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Individuals homozygous for linked ‘p’ and 'q' genes were crossed with wild type (+ +). The $F_{1}$ hybrid thus produced was test crossed. It produced the progeny in the following proportion:
++=800
pq= 900
+p= 100
+q= 200
Calculate the distance between p and q genes present on the same chromosome.

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30 map units

10 map units

20 map units

15 map units

answer is B.

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Detailed Solution

Number of Recombinant= 100+200=300
Total number of offspring= 800+900+100+200= 2000
Distance between p and q genes = (300/2000) x100 = 15 map units

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