An AAbb strain is crossed to an aaBB strain and the resulting progeny self-crossed. If A andB are located 10 cM apart from each other on the same chromosome, what percentage of the self-progeny will be of genotype aaBB

20.3%

Explanation:

Given that:

An AAbb strain is crossed to an aaBB strain

i.e.

                   AAbb      ×      aaBB

gametes       Ab                   aB

For F₁

generation;  

If the resulting progeny are now self-crossed.

We know that the genetic distance between these two genes on the same chromosome is said to be 10 cM.

i.e. the recombinant gene = 10%; Hence, the parental gene will be (100 - 10)% = 90%

Ab ×   aB  =   Aa    AB      ab     Bb

From above; the parental gene;

Ab and   aB = 90% with each being 45%

i.e. Ab = 45%   and aB = 45%

while the recombinants genes are:

AB and ab = 10%

i.e. AB = 5%  and ab = 5%

Finally; the percentage of aaBB is = aB% ×  aB% × 100%

the percentage of aaBB is = 0.45 × 0.45 × 100%

the percentage of aaBB is = 20.25%  ≅ 20.3%

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explanation:

jujj

that scientist would be robert hooke