Hardy-Weinberg Equilibrium Problems

Hardy-Weinberg Equilibrium Problems

Problem 1: Cystic Fibrosis in US Caucasian Newborns

  • Given Data: 1 in 1700 newborns have cystic fibrosis (cc).

  • Definitions:

    • C: Normal allele (dominant)

    • c: Cystic fibrosis allele (recessive)

a) Determine Frequencies of Alleles and Genotypes

  • Frequency of genotype cc:

    • Frequency = 1/1700 = 0.000588.

  • Calculate frequency of allele c (q):

    • q^2 = 0.000588 → q = √0.000588 ≈ 0.0242.

  • Calculate frequency of allele C (p):

    • p = 1 - q = 1 - 0.0242 ≈ 0.9758.

  • Genotype frequencies:

    • CC (homozygous normal) = p^2 ≈ (0.9758)^2 ≈ 0.9524.

    • Cc (heterozygous) = 2pq ≈ 2(0.9758)(0.0242) ≈ 0.0473.

    • cc = q^2 = 0.000588.

b) Homozygous Normal Individuals

  • Total individuals: 1700

  • Homozygous normal (CC) = Total population × Frequency of CC

    • Number = 1700 × 0.9524 ≈ 1620.

d) Significance of cc Phenotype

  • Relevance: The presence of cc (cystic fibrosis) indicates individuals are homozygous recessive, which presents critical information for population genetics and understanding genetic diseases.

Problem 2: Albinism in North America

  • Given Data: Average frequency of albinism = 1 in 20,000.

  • Definitions:

    • a: Albinism allele (recessive)

    • A: Normal allele (dominant)

a) Determine Frequencies of Each Allele

  • Frequency of genotype aa (albinism) = 1/20000 = 0.00005.

  • Calculate frequency of allele a (q):

    • q^2 = 0.00005 → q ≈ 0.00707.

  • Calculate frequency of allele A (p):

    • p = 1 - q ≈ 1 - 0.00707 ≈ 0.99293.

b) Carriers of Albinism

  • Frequency of carriers (Aa):

    • Aa frequency = 2pq = 2 (0.99293)(0.00707) ≈ 0.01405.

  • Percentage of population that are carriers:

    • Percentage = 0.01405 × 100 ≈ 1.41%.

c) Homozygous Normal Individuals

  • Frequency of homozygous normal (AA):

    • AA frequency = p^2 ≈ (0.99293)^2 ≈ 0.98593.

  • Percentage:

    • Homozygous normal = 0.98593 × 100 ≈ 98.59%.

Problem 3: Sickle Cell Anemia in Africa

  • Given Data: 9% of the population has the allele for sickle-cell anemia (ss).

Calculation for Heterozygous (Ss) Individuals

  • Frequency of ss = q^2 = 0.09 → q = √0.09 = 0.3.

  • Frequency of allele S (dominant):

    • p = 1 - q = 1 - 0.3 = 0.7.

  • Percentage of population that is heterozygous (resistant to malaria):

    • Frequency of heterozygotes (Ss) = 2pq = 2(0.7)(0.3) = 0.42.

    • Percentage = 0.42 × 100 = 42%.

Problem 4: Wing Coloration in Scarlet Tiger Moth

  • Given Data: Sample size = 1612 individuals. Genotype frequencies:

    • White-spotted (AA) = 0.

    • Intermediate (Aa) = 0.

    • Little spotting (aa) = 5.

  • Frequencies Calculation:

    • Total = AA + Aa + aa = 1612.

      • AA = 0, Aa = 0, aa = 5 → frequencies will vary based on counts.

Problem 5: Building a New Isolated Population

  • Given Data: Two friends are heterozygous (Cc) for cystic fibrosis (c).

A) Instances of Cystic Fibrosis on the Island

  • Total population = 20 (you + 19 friends).

  • Frequency of individuals with cystic fibrosis (cc):

    • Frequencies remain the same under Hardy-Weinberg equilibrium. Calculate proportion as before based on past frequency.

B) Cystic fibrosis births on the island

  • Birth frequency on mainland = 0.059%.

  • Island instances would be calculated from population size and mainland frequency.