Bch2011 The Pre-Practical Exercise Answers Assessment Answer
Answer:
Solution
|
Tube number |
1 |
2 |
3 |
4 |
5 |
6 sample |
|
Volume of standard Hb solution (ml) |
- |
1.0 |
2.0 |
3.0 |
4.0 |
- |
|
Volume of cyanide reagent (ml) |
1.0 |
1.0 |
1.0 |
1.0 |
1.0 |
1.0 |
|
Volume of water (ml) |
4.0 |
3.0 |
2.0 |
1.0 |
|
3.99 |
|
Volume of blood (ml) |
- |
- |
- |
- |
|
0.01 |
|
Absorbance at 540 nm |
0.025 |
0.090 |
0.160 |
0.230 |
0.290 |
0.210 |
|
Amount of Hb (mg/tube) |
0.000 g |
0.0005 g |
0.001 g |
0.003 g |
0.012 g |
|
Determine the amount of Hb in Tubes 1-5 and enter these data in the protocol.
Volume of standard Hb taken = 1.0 ml
Amount of Hb taken = volume of standard Hb taken * standard solution containing Hb
= 1.0 ml * (g/ml) = 0.0005g
Therefore, total mass of Hb in tube 2 = 0.0005g
Tube3
= 2.0 ml * (g/ml) = 0.001g
Therefore, total mass of Hb in tube 3 = 0.001g
Tube 4
= 3.0 ml * (g/ml) = 0.003g
Therefore, total mass of Hb in tube 4 = 0.003g
Tube 5
= 4.0 ml * (g/ml) = 0.012g
Therefore, total mass of Hb in tube 5 = 0.012g
Note: the addition of any amount of water does not affect the total amount of Hb in the tubes because it remains constant, however, it reduces Hb in final solution.
Also the volume of distilled water added to tube 2 = 3 ml
Total volume of tube 2 = 1.0 ml (Volume of standard Hb solution) + 1.0 ml [Volume of cyanide reagent (ml)] + 3.0 ml (Volume of water (ml)] = 5.00 ml
Total volume of tube 3 = 5.0 ml
Total volume of tube 4 = 5.0 ml
Total volume of tube 5 = 5.0 ml
Now, using C1V1 = C2V2 ………..(i)
C1 = concentration
V1 = volume of initial solution 1, standard Hb
C2 = concentration
V2 = volume of final solution 2, final solution
Putting the values in equation 1
Tube 2
1ml * (g/ml) = 5.0 ml * C2
0.0005g = 5.0 * C2
C2 = 0.0001 g/ml
Tube 3
2ml * (g/ml) = 5.0 ml * C2
0.001g = 5.0 * C2
C2 = 0.0002 g/ml
Tube 4
3ml * (g/ml) = 5.0 ml * C2
0.003g = 5.0 * C2
C2 = 0.0006 g/ml
Tube 5
4ml * (g/ml) = 5.0 ml * C2
0.012g = 5.0 * C2
C2 = 0.0024 g/ml
Amount of Hb
= C2V2
Tube 2
0.0001* 5= 0.0005 g
Tube 3
0.0002 * 5 = 0.001 g
Tube 4
0.0006 * 5 = 0.003 g
Tube 5
0.0024 * 5 = 0.012 g
From the graph
- The amount of Hb in tube 6 = 1.6 mg/tube
- The amount of Hb in the 0.01 ml blood sample
Note that 1.6 mg = 5.0 ml of solution
? = 0.01 ml of the blood in the solution
= 0.0032 mg
- The concentration of Hb in the blood sample
- =
= 0.32 g/l
- Is the Hb content normal
No, it is not within the ranges given.
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