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Answer :
a) The frequency of the dominant allele is 0.95.
b) The frequency of a1a1 is 0.9025, the frequency of a1a2 is 0.095, and the frequency of a2a2 is 0.0025.
c) The fitness of the a2a2 genotype is 0.
d) The mean fitness of the population is 0.9975.
e) the frequency of a2 is expected to decrease slightly to 0.095.
f), the frequency of a1a1 is 0.995
a) Since there are only two alleles at the locus, the frequency of the dominant allele (a1) can be calculated as follows:
p + q = 1, where p is the frequency of a1
0.05 + p = 1
p = 0.95
Therefore, the frequency of the dominant allele is 0.95.
b) The three possible genotypes are: a1a1, a1a2, and a2a2.
The frequency of a1a1 can be calculated as p^2 = 0.95^2 = 0.9025
The frequency of a1a2 can be calculated as 2pq = 2(0.95)(0.05) = 0.095
The frequency of a2a2 can be calculated as q^2 = 0.05^2 = 0.0025
Therefore, the frequency of a1a1 is 0.9025, the frequency of a1a2 is 0.095, and the frequency of a2a2 is 0.0025.
c) A recessive lethal allele is lethal only in the homozygous state, so the fitness of the a1a1 genotype is 1, the fitness of the a1a2 genotype is 1, and the fitness of the a2a2 genotype is 0.
d) The mean fitness of the population can be calculated as follows:
w = (p^2)(1) + (2pq)(1) + (q^2)(0)
w = 0.9025 + 0.095 + 0
w = 0.9975
Therefore, the mean fitness of the population is 0.9975.
e) Since the population is in Hardy-Weinberg equilibrium at the start, we can use the Hardy-Weinberg equation to calculate the expected frequencies of a1 and a2 after one generation:
p' = p^2w11 + pqw12 = (0.95)^2(1) + 2(0.95)(0.05)(1) = 0.9025 + 0.095 = 0.9975
q' = q^2w22 + pqw12 = (0.05)^2(0) + 2(0.05)(0.95)(1) = 0.095
Therefore, the frequency of a1 is expected to increase slightly to 0.9975, and the frequency of a2 is expected to decrease slightly to 0.095.
f) After one generation, the new allele frequencies are p' = 0.9975 and q' = 0.095. The new genotype frequencies can be calculated as follows:
a1a1: p'^2 = (0.9975)^2 = 0.995
a1a2: 2p'q' = 2(0.9975)(0.095) = 0.189
a2a2: q'^2 = (0.095)^2 = 0.009
Therefore, the frequency of a1a1 is 0.995, the frequency of a1a2 is 0.189, and the frequency of a2a2 is 0.009.
Learn more about hardy-weinberg equilibrium
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