The degree of freedom associated with this chi-square value also is 2 — 1 = 1, and the associated probability is between .5 and .3. We again assume that there is no significant difference between what we observed and what we expected at this locus in the testcross.
Testing ratios for independent assortment We are now ready to test for the independent assortment of genes at the two loci. If the genes are assorting independently, we can use the multiplication rule to obtain the probabilities and numbers of progeny inheriting different combinations of phenotypes:
Expected Expected Geno- pheno- propor- Expected Observed types types tions numbers numbers y+y cv+ cv brown, 1/2 X 1/2 = 50 63
The observed and expected numbers of progeny can now be compared by using the chi-square test:
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