Question

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

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

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 ++\(+\).

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 \(c\)

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 ++\(+\).

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 \(c\).

Key concepts

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.