Competing analytical strategies of combining associated SNPs for estimating genetic risks

Abstract

In genomewide association study (GWAS) of a complex phenotype, a large number of variants, many with small effect sizes, are found to contribute to the variability of the phenotype. Subsequent to the identification of such variants in a GWAS, it is of interest to estimate the risk jointly conferred by the variants. We propose three different strategies of combining the risk SNPs to calculate an allele dosage score. Using simulations, we evaluate the different measures of allele dosage score with respect to the risk prediction accuracy of a binary trait and the proportion of variance explained for a quantitative trait. For a binary trait, an allele dosage score defined based on log odds ratio performs marginally better than the other two measures. For a quantitative trait, the measure based on the standardized slope coefficient in linear regression of the trait on single-nucleotide polymorphism (SNP) genotypes performs better than the measures using the weights proportional to log P-value and the proportion of variance explained. We demonstrate the utility of these measures using a real data on type 2 diabetes and fasting blood sugar level in a south Indian population.

Appendix

Risk prediction for a binary end-point trait

The popular case–control study design usually involves a prefixed number of cases and controls. We demonstrate that such a design is inappropriate for estimating risks conferred by individual SNPs. The prevalence of a disease is a function of the allele frequencies and the penetrances at the different causal loci. Since the prevalence of a disease in a population is usually much lower than $50\%$ (or the proportion of cases in GWAS data), there is a gross over-representation of cases in the usual study design comprising equal number of cases and controls (or with substantially larger case control ratio compared to the population) leading to overestimation of the penetrances. On the other hand, a large sample of individuals chosen at random from a population irrespective of their affection status is expected to carry proper information on the disease prevalence in the population.

For a disease trait, define a variable Y for an individual, such that $Y=1$, if the individual is a case, and $Y=0$ if the individual is a control. We assume that the disease locus is biallelic with alleles D and d having frequencies p and ($1-p$), respectively in the population. Suppose we have genotype data at the disease locus on 2n individuals comprising $n_1$ cases and $n_2$ $(=2n-n_1)$ controls. The penetrances at the disease locus are defined as: $f_2=P($case|DD),  $f_1=P($case|Dd),  $f_0=P($case|dd). The overall disease prevalence is given by: P(case$)=f_2p^2 + 2f_1 p (1-p) + f_0 (1-p)^2$. Using a logistic link function between the disease status and SNP genotype, we can estimate the penetrance parameters from the data.

$P(DD|\text{case})=\frac{f_2p^2}{P(\text{case})},$ $P(Dd|\text{case})=\frac{2 f_1 p(1-p)}{P(\text{case})},$ $P(dd|\text{case})=\frac{f_0(1-p)^2}{P(\text{case})},$ and, $P(DD|\text{control})=\frac{(1-f_2)p^2}{P(\text{control})},$ $P(Dd|\text{control})=\frac{2 (1-f_1) p(1-p)}{P(\text{control})},$ $P(dd|\text{control})=\frac{(1-f_0)(1-p)^2}{P(\text{control})},$ where $P(\text{control}) = 1-P(\text{case})$. Let $n_{case}^{DD}$ and $n_{control}^{DD}$ denote the number of case and control individuals who have the genotype DD, respectively. Similarly, $n_{case}^{Dd},$ $n_{case}^{dd},$ $n_{control}^{Dd},$ $n_{control}^{dd}$ are defined. Hence, $E(n_{case}^{G}) = n_1 \times P(G|\text{case})$ and $E(n_{control}^{G}) = n_2 \times P(G|\text{control})$, $G=DD,Dd,dd.$ Since for a given G, ${n_{case}^{G}}/{n_1}$ is a consistent estimator of $P(G|\text{case})$, we assume that, $n_{case}^{G} \approx n_1 \times P(G|\text{case}),$ for large $n_1$. Using the same argument, for a given G,  we assume that $n_{control}^{G} \approx n_2 \times P(G|\text{control})$, for large $n_2$.

Let X denote the number of D alleles in a genotype and hence, $X=0,1,2$. We model the probability of a case conditioned on the genotype at the disease locus via the logistic link as follows: $P(\text{case}|X=x) = \frac{\text{ exp }\{\alpha +\beta x\}}{1+\text{exp}\{\alpha +\beta x\}}.$ So, $P(\text{control}|X=x) = \frac{1}{1+\text{exp}\{\alpha +\beta x \}}.$

Following the above discussion, for a given choice of $n_1,n_2,$ and $f_2,f_1,f_0,$ and p,  we assume the values of $n_{case}^{G}$ and $n_{control}^{G}$ to be $n_1 P(G|\text{case})$ and $n_2 P(G|\text{control}),$ respectively, $G=DD,Dd,dd$. Suppose, we model the likelihood L of the genotype data of the cases and controls: $n_{case}^{G} ( = n_{1}^{G})$ and $n_{control}^{G} ( = n_{2}^{G}),$ $G=DD,Dd,dd,$ based on the logistic model presented above. Then the log-likelihood function denoted by $\text{ log }(L)$ is given by:

$$\begin{aligned} \text{ log }(L)= & {} n_{1}^{DD} (\alpha + 2\beta ) + n_{1}^{Dd} (\alpha + \beta ) \\&+ n_{1}^{dd} \alpha - (n_{1}^{DD}+n_{2}^{DD}) log(1+exp\{\alpha +2\beta \})\\&- (n_{1}^{Dd}+n_{2}^{Dd}) log(1+exp\{\alpha +\beta \}) \\&- (n_{1}^{dd}+n_{2}^{dd}) log(1+exp\{\alpha \}). \end{aligned}$$

For ease of exposition, we define the following quantities: $\text{ prob}_2 = \frac{\text{exp}\{\alpha +2\beta \}}{1+\text{exp}\{\alpha +2\beta \}},$ $\text{ prob}_1 = \frac{\text{exp}\{\alpha +\beta \}}{1+\text{exp}\{\alpha +\beta \}},$ $\text{ prob}_0 = \frac{\text{exp}\{\alpha \}}{1+\text{exp}\{\alpha \}}.$ Then the $1{\mathrm{st}}$ and $2{\mathrm{nd}}$ order partial derivatives of the log data likelihood are given by:

$$\begin{aligned}\frac{\partial \text{ log }(L)}{\partial \alpha } &= (n_{1}^{DD}+n_{1}^{Dd}+n_{1}^{dd}) - (n_{1}^{DD}+n_{2}^{DD})\text{ prob}_2 \\&\quad - (n_{1}^{Dd}+n_{2}^{Dd})\text{ prob}_1 - (n_{1}^{dd}+n_{2}^{dd})\text{ prob}_0\\\frac{\partial \text{ log }(L)}{\partial \beta } &= 2 n_{1}^{DD} + n_{1}^{Dd} - 2 (n_{1}^{DD}+n_{2}^{DD})\text{ prob}_2 \\&\quad - (n_{1}^{Dd}+n_{2}^{Dd})\text{ prob}_1 \\\frac{{\partial }^2 \text{ log }(L)}{\partial {\alpha }^2} &= - (n_{1}^{DD}+n_{2}^{DD})\text{ prob}_2(1-\text{ prob}_2)\\&\quad - (n_{1}^{Dd}+n_{2}^{Dd})\text{ prob}_1(1-\text{ prob}_1) - (n_{1}^{dd}+n_{2}^{dd})\text{ prob}_0(1-\text{ prob}_0)\\\frac{{\partial }^2 \text{ log }(L)}{\partial {\beta }^2} &= - 2 (n_{1}^{DD}+n_{2}^{DD})\text{ prob}_2(1-\text{ prob}_2) \\&\quad - (n_{1}^{Dd}+n_{2}^{Dd})\text{ prob}_1(1-\text{ prob}_1). \end{aligned}$$

Using the above equations in the Fisher’s scoring method, we obtain m.l.e. of $\alpha$ and $\beta$. Alternatively, we could also use the Newton Raphson method or the iteratively reweighted least squares method to obtain the m.l.e. Based on the m.l.e. of $\alpha$ and $\beta$, we estimate the penetrance parameters $f_2,f_1,f_0,$ as $P(\text{case}|X=2), P(\text{case}|X=1), P(\text{case}|X=0),$ respectively, using the logistic model discussed above.

For a given choice of $n_1,n_2,$ and $f_2,f_1,f_0,$ and p,  (and hence, $n_{1}^{G},n_{2}^{G},$ $G = DD,Dd,dd$), it is expected that the estimated penetrances would be close to the true choice of $f_2,f_1,f_0$. For example, we consider a choice of the penetrance parameters: $f_2 = 0.1, f_1 = 0.05, f_0 = 0.01,$ and $p = 0.1,$ that induces an overall prevalence of 0.018. Suppose we consider a total sample of 10,000 individuals. If we consider a design with equal number of cases and controls (i.e., $n_1=n_2=5000$) and estimate $f_2,f_1,f_0$ using the above model, the estimates are found to be $f_2 = 0.93, f_1 = 0.73, f_0 = 0.36$. Hence, it is clear that the true penetrances are grossly overestimated as pointed out earlier. On the other hand, if we consider the number of cases to be proportional to the overall prevalence (i.e., $n_1 = 181,$ $n_2 = 9819)$ we obtain the estimates of the penetrances to be $f_2 = 0.16, f_1 = 0.04, f_0 = 0.01$. Thus, these estimates are much closer to the true penetrances compared to the previous sampling design comprising equal number of cases and controls.