Question

The table shows a population consisting of 80 individuals for which genotypes AA, Aa, and aa have been identified and counted. This organism is a diploid, and A and a are the only alleles at the locus. Using the data from this table as a starting point, categorize each statement below as true or false. Genotype Number of individuals АА 40 Aa 30 aa 10 Total 80 The a allele frequency is 1-0.6875 = 0.3125. The number of a alleles is calculated 2(10) + 30 = 50. The frequencies of A and a must add up to 1. The number of A alleles in this population is 55. The A allele frequency is the number of A alleles divided by the total number of individuals. The A allele frequency is the number of A alleles divided by the total of all chromosomes sampled The frequency of the A allele is 0.6875 The frequency of the A allele is 0.3438 The number of a alleles is calculated 10 + 12 (30) = 25. The number of A alleles in this population is 110. The gene pool in this case is 80 alleles of gene A. In this population, the number of chromosomes with A gene alleles is 160. True False

Answer 1

Answer:

True	False
The a allele frequency is 1-0.6875 = 0.3125 The number of a alleles is calculated 2(10) + 30 = 50. The frequency of the A allele is 0.6875 The number of A alleles in this population is 110. The A allele frequency is the number of A alleles divided by the total of all chromosomes sampled. The frequencies of A and a must add up to 1	The number of a alleles is calculated 10+½(30) = 25 The frequency of the A allele is 0.3438 The gene pool in this case is 80 alleles of gene A. The number of A alleles in this population is 55 In this population, the number of chromosomes with A gene alleles is 160 The A allele frequency is the number of A alleles divided by the total number of individuals.

Explanation:

The table shows a population consisting of 80 individuals for which genotypes AA, Aa, and aa have been identified and counted. This organism is a diploid, and A and a are the only alleles at the locus. Using the data from this table as a starting point, categorize each statement below as true or false.

Genotype	AA	Aa	aa	Total
Number of individuals	40	30	10	80

Frequency of A allele = [{(2×number of individual with genotype AA) + number of individual with genotype Aa} ÷ (2×total number of individual)] = [{(2×40)+30} ÷ (2×80)] = [{80+30} ÷ 160] = [110 ÷ 160] = 0.6875

We known that, sum of the frequency of allele A and allele a is 1, i.e., A+a = 1

So, a = 1-A = 1-0.6875 = 0.3125

Number of A alleles = {(2×number of individual with genotype AA) + number of individual with genotype Aa} = {(2×40)+30} = {80+30} = 110

Number of a alleles = {(2×number of individual with genotype aa) + number of individual with genotype Aa} = {(2×10)+30} = {20+30} = 50

So, total number of allele in the gene pool = (number of allele A + number of allele a) = (110+50) = 160