/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 An al-lele \(A\) that is not let... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 43

An al-lele \(A\) that is not lethal when homozygous causes rats to have yellow coats. The allele \(R\) of a separate gene that assorts independently produces a black coat. Together, \(A\) and \(R\) produce a grayish coat, whereas \(a\) and \(r\) produce a white coat. A gray male is crossed with a yellow female, and the \(\mathrm{F}_{1}\) is \(\frac{3}{8}\) yellow, \(\frac{3}{8}\) gray, \(\frac{1}{8}\) black, and \(\frac{1}{8}\) white. Determine the genotypes of the parents.

Short Answer

Expert verified
The genotypes of the parents are: male (AaRr) and female (Aarr).

Step by step solution

01

Identify Phenotypes and Genotypes

First, we identify the genotypes associated with each phenotype. - Yellow coat: Homozygous for the allele \( A \) and lack the influence of \( R \). This is \( AA \) or \( Aa \), depending on whether \( R \) is present.- Black coat: The presence of \( R \) without \( A \). This is genotype \( rr \) with \( RR \) or \( Rr \).- Gray coat: Combination of \( A \) and \( R \). This can only be \( AaRr \), \( AARr \), \( AaRR \), or \( AARR \).- White coat: Absence of \( A \) and \( R \). This is \( aarr \).
02

Understand the Given Ratios

The offspring ratios are \(\frac{3}{8}\) yellow, \(\frac{3}{8}\) gray, \(\frac{1}{8}\) black, and \(\frac{1}{8}\) white. This suggests the cross impacts both A in color expression and R in modifying the color.
03

Determine Genotypes of Offspring

- \(\frac{3}{8}\) yellow suggests a 3:1 ratio typical of a single-gene dominant scenario in heterozygous form.- \(\frac{3}{8}\) gray suggests the effect of both dominant alleles \( A \) and \( R \).- \(\frac{1}{8}\) black and \(\frac{1}{8}\) white indicate recessive alleles \( a \) and \( r \). This suggests the \( F_1 \) phenotype ratio aligns with a double heterozygote parent.
04

Determine Genotypes of Parents

The gray male is heterozygous for both traits since he produces all phenotypes. - His genotype is \( AaRr \).The yellow female cannot have \( R \) since her coat is not gray, and must be homozygous for \( A \) but missing \( r \). - Her genotype is \( AaRr \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phenotypic Ratios
Phenotypic ratios in genetics help us understand the distribution of observable traits in the offspring of a genetic cross. In the provided problem, there are four distinct phenotypes鈥攜ellow, gray, black, and white. Each phenotype corresponds to specific genetic combinations.

Here's how these phenotypes map to genetic ratios:
  • Yellow coat appears \( \frac{3}{8} \) in the offspring, suggesting a significant presence of the dominant allele \( A \) without the influence of dominant \( R \).
  • Gray coat also appears \( \frac{3}{8} \) of the time, implying the presence of both \( A \) and \( R \) alleles.
  • Black coat appears \( \frac{1}{8} \), requiring the presence of dominant \( R \) alone.
  • White coat appears \( \frac{1}{8} \), indicating the absence of both \( A \) and \( R \) alleles.
By analyzing these ratios, it is easier to deduce how the alleles interact to produce different traits.
Genotype Determination
Genotype determination involves identifying the genetic makeup that leads to particular phenotypes. In this case, we are given a cross between gray and yellow-coated rats, which allows us to predict parent and offspring genotypes.

For genotype determination:
  • Gray male: To produce all phenotypes in the \( F_1 \) generation, the gray male must be heterozygous for both traits. This means his genotype is \( AaRr \), having both dominant alleles \( A \) and \( R \).
  • Yellow female: As she has a yellow coat, the genotype must lack \( R \), which is why her potential genotype is \( aa \), and having either \( Aa \) or \( AA \) since yellow is dominant. The given cross results suggest a heterozygous \( Aa \) with \( rr \) to produce the observed phenotype ratios.
Thus, the combined genotype of the offspring helps determine the parental genotypes needed to arrive at the observed phenotypic pattern.
Mendelian Genetics
Mendelian genetics is the foundation for understanding inheritance patterns and involves principles derived from Gregor Mendel鈥檚 experiments. His key principles include the concepts of dominant and recessive alleles, segregation, and independent assortment, which are reflected in this exercise.

Mendel observed:
  • Each characteristic is controlled by two alleles, one inherited from each parent.
  • Dominant alleles mask the presence of recessive ones.
  • Different genes assort independently if they are not linked.
In our example, independent assortment is observed where two separate genes \( A \) and \( R \) assort independently to produce the observed color phenotypes. The segregation of the alleles results in distinct phenotype ratios aligning with Mendel鈥檚 laws, helping predict offspring outcomes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

After irradiating wild-type cells of Neurospora (a haploid fungus), a geneticist finds two leucine-requiring auxotrophic mutants. He combines the two mutants in a heterokaryon and discovers that the heterokaryon is prototrophic. a. Were the mutations in the two auxotrophs in the same gene in the pathway for synthesizing leucine or in two different genes in that pathway? Explain. b. Write the genotype of the two strains according to your model. c. What progeny and in what proportions would you predict from crossing the two auxotrophic mutants? (Assume independent assortment.)

A woman who owned a purebred albino poodle (an autosomal recessive phenotype) wanted white puppies; so she took the dog to a breeder, who said he would mate the female with an albino stud male, also from a pure stock. When six puppies were born, all of them were black; so the woman sued the breeder, claiming that he replaced the stud male with a black dog, giving her six unwanted puppies. You are called in as an expert witness, and the defense asks you if it is possible to produce black offspring from two pure-breeding recessive albino parents. What testimony do you give?

A plant believed to be heterozygous for a pair of alleles \(B / b\) (where \(B\) encodes yellow and \(b\) encodes bronze) was selfed, and, in the progeny, there were 280 yellow and 120 bronze plants. Do these results support the hypothesis that the plant is \(B / b ?\)

A dominant allele \(H\) reduces the number of body bristles that Drosophila flies have, giving rise to a "hairless" phenotype. In the homozygous condition, \(H\) is lethal. An independently assorting dominant allele \(S\) has no effect on bristle number except in the presence of \(H\) in which case a single dose of \(S\) suppresses the hairless phenotype, thus restoring the hairy phenotype. However, \(S\) also is lethal in the homozygous (S/S) condition. a. What ratio of hairy to hairless flies would you find in the live progeny of a cross between two hairy flies both carrying \(H\) in the suppressed condition? b. When the hairless progeny are backcrossed with a parental hairy fly, what phenotypic ratio would you expect to find among their live progeny?

Most of the feathers of erminette fowl are light colored, with an occasional black one, giving a flecked appearance. A cross of two erminettes produced a total of 48 progeny, consisting of 22 erminettes, 14 blacks, and 12 pure whites. What genetic basis of the erminette pattern is suggested? How would you test your hypotheses?
See all solutions

Recommended explanations on Biology Textbooks

Reproduction

Read Explanation

Substance Exchange

Read Explanation

Control of Gene Expression

Read Explanation

Energy Transfers

Read Explanation

Biology Experiments

Read Explanation

Genetic Information

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.