Using Punnett Squares to Prove Segregation
We can use the letter symbol ‘D’ to represent the normal leaf growth allele and ‘d’ to represent that alternative version causing disease lesion mimics.
The homozygous normal parent had two copies of the ‘D’ allele. Therefore it had the genotype DD. The homozygous disease lesion mutant was the genotype dd. Therefore the first cross could be depicted as follows:
True breeding parents are homozygous. Paired genes are separated to form gametes. When crossed, all the F1s produced were the same. (Image by P. Hain)
The paired genes separated when the parents made gametes. Since both parents were homozygous at this gene pair, they each made only one kind of gamete. When these gametes came together only one genotype combination was possible, Dd or the heterozygous genotype.
The F1 plants had the same appearance or phenotype as the normal parent but our hypothesis predicts that they carry the ‘d’ allele. If this is true they must mask its presence with the ‘D’ allele. Thus we are proposing that the normal leaf allele, ‘D’ is dominant over the recessive ‘d’ allele. The only way to test the hypothesis that the ‘d’ allele is present but masked in the F1 is to obtain evidence that the F1s could pass the ‘d’ allele on to their progeny. The F2 data confirmed the hypothesis.
The occurrence of the mutant phenotype among the F2 progeny is evidence that the F1 plants had the ‘d’ allele and passed it on. The 3:1 phenotype ratio observed in the F2 is consistent with the principle of segregation.
Segregation of paired genes in heterozygous F1s to produce a 3:1 phenotype and 1:2:1 genotype ratio. (Image by P. Hain)
According to the hypothesis of segregation, all three possible genotypes should be produced in the F2 generation. What should the genotypic ratio be? How can we determine a plants genotype? The Punnet square predicts a 1:2:1 genotype ratio for the DD,Dd and dd F2s. Because we can observe phenotype and not genotype in this experiment, we need to attempt to indirectly verify this ratio. Again this can be accomplished by observing the F3 progeny.
The data table gives the results of selfing the F2s with the normal phenotype. Two different types of normal F2s were again observed. Homozygous F2s would give rows of all normal F3s and heterozygous F2s gave mixed rows of both normal and mutant F3s. What ratio would we expect for these two types of rows? Again, the Punnet square predicts a 1:2 ratio. Do the results support this? Yes, the mixed rows were in the majority and if the numbers are combined from both reciprocal crosses, normal F2 plants gave 37 F3 rows that were all normal and 62 rows that were mixed. These numbers agree with what the principle of segregation predicts.