Chapter 26.

Multiple Choice Questions

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

Case History A

1.       Note that the rate constant k equals the amount secreted per unit time divided by the total amount of TSH (mol) present in the total blood volume (TBV). The amount of TSH secreted per unit time is a flux (J) or mol per min per area unit. The total amount of TSH in the blood is TBV multiplied by the concentration of TSH in the blood (C). The rate constant (k) is equal to ln2/T½ or 0.693/T½.

           Hence, J = (C × TBV × k) = (C × TBV × 0.693)/T½.

 2.        From the plasma concentration of TSH, the whole-blood concentration can be calculated: (100 × 0.5) = 50 pmol per l of  blood.

J = (50 × 4 × 0.693)/55 = 2.52 pmol per min.

 3.        24 hours equals 1440 min. The 24 hour secretion flux is: (2.52 × 1440) = 3628.8 pmol daily.

1 pmol TSH = (10-12 × 31 000) g.

J = (3628.8 × 31,000 × 10-12) g = 112.5 mg daily.

The daily fractional secretion of TSH is (112.5/300 mg) = 0.375 or 1/3 of the hormone storage.

Case History B

 1.        1 fmol of vasopressin is equal to (10-15 × 1084)* 10-3 mol ADH = 100 fmol.

A normal 60 kg person secretes the following  flux of vasopressin:

(100 × 60 × 60)/(10-15 × 1084) g  per hour or 0.33 mg per hour. The vasopressin secretion of this patient is only 5% or 0.0166 mg per hour.

 2.        The ECV must be 12 kg or litres and the distribution volume for vasopressin 14.4 l.

3.         1084 ug vasopressin equals 1mmol. Thus, 3 mg equals (3/1084) mmol or 2.768 nmol. The rise in [vasopressin] must be: (2.768/14.4) = 0.171 nM (171 pmol per l).

 4.        The new concentration is 171.01 pmol per l, and the ratio is 171.01/2 = 85.5.

 5.        The plasma [vasopressin] necessary to normalise the diuresis of this patient is much higher than normal.

Insensitive or destroyed receptor proteins for vasopressin on the target cells in the kidney tubules can explain this type of diabetes insipidus. However, abnormal receptor proteins can be secondary to abnormal ATPase mechanisms.

6.         The urine loss is high, namely (10/60) or 1/6 of the patients body weight. Such a high water loss is only possible, if she continues to drink. Any uncompensated water loss comprising more than 1/5 of the total body water can be fatal.  

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