GWAS and Complex Traits
Complex Traits and GWAS
Complex traits are influenced by multiple genetic and environmental factors, unlike Mendelian traits. Genome-wide association studies (GWAS) are used to identify genetic variants associated with these complex traits.
GWAS Applications
GWAS can be applied to study any phenotype. For example, a genetic testing company provided results that showed genetic variants are linked to cilantro preference.
Association Mapping of QTLs
Quantitative trait loci (QTLs) are genes contributing to complex traits.
Association mapping is useful when controlled crosses are not possible (e.g., in human studies).
It relies on recombination events that have occurred over generations.
Association Mapping in Case vs. Control Studies
Recombination shuffles variants linked on the same chromosome over many generations.
Variants that are physically closer are less likely to undergo recombination between them.
Haplotypes and Linkage
Haplotype: A combination of alleles at multiple loci that are transmitted together.
Example: Consider two sites:
Site 1: A or T
Site 2: G or C
Possible haplotypes: AG, AC, TG, TC
Linkage Equilibrium vs. Linkage Disequilibrium
Linkage Equilibrium: Haplotypes are present at equal frequencies, and knowing the sequence at one site provides no information about the sequence at another site.
Linkage Disequilibrium (LD): Variants at two loci are correlated, and the sequence at one site can predict the sequence at another site.
Linkage Disequilibrium Representation
LD is represented as a triangle diagram.
Block color indicates the strength of the statistical correlation for pairwise comparisons of SNPs.
Darker color indicates stronger correlation.
Utility of Linkage Disequilibrium
LD makes it more efficient to identify SNPs associated with complex traits using SNP microarrays.
Only tag SNPs need to be surveyed on a microarray to infer the full haplotype.
GWAS to Identify SNPs
GWAS is used to identify SNPs associated with complex traits such as:
Type 2 diabetes
Autism spectrum disorder
Case-Control Comparison in GWAS
In a GWAS study:
Cases: Individuals with the disease.
Controls: Individuals without the disease.
SNPs are analyzed for each group.
Example data:
SNP1: Cases have a higher frequency of C allele (55.4%) compared to controls (47.4%), p-value = .
SNP2: Cases and controls have similar frequencies of C allele (42.8% vs 43.2%), p-value = 0.80.
P-value in GWAS
The null hypothesis is that there is no association between a SNP and the disease.
The p-value is the probability of observing the extreme bias by random chance if the null hypothesis is true.
Typically, the null hypothesis is rejected when p < 0.05.
A chi-squared test is commonly used to test this association.
Chi-Squared Test for Independence
The chi-squared test assesses the independence of SNP alleles and case/control status.
Formulae:
Observed (O) values: OCaG, OCoG, OCaC, OCoC (Cases with G, Controls with G, Cases with C, Controls with C).
Expected (E) values: ECaG, ECoG, ECaC, ECoC, calculated based on total allele frequencies.
values:
Total =
Degrees of freedom (df) = 1.
Degrees of Freedom
With the row and column totals given, calculating one cell value allows you to determine all other values; hence, df = 1.
P-value Determination
The p-value can be obtained from a value and degrees of freedom using:
Statistical tables
Software like Excel or Google Sheets using the
CHIDISTfunction:CHIDIST( value, df)
Example Chi-Squared Test
Given data for SNP1:
Cases:
O(G) = 1716, E(G) = 1902, = 18.2
O(C) = 2132, E(C) = 1946, = 17.8
Controls:
O(G) = 3089, E(G) = 2903, = 11.9
O(C) = 2783, E(C) = 2969, = 11.7
Total = 59.6, df = 1, p =
Manhattan Plot
Manhattan plots are used to visualize the results of GWAS, showing the association of each SNP with the phenotype.
Multiple Testing Correction
In GWAS, many SNPs are tested, necessitating correction for multiple testing.
To maintain overall significance, the p-value threshold is adjusted:
Calculating Allelic Odds Ratio
Odds of C allele given Case status =
Odds of C allele given Control status =
Odds Ratio = \frac{2132/1716}{2783/3089} = 1.38$$
Interpretation: The case group has 1.38 times the odds of having the disease/phenotype compared to the control group.