The best way to approach the 2nd question is by way of drawing a pedigree. Well...I don't think one is really needed for this question since there are
ONLY two generations.
Question 2So, let's define our alleles:
A = dominant (seborrheic keratosis)
a = recessive (normal)
So, there can be either one of two crosses:
Aa (man) x aa (woman)
or AA (man) x aa (woman)
We know that a man has the disease, so he must have one of the following genotypes:
Aa or
AA. Were not sure which one because the question doesn't specify whether he's heterozygous or homozygous. So, the man's genotype would then be
A_. The _ could either be the "
A" allele or the "
a" allele. The woman's genotype would have to be
aa since she is normal.
If the man's genotype were to be
Aa, the children would either have the genotypes
Aa or
aa. So,
1/2 normal and
1/2 carriers.
If the man's genotype were to be
AA, then ALL the children would have the genotype
Aa and would all be carriers.
a) The man could either be AA or Aa, so 1/2. Since this question is asking for the probability that all three children will be normal, you would only have to consider the genotype Aa, since the "
a" allele is normal. So, the probability that all three children would be normal would be (1/2)^3 =
1/8b) The answer to this question would be the same as question 2a. This is because the question is asking for the chance that all three children would be affected, and therefore you would have to again consider AA or Aa, so 1/2. For a child to be affected, he can either have the genotypes AA or Aa, so the answer should be
1/8.
I know this is a little confusing, but if you still have any problems understanding this question, please feel free to send me a personal message and I will be more than willing to explain it to you.
Hope this helps!
I will post the 3rd question next.
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