I II III II-1 III-1 11-2 Aa 1-1 11-3 Aa 1-2 11-4 Example: Given the pedigree above, what is the probability that the granddaughter (III-1) will be heterozygous?

Human Anatomy & Physiology (11th Edition)
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Chapter1: The Human Body: An Orientation
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I
II
III
II-1
11-2
Aa
1-1
11-3
Aa
1-2
II-4
III-1
Example: Given the pedigree above, what is the probability that the granddaughter (III-1) will be
heterozygous?
In order to determine the probability that III-1 is heterozygous, we need to determine the
probability of the possible genotypes for her mother (II-2). Because the grandparents are both
heterozygous, we would expect the following genotypic ratios in their offspring: ¼ AA, ½ Aa,
and 14 aa. However, we have to use all possible information, and the circle representing III-1 is
not shaded in, so she cannot be aa. Therefore, we eliminate the probability that II-2 is aa, and the
final probability for II-2 is 1/3 AA and 2/3 Aa. We assume II-1 is AA because he is marrying
into the family and we assume everyone marrying into a family is homozygous wild-type, unless
proven otherwise.
If II-2 is AA, and we assume II-1 is AA, then there is no chance that their daughter (III-1)
is Aa
If II-2 is Aa, and we assume II-1 is AA, then there is a 50% chance that their daughter
(III-1) is Aa
We add these two probabilities to get the final probability that III-1 is Aa
E. Through genetic analysis, we find that II-1 is actually heterozygous. Given this new
information, what is the probability that his daughter (III-1) is heterozygous?
F. If we change the genotype of the grandfather (I-1) to AA, what is the probability
that his granddaughter (III-1) is heterozygous? Continue to assume that II-1 is
heterozygous.
G. Let's go back to assuming that II-1 is AA. However, let's continue to say that I-1 is
also AA. What is the probability that III-1 is heterozygous?
Transcribed Image Text:I II III II-1 11-2 Aa 1-1 11-3 Aa 1-2 II-4 III-1 Example: Given the pedigree above, what is the probability that the granddaughter (III-1) will be heterozygous? In order to determine the probability that III-1 is heterozygous, we need to determine the probability of the possible genotypes for her mother (II-2). Because the grandparents are both heterozygous, we would expect the following genotypic ratios in their offspring: ¼ AA, ½ Aa, and 14 aa. However, we have to use all possible information, and the circle representing III-1 is not shaded in, so she cannot be aa. Therefore, we eliminate the probability that II-2 is aa, and the final probability for II-2 is 1/3 AA and 2/3 Aa. We assume II-1 is AA because he is marrying into the family and we assume everyone marrying into a family is homozygous wild-type, unless proven otherwise. If II-2 is AA, and we assume II-1 is AA, then there is no chance that their daughter (III-1) is Aa If II-2 is Aa, and we assume II-1 is AA, then there is a 50% chance that their daughter (III-1) is Aa We add these two probabilities to get the final probability that III-1 is Aa E. Through genetic analysis, we find that II-1 is actually heterozygous. Given this new information, what is the probability that his daughter (III-1) is heterozygous? F. If we change the genotype of the grandfather (I-1) to AA, what is the probability that his granddaughter (III-1) is heterozygous? Continue to assume that II-1 is heterozygous. G. Let's go back to assuming that II-1 is AA. However, let's continue to say that I-1 is also AA. What is the probability that III-1 is heterozygous?
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