In a population of cats, the allele for short tails (T) is dominant over the allele for long tails (t). If the frequency of the T allele is 0.6, what is the expected frequency of cats with short tails in the population, assuming Hardy-Weinberg equilibrium?

• A. 0.84
• B. 0.16
• C. 0.36
• D. 0.48

Answer: (A)

Verified step by step guidance

1. Identify the allele frequencies given: the frequency of the dominant allele T is \(p = 0.6\). Since there are only two alleles, the frequency of the recessive allele t is \(q = 1 - p = 1 - 0.6\).
2. Recall the Hardy-Weinberg equilibrium principle, which states that genotype frequencies can be predicted by the equation \(p^2 + 2pq + q^2 = 1\), where \(p^2\) is the frequency of homozygous dominant (TT), \$2pq\( is the frequency of heterozygous (Tt), and \)q^2$ is the frequency of homozygous recessive (tt).
3. Calculate the expected frequency of cats with short tails. Since short tails are dominant, cats with genotypes TT and Tt will have short tails. Therefore, the frequency of short-tailed cats is \(p^2 + 2pq\).
4. Substitute the values of \(p\) and \(q\) into the expression \(p^2 + 2pq\) to find the expected frequency of short-tailed cats.
5. Interpret the result as the proportion of the population expected to have short tails under Hardy-Weinberg equilibrium.