As in Problem \(12,\) flower color may be red, white, or pink, and flower shape may be personate or peloric. For the following crosses, determine the \(P_{1}\) and \(F_{1}\) genotypes: (a) red, peloric \(\times\) white, personate 1 \(\mathrm{F}_{1}:\) all pink, personate (b) red, personate \(\times\) white, peloric 1 \(\mathrm{F}_{1}:\) all pink, personate (c) pink, personate \(\times\) red, peloric $\rightarrow \mathrm{F}_{1} \quad\left\\{\begin{array}{l}1 / 4 \mathrm{red}, \text { personate } \\ 1 / 4 \mathrm{red}, \text { peloric } \\ 1 / 4 \mathrm{pink}, \text { peloric } \\\ 1 / 4 \mathrm{pink}, \text { personate }\end{array}\right.$ (d) pink, personate \(\times\) white, peloric $\rightarrow \mathrm{F}_{1}\left\\{\begin{array}{l}1 / 4 \text { white, personate } \\ 1 / 4 \text { white, peloric } \\ 1 / 4 \text { pink, personate } \\ 1 / 4 \text { pink, peloric }\end{array}\right.$ (c) What phenotypic ratios would result from crossing the \(\mathrm{F}_{1}\) of (a) to the \(F_{1}\) of \((b) ?\)

Step by step solution

01

Assuming dominant and recessive traits

Let's assume that red (R) and personate (P) are dominant traits, while white (r) and peloric (p) are recessive traits. So, the red peloric P1 genotype would be R_pp, and the white personate P1 genotype would be rrP_. F1 genotypes: --------------------------------------------------------------------------

02

Crossing the P1 genotypes

Crossing R_pp (red, peloric) and rrP_ (white, personate) results in F1 generation: Rp x rP = RrPp

03

Phenotype of F1 generation

So, the F1 genotype is RrPp, which results in pink (due to incomplete dominance) and personate phenotype. This matches with the given F1 phenotypes: all pink, personate. (b) red, personate × white, peloric P1 genotypes: --------------------------------------------------------------------------

04

Assuming dominant and recessive traits

Following the same assumptions as before, red personate P1 genotype would be R_PP, and white peloric P1 genotype would be rrpp. F1 genotypes: --------------------------------------------------------------------------

05

Crossing the P1 genotypes

Crossing R_PP (red, personate) and rrpp (white, peloric) results in F1 generation: RP x rp = RrPp

06

Phenotype of F1 generation

So, the F1 genotype is RrPp, which results in pink (due to incomplete dominance) and personate phenotype. This matches with the given F1 phenotypes: all pink, personate. Now, let's find the phenotypic ratios resulting from crossing the F1 of (a) to the F1 of (b). (c) Crossing F1 of (a) and F1 of (b) --------------------------------------------------------------------------

07

Crossing F1 generations

Both the F1 generations of (a) and (b) have the same genotype, RrPp (pink, personate). So, we need to find the result of crossing RrPp × RrPp.

08

Determine all possible offspring genotypes

We can use a Punnett square to determine the possible offspring genotypes. The square will have 16 possible results which include: RRPP, RRPp, RrPP, RrPp (1 red, personate; 1 red, peloric; 1 pink, personate; 1 pink, peloric) rrPP, rrPp (1 white, personate; 1 white, peloric)

09

Phenotypic ratios

Based on these results, the phenotypic ratios for a cross between F1 of (a) and F1 of (b) are: 1 red personate : 1 red peloric : 1 pink personate : 1 pink peloric : 1 white personate : 1 white peloric

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