A female of genotype \\[ \frac{a}{+}+b^{c} \\] 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?

We are given a female of genotype \(a/++b^c\). The given information about crossover events is as follows: - 68 meiotic tetrads show no crossover events. - 20 show a crossover between loci \(a\) and \(b\). - 10 show a crossover between loci \(b\) and \(c\). - 2 show a double crossover between loci \(a\) and \(b\) and between loci \(b\) and \(c\). For each meiotic tetrad, there are 4 gametes produced, making a total of 400 gametes.

There are 8 different genotypes possible: \(a+b^c, a-b^c, +b^c, +b-c, a+b-c, a-b-c,+b+c, +b-c\). Based on the given crossover events, we calculate the number of gametes produced for each genotype as follows: - \(a+b^c\): 4 tetrads without crossover * 68 + 2 tetrads from double crossover * 2 = 276 - \(a-b^c\): 2 tetrads from crossover between \(a\) and \(b\) * 20 = 40 - \(+b^c\): 2 tetrads from crossover between \(b\) and \(c\) * 10 = 20 - \(+b-c\): 2 tetrads from crossover between \(a\) and \(b\) * 20 = 40 - \(a+b-c\): 2 tetrads from crossover between \(a\) and \(b\) * 20 = 40 - \(a-b-c\): 2 tetrads from crossover between \(b\) and \(c\) * 10 = 20 - \(+b+c\): 2 tetrads from crossover between \(a\) and \(b\) * 20 = 40 - \(+b-c\): 2 tetrads from crossover between \(b\) and \(c\) * 10 = 20

To calculate the map distance between the loci \(a\) and \(b\), we need to find the proportion of tetrads showing a crossover event between these loci. Number of tetrads with crossover event between \(a\) and \(b\) = \(\frac{20}{100}\) We can also calculate the map distance between the loci \(b\) and \(c\) in the same way: Number of tetrads with crossover event between \(b\) and \(c\) = \(\frac{10}{100}\) Map distance between loci, in centiMorgans (cM), is given by the formula: Map distance (cM) = Proportion of crossover events * 100 Map distance between the loci \(a\) and \(b\) = \(\frac{20}{100} * 100 = 20\)cM Map distance between the loci \(b\) and \(c\) = \(\frac{10}{100} * 100 = 10\)cM Assuming the order \(a-b-c\), the map distance between these loci can be represented as: \(a--20--b--10--c\)