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Answer :
a) The frequency of the b allele in the next generation will be 0. b) The frequency of the b allele in two generations will remain 0. c) The frequency of the b allele in two generations will remain 0. d) The difference between the answers in questions 6b and 6c is due to the fact that in 6b, the b allele can still be carried by heterozygous individuals (Bb).
a) To calculate the frequency of the b allele in the next generation, we need to consider the genotypic frequencies and fitnesses.
- The frequency of the b allele in the current generation is 0.2. The genotypic fitnesses are given as WBB = 1.0, WBb = 1.0, and wbb = 0.0.
This means that individuals with the BB and Bb genotypes are fully viable, while individuals with the bb genotype are not viable (have a fitness of 0). Therefore, the b allele can only be passed on by Bb individuals.
- The frequency of the B allele in the current generation is 0.8, so the frequency of the Bb genotype is 2 times the product of the frequencies of B and b, which is [tex]2 \times 0.8 \times 0.2 = 0.32[/tex]. Since the BB genotype has a frequency of [tex]0.8^2 = 0.64[/tex] and the bb genotype cannot survive, the frequency of the b allele in the next generation will only come from the Bb genotype.
The contribution of the b allele from Bb individuals is 0.32 / 2 = 0.16. Since all BB individuals will survive and no bb individuals will contribute, the frequency of the b allele in the next generation will be 0.16 / (0.64 + 0.16) = 0.16 / 0.8 = 0.2.
However, since the bb genotype is lethal, any b alleles that end up in the bb genotype will not be passed on. This means that the effective frequency of the b allele that is actually passed on is reduced. The frequency of the b allele in the next generation will be:
[tex]p' = (1/2) * (frequency of Bb genotype) / (sum of viable genotypes)[/tex]
[tex]p' = (1/2) * (0.32) / (0.64 + 0.32)[/tex]
[tex]p' = (1/2) * (0.32) / (0.96)[/tex]
[tex]p' = 0.16 / 0.96 p' = 0.1667[/tex]
Since the bb genotype is lethal, the b allele frequency is halved each generation. Therefore, in one generation, it goes from 0.2 to 0.1667, and in the next generation, it will be halved again:
[tex]p'' = 0.1667 / 2 = 0.08335[/tex]
However, since the bb genotype cannot survive, the b allele will eventually be eliminated from the population. This process will continue until the frequency of the b allele is effectively 0.
b) Since the b allele is lethal in the homozygous state (bb), it will be eliminated from the population over time. After one generation, the frequency of the b allele is 0.08335. In the next generation, it will be halved again:
[tex]p''' = 0.08335 / 20[/tex]
Effectively 0, as it will continue to decrease towards 0 with each generation
c) If the fitnesses are WBB = 1.0, WBb = 0.0, and Wbb = 0.0, the b allele is lethal in both the heterozygous and homozygous states.
This means that no individuals carrying the b allele will survive to reproduce. Therefore, the frequency of the b allele will immediately drop to 0 in the next generation and remain 0 in all subsequent generations.
d) The difference between the answers in questions 6b and 6c is due to the viability of the heterozygote (Bb). In 6b, the Bb genotype is viable, which means the b allele can be maintained in the population at a low frequency by being carried by heterozygous individuals.
However, in 6c, the Bb genotype is not viable, so the b allele cannot be maintained in the population and is immediately eliminated. This illustrates the concept of genetic drift and the impact of genotypic fitness on allele frequencies in a population.
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The frequency of the b allele in the next generation will be 0.267 ,the frequency of the b allele in two generations will be 0.071, the frequency of the b allele in two generations with given fitnesses will be 0.4 and the difference between answers in 6b and 6c is large due to the change in fitness values for the heterozygous genotype (WBb).
We can use the Hardy-Weinberg equation and selection to find the allele frequencies in the next generations. First, we calculate the average fitness (w) of the population using the given fitness values and allele frequencies. Then, we apply the selection and find the new allele frequencies for the next generation.
For parts a and b, we follow the same process with the same fitness values for both generations. However, for part c, we use the new fitness values for the heterozygous genotype (WBb = 0.0), which dramatically changes the results.
The frequency of the b allele in future generations depends on the fitness values of the different genotypes. The difference between the two scenarios (6b and 6c) highlights the importance of considering selection and fitness when predicting allele frequencies in a population.
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