Chi-Square Test of Independent Assortment in Drosophila
In a laboratory exercise, students cross two fruit flies that are heterozygous for two independently assorting genes: body colour (wild-type dominant to black) and wing shape (wild-type dominant to vestigial). Both parents have the wild-type phenotype. The students classify 440 offspring and obtain: 240 wild-type for both traits, 80 black body with normal wings, 85 wild-type body with vestigial wings, and 35 black body with vestigial wings. The instructor asks them to test the null hypothesis of independent assortment using a chi-square goodness-of-fit test at the 5% significance level. Critical values are 3.84 for 1 degree of freedom, 5.99 for 2, and 7.81 for 3. A classmate looks only at the double-mutant count and says, '35 is noticeably higher than the expected 27.5, so the genes must be linked.' The students must decide whether the data support linkage, and they must also explain what pattern would be expected if the genes were linked in coupling phase.
1. Calculate the expected counts under independent assortment and the chi-square statistic. State the degrees of freedom and the conclusion at the 5% level.
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For a dihybrid cross with independent assortment, the expected phenotypic ratio is 9:3:3:1. With 440 offspring:
[\text{wild both}=\frac{9}{16}(440)=247.5,] [\text{black normal}=\frac{3}{16}(440)=82.5,] [\text{wild vestigial}=\frac{3}{16}(440)=82.5,] [\text{black vestigial}=\frac{1}{16}(440)=27.5.]
The chi-square statistic is
[\chi^2=\sum\frac{(O-E)^2}{E}.]
[\chi^2=\frac{(240-247.5)^2}{247.5} +\frac{(80-82.5)^2}{82.5} +\frac{(85-82.5)^2}{82.5} +\frac{(35-27.5)^2}{27.5}.]
[\chi^2\approx0.227+0.076+0.076+2.045=2.42.]
There are 4 phenotypic classes, so degrees of freedom (=4-1=3). The critical value is 7.81. Since (2.42<7.81), we fail to reject the null hypothesis. The data are compatible with independent assortment at the 5% significance level.
2. Evaluate the classmate's claim that the elevated double-mutant count proves linkage.
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The classmate is overinterpreting one cell. The double-mutant count is 35 versus an expected 27.5, so it is indeed higher, but its contribution to chi-square is only
[\frac{(35-27.5)^2}{27.5}\approx2.05.]
That alone does not reach the critical value of 7.81 with 3 degrees of freedom, and the total chi-square is only 2.42. With an expected count of 27.5, random sampling noise is on the order of (\sqrt{27.5}\approx5.2), so an observed 35 is about 1.4 standard deviations above expectation-not strong evidence by itself. Also, a linkage explanation should fit the whole table, not just one class. Under coupling-phase linkage, the parental combinations (wild-type both and double mutant) would be overrepresented and the recombinant single-mutant classes underrepresented. Here the double mutant is somewhat high, but the wild-type class is slightly low and the single-mutant classes are only slightly low; the pattern is not a clean coupling signature. The correct conclusion is that the data do not provide significant evidence for linkage.
3. Suppose the double-mutant count had been 10 instead of 35, with the wild-type count 260 and the two single-mutant counts still 85 each. Recalculate chi-square and judge whether this new table would support linkage.
Read the worked answer
The new observed counts still total 440, so the expected counts are unchanged: 247.5, 82.5, 82.5, and 27.5. The statistic becomes
[\chi^2=\frac{(260-247.5)^2}{247.5} +\frac{(85-82.5)^2}{82.5} +\frac{(85-82.5)^2}{82.5} +\frac{(10-27.5)^2}{27.5}.]
[\chi^2\approx0.631+0.076+0.076+11.14=11.92.]
With 3 degrees of freedom, 11.92 exceeds the 5% critical value of 7.81, so we would reject independent assortment. This table would show a significant excess of wild-type offspring and a deficit of double mutants, with the single-mutant classes near expectation. That pattern is not the simple coupling-phase pattern, which would enrich both parental classes (wild-type and double mutant) and deplete the single mutants. A significant result would therefore justify further analysis-such as testing for viability differences or epistasis-rather than immediately concluding simple coupling linkage.