In Drosophila , a cross was made between females expressing the three X-linked r
ID: 35359 • Letter: I
Question
In Drosophila, a cross was made between females expressing the three X-linked recessive traits, scute bristles (sc), sable body (sa), and vermillion eyes (v) and wild-type males. All females were wild type in the F1, while all males expressed all three mutant traits. The F1 flies were crossed and 1000 of the F2 males were analyzed, with the results shown below. (2 points)
scute, sable, vermillion
314
wildtype
280
sable, vermillion
150
scute
156
scute, vermillion
46
sable
30
scute, sable
14
vermillion
10
a. Write the genotypes of the parents in the original cross and the F1 male and female flies.
b. What is the order of and map distance between the three genes? Show work.
c. What is the coefficient of coincidence? Is there any interference and if so, what is it?
scute, sable, vermillion
314
wildtype
280
sable, vermillion
150
scute
156
scute, vermillion
46
sable
30
scute, sable
14
vermillion
10
Explanation / Answer
A.
Genotype of the original cross would be
Female: X+++ Xabc where a, b and c are the representatives of scute bristles, sable body and vermillion eyes respectively and + shows wild type.
Male: X+++ Y
In F1 generation
Female: X+++ Xabc
Male: Xabc Y
B.
As seen from the table provided the parental types are the maximum, i.e wild type and scute, sable and vermillion; and double crossovers are least in number, i.e. scute, sable and vermillion. Studying this, it is clearly seen that the vermillion has switched sides from parental to double crossover, which concludes that vermillion (c) is in middle.
So the map order of these three genes on X chromosome is:
a---------c---------b
Now,
For a----c distance calculation:
Ignore b for now. In the parental, a and c are on same chromosome i.e. are coupling. So, recombinant offspring, which come from a crossover between a and c. will be having ab and bc i.e. either scute and sable or sable and vermillion.
There are 156 + 150 + 14 + 10 = 330 offspring that have had a crossover betweena andc.
Total number of offspring = 1000
The map distance is the number of recombinant offspring divided by the total offspring, times 100. Here: (330/1000) * 100 = 33 map units between a and c.
For c----b distance calculation:
Ignore a for now. In the parental, c and b are on same chromosome i.e. are coupling. So, recombinant offspring, which come from a crossover between c and b. will be having ac and ab i.e. either scute and vermillion or scute and sable.
There are 46 + 30 + 14 + 10 = offspring that have had a crossover between and.
Total number of offspring = 1000
The map distance is the number of recombinant offspring divided by the total offspring, times 100. Here: (100/1000) * 100 = 10 map units between c and b.
C.
As we can see, there are 330 progeny showing recombination between genes A and C. And there are 100 progeny showing recombination between genes B and C. Thus the expected rate of double recombination is (330/ 1000) * (100 / 1000) = 0.033, or 33 per 1000.
However, we can see from the data that 14 + 10 = 24 double recombinants. The coefficient of coincidence is therefore 24 /33 = 0.73.
Interference is 1 0.73 = 0.27.