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.