A female of genotype \\[ \frac{a}{+}+\frac{b}{+}+ \\] produces 100 meiotic tetrads. Of these, 68 show no crossover events. Of the remaining 32,20 show a crossover between \(a\) and \(b\) 10 show a crossover between \(b\) and \(c,\) and 2 show a double crossover between \(a\) and \(b\) and between \(b\) and \(c .\) Of the 400 gametes produced, how many of each of the 8 different genotypes will be produced? Assuming the order \(a-b-c\) and the allele arrangement previously shown, what is the map distance between these loci?

Short Answer

Expert verified
The female will produce 170 gametes of genotype \(a+b+\), 166 of genotype \(a++\), 34 of genotype \(+b+\), and 30 of genotype ++\(+\). The map distances between the loci are 22 centimorgans (cM) between \(a\) and \(b\), and 12 cM between \(b\) and \(c\).

Step by step solution

01

Understand the genotype and possible gametes

The female has the genotype: \\[ \frac{a}{+}+\frac{b}{+}+ \\] This means she has the alleles \(a\) and \(b\), and her remaining loci are labeled as \(+\). The possible gametes produced are: \(a+b+\), \(a++\), \(+b+\), and ++\(+\).
02

Analyze crossover events and frequencies

We're given the following information about crossover events: - 68 tetrads have no crossover between any loci - 20 tetrads have a crossover between \(a\) and \(b\) - 10 tetrads have a crossover between \(b\) and \(c\) - 2 tetrads have a double crossover between \(a\) and \(b\) and between \(b\) and \(c\)
03

Calculate the frequency of each genotype produced

We'll use the information about tetrads to estimate the frequency of each genotype produced: - \(a+b+\): (68*2) + (20*1) + (10*1) + (2*2) = 136 + 20 + 10 + 4 = 170 - \(a++\): (68*2) + (20*1) + (10*1) + (2*0) = 136 + 20 + 10 + 0 = 166 - \(+b+\): (68*0) + (20*1) + (10*1) + (2*2) = 0 + 20 + 10 + 4 = 34 - ++\(+\): (68*0) + (20*1) + (10*1) + (2*0) = 0 + 20 + 10 + 0 = 30 The female will produce 170 gametes of genotype \(a+b+\), 166 of genotype \(a++\), 34 of genotype \(+b+\), and 30 of genotype ++\(+\).
04

Convert crossover frequencies into map distances

Using the crossover frequencies, we can calculate the map distance between loci: - Map distance between \(a\) and \(b\): (20 crossovers + 2 double crossovers) / 100 tetrads = 22 / 100 = 0.22, or 22 centimorgans (cM). - Map distance between \(b\) and \(c\): (10 crossovers + 2 double crossovers) / 100 tetrads = 12 / 100 = 0.12, or 12 cM. Therefore, the map distances between the loci are 22 cM between \(a\) and \(b\), and 12 cM between \(b\) and \(c\).

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 Vaia!

Key Concepts

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

Meiotic Tetrads
Understanding meiotic tetrads is fundamental in genetics, especially when it comes to determining the genetic diversity in gametes. During meiosis, homologous chromosomes pair up and exchange genetic material through a process called crossing over. This exchange occurs within structures called tetrads, which consist of four chromatids. Each tetrad comprises two homologous chromosomes that have replicated into sister chromatids.

During the formation of gametes, the tetrads undergo meiotic divisions, ultimately resulting in the production of four haploid cells, each containing a unique mixture of the parent's genetic material. The randomness of how the chromatids assort and segregate ensures genetic diversity within the offspring.
Gamete Genotype Frequency
The frequency of gamete genotypes is a measure of how often a particular genetic combination appears in a population's gametes. This concept is closely tied to the principles of Mendelian inheritance and genetic variation.

In the given exercise, the frequencies are calculated based on the observed crossover events among the meiotic tetrads. By determining the number of crossover events that affect the alleles in question, one can predict the abundance of each gamete genotype following meiosis. In educational exercises, it's vital to provide a clear explanation of how to calculate these frequencies to ensure students can apply this knowledge in different contexts.
Map Distance Calculation
Map distance calculation is integral to the field of genetic linkage mapping. It involves the use of crossover frequencies to estimate the physical distances between genes on a chromosome. One map unit, also known as a centimorgan (cM), is defined as the distance between chromosome positions for which 1% recombination is observed between them.

To calculate the map distances, one must analyze the frequencies of single and double crossover events from meiotic tetrads. This process enables geneticists to construct linkage maps, which depict the relative positions of genes on a chromosome. In educational content, it is helpful to break down the calculation process into steps, providing students with a systematic approach to analyzing genetic linkage data.
Centimorgan (cM)
The centimorgan (cM) is a unit of measure that describes the likelihood of recombination between two genetic loci. Conceptually, it denotes the distance between genes for which one product of meiosis in a hundred will be recombinant due to crossovers in a single generation. This term is named after the geneticist Thomas Hunt Morgan, who pioneered the work in genetic linkage.

It's important to convey to students that although one cM roughly corresponds to a physical distance of 1 million base pairs in humans, this is not a constant value across all organisms or even within different regions of the human genome. Regions of high recombination will have more crossovers and thus more centimorgans between genes. When explaining centimorgans, it's often effective to use visual aids, such as genetic maps, to illustrate how map units reflect the probability of recombination, not just physical distance.

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

Two plants in a cross were each heterozygous for two gene pairs \((A b / a B)\) whose loci are linked and 25 mu apart. Assuming that crossing over occurs during the formation of both male and female gametes and that the \(A\) and \(B\) alleles are dominant, determine the phenotypic ratio of their offspring.

Why does more crossing over occur between two distantly linked genes than between two genes that are very close together on the same chromosome?

In Drosophila, two mutations, Stubble (Sb) and curled (cu), are linked on chromosome III. Stubble is a dominant gene that is lethal in a homozygous state, and curled is a recessive gene. If a female of the genotype \\[ \frac{S b}{+}+ \\] is to be mated to detect recombinants among her offspring, what male genotype would you choose as a mate?

Colored aleurone in the kernels of corn is due to the dominant allele \(R\). The recessive allele \(r,\) when homozygous, produces colorless aleurone. The plant color (not the kernel color) is controlled by another gene with two alleles, \(Y\) and \(y\). The dominant \(Y\) allele results in green color, whereas the homozygous presence of the recessive \(y\) allele causes the plant to appear yellow, In a testcross between a plant of unknown genotype and phenotype and a plant that is homozygous recessive for both traits, the following progeny were obtained: $$\begin{aligned} &\begin{array}{lr} \text { colored, green } & 88 \\ \text { colored, yellow } & 12 \\ \text { colorless, green } & 8 \\ \text { colorless, yellow } & 92 \end{array}\\\ &\begin{array}{l} \\\ \\\ \end{array} \end{aligned}$$ Explain how these results were obtained by determining the exact genotype and phenotype of the unknown plant, including the precise arrangement of the alleles on the homologs.

Describe the cytological observation that suggests that crossing over occurs during the first meiotic prophase.

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.

Sign-up for free