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 you must always start your calculations by determining the proportion of individuals that display the recessive phenotype; this is the only phenotype from which you can immediately work out its genotype as it will always be homozygous recessive (the dominant phenotype is seen in both homozygous dominant and heterozygous individuals).
In Hardy-Weinberg questions it is a good idea to begin by establishing what information you have been given in the question (i.e. do you know q2, or do you know p?), and then establishing what the question wants you to work out (i.e. are you calculating 2pq?). You can then work out how to get from one to the other.
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 that there is no change in this between generations.
转载自savemyexams
以上就是关于【CIE A Level Biology复习笔记17.2.5 Natural Selection: Hardy-Weinberg Principle】的解答,如需了解学校/赛事/课程动态,可至翰林教育官网获取更多信息。
往期文章阅读推荐:
全网破防!ALevel CIE数学M1疑似错题?经济P2难度飙升?5月6日大考考情分析必看!
A-Level CIE就大规模泄题发布最严处罚!哪些考生必须重考?你的成绩怎么办?

© 2026. All Rights Reserved. 沪ICP备2023009024号-1