Why is linkage disequilibrium high on either side of a beneficial mutation, but not between sites across the mutation, during a selective sweep?

Question

I was reading the paper "A survey of methods and tools to detect recent and strong positive selection" (2017) and came across this:

Upon fixation of the beneficial mutation, elevated levels of LD emerge on each side of the selected site, whereas a decreased LD level is observed between sites found on different sides of the selected site. The high LD levels on the different sides of the selected locus are due to the fact that a single recombination event allows existing polymorphisms on the same side of the sweep to escape the sweep. On the other hand, polymorphisms that reside on different sides of the selected locus need a minimum of two recombination events in order to escape the sweep. Given that recombination events are independent, the level of LD between SNPs that are located on different sides of the positively selected mutation decreases.

I am having trouble understanding how this works. If I am understanding this correctly, if you took two sites on the left side of the beneficial mutation during a selective sweep, you would see high LD between them. The same goes for taking two sites on the right side of the beneficial mutation. (Provided they are close enough to the mutation site of course.) However, if you took one site 300 bp to the left of the mutation, and another 300 bp to the right of it, you may not see the same rise in LD. I am not sure why this would be the case: wouldn't the entire region linked to the beneficial mutation, regardless of which side of the mutation it occurs to, be co-inherited and thus display similar high LD across the board, provided it is overall close enough to the site of beneficial mutation?

Answer 1

High-level answer

Obviously, if the positively selected site occurs on a single haplotype, then all the polymorphisms on that haplotype "win" due to hitchhiking and will have some degree of LD, regardless of whether they are left or right. All that this scenario is asserting is that the left and right flanks have a higher degree of within-flank dependence in their fate than they have inter-flank dependence.

In other words, the right flank will be preserved by recombinations left of the selected site, and the left flank will be preserved by recombinations right of the selected site. The flank blocks' interests are to some degree independent or antagonistic, whereas the polymorphisms within each flank have a common interest.

More detail

For the purpose of discussion, I think it is worthwhile to include the figure that they show here (Figure 2 in the paper):

It is less that there is no LD across the site, but rather that the LD is less marked than between sites on either side and the focal positively selected site (in black, in this figure). Negative selection is removing left or right flank polymorphisms separated from the selected site by recombination. As your quote notes, you would need a compensatory recombination to preserve LD across the site, which is relatively rare. So negative selection is not acting to preserve LD from the left to the right flank.

Everyone on the left is co-inherited due to the selection against recombination in that window. Everyone on the right is co-inherited similarly. However right and left are not co-inherited.

Note: since all polymorphisms on the haplotype are hitchhiking along with the selected site, when I say "selection against" what I mean is that recombination will mean that sites on the other side of the recombination event from the selected site cannot hitchhike at the same time. That hitchhiking is (as expected) the ultimate source of this pattern. And all of these LD blocks will decay with distance as expected.

Omega calculation clarifies within-flank dependence

If you go on to look at the equation 2 that immediately follows in the paper, this becomes maybe a little clearer:

\begin{aligned} \omega = \frac{\left({l \atopwithdelims ()2} + {W-l \atopwithdelims ()2}\right)^{-1}\left(\sum _{i,j\in L}r_{ij}^2 + \sum _{i,j\in R}r_{ij}^2\right)}{\left(l(W-l)\right)^{-1}\sum _{i\in L, j\in R}r_{ij}^2}. \end{aligned}

$\omega$ is looking for two flanks of fairly high intra-flank LD within themselves that have relatively low inter-flank LD with each other.

Answer 2

In addition to the other great answer, I would like to revisit my question because I came across a stellar explanation in another paper I am now reading that also answers the question very succinctly and which I share here:

The LD within both flanking regions of the selected site is compared with the LD between flanking regions. Selective sweeps lead to high LD within but not between the flanking regions.

...

Selective sweeps generate a very specific LD pattern at sites neighbouring the beneficial allele, which can be captured by the ω statistic (Kim & Nielsen, 2004). Owing to the fast increase in frequency of the beneficial allele, genetic variation in the region under selection is only maintained by recombination during the selective sweep. When conceptually regarding a chromosome as two flanking regions adjacent to the selected site, a recombination event divides the chromosome into two parts: (1) the original genetic background of the selected site including one flanking region, the selected site and the second flanking region until the recombination breakpoint; and (2) the alternative genetic background, starting from the recombination breakpoint at the second flanking region. Hence, recombination will always affect only one side of the genomic region separated by the selected site per recombination event, leading to independent recombination patterns at both flanking sites. Therefore, LD should be high when comparing sites within one flanking region but low when comparing sites between the two flanking regions.

https://doi.org/10.1093/zoolinnean/zly007