Consider a locus with two alleles - B and b. B is dominant, while b is recessive. There is no mutation. B has a selective advantage relative to b, so that the fitnesses of the three genotypes are BB = 1, Bb = 1, and bb = 1-s. In this case, s = 0.50, so that bb homozygotes have 50% fitness of heterozygotes and BB homozygotes. If the population has the following genotypic counts prior to selection of BB = 500, Bb = 250, and bb = 250, what is the expected change in the frequency of B after one generation with selection? Please give your answer to two decimal places.

Respuesta :

KerryM

Answer:

0.09

Explanation:

Before selection

Total number in population = 1000

Genotype frequencies

Genotype frequency of BB = 500/1000 = 0.5

Genotype frequency of Bb = 250/1000 = 0.25

Genotype frequency of bb = 250/1000 = 0.25

Allele frequencies

Allele frequency of B = BB genotype frequency + half of the Bb genotype frequency = 0.5 + (0.25/2) = 0.625

Allele frequency of b = bb genotype frequency + half of the Bb genotype frequency = 0.25 + (0.25/2) = 0.375

After selection

We are told that after selection, the genotype frequency of bb is changed as they become 50% less fit. This means the frequency of bb individuals changes from 250 to 125 individuals (50% reduction).

Now the total number of individuals is 500 + 125 + 250 = 875.

Genotype frequencies

Genotype frequency of BB = 500/875 = 0.57

Genotype frequency of Bb = 250/875 = 0.29

Genotype frequency of bb = 125/875 = 0.14

Allele frequencies

Allele frequency of B = BB genotype frequency + half of the Bb genotype frequency = 0.57 + (0.29/2) = 0.715

Allele frequency of b = bb genotype frequency + half of the Bb genotype frequency = 0.14 + (0.29/2) = 0.285

Change in frequency of B after 1 generation

0.715 - 0.625 = 0.09

Genes have alleles that are either dominant or recessive. the genotype that is expressed in physical attributes are dominant traits and the recessive traits are those that get masked in the presence of a dominant allele.

The expected change in the frequency is:

0.09

The frequency can be estimated as:

Data before selection:

  • The total population = 1000

  • Frequency of BB = [tex]\dfrac {500 }{1000} = 0.5[/tex]  

  • Frequency of Bb = [tex]\dfrac{250}{1000} = 0.25[/tex]  

  • Frequency of bb = [tex]\dfrac{250}{1000} = 0.25[/tex]

Allele frequencies can be calculated as:

  • Allele frequency of B

[tex]\begin{aligned}\text{BB genotype frequency + half of the Bb genotype frequency} &= 0.5 + (\dfrac{0.25}{2}) \\&= 0.625\end{aligned}[/tex]

  • Allele frequency of b:

[tex]\begin{aligned}\text{bb genotype frequency + half of the Bb genotype frequency} &= 0.25 + (\dfrac{0.25}{2}) \\&= 0.375 \end{aligned}[/tex]

Data after selection:

bb trait becomes 50 % less fit which means that the frequency of bb trait got reduced to 125 individuals.

Now the total population will be:

[tex]\begin{aligned} &= 500 + 125 + 250 \\&= 875\end{aligned}[/tex]

  • Frequency of BB = [tex]\dfrac{500}{875} = 0.57[/tex]

  • Frequency of Bb = [tex]\dfrac{250}{875} = 0.29[/tex]

  • Frequency of bb = [tex]\dfrac{125}{875 }= 0.14[/tex]

Now allele frequency after selection will be:

  • Allele frequency of B:

[tex]\begin{aligned} \text{BB genotype frequency + half of the Bb genotype frequency} &= 0.57 + (\dfrac{0.29}{2}) \\&= 0.715\end{aligned}[/tex]

  • Allele frequency of b:

[tex]\begin{aligned} \text{bb genotype frequency + half of the Bb genotype frequency}& = 0.14 + (\dfrac{0.29}{2}) \\&= 0.285\end{aligned}[/tex]

Change in frequency of B allele after one generation:

[tex]\begin{aligned} & = 0.715 - 0.625 \\&= 0.09\end{aligned}[/tex]

Therefore, 0.09 is the frequency change.

To learn more about frequency change follow the link:

https://brainly.com/question/6895892