/*! 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 11 In the genetic cross \(A a B b C... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In the genetic cross \(A a B b C c D d E E \times A a B B C c D d E e,\) what fraction of the offspring will be heterozygous for all of these genes \((A a B b C c D d E e) ?\) Assume all genes are unlinked.

Short Answer

Expert verified
The fraction of offspring that will be heterozygous for all genes is \(\frac{1}{32}\).

Step by step solution

01

Identify Possible Gametes

Determine the possible gametes for each parent by considering each gene independently. Both parents have two alleles per gene locus due to being diploid organisms. For each heterozygous locus, each parent can contribute two different alleles, resulting in different gamete combinations. Parent 1 can produce the gametes: \(A B C D E\), \(A B C d E\), \(A b C D E\), \(A b C d E\), \(a B C D E\), \(a B C d E\), \(a b C D E\), \(a b C d E\), each with frequency \(\frac{1}{8}\). Parent 2 can produce: \(A B C D E\), \(A B C d E\), \(A B C D e\), \(A B C d e\) with frequency \(\frac{1}{4}\), because only four combinations need consideration (different allele for only one locus).
02

Determine Offspring Genotypes

Identify how the gametes from both parents combine to affect the genotype of the offspring. The offspring will be heterozygous for a particular gene if it inherits different alleles from each parent at that locus.
03

Calculate Heterozygosity for Each Gene

Since all genes are unlinked, calculate the probability of heterozygosity for each gene separately.- **Gene A**: Offspring heterozygous probability is \(\frac{1}{2}\) as both parents are \(A a\).- **Gene B**: Offspring heterozygous probability is \(\frac{1}{2}\) as one parent is \(B b\) (binary choice) and the other parent is \(B B\).- **Gene C**: Offspring heterozygous probability is \(\frac{1}{2}\) since both parents are \(C c\).- **Gene D**: Offspring heterozygous is probability \(\frac{1}{2}\) because both parents are \(D d\).- **Gene E**: Offspring heterozygous is probability \(\frac{1}{2}\) (one parent is \(E e\) and the other is \(E E\)).
04

Calculate Combined Probability

To find the probability of the offspring being heterozygous for all genes simultaneously, multiply the individual probabilities since each allele segregation is an independent event:\[P = \left( \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \right) = \frac{1}{32}\]
05

Interpret the Results

The probability that the offspring will be heterozygous for all five gene loci \((A a B b C c D d E e)\) is \(\frac{1}{32}\). This means that, on average, 3.125% of the offspring will be heterozygous for all these genes.

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 91Ó°ÊÓ!

Key Concepts

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

Heterozygosity
Heterozygosity refers to the presence of different alleles at a particular gene locus within an organism. Essentially, for a gene to be heterozygous, there needs to be two different alleles, such as having an allele for brown eyes and an allele for blue eyes. In genetics, this is denoted as, for example, Aa where 'A' is one allele and 'a' is another distinct allele. This is important in genetic crosses as it affects the traits the offspring will express.

In the context of genetic crosses, particularly like the exercise here, the interest often lies in determining whether offspring will be heterozygous across multiple gene loci. Practically, given parental genotypes, we can predict heterozygosity by examining how alleles segregate during gamete formation and combine during fertilization. For instance, if both parents possess a mix of alleles (such as Aa and Bb), there is a certain probability that the offspring will be heterozygous, inheriting different alleles from each parent for these loci.
  • Key Idea: Heterozygosity implies genetic variation, vital for diverse expression and adaptation.
  • Example: In an organism with genotype Aa Bb, 'Aa' and 'Bb' are heterozygous for genes A and B respectively.
Independent Assortment
Independent assortment is a fundamental principle of Mendelian genetics. It states that alleles for different genes usually segregate independently of one another into gametes. This means that the inheritance of an allele for one trait generally does not affect the inheritance of an allele for another trait, provided the genes are unlinked (not located close together on the same chromosome).

This principle is key in genetic crosses as it allows for the calculation of probabilities related to different gene combinations in offspring. For example, when considering multiple genes as in the exercise, each gene's alleles separate and combine independently, leading to a variety of potential genetic combinations in the offspring. If considering a dihybrid cross, for example, there are multiple combinations of how alleles can assort independently.
  • Key Idea: Independent assortment augments genetic diversity and is crucial for calculating genetic probabilities.
  • Example: In a cross AaBb x AaBb, independent assortment leads to the classic 9:3:3:1 phenotypic ratio.
Probability Calculation
Calculating probabilities in genetics involves determining the likelihood of specific genetic outcomes based on the potential combination of alleles from the parents. It relies on both Mendel's laws and foundational probability theory. In genetic crosses, independent events are calculated using the multiplication rule of probability. Each gene locus is treated as an independent event, meaning the probability of a complex genotype (like Aa Bb Cc Dd Ee being heterozygous for all genes) is the product of the probabilities of being heterozygous at each locus. In our exercise, because each gene is heterozygous with a probability of 1/2, the overall likelihood is a product of these probabilities: \[P = \left( \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \right)\] This equals \(\frac{1}{32}\) or 3.125%, meaning this is the chance for offspring to inherit heterozygosity for all genes involved.
  • Key Idea: Probability calculations allow prediction of genotype frequencies.
  • Example: Using the multiplication rule, predict combined genotype probabilities for offspring.

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

Linked genes a. must be immediately adjacent to one another on a chromosome. b. have alleles that assort independently of one another. c. never show crossing over d. are on the same chromosome. e. always have multiple alleles.

In epistasis a. the phenotype does not change from generation to generation. b. one gene affects the expression of another. c. a portion of a chromosome is deleted. d. a portion of a chromosome is inverted. e. the behavior of two genes is entirely independent.

In cats, black coat \((B)\) is codominant with yellow (b). The coat color gene is on the X chromosome. Calico cats, which have coats with black and yellow patches, are heterozygous for the coat color alleles. a. Why are most calico cats females? b. A calico female, Pickle, had a litter with one yellow male, two black males, two yellow females, and three calico females. What were the genotype and phenotype of the father?

In Drosophila, white \((w), \operatorname{eosin}\left(w^{e}\right),\) and wild- type red \(\left(w^{+}\right)\) are multiple alleles at a single locus for eye color. This locus is on the \(X\) chromosome. A female that has eosin (pale orange) eyes is crossed with a male that has wild-type eyes. All the female progeny are redeyed; half the male progeny have eosin eyes, and half have white eyes. Assume the female has two \(x\) chromosomes and the male has one \(X\) and one \(Y\) a. What is the order of dominance of these alleles? b. What are the genotypes of the parents and progeny?

The disease Leber's optic neuropathy is caused by a mutation in a gene carried on mitochondrial DNA. What would be the phenotype of their first child if a man with this disease married a woman who did not have the disease? What would be the result if the wife had the disease and the husband did not?

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.