BICD 100
Week 4 Handout: Pedigrees, Isolating Mutants
This week’s section will cover some of the more challenging types of pedigree problems, along
with an introduction to isolating mutants.
There is less material this week, but spend some time carefully going over what has been
covered in class so far.
Some of the upcoming topics (recombination, linkage, chromosome
mapping) will require a solid understanding of the previous material!
Advice
:
• There are three kinds of pedigree problems that involve probability calculations:
1)
Carrier Probability
This kind of problem will ask for the probability of a specific individual on a pedigree of
being a carrier.
To solve this problem (1) determine the mode of inheritance of the allele
(this might be indicated in the problem), (2) calculate the probability of each prior relative
inheriting the allele, starting at the affected relative.
(3) Then, use the product rule to
determine P(carrier), since each inheritance of each allele is an independent event.
Example:
In this pedigree, individual a has a rare autosomal recessive disorder (genotype
tt
).
What is the probability that individual i will be a carrier?
Answer: In order for individual i to be a carrier, she must
inherit a
t
allele from individual g, who inherits the
t
allele
from individual d, who inherits the
t
allele from individual a.
So to solve this problem, calculate the probability of each
prior individual inheriting the
t
allale, then use the product
rule.
P(individual d inheriting the
t
allele) = 1 (she must inherit it, since individual a is
homozygous recessive)
P(individual g inheriting the
t
allele) = ½
P(individual i inheriting the
t
allele) = 1 x ½ x ½ =
¼
*a more difficult question will require probability calculations of two parents being carriers,
then finding the probability of their child having the disease*
2)
Probability of Multiple Cases:
This kind of problem asks for multiple scenerios to be analyzed, then the probabilities are
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Example:
This pedigree shows inheritance of a rare
autosomal dominant disorder.
Those individuals
suffering from the disease are shown in black, and
they are heterozygous for the disease (genotype
Aa).
However, some of the kids in generation III
will have the mutant allele and will develop the
disease, but they have not yet shown the
symptoms.
If individuals IIIe and IIIk have a child
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 Spring '08
 SOOWAL
 Genetics, Probability, Zygosity, Phenylketonuria

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