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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Biology
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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?
I think it is b, right?
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Let me think about this one lol
Cool, take your time, let me know if you would like help. :0
1/16

Not the answer you are looking for?

Search for more explanations.

Ask your own question