Single Coupling

Simplex Repulsion

One Marker In Duplex Coupling And One Marker In Simplex Coupling

Duplex-duplex Linkage In Coupling Phase

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Single Coupling

Hacket, C.A. et al. 1998. Linkage analysis in tetraploid species: a simulation study. Genet. Res. Comb. 71:143-154.

Let X be the dominant allele at one locus and Y be the dominant allele at a second locus. The four homologous chromosomes have the following genotypes:

 XY
xy
xy
xy

The XY chromosome must pair with an xy chromosome. Double reduction gametes are not considered. Only bivalent pairing is assumed. The probability of a crossover event for one bivalent is r.

The only observable recombination is between the xy chromosome that pairs with the XY chromosome. If we consider recombination for the bivalent in coupling phase linkage, we have the following gametic series.
(1-r)/2 XY; (1-r)/2 xy; r/2Xy:r/2XY

Let a, b, c, d, be the observed numbers of progeny in the XY, Xy, xY, and xy classes, respectively. The solution to the proportion of observed recombination, 'r' is solved by maximum likelihood. The probability of observing a, b, c, and d of each class is:

P ( a,b,c,d; 1-r , r , r , 1-r )
2   2   2   2

=    n!    ( 1-r ) a ( r ) b ( r ) c ( 1-r ) d
a!b!c!d! 2   2   2   2  

We can solve for r by taking the derivative of the log of this function with respect to r and setting the derivative equal to zero.

The log - likelihood, L, is given by:

L(r)=log [    n!    ] +a log (1-r) +b log ( r ) + c log ( r ) + d log (l-r)
a!b!c!d! 2 2 2 2

d[L(r)] = 0 +    -a    +  b  +   c   +    -d     
dr (1-r)/2 r/2 r/2 (1-r)/2  

0= b + c - (a + d)
r/2 (1 - r)/2    

(b + c)(1 - r) = (a + d)r
2 2  

  a + d   = b + c
(1 - r)/2 r/2

(b +c)(1 - r) = (a + d)r
2 2

b + c - br - cr = ar + dr
b + c = ar + br + cr + dr
b + c = r(a+b+c+d)
b + c = rn, because a+b+c+d = n

b + c = r; which is the solution.
n

Copyright 2000©, Ted Helms

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