Question

A female of genotype \\[ \frac{a}{+++} \\] 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 eight different genotypes will be produced? Assuming the order \(a-b-c\) and the allele arrangement shown above, what is the map distance between these loci?

Short answer

Question: Calculate the number of gametes produced for each of the 8 different genotypes and the map distance between the loci. Answer: The number of gametes produced for each genotype is as follows: a/+++: 144 +++/a: 144 a/b++: 44 +/+a+: 44 a/ +c+: 24 +/+++: 24 Total = 400 The map distance between the loci a, b, and c is 20 cM between a-b and 10 cM between b-c.

Step-by-step solution

Calculate the frequencies of crossover events

We are given the following data: Total number of tetrads (T) = 100 No crossovers (N) = 68 Crossovers between a-b (Ca-b) = 20 Crossovers between b-c (Cb-c) = 10 Double crossovers between a-b and b-c (CDa-b-b-c) = 2 Now we can calculate the frequencies of each crossover event. Frequency of crossovers between a-b (Fa-b) = Ca-b/T = 20/100 = 0.20 Frequency of crossovers between b-c (Fb-c) = Cb-c/T = 10/100 = 0.10 Frequency of double crossovers between a-b and b-c (FDa-b-b-c) = CDa-b-b-c/T = 2/100 = 0.02

Determine gamete proportions

To determine the proportion of different gametes produced, we need to consider these frequencies. For a single crossover event, half of the resulting gametes will have the crossover genotypes. So, we can find the proportions of the gametes as follows: 1. Non-crossover (N) gametes: This will include both the original genotype (a/+++) and its opposite (+++ /a). The proportion of non-crossover gametes is (1 - Fa-b - Fb-c + FDa-b-b-c) = (1 - 0.20 - 0.10 + 0.02) = 0.72. 2. Crossover a-b (Ca-b) gametes: Gametes will split 50%-50% between genotypes a/b++ and +/+a+. The proportion of crossover a-b gametes is (Fa-b + FDa-b-b-c)/2 = (0.20 + 0.02)/2 = 0.11. 3. Crossover b-c (Cb-c) gametes: Gametes will split 50%-50% between genotypes a/ +c+ and +/+++. The proportion of crossover b-c gametes is (Fb-c + FDa-b-b-c)/2 = (0.10 + 0.02)/2 = 0.06. 4. Double crossover (CDa-b-b-c) gametes: Will be split between genotypes a/ +c+ and +/++a. Double crossovers are already counted in single crossovers so no need to add them now.

Calculate the number of gametes produced for each genotype

Total gametes produced (TG) = 400 Non-crossover gametes (N): The original genotype (a/+++) and its opposite (+++ /a) will be produced in TG * 0.72/2 = 144 of each type. Crossover a-b (Ca-b) gametes: There will be TG * 0.11 = 44 of each of the types a/b++ and +/+a+. Crossover b-c (Cb-c) gametes: There will be TG * 0.06 = 24 of each of the types a/ +c+ and +/+++. The total number of gametes produced for each of the 8 different genotypes is: a/+++: 144 +++/a: 144 a/b++: 44 +/+a+: 44 a/ +c+: 24 +/+++: 24 Total = 400

Calculate the map distance between loci

The map distance is given by the recombination frequency between two loci, expressed in centimorgans (cM). Map distance a-b: Frequency is Fa-b = 0.20. The map distance is 0.20 * 100 = 20 cM. Map distance b-c: Frequency is Fb-c = 0.10. The map distance is 0.10 * 100 = 10 cM. The map distance between the loci a, b, and c is 20 cM between a-b and 10 cM between b-c.

Key concepts

Meiotic Tetrads

In the context of genetic mapping, meiotic tetrads play a crucial role in understanding the recombination of genes. A meiotic tetrad consists of a pair of homologous chromosomes, each consisting of two sister chromatids, that form during meiosis.
This formation is essential for the exchange of genetic material between homologous chromosomes during a process called crossing over.
Each tetrad can potentially undergo crossing over at various points, leading to variations in the combinations of alleles that are present in the resulting gametes.
This genetic variation contributes to the diversity of traits passed on to offspring. In this exercise, we observed 100 meiotic tetrads, where each tetrad can independently undergo crossover events.

Crossover Events

Crossover events are pivotal in reshuffling genetic material between homologous chromosomes. During meiosis, segments of DNA are exchanged between non-sister chromatids of homologous chromosomes.
This process results in new allele combinations, which are crucial for genetic diversity in sexually reproducing organisms.
In the provided problem, there are different types of crossover events:

• Crossovers between loci \(a\) and \(b\) occurred in 20 out of the 100 tetrads.
• Crossovers between \(b\) and \(c\) appeared in 10 tetrads.
• Double crossovers, involving both \(a-b\) and \(b-c\), were observed in 2 tetrads.
  

The frequencies of these events help calculate the possible genetic outcomes and the genetic map distances between the loci involved.

Gamete Genotypes

Gamete genotypes result from the various combinations of alleles due to crossover events. These genotypes are essentially four different types that could arise: non-crossover and crossover products.
Non-crossover gametes retain the parental combinations, while crossover gametes exhibit the rearranged genetic makeup due to recombination.
In this genetic mapping exercise:

• 144 non-crossover gametes of type \(a/+++\) and \(+/a+++\) are produced.
• 44 crossover gametes were of type \(a/b++\) and \(+/+a+\), from crossovers between \(a\) and \(b\).
• 24 crossover gametes of type \(a/+c+\) and \(+/+++\) from crossovers between \(b\) and \(c\).

Studying these results helps researchers understand the relationship of alleles on the chromosome and subsequently predict patterns of inheritance.

Map Distance

Map distance, measured in centimorgans (cM), reflects the frequency of recombination between genetic loci during meiosis. It is an essential element for constructing linkage maps, which visualize gene positions on chromosomes based on recombination frequencies.
The closer together two loci are, the less likely they are to be separated by a crossover event.
In the exercise, two map distances were calculated:

• The map distance between \(a\) and \(b\) was determined to be 20 cM, based on a recombination frequency of 0.20.
• Between \(b\) and \(c\), the map distance was 10 cM, given a recombination frequency of 0.10.
  

These distances provide insights into the physical positioning of genes on chromosomes, enhancing our understanding of genetic relationships.