A nurse is assessing a client's arteriovenous (AV) fistula. Which finding provides evidence of its normal function?

Correct Answer: A

Rationale: An enlarged vein is a normal finding in a functioning arteriovenous (AV) fistula, indicating adequate blood flow and patency. An AV fistula, used for dialysis, relies on arterialization of the vein, which causes it to dilate and thicken over time. Placelessness (likely a typo for 'pulselessness') suggests occlusion, redness may indicate infection, and an ecchymotic area points to trauma or bleeding—none of which are normal. The enlarged vein, often accompanied by a palpable thrill or audible bruit, confirms the fistula is working properly and ready for dialysis access.

Correct Answer: D

Rationale: A serum sodium of 185 mEq/L in diabetes insipidus (DI) warrants immediate intervention. DI causes excessive water loss via dilute urine, leading to hypernatremia (normal 135-145 mEq/L). A level of 185 indicates severe dehydration, risking neurologic damage (e.g., seizures) or death, requiring urgent fluid replacement (e.g., hypotonic IV). Tachycardia, polyuria/thirst, and dry skin are expected but less acute, making hypernatremia the priority for rapid correction.

Correct Answer: B

Rationale: Glomerular filtrate is the fluid filtered from blood plasma into the glomerular capsule. It contains water, electrolytes, glucose, and small molecules but has far fewer proteins due to the filtration membrane's selective barrier (fenestrated endothelium, basement membrane, podocytes), which restricts large proteins like albumin. Plasma retains these proteins, and filtrate includes glucose and sodium in similar concentrations to plasma initially, making B the key difference.

Correct Answer: D

Rationale: The glomerular filtration rate (GFR) is about 125 ml/min, producing 180 L/day of filtrate, but only 1-2 L of urine is excreted daily. Thus, approximately 1% of the GFR is excreted, with 99% reabsorbed by the tubules. Option A reverses this, B underestimates reabsorption, and C is too low, making D the correct statement.

Correct Answer: C

Rationale: Bicarbonate (HCO₃⁻) reabsorption in the proximal tubule involves H⁺ secretion combining with HCO₃⁻ to form H₂CO₃, which breaks into CO₂ and H₂O via luminal carbonic anhydrase (not extracellular). CO₂ diffuses into cells, reforming HCO₃⁻ intracellularly with intracellular carbonic anhydrase. Tubules are impermeable to HCO₃⁻ (D is true), making C false due to the enzyme location error.