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Among the 1000 people, 360 people belong to genotype AA, 480 people to Aa, and the remaining 160 people to aa. Based on this data, the frequency of allele A in the community is:
A. 0.4
B. 0.5
C. 0.6
D. 0.7

Last updated date: 20th Jun 2024
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Hint: The proportion of an allele's overall repetition on a genetic locus in a population is known to be Allele frequency. In general, it is communicated to an extent or a rate. The allele frequencies show the hereditary variety of an animal group's population or proportionally the lavishness of its genetic stock, in the community hereditary qualities.

Complete answer:
The total number of people = 1000
Genotype recurrence of AA = $\dfrac{360}{1000}$ = 0.36
Genotype recurrence of aa = $\dfrac{160}{1000}$ = 0.16
Genotype recurrence of Aa = 2x $\dfrac{240}{1000}$ = 0.48
Presently, as per the Hardy Weinberg balance,
The square of the whole of allele frequencies gives the individual genotype frequencies in a non-developing stable populace.
(A+a)x2= AA + 2Aa +aa
Consequently, allele recurrence of A = $\sqrt{0.36}$ = 0.6
Allele recurrence of a = $\sqrt{0.16}$ = 0.4
Hence, Option (c) 0.6 is the correct answer.

Additional Information:
According to the standard of Hardy-Weinberg, the population of alleles must meet five guidelines to be considered "in balance":
-No quality transformations may happen, and hence allele changes do not happen.
-There must be no relocation of people either into or out of the population.
-Random mating must happen, which means people mate by some coincidence.
-No hereditary float, a possible change in allele recurrence, may happen.
-No regular determination, an adjustment in allele recurrence because of climate, may happen.

In population hereditary qualities, the Hardy–Weinberg standard, otherwise called the Hardy–Weinberg balance, model, hypothesis, or law, expresses that allele and genotype frequencies in a populace will stay steady age to age without other transformative impacts. According to this rule, genotype of AA=0.36, aa=0.16 and Aa= 0.48. For a hereditary balance population: p + q = 1.0 (The total of the frequencies of the two alleles is 100%). In this way allele frequency of A = √0.36 = 0.6.