Question

In a population under Hardy Weinberg Equilibrium, out of a total of 592 people, 44 people were affected by a rare autosomal recessive disorder whereas the rest of the population was normal. How many individuals in this population have the heterozygous and homozygous dominant genotypes for this trait?

• A. 233 and 315, respectively.
• B. 315 and 233, respectively.
• C. 0.27 and 0.73, respectively.
• D. 0.73 and 0.27, respectively.

Hint:

In HWE problems, always start with the information you have about the homozygous recessive group (\(q^2\)), as their genotype is unambiguously known from their phenotype. From \(q^2\), find \(q\), then \(p\), and then you can calculate all other genotype frequencies (\(p^2\) and \(2pq\)).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem applies the principles of Hardy-Weinberg Equilibrium (HWE) to a population. HWE describes the relationship between allele frequencies and genotype frequencies in a population that is not evolving. The key equations are \( p + q = 1 \) and \( p^2 + 2pq + q^2 = 1 \).

Step 2: Key Formula or Approach:
- Let \(p\) be the frequency of the dominant allele (A) and \(q\) be the frequency of the recessive allele (a).
- \(q^2\) = frequency of the homozygous recessive genotype (aa), which corresponds to the affected individuals.
- \(p^2\) = frequency of the homozygous dominant genotype (AA).
- \(2pq\) = frequency of the heterozygous genotype (Aa).
- Total population size (N) = 592.
- Number of affected individuals (aa) = 44.

Step 3: Detailed Calculation:
1. Calculate the frequency of the homozygous recessive genotype (\(q^2\)):
\[ q^2 = \frac{\text{Number of affected individuals}}{\text{Total population}} = \frac{44}{592} \approx 0.07432 \] 2. Calculate the frequency of the recessive allele (\(q\)):
\[ q = \sqrt{q^2} = \sqrt{\frac{44}{592}} \approx \sqrt{0.07432} \approx 0.2726 \] 3. Calculate the frequency of the dominant allele (\(p\)):
\[ p = 1 - q = 1 - 0.2726 \approx 0.7274 \] 4. Calculate the expected number of heterozygous individuals (Aa):
The frequency of heterozygotes is \(2pq\).
\[ \text{Number of Aa} = 2pq \times N = (2 \times 0.7274 \times 0.2726) \times 592 \approx 0.3965 \times 592 \approx 234.7 \] This rounds to 235 individuals. 5. Calculate the expected number of homozygous dominant individuals (AA):
The frequency of homozygous dominants is \(p^2\).
\[ \text{Number of AA} = p^2 \times N = (0.7274)^2 \times 592 \approx 0.5291 \times 592 \approx 313.2 \] This rounds to 313 individuals.
Step 4: Final Answer:
The calculated number of heterozygous individuals is approximately 235, and the number of homozygous dominant individuals is approximately 313. The question asks for heterozygous and homozygous dominant, respectively. The option that most closely matches our calculation is (A) 233 and 315. We can verify this option: \(233 (\text{Aa}) + 315 (\text{AA}) + 44 (\text{aa}) = 592\), which is the correct total. The minor discrepancy is likely due to rounding or the use of slightly non-ideal numbers in the problem statement. Option (A) is the intended answer.