The standard genetic test for linkage is a dihybrid testcross. Consider a general cross of that type, in which it is not known if the genes are linked or not:
A/a · B/b × a/a · b/b
If there is no linkage, that is, the genes assort independently, we have seen from the discussions in this chapter and Chapter 3 that the following phenotypic proportions are expected in progeny:
|
A B |
0.25 |
|
A b |
0.25 |
|
a B |
0.25 |
|
a b |
0.25 |
A cross of this type was made and the following phenotypes obtained in a progeny sample of 200.
|
A B |
60 |
|
A b |
37 |
|
a B |
41 |
|
a b |
62 |
There is clearly a deviation from the prediction of no linkage (which would have given the progeny numbers 50:50:50:50). The results suggest that the dihybrid was a cis configuration of linked genes, A B/a b, because the progeny A B and a b are in the majority. The recombinant frequency would be (37 + 41)/200 = 78/200 = 39 percent, or 39 m.u.
However, we know that chance deviations due to sampling error can provide results that resemble those produced by genetic processes; hence, we need the χ2 (pronounced “chi square”) test to help us calculate the probability of a chance deviation of this magnitude from a 1:1:1:1 ratio.
First, let us examine the allele ratios for both loci. These are 97:103 for A : a, and 101:99 for B : b. Such numbers are close to the 1:1 allele ratios expected from Mendel’s first law, so skewed allele ratios cannot be responsible for the quite large deviations from the expected numbers of 50:50:50:50.
We must apply the χ2 analysis to test a hypothesis of no linkage. If that hypothesis is rejected, we can infer linkage. (We cannot test a hypothesis of linkage directly because we have no way of predicting what recombinant frequency to test.) The calculation for testing lack of linkage is as follows:
|
Observed (O) |
Expected (E) |
O–E |
(O – E)2 |
(O – E)2/E |
|---|---|---|---|---|
|
60 |
50 |
10 |
100 |
2.00 |
|
37 |
50 |
–13 |
169 |
3.38 |
|
41 |
50 |
–9 |
81 |
1.62 |
|
62 |
50 |
12 |
144 |
2.88 |
|
|
χ2 = Σ (O – E)2/E for all classes = 9.88 |
|||
151
Since there are four genotypic classes, we must use 4 − 1 = 3 degrees of freedom. Consulting the chi-
Notice, in retrospect, that it was important to make sure alleles were segregating 1:1 to avoid a compound hypothesis of 1:1 allele ratios and no linkage. If we rejected such a compound hypothesis, we would not know which part of it was responsible for the rejection.