In a random mating population frequency of disease causing recessive allele is 80%. What would be the frequency of carrier individual in population :-a)64%b)32%c)16%d)100%Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Given, frequency of disease causing recessive allele is 80% i.e., q=0.8..then from p+q=1..we get p=0.2..then frequency of carrier individual in the population will be 2pq = 2×0.2×0.8= 0.32...which is 32%

Answer

Frequency of Disease Causing Recessive Allele: 80%

Carrier individuals are those who carry one copy of the disease-causing recessive allele and one copy of the normal allele. They do not exhibit the disease but can pass on the disease-causing allele to their offspring.

To determine the frequency of carrier individuals in the population, we need to consider the Hardy-Weinberg equilibrium equation. According to this equation, in a random mating population, the frequencies of alleles and genotypes remain constant from generation to generation if certain conditions are met.

The equation for the Hardy-Weinberg equilibrium is:

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

Where:
- p^2 represents the frequency of the homozygous dominant genotype (AA)
- 2pq represents the frequency of the heterozygous genotype (Aa)
- q^2 represents the frequency of the homozygous recessive genotype (aa)
- p represents the frequency of the dominant allele (A)
- q represents the frequency of the recessive allele (a)

In this case, we are given that the frequency of the disease-causing recessive allele (a) is 80%, which means q = 0.8. We can calculate p using the equation p = 1 - q.

Calculations:
p = 1 - q
p = 1 - 0.8
p = 0.2

Now, we can use the equation to calculate the frequency of carrier individuals (2pq).

2pq = 2 * 0.2 * 0.8
2pq = 0.32

So, the frequency of carrier individuals in the population is 32% (option B).

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