1Meiosis, recombination fractionsand genetic distanceStatistics 246, Spring 2004Lecture 2A, January 22Initially: pages 1-11.Later: pages 12-18.2Source: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.- 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 ones3Four-strand bundle and exchanges(one chromosome arm depicted)sisterchromatidssisterchromatids4-strand bundle (bivalent)Two exchanges4 meiotic products2 parental chromosomes4Chance 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.5A stochastic model for meiosisA point process X for exchanges along the 4-strandbundleA model for determining strand involvement inexchangesA 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-interferenceor Poisson model.67A model for strand involvement The standard “random” assumption here is No Chromatid Interference (NCI): each non-sister pair of chromatids is equally likelyto be involved in each exchange, independently ofthe strands involved in other exchanges. NCI fits the available data pretty well, but there arebroader models.8The crossover processon meiotic products Changes of (grand)parental origin along meiotic products arecalled crossovers. They form the crossover point process Calong the single chromosomes. Under NCI, C is a Bernoulli thinning of X with p=0.5, that is,each exchange has a probability of 1/2 of involving a givenchromatid, independently of the involvement of otherexchanges.1 change2 changes1 change no change9From exchanges to crossovers Usually we can’t observe exchanges, but on suitablymarked chromosomes we can track crossovers. Call a meiotic product recombinant across an intervalJ, and write R(J), if the (grand)parental origins of itsendpoints differ, i.e. if an odd number of crossovershave occurred along J. Assays exist for determiningwhether this is so. We usually write pr(R(J))=r, andcall r the recombination fraction.Recombination across the intervalNo recombinationRecombinationNo recombination104 NR 2R, 2NR4NR2R, 2NR2R, 2NR4RCounting recombinants R and non-recombinants NRacross the interval AB11 Mather’s formulaUnder NCI, if n>0, pr(R(J) | X(J) = n ) = 1/2. Proof. Suppose that n>0. Consider a particular chromatid. Ithas a probability of 1/2 of being involved in any givenexchange, and its involvement in any of the n separateexchanges are independent events. Thus the chance that it isinvolved in an odd number of exchanges is the sum over allodd k of the binomial probabilities b(k; n, 1/2), which equals1/2 (check). Corollary (Mather): pr(R(J)) = 1/2 × pr( X(J) > 0). It follows that under NCI, the recombination fraction r = pr(R(J))is monotone increasing in the size of J, and ≤ 1/2.12The Poisson model Suppose that the exchange process X is a Poisson process,i.e. that the numbers of exchanges in any pairwise disjointset of intervals are mutually independent Poisson randomvariables. Denoting the mean number of exchanges in intervalJ by λ(J), we can make a monotone change of thechromosome length scale to convert this mean to λ|J|, where|J| is the length of J. This foreshadows the important notion ofgenetic or map distance, where rate = length. Exercise: Prove that if X is a Poisson process, so is thecrossover process C.131 2 3r12r23r13r13 ≠ r12 + r23 Recombination and mapping Sturtevant (1913) first used recombination fractions toorder (i.e. map) genes. Problem: the recombinationfraction does not define a metric. Let’s consider 3 loci, denoted by 1, 2 and 3, and putrij = pr(R(i--j)).In general,14Triangle inequality We will prove that under NCI, r13 ≤ r12 + r23 . To see this, define p00 = pr(R(1--2)&R(2--3)), p01 = pr(R(1--2)&R(2--3)) p10 = pr(R(1--2)&R(2--3)), p11 = pr(R(1--2)&R(2--3)), where the denotes the complement (negation) of the event. Now notice that R(1--2)&R(2--3) + R(1--2)&R(2--3) = R(1--2), R(1--2)&R(2--3) + R(1--2)&R(2--3) = R(2--3), and R(1--2)&R(2--3) + R(1--2)&R(2--3) = R(1--3) (think about this one). Thus we have p10 + p11 = r12 , p01 + p11 = r23 , and p00 + p11 = 1-r13 . Adding the three equations, and using the fact that the pij sum to 1 gives r12 + r23 - r13 = 2p11 ≥ 0. In general this inequality is strict. Under the Poisson model, p11 = r12r23 .15 Map distance: d12 = E{C(1--2)} = av # COs in 1--2 Unit: Morgan, or centiMorgan.1 2 3d12 d23 d13d13 = d12 + d23 Genetic mapping or applied meiosis: a BIG business• Placing genes and other markers along chromosomes;•Ordering them in relation to one another;•Assigning map distances to pairs, and then globally.Map distance and mapping16Haldane’s map function Suppose that X is a Poisson process, and that themap length of an interval J is d. Then the mean number λ(J) of exchanges acrossJ is 2d, and by Mather, the recombination fractionacross J isMore generally, map functions relate recombinationfraction to genetic distance; r ~ d for r small.€ r =12(1− e−2d).17The program from now on With these preliminaries, we turn now to thedata and models in the literature which throwlight on the chance aspects of meiosis. Mendel’s law of segregation: a result ofrandom sampling of meiotic products, withallele (variant) pairs generally segregating inprecisely equal numbers. As usual in biology, there are exceptions.1819Random spindle-centromere attachment at 1st meiotic divisionIn 300 meioses in angrasshopper heterozygousfor an inequality in the size ofone of its chromosomes,the smaller of the twochromosomes moved withthe single X 146 times, whilethe larger did so 154 times.Carothers, 1913.xsmallerlarger20Tetrads In some organisms - fungi, molds, yeasts - allfour products of an individual meiosis can berecovered
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