Question

4. In a large, randomly mating population, the frequency of the allele (q) for a particular disease is 0.057. The results of studies have shown that people with the following genotypes at the beta-chain locus produce the following average numbers of offspring: Genotype Average number of offspring produced QQ Og ga a. What will the frequency of the q allele be in the next generation? b. What will the frequency of the q allele be at equilibrium?

Answer 1

Ans.

By randomly particular diesese is 0.057.

From that population results in double recessive genotype (qq)

Here, frequency of recessive allele (q) = 0.057

And Frequency of recessive qq genotype (q²) = 2

Frequency of dominant QQ genotype ( p²) = 4

Frequency of Qq genotype ( 2pq) = 6

The Hardy- Weinberg equation,

p² + 2pq + q² =1

Frequency of q allele be in the next generation

For q allele frequency ( 6 + 2 + 2 )= 10

For p allele frequency ( 4 + 4 + 6 )= 14

So the allele q frequency = 10/ 24 = 0.42

The frequency of q allele is

q² + pq = q² + q(1 - q)

q = q² + q - q2

q = 2² + 2 - 2²

q = 1.4 + ( 2 - 1.4)

q = 1.4 + 0.6

q = 2

​​​​​​While the q allele frequency stay the same from one generation to the next generation.

Ans b

What will the frequency of q allele be at equilibrium?

By the Hardy wienberg equilibrium

q² = Individual with recessive genotype ( qq )/Total population

q² = 2/ 12 = 0.16

By the using q2 we can calculate q allele frequency

q2 = 0.16

√q2 = √ 0.16

q = 0.4

Therefore, the recessive genotype allele frequency (q) has remained same in the population is in state of Hardy wienberg equilibrium.