Question

an allele w for white wool is dominant over allele w for black wool. in a sample of 500 sheep it is observed that 350 are white and 150 are black. calculate the allele frequencies within this population and also calculate the genotypic frequencies. finally calculate the number of sheep that you would expect to have each genotype. assume that the population is in the hardy weinberg equilibrium

Answer 1

@oliver69

Answer 2

Screw Oliver I got this.



Here you go:

1. Calculate allele frequencies:

Let \( p \) be the frequency of the ascendant allele (\( W \) for white wool), and let \( q \) be the frequency of the recessive allele (\( w \) for ebony wool).

\[
\commence{align*}
p & = \frac{2 \cdot (\text{number of WW sheep}) + (\text{number of Ww sheep})}{2 \cdot (\text{total number of sheep})} \\
p & = \frac{2 \cdot 350 + 150}{2 \cdot 500} = \frac{850}{1000} = 0.85 \\
q & = 1 - p = 1 - 0.85 = 0.15
\end{align*}
\]

2. Calculate genotypic frequencies:

The genotypic frequencies are as follows:
- WW genotype frequency (\( p^2 \)): \( (0.85)^2 = 0.7225 \)
- Ww genotype frequency (\( 2pq \)): \( 2 \cdot 0.85 \cdot 0.15 = 0.255 \)
- ww genotype frequency (\( q^2 \)): \( (0.15)^2 = 0.0225 \)

3. Calculate the number of sheep with each genotype:

- Number of WW sheep: \( 0.7225 \cdot 500 \approx 361 \) WW sheep
- Number of Ww sheep: \( 0.255 \cdot 500 \approx 128 \) Ww sheep
- Number of ww sheep: \( 0.0225 \cdot 500 \approx 11 \) ww sheep

So, in this population, you would expect approximately 361 WW sheep, 128 Ww sheep, and 11 ww sheep under Hardy-Weinberg equilibrium.

You're welcome 😎

Answer 3

Here you go:

1. Calculate allele frequencies:

Let \( p \) be the frequency of the ascendant allele (\( W \) for white wool), and let \( q \) be the frequency of the recessive allele (\( w \) for black wool).

\[
\commence{align*}
p & = \frac{2 \cdot (\text{number of WW sheep}) + (\text{number of Ww sheep})}{2 \cdot (\text{total number of sheep})} \\
p & = \frac{2 \cdot 350 + 150}{2 \cdot 500} = \frac{850}{1000} = 0.85 \\
q & = 1 - p = 1 - 0.85 = 0.15
\end{align*}
\]

2. Calculate genotypic frequencies:

The genotypic frequencies are as follows:
- WW genotype frequency (\( p^2 \)): \( (0.85)^2 = 0.7225 \)
- Ww genotype frequency (\( 2pq \)): \( 2 \cdot 0.85 \cdot 0.15 = 0.255 \)
- ww genotype frequency (\( q^2 \)): \( (0.15)^2 = 0.0225 \)

3. Calculate the number of sheep with each genotype:

- Number of WW sheep: \( 0.7225 \cdot 500 \approx 361 \) WW sheep
- Number of Ww sheep: \( 0.255 \cdot 500 \approx 128 \) Ww sheep
- Number of ww sheep: \( 0.0225 \cdot 500 \approx 11 \) ww sheep

So, in this population, you would expect approximately 361 WW sheep, 128 Ww sheep, and 11 ww sheep under Hardy-Weinberg equilibrium.

You're welcome 😎

Answer 4

thanks