Which of the following is not in the sequence of proper kidney blood flow? The starting point is the renal artery and the finishing point is the renal vein.

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Multiple Choice Questions On Urinary System Questions

Question 1 of 5

Which of the following is not in the sequence of proper kidney blood flow? The starting point is the renal artery and the finishing point is the renal vein.

Correct Answer: C

Rationale: Blood flows: renal artery → interlobar → arcuate (arciform) → afferent → glomerulus → efferent → arcuate vein → interlobar vein → renal vein; interlobar vein isn't arterial. This corrects flow, key for perfusion, contrasting with venous misplacement.

Question 2 of 5

This substance cannot pass through semipermeable walls of glomerulus

Correct Answer: D

Rationale: Globin, albumin, and blood cells are too large for glomerular filtration all apply. This defines filtration barrier, critical for selectivity, contrasting with small solutes.

Question 3 of 5

This artery passes blood to the kidney

Correct Answer: C

Rationale: Renal artery supplies kidney not iliac (pelvis), cystic (bladder), or coeliac (gut). This identifies perfusion, critical for function, contrasting with other arteries.

Question 4 of 5

What is the filtration membrane, where is it located, and what is it made of?

Correct Answer: A

Rationale: The filtration membrane lies between blood and glomerular capsule comprises fenestrated endothelium (capillaries), visceral membrane (podocytes), and basement membrane (fused laminae), filtering plasma. The proximal tubule reabsorbs no filtration role. The Loop of Henle concentrates lacks this structure. The collecting duct adjusts via aquaporins not filtration. Its three-layered design distinguishes it, critical for selective filtration, unlike reabsorption or concentration sites.

Question 5 of 5

A 37-year-old man has blood glucose of 900 mg/dL and Tmax of 380 mg/min. What is the expected amount of glucose in the urine?

Correct Answer: B

Rationale: Glucose in urine = filtration rate - Tmax: 900 mg/dL × 1.25 (GFR 125 mL/min) = 1125 mg/min - 380 = 745 mg/min excess spills (e.g., diabetes). Zero assumes full reabsorption false above Tmax. 380 is Tmax not excreted. 1125 is filtered ignores reabsorption. 745 mg/min distinguishes it, critical for glucosuria calculation, unlike total, max, or nil values.

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