Determination of calcium in serum by the method of de Waard.

In the centrifuge tube measure 2 ml of distilled water. Add 1 ml of serum. Then add 1 ml of saturated ammonium oxalate solution, and leave for 10 minutes, then centrifuge it for 5 minutes. Calcium oxalate forms a dense white precipitate at the bottom of the test tube. Supernatant is decanted. The precipitate is washed by adding 4 ml of 2% ammonium hydroxide solution in the test tube. It is again centrifuged for 5 minutes. Pour solution from the sediment and add into a test tube 2 ml of 1 n. sulphuric acid. Stir the sediment with a glass stick. Take another test tube and pour 2 ml of 1 n. sulphuric acid (control sample). Immerse test tubes for 1-2 minutes in boiling water bath. The hot solution is titrated from microburette with 0.01 n. of potassium permanganate solution until a pale pink color, disappearing within minutes. The amount of permanganate that went on the titration of control sample is subtracted from consumed amount of potassium permanganate in the test sample.

Example of calculation: 1 ml of 0.01 n. potassium permanganate solution is equivalent to 0.0499 mmol of calcium. For titration of serum 0.54 ml of potassium permanganate was spent; for the titration of control experiment – 0.05 ml of potassium permanganate, hence the amount of calcium in serum is:

(0.54-0.05)*0.04990*100 = 2.044 mmol/l.

Test Questions

1. What is the mineral composition of blood?

2. Give the equations describing the mechanism of action of the main blood buffer systems.

3. When do hypocalcemia and hypercalcemia develop?

4. What is the difference between serum and plasma?

5. Specify causes of hypo-and hyperproteinemia.

Laboratory work 9.

Detection of bile pigments in urine

Bilirubin in normal urine cannot be open by the usual methods, although it may be in it in trace amounts. Excretion of bilirubin in the urine is increased with hyperbilirubinemia. In these cases, qualitative tests for bilirubin become positive. Reactions for the bile pigments are based on the formation of colored products of bilirubin oxidation: biliverdin (green), bilitsianin (blue), holetelin (yellow), etc.

In the test tube put 1 ml of concentrated nitric acid and gently layer urine from pipette. The reaction is called "Gmelin’s test". It is based on the oxidation of urine bilirubin under the influence of nitric acid with a mixture of nitrous acid in a number of differently colored oxidation products. In the presence of bile pigments in the boundary of two fluids green, blue, purple, red and yellow rings appear in the sequence that corresponds to different degrees of oxidation of bilirubin. Furthest from the nitric acid is a green ring of biliverdin (least oxidized product), the closest to the nitric acid is a yellow ring of holetelin (the most oxidized product).

Test Questions

1. Give a scheme for the formation of bilirubin in the body.

2. What is a "direct" and "indirect" bilirubin?

3. What types of jaundice cause hyperbilirubinuriya?

4. How to distinguish obstructive jaundice from hemolytic jaundice using the blood test?

Laboratory work 10.

Date: 2016-04-22; view: 819