A Testcross That Refuses to Give 1:1:1:1
In a maize genetics practical, a student testcrosses a double heterozygote AaBb with aabb. A is the dominant allele for coloured kernels, a gives colourless; B is dominant for smooth kernels, b gives wrinkled. The student counts 100 offspring: coloured, smooth 42; colourless, wrinkled 38; coloured, wrinkled 10; colourless, smooth 10. The student expected a 1:1:1:1 ratio and runs a chi-square test with three degrees of freedom; the 5% critical value is 7.815. A lab partner says the deviation is probably just chance because the sample size is only 100, and that the two genes must assort independently. The student also notices that the two largest classes are the parental combinations coloured-smooth and colourless-wrinkled. The instructor asks for the recombination frequency and whether the heterozygous parent carried the dominant alleles in coupling or repulsion.
1. Carry out the chi-square test and decide whether the data support independent assortment. Evaluate the lab partner’s “small sample” objection.
Read the worked answer
Expected count for each of the four classes under independent assortment = 100/4 = 25.
χ2 = (42 − 25)2/25 + (38 − 25)2/25 + (10 − 25)2/25 + (10 − 25)2/25 = 289/25 + 169/25 + 225/25 + 225/25 = 11.56 + 6.76 + 9.00 + 9.00 = 36.32.
With df = 3, the 5% critical value is 7.815. Since 36.32 ≫ 7.815, the null hypothesis of independent assortment is rejected; the deviation is far too large to be explained by ordinary sampling error.
The partner’s objection is weak for two reasons. First, the chi-square test already accounts for sample size: a sample of 100 is not automatically unreliable if the effect is large. Second, the pattern is not random scatter. Two complementary classes are large and two are small, which is the signature of linkage rather than chance. A small sample would produce noisy counts around 25:25:25:25, not a consistent two-large/two-small pattern.
2. Calculate the recombination frequency and determine whether the dominant alleles are in coupling or repulsion.
Read the worked answer
The recombinant classes are the two small classes: coloured-wrinkled (A−bb) and colourless-smooth (aaB−), each with 10 offspring. Recombination frequency = (10 + 10)/100 = 20%, so the loci are about 20 map units apart.
The large parental classes are coloured-smooth (A−B−) and colourless-wrinkled (aabb). In the testcross parent AaBb, the gametes that produced these parental classes carried AB and ab. Therefore the dominant alleles A and B are on the same homolog, and the recessive alleles a and b are on the other. This is coupling, or cis configuration, not repulsion. In repulsion, the parental classes would be Ab and aB, and the large offspring classes would be coloured-wrinkled and colourless-smooth.
3. If the genes were 20 cM apart, predict the expected counts in 200 offspring and explain why the observed 42:38:10:10 in 100 offspring is consistent with that prediction.
Read the worked answer
At 20 cM, the recombinant frequency is 0.20, so each recombinant class has frequency 0.10 and each parental class has frequency 0.40.
In 200 offspring, expected counts are: parental coloured-smooth = 0.40 × 200 = 80; parental colourless-wrinkled = 80; recombinant coloured-wrinkled = 0.10 × 200 = 20; recombinant colourless-smooth = 20.
Doubling the observed 100-offspring counts gives 84, 76, 20, and 20. These are close to 80, 80, 20, and 20, with the differences attributable to sampling. The observed data therefore fit a 20 cM linkage model much better than the 1:1:1:1 independent-assortment model, whose expected counts would be 50, 50, 50, 50 in 200 offspring. The recombination frequency estimate is 20%, and the parental classes identify the coupling phase.