Numerical: Hardy-weinberg
In a population, frequency of recessive phenotype (aa) is 0.16. Assuming H-W equilibrium, what is frequency of heterozygotes?
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About This Question
- Subject
- biology
- Chapter
- genetics and evolution
- Topic
- numerical: hardy-weinberg
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
0.48
- Identify Given Value: Recessive phenotype frequency = q2=0.16.
- Find Recessive Allele Frequency (q):
- q=0.16=0.4.
- Find Dominant Allele Frequency (p):
- Since p+q=1, p=1−0.4=0.6.
- Calculate Heterozygote Frequency (2pq):
- Frequency=2×p×q=2×0.6×0.4
- Frequency=2×0.24=0.48.
- Check: p2(0.36)+2pq(0.48)+q2(0.16)=1.00.
This hard difficulty biology question is from the chapter genetics and evolution, covering the topic of numerical: hardy-weinberg. It appeared in the 2025 exam.
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