Human genetics
Basic Mendelian genetics apply
But not a model organism like Mendel’s peas
Most traits are affected by more than one
gene and cannot be analyzed using simple
Mendelian genetics.
Many single gene traits that can be
followed cause rare diseases, rather than
common phenotypes.
Genetics is not straight forward:
No pure breeding humans;
Generation time is long;
No controlled matings;
Siblings rarely mate to generate F2.
Small family sizes
How can one study human genes, and map genes
that cause human diseases?
• Geneticists make use of the variation that already
exists in the population.
• Pedigree analysis. Rare traits can be studied by
carefully analyzing families that are affected. In other
words, the trait may be rare in the population but
common in a particular family.
Important assumption: When a trait is rare, and there is a
random mating, you can always assume the person is not a
carrier (heterozygotes).
Pedigree Analysis
male
affected
individual
female
mating
offspring
Fetus or unspecified progeny
Filled, means affected, open means unaffected (normal).
(brachydactyly, 1906).
 Dominant single gene inheritance.
 Heterozygotes (carriers) show phenotype.
aa
aa
aa
aa aa
aa
Aa
aa
Aa Aa aa aa
Aa Aa Aa aa
?
1/2
1. Assign the genotypes to all individuals,
left top to bottom right, with Aa, etc.
2. Probability that diamond is affected?
Recessive (rare) Traits
- affected individuals appear in
progeny of unaffected parents
- frequently show up in
consanguineous matings
(partners are blood related)
- all progeny from two affected
individuals are affected
A-
aa
Aa
A-
Aa
A- Aa
Aa
Aa
Aa
aa
aa
Fill in the genotypes, left to right, top to bottom, with Aa etc.
What is the chance the child will be affected by this rare disease?
A-
aa
Aa
Assume AA
=rare trait
A-
1/2 Aa A-
A- Aa
1/2 Aa A-
Aa
A-2/3 Aa aa
A
a
A
AA
Aa
a
Aa
aa
1/4 aa
Dad carrier
(1/2 x 1/2)
Mom carrier
x (2/3)
Child aa
x 1/4 =
1/24 Law of the product
Tips for pedigree analysis
• For an individual to be affected (homozygous), both
parents need to be carriers (heterozygous). Therefore,
to calculate the chance that an unknown individual will be
affected, calculate the chances that the parents will be
heterozygous, then multiply these by the chance that the
offspring will get both the the mutant alleles.
• For rare traits, make the simplifying assumption that
unrelated, unaffected individuals are not carriers.
Cystic fibrosis is a rare recessive disorder. Jack whose brother
has the disease marries Jill whose aunt has the disease.
a) If they have one child, what is the probability that the
child will have cystic fibrosis?
b) If their first child has cystic fibrosis, what is the
probability that their second child will have cystic fibrosis?
c) If their second child has cystic fibrosis, what is the
probability that their third child will have cystic fibrosis?
d) If their first child has cystic fibrosis, what is the probability
that of their next three children, two will be carriers and the
other affected?
Cystic fibrosis is a rare recessive disorder. Jack whose brother
has the disease marries Jill whose aunt has the disease.
a) If they have one child, what is the probability that the
child will have cystic fibrosis?
Aa
Aa
Aa
Jack
aa
AA
Jill
A-
A2/3 Aa
A- 1/2 Aa
2/3 Aa
1/4 aa
2/3 X 2/3 X 1/2 X 1/4 = 1/18
Aa
aa
b) If their first child has cystic fibrosis, what is the
probability that their second child will have cystic fibrosis?
Aa
Aa
Aa
Jack
aa
A2/3 Aa
AA
A-
Aa
aa
2/3 Aa
Jill
A- 1/2 Aa
Aa
1/4 aa
c) If their second child has cystic fibrosis, what is the
probability that their third child will have cystic fibrosis? 1/4
d) If their first child has cystic fibrosis, what is the probability
that of their next three children, two will be carriers and the
other affected?
Jill
Jack
Aa
Aa
?
How many ways?
het het
het
het
het het
3!
1! 2!
=
Het Aa = 1/2
Ho aa = 1/4
Child 1
Child 2
Child 3
1/2
X 1/2
X 1/4 = 1/16
=3
1X2X3
1X1X2
= 1/16 + 1/16 + 1/16 = 3/16
=3
Genotype/Phenotype
Chapter 3
So far we have looked at traits that are clearly dominant or recessive.
X
Purple
White
F1
Precursor
Purple pigment
Purple pigment forming enzyme
Purple
P = Dominant allele makes a normal enzyme
p = Recessive allele makes a defective one
But nature is usually more complicated that this, and Mendel was successful
because he chose simple examples to start with (and he got lucky).
Unifying theme of Chapter 3: examples of variations of
Mendelian phenotypic ratios.
(variations of 3:1 or 9:3:3:1).
The underlying genetic ratios are unchanged.
In a Punnett square, ratios of the gamete types are the same as
before.
In this way, Mendel’s laws are not broken at all.
Extensions to Mendelian Analysis: relating genotype and phenotype.
Variations with single genes
• Incomplete dominance
• Co-dominance
• Multiple alleles • Lethal Alleles
• Pleiotropic Effects
Variations with multiple genes
 Two genes affecting the same trait
 Complementation
 Duplicate genes
 Epistasis
Variations of genes with the environment
 Penetrance and Expressivity
 Effect of the environment on phenotype