In an experiment, individual homozygous for 'ab' gene were crossed with wild type. The F¹ hybrid was test crossed and progenies produced in
following ratio :
++/ab = 896
ab/ab = 880
+a/ab = 108
+b/ab = 116
calculate the distance b/w 'a' and 'b' gene.
Answers
Answered by
0
Answer:
= 11.2 %
Explanation:
The recombination frequency for this crossbreed is
= total number of progenynumber of recombinant progeny ×100
= 548+340+44+6844+68 ×100
= 11.2 %
1% recombinant frequency is equivalent to 1 centimorgan or 1 map unit or m.u.
Therefore the distance between a and b genes is 11. 2 m.u.
Answered by
0
Answer:
Explanation:This question does not make sense , homozygous for a gene indicates the two genes that make a characteristic are identical for example ( aa or BB)\
In your question the individual with 'ab' is heterozygous not homozygous.
Similar questions
Science,
4 months ago
Biology,
4 months ago
Accountancy,
8 months ago
Chemistry,
11 months ago
Geography,
11 months ago