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:

hey, i believe your answer would be d. sugar ribose and the base uracil. i believe this is the answer, because rna consists of sugar ribose, adenine, guanine, cytosine, and uracil.

your first option is incorrect, because that's dna. where it consists of sugar deoxyribose, adenine, guanine, cytosine and thymine.

your second option is also incorrect, this is because sugar deoxyribose is in dna, and uracil is in rna. these two do not make sense together.

your third option being incorrect, because sugar ribose is in rna, but not the base of thymine. rna consists of uracil, not thymine.

lastly your fourth option being correct! this answer is correct, because both sugar ribose, and the base uracil are in rna! ~

answer is d. sugar ribose and the base uracil.