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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 × 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 × 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 × 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 Drosophila, an \(\mathrm{X}\) -linked recessive mutation, scalloped (sd), causes irregular wing margins. Diagram the \(F_{1}\) and \(F_{2}\) results if (a) a scalloped female is crossed with a normal male; (b) a scalloped male is crossed with a normal female. Compare these results to those that would be obtained if the scalloped gene were autosomal.

In humans, the ABO blood type is under the control of autosomal multiple alleles. Red-green color blindness is a recessive X-linked trait. If two parents who are both type A and have normal vision produce a son who is color- blind and type \(0,\) what is the probability that their next child will be a female who has normal vision and is type \(0 ?\)

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?

Pigment in the mouse is produced only when the \(C\) allele is present. Individuals of the \(c c\) genotype have no color. If color is present, it may be determined by the \(A\) and \(a\) alleles. AA or Aa results in agouti color, whereas aa results in black coats. (a) What \(F_{1}\) and \(F_{2}\) genotypic and phenotypic ratios are obtained from a cross between \(A A C C\) and aace mice? (b) In the three crosses shown here between agouti females whose genotypes were unknown and males of the aacc genotype, what are the genotypes of the female parents for each of the following phenotypic ratios? (1) 8 agouti (2) 9 agouti (3) 4 agouti 8 colorless 10 black \(\quad 5\) black 10 colorless

In goats, development of the beard is due to a recessive gene. The following cross involving true-breeding goats was made and carried to the \(\mathrm{F}_{2}\) generation: \(P_{1}:\) bearded female \(\times\) beardless male \(\mathrm{F}_{1}:\) all bearded males and beardless females \\[ \mathrm{P}_{1} \times \mathrm{F}_{1} \longrightarrow\left\\{\begin{array}{l} 1 / 8 \text { beardless males } \\ 3 / 8 \text { bearded males } \\ 3 / 8 \text { beardless females } \\ 1 / 8 \text { bearded females }\end{array}\right.\\] Offer an explanation for the inheritance and expression of this trait, diagramming the cross. Propose one or more crosses to test your hypothesis.
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