Question 1. For a locus with two alleles (A, a), imagine that there are three po
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Question
Question 1. For a locus with two alleles (A, a), imagine that there are three populations with the following genotype compositions:
number of individuals with:
AA Aa aa
first pop. 16 32 32
second pop. 0 64 16
third pop. 32 0 48
What are the expected Hardy-Weinberg genotype frequencies for each population? Which of the three populations is closest to the composition expected at Hardy-Weinberg equilibrium?
Question 2. Suppose that there are two loci that both influence the body size of an organism. At the first locus, allele "a" has no effect on growth, while each copy of allele "A" causes body size to increase by one unit:
genotype: aa aA AA
phenotype: 0 +1 +2
At the second locus, alleles influence growth in the same fashion, but have a stronger effect:
genotype: bb bB BB
phenotype: 0 +2 +4
Every diploid individual must have two alleles at the (A, a) locus, along with two alleles at the (B, b) locus. How many possible diploid genotypes are there? If phenotypes combine additively across loci, then the phenotype of a two-locus genotype will be the sum of phenotypic effects of the alleles it carries at the first locus plus the phenotypic effects conferred by alleles at the second locus. For example, genotype "AABB" has a phenotype of 6 (= 2+4) according to this model. Work out the predicted phenotype for each possible genotype. How many distinct phenotypes are there? Are there any cases where 2 different genotypes have an identical predicted phenotype?
Explanation / Answer
Q1). Hardy and Weinberg also described all the possible genotypes for a gene with two alleles. The binomial expansion representing this is, p2 + 2pq + q2 = 1.0
Where,
p2 = proportion of homozygous dominant individuals
q2 = proportion of homozygous recessive individuals
2pq = proportion of heterozygotes.
The genotypic frequency for first population: Given that AA = 16. Aa = 32 and aa = 32.
We can oberve that the total number of individuals = 80.
Thus, p2 = 16/ 80 = 0.2
q2 = 32/80 = 0.4
2pq = 32/ 80 = 0.4
According to Hardy-Weinberg equilibrium, p = p2 + 1/2 (2pq) = 0.2 + 1/2 (0.4) = 0.4
q = q2 + 1/2 (2pq) = 0.4 + 1/2 (0.4) = 0.6
Thus, for the given population, the frequency of p allele = 0.4 (dominant), and the frequency of p allele = 0.6 (recessive).
The genotypic frequency for second population: Given that AA = 0. Aa = 64 and aa = 16.
We can oberve that the total number of individuals = 80.
Thus, p2 = 0/ 80 = 0
q2 = 16/80 = 0.2
2pq = 64/ 80 = 0.8
According to Hardy-Weinberg equilibrium, p = p2 + 1/2 (2pq) = 0 + 1/2 (0.8) = 0.4
q = q2 + 1/2 (2pq) = 0.2 + 1/2 (0.8) = 0.6
Thus, for the given population, the frequency of p allele = 0.4 (dominant), and the frequency of p allele = 0.6 (recessive). But, it is given that AA = 0, the population does not obey Hardy-Weinberg equilibrium (because if we back calculate, p2 (AA) must be equal to 0.36, but it is given zero).
The genotypic frequency for Third population: Given that AA = 32. Aa = 0 and aa = 48.
We can oberve that the total number of individuals = 80.
Thus, p2 = 32/ 80 = 0.4
q2 = 48/80 = 0.6
2pq = 0
According to Hardy-Weinberg equilibrium, p = p2 + 1/2 (2pq) = 0.4 + 0 = 0.4
q = q2 + 1/2 (2pq) = 0.6 + 0 = 0.6
Thus, for the given population, the frequency of p allele = 0.4 (dominant), and the frequency of p allele = 0.6 (recessive). But, it is given that Aa = 0, the population does not obey Hardy-Weinberg equilibrium (because if we back calculate, 2pq must be equal to 0.48, but it is given zero).
From the above observations we can say that the population -I is closest to the composition expected at Hardy-Weinberg equilibrium (but not exactly meeting).
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