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: Concept Overview:
This problem utilizes the Hardy-Weinberg Equilibrium (HWE) principles to analyze a population's genetic makeup. HWE relates allele and genotype frequencies in non-evolving populations, using the equations \( p + q = 1 \) and \( p^2 + 2pq + q^2 = 1 \).

Step 2: Formulas and Given Data:
- \(p\) = frequency of the dominant allele (A), and \(q\) = frequency of the recessive allele (a).
- \(q^2\) = frequency of homozygous recessive genotype (aa), representing affected individuals.
- \(p^2\) = frequency of homozygous dominant genotype (AA).
- \(2pq\) = frequency of heterozygous genotype (Aa).
- Total population size (N) = 592.
- Number of affected individuals (aa) = 44.

Step 3: Detailed Calculations:
1. Calculate \(q^2\):
\[ q^2 = \frac{\text{Number of affected individuals}}{\text{Total population}} = \frac{44}{592} \approx 0.07432 \]\ 2. Calculate \(q\):
\[ q = \sqrt{q^2} = \sqrt{\frac{44}{592}} \approx \sqrt{0.07432} \approx 0.2726 \]\ 3. Calculate \(p\):
\[ p = 1 - q = 1 - 0.2726 \approx 0.7274 \]\ 4. Calculate the expected number of Aa individuals:
Heterozygote frequency is \(2pq\).
\[ \text{Number of Aa} = 2pq \times N = (2 \times 0.7274 \times 0.2726) \times 592 \approx 0.3965 \times 592 \approx 234.7 \]\ Rounded, this is 235 individuals. 5. Calculate the expected number of AA individuals:
Homozygous dominant frequency is \(p^2\).
\[ \text{Number of AA} = p^2 \times N = (0.7274)^2 \times 592 \approx 0.5291 \times 592 \approx 313.2 \]\ Rounded, this is 313 individuals.
Step 4: Conclusion:
The estimated number of heterozygotes (Aa) is 235, and homozygous dominants (AA) is 313. The closest answer choice is (A) 233 and 315. Verification: \(233 (\text{Aa}) + 315 (\text{AA}) + 44 (\text{aa}) = 592\). The slight difference likely stems from rounding. Option (A) is the correct answer.