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Single Gene Mutations and
Inheritance II
April 4, 2008
Lisa Schimmenti, M.D.
Objectives
•
•
•
•
Understand autosomal recessive inheritance.
Know the concept of Lyonization.
Learn and understand the various forms of X
linked inheritance.
Know how to calculate the risk of affected
offspring given a family history and some
facts about carrier frequency using the HardyWeinberg Calculation.
Autosomal Recessive
pattern of inheritance
• Males and females equally affected
• Condition seen in sibs but not
•
•
•
usually in other relatives
Risk of recurrence in sibs 25%
2/3 of unaffected sibs are carriers
Consanguinity increases risk
Autosomal Recessive
Probabilities
Mother
A
A
AA
a
Aa
a
Aa
aa
25% affected
2/3 of unaffected are
carriers
Loss of function alleles
• Likely when point mutations give
•
•
the same phenotype as deletion of
an allele
Inherited as recessive traits when
50% of expression is sufficient for
normal phenotype
Loss of function and gain of
function changes in the same gene
will cause different diseases
Effect of consanguinity
DD
DD
Dd
Dd
DD
Dd
Dd
DD
Dd
DD
Dd
DD
dd
Consanguinity increases the risk of sharing a common ancestral mutation
Sweeping generalizations
about genes inherited as AR
traits
•
•
•
Transcription from one allele is sufficient
Severity dependent on nature and location of
mutation
What we describe in inheritance pattern as
homozygosity is USUALLY compound
heterozygosity at molecular level. Allelic
heterogeneity is COMMON
A common type of AR trait
Mutation in enzyme gene
•
Example:
–degradation of mucopolysaccharides
requires a series of lysosomal
enzymes.
–abnormalities in any of these
enzymes can cause a similar
phenotype:
• coarse facies, enlarged organs, skeletal
One MPS disease:
a-L-iduronidase deficiency
• Severe mutations =
•
Hurler
Milder mutations =
Scheie
http://medgen.genetics.utah.edu
/
http://www.mpssociety.ca/gallery/ReaganKnight.html
Phenotype is dependent on nature of mutation in alleles
A family whose child is affected with maple syrup urine disease (MSUD) has sought
analysis of their child’s MSUD mutations. MSUD is an enzyme deficiency (inborn error of
metabolism) that is inherited in an autosomal recessive pattern.
Two mutations are identified in the child’s DNA. The first deletes two base pairs in the
coding sequence in exon 1 o f the gene, the second is a T-to-A transversion that alteres a
tyrosine to an asparagine at residue 394. W hich of the following statements are TRUE.
a.
b.
c.
d.
e.
The first mutation is likely to be mild because it is “in frame.”
Given the nature of the mutations the family is unlikely to have another child that
is similarly affected.
Neither mutation is likely disease-causing as one yields a frame shift and the other
a simple aminoacid substitution.
The child is a compound heterozygote for this gene locus.
The second mutation is likely to be paternally derived because it is a transversion.
Genes in populations
How recessive conditions
occur in population groups.
Deafness
•
•
•
•
1 in 1000 infants is born deaf.
Infants are screened for deafness/hearing
loss at birth by automated hearing screening
methods.
There are over 400 genes that cause hearing
loss
Autosomal recessive mutations in GJB2,
encoding Connexin 26, will be causative in
nearly half of all deaf individuals.
Photo credit: Josie Helmbrecht
Hearing loss is the most
common condition found at
birth
Gene
Mutations in Connexin26
are a common cause of
hearing
loss
Start codon
Exon 1
158 bp
Exon2
681 bp
Protein
Six connexons form a hemichannel in one cell membrane.
When two cell membranes meet, a gap junction is formed.
Connexin 26
•
Common deafness alleles
– One method for determining carrier frequency
• Screen hearing individuals for the presence of the gene
•
•
•
mutation.
Remember, we have two of each gene
Carriers are heterozygotes
– one wildtype and one mutant allele
35delG
– This is the most common allele in European
populations.
– How do you determine the carrier frequency
without testing hundreds of individuals?
Hardy-Weinberg Law
•
•
A mathematical model for calculating allele
and gene frequencies in a population
Assumes
– Large population with random mating
– Allele frequencies remain constant
• no new mutations
• no reproductive selection bias
• no significant immigration for populations with different
allele frequencies
Allele frequencies
• p = the frequency of allele 1
• q = the frequency of allele 2
The sum of all alleles for a population is 1
p + q =1
Hardy Weinberg Calculation
• Binomial expansion
(p+q)2 = p2 + 2pq +q2
p2 = the number of homozygous wildtype individuals in the
population
2pq = the number heterozygous carriers in the population
q2 = the number of homozygous affected individuals in the
population
Using the Hardy Weinberg
calculation to determine the
carrier frequency in a
population
q2 = affected individuals in the population
for Connexin 26 related deafness:
q2 =1 in 2000
(an approximated number for this exercise)
p2 @ 1
Solve for 2pq
Using the Hardy Weinberg
calculation to determine the
carrier frequency in a population
q2 =1 in 2000 = 0.0005
q @ 0.02
2pq = 0.04
The carrier frequency is 0.04 or 1 in 25.
•
Hardy Weinberg
calculations are used in
clinic every day
Clinical situation:
– A 23 year old hearing woman named Marie comes
to your genetics clinic. Her family is of European
descent Marie has a deaf sister, Sally, who has
deafness caused by mutations in the gene
encoding Connexin 26. Her sister's genotype is
35delG/35delG.
– Marie would like to know her chance of having
deaf children caused by mutations in Connexin 26
if the father of the child is of European descent.
How do you answer
Marie's question?
•
What is Marie's chance of being a carrier of
35delG?
•
What is the chance that the father of Marie's
child will be a carrier of 35delG?
•
What is the chance that Marie will have a
child with hearing loss?
Draw a pedigree
Marie
Sally
What is Marie's chance of
being a carrier of 35delG?
• Marie's parents are obligate carriers.
• Marie is unaffected.
• The chance of being a carrier
if unaffected is 2/3.
Use a Punnett square if in doubt. Wt/35delG
Marie
Wt/35delG
Sally
35delG/35delG
What is the chance that the father of
Marie's child will be a carrier of
35delG?
q2 =1 in 2000 = 0.0005
q @ 0.02
2pq = 0.04
The carrier frequency is 0.04 or 1 in 25.
The father's chance of being a carrier is 1/25.
What is the chance that Marie will
have a child with hearing loss?
•
The chance that an affected child will be born
to a carrier couple is 1 in 4 with each
pregnancy.
•
Calculation:
Father of baby's chance of being a carrier
• 2/3 x 1/25 x 1/4 = 1/150
Marie's chance of being a carrier
Inheritance patterns and genes
• Single allele change gives disease
– Genes on X (X - linked)
– Genes on autosome (AD)
• Mutation in both alleles required for
disease
– Genes on autosome (AR)
– (rare) female homozygote for XL
X-linked patterns of
inheritance
• X-linked recessive
– Primarily males affected
– Females typically unaffected, but there are
exceptions
• X-linked dominant
– Male surviving
– Either sex affected, males more severely than females
– Male lethal
– Only females seen with disease, mosaic patterns of
expression
X-linkage: already exceptions
to the “rules”
• The lines of dominant and recessive are
blurred
• X-linked conditions are sometimes
called “semi-dominant” in women
• Need dosage compensation
Lyonization
• Only one X is active in each cell. All
others are inactivated.
– X inactivation is random
– X inactivation is fixed
– X inactivation occurs early in development
(late blastocyst)
XX
Lyonization
XX XX XX
XX
XX
XX
XX
XX XX
XX
X = active X
I = inactive X
X Inactivation
XI IX IX
IX
IX
XI
XI
IX
IX
XI
Consequences of
“Lyonization”
• Females are mosaic for their X-linked gene
manifestations
– should be roughly equal
• All are functionally hemizygous
– dosage compensation: have the same number of
copies as males
• Inactive X may be evident in cells
– The Barr body
• Shifting from random distribution may result in
manifestation of disease in females
– skewed lyonization
X-linked recessive “rules”
• Affects mainly males
– Affected males are usually born to unaffected
parents
– Females can be affected if born to an affected
father and carrier mother
– Females can be affected if they have very skewed
lyonization
• No male to male transmission
• Males born to carrier mother have 50% risk of
inheriting altered gene
X-linked recessive
What does the pattern look like?
Carrier female
Affected male
Normal male
Normal female
X-linked recessive recurrence
risks
Mother
X1
X1
Y
X1X1
X1Y
X2
X1X2
Daughters
50% normal
50% carriers
X2Y
Sons
50% normal
50% affected
Duchenne/Becker
Muscular Dystrophy
• Mutations in dystrophin gene
• Duchenne
– Typically del/dup but frameshift
– Severe
– No protein product
• Becker
– Also del/dup in frame so milder
– Shortened protein
Clinical manifestations of
Duchenne MD
http://www.mdausa.org/publications/fa-md-9.html
Onset before age 5
Calf pseudohypertrophy
1/3500 males
Elevated creatine kinase
Severe muscle wasting
Affects resp and cardiac mm
Females 8-10% some weakness
CK > 95%tile in 2/3 of carriers
http://www.neuro.wustl.edu/neuromuscular/musdist/lg.html
Dystrophin gene and molecule
Huge gene (2.3 Mb)
14kb transcript
3685 AAs
High mutation rate (10-4)
http://www.ncbi.nlm.nih.gov/cgi-bin/SCIENCE96/gene?DMD
Using dystrophin antibody
normal
Becker
Duchenne
B B n D D
http://www.neuro.wustl.edu/neuromuscular/musdist/lg.html
X-linked dominant “rules”
• Males and females both affected , but
typically males are worse than females.
• If male lethal, only affected females observed.
• Affected males will only have affected
daughters and unaffected sons.
• No male -> male transmission seen
X-linked dominant
What does the pattern look like?
Affected female
Affected male
Normal male
Normal female
Example with male survival
X-linked dominant, male lethal
What does the pattern look like?
Affected female
Normal male
Normal female
X-linked dominant recurrence
risks
Mother
X1
X1
Y
X1X1
X1Y
X2
X1X2
Daughters
50% normal
50% affected
X2Y
Sons
50% normal
50% affected
Incontinentia pigmenti
X-linked dominant
(male lethal condition)
•
•
•
•
•
•
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Linear blisters in newborn girls
“Crops” followed by scarring
Small teeth
Eye abnormalities
Patchy hair loss
Only girls are affected
Males are typically not affected
Infant with IP
Linear blisters on leg
Linear erosions on soles
http://dermatology.cdlib.org/DOJvol4num1/path/incont.html
Potentially confusing in
X-linked pedigree analysis
• Male lethal X-linked conditions
• New mutations may be hard to
recognize as X-linked
• Sex-limited conditions may look Xlinked
• With carrier mother and affected father
can see “male to male” inheritance
Fred Jones is red-green color blind as i s his son, Frank, but Frank’s mom , Freida, is not.
However, as an expert in genetics, you know that red-green color blindness is inherited in
an X-linked pattern. What is the most likely explanation for the apparent inheritance of
this trait in this family?
a.
Color blindness is a sex-limited trait occurring only in males
b.
Frank is probably not Fred’s biologic offspring
c.
Frank has a new mutation for red-green color blindness
d.
Freida is a carrier (heterozygote) for red-green color blindness
e.
Males homozygous for X-linked traits usually don’t survive.
What is the chance that the next BOY born to this couple will also be red-green color blind?
a.
100% because Frank is affected.
b.
100% because Freida is a carrier.
c.
50% because Fred is affected.
d.
50% because Frieda is a carrier.
e.
50% because Frank is affected.