Practice Test 2 – PCB3063
Page 1 of 6
1.
Consider the following pedigree, which has individuals that are hypersensitive to UV radiation
indicated by solid black:
a.
What is the mode of inheritance for the gene that causes the hypersensitivity? (dominant
or recessive)
Recessive – because affected children can have unaffected parents.
b.
What is the probability that individual 4 in generation III would have a hypersensitive
child if he married a woman that was hypersensitive?
Individual 4 in generation III is not hypersensitive, so they must be heterozygous or homozygous for
the dominant allele. Call the dominant (nonhypersensitive) allele
H
and the recessive (hypersensitive)
allele
h.
His parents must both be heterozygotes, so (in general) the probability they would have an
HH
child is
¼
, the probability of a
Hh
heterozygote is
½
, and the probability of a
hh
child is
¼
. However,
individual 4 of generation 3 cannot
be
hh
– this means he has a 1/3 probability of being
HH
and a 2/3
probability of being
Hh
. If he had a child with a hypersensitive woman, there is a
½
probability of the
child being hypersensitive if
he is heterozygous (
Hh
). If he is homozygous dominant, the child cannot
be hypersensitive. Thus, the probability that individual 4 in generation III would have a hypersensitive
child if he married a woman that was hypersensitive is 2/3 times 1/2 = 1/3
.
c.
What is the probability that individual 6 in generation III would have a hypersensitive
child if she married a nonhypersensitive man who has a son from a previous
relationship that is hypersensitive? Explain your answer.
The genotypic probabilities for individual 6 in generation III are identical to those for individual 4 in
generation III (see part b – the probabilities are 1/3 probability of being
HH
and a 2/3 probability of
being
Hh
). If a nonhypersensitive man had a hypersensitive child from a previous relationship, he
must
be heterozygous. If individual 6 in generation III is heterozygous, the probability that they will
have a hypersensitive child is 1/4. So the probability that individual 4 in generation III would have a
hypersensitive child if he married a woman that was hypersensitive is 2/3 times 1/4 = 1/6
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Practice Test 2 – PCB3063
Page 2 of 6
2.
In a family of five children what is the probability that:
a)
all are males?
0.5
5
b)
three are males and two are females?
0.5
5
– Note that this is the probability of any specific
birth order (e.g., M M F M F). In previous years I
have taught the use of binomial probability formula which would allow you to calculate the probability
of any
birth order (e.g., three males and two females might be born in the order FFMMM or FMFMM
or FMMFM or etc.). You don’t have to know how to calculate this.
c)
two are males and three are females?
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 MARTA
 Probability

Click to edit the document details