Statistics 246 Spring 2006

Size: px

Start display at page:

Download "Statistics 246 Spring 2006"

Egbert Lester
6 years ago
Views:

1 Statistics 246 Spring 2006 Meiosis and Recombination Week 3, Lecture 1 1

2 - the process which starts with a diploid cell having one set of maternal and one of paternal chromosomes, and ends up with four haploid cells, each of which has a single set of chromosomes, these being mosaics of the parental ones Source: 2

3 The action of interest to us happens around here : Chromosomes replicate, but stay joined at their centromeres Bivalents form Chiasmata appear Bivalents separate by attachment of centromeres to spindles. Source: 3

4 Four-strand bundle and exchanges (one chromosome arm depicted) sister chromatids sister chromatids 2 parental chromosomes 4-strand bundle (bivalent) Two exchanges 4 meiotic products 4

5 Chance aspects of meiosis Number of exchanges along the 4-strand bundle Positions of the exchanges Strands involved in the exchanges Spindle-centromere attachment at the 1st meiotic division Spindle-centromere attachment at the 2nd meiotic division Sampling of meiotic products Deviations from randomness called interference. 5

6 A stochastic model for meiosis A stochastic point process X for exchanges along the 4-strand bundle A stochastic model for determining strand involvement in exchanges A model for determining the outcomes of spindlecentromere attachments at both meiotic divisions A sampling model for meiotic products Random at all stages defines the no-interference or Poisson model. 6

7 Point process for exchanges We ll come to this in a moment, going first to strand involvement in exchanges, to derive an important result that is independent of the model for exchanges. Usually we can t observe exchanges - they occur inside meioses, and we usually observe only single meiotic products - but on suitably marked meiotic products (chromosomes) we can track crossovers, see next slide. These give us indirect information about the exchange process. We can make inferences about exchanges when all the products of a single meiosis can be recovered together, and we ll see cases of this. 7

8 From exchanges to crossovers Changes of parental origin along meiotic products are called crossovers. They form the crossover point process C along the 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 its endpoints differ, i.e. if an odd number of crossovers have occurred along J, and parental otherwise. Assays exist for determining whether this is so, usually simply genotyping segregating markers at the endpoints of J. 8

9 A model for strand involvement The standard assumption here is No Chromatid Interference (NCI): each non-sister pair of chromatids is equally likely to be involved in each exchange, independently of the strands involved in other exchanges. NCI fits the available pretty well, but of course not perfectly. There are broader models, though not in general use. It is interesting to make up models with different strand involvement rules and explore their consequences. Formally, NCI says that the crossover process C is a Bernoulli thinning of the exchange process X with p=0.5, that is, we copy an exchange point to a crossover point on a meiotic product with probability 1/2, independently for all exchanges. 9

10 The relation between exchanges and crossovers Key to making inferences about unobserved exchanges from observable crossovers are the assumptions connecting them. NCI is our principal one, but before we look carefully at that, let s do draw some pictures and get an idea of the combinatorics involved. We ll begin by defining an interval J by segregating markers at its endpoints, and looking at possible meiotic products. Our interest is in whether a meiotic product is recombinant or parental across the interval, and how often this occurs. 10

11 Two exchanges (4/16 cases) 11

12 Exercise With Emerson & Rhodes, The American Naturalist, Vol 67 No 711 (Jul-Aug 1933): , extend this table to exchanges with 3, 4 or more exchanges in the interval, and conclude that in all cases apart from 0 exchanges, exactly one half of the meiotic products are parental, and one half recombinant. Develop a proof of their conclusion that under NCI, for n>0, pr(r(j) X(J) = n ) = 1/2, and so pr(r(j)) = 1/2 pr( X(J) > 0 ) (*) 12

13 Recombination fractions The recombination fraction across an interval J is defined to be r = pr(r(j)). Under NCI, (*) tells us that 0 r 1/2. In a sense it gives an indication of the chromosomal length of the interval J. Exercise. Prove that under NCI, r = r(j) is monotone in the length of J. However, the use of r as a metric must be limited, as it is bounded by 1/2. 13

14 The Poisson (no interference) model for exchanges Suppose that the exchange process X is a Poisson point process with mean measure λ. That is, for a chromosomal interval J, the distribution of X(J) is Poisson with mean λ(j), and exchanges across disjoint intervals are mutually independent. The crossover process C, the thinned version of X, is thus also a Poisson process, with mean measure d = λ/2. Further, under NCI, we have the formula r(j) = 1 2 (1 e λ(j ) ) = 1 2 (1 e 2d (J ) ). 14

15 Recombination fractions and mapping r 1 12 r r 13 r 13 r 12 + r 23 The recombination fraction does not define a metric. However,Sturtevant (1913) used recombination fractions to order genes. Exercise: Can we determine the order of three loci on a chromosome from their 3 pairwise recombination fractions. Can you extend this to any number of loci along a chromosome? 15

16 More on the Poisson model for exchanges Exercise: Prove that for the Poisson model, we have the following formula for loci in order 1-2-3: r 13 = r 12 + r 23-2r 12 r 23. Extend to 4 loci. Relations such as r = 1 2 (1 e 2d ) between observable recombination fractions and parameters d of unobservable exchange processes are called map functions. This one is known as Haldane s, after JBS Haldane who introduced it in

17 Multilocus recombination probabilities Let s forget the Poisson model for a moment. Suppose that we have 3 loci, along a chromosome. Write E ij for the event of at least one exchange between loci i and j during meiosis, and R ij for the event of recombination between loci i and j on a meiotic product. Now define probs p and q as follows, where - denotes the complementary event: q 00 = pr(e 12 & E 23 ) p 00 = pr(r 12 & R 23 ) q 01 = pr(e 12 & E 23 ) p 01 = pr(r 12 & R 23 ) q 10 = pr(e 12 & E 23 ) p 10 = pr(r 12 & R 23 ) q 11 = pr(e 12 & E 23 ) p 11 = pr(r 12 & R 23 ) 17

18 Exercise: Prove the following formulae under NCI, and extend to n> 3 loci. p 11 = 1 4 q 11 p 10 = 1 4 q q 10 p 01 = 1 4 q q 01 p 00 = 1 4 q q q 01 + q 00 Infer that we can order any number of loci under NCI from multilocus recombination fractions.can we order loci with these recombination fractions without NCI? 18

19 Map distance and mapping Map length: d(j) = E{C(J)} = av # crossovers in J Also called genetic map distance, the unit is the Morgan, after TH Morgan, but centimorgans are more commonly used. d 1 12 d d 13 Map distances are automatically additive (why?), i.e. d 13 = d 12 + d

20 Map distance and mapping d 1 12 d d 13 d 13 = d 12 + d 23 Genetic mapping or applied meiosis is 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. 20

21 The program from now on With these preliminaries, we turn now to the data and models in the literature which throw light on the chance aspects of meiosis. Mendel s law of segregation: a result of random sampling of meiotic products, with allele (variant) pairs generally segregating in precisely equal numbers. As usual in biology, there are exceptions. 21

22 22

Random spindle-centromere attachment at 1st meiotic division x smaller larger In 300 meioses in an grasshopper heterozygous for an inequality in the

23 Random spindle-centromere attachment at 1st meiotic division x smaller larger In 300 meioses in an grasshopper heterozygous for an inequality in the size of one of its chromosomes, the smaller of the two chromosomes moved with the single X 146 times, while the larger did so 154 times. Carothers,

24 Tetrads In some organisms - fungi, molds, yeasts - all four products of an individual meiosis can be recovered together in what is known as an ascus. These are called tetrads. The four ascospores can be typed individually. In some cases - e.g. N. crassa, the red bread mold - there has been one further mitotic division, but the resulting octads are ordered. 24

25 25

26 Meiosis data in which all exchanges are precisely located (from microarrays) Yeast Figure courtesy of J Derisi 26

27 Using ordered tetrads to study meiosis Data from ordered tetrads tell us a lot about meiosis. For example, we can see clear evidence of 1st and 2nd division segregation. We first learned definitively that normal exchanges occur at the 4-stand stage using data from N. crassa, and we can also see that random spindle-centromere attachment is the case for this organism. Finally, aberrant segregations can occasionally be observed in octads. 27

28 Meiosis in N.crassa 28

29 First-division segregation patterns 29

30 Second-division segregation patterns 30

31 Different 2nd division segregation patterns Under random spindle-centromere attachment, all four patterns should be equally frequent. 31

32 Lindegren s 1932 N. crassa data 32

33 2-strand double exchanges lead to FDS There is a nice connexion between the frequencies of multiple exchanges between a locus and its centromere and the frequency of 2nd division segregations at that locus. 33

34 A simple calculation and result Let F k (resp. S k ) denote the number of strand-choice configurations for k exchanges leading to first (resp. second) division segregation at a segregating locus. By simple counting (Exercise!) we find that F 0 =1 and S o = 0, while for k>0, F k+1 = 2S k, and S k+1 = 4F k + 2S k. Assuming NCI, the proportion s k of second-division segregants among meioses having k exchanges between our locus and the centromere is 34

35 s k = 2 3 [1 ( 1 2 ) k ], k > 0. If the distribution of the # of exchanges is (x k ), then the frequency of SDSs is s = x x x If the distribution is Poisson (2d) then we find s = 2 3 (1 e 3d ). This is a map-function: between the unobservable 35 map distance d and the observable SDS frequency s.

Advanced Algorithms and Models for Computational Biology -- a machine learning approach

Advanced Algorithms and Models for Computational Biology -- a machine learning approach Population Genetics: meiosis and recombination Eric Xing Lecture 15, March 8, 2006 Reading: DTW book, Meiosis Meiosis