Question

Hemoglobin is a large protein molecule that is responsible for carrying oxygen through the blood. Iron ions are a relatively small component of hemoglobin. There are four Fe2+ions that are part of the much larger hemoglobin structure. In a single red blood cell there are 2.50x108molecules of hemoglobin. If a single Fe2+ion has an atomic radius of 75.1 pm and a redblood cell has a volume of 95 μm3, what percentage of the total red blood cell volume is taken up by Fe2+ions?

Answer 1

Answer:

The percentage of volume taken by iron(II) in the blood cell is 0.0019%.

Explanation:

Radius of iron(II) ions ,r= 75.1 pm =$7.51\times 10^{-5} \mu m$

1 pm = 10^{-6} μm

Volume of sphere  =$\frac{4}{3}\pi r^3$

Volume of single iron(II) ion = V

$V=\frac{4}{3}\times 3.14\times (7.51\times 10^{-5} \mu m)^3$

$V=1.7742\times 10^{-12} \mu m^3$

Number of iron(II) ions in one hemoglobin structure = 4

Number of hemoglobin structure in blood cell = $2.50\times 10^8$ molecules

Then number of iron (II) ions in $2.50\times 10^8$ molecules of hemoglobin:

$4\times 2.50\times 10^8 ions=10^9 ions$

Volume of $10^9$ ions of iron = $V\times 10^9$

Volume of the  hemoglobin structure,V' = $95 \mu m^3$

Percentage volume of iron (II) ions in a single blood cell:

$\frac{10^9\times V}{V'}\times 100$

$\frac{10^9\times 1.7742\times 10^{-12} \mu m^3}{95 \mu m^3}\times 100$

=$0.0019\%$