1. AA CC AC
Mean observed IL-6R concentration of each genotype:
CC: / CA: / AA: (10-8 g/mL)
Total Variance of IL-6R concentration = 1.35
Frequencies: C, frequency: p=0.39 / A, frequency: q =0.61 -a AA AC CC a d
mean 2a
QUESTIONS (Falconer & MacKay; 1996: Introduction to quantitative genetics)
Calculate genotypic values (a and d) (page 109)
Calculate the average effect of the alleles (page 113)
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 squared deviations from the mean)
6. Calculate heritability

2. QUESTION Calculate genotypic values (a and d)
midpoint= 4.47 -a AA AC CC
5.698
4.418
3.238 a d
Midpoint = ( )/2 = 4.47

3. a= 0.5*(sIL6RCC - sIL6RAA) = 0.5*(5.698-3.238) = 1.23
QUESTION Calculate genotypic values (a and d)
= 4.47 -a AA AC CC
5.698
4.418
3.238 a d
= 2.46
a= 2.46/2=1.23
d= = -0.05
a= 0.5*(sIL6RCC - sIL6RAA) = 0.5*( ) = 1.23
d= sIL6RAC - (sIL6RAA + a ) = ( ) = -0.05

4. 2. Calculate the average effect of the two alleles
Minor allele: C, frequency: p= 0.39
Major Allele: A, frequency: q=0.61
(Allele frequencies from Sib-pair ( ) using the best linear unbiased estimator (BLUE) option, which accounts for familial relatedness)
a=1.23, d= -0.05
Average effect C= q[a+d(q-p)] = 0.61[ ( )] = 0.74
Average effect A= -p[a+d(q-p)]= [ ( )] = -0.48

5. Calculate the genotype frequencies
Calculate the mean IL6-R concentration in the population
Calculate how much of the total variance is explained by this SNP
Minor allele: C, frequency: p=0.39
Major Allele: A, frequency: q =0.61

6. CC CA AA 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
CC CA AA
Genotype frequencies = p pq q2 
= *0.39* =
mean IL6R
2. Population mean p2 * pq * q2 *
0.15 * * * = 4.17
3. Deviation from mean = =
4. Squared deviation = = = =0.88

7. Calculate the genotype frequencies
Calculate the mean IL6-R concentration in the population
Calculate how much of the total variance is explained by this SNP
CC CA AA
4. Squared deviation = = = =0.88
5. Variance (SNP) = p2 * pq * q2 * 0.88
= 0.15* * *0.88 = 0.71 (SD = 0.84)

8. Alternative way of estimating the mean: Mean = a(p-q) + 2pqd
a = and d=
a(p-q) = 1.23( ) = 1.23 *( -0.22) = -0.27
2pqd = 2*0.39*0.61*-0.05 =
This value of the mean, however, is measured from the mid-homozygote point, which was 4.47 for these data.
Thus: (-0.27 – 0.012) = 4.19

9. d= sIL6RAC - (sIL-6RAA + a ) = -0.05
Alternative way of estimating genetic variance (due to this SNP)
a= 0.5*(sIL6RCC - sIL6RAA) = 1.23
d= sIL6RAC - (sIL-6RAA + a ) = -0.05
VA = 2pq [a + d (q-p)]2 =
2*0.39*0.61[ *( )]2= 0.71
VD = (2pqd)2 =
(2*0.39*0.61*-0.05)2 = 5.66 x 10-4

10. Total variance of IL-6R concentration= 1.35
Calculate how much of the total variance is explained by this SNP
Total variance of IL-6R concentration= 1.35
% variance explained by SNP= 0.71/1.35 = 53%

11. Announcement: Behavior Genetics in Amsterdam
5 to 19 August 2017: Summer school Twin Research and Human Genetics
Starting august 2018: New Research Master Genes, Behaviour and Health
Period 1
Period 2
Period 3
Period 4
Period 5
Period 6
Year 1
Introduction to omics
Behavioural Genetics
Epigenomics and Sequencing in Behaviour and Health
Imaging and Cardiovascular Genetics
Gene finding: GWA studies and beyond
Elective 1
Internship I  (24EC)
Year 2
Exposome and gene-environment interaction
Personalised medicine and health
Grant writing and science communication
Internship II – MA thesis (30 EC)
Complex trait genetics
Elective 2