Single Coupling

Simplex Repulsion

One Marker In Duplex Coupling And One Marker In Simplex Coupling

Duplex-duplex Linkage In Coupling Phase

Duplex-duplex Linkage In Coupling Phase

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

 X          Y(A)   X           Y(C)  
o                   o                Case 1
                                     A+B
o            (B)   o             (D) C+D
 x             y   x              Y  

 

 X          Y(A)   x           y(B)  
o                   o                 Case 2
                                      A+C
o                   o                 B+D
 X         Y (C)   x           y(D)  

 

 X          Y(A)   X          Y(C)  
o                   o                Case 3
                                      A+D
o                   o                 B+C
 x           y(D)   x           Y(B)    

Case 1 - gametes and Case 3-gametes.


 

Case 2 - gametes

  ½ XY ½ XY
½ xy
¼ XxYy ¼ XxYy
¼ XxYy ¼ XxYy
½ xy

We can now summarize the probabilities of each type of gamete, using probabilities of 1/3 for each of the three cases.

Gametes Probability
XXYY
XXYy
XXyy
XxYY
XxYy
Xxyy
xxYY
xxYy
xxyy

Now we must collect the nine gametic classes into four classes due to dominance gene action. With dominance gene action we can only distinguish between four classes.

Gametes Probability Observed Number
X_Y_ 1/12 [10 - 4r + 2r2] a
X_yy 1/12 [4r - 2r2] b
xxY_ 1/12 [4r - 2r2] c
xxyy 1/12 (2 - 4r + 2r2) d

Now we need to double check the results because the four probabilities should sum to unity.

Next we find the likelihood for r.

Now we find the multinomial expression for the likelihood for the observed data and solve to find the maximum likelihood estimate of r.

The quadratic equation is in the form

and has the solution

In this case a = bcd + abc + acd + abd
In this case b = -2(bcd + abc + acd + abd)
In this case c = 5bcd + bca

To derive the same results as Hackett et al. We substitute

Y is an intermediate variable
Y is an5-2r + r2 = 4 + y
Y is an2r - r2 = 1 - y
Y is an1 - 2r2 + r = y

so we can substitute

n = a + b + c + d;

The quadratic equation is in the form

with the solution

In our case a = -n, b = (a - 4b - 4c - 3d), c = 4d

Copyright 2000©, Ted Helms

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