Analysis of an $F_2$ derived from inbred lines

An analysis of inbred lines uses the same basic design as Thoday, but takes advantage of more information.⁸ We start with two inbred lines $M_1QM_2/M_1QM_2$ and $m_1qm_2/m_1qm_2$, make an $F_1$, intercross them, and score the phenotype and marker genotype of each individual. Analysis of the data is based on calculating the frequency of each genotype at the $Q$ locus as a function of the genotype at the marker loci and the recombination fractions between the marker loci and $Q$.⁹ For example,

$$\begin{eqnarray*} P(M_1QM_2/M_1QM_2) &=& \left((1-r_{1Q})(1-r_{Q2})/2\right)^2 \... ...\ P(M_1qM_2/M_1qM_2) &=& \left(r_{1Q}r_{Q2}/2\right)^2 \qquad . \end{eqnarray*}$$

Because the frequency of $M_1M_2/M_1M_2 = \left((1-r_{12})/2\right)^2$, we can use Bayes' Theorem to write the conditional probabilities of getting each genotype as

$$\begin{eqnarray*} P(QQ\vert M_1M_2/M_1M_2) &= {(1-r_{1Q})^2(1-r_{Q2})^2 \over (1... ...M_2/M_1M_2) &= {r_{1Q}^2 r_{Q2}^2 \over (1-r_{12})^2} \qquad . \end{eqnarray*}$$

Clearly, if we wanted to we could right down similar expressions for the nine remaing marker genotype classes, but we'll stop here. You get the point.¹⁰

Now that we've got this we can write down the likelihood of getting our data, namely

$$L(x\vert M_j) = \sum_{k=1}^N \phi (x\vert\mu_{Q_k}, \sigma^2) P(Q_k\vert M_k) \qquad ,$$

where $N$ is the number of QTL genotypes considered, $\phi (x\vert\mu_{Q_k}, \sigma^2)$ is the probability of getting phenotype $x$ given the mean phenotype, $\mu_{Q_k}$, and variance, $\sigma^2$, associated with $Q_k$, and $P(Q_k\vert M_k)$ is the probability of getting $Q_k$ given the observed marker genotype. Fortunately, we don't have to do any of these calculations, all we do is to ask our good friend (QTL Cartographer) to do the calculations for us. It will scan the genome, and tell us how many QTL loci we are likely to have, where they are located relative to our known markers, and what the additive and dominance effects of the alleles are.