Question

Gaucher disease type 1 (GD1 ) is a recessive genetic disease that affects 1 in 900 individuals in a particular population. GD1 is caused by a mutation in the enzyme glucocerebrosidase. Assuming the population is in Hardy-Weinberg equilibrium, calculate the frequency of the wild-type (nonmutant) allele for the enzyme glucocerebrosidase.

Answer 1

Answer:

Using the Hardy-Weinberg equilibrium equation, the frequency for the wild-type allele (referred to as p, see explanation below) comes out to be $\frac{2}{3}$

Explanation:

An individual will have to be homozygous recessive for the mutation in the enzyme glucocerebrosidase to have the Gaucher disease type 1 (GD1). This percentage will be $q^{2}$

$q^{2}$ = $\frac{1}{900}$ × 100 = $\frac{1}{9}$

$q$ = $\sqrt{\frac{1}{9} }$ = $\frac{1}{3}$

We know that $\\ p + q = 1$ because the sum of both allele frequencies has to be 100%

So $\\ 1 - \frac{1}{3} = \frac{2}{3}$

Hence, $\\ p = \frac{2}{3}$

Hope that answers the question, have a great day!