Hardy-Weinberg Practice Problems Study Notes

Hardy-Weinberg Practice Problems

Problem 1: Sickle-cell Anemia and Malaria Resistance

• Given: 9% of an African population has sickle-cell anemia (genotype ss).
• Question: What percentage of the population will be more resistant to malaria because they are heterozygous (genotype Ss)?
• Key Concept: Heterozygous individuals (Ss) are favored in malaria-endemic areas due to the selective advantage they hold, possessing some protection against malaria while being less susceptible to sickle-cell anemia.

Problem 2: Hardy-Weinberg Equilibrium and Homozygous Recessive Genotype

• Given: 81% of individuals are homozygous recessive for a gene.
• Task: Determine the expected percentage of heterozygous (genotype Aa) individuals in the next generation.
• Use the Hardy-Weinberg principle: If the proportion of homozygous recessive individuals (aa) is denoted as $q^2 = 0.81$, then:
  • $q = ext{sqrt}(0.81) = 0.9$ (frequency of allele a)
  • Since $p + q = 1$, then $p = 1 - 0.9 = 0.1$ (frequency of allele A)
  • The frequency of heterozygous genotype (Aa) is given by $2pq = 2(0.1)(0.9) = 0.18$ or 18%.

Problem 3: Genotype Frequencies Calculation from Homozygous Recessive Genotype

• Given: Population has 36% homozygous recessive genotype (aa).
• Calculate:
  • A. Frequency of the aa genotype:
  • Given as 36% or 0.36.
  • B. Frequency of the a allele:
  • $q^2 = 0.36
    ightarrow q = ext{sqrt}(0.36) = 0.6$.
  • C. Frequency of the A allele:
  • $p = 1 - q = 1 - 0.6 = 0.4$.
  • D. Frequencies of the genotypes AA and Aa:
  • Frequency of AA = $p^2 = (0.4)^2 = 0.16$.
  • Frequency of Aa = $2pq = 2(0.4)(0.6) = 0.48$.
  • E. Frequency of two phenotypes (A is dominant):
  • Phenotype A (dominant): $p^2 + 2pq = 0.16 + 0.48 = 0.64$ (64% dominant phenotype).
  • Phenotype a (recessive): $q^2 = 0.36$ (36% recessive phenotype).

Problem 4: Classroom Grade Frequencies and Genetics

• Given: 100 students in class; 96 did well (dominant phenotype) and 4 received an F (homozygous recessive condition).
• Assume 4% represents the frequency of the homozygous recessive genotype (ff).
• Calculate:
  • A. Frequency of recessive allele:
  • $q^2 = 0.04$ therefore $q = ext{sqrt}(0.04) = 0.2$.
  • B. Frequency of dominant allele:
  • $p = 1 - q = 1 - 0.2 = 0.8$.
  • C. Frequency of heterozygous individuals:
  • Frequency of heterozygous (Ff) = $2pq = 2(0.8)(0.2) = 0.32$ or 32%.

Problem 5: Butterfly Color Allele Frequencies

• Given: Color brown (B) is dominant over white (b); 40% of butterflies are white (bb).
• Calculate:
  • A. Percentage of butterflies that are heterozygous (Bb):
  • $q^2 = 0.4
    ightarrow q = ext{sqrt}(0.4) = 0.632$. Thus, frequency of allele b is 0.632.
  • Therefore, frequency of allele B is $p = 1 - q = 1 - 0.632 = 0.368$.
  • Heterozygous (Bb) frequency: $2pq = 2(0.368)(0.632) = 0.4656$. 46.56% are heterozygous.
  • B. Frequency of homozygous dominant individuals (BB):
  • $p^2 = (0.368)^2 = 0.135424$, or approximately 13.54%.

Problem 6: Beetle Population Genotypes and Phenotypes

• Given: 396 red-sided and 557 tan-sided beetles.
• Red is recessive: denote red as rr and tan as Rr or RR.
• Calculate:
  • A. Allele frequencies of each allele (R and r):
  • Total beetle count = 396 + 557 = 953.
  • Frequency of rr: $q^2 = rac{396}{953}
    ightarrow q = ext{sqrt}(rac{396}{953})$.
  • Using total alleles, determine p and q respectively.
  • B. Expected genotype frequencies:
  • Using derived frequencies of R and r, calculate expected genotype frequencies: $p^2$, $2pq$, and $q^2$.
  • C. Number of expected heterozygous (Rr) individuals:
  • Use previously calculated $2pq$.
  • D. Expected phenotype frequencies based on calculations above.
  • E. Population increase: If count goes to 1,245, calculate expected red-sided and tan-sided beetles using Hardy-Weinberg principles.

Problem 7: Laboratory Mice Color Frequencies

• Given: 35% are white (aa), thus homozygous recessive prevalence is marked.
• Calculate allelic and genotypic frequencies for this population:
  • $q^2 = 0.35
    ightarrow q = ext{sqrt}(0.35)$ generates the allele frequencies and then isolates alleles A and a respectively.

Problem 8: Cystic Fibrosis on Isolated Island

• Adventure scenario with 20 individuals, 2 heterozygous for cystic fibrosis (allele c).
• Calculate the incidence of cystic fibrosis on the island:
  • Assuming allele frequency remains steady, apply Hardy-Weinberg to predict incidence rate of condition derived from initial frequencies established.

Problem 9: MN Blood Group Allele Frequencies

• Given: Sample of 1,000 individuals for MN blood group; MM = 0.49.
• Calculate:
  • A. Allele frequencies of M and N:
  • Determine $p$ and $q$ from genotype proportions.
  • B. Predicted heterozygous individuals:
  • Use established frequencies to calculate expected heterozygous representation based on relevant frequencies.

Additional Problems from Old Textbook

Problem 2: Frequency of Heterozygotes at Hardy-Weinberg Equilibrium

• Given: Two alleles B and b; frequency of B is 0.7.
• Determine frequency of heterozygotes in equilibrium:
  • $p = 0.7$, thus $q = 1 - 0.7 = 0.3$; Calculate $2pq$.

Problem 3: Recessive Trait Frequency Calculation

• Given: 16% show recessive trait.
• Determine dominant allele frequency:
  • $q^2 = 0.16
    ightarrow q = 0.4$, hence dominant frequency is calculated as 1 - q.

Problem 4: Frequency of Heterozygotes with Given Allele Frequencies

• Given: Allele frequency of b = 0.4; use Hardy-Weinberg to infer.

Problem 5: Dominance Estimated Percent Representation

• Given: 16% showing recessive trait, calculate the dominant trait representation in the population based on derived allele frequencies.

Problem 11: Field Mice Coat Color Genetic Study

• Given: 48% indicate heterozygosity; determine dominant allele frequency.

Problem 12: Shorthorn Cattle Allele Frequencies in Random Sample

• Given: Color frequencies and proportions of CR and Cr genotypes; use established formulas for estimation and determine expressed equilibrium based on calculated values.