Answer:
A. Volume of blood = 5308.06 mL
B. Plasma volume = 2919.43 mL
C. Volume of packed red cells destroyed daily = 19.91 mL
This corresponds to 0.37% of whole blood volume
Volume of whole blood made daily = 44.24 mL
D. Amount of hemoglobin degraded daily = 7.08 g/day
E. Amount of iron liberated from hemoglobin daily = 24 mg/day or 0.00043 mol/day
F. Amount of bilirubin liberated = 62 mg/day
G. Total amount of hemoglobin in blood = 849.3 g of hemoglobin
Total amount of iron in blood = 2.89 g of iron
Explanation:
A. Volume = mass/density
mass = 0.08 * 70 kg * 1000g/1 kg = 5600 g; density of blood = 1.055 g/mL
volume of blood = 5600 g/1.055 g/mL = 5308.06 mL
B. Plasma volume = total blood volume * (1 - hematocrit)
Plasma volume = 5308.06 * (1 - 0.45) = 2919.43 mL
C. Volume of packed red cells = 0.45 * 5308.06 = 2388.63 mL
Since average lifespan is 120 days, volume of red blood cells destroyed daily = 2388.63 mL / 120 = 19.91 mL
This corresponds to (19.91 mL/5308.06 mL) * 100% = 0.37% of whole blood volume
Since volume of red cells destroyed daily = volume produced daily
Ratio of whole blood to red cells = 1 : 0.45
Volume of whole blood made daily = 19.91mL * 1/0.45 = 44.24 mL of whole blood daily.
D. Hemoglobin concentration = 16 g/dL = 16 g/100 mL = 0.16 g/mL
Amount of hemoglobin degraded daily = 0.16 g/mL * 44.24 mL = 7.08 g/day
E. molar mass of iron atom, Fe = 56 g/mol
Four iron atoms will have a mass of 56 * 4 = 224g
percentage mass of four Fe atoms in hemoglobin = 224/66600 * 100% = 0.34%
amount of iron liberated from hemoglobin daily = 7.08 * 0.34% = 0.024 g/day = 24 mg/day
moles of iron per day = 0.024/56 g/mol = 0.00042 mol/day
F. Amount of bilirubin liberated = 7.08 g * 584.7/66600 = 0.062 g/day = 62 mg/day
G. Total amount of hemoglobin in blood = 0.16 g/mL * 5308.06 mL = 849.3 g of hemoglobin
Total amount of iron in blood = 0.34% * 849.3 = 2.89 g of iron
