Computational Genetics Lecture 1

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Computational GeneticsLecture 1
Background Readings Chapter 23 of An
introduction to Genetics, Griffiths et al. 2000,
Seventh Edition (CS/Fishbach/Other libraries).
This class has been edited from several sources.
Primarily from Terry Speeds homepage at Stanford
and the Technion course Introduction to
Genetics. Changes made by Dan Geiger.
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Course Goals

• Learning about computational and mathematical
  methods for genetic analysis.
• We will focus on Gene hunting finding genes for
  simple human diseases.
• Methods covered in depth linkage analysis (using
  pedigree data), association analysis (using
  random samples).
• Another goal is to learn more about Bayesian
  networks usage for genetic linkage analysis.

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Human Genome

• Most human cells contain
• 46 chromosomes
• 2 sex chromosomes (X,Y)
• XY in males.
• XX in females.
• 22 pairs of chromosomes, named autosomes.

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Genetic Information

• Gene basic unit of genetic information. They
  determine the inherited characters.
• Genome the collection of genetic information.
• Chromosomes storage units of genes.

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Sexual Reproduction
Meiosis
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The Double Helix
Source Alberts et al
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Central Dogma
?????
?????
cells express different subset of the genes In
different tissues and under different conditions
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Chromosome Logical Structure

• Marker Genes, SNP, Tandem repeats.
• Locus location of markers.
• Allele one variant form of a marker.

Locus1 Possible Alleles A1,A2
Locus2 Possible Alleles B1,B2,B3
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Alleles - the ABO locus example
Genotype Phenotype
A/A, A/O A
B/B, B/O B
A/B AB
O/O O

• O is recessive to A.
• A is dominant over O.
• A and B are codominant.
• Multiple alleles A,B,O.

Trait Character Phenotype
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??????

• ??? ?????? ?????????. ???? ???? ??? ?? ????
  ??????? ???
• ?????????, ??????? ????? ???? ?????????
  ?????.

1. AA ?- aa ?? ??????????? (Homozygote) ????
   ????????? ????????, ??????. Aa ??? ?????????
   (Hetrozygote).

3. ????? ?????? (A,B,O),
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X-linked
genotype
phenotype

• b - dominant allele. Namely, (b,b), (b,w) is
  Black.
• w - recessive allele. Namely, only (w,w) is
  White.
• This is an example of an X-linked
• trait/character.
• For males b alone is Black and w alone is white.
• There is no homolog gene on the Y chromose.

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Mendels Work
Modern genetics began with Mendels experiments
on garden peas (Although, the ramification of his
work were not realized during his life time). He
studied seven contrasting pairs of characters,
including The form of ripe seeds
round, wrinkled The color of the seed
albumen yellow, green The length of
the stem long, short
Mendel Gregor. 1866. Experiments on Plant
Hybridization. Transactions of the Brünn Natural
History Society.
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Mendels first law Characters are controlled by
pairs of genes which separate during the
formation of the reproductive cells (meiosis)
A a
a
A
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P AA X aa
F1 Aa
F1 X F1 Aa X Aa
test cross Aa X aa
Gametes A a A AA
Aa a Aa aa
Gametes A a a Aa
aa

Phenotype
1A 1 a
F2 1 AA 2 Aa 1 aa

Phenotype
A a
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??????
1. ????? ?? 1F ?? ???? ???? 2F ???? ??? ???????
?????? ??????? ????????? ???? ?????? ???????
??????? ??? 31.
2. ????? ???? ????? ????? 1F ?? ????? ???
??????? ???????. ???? ??? ??????? ??????
??????? ????????? ???? ?????? ??????? ???????
??? 11
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Mendel's First low.
Results of crosses in which parents differed for
one character
F2 ratio F2 F1 Parental Phenotype
2.961 5474 round 1850 wrinkled Round 1. Round X wrinkled seeds
3.011 6022 yellow 2001 green yellow 2. Yellow X green seeds
3.151 705 purple 224 white purple 3. Purple X white petals
2.951 882 inflated 299 pinched inflated 4. Inflated X pinched pods
2.821 428 green 152 yellow green 5. Green X yellow pods
3.141 651 axial 207 terminal axial 6. Axial X terminal flowers
2.841 787 lon 277 short long 7. Long X short stems
Conclusion, First low The two members of a gene
pair segregate from each other into the gametes.
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????? ?????? ?? ?????? ??????? (??????? ?? ???
?????).
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Polydactyly A dominant mutation
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Brachydactyly A dominant mutation
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Mendels second law When two or more pairs of
genes segregate simultaneously, they do so
independently.
A a B b
A B
A b
a B
a b
PAB PA ? PB PAbPA ? Pb PaBPa ? PB
PabPa ? Pb
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Mendel's second low.


A d
ihybrid cross for color and shape of pea seeds



P wrinkled and yellow X round and green






rrYY
RRyy

F1 round yellow



Rr Yy
X Rr Yy


F2 round yellow
315


round green 108

wrinkled yellow 101
wrinkled green 32

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a. Check segregation pattern for each allele in
F2

416 yellow 140 green (2.971)

423 round 133 wrinkled (3.181)


Conclusion both trai
ts behave as single genes
, each carrying


two different alleles.
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Question Is there independent assortment of
alleles of the different genes?

Probability to get yellow is 3/4 probability to
get round is 3/4
v
v

Probability to get yellow is 3/4 probability to
get wrinkled is 1/4
v

Probability to get green is 3/4 probability to
get round is 3/4
probability to get green round is
X 3/4, namely 3/16

1/4
v

Probability to get green is 1/4 probability to
get wrinkled is 1/4
.
probability to get
green

wrinkled
is
1
/4 X 1/4, namely
1
/16

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A standard presentation in terms of counts
expected expected observed
yellow round 9 312.75
315 yellow wrinkled 3
104.25 101 green round 3
104.25 108 green wrinkled
1 34.75 32 Total
16 556
556
Conclusion, second law Different gene pairs
assort independently in gamete formation


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Exceptions to Mendels Second Law
Morgans fruit fly data (1909) 2,839 flies Eye
color A red a purple Wing length B normal b
vestigial
AABB x aabb
AaBb x aabb
AaBb Aabb
aaBb aabb Expected 710
710 710 710 Observed 1,339
151 154 1,195
The pair AB stick together more than expected
from Mendels law.
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Morgans explanation
F1
F2
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Parental types AaBb, aabb Recombinants Aabb,
aaBb
The proportion of recombinants between the two
genes (or characters) is called the
recombination fraction between these two genes.
It is usually denoted by r or ?. For
Morgans traits r (151
154)/2839 0.107 If r lt 1/2 two
genes are said to be linked. If r
1/2 independent segregation
(Mendels second law).
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Recombination Phenomenon(Happens during Meiosis)
Male or female

• Recombination
• Haplotype

The recombination fraction Between two loci on
the same chromosome Is the probability that they
end up in regions Of different colors
??? ??? ?????, ?? ???
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?????????? ??????? ?????? ???????
???????? ??? ?????? ????????? ??????.
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Example ABO, AK1 on Chromosome 9
Phase inferred
Hardy-Weinberg law of population genetics permits
calculation of genotype frequencies from allele
frequencies P(a) frequency of a in the
population P(ab) 2P(a)P(b) Hardy-Weinberg
equilibrium corresponds to a random union of two
gamets, called zygote.
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Example ABO, AK1 on Chromosome 9
Phase inferred
Recombination fraction is 12/100 in males and
20/100 in females. One centi-morgan means one
recombination every 100 meiosis. One centi-morgan
corresponds to approx 1M nucleotides (with large
variance) depending on location and sex.
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Conventions
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Maximum Likelihood Principle
What is the probability of data for this
pedigree, assuming a recessive mutation ?
What is the probability of data for this
pedigree, assuming a dominant mutation ?
Maximum likelihood principle Choose the model
that maximizes the probability of the data.
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Linkage Equilibrium

• Linkage Equilibrium haplotype frequency is the
  product of the underlying alleles frequencies
  independence.
• Exceptions occur for tightly linked loci.

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One locus founder probabilities
Founders are individuals whose parents are not
in the pedigree. They may of may not be typed
(namely, their genotype measured). Either way, we
need to assign probabilities to their actual or
possible genotypes. This is usually done by
assuming Hardy-Weinberg equilibrium (H-W). If the
frequency of D is .01, then H-W says

pr(Dd )
2x.01x.99 Genotypes of founder couples are
(usually) treated as independent.


pr(pop Dd , mom dd )
(2x.01x.99)x(.99)2
D d
1
2
1
D d
dd
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One locus transmission probabilities
Children get their genes from their parents
genes, independently, according to Mendels laws
also independently for different children.
D d
D d
2
1
d d
3
pr(kid 3 dd pop 1 Dd mom 2 Dd ) 1/2
x 1/2
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One locus transmission probabilities - II
D d
D d
1
2
4
3
5
D d
d d
D D
pr(3 dd 4 Dd 5 DD 1 Dd 2 Dd )
(1/2 x 1/2)x(2 x 1/2 x 1/2) x (1/2 x 1/2). The
factor 2 comes from summing over the two mutually
exclusive and equiprobable ways 4 can get a D
and a d.
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One locus penetrance probabilities
Pedigree analyses usually suppose that, given the
genotype at all loci, and in some cases age and
sex, the chance of having a particular phenotype
depends only on genotype at one locus, and is
independent of all other factors genotypes at
other loci, environment, genotypes and phenotypes
of relatives, etc. Complete penetrance
pr(affected DD
) 1 Incomplete penetrance)
pr(affected DD ) .8
DD
DD
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One locus penetrance - II
Age and sex-dependent penetrance (liability
classes)

pr(
affected DD , male, 45 y.o. ) .6
D D (45)
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?????? ?????
????? ??????? ????????? ?? ??????? ??????? ??
???? ?????
A healthy daughter transmits The mutation to her
daughter. .
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One locus putting it all together
Assume penetrances pr(affected dd ) .1,
pr(affected Dd ) .3 pr(affected DD ) .8,
and that allele D has frequency .01. The
probability of data for this pedigree assuming
penetrances of ?10.1 and ?20.3 is the product
(2 x .01 x .99 x .7) x (2 x .01 x .99 x .3) x
(1/2 x 1/2 x .9) x (2 x 1/2 x 1/2 x .7) x (1/2 x
1/2 x .8)
This is a function of the penetrances. By the
maximum likelihood principle, the values for ?1
and ?1 that maximize this probability are the ML
estimates.
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Fully penetrant Recessive Disease
2
1
5
3
4
Let q be the probability of the disease allele.
The probability of data for this pedigree
assuming full penetrance is the product
L (1-q) x q x (1-q) x q (3/4)(3/4)(1/4)
Exercise write the likelihood for a fully
penetrant dominant disease.
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Linkage Recombination

• Tutorial 2
• by Maayan Fishelson

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Crossing Over

• Sometimes in meiosis, homologous chromosomes
  exchange parts in a process called crossing-over.
• New combinations are obtained, called the
  crossover products.

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Recombination During Meiosis
Recombinant gametes
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Linkage

• 2 genes on separate chromosomes assort
  independently at meiosis.
• 2 genes far apart on the same chromosome can also
  assort independently at meiosis.
• 2 genes close together on the same chromosome
  pair do not assort independently at meiosis.
• A recombination frequency ltlt 50 between 2 genes
  shows that they are linked.

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Two Loci Inheritance
Recombinant
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Linkage Maps

• Let U and V be 2 genes on the same chromosome.
• In every meiosis, chromatids cross over at random
  along the chromosome.
• If the chromatids cross over between U V, then
  a recombinant is produced.

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Recombination Fraction

• The recombination fraction ? between two loci
• is the percentage of times a recombination
• occurs between the two loci.
• ? is a monotone, nonlinear function of the
• physical distance separating between the loci
• on the chromosome.

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Centimorgan (cM)

• 1 cM (or 1 genetic map unit, m.u.) is the
  distance between genes for which the
  recombination frequency is 1.

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Interference

• Crossovers in adjacent chromosome regions are
  usually not independent. This interaction is
  called interference.
• A crossover in one region usually decreases the
  probability of a crossover in an adjacent region.

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Building Genetic Maps

• At first only genes with variant alleles
  producing detectably different phenotypes were
  used as markers for mapping.
• Problem the chromosomal intervals between the
  genes were too large ? the resolution of the maps
  wasnt high enough.
• Solution use of molecular markers (a site of
  heterozygosity for some type of silent DNA
  variation not associated with any measurable
  phenotypic DNA variation).

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Linkage Mapping by Recombination in Humans.

• Problems
• Its impossible to make controlled crosses in
  humans.
• Human progenies are rather small.
• The human genome is immense. The distances
  between genes are large on average.

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Lod Score for Linkage Testing by Pedigrees

• The results of many identical matings are
  combined to get
• a more reliable estimate of the recombination
  fraction.
• Calculate the probabilities of obtaining a set of
  results in a family on the basis of (a)
  independent assortment and (b) a specific degree
  of linkage.
• Calculate the Lod score log(b/a).

A Lod score of 3 is considered convincing
support for a specific recombination fraction.