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MCDB 1041 Class 7
Visualizing inheritance
patterns: Pedigrees
Review session for Quiz 1
Friday 3:30 pm Porter B121—down the hall
from coffee shop
Learning goals:
1.  Construct a pedigree from given information
2. Determine which mode of inheritance is
most likely based on information in a pedigree
3. Calculate the likelihood of a genetic event
based on a pedigree
4. Take into account information presented on
a pedigree to exclude certain genotypes
Dylan is genotype AaBB for two genes that
cause recessive diseases. If his wife is
aaBb, what is the chance his child will be a
carrier for both diseases (AaBb)?
a.  1 (100%)
b.  3/4 4 possible different genotypes of offspring
c. 1/2 and only one way to get AaBb, so ¼;
d. 1/4 Or, think about it this way:
½ chance of getting Aa x ½ chance of
e. 0
getting Bb, multiply togeter to get ¼
Carol is a carrier for two recessive disease
alleles DdFf. Her husband is DDFF. What is
the chance that their child will be a carrier
of either the d or the f allele?
a.  0
b. ¼
c. ½ 1/2
d. ¾
Because there are two possible different
e. 1 Genotypes that lead to this pheontype:
DdFF or DDFf (1/4 chance of each)
Modes of inheritance
Tells us about HOW traits are inherited
Autosomal recessive
Autosomal dominant
X-linked recessive
X-linked dominant
Y-linked
mitochondrial
Most common
An example of why pedigrees are used:
Huntington s disease is a deadly genetic neurological and
muscular disease that progresses over a course of 10-20
years (onset at 30-50 years of age)
Alan s mother died before he was 5 years old, of
Huntington s. His grandmother also died from the disease.
Alan has a sister and a brother who are each already married.
His sister has a son.
Alan wonders if he is at risk. He finds out that there is a
test to determine whether or not he will develop Huntington s
later in life.
Draw the pedigree of what you know about Alan’s family.
What is his chance of developing Huntington’s?
Autosomal dominant inheritance
(examples: polydactyly, neurofibromatosis, Huntington Disease)
Heterozygotes exhibit the phenotype.
Autosomal recessive inheritance
(examples: albinism, cystic fibrosis, sickle cell anemia, etc.)
Heterozygotes have one copy of the recessive allele
but exhibit the dominant (normal) phenotype
Note: in many autosomal recessive
pedigrees, carriers are not known (or
shown) since they have a normal
phenotype
Practice!
In the pedigree below, there could be carriers
that are unmarked. What mode of inheritance is
possible in the pedigree shown below?
a) autosomal recessive
b) autosomal dominant
c) either
I.
II.
1
1
2
Dd, DD = normal
dd = deaf
2
3
III
1
This pedigree shows a family with a form of deafness that is inherited in
an autosomal recessive manner. Members of the family with filled
symbols are deaf. Which members of this family are definitely
heterozygous (Dd)?
A.  I-1 and I-2
B.  I-1, I-2, and II-1
C.  I-1, II-1, and II-3
D.  I-1, I-2, and II-3
E.  I-1, I-2, II-1, and II-3
Up until now, we ve really only thought
about autosomal chromosomes.
What about the X and Y chromosomes?
Sex chromosomes
The sex of an individual is determined
genetically by the sex chromosomes.
X and Y -> male
X and X -> female
•  Men only have 1 X chromosome
•  Men pass on an X to daughter,
Y to son
X
© Biophoto Associates/Photo Researchers
Y
X-linked recessive traits
•  Males inherit an X only
from mom while females
inherit an X from each
parent
•  Because males inherit
their only X from their
mother, an X-linked
pedigree often has carrier
females with affected sons.
Affected son
X-linked recessive inheritance
hemophilia
X-linked dominant inheritance:
As with an autosomal dominant trait,
inheriting a single allele gives you the trait
example: congenital generalized hypertrichosis
(probably where werewolf legends came
from)
What is the most likely outcome if you can observe the
children of the affected women in generation III
(assume they all have children with normal, noncarrier males)?
a.  All male children (and only males) will have the trait
b.  All females (and only females) will have the trait
c.  Some males and no females will have the trait
d.  All males and females will have the trait
e.  Some males and some females will have the trait
Matt and Sue have 14 children, 7 boys and 7 girls.
All 7 girls have X-linked dominant hypophosphatemia
and all 7 boys are normal. People with
hypophosphatemia typically show symptoms when
they eat a lot of carbohydrates, so one can have the
disease and not know it.
X- = the X with the mutation
X+ = the normal X
The genotype for Sue that is most consistent with
this information is:
A). X+X+
B). X+XC). X-X-
Another practice question
Mary is a carrier of an X-linked recessive disease
and a carrier of an autosomal disease. If Mary
has a child with a man who has only non-disease
alleles, what is the chance their child will have the
X-linked disease?
a.  0
b. ¼
c. ½
d. ¾
e. 1