Question

In humans, red blood cells have a number of proteins embedded in the cell membrane. One type of protein, the Rh factor, is controlled by a single gene and is either present or missing from the red blood cells. If present, the individual has the Rh phenotype. If missing, the individual has the Rh– phenotype. Rh is the dominant to Rh–. Suppose that, in the Yoruban population, the frequency of the Rh– phenotype is 0.01. Using the Hardy–Weinberg equations, calculate the frequency of the Rh allele to at least two decimal places.

Answer 1

Answer:

The frequency of the Rh allele $= 0.90$

Explanation:

Given -

Rh is the dominant to Rh–

Therefore alleles for Rh will be represented by "p" and allele for Rh- will be represented by "q"

So  , the frequency of the Rh– phenotype is 0.01.

Which means $q^{2} = 0.01$

Hence, frequency for allele of Rh–

$=\sqrt{q^2} \\= \sqrt{0.01} \\= 0.1$

As per Hardy Weinberg's first equation-

$p+q=1$

Substituting the value of "q" in above equation, we get -

$p+0.10=1\\p=1-0.1\\p=0.90$

So the frequency of the Rh allele $= 0.90$