1Statistics 246 Spring 2006Meiosis and RecombinationWeek 3, Lecture 12Source:http://www.accessexcellence.org- the process which starts witha diploid cell having one set ofmaternal and one of paternalchromosomes, and ends upwith four haploid cells, each ofwhich has a single set ofchromosomes, these beingmosaics of the parental ones3Source:http://www.accessexcellence.orgThe action of interest to ushappens around here :•Chromosomes replicate, butstay joined at their centromeres•Bivalents form•Chiasmata appear•Bivalents separate by attachmentof centromeres to spindles.4Four-strand bundle and exchanges(one chromosome arm depicted)sisterchromatidssisterchromatids4-strand bundle (bivalent)Two exchanges4 meiotic products2 parental chromosomes5Chance aspects of meiosisNumber of exchanges along the 4-strand bundlePositions of the exchangesStrands involved in the exchangesSpindle-centromere attachment at the 1st meioticdivisionSpindle-centromere attachment at the 2nd meioticdivisionSampling of meiotic products Deviations from randomness called interference.6A stochastic model for meiosisA stochastic point process X for exchanges along the4-strand bundleA stochastic model for determining strand involvementin exchangesA model for determining the outcomes of spindle-centromere attachments at both meiotic divisionsA sampling model for meiotic products Random at all stages defines the no-interference orPoisson model.7Point process for exchanges We’ll come to this in a moment, going first to strandinvolvement in exchanges, to derive an important resultthat is independent of the model for exchanges. Usually we can’t observe exchanges - they occur “inside”meioses, and we usually observe only single meioticproducts - but on suitably marked meiotic products(chromosomes) we can track crossovers, see next slide.These give us indirect information about the exchangeprocess. We can make inferences about exchanges when all theproducts of a single meiosis can be recovered together,and we’ll see cases of this.8From exchanges to crossovers Changes of parental origin along meiotic products are calledcrossovers. They form the crossover point process C alongthe single chromosomes. A meiotic product is called recombinant across an interval J -we’ll denote the event by R(J) - if the parental origins of itsendpoints differ, i.e. if an odd number of crossovers haveoccurred along J, and parental otherwise. Assays exist for determining whether this is so, usually simplygenotyping segregating markers at the endpoints of J.9A model for strand involvement The standard assumption here is No Chromatid Interference(NCI): each non-sister pair of chromatids is equally likely to beinvolved in each exchange, independently of the strandsinvolved in other exchanges. NCI fits the available pretty well, but of course not perfectly.There are broader models, though not in general use. It isinteresting to make up models with different strand involvementrules and explore their consequences. Formally, NCI says that the crossover process C is a Bernoullithinning of the exchange process X with p=0.5, that is, wecopy an exchange point to a crossover point on a meioticproduct with probability 1/2, independently for all exchanges.10The relation betweenexchanges and crossovers Key to making inferences about unobserved exchangesfrom observable crossovers are the assumptionsconnecting them. NCI is our principal one, but before welook carefully at that, let’s do draw some pictures andget an idea of the combinatorics involved. We’ll begin by defining an interval J by segregatingmarkers at its endpoints, and looking at possible meioticproducts. Our interest is in whether a meiotic product isrecombinant or parental across the interval, and howoften this occurs.11Two exchanges (4/16 cases)12Exercise With Emerson & Rhodes, The American Naturalist, Vol67 No 711 (Jul-Aug 1933): 374-377, extend this tableto exchanges with 3, 4 or more exchanges in theinterval, and conclude that in all cases apart from 0exchanges, exactly one half of the meiotic products areparental, and one half recombinant. Develop a proof of their conclusion that under NCI, forn>0, pr(R(J) | X(J) = n ) = 1/2,and so pr(R(J)) = 1/2 × pr( X(J) > 0 )………(*)13Recombination fractions The recombination fraction across an interval J isdefined to be r = pr(R(J)). Under NCI, (*) tells usthat 0 ≤ r ≤ 1/2. In a sense it gives an indication of thechromosomal length of the interval J. Exercise. Prove that under NCI, r = r(J) ismonotone in the length of J. However, the use of r as a metric must be limited,as it is bounded by 1/2.14The Poisson (no interference)model for exchangesSuppose that the exchange process X is a Poisson pointprocess with mean measure λ. That is, for achromosomal interval J, the distribution of X(J) isPoisson with mean λ(J), and exchanges across disjointintervals are mutually independent.The crossover process C, the thinned version of X, is thusalso a Poisson process, with mean measure d = λ/2.Further, under NCI, we have the formula€ r(J) =12(1− e−λ(J )) =12(1− e−2d (J )).151 2 3r12r23r13r13 ≠ r12 + r23 Recombination fractions and mapping The recombination fraction does not define a metric.However,Sturtevant (1913) used recombination fractions toorder genes. Exercise: Can we determine the order of three loci on achromosome from their 3 pairwise recombination fractions. Canyou extend this to any number of loci along a chromosome?16More on the Poissonmodel for exchangesExercise: Prove that for the Poisson model, we havethe following formula for loci in order 1-2-3: r13 = r12 + r23 - 2r12r23 .Extend to 4 loci.Relations such asbetween observable recombination fractions andparameters d of unobservable exchange processes arecalled map functions. This one is known as Haldane’s,after JBS Haldane who introduced it in 1919.€ r =12(1− e−2d)17Multilocus recombination probabilities Let’s forget the Poisson model for a moment. Suppose thatwe have 3 loci, 1-2-3 along a chromosome. Write Eij for theevent of at least one exchange between loci i and j duringmeiosis, and Rij for the event of recombination between loci iand j on a meiotic product. Now define probs p and q asfollows, where - denotes the complementary event:€ € q00= pr(E 12& E 23) p00= pr(R 12& R 23)q01= pr(E 12& E23) p01= pr(R 12& R23)q10= pr(E12& E 23) p10= pr(R12& R 23)q11=
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