For this case we have the following functions:
We must find the product of both functions:
We apply distributive property term to term:
We add similar terms:
Answer:
Option B
I need help on questions 3 and 4
1 answer:
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I added a picture of a badly-drawn cross as an example to help me explain. Let's say h is the recessive allele, in this case, that would cause a recessive condition. The heterozygous Hh parent is the carrier and they are being crossed with a homozygous dominant HH parent.
As you can see, all of the children are either homozygous dominants HH or heterozygous carriers Hh. None of the children can be homozygous recessive hh because there is not a recessive allele h to inherit from the other parent. Because H is dominant over h, the recessive condition won't be displayed.
As you can see, all of the children are either homozygous dominants HH or heterozygous carriers Hh. None of the children can be homozygous recessive hh because there is not a recessive allele h to inherit from the other parent. Because H is dominant over h, the recessive condition won't be displayed.