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The probability that a dihybrid cross of heterozygotes will produce an individual exhibiting BOTH recessive traits is:
9/16
1/4
3/16
1/16
 one year ago
 one year ago
The probability that a dihybrid cross of heterozygotes will produce an individual exhibiting BOTH recessive traits is: 9/16 1/4 3/16 1/16
 one year ago
 one year ago

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bluesBest ResponseYou've already chosen the best response.0
The best way to do a question like this is to set up a dihybrid cross and do it. It tells you that both parents are heterozygotes for both traits. So you could call their genotpyes AaBb. What you want to do next is set up the actual Punnett Square for AaBb X AaBb, and count the number of offspring that are AaBb themselves. Is that enough guidance, or would you like help setting up the Punnett Square?
 one year ago

churrityBest ResponseYou've already chosen the best response.0
I think it is b, right?
 one year ago

bluesBest ResponseYou've already chosen the best response.0
Oh, I read the question wrong. It wants the offspring that will be aabb. What you should do is actually set up the square, fill in the offspring's genotypes, and see how many of them are aabb.
 one year ago

churrityBest ResponseYou've already chosen the best response.0
Let me think about this one lol
 one year ago

bluesBest ResponseYou've already chosen the best response.0
Cool, take your time, let me know if you would like help. :0
 one year ago
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