Genes A and B are on different chromosomes. What fraction of the offspring of the cross
AaBb x AaBb will be homozygous for all the dominant alleles?
A. 1/64
B 3/9
C 1/32
D.1/16
B 1/9
The answer is D. May I know the explanation?
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is the answer 1/16? i think you made a small typo
here how I solved this question:
from the first Aa you have 1/2 chance of getting either A or the a.
from Bb you again have a 1/2 chance of getting the either B or b.
for the second one in the cross we do this again
we have 1/2 chance of getting one of the A or a from this one and the same for B or b.
at the end the probability we need is the multiplication of all which means:
1/2x1/2x1/2/x1/2= 1/16
let me know if you got confused in any way
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I would agree with @nihantokar
I have compared it with F2 gen of mendelian dihybrid cross. Or it could also be solved by product rule as mentioned above by Nihan.
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@Ibraheem_Saeed that’s the best way to solve this question in order to not get confused but as in the exam we have limited time I wanted to use the product rule.
thanks for sharing!
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Isn’t is 1/16?
Because like homozygous for DOMINANT Alleles is essentially AABB. Which only happens once in a dihybrid cross.
Same I also think it is a typographical error
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Yes it was a typo. I get it. Thank you.
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