In Drosophila, kidney-shaped eye (k), cardinal eye(cd), and body (e) are three r
ID: 80694 • Letter: I
Question
In Drosophila, kidney-shaped eye (k), cardinal eye(cd), and body (e) are three recessive genes. If homozygous kidney, cardinal females are crossed with homozygous ebony males, the F1 offspring are all wild-type. If heterozygous F1 females are mated with kidney, cardinal, ebony males, the following 2,000 progeny appear 880 kidney, cardinal 887 ebony 64 kidney, ebony 67 cardinal 49 kidney 46 ebony, cardinal 3 kidney, ebony, cardinal 4 wild-type a. Determine the chromosomal composition of the F1 females. b. Derive a map of the three genes. c. Calculate the interference.Explanation / Answer
a. Determine the chromosomal composition of the F1 females.
From the flies in the P generation ("kidney, cardinal" and "ebony"), we know that in the F1 one chromosome will carry the alleles k, cd,+ while the other chromosome will carry the alleles +,+,e.
b. Derive a map of the three genes.
In this case, only two classes are in highest frequency, which indicates that all three loci are linked. That next step is to determine the gene order. To do this we must compare the parental classes (highest frequency) to the double crossover classes (lowest frequency).
For the parental classes these are:
kidney, cardinal and ebony
For the double crossover classes these are:
kidney, ebony, cardinal and wildtype
A comparison of the phenotypes from the parental classes (for example, "kidney, cardinal") with the double crossover classes (for example "kidney, ebony, cardinal") shows that the locus ebony is in the middle because it is the one locus that does not match.
The next step is to calculate the map distance among the loci. To do this we need a table as given below number of recombinants between
kidney-ebony ebony-cardinal kidney-cardinal
------------------------------------------------------------------------------------------------
64 49 49
67 46 46
3 3 64
4 4 67
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Totals 138 102 136
To calculate the distance between each locus, we use the same formula as before: the number of recombinants between the loci divided by the total number of flies and then times 100.
distance (kidney-ebony) = 138/2000 x 100 = 6.9 m.u.
distance (ebony-cardinal) = 102/2000 x 100 = 5.1 m.u.
distance (kidney-cardinal) = 6.9 + 5.1 = 12.0
The map is this.
k e cd
---+------------+------------+-------
6.9 m.u. 5.1 m.u.
C. Calculate the interference
Interference = 1 – C.O.C
C.O.C = Observed # of double recombinant / Expected # of double recombinant
Observed # of double recombinant = 7
Expected # of double recombinant = (0.069 x 0.051) X Total population = 0.003519 X 2000 = 7.038
C.O.C = 7/7.038 = 0.9946
Interference = 1 – C.O.C = 1 – 0.99 * 100
=99%