In a test-cross between an individual who is AaBb, and an individual who is aabb, the following types and numbers of offspring were obtained:
22 AaBb
67 Aabb
70 aaBb
21 aabb
Which of the following is true?
| a. |
The A and B genes are linked to each other, and are 43 cM apart. |
|
| b. |
The recombinant gametes from the double heterozygote are Ab and aB |
|
| c. |
The A and B alleles in the AaBb parent are present in repulsion |
|
| d. |
It is possible that the two genes are present on different chromosomes |
AaBb X aabb
Test cross AB/ab X ab/ab
1. AB/ab =22
2. ab/ab =21
3. Ab/ab = 67
4. aB/ab = 70
Here 1 & 2 are recombinant and frequency is 22+21= 43 and parental combination Ab & aB
a) so map distance between a & b is 43 map unit
b) No recombinant gamate is not Ab & aB. In this test cross recombinant gamate is AB & ab.
c) a & b allele is this parent present in trans position and they are in repulsion
d) As the genes are linked and their distance is below 50% so they are present on same chromosome.
In a test-cross between an individual who is AaBb, and an individual who is aabb, the...
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explain
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