Chi-Square, Testcross and a Three-Season Chili Breeding Plan
A plant breeder crosses two true-breeding chili lines. Line A is powdery-mildew resistant but low-yielding; Line B is susceptible but high-yielding. All F1 plants are resistant and high-yielding. Selfing the F1 produces 240 F2 plants scored as 128 resistant high-yielding, 52 resistant low-yielding, 44 susceptible high-yielding, and 16 susceptible low-yielding. The breeder expected a 9:3:3:1 ratio and wants to know whether the deviation is meaningful. A testcross of the same F1 to a susceptible low-yielding tester gives 96 progeny: 24 resistant high-yielding, 26 resistant low-yielding, 23 susceptible high-yielding, and 23 susceptible low-yielding. The breeder has no molecular markers, one growing season per year, greenhouse space for 300 F2 plants next season, and wants a true-breeding resistant high-yielding line within three seasons.
1. Carry out a chi-square test for the F2 data and state whether the 9:3:3:1 expectation is rejected. Show expected counts, the statistic, degrees of freedom, and your decision at p = 0.05.
Read the worked answer
Total F2 = 240. For 9:3:3:1, expected counts are:
- resistant high: 240 × 9/16 = 135
- resistant low: 240 × 3/16 = 45
- susceptible high: 240 × 3/16 = 45
- susceptible low: 240 × 1/16 = 15
Chi-square = (128−135)²/135 + (52−45)²/45 + (44−45)²/45 + (16−15)²/15 = 49/135 + 49/45 + 1/45 + 1/15 = 0.363 + 1.089 + 0.022 + 0.067 = 1.54.
Degrees of freedom = 4 classes − 1 = 3. The critical value at p = 0.05 is 7.815. Since 1.54 < 7.815, the 9:3:3:1 expectation is not rejected. The observed deviation is compatible with chance. Note that failure to reject does not prove independent assortment; it only means this sample does not provide evidence against it.
2. The testcross gave 24, 26, 23 and 23 progeny in the four classes. Test the 1:1:1:1 expectation and explain what the testcross adds beyond the F2 test.
Read the worked answer
Total testcross progeny = 96, so the expected number in each class is 24. Chi-square = (24−24)²/24 + (26−24)²/24 + (23−24)²/24 + (23−24)²/24 = 0 + 4/24 + 1/24 + 1/24 = 0.25. Degrees of freedom = 3, and 0.25 is far below 7.815, so the 1:1:1:1 expectation is not rejected. The testcross directly samples the gametes produced by the F1, because the tester contributes only recessive alleles. It is therefore a more direct check for linkage than the F2 ratio, which depends on both gamete frequencies and fertilisation combinations. Here both tests agree with independent assortment. If the testcross had shown an excess of parental classes and a deficit of recombinant classes, linkage would be indicated even if the F2 chi-square had not rejected 9:3:3:1.
3. Design a breeding plan to obtain a true-breeding resistant high-yielding line within three seasons, using 300 F2 plants and no molecular markers. Include expected numbers and the role of progeny testing.
Read the worked answer
Grow the 300 F2 plants and select resistant high-yielding individuals. Expect about 300 × 9/16 = 169 such plants. Of these, only 1/9 are expected to be homozygous at both loci (RRHH), about 19 plants; the rest carry at least one heterozygous locus. Because resistant high-yielding F2 plants cannot be distinguished by phenotype alone, self each selected plant and keep its seed separately. In the next season, grow F3 families from the selections. A family that does not segregate for either resistance or yield is true-breeding; discard families that segregate. Keep several non-segregating families to guard against loss and to maintain genetic diversity. In the third season, increase seed from the best non-segregating family. Progeny testing is essential because about 8/9 of the selected F2 plants are not fully homozygous. Keeping roughly 25–30 F2 selections gives a high chance of recovering at least one RRHH line.