Individuals homozygous for 'xy' genes were crossed with wild type '++' . The `F_1` dihybrid thus produced was test crossed . It produced progeny in the following in the following ratio `'+ + ' 900, ' +y ' 115, 'xy' 880, 'x+' 105. Find out the recombination frequency .
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Answer:
11 units
Explanation:
The distance between c and d genes = ( Recombinants / Total individuals ) x 100
Total individuals = 2000
Recombinants = 220
= ( 220 / 2000 ) x 100
= 22 / 2
= 11
So correct answer is ' C '.
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