7. Three Drosophila X-linked recessive genes are y, w, and ct for yellow body, w
ID: 34001 • Letter: 7
Question
7. Three Drosophila X-linked recessive genes are y, w, and ct for yellow body, white eyes and cut wing. A yellow, white-eyed, normal-winged female was crossed with a male with cut wings but wild-type for the other two phenotypes. F1 females were wild-type for all three traits, and males were yellow with white eyes and normal wings. These flies were crossed, but only the MALES were tallied among the F2 progeny as follows:
y+c
9
+w+
6
ywc
90
+++
95
++c
424
yw+
376
y++
0
+wc
0
Diagram the genotypes of the F1 parents.
Construct a map of the genes.
Were any double crossovers expected?
Could the F2 female offspring be used to construct a map? Why or why not?
y+c
9
+w+
6
ywc
90
+++
95
++c
424
yw+
376
y++
0
+wc
0
Explanation / Answer
a)
Recombinant between ++, yw
= 9+6 =15
Recombinant between +c, y+
= 9+6+ 90+95 = 200
Recombinant between +c, w+
= 90+95 = 185
Total progeny = 1000
Recombinant frequency between y and w = 15/1000*100 = 1.5%
Recombinant frequency between y and c = 200/1000*100 = 20%
Recombinant frequency between w and c = 185/1000*100 = 18.5%
So map would be
y 1.5 w 18.5 c+
b)
y++, +wc are double crossover products. These products did not appear.
c)
F2 female offspring could not be used to construct a map because it is not recessive. Mapping can be only done while cross with recessive.
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