Question

A pea plant homozygous for yellow round seed is crossed with its recessive parents. Calculate the phenotypic and genotypic ratio with the help of a checkerboard.

Answer 1

Mendel started with pure lines. Therefore, the genotype of a parent with yellow round seeds in YYRR and that of a parent with green wrinkled seeds is yyrr. Both the parents are homozygous. Therefore, they would produce only one type of gamete i.e. YR and yr respectively. All F1 dihybrid seeds resulting from the cross will be heterozygous for both the traits and with genotype YyRr. Due to dominance, all the seeds of the F1 generation will be yellow round.

Mendel allowed the selfing of F1 dihybrids. During gamete formation by the dihybrids, the alleles in both pairs separate (law of segregation). Each gamete will receive only one allele from each pair. A gamete that receives ‘Y’ for colour may receive ‘R’ for shape or ‘r’ for shape. This would result in the formation of YR and yr types of gametes. Similarly, a gamete that receives ‘y’ for colour may receive ‘R’ or ‘r’ for shape. This would result in the formation of yR and yr types of gametes (Independent assortment). Thus F1 dihybrid would produce four different types of gametes in equal proportion. (Each type will be 25%).
There would be four types of male gametes and four types of female gametes and random fusion will take place during selfing. Due to this chance fusion (4 x 4 = 16) sixteen combinations which fall into nine categories as shown in Punnett square are possible.
The nine different genotypes are; YYRR, YYRr, YyRR, YyRr, Yyrr, yyRR, yyRr and yyrr.
The genotypic ratio is 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1. Due to dominance there are only four phenotypes. The phenotypic ratio 9 : 3 : 3 : 1, is called the dihybrid ratio.