A true-breeding stock of Drosophila melanogaster shows phenotypes pink-eyes (pk)
ID: 53272 • Letter: A
Question
A true-breeding stock of Drosophila melanogaster shows phenotypes pink-eyes (pk), javelin-bristles (jv), and striped-body (sr). Whether the genes are sex-linked or autosomal, what their chromosomal order is, and what the distances among them are, are all as yet unknown. Females of this stock are crossed to males of a true-breeding stock for the wild-type, and both sexes of their F1 have wild-type phenotypes, +pk + jv + sr. F1 females are then crossed back to males of the true-breeding pk jv sr stock. Why do we expect that both sexes of the progeny of this test cross will yield equivalent information about recombination events occurring in their F1 mothers? Explain in terms of genotypes and phenotypes of the original P-generation cross and the resulting F1 . Now consider the test cross phenotypes, hence the genetics of the F1 female gametes, seen in the test cross progeny:
pk jv sr 298
+ pk jv sr 17
pk +jv + sr 22
pk +jv sr 120
pk jv + sr 56
+ pk + jv + sr 310
+ pk + jv sr 49
+ pk jv +sr 128
What is the order of the three genes? How do you know? What are the distances among the 3 genes, in map units? How have you determined this?
Explanation / Answer
Based on the given data,
S.no
Genotype
Number
Of progeny
Type
1
pk jv sr
298
Parental
2
+ pk jv sr
17
Double crossover (DCO)
3
pk +jv + sr
22
Double crossover (DCO)
4
pk +jv sr
120
Single crossover (DCO)
5
pk jv + sr
56
Single crossover (SCO)
6
+ pk + jv + sr
310
Parental
7
+ pk + jv sr
49
Single crossover (SCO)
8
+ pk jv +sr
128
Single crossover (SCO)
Total = 1000
So, the parental genotypes are as follows:
pk jv sr: pink-eyes (pk), javelin-bristles (jv), and striped-body (sr)
+pk + jv + sr: (wild-type)
There are 17 + 120 + 128 + 22 = 287 progeny show recombination between genes m and r.
Thus, the gene distance between pk and jv is 28.7 map units (m.u).
There are 17 + 56 + 49 + 22 = 242 progeny show recombination between genes r and t.
Thus, the gene distance between jv and sr is 24.2 (m.u), and the gene distance between pk and sr is (28.7+ 24.2 = 52.9 m.u).
From these sums it is clear that the pk and jv have the largest RF values than jv and sr; therefore the sr locus must lie between them.
The genetic map of the three genes is as follows:
S.no
Genotype
Number
Of progeny
Type
1
pk jv sr
298
Parental
2
+ pk jv sr
17
Double crossover (DCO)
3
pk +jv + sr
22
Double crossover (DCO)
4
pk +jv sr
120
Single crossover (DCO)
5
pk jv + sr
56
Single crossover (SCO)
6
+ pk + jv + sr
310
Parental
7
+ pk + jv sr
49
Single crossover (SCO)
8
+ pk jv +sr
128
Single crossover (SCO)
Total = 1000
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