Testing Independent Assortment with Too Few F2 Plants
A breeder working with a self-pollinated pulse crosses two true-breeding parents differing for plant height (tall vs dwarf) and flower colour (purple vs white). All F1 plants are tall with purple flowers. In the F2, 200 plants are scored: 118 tall-purple, 38 tall-white, 34 dwarf-purple and 10 dwarf-white. A chi-square test is run and the breeder concludes that the two loci assort independently, so she abandons the linkage hypothesis and starts selecting dwarf-purple plants for release. Her target is a true-breeding dwarf-purple line. The crop shows 3-5% natural outcrossing, and the F1 seed was produced in a block only 30 m from another trial of the same species. One dwarf-purple F2 plant she likes is being multiplied directly from open-pollinated seed.
1. Compute the chi-square test for a 9:3:3:1 ratio, state the degrees of freedom, and evaluate whether the breeder's conclusion of independent assortment is justified.
Read the worked answer
Total = 200, so the expected numbers are 9/16 x 200 = 112.5, 3/16 x 200 = 37.5, 37.5 and 12.5.
chi-square = (118-112.5)^2/112.5 + (38-37.5)^2/37.5 + (34-37.5)^2/37.5 + (10-12.5)^2/12.5 = 30.25/112.5 + 0.25/37.5 + 12.25/37.5 + 6.25/12.5 = 0.269 + 0.007 + 0.327 + 0.500 = 1.10.
Four phenotypic classes minus one = 3 degrees of freedom; the 5% critical value is 7.81, and the 1% value is 11.34. Since 1.10 is far below 7.81, the data are compatible with 9:3:3:1 and the null hypothesis is not rejected.
But not rejecting is not the same as proving independence. The test has low power at n = 200: expected frequencies under moderate linkage differ from 9:3:3:1 by only a few plants, which is well inside sampling error. With 200 plants, the standard error of an F2-based estimate of recombination is large, and the F2 is a weak instrument for linkage. The correct move is to run a testcross, crossing F1 plants to a dwarf-white parent; with 200 testcross progeny the standard error of the recombination fraction is sqrt(r(1-r)/n), about 2.8% for r = 0.20, which is precise enough to detect or exclude moderate linkage. A chi-square test on pooled F2 data is not, by itself, a linkage test.
2. How many F2 plants must be scored to be 99% sure of finding at least one true-breeding dwarf-purple individual, and how would the breeder confirm which of the 34 dwarf-purple plants are homozygous at the flower-colour locus?
Read the worked answer
A true-breeding dwarf-purple plant must be homozygous recessive at the height locus and homozygous dominant at the colour locus, i.e. ddPP. Its frequency in the F2 is 1/16 = 0.0625. The probability of finding at least one in n plants is 1 - (15/16)^n. Setting this to 0.99: (15/16)^n = 0.01; n = ln(0.01)/ln(0.9375) = (-4.605)/(-0.0645) = 71.4, so 72 plants is the theoretical minimum.
That figure is a floor, not a plan. It assumes full germination, no outcrossing, no misclassification and no seedling loss. With 3-5% outcrossing the effective frequency of genuine selfed homozygotes falls, so a working population of 100-150 F2 plants is prudent; her 200 plants already provide a comfortable margin.
The 34 dwarf-purple plants are a mixture: on average 1/3 (about 11) are ddPP and 2/3 (about 23) are ddPp. Phenotype cannot separate them. Confirm each by a progeny test: cross or self each plant and score 15-20 progeny for white flowers. A ddPP plant gives no white progeny; a ddPp plant gives roughly 1 in 4 white, so 15 progeny give a 98.7% chance of detecting a heterozygote (1 - 0.75^15). If the breeder instead selfs the plant directly, an open-pollinated bulk is unacceptable because of outcrossing - bag the inflorescences. The 10 dwarf-white plants are the check on the population: close to the expected 12.5, they give no evidence of contamination, but they do confirm that both recessive alleles are present in the block.
3. What are the practical consequences of the 30 m isolation distance and 3-5% outcrossing for the F1 seed lot and for the line the breeder is about to release?
Read the worked answer
For the F1 seed lot: the cross was made in a block only 30 m from another trial of the same species, with 3-5% outcrossing in the crop. Pollen from the neighbouring trial could have fertilized some flowers on the female parent. The consequence is a low frequency of plants in the F1 that are not true F1s. Because the F1 generation is uniform for the dominant alleles, such contamination stays hidden until segregation in the F2, where it shows up as an excess of parental types and distorted ratios - which would also bias any linkage conclusion drawn later.
For the release: the breeder is multiplying a selected F2 plant from open-pollinated seed. At 4% outcrossing, roughly 4 in every 100 progeny have a foreign pollen parent, and the line will not breed true. Any variety released must be built from progeny-tested, selfed or bagged families, with rogueing of off-types over at least two generations.
Recommendations: bag or cage the inflorescences at the F1 and F2 multiplication stages; increase the isolation distance to the norm for breeder seed of the crop, or use a temporal barrier so the neighbouring trial is not flowering at the same time; keep a voucher sample of the F1 seed to check later; and recognise that the 30 m block tells you the F1 was probably contaminated at a low rate, so the F2 segregation data should be verified in a properly isolated repeat before any linkage claim is accepted.