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12. Models of Recombination
                                                                                            259
                              distribution with mean λ. Then a gamete is recombinant with probability
                                                                            1
                               1
                                     −λ
                                                                                  −λ
                                (1 − e
                                                                             (1 + e
                                                                                    ). Haldane’s
                                       ) and nonrecombinant with probability
                               2
                                                                            2
                              model, which postulates that the chiasma process is Poisson, turns out to
                              be crucial in generalizing Mather’s formula [17, 24]. To derive this gener-
                              alization, we employ a randomization technique pioneered by Schr¨odinger.
                                Consider a sequence of k+1 loci along a chromosome. The k+1 loci define
                              k adjacent intervals I 1 ,... ,I k . A gamete can be recombinant on some of
                              these intervals and nonrecombinant on others. For a subset S ⊂{1,...,k},
                              let y S denote the probability that the gamete is recombinant on each of the
                              intervals I i indexed by i ∈ S and nonrecombinant on each of the remaining
                              intervals I i indexed by i/∈ S. Under Haldane’s model, the numbers of
                              chiasmata falling on disjoint intervals are independent. Therefore,
                                                        1             1
                                             y S  =      (1 − e −λ i )  (1 + e −λ i ),    (12.3)
                                                        2             2
                                                     i∈S          i/∈S
                              where λ i denotes the expected number of chiasmata on interval I i .
                                                                                   = n i of chias-
                                Next consider what happens when we fix the number N I i
                              mata occurring on each interval I i . The probability of the recombination
                              pattern dictated by S now changes to the conditional probability y n S ,
                              where n is the multi-index (n 1 ,...,n k ). We recover y S via
                                                                  k
                                                          	         λ n i
                                                                     i
                                                  y S  =     y n S     e −λ i .           (12.4)
                                                                    n i !
                                                           n     i=1
                                                                                  k
                              Equating formulas (12.3) and (12.4) and multiplying by  e , we de-

                                                                                      λ i
                                                                                  i=1
                              duce that
                                              k                 k
                                                                   i
                                                        	         λ n i
                                                e λ i  =
                                           y S             y n S
                                                                   n i !
                                             i=1         n     i=1
                                                          1  k

                                                     =          (e λ i  − 1)  (e λ i  +1).  (12.5)
                                                          2
                                                             i∈S        i/∈S
                              We extract the coefficient y nS by evaluating the partial derivative
                                                       k          k

                                                    ∂  i=1  n i
                                                               y S  e λ i
                                                  ∂λ n 1  ··· ∂λ n k
                                                    1      k     i=1
                              at λ 1 = ··· = λ k = 0. Equating the two results from identity (12.5) yields
                                                    1  k

                                         y n S  =         (1 − 1 {n i =0} )  (1+1 {n i =0} )
                                                    2
                                                        i∈S           i/∈S
                                                    1  k

                                               =          (−1)  |S∩T |        ,
                                                                   1
                                                    2               {  i∈T  n i =0}
                                                        T
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