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Mary and Tom are a young couple from Montreal. They have plans to start a family but
are considering moving to Costa Rica first because they are both fed up of winter. However,
they are both aware that albinism runs in their family even though neither of them is albino.
They think it might be cruel for an albino child to live in a country where most of the social
gatherings take place in the sunny outdoors year-round.
They decide to come to you, a reputable genetic counselor for guidance. If they are not
likely to have children afflicted with albinism, they pack their bags tonight. If they are, they will
move to Laval and try and get a job at the Biodome to experience simulated tropical weather
during winters (hopefully, they won’t be made to work in the Arctic sector during summers).
As requested, they
each provide you with
enough
familial
information to construct
these pedigrees:
1) Is the inheritance
pattern autosomal or
sex-linked, dominant or
recessive?
2)
What
is
the
likelihood of affected
children?
Question 2:
The following pedigree shows a family with a rare disease that is disfiguring but not fatal.
Individuals III-5 and III-6 have been betrothed to each other. What is the probability of them
having an affected child?
Question 3
Familial hypercholesterolemia is a genetic disorder that is caused by a mutation in the gene
for the Ldl receptor. Patients develop coronary artery disease at an early age (age 35). This
disease is usually rare, but it occurs at a higher frequency in Northeastern Quebec.
You work as a genetic counselor in Northeastern Quebec. You meet with a young couple,
Pierre and Jane (ages 25 and 26) because they wish to start a family. Although neither
individual shows symptoms of coronary artery disease, both have a family history of the
disease and they are concerned that their future children may be affected as well. They
provide you with family information and you are able to reconstruct pedigrees.
a) Based on the above pedigrees, what appears to be the mode of inheritance of this
disease?
b) Based on the pedigrees, and your knowledge of the disease, what is the probability that
Jane will develop hypercholesterolemia? And Pierre?
c) What is the probability that their child will inherit the disease?
Question 4
You are at your family reunion in Oklahoma when out of nowhere a giant tornado appears.
You are a quite heroic person and need to make a snap decision about whom to save. Being
the able geneticist that you are how many of each type of relative would you save given that
you will certainly perish?
A) Siblings
B) Cousins
C) Nieces and Nephews
You have been given the task of determining the genetic basis of leaf number in clover. You
decide to cross a pure-breeding four-leaf plant to two separate, randomly chosen threeleaved plants. The results are as follows:
Cross A - 4-leaf X 3-leaf
Cross B - 4-leaf X 3-leaf
Progeny phenotype
counts
3-leaf
4-leaf
179
0
75
81
- Assuming leaf number in clover is controlled by a single gene, how would you explain your
results?
- The progeny phenotype counts in Cross B are slightly different from what you would expect.
Why?
Answer CC:
1. Autosomal, recessive.
2. Mary’s genotype must be A/a; Tom’s father has 2/3 probability of being a carrier, so
Tom has 1/3 chance of being a carrier (2/3 x 1/2).
Each child will have 1/12 chance of albinism (1/4 x 1/3).
Question 2:
III-5 must be a carrier as she got an x from dad, and III-6 can only give one X-chromosome
that has the mutation…It is X linked, so for an affected child it is probably best to look at the
probability of a girl and a boy separately and then add them up: p(girl) ½ * ½ + p(boy) ½ * ½=
½ total
Answers:
a) autosomal dominant
b) Jane has a ½ probability of developing the disease. Pierre has no probability of developing
the disease.
c) Their child has a ¼ chance of inheriting the disease. This is a function of the probability
that Jane will develop the disease (1/2) and the probability that the child will inherit this allele
from her (1/2 ).
Answers
The question is: on average, how many people would you have to save to account for the
loss of your life.
Siblings – 2
Cousins – 8
Nieces and Nephews – 4
The results are easily explained if the allele for four leaves is recessive to that for three
leaves, say L/- = three leaves and l/l = four leaves. As such, Cross A must have been l/l x L/L
and Cross B must have been l/l x L/l.
Despite the equal probability of drawing each type of gamete, in this case from the threeleaved plant (only one type of gamete from the four-leaved plant), even with a sample size
like 156, sampling error will play a role in dictating the observed numbers of progeny in each
phenotypic class.