Transcript Identity by descent

TOPICS IN QUANTITATIVE AND
POPULATION GENETICS
GENETICS 202
Jon Bernstein
Department of Pediatrics
October 6, 2015
Session Goals

Understand and demonstrate how the Hardy-Weinberg
distribution can be used to estimate carrier frequencies from
disease incidence for autosomal recessive conditions.

Recognize that different individuals have different
perceptions of quantitatively similar risks.

Understand how and to what extent consanguinity can
increase the risk of autosomal recessive conditions.
Session Goals

Understand how Bayes Theorem can be used to revise a
risk estimate based on new clinical information
◦ (short video available)
Lecture Outline

Clinical Case
◦ Hardy-Weinberg Distribution
◦ Consanguinity – Identity by descent
◦ Bayes Theorem
 Updating a risk estimate using new information
Clinical case:

A father in a first cousin
marriage is a carrier for
ARPKD. They have had
had 1 healthy child.
◦ How does the presence of
consanguinity and the
health child affect the risk
estimate for the mother
to be a carrier?
Aa
??
?
??
Autosomal Recessive Polycystic Kidney Disease

Incidence
◦ Approximately 1 in 15,000
Mutations in PKHD1 on
chromosome 6p
 Classically features multiple
bilateral kidney cysts at a
young age and hepatic
fibrosis

Avni, 2002, Pediatric Radiology, Hereditary polycystic kidney diseases in children: changing
sonographic appearances through childhood, PMID: 12164348
Autosomal Recessive – Recurrence Risk – The
generic problem

What is the chance that
each parent is a carrier and
passes on a mutation?
Aa
(_ x _) x( _ x _)
Aa
?
aa
(1 x 1/2) x( 1 x 1/2)=1/4
??
Autosomal Recessive – Recurrence Risk

What if one parent is
known to be a carrier and
the genetic status of the
other is unknown?
Aa
??
?
(_ x _) x ( _ x _)
(1 x 1/2) x( ? x 1/2)=?
??
Population Frequency and Risk to be a Carrier

If the incidence of a
recessive disease is 1 in
5,000; is the risk for the
circled family member to
be a carrier higher or
lower than if the
incidence were 1 in
15,000?
Aa
??
?
??
How can we estimate the risk that an individual is
a carrier?
In general, each person
has two copies (alleles) of
each autosomal gene.
A=dominant allele
a=recessive allele
Possible genotypes at a
given locus
AA
Aa
aA
aa
These can be considered
together
How can we estimate the risk that an individual is
a carrier?
The population of people
has a population of
alleles – the gene pool.
 Under certain conditions
one can assume that the
two alleles at a locus in
an individual are selected
at random from the gene
pool.

AA
a a
a A A
The Gene Pool
How can we estimate the risk that an individual is
a carrier?

If we select one allele at
random the chance that
it will be “A” or “a” is
dependent on the
relative frequency of
these alleles in the gene
pool.
_
AA
a a
a A A
The Gene Pool
How can we estimate the risk that an individual is
a carrier?






What if we are selecting
two alleles from the gene
pool?
There are four possible
combinations (3 genotypes)
AA
Aa
Can be considered together
aA
aa
__
AA
a a
a A A
The Gene Pool
How can we estimate the risk that an individual is
a carrier?
What are the chances of
having each of these
genotypes?
AA
Aa
aA
aa
Chance of AA is:
Probability of drawing A twice
in a row.
If “p” is the probability of
drawing an “A” then the
chance of drawing two “A”’s in
a row (AA) is p x p=p2
How can we estimate the risk that an individual is
a carrier?
What are the chances of
having each of these
genotypes?
AA
Aa
aA
aa
Chance of aa is:
Probability of drawing “a”
twice in a row.
If q is the probability of
drawing an “a” then the
chance of drawing two “a”’s in
a row (aa) is q x q=q2
How can we estimate the risk that an individual is
a carrier?
What are the chances of
having each of these
genotypes?
AA
Aa
aA
aa
Can be considered together
Chance of drawing Aa or aA is:
Probability of drawing “A” and then
“a” OR “a” and then “A”
Chances of drawing Aa is p x q=pq
Chances of drawing aA is q x p=pq
Chances of Aa or aA is pq + pq=2pq
How can we estimate the risk that an individual is
a carrier?

At a given locus, in the case of two alleles
◦ p=frequency the dominant allele, A
◦ q=frequency of the recessive allele, a
 p2 is the frequency of homozygotes, AA
 q2 is the frequency of homozygotes, aa
 2pq=frequency of heterozygotes, Aa or aA, (collectively referred to as Aa)

The Hardy-Weinberg Distribution
◦ p2+2pq+q2
◦ (p+q)(p+q)
Autosomal Recessive – Recurrence Risk

Carrier frequency in population = 2pq

◦ This is also the risk for someone with no family history of the
condition to be a carrier
If (p+q)(p+q)=p2+2pq+q2=1 and the recessive condition is rare

2pq  2 X 1 x 𝐼𝑛𝑐𝑖𝑑𝑒𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛
Autosomal Recessive – Recurrence Risk

What is the risk to the next
child if one parent is known
to be a carrier and the
genetic status of the other
is unknown?
(_ x _) x ( _ x _)
(1 x 1/2) x( ? x 1/2)=?
Aa
??
?
??
Carrier frequency for ARPKD

Incidence = 1 in 15,000

2pq2(1)(
1 / 15,000 )=1/61
Autosomal Recessive – Recurrence Risk - ARPKD

What is the risk to the next
child if one parent is known
to be a carrier and the
genetic status of the other
is unknown?
Aa
(_ x _) x ( _ x _)
(1 x 1/2) x( 1/61 x 1/2)=1/244
??
?
??
1 in 5,000 versus 1 in 15,000

If the population frequency of a recessive disease is 1
in 5,000
◦ 2pq  2(1)( 1/5,000)= 1/35 = Carrier Frequency

If the population frequency of a recessive disease is 1
in 10,000
◦ 2pq  2(1)( 1/15,000)=1/61 = Carrier Frequency
We advise the couple that they have a risk of
1/244 to have a child with ARPKD

How might they react?

What factors might influence this reaction?
We advise the couple that they have a risk of
1/244 to have a child with ARPKD

What factors might influence this reaction?
◦ Severity of the disease
 Individual’s impression of the severity and impact of the disease
◦ Prior notion of risk was higher or lower
◦ Patient’s understanding of probability and how the
information is delivered
 1/244 risk of affected, 0.4% risk (Negative framing)
 243/244 chance of unaffected, 99.6% chance (Positive framing)
What if there are three alleles (A, B and O)










There are nine possible
combinations (6 genotypes)
AA
AO
Can be considered together
OA
OO
AB
Can be considered together
BA
BB
BO
Can be considered together
OB
__
AB
A O
O B A
The Gene Pool
What if there are three alleles?


The allele frequencies can be
considered as p, q and r for A, B
and O respectively
__
A OA B
O B A
The distribution of genotypes
will be p2+2pq+2pr+2qr+q2+r2
 (p+q+r)(p+q+r)
BB
AA
AB
AO BO
OO
The Gene Pool
What are assumptions in using the Hardy-Weinberg
distribution?

The populations of alleles is stable, the population is
very large and mating is random
◦
◦
◦
◦
Assortative (random) mating
No migration
No new mutations
No selection
Why is it ok to assume no selection for autosomal recessive
traits?
Fraction of alleles
2(q2)/(1(2pq)+2(q2)) = q
Derivations of fraction of
alleles exposed to selection
q/1(2pq)+2(q2)+q = 1/2p+2q+1= 1/3
New Clinical Genetics 2e
Andrew Read and Dian Donnai
ISBN: 9781904842804
© Scion Publishing Ltd, 2011
When can selection operate effectively on a
recessive trait?

Heterozygote advantage
◦ When heterozygous carriers of a recessive allele have greater
relative fitness, than either class of homozygotes
 Relative Fitness
 The probability that individuals with a particular genotype will successfully
reproduce relative to those of another genotype
Autosomal Recessive – Recurrence Risk with
Consanguinity

If the parents are related
does that increase the risk
their child will be affected
by a recessive disease?
(_ x _) x ( _ x _)
(1 x 1/2) x( ? x 1/2)=?
Aa
??
?
??
Consanguinity
• Clinical definition of consanguinity
• The union of individuals who are related to each other as
second cousins (5th degree relatives) or closer.
http://www.consang.net
First, Second and Third Degree Relatives
2
2
2
Grandmother
Grandfather
2
Aunt
3
First
cousin
2
Grandfather
1
1
Father
Mother
1
Patient
2
Uncle
1
Brother
Sister
2
1
Niece
Son
1
Daughter
Grandmother
2
Nephew
3
First
cousin
Consanguinity – and the risk of recessive disease

The more closely related the parents, the greater the
increase in risk
◦ The risk decreases the more distantly related the parents are
◦ The risk is proportional to the fraction of genes which are
shared by the parents
 Identity by descent
Autosomal Recessive – Recurrence Risk

If the parents are related
does that increase the risk
their child will be affected?
(_ x _) x ( _ x _)
(1 x 1/2) x( ? x 1/2)=?
How closely related are the parents?
Aa
??
?
??
If the parents are first cousins….
(1 x 1/2)x(1/8 x 1/2)=1/32
1/2
1/2
1/2
Compare risk of affected
child to that estimated by
H-W
??
Aa
?
??
Identity by Descent and the Coefficient of
inbreeding


Identity by descent
◦ When both copies of an allele in an
individual are identical because they
were inherited from a common
ancestor
Coefficient of inbreeding
◦ The fraction of a person’s genes that
are identical by descent
◦ The probability that a person is
identical by descent at a particular
locus
Cc
??
?
cc
Identity by Descent – First Cousins

The chance that allele A1 will
be inherited by individual IV1 in a homozygous state is:
◦ (1/2x1/2x1/2)x(1/2x1/2x1/2)=1/64

The chance that allele A1, A2,
A3 or A4 will be inherited by
individual IV-1 in a
homozygous state is
4x1/64=1/16
1/2
1/2
1/2
1/2
1/2
1/2
Thompson and Thompson, Genetics in Medicine, 7th Edition, 2007
If the parents are second cousins….
(1 x 1/2)x(1/32 x 1/2)=1/128
1/2
1/2
1/2
1/2
1/2
??
Aa
?
??
Fraction of genes shared in consanguineous
unions
Thompson and Thompson, Genetics in Medicine, 7th Edition, 2007
What if there is more than one consanguineous
union in the family?
New Clinical Genetics 2e
Andrew Read and Dian Donnai
ISBN: 9781904842804
© Scion Publishing Ltd, 2011
Clinical case:

A father in a first cousin
marriage is a carrier for
ARPKD. They have had
had 1 healthy child.
◦ How does the presence of
consanguinity and the
healthy child affect the
risk estimate for the
mother to be a carrier?
Aa
??
?
??
Modifying a risk using additional information

Given a risk of a particular event, this risk can be
modified using Bayes’ theorem to take into account
additional information
◦ Risk of developing a disease can be modified by
 Family History
 Test Results
 Age (with age dependent penetrance)
Autosomal Recessive Recurrence Risk


What is the chance that the
patient’s unaffected brother is a
carrier?
At conception there are four
possible pairs of alleles he could
inherit with equal probability.
◦ AA, Aa, aA, aa

But we know that he is not aa
(conditional probability of zero),
so 2/3rds risk of being a carrier
Unaffected
AA
Aa
Aa
Aa
?
aa
??
Affected
aA
aa
= brother is a carrier
Given genotype what is the chance of no symptoms in first
child, and given no symptoms what is chance of each
genotype
No symptoms
Genotype
1/4
0
1
aa
1
Aa
0/4
aa and symptoms
1/4
Aa and no symptoms
2/4
Aa and symptoms
0/4
AA and no symptoms
1/4
AA and symptoms
0/4
0
1/2
1/4
aa and no symptoms
1
AA
0
Incorporating additional information into risk
estimates

If the couple (first
cousin example) has
had 1 healthy child.
◦ Does this increase or
decrease the
probability that the
mother is a carrier?
Aa
??
?
??
Given mother a carrier, chance for first child to be
unaffected
Affected
1/32
Mother carrier and child affected
1/4
C
3/4
1/8
7/8
NC
3/32
Mother carrier and child unaffected
Not affected
Affected
0/32
Mother not a carrier and child affected
0
1
Not affected
28/32
Mother not a carrier and child unaffected
Given first child unaffected, chance for mother to
be a carrier
Affected
1/32
Mother carrier and child affected
1/4
C
3/4
1/8
7/8
NC
3/32
Mother carrier and child unaffected
Not affected
Affected
0/32
Mother not a carrier and child affected
0
1
Not affected
28/32
Mother not a carrier and child unaffected
Example of Risk Adjustment Using Bayes Method
Mother is a carrier
Mother is not a
carrier
Prior Probability
1/8=.125
7/8=.875
Conditional
Probability
3/4
1/1
Joint Probability
3/32
7/8=28/32
Final/Posterior
Probability
(3/32) / (3/32+28/32)
= 3/31 =.1
(28/32)/(3/32+28/32)
= 28/31=.90
Aa
??
?
aa
Students will not be asked to perform this calculation on the final exam
??
Visual Depiction of Conditional Probabilities
Mother is a Carrier
Mother is Not a Carrier
A
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
U
+
1/4
-
3/4
Aa
??
?
aa
??
= brother is a unaffected
¾ of children will be unaffected
-
28/28
All children will be unaffected
Incorporating additional information into risk
estimates

If the couple has had
2 healthy children.
Aa
◦ Does this increase or
decrease the
probability that the
mother is a carrier?
??
?
aa
??
What if there are two unaffected children?
Mother is a carrier
Mother is not a
carrier
Prior Probability
1/8=.125
7/8=.875
Conditional
Probability
3/4
1/1
Conditional
Probability #2
3/4
1/1
Joint Probability
9/128
7/8=112/128
Final/Posterior
Probability
(9/128) / (9/128+112/128) =
9/121 =.07
(112/128)/(9/1+112/128)=
112/121=.93
Aa
??
?
aa
Students will not be asked to perform this calculation on the final exam
??
Lecture Summary

The Hardy-Weinberg distribution can be used to estimate carrier
frequencies for autosomal recessive conditions given the
incidence of the condition

Consanguinity can increase the risk of autosomal recessive
conditions
◦ The degree of increased risk is related to the degree of consanguinity of
the couple
◦ The increase in risk falls off quickly for relationships more distant than
second cousins
Lecture Summary

Bayes’ Theorem can be used to update risk estimates
to take into account new clinical information
◦ Outcomes that are unlikely generally require stronger
evidence than those that are likely
Review Question

A severe autosomal recessive condition is less subject to
selection than an equally severe autosomal dominant
condition because
◦ A) Dominant conditions are more severe than recessive
conditions
◦ B) Both parents need to be carriers for a recessive condition to
occur
◦ C) Most of the recessive alleles are present in healthy carriers
◦ D) There can be variable expressivity in a dominant condition
Review Question

The fraction of parental
alleles inherited in
common by siblings is
◦
◦
◦
◦
A) 0
B)1/4
C)1/2
D)3/4
Synthesis Question

If a patient is suspected to have a genetic disease and
undergoes a DNA test that is negative. The chances that he is
affected by the disease are?
◦ A) 100%
◦ B) 25%
◦ C) 0%
◦ D) It depends on the characteristics of the test
◦ E) It depends on the characteristics of the test and the
probability that the patient might have the disease before
having the test