Question

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

Answer 1

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.

Note:
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.