Interval Mapping

Consider a QTL flanked by two markers segregating in an F2 population.

Basics for interval mapping

To fully understand interval mapping, first we need to cover some basics:

• Conditional probabilities
• Likehood inference and maximum likelihood

Conditional probabilities of QTL genotypes

Consider a QTL flanked by two markers segregating in an F2 population.

The recombination frequency between M1 and Q is 0.05 or 5cM and between Q and M2 is 0.20 or 20cM. Assume no interference.

• What are the probabilities of the three QTL genotypes given a marker genotype of $M_1M_1M_2M_2$?

$$\begin{align*} & P(QQ|M_1M_1M_2M_2) \\ & P(Qq|M_1M_1M_2M_2) \\ & P(qq|M_1M_1M_2M_2) \\ \end{align*}$$

Conditional probabilities of QTL genotypes

Consider a QTL flanked by two markers segregating in an F2 population.

$$\begin{align*} & P(QQ | M_1M_1M_2M_2) = \frac{P(M_1M_1QQM_2M_2)}{P(M_1M_1M_2M_2)} \\ \end{align*}$$

$$\begin{align*} & P(M_1M_1M_2M_2) = P(M_1M_2)^2 = (\frac{1- c_{12}}{2})^2 \\ \end{align*}$$

$$\begin{align*} & P(M_1M_1QQM_2M_2) = P(M_1QM_2)^2 = (\frac{(1- c_{1Q})(1 - c_{2Q})}{2})^2 \\ \end{align*}$$

$$\begin{align*} & P(QQ | M_1M_1M_2M_2) = \frac{(\frac{(1- c_{1Q})(1 - c_{2Q})}{2})^2}{(\frac{1- c_{12}}{2})^2} = 0.973\\ \end{align*}$$

Conditional probabilities

Consider a QTL flanked by two markers segregating in an F2 population.

$$\begin{align*} & P(QQ | M_1M_1M_2M_2) = \frac{(\frac{(1- c_{1Q})(1 - c_{2Q})}{2})^2}{(\frac{1- c_{12}}{2})^2} = 0.973\\ \end{align*}$$

$$\begin{align*} P(Qq | M_1M_1M_2M_2) & = \frac{P(M_1M_1QqM_2M_2)}{P(M_1M_1M_2M_2)} \\ P(M_1M_1QqM_2M_2) & = P(M_1 Q M_2)P(M_1 q M_2) + P(M_1 q M_2)P(M_1 Q M_2) \\ & = 2(\frac{(1- c_{1Q})(1 - c_{2Q})}{2})(\frac{c_{1Q}c_{2Q}}{2}) \\& = 0.0038 \end{align*}$$

Here, $c_{1Q} = 0.05$ and $c_{2Q} = 0.2$

Conditional probabilities

Consider a QTL flanked by two markers segregating in an F2 population.

$$\begin{align*} & P(QQ | M_1M_1M_2M_2) = \frac{(\frac{(1- c_{1Q})(1 - c_{2Q})}{2})^2}{(\frac{1- c_{12}}{2})^2} = 0.973\\ & P(Qq | M_1M_1M_2M_2) = \frac{2(\frac{(1- c_{1Q})(1 - c_{2Q})}{2})(\frac{c_{1Q}c_{2Q}}{2})}{(\frac{1- c_{12}}{2})^2} = 0.026\\ & P(qq | M_1M_1M_2M_2) = \frac{(\frac{c_{1Q}c_{2Q}}{2})^2}{(\frac{1- c_{12}}{2})^2} = 1.7 \times 10^{-4}\\ \end{align*}$$

Conditional probabilities

If the genotypic values for each of the QTL genotypes were given as below:

• What is the expected value of individuals with the $M_1M_1M_2M_2$?

$$\begin{align*} & E(M_1M_1M_2M_2) = 0.973 \times 7 + 0.026 \times 5 + 1.7\times 10^{-4} \times 0 = 6.94 \\ \end{align*}$$

Maximum likelihood

What does it mean to calculate the likelihood of somethings?

The likelihood function is represented as $L(\theta|s) = f_{\theta}(s)$ .

This function represents the likelihood of a certain parameter value ( $\theta$ ) given a data vector ( $s$ ).

Interpretation of the likelihood function

• The $f_{\theta}(s)$ represents the probability density function with $\theta$ set as the parameter and $s$ set as the observations.

• The value of $L(\theta|s)$ is called the likelihood of $\theta$.

Maximum likelihood

What does it mean to calculate the likelihood of somethings?

The likelihood function is represented as $L(\theta|s) = f_{\theta}(s)$ .

This function represents the likelihood of a certain parameter value ( $\theta$ ) given a data vector ( $s$ ).

Estimation

To find the value of $\theta$ with the maximum likelihood, a range of theta values is tested against the observed data, and the $\theta$ giving the highest likelihood is determined to be the maximum likelihood estimator of $\theta$.

Note: we are fixing the data and varying the parameter.

Example: Binomial distribution

You tossed a coin ten times and observed four heads.

What is the maximum likelihood estimator of $p$, the probability of obtaining a head?

• Substitute the observation into the binomial probability density function (pdf) and vary the value of $p$.

$$\begin{align*} & L(p | k) = \binom{n}{k}p^kq^{n-k} \\ & L(p | 4) = \binom{10}{4}p^4q^{10-4} \\ \end{align*}$$

• Vary $p$ from 0 to 1, with step size 0.1.
• Each $p$ is assumed to be the true value, then the likelihood of the data is calculated.

$$\begin{align*} & L(0 | 4) = 0; L(0.1 | 4) = 0.01; L(0.2 | 4) = 0.09; ... \\ & L(0.4 | 4) = 0.25; ... \\ & L(0.7 | 4) = 0.04; ... ; L(1.0 | 4) = 0 \\ \end{align*}$$

$p=0.4$ is our maximum likelihood (ML) estimator for $p$.

Construction of QTL likelihood functions?

When a major bi-allelic locus is segregating in a population. The distribution of the entire population can be broken into three underlying distributions:

• The distribution of the $QQ$ individuals,
• $Qq$ individuals,
• $qq$ individuals.

The likelihood of the genotypic parameters given phenotypic value z is:

$$\begin{align*} L(z) & = L(\mu_{QQ}, \mu_{Qq}, \mu_{qq}, \sigma^2 | z) \\ & = P(QQ)f(z, \mu_{QQ}, \sigma^2) + P(Qq)f(z, \mu_{Qq}, \sigma^2) + P(qq)f(z, \mu_{qq}, \sigma^2)\\ \end{align*}$$

• Where $P(Q_k)$ equals the probability of a particular genotype

  • e.g. 1/4 in an F2 population for $QQ$

• $f(z, \mu_k, \sigma^2)$ is the probability density function for a normally distributed random variable with mean $\mu_k$ and variance $\sigma^2$.

  • The mean value of $QQ = a$, $Qq=d$ and $qq = -a$.

Construction of QTL likelihood functions?

When a major bi-allelic locus is segregating in a population. The distribution of the entire population can be broken into three underlying distributions:

• The distribution of the $QQ$ individuals,
• $Qq$ individuals,
• $qq$ individuals.

The likelihood of the genotypic parameters given phenotypic value z is:

$$\begin{align*} L(z) & = L(\mu_{QQ}, \mu_{Qq}, \mu_{qq}, \sigma^2 | z) \\ & = P(QQ)f(z, \mu_{QQ}, \sigma^2) + P(Qq)f(z, \mu_{Qq}, \sigma^2) + P(qq)f(z, \mu_{qq}, \sigma^2)\\ \end{align*}$$

For $n$ random (unrelated) individuals, the overall likelihood is the product of the $n$ individual likelihoods

$$\begin{align*} L(z_1, z_2, .., z_n) & = L(\mathbf{z}) = \prod_{j=1}^{n}{L(z_j)}\\ \end{align*}$$

Construction of QTL likelihood functions?

Now, let's return to our conditional probabilities, specifically the probability of a QTL genotype given a marker genotype.

The likelihood of an individual with phenotypic value $z$ given a marker genotype $M_i$ is represented as:

$$\begin{align*} L(z|M_i) & = P(QQ|M_i)f(z, \mu_{QQ}, \sigma^2) + P(Qq|M_i)f(z, \mu_{Qq}, \sigma^2) + P(qq|M_i)f(z, \mu_{qq}, \sigma^2)\\ \end{align*}$$

For example, the likelihood for genotype MM is: $$\begin{align*} L(z|MM) & = P(QQ|MM)f(z, \mu_{QQ}, \sigma^2) + P(Qq|MM)f(z, \mu_{Qq}, \sigma^2) + P(qq|MM)f(z, \mu_{qq}, \sigma^2)\\ \end{align*}$$

• The $P(Q_k|M_j)$ parts are a function of the map positions and experimental design, so that

$$\begin{align*} L(z|MM) & = (1-c)^2f(z, \mu_{QQ}, \sigma^2) + 2c(1-c)f(z, \mu_{Qq}, \sigma^2) + c^2f(z, \mu_{qq}, \sigma^2)\\ \end{align*}$$

• And the QTL effects enter through the means and variances of the underlying normal distributions $f_{\theta}(z)$ or $f(z, \mu_{Q_k}, \sigma^2)$.

Back to interval mapping

To calculate the likelihoods for an interval, we simply insert the probabilities of a QTL genotype given a marker interval genotype.

For example,

$$\begin{align*} L(z|M_1M_1M_2M_2) = & \frac{(1-c_1)^2(1-c_2)^2}{(1-c_{12})^2}f(z, \mu_{QQ}, \sigma^2) \\ & + \frac{2c_1c_2(1-c_1)(1-c_2)}{(1-c_{12})^2}f(z, \mu_{Qq}, \sigma^2) \\ & + \frac{c_1^2c_2^2}{(1-c_{12})^2}f(z, \mu_{qq}, \sigma^2)\\ \end{align*}$$

• This likelihood value is calculated for each genetic position in between the two flanking markers by varying value of recombination rate (c).

• The span of the entire interval ( $c_{12}$ ) is calculated using a mapping function.

• The values of $\mu_{QQ}, \mu_{Qq}, \mu_{qq}, \sigma^2$ are estimated at each genetic position.

Back to interval mapping

Step 1: Calculate the likelihood with a QTL

• Given $c_1, c_2, c_{12}$ and the phenotypic data

Step 2: Calculate the likelihood that no underlying QTL exists (the data arose in the absence of a QTL).

• That is, there are no underlying QTL genotypes and the distribution of individuals with the $M_1M_1M_2M_2$ genotype consists of a single distribution.

Step 3: Finally, compute the ratio.

• This ratio is what provides the likelihood ratio.

LOD score

The ratio is converted to the famous logarithm of odds (LOD) score

$$\begin{align*} LOD = log_{10}(\frac{L_{full}}{L_{reduced}}) \end{align*}$$

• Where $L_{full}$ is the likelihood of a QTL at assumed genetic position given the data.
• $L_{reduced}$ is the likelihood of no QTL present given the data.

If the LOD score is 3, for example, this means that the likelihood for a model including a QTL at the given genetic position is 1,000 times higher than no QTL at that position!

Statistical significance

QTL mapping involves a large number of tests, which requires adjustments for multiple testing to keep the experiment-wise error rate low.

Bonferroni correction

• Assume all tests are independent, which is not the case in QTL mapping because markers are linked.
• Overly conservative for QTL mapping.

Statistical significance

QTL mapping involves a large number of tests, which requires adjustments for multiple testing to keep the experiment-wise error rate low.

Permutation test

A commonly used technique for QTL mapping.

• Basically, the phenotypic data is randomized relative to the marker data so that the null hypothesis is established.

• Then, the test statistic for each QTL is calculated and the largest test statistic across the genome is tabulated.

• This is repeated 1,000 or more times in order to establish an empirical distribution of the test statistic under the null hypothesis.
• The test statistics calculated for the real data are compared to this distribution to determine the significance level.

Simulating a QTL mapping experiment

library(qtl)
set.seed(12347)
# Five autosomes of cM length 50, 75, 100, 125, 60
L <- c(50, 75, 100, 125, 60)
map <- sim.map(L, L/5+1, eq.spacing=FALSE, include.x=FALSE)
# Simulate a backcross with two QTL
a <- 0.7
mymodel <- rbind(c(1, 40, a), c(4, 100, a))
pop <- sim.cross(map, type="bc", n.ind=200, model=mymodel)
plot.map(pop)

Simulating a QTL mapping experiment

Checking phenotypic distribution

hist(pop$pheno$phenotype, main="simulated phenotype", 
     breaks=50, xlab="Phenotype", col="#cdb79e")

Single-marker analysis

# single-QTL scan by marker regression with the simulated data
out.mr <- scanone(pop, method="mr")
# plot of marker regression results for chr 4 and 12
plot(out.mr, chr=c(1,2,3,4,5), ylab="LOD Score")

Haley-knott Regression

This is a version of interval mapping which is a very good approximation to interval mapping via maximum likelihood.

# single-QTL scan using Haley-knott Regression approach
out.hk <- scanone(pop, method="hk")
# plot of marker regression results for chr 4 and 12
plot(out.hk, chr=c(1,2,3,4,5), ylab="LOD score")

Plot QTL effect

# summary of out.mr
summary(out.mr, threshold=3)

##      chr pos  lod
## D1M4   1  32 3.77

effectplot(pop, mname1="D1M4", main="Chr1")