RL Lectures 47-48: Genetics of Complex Diseases

Question 1

What is the heritability component and how can it be determined?

Answer 1

the extent to which the expression of a common disease is dependent on genetic variation
determined by family studies

Question 2

What is the equation for calculating heritability?

Answer 2

h^2 = 2 ( r(mz) - r(dz) )
where r(mz)  is the correlation for monozygotic twins and r(dz) is the correlation for dizygotic twins

Question 3

What is the range of heritability as per the equation?

Answer 3

0 (no heritability component)

1 (completely heritable with no measurable environmental impact)

Question 4

What study methods are available to identify variants that are associated with increased risk for disease?

Answer 4

case-control association study

GWAS

Question 5

Describe the steps of a case-control association study.

Answer 5

1- obtain genotypes at genomic positions of interest in a group of individuals with the disease and in a matched cohort of unaffected
2- count alleles in each group to determine differences in frequency of specific allele between the two groups
3- use the odds ratio to calculate the effect size (impact)

Question 6

What is required to perform a case-control association study?

Answer 6

a candidate gene that you believe to be associated with a phenotype

Question 7

How do you calculate the odds ratio (OR)?

Answer 7

= (Allele1(ds) / Allele2(ds) ) / (Allele1(c) / Allele2(c) )

Question 8

How do you interpret the odds ratio?

Answer 8

OR <1 protective mutation
OR = 1 neutral (no relationship)
OR > 1 predisposition mutation

Question 9

List some of the limitations of GWAS.

Answer 9

requires large cohort size
missing heritability
misinterpretation of results

Question 10

What is a haplotype?

Answer 10

a series of alleles along the same segment of a chromosome strand can be linked (alleles that are closer together tend to co-segregate and are inherited together) or unlinked (alleles further apart with higher liklihood they will be separated during recombination => lower liklihood they will be inherited together)

Question 11

What is lineage equilibrium

Answer 11

haplotype frequencies in a population are different from allele (SNP) frequencies due to the loci that are not linked

Question 12

What is linkage disequilibrium?

Answer 12

haplotype frequencies in a population are same or close to allele (SNP) frequencies due to the loci being linked

Question 13

Describe a LD score.

Answer 13

reflects how linked or not linked the loci are in a haplotype
if score is 0% it is not linked (haplotypes at equilibrium)
if score is at 100% they are linked (haplotypes are at complete disqeuilibrium)