p + q = 1
p2 + q2 + 2pq = 1
In a population of birds, 10% of the individuals exhibit the recessive phenotype of white feathers. Calculate the frequencies of all genotypes.
Solution:
To calculate the frequencies of the homozygous dominant ( p2 ) and heterozygous ( 2pq ):
Step 1: Find q
Step 2: Find p (the frequency of the dominant allele F). If q = 0.32, and p + q = 1
p + q = 1
p = 1 - 0.32
p = 0.68
Step 3: Find p2 (the frequency of homozygous dominant genotype)
0.682 = 0.46
p2 = 0.46
Step 4: Find 2pq = 2 x (p) x (q)
2 x (0.68) x (0.32)
= 0.44
Step 5: Check calculations by substituting the values for the three frequencies into the equation; they should add up to 1
p2 + 2pq + q2 = 1
0.46 + 0.44 + 0.10 = 1
In summary:
When you are using Hardy-Weinberg equations, start your calculations by determining the proportion of individuals that display the recessive phenotype - you will always know the genotype for this: homozygous recessive. Remember that the dominant phenotype is seen in both homozygous dominant, and heterozygous individuals. Also, don’t mix up the Hardy-Weinberg equations with the Hardy-Weinberg principle. The equations are used to estimate the allele and genotype frequencies in a population. The principle suggests that there is an equilibrium between allele frequencies and there is no change in this between generations.
转载自savemyexams
© 2024. All Rights Reserved. 沪ICP备2023009024号-1