F2 Linkage Analysis
Analysis of F2 results
Estimation of recombination frequency
χ2 test of existence of linkage
Standard error of recombination freq.
Standard error of recombination freq.
F2 Linkage data
Coupling vs. repulsion:
Coupling - two dominant alleles are contributed by the same parent
(eg, AABB x aabb).
Repulsion - two dominant alleles are contributed from the opposite
parents (eg, AAbb x aaBB).
Coupling and repulsion have different ratios and different error rates in
the F2, so they have to be dealt with separately. Coupling will be first:
Genotype class
Symbol
A_B_
A_bb
aaB_
aabb
Number
128
20
16
36
Expected (no linkage)
112.5
37.5
37.5
12.5
a1
a2
a3
a4
Note there appear be deviations from 9:3:3:1. Can this be proven and
the recombination frequency quantified?
Calculation of Recomb. Freq. for AB
The formula for this calculation is somewhat complex. The method that
has been used is called “product ratio” method.
This has two steps 1) calculation of a ratio based on observed numbers,
then, 2) consult a table to translate the ratio to a recombination
frequency. The ratio, z, is calculated as follows for coupling:
Zc = (a2 x a3)/(a1 x a4) = (20 x 16)/(128 x 36) = 0.0694
From the table the value 19.2 appear under coupling next to Z = 0.07, the
closest to our value. Hence, r(%) = 19.2.
Repulsion F2 data: same loci as
previous example, A & B
Below is a repulsion linkage example:
Genotype class
Symbol
A_B_
A_bb
aaB_
aabb
Number
100
50
46
4
Expected (no linkage)
112.5
37.5
37.5
12.5
Zr = (a1xa4)/(a2xa3) = (100 x 4)/(50 x 46) = 0.174
From table, under coupling, r% = 26.6
a1
a2
a3
a4
Second example, F2 data: coupling
The complete analysis will be conducted on these data:
Genotype class
Symbol
A_B_
A_bb
aaB_
aabb
Number
1405
100
90
405
Expected (no linkage)
1125
375
375
125
Zc = (a2xa3)/(a1xa4) = (100 x 90)/(1405 x 405) = 0.0158
r = 0.099
a1
a2
a3
a4
Chi-square test for linkage
This is set up and can be calculated the
same way as for the test cross:
Observed data:
G enotype
B_
bb
Sum
A_
1405
100
1505
aa
90
405
495
1495
505
2000
Sum
Chi-square test; continued:
Expected numbers. E.g., For A_B_ the value is 1495x1505/2000 = 1125:
Genotype
B_
bb
Sum
A_
1125
380
1505
aa
370
125
495
1495
505
2000
B_
bb
Sum
Observed - expected:
Genotype
Sum
A_
280
-280
0
aa
-280
280
0
0
0
Sum
Chi-square (1 d.f.) > 1000; there is linkage.
Standard error of the recombination freq.
The table on the next slide can be used to obtain SEs of
recombination frequencies In the table the recombination
frequency, usually represented as r, is p.
Using the p of 0.099, the table value is 0.10189. SE =
sqrt(0.10189/2000) = 0.007.
Recomb. Freq. is estimated as 0.099 +/- 0.014 (95% confidence
interval).
Statistics for the previous F2 population
The standard errors and confidence limits for these data are:
Repulsion example - 200 individuals:
The chi-square test can be performed exactly as for the coupling example,
using the same method for calculating expected numbers, etc.
The SE of r utilizes p = 0.266 yielding a table value of 0.8386.
From this, SE = sqrt(0.8386/200) = 0.065.
r = 0.266 +/- 0.127 (95% confidence interval statement).
Coupling example - 200 individuals:
Here, p = 0.192 for a table (variance) value of 0.1975.
The SE = sqrt (0.1975/200) = 0.0009875/200 = .031.
r = 0.192 +/- 0.063
Note: For the coupling case, the confidence interval is 1/2 the size.
Expected class frequencies for linkage in
test cross and F2 populations
Genotype
Test cross:
AB/ab
Ab/ab
aB/ab
ab/ab
Coupling
Repulsion
0.5(1 – r)
0.5(r)
0.5(r)
0,5(1 – r)
0.5(r)
0.5(1 – r)
0.5(1 – r)
0.5(r)
0.25(3-2r-r2)
0.25(2r-r2)
0.25(2r-r2)
0.25(1-r)2
0.25(2-r2)
0.25(1-r2)
0.25(1-r2)
0.25r2
F2:
A_B_
A_bb
aaB_
aabb