Helen Stuart, Cardiff

1. How can we use Hardy-Weinberg equilibrium to calculate carrier frequencies of recessive diseases if we know the proportion of people who are affected? Helen Stuart, Cardiff

2. Essay plan • Description of Hardy-Weinberg principle (HWP) • Allele frequencies: p+q=1 • Genotype frequencies: 2pq + q2 + p2 • Assumptions of HWP including single locus with 2 alleles, random mating, no selection, limitless population • Know proportion of people affected = q2 • Deduce q value • Use HWP (q+p=1) to calculate p value • Use HWP to calculate carrier frequency = 2pq • Give autosomal (Tay Sachs or CF) and X-linked (red-green colour blindness) HW calculation examples

3. Hardy-Weinberg equilibrium • Hardy-Weinberg equilibrium states that both allele and genotype frequencies in a population remain constant (i.e. they are in equilibrium) from generation to generation unless specific disturbing influences are introduced. wikipedia

4. Simplest scenario • In the simplest case of a single locus with two alleles: the dominant allele is denoted A and the recessive a and their frequencies are denoted by p and q • freq(A) = p and freq(a) = q • So, p + q = 1 • If the population is in equilibrium, then we will have: • freq(AA) = p2 for the AA homozygotes in the population • freq(aa) =  q2 for the aa homozygotes • and freq(Aa) = 2pq for the heterozygotes.

5. How to calculate carrier frequency of a recessive disorder…. • If you know the proportion of people who are affected with an autosomal recessive condition, you know q2 • Remember: q2 is the genotype frequency of aa homozygotes (i.e. affected individuals) • Use this value to calculate q, i.e. square root of q2 • Use HWP to calculate p (p+q=1) • Use HWP to calculate carrier frequency = 2pq

6. Example 1: Autosomal recessive use of Hardy-Weinberg principle Tay Sachs disease affects 1 in 3,900 Ashkenazi Jews. What is the expected frequency of carriers? q2 = 1/3,900 q = square root of 1/3,900 = 1/62 p+q=1 so: p+1/62=1 so: p=61/62 Carrier frequency (2pq) = 2 x 61/62 x 1/62 = approx 1/31 in Ashkenazi Jewish population

7. Example 2: X-linked use of Hardy-Weinberg principle Red-green colour blindness has an incidence among males of 1/12 (8%). What is the expected frequency of female carriers? For X-linked disorders, the affected genotype frequency is q (not q2 as for recessive conditions). So, q= 1/12 p+q=1 so: p+1/12=1 so: p=11/12 Female carrier frequency (2pq) = 2 x 11/12 x 1/12 = approx 22/144 (approx 1/8 or 15%)

8. Example 3: autosomal recessive familial case A pregnant African-American couple presents concerned about their risk of having a child with sickle cell anemia (SSA). The husband has a sister with SSA, and the wife has no family history for SSA. Neither have been tested for SS-trait (SST). The disease occurs in 1 in 576 live born African-American children. Calculate this couple’s risk of delivering a baby with SSA. Hint: The square root of 576 = 24

9. Husband’s risk of being a carrier is based on family history = 2/3 Example 3: autosomal recessive familial case

10. Example 3: autosomal recessive familial case • Calculate wife’s SST carrier risk using Hardy-Weinberg equilibrium calculation: • q2 = 1/576 • So, q = 1/24 • p+q=1 so: p+1/24=1 so: p=23/24 • Carrier frequency (2pq) = 2 x 1/24 x 23/24 = approx 1/12 • Couple’s risk of SSA = 2/3 x 1/12 /1/4 = 2/144 = 1/72 http://www.medschool.lsuhsc.edu/ob_gyn/study_guide/D%20Basic%20genetics%20review.doc

11. References • Strachan and Read chapter 3 • http://www.medschool.lsuhsc.edu/ob_gyn/study_guide/D%20Basic%20genetics%20review.doc