contestada

In a population with two alleles, B and b, the allele frequency of b is 0.4. B is dominant to b. What is the frequency of individuals with the dominant phenotype if the population is in Hardy-Weinberg equilibrium

Respuesta :

koishinobi18

Answer:

The frequency of the dominant (normal) allele in the population (p) is simply 1 - 0.02 = 0.98 (or 98%). The percentage of heterozygous individuals (carriers) in the population.

Explanation:

(i found it here if you have questions

Hardy-Weinberg - Kansas State Universitywww.k-state.edu › parasitology › biology198 › answers1)

akindeleot

Answer:

The frequency of the number of individuals with the dominant phenotype is 0.84

Explanation:

in Hardy-Weinberg equilibrium equation, p & q are used to represent the frequency of alleles in a population.

Frequency = Total No of 1 allele / total alleles

if frequency of p allele = 0.6

therefore p + q = 1        ==> q = 0.4

To get these frequencies, square (p + q)

p² + 2pq + q² = 1

The proportion of individuals that are homozygous dominant for B allele is p²

P² = 0.6 x 0.6 = 0.36

The proportion of individuals that are heterozygous for B and b alleles is 2pq

2pq = 2 x 0.6 x 0.4 = 0.48

The frequency of number of individuals with dominant female type

= 0.36 + 0.48 = 0.84

Otras preguntas