College

Assuming the conditions set by Hardy-Weinberg equilibrium, if the frequency of phenotype A is 96%, what would be the frequency of A in the next generation?

Answer :

The frequency of phenotype A in the next generation would remain 96% (0.96) under Hardy-Weinberg equilibrium.

To determine the frequency of the allele A in the next generation under Hardy-Weinberg equilibrium, we first need to find the allele frequencies from the phenotype frequencies.

Step 1: Understand the Phenotype Frequencies

  • Let p be the frequency of allele A.
  • Let q be the frequency of allele a.

Given that the frequency of phenotype A is 96%, this typically refers to the dominant phenotype. We can denote:

  • Frequency of phenotype A (AA + Aa) = 0.96
  • Frequency of phenotype a (aa) = 1 - 0.96 = 0.04

Step 2: Apply the Hardy-Weinberg Equation

According to the Hardy-Weinberg principle:

[tex]p^2 + 2pq + q^2 = 1[/tex]

Where:

  • [tex]p^2[/tex] = frequency of homozygous dominant (AA)
  • 2pq = frequency of heterozygous (Aa)
  • [tex]q^2[/tex] = frequency of homozygous recessive (aa)

Since we have [tex]q^2[/tex] (the frequency of aa) = 0.04:

Step 3: Calculate q

To find q, we take the square root of [tex]q^2[/tex]:

  • q = [tex]\sqrt{(0.04)[/tex] = 0.2

Step 4: Calculate p

Now, using the relationship p + q = 1:

  • p = 1 - q = 1 - 0.2 = 0.8

Step 5: Calculate the Frequency of Phenotype A in the Next Generation

Under Hardy-Weinberg equilibrium, the frequencies of the alleles remain constant from generation to generation. Thus, the frequency of phenotype A in the next generation would still reflect the current allele frequencies.

Using the calculated frequency of A (p):

Frequency of phenotype A in the next generation = [tex]p^2[/tex] + 2pq

Calculating that:

  • [tex]p^2 = (0.8)^2[/tex] = 0.64 (frequency of AA)
  • 2pq = 2 * (0.8) * (0.2) = 0.32 (frequency of Aa)

Adding them together gives:

Frequency of phenotype A in the next generation = 0.64 + 0.32 = 0.96

Frequency of phenotype A in current generation: 96% Frequency of phenotype A in the next generation (under Hardy-Weinberg equilibrium): 96%

To understand the frequency of phenotype A in the next generation under Hardy-Weinberg equilibrium, let's first review what Hardy-Weinberg equilibrium means.

Hardy-Weinberg Equilibrium

Hardy-Weinberg equilibrium describes a population that is not evolving, where allele and genotype frequencies remain constant from generation to generation. For two alleles (A and a) in a population, the frequencies can be represented as:

[tex]p + q = 1[/tex]

where:

  • [tex]p[/tex] is the frequency of allele A (dominant)
  • [tex]q[/tex] is the frequency of allele a (recessive)

In this scenario, we also use the following formula:

[tex]p^2 + 2pq + q^2 = 1[/tex]

where:

  • [tex]p^2[/tex] represents the frequency of homozygous dominant (AA) individuals
  • [tex]2pq[/tex] represents the frequency of heterozygous (Aa) individuals
  • [tex]q^2[/tex] represents the frequency of homozygous recessive (aa) individuals

Given Information

You mentioned that the frequency of phenotype A is 96%. This means we need to calculate the allele frequencies based on this information.

Assuming phenotype A includes both homozygous dominant (AA) and heterozygous (Aa), we have:
[tex]p^2 + 2pq = 0.96[/tex]

This equates to 96%, which implies that the rest of the population would consist of phenotype a:
[tex]q^2 = 0.04[/tex]

Calculating Frequencies

From the above equation, we can find [tex]q[/tex]:
[tex]q = ext{sqrt}(0.04) = 0.2[/tex]

Now we can find [tex]p[/tex]:
[tex]p = 1 - q = 1 - 0.2 = 0.8[/tex]

Next Generation Frequencies

Assuming the population is in Hardy-Weinberg equilibrium, these allele frequencies will remain constant. Hence, the frequency of phenotype A in the next generation will still be:
[tex]p^2 + 2pq = 0.96[/tex]