Exercise: Effect of the IL6R gene on IL-6R concentration

1. Exercise: Effect of the IL6R gene on IL-6R concentration

2. Excercise: Effect of the IL6R gene on IL-6R concentration • Interleukin-6 receptor (IL-6R) is involved in the immune system • We measured IL-6R concentration in blood in ~5000 individuals • IL-6R concentration in blood is a quantitativetrait IL-6R concentration (pg/mL)

3. Genetics  IL-6R concentration  common disease • IL-6R protein is encodedby the IL6R gene (chromosome 1) • IL6R gene important forseveral common diseases • A SNP (single nucleotide polymorphism) in the IL6R gene (rs2228145) is associatedwith: • Asthma1 • Coronaryheart disease2 • Type 1 diabetes3 1 Ferreira M.A. et al Lancet 2011; 2 IL6R consortium Lancet 2012; 3Ferreira R.C. et al PLoS Genetics 2013

4. Genetics  IL-6R concentration  common disease • Functional studies showedthat the SNP affects IL-6R production • Rs2228145: effect? Mean=4.17 Variance=1.35 Effect of SNPs=? IL-6R concentration (pg/mL) in 5000 people

5. Aim:to estimate how much variance in IL-6R levels is explained by the known SNP and how much by other genetic variants? Methods • We measured IL-6R concentration in ~5000 individuals (Twins & parents & siblings) • We estimated theHeritability: Variance of sIL-6R level explainedbytotalgeneticeffects • We measuredgenome-wide SNP genotypes of the same subjects: • How muchvariance is explainedbyallSNPs in the genome(Genomewide-complex trait analysis, GCTA) • How muchvariance is explainedbyallgeneticvariation in the IL6R gene (linkage analysis) • How muchvariance is explainedbythe SNP rs2228145 (excercise)

6. Heritability of sIL-6R level H2 = Broad-senseheritability a2 = Additivegeneticeffects d2 = Non-additive geneticeffects c2 = Sibling-shared environment e2 = Unique environment H2 a2 d2 c2 e2

7. Variance explained by chromosome-wide SNPs SNPs in the IL6R gene on Chromosome 1 (+/- 10MB): 54.7 % (SE=2.5%)

8. Combined linkage and association analysis Chi-squared from linkage test Chi-squared from linkage test – while modeling association for individual SNPs Chr1: (q21.3) IL6R region: • Variance explained by linkage (VA/Vtotal): 69 % • Variance explained by linkage after correction for rs2228145: 19%

9. Exercise: Effect of the IL6R gene on IL-6R concentration INFORMATION • The SNP (single nucleotidepolymorphism) has 2 alleles: • Minor allele: C, frequency: p=0.39 • Major Allele: A, frequency: q =0.61 • Mean IL-6R concentration of each genotype: • CC: 5.698 (10-8 g/mL) • CA: 4.418 (10-8 g/mL) • AA: 3.238 (10-8 g/mL) • Variance of IL-6R concentration=1.35 QUESTIONS Calculate genotypic values (a and d) (Falconer page 109) Calculate the genotype frequencies (Falconer page 7) Calculate the mean IL6-R concentration in the population (Falconer page 110) Calculate how much of the variance is explained by this SNP (Variance= Sum of squareddeviationsfrom the mean)

10. m=0 -a a Model: gene with 2 alleles A and a and 3 genotypes AA, Aa and aa aa AA Aa The difference on a quantitative scale between AA and aa is 2a. The middle (m) is zero and the value of Aa is 0 (no dominance).

11. d m -a a Model: gene with 2 alleles A and a and 3 genotypes AA, Aa and aa aa Aa AA The deviation from m (middle) of the heterozygote Aa is d: partial dominance.

12. Mean?

13. (recall p+q = 1) mean: p2 a + 2pqd - q2 a = a(p2 – q2) + 2pqd = a(p-q)(p+q) + 2pqd = Mean = a(p-q) + 2pqd a(p-q) : attributable to homozygotes 2pqd : attributable to heterozygotes

14. Exercise: Effect of the IL6R gene on IL-6R concentration INFORMATION • The SNP (single nucleotidepolymorphism) has 2 alleles: • Minor allele: C, frequency: p=0.39 • Major Allele: A, frequency: q =0.61 • Mean IL-6R concentration of each genotype: • CC: 5.698 (10-8 g/mL) • CA: 4.418 (10-8 g/mL) • AA: 3.238 (10-8 g/mL) • Variance of IL-6R concentration=1.35 QUESTIONS Calculate genotypic values (a and d) (page 109) Calculate the genotype frequencies (page 7) Calculate the mean IL6-R concentration in the population (page 110) Calculate how much of the variance is explained by this SNP (Variance= Sum of squareddeviationsfrom the mean)

15. AA AC CC 3.238+1.23= 4.47 QUESTIONS • Calculate genotypic values (a and d) (Falconer, page 109) 3.238 4.418 5.698 AA AC CC -a d 0 a 5.698-3.238 = 2.46 a= 2.46/2=1.23 d= 4.418-4.47= -0.05

16. CC CA AA • Genotype frequencies = p2 + 2pq + q2 = 0.392 + 2*0.39*0.61 + 0.61 2 = 0.15 + 0.48 + 0.37 mean IL6R 5.6984.4183.238 2. Population mean p2 *5.698 + 2pq * 4.418 + q2 * 3.238 0.15 *5.698 + 0.48 * 4.418+ 0.37 * 3.238= 4.17 3. Deviation from mean = 5.698-4.174.418 - 4.17 3.238 - 4.17 = 1.52 0.24 -0.94 3. Squared deviation = 1.522 =2.32 0.242 =0.06 -0.942 =0.88 3. Variance (SNP) = p2 * 2.32 + 2pq * 0.06 + q2 * 0.88 = 0.15* 2.32 + 0.48* 0.06 + 0.37*0.88 = 0.71 (SD = 0.84) • Calculate the genotype frequencies • Calculate the mean IL6-R concentration in the population (page 110) • Calculate how much of the variance is explained by this SNP Minor allele: C, frequency: p=0.39 Major Allele: A, frequency: q =0.61 Total variance of IL-6R concentration= 1.35 % variance explained by SNP= 0.71/1.35 = 53%

17. mean: p2 a + 2pqd - q2 a = a(p-q) + 2pqd Variation: 2pq[a+d(q-p)]2 + (2pqd)2 Population variation depends on ‘a’ (difference between homozygote individuals), ‘d’ (deviation of heterozygote persons from zero) and on allele frequency (p & q).

21. Observed and predicted height and BMI in Europe. (using within family genetic variance)