Hardey Weinberg-Taste Frequency problem (please help walk me through)

Question

The ability to taste PTC is due to a single dominate allele "T". You sampled 245 individuals in a biology class and determined that 150 could detect the bitter taste of PTC and 95 could not.

What is the predicted frequency of the recessive allele (t)?

What is the predicted frequency of dominant allele (T)?

In a population of 10,000 people, how many would be heterozygous (assuming Hardy-Weinberg equilibrium)? Homozygous dominant? Homozygous recessive? Calculate all of the potential frequencies.

Answer 1

Let's break down the problem step by step.

*Given data:*

- Total sampled individuals: 245

- Individuals who can taste PTC (genotypes TT or Tt): 150

- Individuals who cannot taste PTC (genotype tt): 95

*Assumptions:*

- Single locus with two alleles: T (dominant) and t (recessive)

- Hardy-Weinberg equilibrium (HWE) applies

*Allele frequencies:*

Let p = frequency of dominant allele T

Let q = frequency of recessive allele t

We know that p + q = 1 (since these are the only two alleles)

*Genotype frequencies:*

- TT (homozygous dominant)

- Tt (heterozygous)

- tt (homozygous recessive)

*Calculating allele frequencies:*

Since 95 individuals cannot taste PTC, they must be tt (homozygous recessive).

q^2 = 95/245 ≈ 0.387 (frequency of tt genotype)

q = √(0.387) ≈ 0.622 (frequency of recessive allele t)

p = 1 - q ≈ 1 - 0.622 ≈ 0.378 (frequency of dominant allele T)

*Predicted frequencies:*

- Recessive allele (t): q ≈ 0.622

- Dominant allele (T): p ≈ 0.378

*Genotype frequencies in the population:*

Under HWE, genotype frequencies can be calculated using the allele frequencies:

- TT (homozygous dominant): p^2 ≈ 0.378^2 ≈ 0.143

- Tt (heterozygous): 2pq ≈ 2(0.378)(0.622) ≈ 0.469

- tt (homozygous recessive): q^2 ≈ 0.622^2 ≈ 0.387

*Number of individuals in a population of 10,000:*

- Heterozygous (Tt): 0.469 × 10,000 ≈ 4,690

- Homozygous dominant (TT): 0.143 × 10,000 ≈ 1,430

- Homozygous recessive (tt): 0.387 × 10,000 ≈ 3,870

*Additional frequencies:*

- Carriers of the recessive allele (Tt and tt): q(p + q) = 0.622(0.378 + 0.622) ≈ 0.856 or 8,560/10,000

- Total individuals who can taste PTC (TT and Tt): p(p + q) = 0.378(0.378 + 0.622) ≈ 0.613 or 6,130/10,000

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