In a laboratory population of diploid, sexually reproducing organisms a certain trait is determined by a single autosomal gene and is expressed as two phenotypes. A new population was created by crossing 102 pure–breeding (homozygous) dominant individuals with 98 pure breeding (homozygous) recessive individuals. After four more generations, the following results were obtained.
Evaluate if the population is in Hardy–Weinberg equilibrium and calculate the allele frequencies (p and q) for each generation.
Where are the results after four generations? Repost later with the information and maybe someone will help you solve this problem.
To determine if a population is in Hardy-Weinberg equilibrium, we need to compare the observed genotype frequencies in the population to the expected genotype frequencies under the assumptions of the Hardy-Weinberg principle.
The Hardy-Weinberg equilibrium states that in a large, randomly mating population, the frequencies of alleles and genotypes will remain constant from one generation to the next if certain conditions are met:
1. No mutation occurs
2. No migration (gene flow) occurs
3. The population is large and randomly mating
4. Natural selection is not acting on the trait
5. Random mating is occurring with respect to the trait (no mate preference)
Let's start by calculating the allele frequencies (p and q) for the initial generation.
Given information:
- Pure-breeding (homozygous) dominant individuals = 102
- Pure-breeding (homozygous) recessive individuals = 98
Let's assume the dominant allele is represented by "A" and the recessive allele is represented by "a".
To calculate the allele frequencies:
- Total number of individuals in the population = 102 + 98 = 200
- The total number of dominant alleles (A) = 2 * number of dominant individuals = 2 * 102 = 204
- The total number of recessive alleles (a) = 2 * number of recessive individuals = 2 * 98 = 196
The allele frequency of A (p) = (number of A alleles) / (total number of alleles) = 204 / 400 = 0.51
The allele frequency of a (q) = (number of a alleles) / (total number of alleles) = 196 / 400 = 0.49
Now, let's move on to calculate the genotype frequencies in the subsequent generations and compare them to the expected frequencies under Hardy-Weinberg equilibrium.
After each generation, we can calculate the expected genotype frequencies using the Hardy-Weinberg equation:
p^2 + 2pq + q^2 = 1
where p^2 represents the frequency of the homozygous dominant genotype (AA), 2pq represents the frequency of the heterozygous genotype (Aa), and q^2 represents the frequency of the homozygous recessive genotype (aa).
For generation 1, we can calculate the expected genotype frequencies using the allele frequencies calculated earlier:
Expected frequency of AA = p^2 = (0.51)^2 = 0.2601
Expected frequency of Aa = 2pq = 2 * 0.51 * 0.49 = 0.4998
Expected frequency of aa = q^2 = (0.49)^2 = 0.2401
Now, compare the expected genotype frequencies with the observed genotype frequencies in generation 1. If they are similar, the population is in Hardy-Weinberg equilibrium. If not, the population is not in equilibrium.
Repeat this process for subsequent generations to evaluate if the population remains in Hardy-Weinberg equilibrium.