Hardy-Weinberg Equilibrium Practice Problems with Answers
Understanding the Hardy-Weinberg equilibrium is a cornerstone of population genetics, providing a framework to analyze genetic variation and evolutionary forces in populations. This principle helps determine whether a population is evolving or remains stable in terms of genetic structure. So to master this concept, practicing problems is essential. Below, we present a series of Hardy-Weinberg equilibrium practice problems with answers, designed to reinforce your understanding of allele frequencies, genotype distributions, and evolutionary principles Took long enough..
Problem 1: Calculating Allele Frequencies from Genotype Frequencies
Question:
In a population of 1000 individuals, 360 are homozygous dominant (AA), 480 are heterozygous (Aa), and 160 are homozygous recessive (aa). Calculate the allele frequencies ((p) and (q)) for the dominant (A) and recessive (a) alleles Practical, not theoretical..
Solution:
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Genotype frequencies:
- ( \text{freq}(AA) = \frac{360}{1000} = 0.36 )
- ( \text{freq}(Aa) = \frac{480}{1000} = 0.48 )
- ( \text{freq}(aa) = \frac{160}{1000} = 0.16 )
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Allele frequencies:
- ( p = \sqrt{\text{freq}(AA)} = \sqrt{0.36} = 0.6 )
- ( q = \sqrt{\text{freq}(aa)} = \sqrt{0.16} = 0.4 )
Alternatively, ( p = \text{freq}(A) = \frac{2 \times \text{AA} + \text{Aa}}{2 \times \text{total individuals}} = \frac{2(360) + 480}{2000} = 0.6 ) That's the part that actually makes a difference..
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Verification:
( p + q = 0.6 + 0.4 = 1 ) (valid). -
Check Hardy-Weinberg equation:
( p^2 + 2pq + q^2 = (0.6)^2 + 2(0.6)(0.4) + (0.4)^2 = 0.36 + 0.48 + 0.16 = 1 ).
Answer:
The allele frequencies are ( p = 0.6 ) (dominant allele) and ( q = 0.4 ) (recessive allele). The population is in Hardy-Weinberg equilibrium.
Problem 2: Predicting Genotype Frequencies in the Next Generation
Question:
In a population, the frequency of the recessive allele ((q)) is 0.3. What are the expected genotype frequencies in the next generation under Hardy-Weinberg equilibrium?
Solution:
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Allele frequencies:
- ( q = 0.3 )
- ( p = 1 - q = 1 - 0.3 = 0.7 )
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Genotype frequencies:
- ( p^2 = (0.7)^2 = 0.49 ) (homozygous dominant, AA)
- ( 2pq = 2(0.7)(0.3) = 0.42 ) (heterozygous, Aa)
- ( q^2 = (0.3)^2 = 0.09 ) (homozygous recessive, aa)
Answer:
The expected genotype frequencies are 49% AA, 42% Aa, and 9% aa in the next generation Worth keeping that in mind. That's the whole idea..
Problem 3: Determining if a Population is in Equilibrium
Question:
