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.
| |
( |
1-r |
) |
XY |
( |
1-r |
) |
xy |
( |
r |
) |
Xy |
( |
r |
) |
xY |
| 2 |
2 |
2 |
2 |
| ( |
1-r |
) |
XY |
|
| 2 |
| ( |
1-r |
) |
xy |
| 2 |
| ( |
r |
) |
Xy |
| 2 |
| ( |
r |
) |
xY |
| 2 |
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 |
2 |
( |
1-r |
) |
2 |
|
|
|
|
|
|
|
|
|
|
| 3 |
2 |
|
|
|
|
|
|
|
|
| XXYy |
2 |
( |
1-r |
) |
( |
r |
) |
+ |
2 |
( |
1-r |
) |
( |
r |
) |
| 3 |
2 |
2 |
3 |
2 |
2 |
| XXyy |
2 |
( |
r |
) |
2 |
|
|
|
|
|
|
|
|
|
|
| 3 |
2 |
| XxYY |
2 |
( |
1-r |
) |
( |
r |
) |
+ |
2 |
( |
1-r |
) |
( |
r |
) |
| 3 |
2 |
2 |
3 |
2 |
2 |
| XxYy |
2 |
( |
1-r |
) |
2 |
+ |
2 |
( |
1-r |
) |
2 |
+ |
2 |
( |
r |
) |
2 |
+ |
2 |
( |
r |
) |
2 |
+ |
1 |
(1) |
| 3 |
2 |
3 |
2 |
3 |
2 |
3 |
2 |
3 |
| Xxyy |
2 |
( |
1-r |
) |
( |
r |
) |
+ |
2 |
( |
1-r |
) |
( |
r |
) |
|
|
|
|
|
|
|
|
|
|
|
| 3 |
2 |
2 |
3 |
2 |
2 |
| xxYY |
2 |
( |
r |
) |
2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
| 3 |
2 |
| xxYy |
2 |
( |
1-r |
) |
( |
r |
) |
+ |
2 |
( |
1-r |
) |
( |
r |
) |
|
|
|
|
|
|
|
|
|
|
|
| 3 |
2 |
2 |
3 |
2 |
2 |
| xxyy |
2 |
( |
1-r |
) |
2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
| 3 |
2 |
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.
| 1 |
[10+2-4r+4r+4r-4r+2r2-2r2-2r2+2r2] = |
12 |
= 1 |
|
|
| 12 |
12 |
|
|
Next we find the likelihood for r.
| L(r)= |
n! |
[ |
5-2r+r2 |
] |
a |
[ |
2r-r2 |
] |
b |
x |
[ |
2r-r2 |
] |
c |
[ |
1-2r+r2 |
] |
d |
| a!b!c!d! |
6 |
|
6 |
|
6 |
|
6 |
|
Now we find the multinomial expression for the likelihood for the observed
data and solve to find the maximum likelihood estimate of r.
| Log[L(r)]=log |
( |
n! |
) |
+a log |
( |
5-2r+r2 |
) |
/6+b log |
( |
2r-r2 |
) |
| a!b!c!d! |
| |
/6+c log |
( |
2r-r2 |
) |
/6+d log |
( |
1-2r+r2 |
) |
/6 |
|
| d |
Log[L(r)]= |
a(2r-2) |
+ |
b(2-2r) |
+ |
c(2-2r) |
+ |
d(2r-2) |
=0 |
| dr |
5-2r+r2 |
2r-r2 |
2r-r2 |
1-2r+r2 |
| r2(bcd+abc+acd+abd)+r(-2bcd-2abc-2acd-2abd)+5bcd+bca=0 |
The quadratic equation is in the form
| ax2+bx+c=0 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 |
| |
5-2r + r2 = 4 + y |
| |
2r - r2 = 1 - y |
| |
1 - 2r + r2 = y |
| L(r)= |
( |
4+y |
) |
a |
( |
1-y |
) |
b |
( |
1-y |
) |
c |
( |
y |
) |
d |
| 6 |
|
6 |
|
6 |
|
6 |
|
| Log[L(r)]=a log |
( |
4+y |
) |
+b log |
( |
1-y |
) |
|
|
|
|
|
|
|
|
| 6 |
6 |
|
|
| +c log |
( |
1-y |
) |
+d log |
( |
y |
) |
|
|
|
|
|
|
|
|
| 6 |
6 |
| d log[L(r)] |
= |
a(1) |
+ |
b(-1) |
+ |
c(-1) |
+ |
d(1) |
=0 |
| dr |
(4+Y)/6 |
(1-y)/6 |
(1-y)/6 |
y/6 |
| a |
- |
b |
- |
c |
+ |
d |
=0 |
|
|
| (4+y) |
(1-y) |
(1-y) |
y |
|
|
so we can substitute
n = a + b + c + d; -ny2 +(a - 4b - 4c -3d)y + 4d = 0
The quadratic equation is in the form
ax2 + bxc =0 with the solution
In our case a = -n, b = (a - 4b - 4c - 3d), c = 4d
