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]
