In a certain population of rabbits, the allele for brown fur is dominant over the allele for white fur. If 10 out of 100 rabbits have white fur, what is the allele frequency for the dominant allele?

The Hardy-Weinberg equation is: [tex]\([ \text{homozygous dominant} ]^2 + 2 [ \text{heterozygous} ] + [ \text{homozygous recessive} ]^2 = 1\)[/tex]

A. 0.90

B. 0.68

C. 0.10

D. 0.95



Answer :

To solve this problem, we'll use the principles of Hardy-Weinberg equilibrium. The Hardy-Weinberg equation is given as [tex]\( p^2 + 2pq + q^2 = 1 \)[/tex], where [tex]\( p \)[/tex] and [tex]\( q \)[/tex] represent the frequencies of the dominant and recessive alleles, respectively.

Step-by-step solution:

1. Identify given data:
- Total population of rabbits ([tex]\( N \)[/tex]) = 100
- Number of white fur rabbits = 10

2. Determine the frequency of white fur rabbits:
Since white fur is recessive, the number of white fur rabbits is equal to the number of rabbits that are homozygous recessive ([tex]\( q^2 \)[/tex]).
[tex]\[ q^2 = \frac{\text{Number of white fur rabbits}}{\text{Total population}} = \frac{10}{100} = 0.1 \][/tex]

3. Calculate the frequency of the recessive allele ([tex]\( q \)[/tex]):
To find [tex]\( q \)[/tex], take the square root of [tex]\( q^2 \)[/tex].
[tex]\[ q = \sqrt{q^2} = \sqrt{0.1} \approx 0.316 \][/tex]

4. Determine the frequency of the dominant allele ([tex]\( p \)[/tex]):
Using the relation [tex]\( p + q = 1 \)[/tex], solve for [tex]\( p \)[/tex].
[tex]\[ p = 1 - q = 1 - 0.316 \approx 0.684 \][/tex]

Hence, the frequency of the dominant allele ([tex]\( p \)[/tex]) is approximately [tex]\( 0.684 \)[/tex] or [tex]\( 68.4\% \)[/tex].

The closest answer choice is:
B. 0.68