Question

As in the plants of Problem \(6,\) color may be red, white, or pink and flower shape may be personate or peloric. Determine the \(\mathrm{P}_{1}\) and \(\mathrm{F}_{1}\) genotypes for the following crosses: (a) red, peloric \(\times\) white, personate \(\mathrm{F}_{1}:\) all pink, personate (b) red, personate \(\times\) white, peloric \(\mathrm{F}_{1}:\) all pink, personate (c) pink, personate \(\times\) red, peloric $F_{1:}\left\\{\begin{array}{l}1 / 4 \text { red, personate } \\ 1 / 4 \text { red, peloric } \\ 1 / 4 \text { pink, personate } \\ 1 / 4 \text { pink, peloric }\end{array}\right.$ (d) pink, personate \(x\) white, peloric $\mathbf{F}_{1:}\left\\{\begin{array}{l}1 / 4 \text { white, personate } \\ 1 / 4 \text { white, peloric } \\ 1 / 4 \text { pink, personate } \\ 1 / 4 \text { pink, peloric }\end{array}\right.$ (e) What phenotypic ratios woud result from crossing the \(\mathrm{F}_{1}\) of (a) to the \(F_{1}\) of \((b) ?\)

Short answer

In conclusion, when crossing plants with different flower shapes and colors, the parental genotypes and first filial genotypes are as follows: (a) P1: RRss & rrSS, F1: RrSs (b) P1: RRSS & rrss, F1: RrSs (c) P1: RrSs & RRss, F1: 1/4 RRSs, 1/4 RRss, 1/4 RrSs, 1/4 Rrss (d) P1: RrSs & rrss, F1: 1/4 rrSs, 1/4 rrss, 1/4 RrSs, 1/4 Rrss (e) When crossing the F1 of (a) to the F1 of (b), the phenotypic ratios are 1 red, personate: 1 red, peloric: 2 pink, personate: 2 pink, peloric: 2 white, personate: 1 white, peloric.

Step-by-step solution

Determine P1 genotypes

The parents are: 1. Red, peloric: RRss (red requires two R alleles, and peloric requires two s alleles) 2. White, personate: rrSS (white requires two r alleles, and personate requires two S alleles)

Determine F1 genotypes

Each offspring will inherit one allele for color and one allele for flower shape from each parent. In this case, F1 offspring will have the genotype RrSs (pink, personate). (b) red, personate × white, peloric F1: all pink, personate

Determine P1 genotypes

The parents are: 1. Red, personate: RRSS (red requires two R alleles, and personate requires two S alleles) 2. White, peloric: rrss (white requires two r alleles, and peloric requires two s alleles)

Determine F1 genotypes

In this case, F1 offspring will also have the genotype RrSs (pink, personate). (c) pink, personate × red, peloric F1: 1/4 red, personate; 1/4 red, peloric; 1/4 pink, personate; 1/4 pink, peloric

Determine P1 genotypes

The parents are: 1. Pink, personate: RrSs 2. Red, peloric: RRss

Determine F1 genotypes

A Punnett square can be used to determine the F1 genotypes and their frequencies: 1/4 RRSs (red, personate); 1/4 RRss (red, peloric); 1/4 RrSs (pink, personate); 1/4 Rrss (pink, peloric) (d) pink, personate × white, peloric F1: 1/4 white, personate; 1/4 white, peloric; 1/4 pink, personate; 1/4 pink, peloric

Determine P1 genotypes

The parents are: 1. Pink, personate: RrSs 2. White, peloric: rrss

Determine F1 genotypes

Using a Punnett square: 1/4 rrSs (white, personate); 1/4 rrss (white, peloric); 1/4 RrSs (pink, personate); 1/4 Rrss (pink, peloric) (e) What phenotypic ratios would result from crossing the F1 of (a) to the F1 of (b)?

Determine the cross

Crossing F1 of (a) and the F1 of (b) involves crossing RrSs (pink, personate) with RrSs (pink, personate).

Determine phenotypic ratios

Using a Punnett square, we can determine the phenotypic ratios of the offspring: - 1/4 RRSS (red, personate) - 1/4 RRSs or RRss (red, personate or red, peloric) - 1/4 RrSs (pink, personate) - 1/4 Rrss or rrSs (pink, peloric or white, personate) - 1/16 rrSS (white, personate) - 1/4 rrSs or rrss (white, personate or white, peloric) The phenotypic ratio is 1 red, personate: 1 red, peloric: 2 pink, personate: 2 pink, peloric: 2 white, personate: 1 white, peloric.