Question

In a population, researchers found three genotypes for the gene A with following frequencies: AA: 0.25 Aa: 0.26 aa: 0.49 Question 2 With the present allele frequencies, what proportions of the dominant phenotype would you expect in the population if it were at Hardy Weinberg equilibrium? (rounded to the nearest digit) O 0.62 O 0.51 O 0.69 O 0.77 O 0.14

Answer 1

Genotype frequencies given :

AA - 0.25

Aa -0.26

aa - 0.49

So, number of each genotype in the population of 100

AA = 25

Aa = 26

aa = 49

Allele number for the population can be calculated by multiplying 2 to the number of homozygous individuals, which is then added to number of heterozygous individuals. Allele frequency can be calculated by dividing no. Of a allele in a population by total number of alleles.

Here, no. Of allele A = [(25 ×2 ) + 26] =76

no. Of allele a = [(49×2)+26] = 124

Total number of alleles (A+a) = 200

Frequency of allele A = 76/200 = 0.38

Frequency of allele a = 124 /200 0.62

These frequencies are now used in hardy Weinberg rule P²+2pq+q², where p is frequency of dominant allele, q is frequency of recessive allele, p² is frequency of homozygous dominant, 2pq is frequency of heterozygous and q² of homozygous recessive individuals.

Dominant phenotype would be expressed by organisms with homozygous dominant genotype and heterozygous genotype.

So , proportion of individuals with dominant phenotype =( p² + 2pq)

= (0.38)² + (2×0.38×0.62)

= 0.1444 + 0.4812

= 0.6256 = 0.62