/*! 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 44 The genotype \(r / r ; p / p\) g... [FREE SOLUTION] | 91Ó°ÊÓ

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The genotype \(r / r ; p / p\) gives fowl a single comb, \(R /-; P /-\) gives a walnut comb, \(r / r ; P /-\) gives a pea comb, and \(R /-; p / p\) gives a rose comb (see the illustrations) Assume independent assortment. a. What comb types will appear in the \(\mathrm{F}_{1}\) and in the \(\mathrm{F}_{2}\) and in what proportions if single-combed birds are crossed with birds of a true-breeding walnut strain? b. What are the genotypes of the parents in a walnut \(x\) rose mating from which the progeny are \(\frac{3}{8}\) rose, \(\frac{3}{8}\) walnut, \(\frac{1}{8}\) pea, and \(\frac{1}{8}\) single? c. What are the genotypes of the parents in a walnut \(x\) rose mating from which all the progeny are walnut? d. How many genotypes produce a walnut phenotype? Write them out.

Short Answer

Expert verified
a. F1: All walnut; F2: 9/16 walnut, 3/16 rose, 3/16 pea, 1/16 single. b. Parents: \(R/r; p/p \times R/r; P/p\). c. Parents: \(R/R; P/p \times R/r; P/P\). d. Four genotypes: \(R/R; P/P\), \(R/R; P/p\), \(R/r; P/P\), \(R/r; P/p\).

Step by step solution

01

Identify Genotypes for F1 Generation

Cross a single-combed bird with genotype \(r/r; p/p\) and a walnut-combed bird, which must have the genotype \(R/R; P/P\) for true-breeding. Since we assume independent assortment, there will be only one genotype in the \(F_1\) generation, which is \(R/r; P/p\). This will all express the walnut comb as both dominant traits are present.
02

Analyze F2 Generation

To find the phenotypes in the \(F_2\) generation, perform a dihybrid cross \(R/r; P/p \times R/r; P/p\). The genotypic ratio will be 1 \(:\) 2 \(:\) 1 for each gene, which can be shown as \(R/R\), \(R/r\), \(r/r\) and \(P/P\), \(P/p\), \(p/p\). You can calculate proportions of different comb types (walnut, rose, pea, single) by multiplying the independent proportions: - Walnut (\(R/-; P/-\)): \(9/16\)- Rose (\(R/-; p/p\)): \(3/16\)- Pea (\(r/r; P/-\)): \(3/16\)- Single (\(r/r; p/p\)): \(1/16\).
03

Derive Genotypes from Given Ratios

Progeny ratios are \(3/8\) rose, \(3/8\) walnut, \(1/8\) pea, and \(1/8\) single. This suggests a parental cross of \(R/r; p/p \times R/r; P/p\). Here, the offspring distribution fits the phenotypic ratio derived from these genotypes.
04

Identify Uniform Walnut Offspring

All progeny being walnut suggests both parents are heterozygous for both traits with one already homozygous dominant for one, like \(R/R; P/p \times R/r; P/P\). This ensures all offspring have at least one dominant allele for both characters, exhibiting the walnut phenotype.
05

List All Genotypes for Walnut Phenotype

The walnut phenotype is expressed by any genotype with at least one dominant allele for both traits (\(R/-; P/-\)). The possible genotypes are:1. \(R/R; P/P\)2. \(R/R; P/p\)3. \(R/r; P/P\)4. \(R/r; P/p\).

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Key Concepts

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

Genotype and Phenotype
Genotype refers to the genetic makeup of an organism, comprising the alleles inherited from both parents. Phenotype, on the other hand, is the observable expression of these genes — essentially, how the organism looks or behaves. For example, in the context of our exercise, the genetic combinations determine the type of comb a fowl will have. When we say that a genotype of \(R/-; P/-\) results in a walnut comb, it means that as long as there is at least one dominant allele for both traits, the bird will have a walnut comb.

Understanding the difference between genotype and phenotype is crucial in genetic studies. The genotype acts as instructions while the phenotype is what you see as a result of those instructions.
  • If a bird's genotype is \(r/r; p/p\), the phenotype is a single comb since no dominant alleles are present for either trait.
  • Combinations like \(R/-; P/p\) will yield a walnut comb, showcasing the dominant effect of alleles \(R\) and \(P\).
  • Specific genotypes such as \(r/r; P/-\) result in a pea comb, despite only having dominant alleles for one trait.
A deeper grasp of how genotype determines phenotype helps clarify not just animal traits, but broader concepts in genetics.
Dihybrid Cross
A dihybrid cross is a breeding experiment examining the inheritance of two traits simultaneously, which can each have two different alleles. In our exercise, these two traits include the presence or absence of the comb characteristics dictated by \(R/r\) and \(P/p\) alleles.

When setting up a dihybrid cross, follow these steps:
  • Select parental genotypes: For example, \(R/r; P/p\) representing the F1 generation from our study.
  • Set up a Punnett Square for each trait independently, then combine these squares in a larger, four-square format to account for all possibilities.
  • Assess the resulting genotypic ratios in the offspring to predict the phenotypic outcome.
The classic Mendelian dihybrid cross, like the one used in the exercise, follows a 9:3:3:1 ratio when both parents are heterozygous for both traits. Here:
  • 9/16 are predicted to be walnut combed, as they hold the dominant gene expression for both traits \(R/-; P/-\).
  • 3/16 will have rose combs with a \(R/-; p/p\) genotype, showing the dominant R gene alone with recessive p.
  • 3/16 develop pea combs as \(r/r; P/-\), with only P dominant gene being expressed.
  • 1/16 are single combed \(r/r; p/p\) expressing only recessive genes.
This process showcases the concept of independent assortment, which highlights how alleles separate and reorganize independently during reproduction.
Mendelian Inheritance
Mendelian inheritance refers to the patterns of heredity first described by Gregor Mendel in the 19th century. It forms the foundation of our understanding of genetic inheritance and was demonstrated by Mendel through his pea plant experiments. These patterns highlight how traits are passed from parents to offspring, governed by distinct units known as genes.

In Mendelian inheritance, several key principles are evident:
  • **Principle of Segregation:** During gamete formation, alleles responsible for a trait separate, ensuring that offspring receives one allele from each parent. This explains how only one plant out of four would end up with both recessive traits, like pea and single combs in our genetic exercise.
  • **Principle of Independent Assortment:** Genes for different traits segregate independently, which is why in a dihybrid cross, you observe various combinations of traits (genotype \(R/r; P/p\) results in a 9:3:3:1 phenotypic ratio).
  • **Dominant and Recessive Traits:** This principle posits that dominant alleles can mask the presence of recessive ones. Hence, in our birds example, walnut phenotypes (\(R/-; P/-\)) are dominant, overshadowing single combs.
Each principle aligns with the classic breeding experiments illustrated, demonstrating Mendel's reliable laws as a predictive mechanism for inheritance. Understanding these underlying guidelines helps make sense of more complex genetic concepts encountered in both basic and advanced genetic studies.

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Most popular questions from this chapter

Several mutants are isolated, all of which require compound G for growth. The compounds (A to E) in the biosynthetic pathway to G are known, but their order in the pathway is not known. Each compound is tested for its ability to support the growth of each mutant (1 to 5). In the following table, a plus sign indicates growth and a minus sign indicates no growth. $$\begin{array}{rcccccc} {5}{c} {\text {compound tested}} \\ \ { 2 - 6 }{1}{c} {} & &\mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{D} & \mathrm{E} & \mathrm{G} \\ \hline \text { Mutant } & 1 & \- & \- & \- & \+ & \- & \+ \\ & 2 & \- & \+ & \- & \+ & \- & \+ \\ & 3 & \- & \- & \- & \- & \- & \+ \\ & 4 & \- & \+ & \+ & \+ & \- & \+ \\ & 5 & \+ & \+ & \+ & \+ & \- & \+ \\ \hline \end{array}$$ a. What is the order of compounds A to Ein the pathway? b. At which point in the pathway is each mutant blocked? c. Would a heterokaryon composed of double mutants 1,3 and 2,4 grow on a minimal medium? Would 1,3 and \(3,4 ?\) Would 1,2 and 2,4 and \(1,4 ?\)

Many kinds of wild animals have the agouti coloring pattern, in which each hair has a yellow band around it. a. Black mice and other black animals do not have the yellow band; each of their hairs is all black. This absence of wild agouti pattern is called nonagouti. When mice of a true-breeding agouti line are crossed with nonagoutis, the \(F_{1}\) is all agouti and the \(F_{2}\) has a 3: 1 ratio of agoutis to nonagoutis. Diagram this cross, letting \(A\) represent the allele responsible for the agouti phenotype and \(a\) nonagouti. Show the phenotypes and genotypes of the parents, their gametes, the \(F_{1}\), their gametes, and the \(F_{2}\) b. Another inherited color deviation in mice substitutes brown for the black color in the wild-type hair. Such brown-agouti mice are called cinnamons. When wildtype mice are crossed with cinnamons, all of the \(\mathrm{F}_{1}\) are wild type and the \(\mathrm{F}_{2}\) has a 3: 1 ratio of wild type to cinnamon. Diagram this cross as in part \(a\), letting \(B\) stand for the wild-type black allele and \(b\) stand for the cinnamon brown allele. c. When mice of a true-breeding cinnamon line are crossed with mice of a true- breeding nonagouti (black) line, all of the \(F_{1}\) are wild type. Use a genetic diagram to explain this result. d. In the \(F_{2}\) of the cross in part \(c,\) a fourth color called chocolate appears in addition to the parental cinnamon and nonagouti and the wild type of the \(\mathrm{F}_{1}\). Chocolate mice have a solid, rich brown color. What is the genetic constitution of the chocolates? e. Assuming that the \(A / a\) and \(B / b\) allelic pairs assort independently of each other, what do you expect to be the relative frequencies of the four color types in the \(\mathrm{F}_{2}\) described in part \(d ?\) Diagram the cross of parts \(c\) and \(d\) showing phenotypes and genotypes (including gametes). f. What phenotypes would be observed in what proportions in the progeny of a backcross of \(\mathrm{F}_{1}\) mice from part \(c\) with the cinnamon parental stock? With the nonagouti (black) parental stock? Diagram these backcrosses. g. Diagram a testcross for the \(\mathrm{F}_{1}\) of part \(c .\) What colors would result and in what proportions? h. Albino (pink-eyed white) mice are homozygous for the recessive member of an allelic pair \(C / c,\) which assorts independently of the \(A / a\) and \(B / b\) pairs. Suppose that you have four different highly inbred (and therefore presumably homozygous) albino lines. You cross each of these lines with a true-breeding wild-type line, and you raise a large \(\mathrm{F}_{2}\) progeny from each cross. What genotypes for the albino lines can you deduce from the following \(\mathrm{F}_{2}\) phenotypes? $$\begin{array}{cccccc} & {4}{c} {\text {Phenotypes of progeny}} \\ { 2 - 5 } \mathrm{F}_{2} \text { of } \text { line } & \begin{array}{c} \text { Wild } \\ \text { type } \end{array} & \text { Black } & \begin{array}{c} \text { Cinna- } \\ \text { mon } \end{array} & \begin{array}{c} \text { Choco- } \\ \text { late } \end{array} & \text { Albino } \\ \hline 1 & 87 & 0 & 32 & 0 & 39 \\ 2 & 62 & 0 & 0 & 0 & 18 \\ 3 & 96 & 30 & 0 & 0 & 41 \\ 4 & 287 & 86 & 92 & 29 & 164 \\ \hline \end{array}$$ (Adapted from A. M. Srb, R. D. Owen, and R. S. Edgar General Genetics, 2nd ed. W. H. Freeman and Company, 1965.)

Because snapdragons (Antirrhinum) possess the pigment anthocyanin, they have reddish purple petals. Two pure anthocyaninless lines of Antirrhinum were developed, one in California and one in Holland. They looked identical in having no red pigment at all, manifested as white (albino) flowers. However, when petals from the two lines were ground up together in buffer in the same test tube, the solution, which appeared colorless at first, gradually turned red. a. What control experiments should an investigator conduct before proceeding with further analysis? b. What could account for the production of the red color in the test tube? c. According to your explanation for part \(b\), what would be the genotypes of the two lines? d. If the two white lines were crossed, what would you predict the phenotypes of the \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) to be?

Wild-type strains of the haploid fungus Neurospora can make their own tryptophan. An abnormal allele \(t d\) renders the fungus incapable of making its own tryptophan. An individual of genotype \(t d\) grows only when its medium supplies tryptophan. The allele \(s u\) assorts independently of \(t d ;\) its only known effect is to suppress the \(t d\) phenotype. Therefore, strains carrying both \(t d\) and \(s u\) do not require tryptophan for growth. a. If a \(t d ; s u\) strain is crossed with a genotypically wildtype strain, what genotypes are expected in the progeny and in what proportions? b. What will be the ratio of tryptophan-dependent to tryptophan-independent progeny in the cross of part \(a ?\)

Two normal-looking fruit flies were crossed, and, in the progeny, there were 202 females and 98 males. a. What is unusual about this result? b. Provide a genetic explanation for this anomaly. c. Provide a test of your hypothesis.

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