You are researching a population of 100 cows, where 91 of them are grey and 9 are black. You know that the black colour is a recessive trait for this type of cows. Using the Hardy-Weinberg Equilibrium equation (p2+ 2pq + q2 = 1), what number would be heterozygous?

Question

BlackTom AI · Accepted Answer

We start by identifying the given information: there are 100 cows, with 9 black cows. Since black is the recessive phenotype, its frequency in the population equals q^2 = 9/100 = 0.09.

From q^2 = 0.09, we take the square root to find q: q = sqrt(0.09) = 0.3. The frequency of the dominant allele is p = 1 - q = 1 - 0.3 = 0.7.

The heterozygous genotype frequency is given by 2pq. Substituting p and q gives 2pq = 2 × 0.7 × 0.3 = 0.42.

To get the actual number of heterozygous individuals in the population of 100 cows, multiply by the total population: 0.42 × 100 = 42 cows.

Now examine each option to see why they are or aren’t correct:
- Option 32: If 2pq were 0.32, that would imply p and q values that do not satisfy q^2 = 0.09 (since p + q = 1 would not hold with these values). This is inconsistent with the given recessive phenotype frequency. 
- Option 42: This matches the calculated heterozygote frequency 2pq = 0.42, yielding 42 cows. It is the correct calculation under Hardy-Weinberg with the provided data. 
- Option 70: A heterozygote proportion of 0.70 would require p to be near 0.7 but would force q to be near 0.3 only if 2pq equals 0.42; here 0.70 is simply the value of p, not 2pq, so this overstates the heterozygote fraction. 
- Option 80: A heterozygote proportion of 0.80 would imply a different combination of p and q that does not align with q^2 = 0.09; this is not compatible with the observed recessive phenotype frequency.

Putting it together, the heterozygous count that aligns with the recessive phenotype frequency and the Hardy-Weinberg relationship is 42 individuals.

AGRI10051_2025_SM2 Exam: Genetics for Agriculture (AGRI10051_2025_SM2)- Requires Respondus LockDown Browser

You are researching a population of 100 cows, where 91 of them are grey and 9 are black. You know that the black colour is a recessive trait for this type of cows. Using the Hardy-Weinberg Equilibrium equation (p2+ 2pq + q2 = 1), what number would be heterozygous?

View Explanation

Log in for full answers

Similar Questions

You are researching a population of 100 cows, where 91 of them are grey and 9 are black. You know that black colour is a recessive trait for this type of cow. Using the Hardy-Weinberg Equilibrium equation (p2 + 2pq + q2 = 1), what number would be heterozygous?

Assuming the population is in Hardy-Weinberg equilibrium, if 16% of the individuals in a population show a recessive trait (q2=16%), what is the allelic frequency for the dominant allele?

What occurs to allele frequencies when a population is at Hardy-Weinberg equilibrium?

What would disrupt the conditions required for Hardy-Weinberg equilibrium?

The allele for brown eyes (B) is found within a population at a frequency of 0.7; the allele for blue eyes (b) at a frequency of 0.3. If the conditions for Hardy-Weinberg equilibrium are satisfied, what will be the expected frequency of heterozygous (Bb) individuals?

Which of these would support a population remaining at genetic equilibrium / NOT evolving? (In other words, which of these would not cause a population to evolve?)

Which of the following factors would cause a population to deviate from Hardy–Weinberg equilibrium?

What occurs to allele frequencies when a population is at Hardy-Weinberg equilibrium?

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler