Question

Data set B presents the results of a testcross using female flies heterozygous for three traits and male flies, which are homozygous recessive for the same three traits. For this part of the report do the following:You have first to determine the gene order. Construct a genetic map for these three genes, including the map distances between them. Clearly indicate the logic you followed and show all your calculations.

91 504 346 342 343 344 476 79 B5 0 + 0 + + 0 0 0 0 + 0 0 + + + 0 + + 0 0 + + 0 +

Answer 1

In a double cross over problems like this, the following steps are used to determine the map distance and gene order:

1. The lowest frequencies will be DCOs

2. The highest frequencies will be Parental types.

3. Order of the gene can be identified by comparing the DCOs and Parental types.

e.g DCOs are: a + c AND + b +

While parental types are : a + + AND + b c,

Comparing these, we can see that the middle gene is c as it is the only one that is reciprocal in the DCO and the parental type. Therefore, order of the gene is a-c-b.

4. The total population = 2525

5. Using the parental type using the gene order and we can determine where the crossovers are taking place. In this case, the single crossover between a and c would mean that the recombinants would be a c b AND + + + and the single crossover between c and b would mean that the recombinants would be a b + AND + + c

6. Map distance can be calculated using the formula:

(Numberofrecombinants + Total population) x 100

Therefore, the map distance between a and c is

(343+346+79+91) / 2525 = 859/ 2525 = 0.34 = 34 mu

Therefore, the map distance between c and b is

(342+344+91+79) / 2525 = 856/2525 = 0.339 =33.9 mu

(Numberofrecombinants + Total population) x 100