Chapter 25. 

Multiple Choice Questions

Answers A, C, D and E are true statements, whereas B is false.

Case History A

1.  No, b-blockers often aggravate the glucose intolerance and worsen the hyperlipidaemia (cholesterolaemia).

2. Carboanhydrase inhibitors promote the outflow of the aqueous humour from the eye and probably diminishes its secretion. Blockage of carboanhydrase inhibits the reabsorption of bicarbonate at the brush borders of the proximal tubule cells, and also the smaller reabsorption at the intercalated cells of the collecting ducts. Thus, more bicarbonate, Na+ and water is excreted in the urine. Therefore, the patient may develop a metabolic acidosis as well as dehydration. However, acetazolamide is a mild diuretic and often preferred to reduce the intraocular pressure in cases of glaucoma.

3. Yes, diuretics  sometimes aggravate diabetes mellitus. Thiazides and loop diuretics aggravate the glucose intolerance.

4. The renal oxygen uptake is (1.2 l/min * 15 ml oxygen/l) = 18 ml oxygen per min. In percent of the total oxygen uptake for an adult person (250 ml/min), this is: 18/250 = 0.072 or 7.2%.

5. The kidney weight in % of the total body weight: (300 *100/70 000) = 0.43%.

6. Is the renal bloodflow redundant compared to the renal oxygen consumption? Yes, the kidneys consume much more oxygen than indicated by their relative weight. The redundant renal bloodflow reflects the need of a high filtering and reabsorption capacity for the regulation of the body fluids.

Case History B

1. The blood [glucose] is normal, since the two values are identical - an ideal fasted blood [glucose]: 1000 mg equal to 1000/180 or 5.56 mmol or per litre 5.56 mM.

2. The creatinine clearance is (minute diuresis*urine creatinine/plasma creatinine): 0.0015 litre of urine per min * 6 mM/0.09 mM = 0.1 litre of plasma per min. 

3. The creatinine clearance is an acceptable estimate of the inulin clearance and thus of the plasma volume passing the glomerular barrier each min. The reabsorption of glucose in the proximal tubules  is thus: Creatinine clearance * blood [glucose] = 0.1 * 5.56 mM or 0.556 mmol glucose each min. This is 0.556/1.78 =0.31 or 31% of the normal reabsorption capacity. The maximal glucose reabsorption capacity is probably somewhat higher being related to the saturation threshold.

4. The appearance threshold in this case is a normal fasting concentration. The saturation threshold is higher, because glucose appears as soon as a few nephrons with a large glomerulus and small tubules are overwhelmed. 

The saturation threshold is the blood glucose concentration, where the reabsorption mechanism of all nephrons are saturated. 

Case History C

1. Rheumatic fever.

2. The acute postinfectious glomerulonephritis occurs typically in a child, who has suffered from streptococcal tonsillitis a few weeks before. Haematuria, albuminuria, and oliguria characterise the condition with salt-water retention causing oedemas and hypertension.

3. The GP arranges immediate admission to hospital for further examination and treatment. 

Case History D

 1.        The nurse turned her intermediary metabolism into ketosis by fasting, and developed a classical hunger diabetes. She had a totally compensated (normal pH) metabolic acidosis.

 2.        The nurse was advised to improve her eating habits (healthy and varied food). Her water without food diet proved to be an insufficient eating pattern.

Case History E

 1.        The clearance for PAH = 100/0.2 = 500 ml of plasma per min. This is the definition of clearance: The excretion flux of PAH divided by its plasma concentration. At least 500 ml of plasma with low plasma concentration must pass the kidneys each minute to accomplish this clearance.

 2.        The tubular secretion flux of PAH equals the excretion flux in the urine minus the filtration flux.

The secretion flux: (100 - (125 × 0.2 × 0.8)) = 80 mg/min of PAH.

 3.        Renal plasma flow (RPF) equals the excretion flux for PAH divided by the renal, arteriovenous concentration difference for PAH (0.2 - 0.02 mg/ml). 

RPF = 100/(0.2 - 0.02) = 556 ml of plasma per min. This is the true RPF, which is 10% higher than the clearance for PAH determined in 1. 

RBF = 556/0.57 = 975 ml of blood per min.

 4.        The filtration flux for PAH: (125 × 1 × 0.8) = 100 mg of PAH per min. The excretion flux for PAH: (80 + 100) = 180 mg of PAH per min.

 5.        The new clearance for PAH: 180/1 = 180 ml of plasma per min.

Return to chapter 25

Return to Content