2023.02.15 | Questions 29-31
Biostats (3/3) - Hardy-Weinberg
Hello,
This is part 3 of 3 of the biostats series. This set of questions focuses on calculations that involve Hardy-Weinberg equilibrium. Find a pen and paper and give these questions a try!
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-Daniel
Question 29
An autosomal recessive disorder has a population frequency of 1:90,000. What is the carrier frequency for this disorder in the general population? Assume the population is in Hardy-Weinberg equilbrium (HWE).
Hint: the question asks about carrier frequency, not allele frequency. Use the Hardy-Weinberg equation: p² + 2pq + q²
Question 30
Red–green color blindness (RGCB) is an X-linked recessive disorder due to mutations in genes that encode opsins. RGCB affects 1 in 12 males. Assuming Hardy-Weinberg equilibrium, what is the carrier frequency for RGCB?
Question 31
An autosomal recessive disease has a carrier frequency of 1/50 in a specific population. Assume the population is in Hardy-Weinberg equilibrium. Which of the following is the expected frequency of individuals with the disease in this population?
Explanations
Question 29 (Ans: 1/150)
Questions on Hardy-Weinberg equilibrium (HWE) are common on board exams. Questions may involve calculating allele or genotype frequencies directly, or they may ask about the assumptions of HWE, which include no mutation, random mating, no gene flow, infinite population size, and no selection. The Hardy-Weinberg equation is p2 + 2pq + q2 = 1. Also, p + q = 1. In this case, assume that there are only two alleles in the population — a major allele (p) and a minor (mutant) allele (q), — though keep in mind that HWE calculations can be performed for multiple alleles at a single locus. When the minor allele frequency (q) is small, the major allele frequency (p) can be assumed to be close to 1. For autosomal recessive disorders, the variables in these equations can be interpreted as follows. Feel free to refer back to this guide for each of the questions that follow.
p = frequency of the more common (major) allele
q = frequency of the minor (disease-associated) allele
p² = frequency of homozygotes ("unaffected, not a carrier")
2 x p x q = frequency of heterozygotes ("unaffected carrier", aka carrier frequency)
q² = frequency of homozygotes ("affected", aka disease frequency)What is the difference between carrier and allele frequencies? For an autosomal recessive disorder, the carrier frequency (2 x p x q) is typically twice the minor allele frequency q (assuming the minor allele frequency is small, and p is close to 1).
For Question 29, we can calculate the carrier frequency as follows. My approach is to first write down the values that are given to us, then to calculate the values that are needed to answer the question.
(Nomenclature: “sqrt” = square root)
# Given to us:
q² = 1/90,000 (disease frequency)
# Derived from q²
q = sqrt(1/90,000) = 1/300 (minor allele frequency)
p = 1 - q = 299/300 ≈ 1 (major allele frequency)
2 x p x q = 2 x 1 x (1/300) = 1/150 (heterozygote frequency, i.e. carrier frequency)Thus, the answer to Question 29 is 1/150. Note that if the question had asked about minor allele frequency, then the answer would have been 1/300.
Question 30 (Ans: 11/72)
This question asks us to calculate the carrier frequency for red–green color blindness (RGCB), an X-linked recessive disorder. For HWE calculations involving X-linked disorders, it is helpful to think in terms of allele frequencies.
Note that only women can be carriers in X-linked recessive disorders (as they have two X chromosomes). Males have only one X chromosome and cannot be carriers for an X-linked disorder — they are either affected or unaffected.
The frequencies of the major and minor (i.e. mutant) alleles can be determined from the incidence of the RGCB phenotype in males (not females). Because males only have one X chromosome, the frequency of affected males is equal to the minor (mutant) allele frequency in the entire population (under HWE), which includes both males and females.
# Given to us:
q = 1/12 (incidence of RGCB in males; also the minor allele frequency in the total population. Think of this as "One in twelve X chromosomes in the total gene pool has a mutation.")
# Derived from q
p + q = 1
p = 1 - q = 1 - 1/12 = 11/12 (major allele frequency)
# Carrier frequency
2 x p x q = 2 x (11/12) x (1/12) = 22/144 = 11/72 (answer!)Now, suppose we had been asked: “What proportion of women would be expected to be affected with RGCB?” This value represented by q² and can be calculated as:
q² = 1/12 x 1/12 = 1/144 (incidence in females)Question 31 (Ans: 1/10,000)
This question asks about the expected frequency of individuals with an autosomal recessive disease in the population (q²). We are given the carrier frequency (1/50), which is represented by 2 x p x q. We also assume that p (the major allele frequency) is ~1. We use the same framework as above to solve this question.
# Given to us:
2 x p x q = 1/50 (carrier frequency)
p ≈ 1 (major allele frequency)
# solve for q
2 x 1 x q = 1/50
q = (1/50) / (2 x 1) #rearranged version of the above
q = 1/100
q² = 1/100 x 1/100 = 1/10,000 The answer is that 1/10,000 individuals would be expected to have the disease in this population.
Additional Resources
Video for X-linked HWE calculations (YouTube)
Autosomal dominant HWE question (Question 5)
Learning objective
Calculations that assume Hardy-Weinberg equilibrium (HWE) can be used to estimate the expected proportion of homozygotes and heterozygotes based on the major and minor allele frequencies in a given population. Calculations of major and minor allele frequencies as well as the number of affected, carrier, and unaffected individuals can be derived from the HWE equations. Equations are valid for autosomal recessive, X-linked, and autosomal dominant disorders.
2023 ABMGG General Exam Blueprint | III. Population Genetics → c. Hardy-Weinberg equilibrium (page 2)