For the following questions use the Hardy-Weinberg Equations: p2+ 2pq + q2 = 1 and p + q = 1 A study was done of the current population of individuals to determine the frequency for the tongue rolling trait. At the conclusion of the study it was determined that 16% of the population was homozygous recessive and were not tongue rollers. Using this information complete the Hardy-Weinberg equations and determine the following: A) What is the percentage of individuals that are heterozygous for tongue rolling. B) What is the frequency of the p allele in the population.
The Correct Answer and Explanation is:
We are given that 16% of the population is homozygous recessive (non-tongue rollers), which means: q2=0.16q^2 = 0.16q2=0.16
To find the allele frequencies and solve for the rest of the Hardy-Weinberg equilibrium components, follow these steps:
Step 1: Find q (frequency of the recessive allele)
q=q2=0.16=0.4q = \sqrt{q^2} = \sqrt{0.16} = 0.4q=q2=0.16=0.4
Step 2: Find p (frequency of the dominant allele)
p=1−q=1−0.4=0.6p = 1 – q = 1 – 0.4 = 0.6p=1−q=1−0.4=0.6
Step 3: Find 2pq (frequency of heterozygous individuals)
2pq=2×0.6×0.4=0.482pq = 2 × 0.6 × 0.4 = 0.482pq=2×0.6×0.4=0.48
Step 4: Convert to percentage
0.48=48%0.48 = 48\%0.48=48%
Final Answers:
A) The percentage of individuals that are heterozygous for tongue rolling is 48%
B) The frequency of the p allele in the population is 0.6
Explanation
The Hardy-Weinberg principle provides a mathematical model to study genetic variation in a population under certain ideal conditions. The conditions include no mutation, random mating, no gene flow, infinite population size, and no selection. It is a useful tool to estimate the frequency of alleles and genotypes when these conditions are approximately met.
In this problem, the trait being analyzed is the ability to roll the tongue. We are told that 16 percent of the population are non-tongue rollers and that this is a homozygous recessive trait. According to Hardy-Weinberg notation, the frequency of the homozygous recessive genotype is represented by q². Therefore, we set q² = 0.16. By taking the square root of 0.16, we find that q = 0.4, which represents the frequency of the recessive allele.
Once we know q, we can find p, the frequency of the dominant allele, because p + q = 1. Substituting q = 0.4, we find p = 0.6. To determine the frequency of heterozygous individuals (those who have one dominant and one recessive allele), we use the formula 2pq, which equals 2 × 0.6 × 0.4 = 0.48, or 48 percent.
Thus, 48 percent of the population are heterozygous tongue rollers, and the dominant allele frequency is 0.6. This application demonstrates how observed phenotype data can be used to infer the underlying genetic structure of a population.
