Biology, asked by Aiba987, 11 months ago

600 insects :200 are homozygous purple, 150 are heterozygous purple, 250 are homozygous green. calculate p (square ) &q. From Hardy Weinberg equilibrium no theoretical answer. Take help from Google or form any elder. hai koi maikalal ya maikalali jo is ka answer de skta hai​

Answers

Answered by dharma2081974
2

Answer:

1. The frequency of two alleles in a gene pool is 0.19 (A) and 0.81(a). Assume that the population is in

Hardy-Weinberg equilibrium.

(a) Calculate the percentage of heterozygous individuals in the population.

According to the Hardy-Weinberg Equilibrium equation, heterozygotes are represented by the 2pq

term. Therefore, the number of heterozygous individuals (Aa) is equal to 2pq which equals

2 × 0.19 × 0.81 = 0.31 or 31%

(b) Calculate the percentage of homozygous recessives in the population.

The homozygous recessive individuals (aa) are represented by the q

2

term in the H-W equilibrium

equation which equals 0.81 × 0.81 = 0.66 or 66%

2. An allele W, for white wool, is dominant over allele w, for black wool. In a sample of 900 sheep, 891 are

white and 9 are black. Calculate the allelic frequencies within this population, assuming that the

population is in H-W equilibrium.

The allelic frequency of w is represented by the q term and the allelic frequency W is represented by the

p term. To calculate the value of q, realize that qq or q

2

represents the homozygous recessive individuals

or the black sheep in this case. Since there are 9 black sheep, the frequency of black sheep

=

# individuals 9 0.01

total individuals 900 = = , thus ww = q

2 = 0.01

∴ 2

q q == = 0.01 0.1

Additionally, p + q = 1 thus p = 1 – q or p = 1 – 0.1 or 0.9 ∴ p = W = 0.9 and q = w = 0.1

3. In a population that is in Hardy-Weinberg equilibrium, the frequency of the recessive homozygote

genotype of a certain trait is 0.09. Calculate the percentage of individuals homozygous for the dominant

allele.

We know that the frequency of the recessive homozygote genotype is q

2 and equal to 0.09.

∴ 2

q q == = 0.09 0.30 AND we also know that p + q = 1

Thus, p = 1 – q ∴ p = 1 – 0.30 = 0.70

∴The homozygote dominants are represented by p

2 = (0.70)2

= 0.49 or 49%

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