Question

In a random-mating population ( p=0,9929 , Q= 0,0071) are 400 heterozygous individuals out of the 20 000 total number of the population
is the population in equilibrium ?
What
should the number of heterozygotes be at equilibrium ?

Answer 1

Answer:

According to the Hardy Weinberg law, the allele and genotype frequencies in a population remain constant under the absence of factors responsible for evolution. It states that the sum of all genotype frequencies can be represented as the binomial expansion of the square of the sum of p and q. This sum is equal to one. (p + q)2 = p2 + 2pq + q2 = 1.

Here, "p" is the frequency of dominant allele and q is the frequency of the recessive allele. The "2pq" in equation shows the frequency of heterozygotes in the population.

In the question, the frequency of dominant homozygous individuals p

2

= 49% or 0.49.

Thus, the frequency of a dominant allele (p)= "0.7".

Since, p+q=1, so q=1-p.

The frequency of recessive allele (q)= 1-p=1-0.7=0.3.

Thus, the frequency of heterozygotes = "2pq= 2 X 0.7 X 0.3 = 0.42 or 42%".

So, option B is correct.

Answer 2

Equilibrium = 4000

Heterozygots = 2000