Calculate the number of each genotype in a Hardy-Weinberg equilibrium population

Question

Calculate the number of each genotype in a Hardy-Weinberg equilibrium population of 4800 individuals with a "T" allele frequency of 0.72, write the number of each genotype in the spaces below: Consider an initial population of 512 TT individuals, 205 Tt individuals, and 83 tt individuals. What are the frequencies of the alleles? Frequency of A = Frequency of a = Given the allele frequencies you calculated above, determine the actual numbers of each of the genotypes you would expect based on the allele frequencies calculated: No. TT = No. Tt = No. Tt =

Explanation / Answer

10 answer

GENOTYPE

frequency

TT

512

P^2

HOMOZYGOUS DOIMNANT

512/800=0.64

Tt

205

2pq

HETEROZYGOUS

205/800=0.2562

tt

83

P^2

HOMOZYGOUS RECESSIVE

83/800=0.1037

total

800

1

By using gene count method

FREQUENCY OF T ALLELE = D+1/2 H

= 0.64 + (0.2562/2)= 0.7681

FREQUENCY OF t ALLELE = R+1/2 H

= 0.1037+ (0.2562/2)= 0.2318

ACCORDING hardy Weinberg

P+q=1

=0.7681 +0.2318 =1

GENOTYPE

frequency

Expected frequencies

Expected number

TT

P^2

0.7681 x 0.7681 =0.5899

0.7681 x 0.7681 x 800=471.98 =472

Tt

2pq

2x 0.7681 x 0.2318 =0.356

2x 0.7681 x 0.2318 x 800284.87 =285

tt

P^2

0.2318 x 0.2318 =0.0537

0.2318 x 0.2318 x 800=42.98 =43

800

GENOTYPE

frequency

TT

512

P^2

HOMOZYGOUS DOIMNANT

512/800=0.64

Tt

205

2pq

HETEROZYGOUS

205/800=0.2562

tt

83

P^2

HOMOZYGOUS RECESSIVE

83/800=0.1037

total

800

1

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