Answers A, B and E are true statements, whereas C, and D are false.
1. All cell membranes are water permeable, so the total body water will be evenly distributed across the cell membranes. There is no loss of solutes, so the total amount is maintained: (300 mOsmol kg-1 * 42 kg) = (x * 39.5 kg). Accordingly, the mean osmolality is increased to: (300* 42/39.5) = 319 mOsmol kg-1.
2. Yes. The hypothalamic osmoreceptors are stimulated, and they release ADH from the neurohypophysis. Increased serum-ADH will increase the water reabsorption and tend to minimise the hyperosmolality.
3. Almost all NaCl of the body is located in the ECV, so the normal content (300* 14) = 4200 mOsmol is reduced to (4200 - 900) = 3300. Accordingly, the new osmolality would be: [300 * (3300/4200)] = 236 mOsmol kg -1, if the cell membranes were impermeable, but the cells swell and a higher osmolality is maintained on the cost of reduced ECV and vascular phase.
4. Following a pure water, loss the person has increased osmolality in the ECV. This is the adequate stimulus for thirst and for ADH release. The person is immediately in antidiuresis in order to compensate for the water loss, water is drawn from the cells into the ECV, and the person tries to satisfy his thirst. The condition is rapidly compensated for and not really dangerous.
5. The solutes of ECV are mainly NaCl, so the combined loss of 6 kg water and (6 * 300) = 1800 mOsmol is basically from the ECV. The normal ECV of 14 l is reduced to 8, which is serious. Osmolality is unchanged, so there is no net transfer across the cell membranes. The seriously reduced vascular space implies that the patient develops shock and haemoconcentration. The threat to the effective circulating blood volume triggers the renin-angiotensin-aldosterone cascade. Aldosterone increases the Na+ reabsorption in the distal renal tubules, and thus minimises the NaCl excretion from 200 mmol daily towards 10. Without saline infusions this natural compensation will take up to 9 days. Such a prolonged shock condition is extremely dangerous to the patient and probably fatal.
1. Two physical half-lives equals the 24 hours following the injection of radio-active K+. Thus the remaining activity of the injected 42K+ is: 555 000 Bq × (½)2 = 138,750 Bq. The specific activity, SA, is 55.5 Bq/mmol. SA is equal to the ratio between the total contents of 42K+/39K+, that is in the urine 2220/40 = 55.5 Bq/mmol after 24 hours. The exchangeable K+ pool of her body:
(Injected- eliminated)/SA = (138 750 - 4144 - 2220)/55.5 Bq/(Bq/mmol) = 2385 mmol. The normal value is (71kg × 41) = 2911 mmol 39K.
2. The elimination flux is 40 mmol/12 hours = 3.333 mmol/hour.
The elimination flux divided by the K+ pool is the rate constant, k:
3.333/2385 (mmol/hour)/mmol = 0.0014 hour-1.
The biological half-life is 0.693/k or 495 hours.
The ratio between physical and biological half-life is 12/495 = 0.02424, and the biological half-life is the main determinant of the total half-life.
3. The diagnosis for the patient is hypokalaemia with K+ depletion of the body. Intensive use of diuretics without compensatory K+ intake is the cause.
4. The most effective diuretics block the NaCl reabsorption including the important one in the thick ascending limb of the Henle loop. Since the tubular fluid is almost isosmotic, large fluxes of NaCl, K+ and water reach the distal tubules. The large Na+ flux is a strong stimulus to the aldosterone controlled distal Na+ reabsorption and K+ secretion, so over the years she has lost K+ and H+. This explains her development of hypokalaemia and of metabolic alkalaemia.
5. The amount of natural 40K is measured in a whole body counter. This amount divided by the specific activity (0.012%) is the total body potassium.
1. We must calculate the elimination rate constant (k) for the excretion of this substance through the kidneys. The definition of k is the fraction of the total amount of substance in the body eliminated per time unit.
The available information suggests that the substance is like inulin, which is eliminated by glomerular ultrafiltration, without being secreted or reabsorbed. For such substances the calculation is:
k1 = GFR/ECV = 120/14 000 = 0.0086 min-1.
2. k2 = RPF/ECV = 700/14 000 = 0.05 min-1.