How many markers will you need?

If markers are randomly placed through the genome, then the average distance between a QTL and the closest marker is

$$E(m) = {L \over 2(n+1)} \qquad ,$$

where $L$ is the total map length and $n$ is the number of markers employed. The upper 95% confidence limit for the distance is

$${L \over 2}\left(1 - 0.05^{(1/n)}\right) \qquad .$$

Since the human genome is 33M (3300cM), 110 random markers give an average distance of 14.9cM and an upper 95% confidence limit of 44.3cM, corresponding to recombination frequencies of 0.13 and 0.29, respectively. Since there are about 30,000 genes in the human genome, there are roughly 10 genes per centimorgan. So if you're QTL is 44cm from the nearest marker, there are probably over 400 genes in the chromosomal segment you've identified.

If $r_{MQ}$ is the recombination fraction between the nearest marker locus and the QTL of interest, the frequency of recombinant genotypes among $F_2$ progeny is $2r_{MQ}(1-r_{MQ}) + r_{MQ}^2$. As you can see from the graph in Figure 1, there's a nearly linear relationship between recombination frequency and the frequency of recombinant phenotypes ($p$ in the graph). Think about what that means. Having a really dense map with a lot of markers is great, because it will allow you to map your QTL very precisely, if you look at enough segregating offspring to have a reasonable chance of picking up recombinants between them. With 3300cM in the human genome and roughly 3 GB of sequence to get within 1 MB of the actual QTL, you'd need one marker per centimorgan. To have 10 recombinants between markers bracketing the QTL, you'd need to analyze 1000 chromosomes.⁸

Figure 1: The relationship between recombination frequency, $r$, and the frequency of recombinant phenotypes, $p$, assuming a Haldane mapping function.