Question

1. suppose a population of lemnings has a dominant gene for white tooth enamel, while the recessive was yellow tooth enamel. if 23 lemnings is a population of 325 have the recessive trait.
2. what is the frequency of the recessive genotype in this population?
3. what is the frequency of the homozygous dominant condition?

Explanation:

Step1: Identify Genotype Counts

Let's assume the dominant allele is \( A \) and the recessive allele is \( a \). The yellow tooth enamel (dominant) individuals can be \( AA \) (homozygous dominant) or \( Aa \) (heterozygous), and white tooth enamel (recessive) individuals are \( aa \) (homozygous recessive). We know the number of recessive individuals (\( aa \)) is 23, and the total population is 325. First, find the number of dominant - phenotype individuals: \( 325 - 23=302 \).

Step2: Calculate Allele Frequencies (Hardy - Weinberg Principle)

The Hardy - Weinberg equation is \( p^{2}+2pq + q^{2}=1 \), where \( q^{2} \) is the frequency of the recessive genotype (\( aa \)), \( p^{2} \) is the frequency of the homozygous dominant genotype (\( AA \)), and \( 2pq \) is the frequency of the heterozygous genotype (\( Aa \)). Also, \( p + q=1 \), where \( p \) is the frequency of the dominant allele (\( A \)) and \( q \) is the frequency of the recessive allele (\( a \)).

First, calculate \( q^{2} \): \( q^{2}=\frac{\text{Number of } aa}{\text{Total population}}=\frac{23}{325}\approx0.0708 \)

Then, find \( q \): \( q = \sqrt{q^{2}}=\sqrt{\frac{23}{325}}\approx\sqrt{0.0708}\approx0.266 \)

To find the frequency of the recessive genotype (\( aa \)) is \( q^{2}\approx0.0708 \) or 7.08%

For the homozygous dominant (\( AA \)) frequency (\( p^{2} \)): Since \( p = 1 - q \), \( p=1 - 0.266 = 0.734 \)

Then \( p^{2}=(0.734)^{2}\approx0.539 \) or 53.9%

(Note: If we want to find the number of homozygous dominant individuals, we can calculate \( p^{2}\times\text{Total population}=0.539\times325\approx175 \), and the number of heterozygous individuals is \( 2pq\times325 = 2\times0.734\times0.266\times325\approx2\times0.195\times325 = 0.39\times325 = 126.75\approx127 \), and \( 175 + 127+23 = 325 \) which checks out)

Answer:

The frequency of the recessive genotype (\( aa \)) is approximately \( \frac{23}{325}\approx0.071 \) (or 7.1%). The frequency of the homozygous dominant genotype (\( AA \)) is approximately \( p^{2}\approx0.539 \) (or 53.9%).