Chapter 3: Evolution - Numerical Questions

Here are some numerical problems based on the principles of evolution, focusing on the Hardy-Weinberg equilibrium.

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Question 1: Calculating Allele Frequencies

In a population of 1000 individuals, 360 have the genotype AA, 480 have the genotype Aa, and 160 have the genotype aa. What are the frequencies of the alleles A and a in this population?

Solution:

• Total individuals: 1000
• Number of A alleles:
  • From AA individuals: 360 x 2 = 720
  • From Aa individuals: 480 x 1 = 480
  • Total A alleles: 720 + 480 = 1200
• Number of a alleles:
  • From aa individuals: 160 x 2 = 320
  • From Aa individuals: 480 x 1 = 480
  • Total a alleles: 320 + 480 = 800
• Total alleles in the population: 1200 (A) + 800 (a) = 2000
• Frequency of allele A (p):
  • p = (Total A alleles) / (Total alleles) = 1200 / 2000 = 0.6
• Frequency of allele a (q):
  • q = (Total a alleles) / (Total alleles) = 800 / 2000 = 0.4
• Check: p + q = 0.6 + 0.4 = 1.0

Therefore, the frequency of allele A is 0.6 and the frequency of allele a is 0.4.

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Question 2: Hardy-Weinberg Equilibrium

The frequency of a recessive allele in a population is 0.2. Assuming the population is in Hardy-Weinberg equilibrium, what is the frequency of the heterozygous genotype?

Solution:

• Frequency of the recessive allele (q): 0.2
• Since p + q = 1, the frequency of the dominant allele (p) is:
  • p = 1 - q = 1 - 0.2 = 0.8
• The frequency of the heterozygous genotype is given by 2pq.
  • Frequency of heterozygotes = 2 * p * q = 2 * 0.8 * 0.2 = 0.32

Therefore, the frequency of the heterozygous genotype is 0.32 or 32%.

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Question 3: Application of Hardy-Weinberg Principle

In a population that is in Hardy-Weinberg equilibrium, the frequency of the homozygous recessive genotype is 0.09. What is the frequency of the dominant allele?

Solution:

• Frequency of the homozygous recessive genotype (q²): 0.09
• The frequency of the recessive allele (q) is the square root of q².
  • q = √0.09 = 0.3
• The frequency of the dominant allele (p) is calculated as:
  • p = 1 - q = 1 - 0.3 = 0.7

Therefore, the frequency of the dominant allele is 0.7.

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Question 4: Predicting Genotype Frequencies

If the frequency of the dominant allele (A) in a population is 0.4, what are the expected frequencies of the genotypes AA, Aa, and aa, assuming the population is in Hardy-Weinberg equilibrium?

Solution:

• Frequency of the dominant allele (p): 0.4
• Frequency of the recessive allele (q):
  • q = 1 - p = 1 - 0.4 = 0.6
• Expected genotype frequencies:
  • Frequency of AA (p²): (0.4)² = 0.16
  • Frequency of Aa (2pq): 2 * 0.4 * 0.6 = 0.48
  • Frequency of aa (q²): (0.6)² = 0.36
• Check: p² + 2pq + q² = 0.16 + 0.48 + 0.36 = 1.0

Therefore, the expected frequencies are:

• AA: 0.16 or 16%
• Aa: 0.48 or 48%
• aa: 0.36 or 36%