Final answer:
In a population of 5150 tadpoles, 5 lack lower limbs due to a homozygous recessive condition. Using the Hardy-Weinberg formulas, the genotypic and allele frequency values can be calculated. The calculated values are p² = 0.9710, 2pq = 0.0285, q² = 0.000210, p = 0.9855, and q = 0.0145.
Step-by-step explanation:
A Hardy-Weinberg equilibrium population refers to a population in which the allele frequencies remain constant over generations. In order to remain in equilibrium, the population must meet certain conditions, including random mating, no mutations, no genetic drift, no gene flow, and no natural selection.
In this scenario, we want to find the genotypic and allele frequency values using the Hardy-Weinberg formulas.
Given that 5 out of 5150 tadpoles lack lower limbs due to a homozygous recessive condition, the frequency of the recessive allele (q) can be calculated as the square root of the proportion of the population affected by the trait (5/5150). The value of q would be approximately 0.0145. Knowing that p + q = 1, we can find the value of p as 1 - 0.0145 = 0.9855.
Now we can calculate:
- p² = (0.9855)²
- 2pq = 2 * (0.9855) * (0.0145)
- q² = (0.0145)²
These calculations will provide the genotypic and allele frequency values.
- p² = 0.9710 (genotypic frequency of homozygous dominant individuals)
- 2pq = 0.0285 (genotypic frequency of heterozygous individuals)
- q² = 0.000210 (genotypic frequency of homozygous recessive individuals)
- p = 0.9855 (allele frequency of the dominant allele)
- q = 0.0145 (allele frequency of the recessive allele)