MBChB MSC PhD DCH UK MRCPCH Definition Population genetics is the study of genetic variation in populations Allow us to understand how and why the prevalence of various genetic diseases differs among populations ID: 933885
Download Presentation The PPT/PDF document "Population Genetics Dr. Mohammed Hussein" is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
Slide1
Population Genetics
Dr. Mohammed Hussein
M.B.Ch.B, MSC, PhD, DCH (UK), MRCPCH
Slide2Definition
Population genetics is the study of genetic variation in populations. Allow us to understand how and why the prevalence of various genetic diseases differs among populations
.
Slide3Hardy-Weinberg Equilibrium
Hardy
: British Mathematician
Weinberg
: German Physician
Slide4Hardy-Weinberg Equilibrium
Slide5Hardy-Weinberg
Equilibrium
If a population is large and if individuals mate at random with respect to their genotypes at a locus, the population should be in
H-W equilibrium.
Slide6Hardy-Weinberg
Equilibrium
This means that there is
a constant
and predictable relationship between genotype frequencies and allele frequencies.
The Hardy Weinberg equation, allows one
to estimate
genotype frequencies if one
knows allele
frequencies, and vice versa.
Slide7Hardy-Weinberg
Equilibrium
P
2
+ 2pq + q
2
= 1
p
= a normal allele
q
= a disease producing allele
Slide8p
= a normal alleleq = a disease-producing allele
pp
= the genotype
pp
(homozygous normal) =
P
2
qq = the genotype
qq
(homozygous affected) =
q
2pq =
heterozygous ( affected or carrier) =
2pq
P2 + 2pq + q2 = 1
Slide9Heterozygous male
pq
Heterozygous female
pq
Slide10q
p
p
q
p
p
p
q
p
q
q
q
pp
pq
pq
qq
Heterozygous
Heterozygous
Homozygous
for the normal allele
Homozygous
for the
diseased allele
pp
+
pq
+
pq
+
qq
=
1
p
2
+
2pq
+
q
2
=
1
1/4
1/4
1/4
1/4
Slide11P
2 + 2pq + q2 = 1
p
= frequency of the normal allele
q
= frequency of the disease-producing allele
p
2
= frequency of the homozygous normal
q
2
= frequency of the homozygous affected
2pq
= frequency of heterozygous
Slide12Simplification
Generally p, the normal allele frequency in the population, is very close to 1 (e.g., most of the alleles of this gene are normal). In this case, we may assume that
p
~ 1, and the equation simplifies
to:
P
2
+ 2pq + q
2
= 1
1 + 2q + q
2
= 1
Slide131 + 2q + q
2 = 1
q
= frequency of the
diseased-producing
allele
2q
= frequency of heterozygous
for diseased
alleleq2= frequency of homozygous for diseased allele
Slide14In
Autosomal Dominant diseasesThe diseased allele (q) is dominant over the normal allele (p), so
pp
=
normal
qq
= homozygous affectedpq
=
heterozygous affected
Slide15In Autosomal Dominant
diseasesq = frequency of the diseased-producing allele2q
= heterozygous affected =
prevalence of the
disease
q
2
= homozygous affected (sever disease
)
1 + 2q + q
2
= 1
P
2
+ 2pq + q
2 = 1
Slide16In Autosomal Recessive diseases
The normal allele (p) is dominant over the diseased allele (q), so pp is normal
p
q
is heterozygous carrier
qq
is homozygous affected
Slide17In Autosomal Recessive
diseasesq = frequency of the diseased-producing allele2q
= heterozygous
carrier
q
2
= homozygous affected
(
prevalence of the disease)
1 + 2q + q
2
= 1
P
2
+ 2pq + q
2 = 1
Slide18In
X-linked Recessive diseasesThe normal allele (p) is dominant over the diseased allele (q)As the diseased allele presented only on the X chromosome
X
p
Y
= normal male
X
q
Y = affected male X
p
X
p
= normal femaleXpXq =
female carrier
Xq Xq
= affected female
Slide19q = frequency
of the diseased-producing alleleq = frequency of the hemizygous affected males (X
q
Y
)
(
prevalence of the disease
)
2q = heterozygous female carrier (Xp
X
q
)q2 = frequency of homozygous affected females (X
q
Xq )
1 + 2q + q2 = 1
P2 + 2pq + q
2 = 1
Slide20Slide21q
2q
q
2
Diseased-producing
allele
Heterozygous
Homozygous
X-L R
AD
AR
X-L R
AD
AR
X-L R
Disease prevalence
Disease prevalence
Carrier
Female carrier
Sever affected
Disease prevalence
Affected females
Slide22AD diseases:q = frequency of the disease-producing allele2q = heterozygous affected =
prevalence of the disease
q
2
= homozygous affected (sever disease)
AR diseases:
q
=
frequency of the disease-producing allele2q
= heterozygous carriers
q
2
= homozygous affected = prevalence of the diseaseX-L - R diseases:q =
frequency of the disease-producing
allele q = prevalence of the disease
2q = heterozygous female carriersq2 = homozygous affected females (sever disease)
Slide23A Practical Application of the H-W Principle
If the prevalence of homocystinuria (AR genetic disease)
is
1/10,000, find out
The
frequency of the
disease-producing
allele
The carrier frequency
Slide24Answer
q2 =Disease prevalenceq2 =
1/10,000
Frequency of the
disease-producing
allele =
1/100
2q
= Carrier
frequency 2 x 1/100 = 2/100 = 1/50
The carrier frequency =
1/50
AR Disease
q
= disease-producing allele
2q = carrier frequency
q2 = disease prevalence
q = 1/100
Disease prevalence 1/10,000Disease-producing allele ?The carrier frequency ?
Slide25Example 2
The prevalence of hyperlipidemia (AD disease) is 1/500, calculate The frequency of the disease-producing allele
The frequency of the homozygous affected
Slide26Answer
2q =Prevalence 2q= 1/500
Frequency of the
disease-producing
allele =
1/1000
q
2
= homozygous affected q
2
=
1/1000 x 1/1000 = 1/1000000 = 1/106 Frequency of the homozygous affected = 1/106
AD Disease
q
= disease-producing allele2q = disease prevalenceq2 = homozygous affected
q= 1/1000
Disease prevalence 1/500Disease-producing
allele ?
The
homozygous affected?
Slide27Example 3
The prevalence of hemophilia in males is 1/10,000, calculate The frequency of the disease-producing allele
The prevalence (frequency) of female carriers
The frequency of the females affected.
Slide28Answer
q = The prevalence of the disease q = 1/10,000
Frequency of the
disease-producing
allele =
1/10,000
2q
= heterozygous female carrier
2q = 2 x 1/10,000 = 2/10,000 = 1/5,000
Frequency of the
female carrier=
1/5,000
q
2 = affected female q2 = 1/10,000 x 1/10,000 = 1/100,000,000 = 1/108
Frequency
of the females affected= 1/10
8X-linked R Diseaseq = disease-producing allele (disease prevalence)
2q = heterozygous female carrier q2 = homozygous affected female (sever disease)
Disease prevalence 1/10,000Disease-producing allele ?The female carrier?
The female affected?
as
q =
frequency of the disease-causing allele
Slide29How to calculate the probability for a family to have a child with an autosomal recessive disease?
Slide30The probability is
based on the joint occurrence of 4 events:The probability that the male carry
the abnormal allele
The
probability that
the
he
will
pass
the allele to the childThe probability that the female carry the abnormal alleleThe probability that the
she
will
pass
the allele to
the child1 * 2 * 3 * 4 = the risk of having a child with the disease
Slide31The probability that the
parent carry the abnormal alleleThen the probability that he/she carries the abnormal gene is
100% which equal to
1
If the parent is
known
affected by the disease or
known
carrier
Slide32If the parent is don’t know whether carrier or not
The probability that the parent carry the abnormal allele
Then we have to find the carrier risk or the carrier frequency
If non of his/her family members is affected or carrier then we need to calculated the
carrier frequency of the population
However if one of his/her brothers is affected then the probability is
67% or 2/3
Slide33This means that the probability to pass
the gene is 100% or 1 The probability that the parent will pass
the abnormal allele
Otherwise
This means that the probability to
pass
the gene is 50% or
1/2
If the parent is known
affected
by the
disease
Slide34A Practical Application of the H-W Principle
A 20-year-old female has an AR genetic disease called Phenylketonuria (PKU). She asks you 2 questions:What is the chance that she had to marry a man with the disease-producing allele.
What is the probability that she will have a child with PKU?
Note
: the
prevalence
of
PKU
in the population is 1/10,000.
Slide35Answer
q2 =Disease prevalenceq2 =
1/10,000
Frequency of the disease-causing allele =
1/100
2q
= Carrier
frequency
2
x 1/100 = 2/100 = 1/50The carrier frequency = 1/50
The chance that she had to marry a man with disease-producing allele is
1/50
AR Disease
q
= disease-causing allele
2q = carrier frequency
q2 = disease prevalence
q = 1/100
Disease prevalence 1/10,000The carrier frequency ?
Slide36The answer to the second question is based on the joint occurrence of 4 events:
The probability that the female carries the abnormal allele is The probability that the she will passes her abnormal allele to the child is
The probability that the
male carries
the abnormal allele is
The probability that the
he will passes
his abnormal allele to the child is
1 * 1 * 1/50*1/2 = 1/100
(
1
) as she is known affected
1/50
as we calculate it before
1/2
as the disease is AR as he is carrier
1
as the disease is AR as she is affected
Slide37Example
A couple come
for genetic
counseling
about the risk that their child will
have cystic fibrosis
.
The husband's
sister has the disease, but he is not affected and there is no history in the wife's family of the disease
.
Note: the prevalence of CF is 4/10000.
Slide38Answer
q2 =Disease prevalenceq2 =
4/10,000
Frequency of the disease-causing allele =
2/100
2q
= Carrier
frequency
2
x 2/100 = 4/100 = 1/25The carrier frequency =
1/25
The carrier
risk of CF in
population,
which is the risk for the wife = 1/25
AR Disease
q = disease-causing allele2q = carrier frequency
q2 = disease prevalence
q = 2/100
Disease prevalence 4/10,000The carrier frequency ?
Slide39Answer
Regarding the husband, as his sister is affected, so both parents must be carriers. Since he is not affected therefore there is a
2/3
chance that he is a carrier.
pp
pq
pq
qq
Slide401/25 *
1/2 *
2/3 *
1/2 *
= 1/150
The probability that
the
female carry
the abnormal allele
The
probability that
the
she
will
pass
her abnormal
allele to the
childThe probability that the male carry the abnormal alleleThe probability that the he will pass
his abnormal allele to the child
Slide41Factors responsible for genetic variation in populations
Slide42Factors responsible for genetic variation in populations
Although
human populations are typically in H-W equilibrium for most loci, deviations from equilibrium can be produced
by:
Mutation
Natural Selection
Genetic Drift
Gene Flow
Consanguinity
Slide43Mutation
Mutation
is ultimately the source of all new genetic variation in populations.
Founder
effect
Slide44Natural
Selection
Natural
selection acts upon genetic variation, increasing the frequencies of alleles that promote survival or fertility (referred to as fitness) and decreasing the frequencies of alleles that reduce fitness.
Slide45Natural
Selection
There is now evidence for heterozygote advantages for several other recessive diseases
that are
relatively common in some populations. Examples include
:
Cystic fibrosis
(heterozygote resistance to typhoid fever)
Hemochromatosis
(heterozygote advantage in iron-poor environments)
Glucose-6-phosphate
dehydrogenase deficiency
(heterozygote resistance to malaria)
Slide46Genetic
Drift
Mutation
rates do not vary significantly from population to population, although they can result in significant differences in allele frequencies when they occur in small populations
Slide47Gene
Flow
Gene
flow refers to the exchange of genes among populations.
Because
of gene flow, populations located close to one another often tend to have similar gene frequencies
Slide48Consanguinity and its Health Consequences
Consanguinity
refers to the mating of individuals who are related to one another (typically, a union is considered to be consanguineous if it occurs between individuals related at the second-cousin level or closer).
Slide49Slide50Consanguinity and its Health Consequences
Dozens
of empirical studies have examined
the health
consequences of consanguinity, particularly first-cousin
matings
.
These
studies
show that the offspring of first-cousin matings are approximately twice as likely to present with a genetic disease as are the offspring of unrelated matings.
Slide51Dr. Mohammed Hussein
M.B.Ch.B, MSC, PhD, DCH (UK), MRCPCH
Thank you
for
Your Attention