The following table gives the number of individuals of each of three genotypes a

Question

The following table gives the number of individuals of each of three genotypes at the same locus in each of three populations: Genotype Pop 1 Pop 2 Pop 3 90 200 50 420 200 500 490 600 450 TC I-1. For each population, calculate the allele frequencies and determine whether the population is currently at Hardy-Weinberg equilibrium 1-2. For each population that is not at Hardy-Weinberg equilibrium, indicate whether there is an excess of homozygous or heterozygous genotypes I-3. For each population that is not at Hardy-Weinberg equilibrium, indicate how many enerations of random the oat equlibrium.

Explanation / Answer

p2 + 2pq + q2 = 1 and p + q = 1

p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p2 = percentage of homozygous dominant individuals
q2 = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals

Since TT =p2,TC =2pq, CC= q2 then p the frequency of the T allele will be equalt to the square root of the 0.09 i.e 0.3 and q= 1-p =1-0.3 =0.7.This popultion is in accordance to Hardy weinberg` rule as here p+q=1(0.3 +0.7 =1)

Now to check whether these datas are according to Hardy Weinberg equation

p+q should be equal to 1. Here p= square root of 0.2 =0.44 and q= square root of 0.6 =0.77.let's check whether p+q=1 or not.

0.44+0.77= 1 .21, Hence this population is not according to the Hardy Weinberg rule.

p2 =0.05,p=square root of p = 0.223

q2 = 0.45 ,q = squareroot of q = 0.67

Now to check whether these datas are according to Hardy Weinberg equation

p+q should be equal to 1. but here 0.223 + 0.67 =0.893 whcih is not equal to 1. hence it is not in accordance to Haedy Weinberg rule.

In second population CC i.e homozygous genotypes are in excess and in third population  TT homozygous genotypes are limited.

genotype number of individuals resulting frequency TT(p2) 90 0.09(90/1000) TC(2pq) 420 0.42(420/1000) CC(q2) 490 0.49(490/1000)

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