distinguishes SIADH from dehydration
- A.
The concentration of ADH in the plasma
- B.
Urine flow rate
- C.
Plasma osmolality
- D.
Free water clearance
- E.
The clearance of Na+
Correct Answer
C. Plasma osmolalityExplanation
Plasma osmolality refers to the concentration of solutes in the blood plasma. In the case of SIADH (syndrome of inappropriate antidiuretic hormone secretion), there is an excessive release of antidiuretic hormone (ADH), leading to water retention and dilution of the blood plasma. This results in a decrease in plasma osmolality. On the other hand, dehydration is characterized by a decrease in total body water, causing an increase in plasma osmolality. Therefore, plasma osmolality is a key parameter that can help differentiate between SIADH and dehydration.Rate this question:
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- 7.
The value that distinguishes central diabetes insipidus from nephrogenic diabetes insipidus
- A.
The concentration of ADH in the plasma
- B.
Urine flow rate
- C.
Plasma osmolality
- D.
Free water clearance
- E.
The clearance of Na+
Correct Answer
A. The concentration of ADH in the plasmaExplanation
The concentration of ADH in the plasma is the key factor that distinguishes central diabetes insipidus from nephrogenic diabetes insipidus. In central diabetes insipidus, there is a deficiency or lack of production of ADH by the hypothalamus or release of ADH by the pituitary gland. This leads to decreased concentration of ADH in the plasma. On the other hand, in nephrogenic diabetes insipidus, there is a problem with the kidneys' response to ADH, leading to decreased sensitivity or resistance to its effects. In this condition, the concentration of ADH in the plasma is usually normal or even elevated.Rate this question:
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- 8.
The increase in peritubular capillary reabsorption that accompanies volume depletion is NOT due to
- A.
A. glomerulotubular balance
- B.
B. increased filtration fraction
- C.
C. increased peritubular capillary plasma oncotic pressure
- D.
D. an increase in angiotensin II
- E.
E. decreased peritubular capillary pressure
Correct Answer
A. A. glomerulotubular balanceExplanation
Glomerulotubular balance refers to the ability of the proximal tubules to regulate reabsorption based on changes in glomerular filtration rate (GFR). In volume depletion, the GFR is decreased, and as a compensatory mechanism, the proximal tubules increase their reabsorption to maintain the balance. Therefore, the increase in peritubular capillary reabsorption during volume depletion is not due to glomerulotubular balance.Rate this question:
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- 9.
H+/K+ ATPase is found in the parietal cells of the stomach and in the
- A.
A. proximal tubule epithelium
- B.
B. thick ascending limb epithelium
- C.
C. thin loop of Henle epithelium
- D.
D. alpha-intercalated cells
- E.
E. principal cells
Correct Answer
D. D. alpha-intercalated cellsExplanation
The correct answer is d. alpha-intercalated cells. H+/K+ ATPase is an enzyme that helps in the secretion of hydrogen ions (H+) into the stomach. It is found in the parietal cells of the stomach, where it plays a crucial role in the production of gastric acid. Alpha-intercalated cells, on the other hand, are found in the collecting ducts of the kidney and are responsible for the secretion of hydrogen ions (H+) and reabsorption of bicarbonate ions (HCO3-). Therefore, H+/K+ ATPase is not found in the proximal tubule epithelium, thick ascending limb epithelium, thin loop of Henle epithelium, or principal cells.Rate this question:
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- 10.
Potassium loss in the urine is increased by
- A.
An increase in intracellular [K+]
- B.
An increase in the activity of the electrogenic H+ pump in intercalated cells
- C.
A decrease in the electrical potential difference across the collecting duct epithelium
- D.
A decrease in the GFR
- E.
Decreased Na+ reabsorption by principal cells
Correct Answer
A. An increase in intracellular [K+]Explanation
An increase in intracellular [K+] can lead to an increase in potassium loss in the urine because when the intracellular concentration of potassium is high, it promotes the movement of potassium out of the cells and into the urine. This can occur through various mechanisms, such as increased potassium secretion by the kidneys or increased potassium excretion in the urine. Therefore, an increase in intracellular [K+] can result in an elevated level of potassium being excreted in the urine.Rate this question:
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- 11.
ADH
- A.
A. is secreted from neurons in the anterior pituitary
- B.
B. secretion is increased by a decrease in plasma osmolarity
- C.
C. secretion is stimulated by decrease in venous return
- D.
D. binds to receptors on the apical membrane of intercalated cells
- E.
E. mediates water reabsorption in the kidneys via V1 receptor
Correct Answer
C. C. secretion is stimulated by decrease in venous returnExplanation
antidiuretic hormone (ADH),
a hormone that decreases the production of urine by increasing the reabsorption of water by the renal tubules. It is secreted by cells of the hypothalamus and stored in the neurohypophysis. ADH is released in response to a decrease in blood volume, an increased concentration of sodium or other substances in plasma, pain, stress, or the action of certain drugs. ADH causes contraction of smooth muscle in the digestive tract and blood vessels, especially capillaries, arterioles, and venules. Acetylcholine, methacholine, nicotine, large doses of barbiturates, anesthetics, epINEPHrine, and norepinephrine stimulate ADH release; ethanol and phenytoin inhibit production of the hormone. Increased intracranial pressure promotes inappropriate increases and decreases in ADH. Synthetic ADH is used in the treatment of diabetes insipidus. Normal values are 1 to 5 pg/mL or less than 1.5 ng/L. Also called vasopressin.Rate this question:
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- 12.
With a loss of blood volume there is both a reduction in GFR and an increase in the reabsorption through the peritubular capillary. The rapid reduction in GFR following volume depletion is primarily due to
- A.
A. tubuloglomerular balance
- B.
B. sympathetic afferent arteriolar vasoconstriction
- C.
C. sympathetic efferent arteriolar vasoconstriction
- D.
D. the drop in systemic arterial pressure
- E.
E. an increase in plasma oncotic pressure
Correct Answer
D. D. the drop in systemic arterial pressureExplanation
afferents are consricted in response to an INCREASE in pressure
efferents are constricted in response to angiotensin 2,Rate this question:
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- 13.
A healthy young man weighs 80 kg. His ECW measures
- A.
A. 4 L.
- B.
B. 16 L.
- C.
C. 24 L.
- D.
D. 32 L.
- E.
E. 48 L.
Correct Answer
B. B. 16 L.Explanation
The correct answer is b. 16 L. The extracellular water (ECW) in a healthy young man typically accounts for about 20% of their body weight. Therefore, if the man weighs 80 kg, his ECW would be approximately 16 L.Rate this question:
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- 14.
When circulating levels of ADH are high, which segment of the nephron has the lowest permeability to H2O?
- A.
A. Proximal tubule
- B.
B. Descending limb of the loop of Henle.
- C.
C. Thick ascending limb of the loop of Henle.
- D.
D. Late distal tubule.
- E.
E. Inner medullary collecting duct
Correct Answer
C. C. Thick ascending limb of the loop of Henle.Explanation
When circulating levels of ADH are high, the thick ascending limb of the loop of Henle has the lowest permeability to water. ADH, also known as antidiuretic hormone, acts on the collecting ducts and the distal tubules to increase water reabsorption. However, the thick ascending limb of the loop of Henle is impermeable to water even in the presence of ADH. This segment actively transports sodium and chloride ions out of the tubular fluid, creating a concentration gradient that allows for the reabsorption of water in the collecting ducts. Therefore, the thick ascending limb of the loop of Henle has the lowest permeability to water.Rate this question:
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- 15.
Which solute is NOT transported by the symporter in the thick ascending limb?
- A.
A. Na+.
- B.
B. K+.
- C.
C. NH4+
- D.
D. Urea.
- E.
E. Cl- .
Correct Answer
D. D. Urea.Explanation
The symporter in the thick ascending limb transports Na+, K+, NH4+, and Cl-. However, urea is not transported by this symporter.Rate this question:
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- 16.
Which one of the components of the plasma listed below is NOT freely filtered?
- A.
A. Urea
- B.
B. Hemoglobin
- C.
C. Creatinine
- D.
D. Glucose
- E.
E. Water
Correct Answer
B. B. HemoglobinExplanation
Hemoglobin is not freely filtered in the plasma because it is a large protein molecule that is too big to pass through the filtration membrane in the kidneys. The filtration membrane only allows small molecules like urea, creatinine, glucose, and water to pass through. Hemoglobin is typically found inside red blood cells and is not normally present in the plasma.Rate this question:
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- 17.
The cells of the inner wall of Bowman’s capsule that share a basal lamina with the endothelial cells of glomerular capillaries are called
- A.
A. Juxtaglomerular cells.
- B.
B. Macula densa cells
- C.
C. Podocytes
- D.
D. Intercalated cells.
- E.
E. Mesangial cells.
Correct Answer
C. C. PodocytesExplanation
Podocytes are cells found in the inner wall of Bowman's capsule in the kidney. They have foot-like projections called pedicels that wrap around the glomerular capillaries, forming filtration slits that allow for the passage of filtered fluid into the Bowman's capsule. These podocytes share a basal lamina with the endothelial cells of the glomerular capillaries, forming the filtration barrier of the glomerulus. Therefore, the correct answer is c. Podocytes.Rate this question:
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- 18.
A kidney stone partially obstructs the ureter of a young man. Which of the following will change as compared to normal?
- A.
A. Pgc
- B.
B. Pt
- C.
C. πgc
- D.
D. πt
- E.
E. Kf
Correct Answer
B. B. PtExplanation
When a kidney stone partially obstructs the ureter, the pressure within the renal pelvis and calyces increases. This increase in pressure is transmitted to the Bowman's capsule, leading to an increase in the glomerular capillary pressure (Pgc). However, the oncotic pressure in the glomerular capillaries (πgc) remains the same because the obstruction does not affect the concentration of proteins in the blood. The filtration coefficient (Kf) also remains unchanged. Therefore, the only parameter that changes as compared to normal is the tubular pressure (Pt), which increases due to the obstruction.Rate this question:
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- 19.
Which of the following would NOT be observed in uncontrolled diabetes mellitus?
- A.
A. Dehydration
- B.
B. Positive glucose clearance
- C.
C. Osmotic diuresis
- D.
D. Hypokalemia
- E.
E. Hyperglycemia
Correct Answer
D. D. HypokalemiaExplanation
In uncontrolled diabetes mellitus, the body is unable to regulate blood sugar levels effectively, leading to hyperglycemia (high blood sugar levels). This high level of glucose in the blood causes osmotic diuresis, where excess glucose is excreted in the urine, leading to increased urine output and dehydration. Since the body is losing fluids through increased urination, dehydration is observed. Positive glucose clearance refers to the kidneys' ability to filter and remove excess glucose from the blood, which is a characteristic of uncontrolled diabetes. However, hypokalemia (low potassium levels) is not typically observed in uncontrolled diabetes mellitus.Rate this question:
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- 20.
Aldosterone secretion from the adrenal cortex is stimulated by
- A.
A. Hypokalemia
- B.
B. ANP
- C.
C. Hypervolemia
- D.
D. Angiotensin II.
- E.
E. Hyperchloremia
Correct Answer
D. D. Angiotensin II.Explanation
Aldosterone secretion from the adrenal cortex is 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 secretion of aldosterone, which promotes the reabsorption of sodium and the excretion of potassium in the kidneys. This helps to increase blood volume and blood pressure. Hypokalemia, ANP, hypervolemia, and hyperchloremia do not directly stimulate aldosterone secretion.Rate this question:
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- 21.
After ingestion of 2 L of water in a 5 minute period, which one of the following occurs?
- A.
A. Thirst
- B.
B. Increased Posm
- C.
C. Decreased ADH secretion.
- D.
D. Decreased RPF.
- E.
E. Decreased free water clearance.
Correct Answer
C. C. Decreased ADH secretion.Explanation
When a large amount of water is ingested in a short period of time, it leads to a decrease in the secretion of antidiuretic hormone (ADH) by the pituitary gland. ADH is responsible for regulating the reabsorption of water in the kidneys. When ADH secretion decreases, it causes an increase in urine production and a decrease in the reabsorption of water, resulting in the excretion of more water from the body.Rate this question:
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- 22.
The plasma concentration of protein Z is 5 mg/ml. Its filterability quotient is 0.2. GFR is 100 ml/min. What is the filtered load of protein Z?
- A.
A. 100 ml/min
- B.
B. 500 ml/min.
- C.
C. 1 L/min.
- D.
D. 100 mg/min
- E.
E. 500 mg/min
Correct Answer
D. D. 100 mg/minExplanation
The filtered load of a substance is calculated by multiplying its plasma concentration by the glomerular filtration rate (GFR). In this case, the plasma concentration of protein Z is 5 mg/ml and the GFR is 100 ml/min. Therefore, the filtered load of protein Z would be 5 mg/ml * 100 ml/min, which equals 500 mg/min. Therefore, the correct answer is d. 100 mg/min.Rate this question:
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- 23.
Which of the following is NOT a step in counter-current multiplication?
- A.
Energy-dependent reabsorption of solute from the tubular fluid of the thick ascending limb.
- B.
Equilibration of the osmolarity of the medullary interstitium and the tubular fluid in the descending limb.
- C.
Slow flow of tubular fluid.
- D.
Reabsorption of water from the tubular fluid of the descending limb.
- E.
Reabsorption of solute and water from the tubular fluid of the proximal tubule.
Correct Answer
E. Reabsorption of solute and water from the tubular fluid of the proximal tubule.Explanation
The correct answer is "Reabsorption of solute and water from the tubular fluid of the proximal tubule." This is not a step in counter-current multiplication because the proximal tubule is located before the loop of Henle, where counter-current multiplication occurs. In counter-current multiplication, the thick ascending limb actively reabsorbs solute from the tubular fluid, creating a concentration gradient in the medullary interstitium. This concentration gradient allows for water reabsorption in the descending limb. The slow flow of tubular fluid also contributes to the establishment of the concentration gradient.Rate this question:
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