Question

assume in a hardy weinberg population of 1600 no of recessive individuals are 256. find the genotypic frequency of heterozygote dominant​

Answer 1

Answer:

From Hardy-Weinberg equilibrium, p + q = 1,

q = 1 – p = 1 – 0.7 = 0.3

q2 =0.3 *0.3=0.09

The homozygous dominant individuals are represented by

1600 *p =1120

1600*q=144

So, the correct option is '144'

Explanation:

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Answer 2

Answer:

The genotypic frequency of heterozygous dominant is 0.48.

Explanation:

• According to the Hardy-Weinberg equilibrium, if no unfavorable influences exist, genetic variation in a population will remain stable from one generation to the next.
• The law predicts that genotype and allele frequencies will remain constant because they are in equilibrium when mating is random in a large population with no disruptive factors.
• Numerous factors, including mutations, natural selection, nonrandom mating, genetic drift, and gene flow, have the potential to upset the Hardy-Weinberg equilibrium.
• p^2 + 2pq + q^2 = 1;

1. p^2 = dominant homozygous frequency (AA)
2. 2pq = heterozygous frequency (Aa)
3. q^2 = recessive homozygous frequency (aa)

• Information given:

1. Total population= 1600
2. No. of recessive individuals (q^2)= 256/1600= 0.16
3. q= 0.4
4. p + q=1; p=0.6
5. 2pq= 2*0.6*0.4= 0.48

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