Question

8\. In a particular Hardy-Weinberg population, there are only two eye colors: brown and blue. Of the population, 36 percent has blue eyes, the recessive trait. What percentage of the population is heterozygous for brown eyes? A. 24% B. 48% C. 60% D. 64%

Short answer

48%

Step-by-step solution

State the Hardy-Weinberg Principle

The Hardy-Weinberg principle states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences. It is given by the equation: \[ p^2 + 2pq + q^2 = 1 \] where p and q are the frequencies of the dominant and recessive alleles respectively, and \[ p + q = 1 \].

Identify given information

The problem states that 36% of the population has blue eyes (the recessive trait). Therefore, \[ q^2 = 0.36 \].

Calculate q

To find the value of q, take the square root of q^2: \[ q = \sqrt{0.36} = 0.6 \].

Calculate p

Since \[ p + q = 1 \], we can find p by: \[ p = 1 - q = 1 - 0.6 = 0.4 \].

Determine heterozygous frequency

The frequency of heterozygous individuals (brown eyes) is given by 2pq: \[ 2pq = 2 \times 0.4 \times 0.6 = 0.48 \]. This represents 48% of the population.

Key concepts

Allele Frequency

Alleles are different forms of a gene that arise by mutation and are found at the same place on a chromosome. In the context of the Hardy-Weinberg equilibrium, allele frequency refers to how often a specific allele appears in a gene pool. It is usually represented by either 'p' or 'q'.

In this exercise, you are dealing with two alleles: one for brown eyes (dominant) and one for blue eyes (recessive). The frequency of the dominant allele is denoted as 'p' and the frequency of the recessive allele as 'q'.

It's important to remember that the sum of the frequencies of these alleles must equal 1. This means: \( p + q = 1\). By knowing one allele frequency, you can easily find the other. For instance, if 'q' (recessive) is 0.6, you can find 'p' (dominant) using the formula:
\[ p = 1 - q = 1 - 0.6 = 0.4 \]

Keeping these relationships in mind will help you solve Hardy-Weinberg problems quickly and accurately.

Genotype Frequency

Genotype frequencies are the proportions of different genotypes in the population. Under the Hardy-Weinberg principle, these can be calculated using the allele frequencies 'p' and 'q'. The most common genotypes in such problems are:

• \( p^2 \): Frequency of homozygous dominant individuals
• \( 2pq \): Frequency of heterozygous individuals
• \( q^2 \): Frequency of homozygous recessive individuals

In the given problem, we know that 36% of the population has blue eyes, which is the recessive trait. This corresponds to \( q^2 = 0.36 \).

To find \( q \), we take the square root of 0.36:
\[ q = \sqrt{0.36} = 0.6 \]

With \( q\) known, we can find \( p\) as follows:

\[ p = 1 - q = 1 - 0.6 = 0.4 \]

Now, by substituting 'p' and 'q' into the formulas for genotype frequencies, you can find out the proportions of each genotype in the population.

Heterozygous

An individual is heterozygous for a trait if they have one dominant and one recessive allele (i.e., 'Bb' in this brown-blue eye example). The combined frequency of heterozygous individuals in a population can be calculated using the Hardy-Weinberg principle.

The formula to find the heterozygous frequency is:
\[ 2pq \]

From previous sections, we know that 'p' (dominant allele frequency) is 0.4 and 'q' (recessive allele frequency) is 0.6. Let's apply these values to the formula:

\[ 2pq = 2 \times 0.4 \times 0.6 = 0.48 \]

Therefore, the heterozygous frequency is 48% in the population. This means 48% of individuals in this Hardy-Weinberg population have one allele for brown eyes and one for blue eyes, making them genetically 'Bb'. This intermediate step is crucial for understanding how genetic variation is maintained in populations.