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?

Question

BlackTom AI · Accepted Answer

In this population genetics problem, we start from the phenotype data: 9% black cows, which are recessive, meaning aa. Therefore q^2 = 0.09, so q = sqrt(0.09) = 0.3. The proportion of the other allele, p, is 1 - q = 0.7. The heterozygous genotype frequency is 2pq = 2 × 0.7 × 0.3 = 0.42.

Option by option analysis:
- Option 42: This matches the calculated heterozygote frequency of 0.42. Since there are 100 cows, 0.42 × 100 = 42 individuals would be heterozygous (Aa). This aligns with Hardy-Weinberg expectations given q = 0.3 and p = 0.7.
- Option 80: If 80 individuals were heterozygous, that would correspond to a frequency of 0.80 for Aa. That would imply p and q values that do not satisfy p^2 + 2pq + q^2 = 1 with q^2 = 0.09; it would require an unrealistically high 2pq value, inconsistent with the observed recessive phenotype frequency.
- Option 70: A heterozygote frequency of 0.70 would also imply 2pq = 0.70, which would force p and q to values that do not produce q^2 = 0.09 when used in p^2 + 2pq + q^2 = 1; this is not mathematically compatible with the given recessive phenotype proportion.
- Option 32: 32 heterozygotes would correspond to a frequency of 0.32, which is lower than the calculated 0.42 from p = 0.7 and q = 0.3, indicating a mismatch with the Hardy-Weinberg proportions based on the observed data.

In summary, the 0.42 heterozygote frequency derived from p = 0.7 and q = 0.3 best fits the given data and the Hardy-Weinberg framework, leading to 42 individuals being heterozygous in a population of 100.

AGRI10051_2025_SM2 Supplementary or Special 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 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?

View Explanation

Log in for full answers

Similar Questions

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?

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?

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