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Chapter 4: Problem 7

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

Expert verified
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

01

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)
02

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 脳 white, peloric F1: all pink, personate
03

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)
04

Determine F1 genotypes

In this case, F1 offspring will also have the genotype RrSs (pink, personate). (c) pink, personate 脳 red, peloric F1: 1/4 red, personate; 1/4 red, peloric; 1/4 pink, personate; 1/4 pink, peloric
05

Determine P1 genotypes

The parents are: 1. Pink, personate: RrSs 2. Red, peloric: RRss
06

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 脳 white, peloric F1: 1/4 white, personate; 1/4 white, peloric; 1/4 pink, personate; 1/4 pink, peloric
07

Determine P1 genotypes

The parents are: 1. Pink, personate: RrSs 2. White, peloric: rrss
08

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)?
09

Determine the cross

Crossing F1 of (a) and the F1 of (b) involves crossing RrSs (pink, personate) with RrSs (pink, personate).
10

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.

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Most popular questions from this chapter

In cats, orange coat color is determined by the \(b\) allele, and black coat color is determined by the \(B\) allele. The heterozygous condition results in a coat pattern known as tortoiseshell, These genes are X-linked. What kinds of offspring would be expected from a cross of a black male and a tortoiseshell female? What are the chances of getting a tortoiseshell male?

In Dexter and Kerry cattle, animals may be polled (hornless) or horned. The Dexter animals have short legs, whereas the Kerry animals have long legs. When many offspring were obtained from matings between polled Kerrys and horned Dexters, half were found to be polled Dexters and half polled Kerrys. When these two types of \(\mathrm{F}_{1}\) cattle were mated to one another, the following \(\mathrm{F}_{2}\) data were obtained: \(3 / 8\) polled Dexters 3/8 polled Kerrys \(1 / 8\) horned Dexters \(1 / 8\) horned Kerrys A geneticist was puzzled by these data and interviewed farmers who had bred these cattle for decades. She learned that Kerrys were true-breeding. Dexters, on the other hand, were not truebreeding and never produced as many offspring as Kerrys. Provide a genetic explanation for these observations.

Students taking a genetics exam were expected to answer the following question by converting data to a "meaningful ratio" and then solving the problem. The instructor assumed that the final ratio would reflect two gene pairs, and most correct answers did. Here is the exam question: "Flowers may be white, orange, or brown. When plants with white flowers are crossed with plants with brown flowers, all the \(\mathrm{F}_{1}\) flow ers are white. For \(\mathrm{F}_{2}\) flowers, the following data were obtained: 48 white 12 orange 4 brown Convert the \(\mathrm{F}_{2}\) data to a meaningful ratio that allows you to explain the inheritance of color. Determine the number of genes involved and the genotypes that yield each phenotype." (a) Solve the problem for two gene pairs. What is the final \(\mathrm{F}_{2}\) ratio? (b) A number of students failed to reduce the ratio for two gene pairs as described above and solved the problem using three gene pairs. When examined carefully, their solution was deemed a valid response by the instructor, Solve the problem using three gene pairs.

Labrador retrievers may be black, brown, or golden in color (see the chapter opening photograph on \(\mathrm{p} .53\) ). Although each color may breed true, many different outcomes occur if numerous litters are examined from a variety of matings, where the parents are not necessarily true-breeding. The following results show some of the possibilities. Propose a mode of inheritance that is consistent with these data, and indicate the corresponding genotypes of the parents in each mating. Indicate as well the genotypes of dogs that breed true for each color. (a) black \(\times\) brown \(\longrightarrow\) all black (b) black \(\times\) brown \(\longrightarrow \quad 1 / 2\) black \(1 / 2\) brown (c) black \(\times\) brown \(\longrightarrow \quad 3 / 4\) black \(1 / 4\) golden (d) black \(\quad \times\) golden \(\longrightarrow \quad\) all black (e) black \(\times\) golden \(\longrightarrow \quad 4 / 8\) golden 318 black \(1 / 8\) brown (f) black \(\times\) golden \(\longrightarrow \quad 2 / 4\) golden \(1 / 4\) black \(1 / 4\) brown (8) brown \(\times\) brown \(\longrightarrow \quad 3 / 4\) brown \(1 / 4\) golden (h) black \(\times\) black \(\longrightarrow 9 / 16\) black \(4 / 16\) golden \(3 / 16\) brown

Two mothers give birth to sons at the same time at a busy urban hospital. The son of mother 1 has hemophilia, a disease caused by an X-linked recessive allele. Neither parent has the disease. Mother 2 has a son without hemophilia, despite the fact that the father has hemophilia. Several years later, couple 1 sues the hospital, claiming that these two newborns were swapped in the nursery following their birth. As a genetic counselor, you are called to testify. What information can you provide the jury concerning the allegation?
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