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Chapter 4: Problem 25

Pigment in the mouse is produced only when the \(C\) allele is pres- ent. Individuals of the ce 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 \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) 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

Short Answer

Expert verified
(a) F1 genotype is AaCe, and F1 phenotype is agouti. (b) F2 genotypic ratio is 1AACC:4AACe:6AaCC:4AaCe:2aaCC:4aaCe:1aaee, and the phenotypic ratio is 15 agouti:6 black:1 colorless. (c) The genotypes of the female parents for different phenotypic ratios are: (1) AaCc for 8 agouti (2) AaCe for 9 agouti (3) Aace for 4 agouti: 8 colorless: 10 black.

Step by step solution

01

Cross the parent generation to obtain F1

Cross the parent genotypes \(A A C C\) and aace to obtain the F1 generation. It's easier to deal with the two alleles separately: \(A A \times aa\) results in all offspring carrying the Aa genotype for the first gene. \(C C \times ce\) results in all offspring carrying the Ce genotype for the second gene. Thus, the F1 genotype for their offspring will be AaCe.
02

Cross the F1 generation to obtain F2

Cross AaCe x AaCe to obtain F2 genotypic ratios: For the first gene (A, a): Aa x Aa produces: 1 AA, 2 Aa, 1 aa (1:2:1 ratio) For the second gene (C, e): Ce x Ce produces: 1 CC, 2 Ce, 1 ee (1:2:1 ratio) Now, we will combine the results from these crosses: 1:2:1 × 1:2:1 = 1:4:6:4:2:4:1:2:1 The genotypic ratio for the F2 offspring is 1AACC:4AACe:6AaCC:4AaCe:2aaCC:4aaCe:1aaee.
03

Determine the F2 phenotypes based on the genotypic ratios

According to the problem: 1. Color is determined by A and a alleles only if the C allele is present. 2. AA or Aa results in agouti color, whereas aa results in black coats. Therefore, we can list the phenotypes using the given information: 1 AACC: Agouti 4 AACe: Agouti 6 AaCC: Agouti 4 AaCe: Agouti 2 aaCC: Black 4 aaCe: Black 1 aaee (Colorless): Colorless The phenotypic ratio for the F2 offspring is 15 agouti:6 black:1 colorless. (a) F1 genotype: AaCe F1 phenotype: Agouti (b) In all the crosses, the male parent has aacc genotype, and we know that the female parents are agouti. Thus, the female parent must have one A allele and at least one C allele.
04

Determine Female Parent Genotypes for the Phenotypic Ratios

(1) 8 agouti: Male genotype: aacc Possible female genotype: AaCc (2) 9 agouti: Male genotype: aacc Possible female genotype: AaCe (Self-cross AaCe x AaCe to verify that this combination can produce the 9 phenotypic ratios) (3) 4 agouti: 8 colorless 10 black: Male genotype: aacc Possible female genotype: Aace (Self-cross AaCe x AaCe to verify that this genotype combination can produce the 4 agouti: 8 colorless: 10 black phenotypic ratio) Therefore, the genotypes of the female parents are: (1) AaCc (2) AaCe (3) Aace

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Most popular questions from this chapter

The trait of medium-sized leaves in iris is determined by the genetic condition \(P P^{\prime}\). Plants with large leaves are \(P P\), whereas plants with small leaves are \(P^{\prime} P^{\prime} .\) A cross is made between two plants each with medium-sized leaves. If they produce 80 seedlings, what would be the expected phenotypes, and in what numbers would they be expected? What is the term for this allelic relationship?

Two different genes, located on two different chromosomes, are responsible for color production in the aleurone layer of com kernels. For color production (either purple or red), the dominant alleles of these two genes \((C \text { and } R\) ) must come together. Furthermore, a third gene, located on a third chromosome, interacts with the \(C\) and \(R\) alleles to determine whether the aleurone will be red or purple. While the dominant allele ( \(P\) ) ensures purple color, the homozygous recessive condition (pp) makes the aleurone red. Determine the \(\mathrm{P}_{1}\) phenotypic ratio of the following crosses: (a) \(C C r r P P \times \operatorname{ccRRp} p\) (b) \(C c R R p p \times C C R r p p\) (c) \(\operatorname{CcRrPp} \times\) CcRrpp.

Three gene pairs located on separate autosomes determine flower color and shape as well as plant height. The first pair exhibits incomplete dominance, where color can be red, pink (the heterozygote), or white, The second pair leads to the dominant personate or recessive peloric flower shape, while the third gene pair produces either the dominant tall trait or the recessive dwarf trait. Homozygous plants that are red, personate, and tall are crossed with those that are white, peloric, and dwarf. Determine the \(P_{1}\) genotype(s) and phenotype(s). If the \(F_{1}\) plants are interbred, what proportion of the offspring will exhibit the same phenotype as the \(F_{1}\) plants?

Duchenne muscular dystrophy (DMD), marked by muscular degeneration, results from an \(\mathrm{X}\) - linked recessive gene. Thus, a female who is heterozygous for this gene and does not have the disease can be a carrier. What kind of offspring can you expect from a DMD-affected male and a carrier female? Can there be a carrier male?

In cattle, coats may be solid white, solid black, or black-andwhite spotted. When true-breeding solid whites are mated with true-breeding solid blacks, the \(\mathrm{F}_{1}\), generation consists of all solid white individuals. After many \(\mathrm{F}_{1} \times \mathrm{F}_{1}\) matings, the following ratio was observed in the \(\mathrm{F}_{2}\) generation: \(12 / 16\) solid white \(3 / 16\) black-and-white spotted \(1 / 16\) solid black Rxplain the mode of inheritance governing coat color by determining how many gene pairs are involved and which genotypes yield which phenotypes. Is it possible to isolate a true-breeding strain of black-and-white spotted cattle? If so, what genotype would they have? If not, explain why not.
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