Question

In genetics, single nucleotide polymorphisms (SNPs) are locations in the (human) genome that exhibit variation across the population. SNPs cause the differences we see in traits such as hair color. Each SNP typically has two possible alleles say A and a and each person's genotype at the SNP is either AA, Aa, or aa, where one allele comes from the person's mother and one from the father. Let X be the number of A alleles at a particular SNP, and suppose we collect a random sample of people from some population. Under some assumptions (such as "random mating" and "no selection") we may assume that where p is called the allele frequency of allele A. What is the maximum likelihood estimator of p? What is the maximum likelihood estimate of the allele frequency of allele A if our sample consists of five people with genotypes AA, aa, Aa, aa, Aa at this particular SNP?

Answer 1

The probability mass function of Xi ~ Binom(2, p) is,

$f(x_i;p) = \binom{2}{x_i} p^{x_i} (1-p)^{2-x_i}$

The likelihood function of X = (X1, X2, ..., Xn) is,

$f(x;p) = \binom{2}{x_1} p^{x_1} (1-p)^{2-x_1} \times \binom{2}{x_2} p^{x_2} (1-p)^{2-x_2} .... \times \binom{2}{x_n} p^{x_n} (1-p)^{2-x_n}$

$f(x;p) = \prod_{i=1}^{n} \binom{2}{x_i} p^{x_i} (1-p)^{2-x_i}$

The log-likelihood function is,

$ln f(x;p) =\sum_{i=1}^{n}ln \binom{2}{x_i} + \sum_{i=1}^{n} x_i ln p + \sum_{i=1}^{n} (2-x_i) ln (1-p)$

For maximum likelihood estimate,

$\frac{\partial }{\partial p}ln f(x;p) = 0$

$\frac{\partial }{\partial p}ln f(x;p) = \sum_{i=1}^{n} x_i / p - \sum_{i=1}^{n} (2-x_i) / (1-p) = 0$

$\Rightarrow \sum_{i=1}^{n} x_i / p - (2n-\sum_{i=1}^{n}x_i) / (1-p) = 0$

$\Rightarrow (1-p)/ p = (2n-\sum_{i=1}^{n}x_i) / \sum_{i=1}^{n} x_i$

$\Rightarrow 1/p - 1 = 2n/\sum_{i=1}^{n}x_i-1$

$\Rightarrow 1/p = 2n/\sum_{i=1}^{n}x_i$

$\Rightarrow p = \sum_{i=1}^{n}x_i /2n$

Thus, the maximum likelihood estimate of the allele frequency is,

$\hat{p} = \sum_{i=1}^{n}x_i /2n$

For a given sample,

AA, aa, Aa, aa, Aa

n = 5 and the frequency of the allele in each of the five people are X = (2, 0, 1, 0, 1) . We get this by counting number of A alleles in each genotype.

$\sum_{i=1}^{n}x_i = 2 + 0 + 1 + 0 + 1 = 4$

The maximum likelihood estimate of the allele frequency is,

$\hat{p} = \sum_{i=1}^{n}x_i /2n = 4 / (2 * 5) = 2/5 = 0.4$