Homework SourseThe following depicts two sets of data obtained from testcro
The following depicts two sets of data obtained from testcross experiment asking whether genes A and B are liked. Conduct a chi square analysis to see if these genes are linked. The null hypothesis is that the two genes are unlinked.
Progeny Experiment 1 Experiment 2
AB 17 34
ab 14 28
Ab 8 16
aB 11 22
Total 50 100
| Class | Observed (exp 1) | Expected (exp 1) | Observed (exp 2) | Expected (exp2) | |
| Parents | 31 | 25 | 62 | 50 | |
| Recombinants | 19 | 25 | 38 | 50 |
Solution
Experiment 1 result: Experiment 2 result:
AB 17 34 ------------ Parents
ab 14 28 ------------ Parents
Ab 8 16 ------------ Recombinant
aB 11 22 ------------ Recombinant
Recombination frequency (expt 1) = Recombinants (expt 1) / Total
= (8+11) /50 = 19 / 50 = 38%
Recombination frequency (expt 2) = Recombinants (expt 2) / Total
= (22+16) /100 = 38 / 100 = 38%
As Recombination frequency is less than 50%, we need to perform Chi Square (X2) test.
Expt 1 Chi square test:
Parents
(AB & ab)
Recombinants
(Ab & aB)
Total X2 = 2.88
Expt 2 Chi square test:
Parents
(AB & ab)
Recombinants
(Ab & aB)
Total X2 = 2.88
The probability is less than 5%, so the hypothesis of independent assortment must be rejected. Thus, having rejected the hypothesis of no Linkage.
Inference: The two loci (A & B) are linked. Null hypothesis is proved to be wrong.
| Class | Observed (O) | Expected (E) | O - E | (O - E)2 | (O - E)2 / E |
| Parents (AB & ab) | 31 | 25 | 31 - 25 = 6 | (6)2 = 36 | 36 / 25 = 1.44 |
| Recombinants (Ab & aB) | 19 | 25 | 19 - 25 = -6 | (-6)2 = 36 | 36 / 25 = 1.44 |
