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
