You perform the following dihybrid cross: A blaBx Ab/aB. The A and B genes control different traits. What phenotypic ratio would you expect to see in the progeny if the two genes are unlinked? What if they are linked?

a. Unlinked: 1:1: 1:1/Linked: 1: 2: 2:1
b. Unlinked: 1:1:11/Linked: 2: 1 1:2
c. Unlinked: 1:1:11/Linked: >9: 3:<3:1
d. Unlinked: 9: 33: 1/Linked: <9:3:3:
e. Unlinked: 9: 3: 3:1 Linked: >9: <3:<3:>1
f. Unlinked: 1:1:1:1Linked: <9:>3:3:<1

Option-D

Explanation:

In the given question the genotype of parental allele is- AbaB X AbaB. The alleles formed will be- AB, Ab,aB, ab in which the crossover allele will be AB and ab.

1. Unlinked genes

Now of these alleles are unlinked then the alleles will assort independently of each other therefore performing the punnett square we will get dihybrid ratio as 9:3:3:1 in which  

1. A_B_ are 9

2. aaB_ are 3

3. A_bb are-3

4. aabb- 1

Since the crossover allele are AB and ab therefore A_B_ are 9 and aabb- 1 represent recombinant genotype.

2. Linked genes

When the genes will be linked then chances of crossing over will be reduced as a result of which only two type of allele will be formed which are-  Ab and aB and the recombinant frequency will become less.

Crossing these allele will  form Aabb , AaBb , AaBb  and aaBB  genotypes. The recombinant ratio will decrease as A_B_ is 2 and aabb is not present.

Thus, option-4 which states that the unlinked ration is  9:3:3:1  and linked ratio is  <9: >3: >3: <1 is correct.

jjrnfnf

explanation:

b r

he may like cookies