Since it is a rare disorder, assume that anyone entering the pedigree (someone w

Question

Since it is a rare disorder, assume that anyone entering the pedigree (someone whose parents are not in the pedigree) is homozygous, unless there is evidence to the contrary. Those who are not shaded show the dominant phenotype. Let A and a represent the dominant and recessive alleles, respectively. Fill in the table below to indicate the genotype or probability function for the genotype of each individual. Give all probabilities as fractions that are simplified to the lowest value. Type your fractions in # format without spaces. For example, one-half should be entered as 1/2 Failure to correctly follow these directions will result in your answer being marked as incorrect. IV Fill in the table below to indicate the genotype or the genotype probability of the individual. Two are completed as examples

Explanation / Answer

For the I-4: Since the individual has a progeny with one child with the disorder (II-3), the parents must both have the recessive allele and the dominant allele (because they are not sick). So the probabilities are 0 for AA, 1 for Aa, and 0 for aa.

For the II-1: one of the parents is healthy and the other has the disease. Since the problem states that we should asume that anyone entering the pedigree (in this case I-1) is homozygous, and that neither I-1 has the disease, nor the progeny, we know that I-1´s genotype is AA, and the parent with the disease is aa. So the test crossing would be:

All the progeny has a genotype Aa, the probability then is 100% or 1 for this genotype and 0 for aa, and AA.

For II-2: We already know that the genotype of the parents (I-3 and I-4) is Aa (solved in I-4, or first question):

We discard aa because we know the individual is not sick, so one out of three options is AA, and two out of three is Aa. So for AA, the probability is 1/3, for Aa is 2/3 .

For III-2: We know that the parent II-1 is AA because of the assumption of anybody entering the pedigree is homozygous. And we know that the parent II-2 could be AA with 1/3 of probability, or Aa with 2/3 of probaility, so this time we make three Punnet Squares, to repeat the option with Aa because it is two times more likely than AA:

So the probability for AA is 8/12 = 2/3 and for Aa is 4/12 = 1/3

For IV-1 is the same, we have a parent AA (entering the pedigree) and a parent (III-2) that has the probability of 2/3 to be AA and 1/3 to be Aa:

The probability for AA is 10/12 = 5/6 and for Aa is 2/12 = 1/6

A A a Aa Aa a Aa Aa

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