Question

In a given population, 1 out of every 400 people has a genetic disorder caused by a completely recessive allele, b. Assuming the population is in Hardy-Weinberg equilibrium, which one of the following is the expected proportion of individuals who are carriers of b allele but are not expected to develop the disorder
A) 380/400
B) 19/400
C) 20/400
D) 38/400​

Answer 1

Explanation:

Correct option is (b) 19/400

Answer 2

The proportion of individuals with the carrier of b allele but not expected to develop the disorder are $\frac{38}{400}$ (Option D).

According to Hardy Weinberg's Law,

$p^2 + 2pq + q^2 = 1\\p + q = 1$

where,  

p - frequency of dominant allele  

q - frequency of recessive allele  

$p^2$ - frequency of dominant homozygous individuals  

$q^2$ - frequency of recessive homozygous individuals  

2pq - frequency of heterozygous individuals

Given - Frequency of recessive homozygous individuals$(q^2) = \frac{1}{400}$  

To find- Proportion of heterozygous individuals in the given population.  

Solution -

Frequency of recessive allele (q)  = $\sqrt \frac{1}{400}$

                                                    q = 0.05

According to the equation, p = 1 - q  

Frequency of the dominant allele (p) = 1 - 0.05  

                                                         p = 0.95

Frequency of heterozygous individual = 2pq = 2 × 0.05 × 0.95  

                                                                  = 0.095

The number of heterozygous individuals = 0.095 × Total number of people in the given population  

                                                                   = 0.095 × 400  

                                                                   = 38