/*! 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 10 In a cross between a black and a... [FREE SOLUTION] | 91影视

91影视

In a cross between a black and a white guinea pig, all members of the \(\mathrm{F}_{1}\) generation are black. The \(\mathrm{F}_{2}\) generation is made up of approximately \(3 / 4\) black and \(1 / 4\) white guinea pigs. (a) Diagram this cross, showing the genotypes and phenotypes. (b) What will the offspring be like if two \(\mathrm{F}_{2}\) white guinea pigs are mated? (c) Two different matings were made between black members of the \(\mathrm{F}_{2}\) generation, with the following results. Cross Cross 1 Cross 2 Offspring All black \(3 / 4\) black, \(1 / 4\) white Diagram each of the crosses.

Short Answer

Expert verified
Answer: In Cross 2, when two black guinea pigs with heterozygous genotypes (Bb) are mated, the ratio of black and white offspring will be 3:1, with 3/4 of the offspring being black and 1/4 being white.

Step by step solution

01

Analyze F1 generation cross

First, note that the black guinea pig is dominant over the white guinea pig, as all members of F1 generation are black. Represent the black guinea pig allele with 'B' and white guinea pig allele with 'b'. Therefore, the initial cross is between a black guinea pig (BB) and a white guinea pig (bb).
02

Create a Punnett square for the F1 cross

A Punnett square can be used to visualize and predict offspring genotypes by pairing alleles from each parent. \[ \begin{array}{c|c c} \times & B & B \\ \hline b & Bb & Bb \\ b & Bb & Bb \\ \end{array} \] All members of F1 generation are black (Bb). #b) What will the offspring be like if two 饾悈2 white guinea pigs are mated?#
03

Analyze the F2 generation

Since there are white guinea pigs present in the F2 generation, their genotype must be (bb). When two F2 white guinea pigs (bb) are mated, we should determine the offspring's genotypes and phenotypes.
04

Create a Punnett square for the white F2 cross

Create a Punnett square, similar to Step 2, to visualize the offsprings' genotypes and phenotypes. \[ \begin{array}{c|c c} \times & b & b \\ \hline b & bb & bb \\ b & bb & bb \\ \end{array} \] All offspring from this cross will be white (bb). #c) Two different matings were made between black members of the 饾悈2 generation, with the following results.#
05

Analyze the different matings in F2 generation

Two different crosses between black members of the F2 generation are given: - Cross 1: All black offspring - Cross 2: \(\frac{3}{4}\) black, \(\frac{1}{4}\) white offspring
06

Diagram Cross 1

Considering dominant nature of black guinea pig, "All black offspring" in the F2 generation suggests that at least one parent must have been homozygous dominant (BB). \[ \begin{array}{c|c c} \times & B & B \\ \hline B & BB & BB \\ b & Bb & Bb \\ \end{array} \] In Cross 1, one parent is homozygous dominant (BB) and the other parent is heterozygous (Bb). All offspring are black: 1/2 (BB) and 1/2 (Bb).
07

Diagram Cross 2

This cross resulted in offspring with the following phenotypes: \(\frac{3}{4}\) black and \(\frac{1}{4}\) white. This indicates that both parents in this cross are heterozygous (Bb). \[ \begin{array}{c|c c} \times & B & b \\ \hline B & BB & Bb \\ b & Bb & bb \\ \end{array} \] In Cross 2, both parents are heterozygous (Bb). The offspring are: 1/4 (BB), 1/2 (Bb), and 1/4 (bb). The phenotypes are \(\frac{3}{4}\) black and \(\frac{1}{4}\) white.

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.

Punnett Square
The Punnett square is an essential tool used in genetics to predict the possible genotypes of offspring from two parents. It provides a simple graphical way of illustrating the cross between the alleles of the parents' genes.

For instance, in our guinea pig exercise, the Punnett square allowed us to visualize how the black and white traits were inherited. By organizing the two possible alleles from each parent along the axes of a grid, we mapped out the potential combinations in the offspring. It's almost like making a chart to see every possible outcome of a genetic pairing.

When interpreting a Punnett square, remember that each cell represents a potential genotype of an offspring. So, it provides a structured and easy way to calculate the probability of each genotype and, by extension, the related phenotype鈥攚hether guinea pigs turn out black or white in this case.
Genotype and Phenotype
Every organism has a set of genetic instructions used to build and maintain its cells, known as its genotype. The physical expression of this genetic information is known as the phenotype. In other words, the genotype can be thought of as the genetic 'blueprint', while the phenotype is like the completed 'construction'.

The exercise showcases how two guinea pigs with different genotypes can have the same phenotype. A black guinea pig might have the genotype 'BB' (homozygous dominant) or 'Bb' (heterozygous), yet in both cases, the phenotype is a black coat color. Thus, the genotype refers to the actual alleles inherited, while the phenotype is the observed trait, such as fur color, in our furry friends.
F1 and F2 Generations
In Mendelian genetics, the terms 'F1' and 'F2' refer to the first and second filial generations, respectively. The F1 generation results from the crossing of two parent (P) organisms, while the F2 generation is produced by crossing two members of the F1 generation.

As seen in the exercise, all the F1 offspring exhibited the dominant phenotype鈥攂lack fur. The F2 generation, born from the F1 individuals, shows the reemergence of the recessive white fur trait. Through careful analysis of these generational patterns, we can determine how traits are transmitted from parents to offspring and predict future occurrences of these traits with remarkable accuracy.
Dominant and Recessive Traits
Traits in Mendelian genetics are determined by alleles which are either dominant or recessive. Dominant traits mask the expression of recessive traits when they are together in a heterozygous genotype. To express a recessive trait, like the white fur in guinea pigs, an organism must inherit two copies of the recessive allele (bb).

It is fascinating how the combination of alleles, BB, Bb, or bb, determines the fur color of guinea pigs. In the exercise, we discovered that two black F2 guinea pigs could produce white offspring because they carried the recessive allele. This exemplifies why it's vital to look beyond the phenotype and consider the underlying genotype for a full understanding of inheritance patterns.

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

Which of Mendel's postulates can only be demonstrated in crosses involving at least two pairs of traits? State the postulate.

For decades scientists have been perplexed by different circumstances surrounding families with rare, early-onset auditory neuropathy (deafness). In some families, parents and grandparents of the proband have normal hearing, while in other families, a number of affected (deaf) family members are scattered throughout the pedigree, appearing in every generation. Assuming a genetic cause for each case, offer a reasonable explanation for the genetic origin of such deafness in the two types of families.

The following are \(\mathrm{F}_{2}\) results of two of Mendel's monohybrid crosses. (a) full pods 882 constricted pods 299 (b) violet flowers 705 white flowers For each cross, state a null hypothesis to be tested using \(\chi^{2}\) analysis. Calculate the \(x^{2}\) value and determine the \(p\) value for both. Interpret the \(p\) values. Can the deviation in each case be attributed to chance or not? Which of the two crosses shows a greater amount of deviation?

Mendel crossed peas having green seeds with peas having yellow seeds. The \(F_{1}\) generation produced only yellow seeds. In the \(F_{2}\) the progeny consisted of 6022 plants with yellow seeds and 2001 plants with green seeds. Of the \(\mathrm{F}_{2}\) yellow-seeded plants, 519 were self-fertilized with the following results: 166 bred true for yellow and 353 produced an \(\mathrm{F}_{3}\) ratio of \(3 / 4\) yellow: \(1 / 4\) green. Explain these results by diagramming the crosses.

How many different types of gametes can be formed by individuals of the following genotypes: (a) \(A a B b\) (b) \(A a B B\) (c) \(A a B b C c\) (d) \(A a B B c c\) (e) \(A a B b c c,\) and (f) \(A a B b C c D d E e ?\) What are the gametes in each case?

See all solutions

Recommended explanations on Biology Textbooks

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.