How many genes? Mapping mouse traits, cont.Inferring linkage and mapping markersThe LOD scoreCalculating the LOD scoreNull probabilities of 2-locus genotypesUsing the LOD scoreLinkage groupsForming linkage groups, cont.Ordering linkage groupsLikelihoods for 3-locus dataMultilocus linkage: #loci >3Multilocus mapping: no detailsChecking the map, after removal of bad markersChecking existing genetic mapsInterplay between error detection and mapsLocations of our markersRR/qtlBenefits of using R/qtlWhy map coat color genes in our C57/BL6 x NOD F2 intercross?Recall our earlier Punnett squareSegregation data at a “random” markerMapping a segregating traitProbabilities of albino-marker genotypes (4)Segregation data at the marker closest to TyrcPowerPoint PresentationSlide 27Honesty in advertising, and LOD thresholdsApproximate probabilities of agouti-marker genotypes (4)Segregation data at the marker closest to the agouti locusSlide 31Slide 32Conclusion: single locus mappingAcknowledgementGeneral exercise1How many genes? How many genes? Mapping mouse traits, cont.Mapping mouse traits, cont.Lecture 3, Statistics 246January 27, 20042Inferring linkage and mapping markersWe now turn to deciding when two marker loci are linked, and if so, estimating the map distance between them. Then we go on and create a full (marker) map of each chromosome, relative to which we can map trait genes. With these preliminaries completed, we can map trait loci.3The LOD score Suppose that we have two marker loci, and we don’t know whether or not they are linked. A natural way to address this question is to carry out a formal test of the null hypothesis H: r=1/2 against the alternative K: r< 1/2, using the marker data from our cross. The test statistic almost always used in this context is log10 of the ratio of the likelihood at the maximum likelihood estimate to that at the null, r=1/2, i.e. € LOD = log10{L(ˆ r )L(1 / 2)}€ ˆ r4Calculating the LOD score Recall that the (log) likelihood here is based on the multinomial distribution for the allocation of n=132 intercross mice into their nine 2-locus genotypic categories. As we saw earlier, it can be written and so we take the difference between this function evaluated at and at r=1/2, which is where qi is 1/16, 1/8 or 1/4, depending on i.€ ˆ r € € log10L(r) = nilog10pi(r)i∑€ LOD = nilog10pi(ˆ r ) /qii∑5Null probabilities of 2-locus genotypes L1 L2A H BA 1/16 1/8 1/16H 1/8 1/4 1/8B 1/16 1/8 1/16 This is just putting r = 1/2 in an earlier table.Exercise: Suggest some different test statistics to discriminate between the null H and the alternative K. How do they perform in comparison to the LOD?6Using the LOD score Normal statistical practice would have us setting a type 1 error in a given context (cross, sample size), and determining the cut-off for the LOD which would achieve approximately the desired error under the null hypothesis. This approach is rarely adopted in genetics, where tradition dictates the use of more stringent thresholds, which take into a account the multiple testing common on linkage mapping. It was originally motivated by a Bayesian argument, and in fact, Bayesian approaches to linkage analysis are increasingly popular. Let us use of Bayes’ formula in the form log10 posterior odds = log10 prior odds + LOD, where the odds are for linkage. With 20 chromosomes, which we might assume approx the same size, and not too long, the prior probability of two random loci being on the same chromosome and hence linked, is about 1/20. In order to overcome these prior odds against linkage, and achieve reasonable posterior odds, say 100:1, we would want a LOD of at least 3.7Linkage groups And so it has come to pass that a LOD must be >3 to get people’s attention. We’ll be a little more precise later. The next step is to define what are called linkage groups. These partition the markers into classes, every pair of markers being either closely linked (i.e. r 0), or being connected by a chain of markers, each consecutive pair of which is closely linked. In practice, we might define closely linked to be something like a) < c1, and b) LOD( ) > c2, where e.g. c1= 0.2, c2 = 3. € ˆ r € ˆ r8Forming linkage groups, cont. When one tries to form linkage groups, it is not unusual to have to vary c1 and c2 a little, until all markers fall into a group of more than just one marker. When this is done, it is hoped that the linkage groups correspond to chromosomes. If the chromosome number of the species is known, and that coincides with the number of linkage groups, this is a reasonable presumption. But much can happen to dash this hope: one may have two linkage groups corresponding to different arms of the same chromosome, and not know that; one can have a marker at the end of one chromosome “linked” to a marker at the end of another chromosome, though this should be rare if there is plenty of data; and so on.9Ordering linkage groups Next we want to order the markers in a linkage group( ideally, on a chromosome). How do we do that? An initial ordering can be done by starting one of the markers, M1 say, on the most distant pair, here distance being recombination fraction, or map distance. Call M2 the closest marker to M1 and continue in this way. Now we want to confirm our ordering. One way is to calculate a (maximized) log likelihood for every ordering, and select the one with the largest log likelihood. But if we have (say) 11 markers on a chromosome, this is 11! = 4107 orders. What people often do is take moving k-tuples of markers, and optimize the order of each, e.g. with k = 3 or 4. Whichever strategy one adopts, multi (i.e. >2) locus methods are needed.10Likelihoods for 3-locus data Suppose that we have 3 markers M1 , M2 and M3 in that order. How do we calculate the log likelihood of the associated 3-locus marker data from our intercross? Recalling the discussion preceding the Punnett square of the last lecture, the parental haplotypes here are a1a2a3 and b1b2b3 while are would no fewer than 6 forms of recombinant haplotypes: the four single recombinants a1a2b3 , a1 b2 b3 , b1b2a3 and b1a2a3 , and the two double recombinants a1b2 a3 and b1a2b3 . Proceeding as before, we calculate the probability of each of these in terms of the recombination fractions r1 and r2 across intervals M1-M2, and M2-M3, respectively.
View Full Document
Unlocking...