A plant of the genotype AaBb is selfed. The two genes are linked and a...
Given that the genes A and B are linked by 50 map unit apart. So the frequency of recombinant is 50%. That is crossing the AaBb with aabb will results 50% recombinants and 50% parental gametes (AB : aB : Ab : ab : : 1 : 1 : 1 : 1). Thus the genes will seggregate by independent assortment as proposed by Mendel.
According to Mendelian dihybrid cross selfing of AaBb will result in aabbprogeneis in the ratio of 1/16.
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A plant of the genotype AaBb is selfed. The two genes are linked and a...
Solution:
Given, AaBb plant is selfed and the two genes are linked and 50 map units apart.
The recombination frequency between the two genes is 50%, which means that 50% of the gametes will be parental and 50% will be recombinant.
The gametes produced by the selfed AaBb plant are AB, Ab, aB, and ab in equal proportions.
The possible genotypes of the progeny and their expected frequencies can be determined using a Punnett square as shown below:
| | AB | Ab | aB | ab |
|---|------|------|------|------|
| AB | AABB | AABb | aaBB | aaBb |
| Ab | AABb | AAbb | aaBb | aabb |
| aB | aaBB | aaBb | AaBB | AaBb |
| ab | aaBb | aabb | AaBb | Aabb |
From the Punnett square, it can be seen that the expected frequency of progeny with the genotype aabb is 1/16 or 6.25%.
Therefore, the correct answer is option D, 1/16.