At a particular locus, frequency of 'A' allele is 0.6 and that of 'a' is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium ?
- 0.16
- 0.48
- 0.36
- 0.24
The Correct Option is B
Solution and Explanation
frequency of heterozygotes (Aa) using the formula:
Frequency of heterozygotes (Aa) = 2 \(\times\) Frequency of 'A' allele (p) \(\times\) Frequency of 'a' allele (q)
Given:
Frequency of 'A' allele (p) = 0.6
Frequency of 'a' allele (q) = 0.4
Frequency of heterozygotes (Aa) = 2 \(\times\) 0.6 \(\times\) 0.4 = 0.48
So, the frequency of heterozygotes in a random mating population at equilibrium is 0.48.
Therefore, the correct answer is (B): 0.48
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Concepts Used:
Hardy-Weinberg Principle
Hardy Weinberg Law:
In a considerable size, random-mating population, the genotype and allele frequencies remain constant in the absence of any evolutionary influences from one generation to another. Influences include a choice of mate, natural selection, genetic drift, mutation, sexual selection, gene flow, genetic hitchhiking, founder effect, meiotic drive, population bottleneck, inbreeding, and assortative mating.
Assumptions for the Hardy Weinberg Principle
Following are a few assumptions for the law:
- Only sexual reproduction can take place.
- The mating process is random.
- The size of the population is indefinitely large.
- Entities are diploid.
- Generations do not overlap.
- Equality of allele frequencies in terms of sexes.
- No gene flow, selection, mutation, migration, or admixture.