Chapter 30.
I. Answers A, B, and E are true statements, whereas C and D are false.
.II. Answers A, C, and D are true statements, whereas B and E are false.
1. 777 ng or 0.777 mg = 1 nmol; Serum-thyroxine: (180 × 777) = 139 860 ng = 140 mg per l.
2. The total thyroxine content of the patient's body: (140 × 8) = 1120 mg.
The elimination rate: (1,120 × 0.14) = 156.8 mg daily or 156.8/0.777 = 201.8 nmol daily.
3. The thyroid uptake of iodide equals the elimination of thyroxine, which is:
(201.8 × 4) nmol/day or (807.2 × 127) = 102 514 ng/24 hours or (102.5/1440) = 0.071 mg per min.
The thyroid plasma clearance for iodide is:
0.071 (mg/min)/0.004 (mg/ml) = 17.75 ml plasma/min.
4. The symptoms and signs suggest hyperthyroidism. The serum thyroxine concentration is almost twice the normal level. The heart attack is probably atrial fibrillation, which frequently occurs in hyperthyroid patients.
1. The renal excretion flux of Ca2+ is reduced due to increased reabsorption of Ca2+ in the distal tubules. The plasma concentration [Ca2+] is increased, due to bone mobilisation, increased intestinal absorption and reduced renal excretion flux of Ca2+. PTH increases the phosphate excretion flux. Hereby the plasma [phosphate] is reduced.
2. The diagnosis is hypoparathyroidism. Patients develop hypoparathyroidism following struma surgery.
3. Hypoparathyroidism leads to a low plasma [Ca2+], and a large plasma [phosphate]. The symptoms and signs of hypoparathyroidism are related to the hypocalcaemia, such as low [Ca2+]; increased neuro-muscular irritability with latent or manifest tetany. Universal tetany can end in respiratory and cardiac arrest.
1. 7.21 = 6.8 + log (HPO4--/H2PO4-); (HPO4--/H2PO4-) = 2.57.
Since HPO4-- + H2PO4- = 0.84 it can be deduced that [HPO4--] is (0.84×2.57)/3.57 = 0.60 and [H2PO4-] is 0.84/3.57 = 0.24, which is different from the normal.
2. [Total calcium] = 2/0.62 = 3.23 mmol per l. The [total calcium] of this patient is much higher than normal (just as her [inorganic phosphate]).
3. The patient has an acidosis (pH = 7.21), probably of metabolic character, due to loss of base (bicarbonate and secondary phosphate).
Ca-Proteins + 2 H+ = Ca2+ + Proteins.
The plasma proteins will bind a smaller and smaller fraction of Ca2+ at increasing H+ concentration in blood plasma, so the free fraction of Ca2+ ([Ca2+]) must increase.
4. Hyperparathyroidism (primary).
5. She has a substantial renal Ca2+ excretion, and small Ca2+ -crystals and stones in her urinary tract. She may have polyuria due to reduced sensitivity to vasopressin.