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Computational GeneticsLecture 1

This class has been edited from several sources. Primarily from Terry Speed’s homepage at Stanford and the Technion course “Introduction to Genetics”. Changes made by Dan Geiger.

Background Readings: Chapter 2&3 of An introduction to Genetics, Griffiths et al. 2000, Seventh Edition (CS/Fishbach/Other libraries).

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Human GenomeMost 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

zygote

gametes

sperm

egg

Meiosis

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The Double HelixS

ourc

e: A

lber

ts e

t al

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Chromosome Logical Structure

Marker – Genes, SNP, Tandem repeats.

Locus – location of markers.

Allele – one variant form of a marker.

Locus1Possible Alleles: A1,A2

Locus2Possible Alleles: B1,B2,B3

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Alleles - the ABO locus example

O is recessive to A.

A is dominant over O.

A and B are codominant.

Multiple alleles: A,B,O.

Phenotype Genotype

A A/A, A/O

B B/B, B/O

AB A/B

O O/O

Trait = Character = Phenotype

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X-linked

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.

genotype

phenotype

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Modern genetics began with Mendel’s 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’s Work

Mendel Gregor. 1866. Experiments on Plant Hybridization. Transactions of the Brünn Natural History Society.

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Mendel’s 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

F2: 1 AA : 2 Aa : 1 aa

~ ~ A aPhenotype

Gametes: A a

a Aa aa

1A : 1 aPhenotype:~ ~

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Mendel’s 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 PAb=PA Pb PaB=Pa PB Pab=Pa Pb

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Recombination Phenomenon(Happens during Meiosis)

Recombination Haplotype

Male or female

:תאי מיןביצית, או זרע

The recombination fraction Between two loci on the same chromosomeIs the probability that they end up in regionsOf different colors

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Example: ABO, AK1 on Chromosome 9

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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A

A1/A1

O

A2/A2

A

A1/A2

O

A1/A2

A

A2/A2

O OA1 A2

A OA1 A2

A | OA2 | A2

O OA2 A2

Recombinant

Phase inferred

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Example: ABO, AK1 on Chromosome 9

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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4

5

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A

A1/A1

O

A2/A2

A

A1/A2

O

A1/A2

A

A2/A2

O OA1 A2

A OA1 A2

A | OA2 | A2

O OA2 A2

Recombinant

Phase inferred

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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 allele’s frequencies: independence.

Exceptions occur for tightly linked loci.

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One locus: founder probabilitiesFounders 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

D d dd

1

21

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One locus: transmission probabilities

Children get their genes from their parents’ genes, independently, according to Mendel’s laws; also independently for different children.

D d D d

d d3

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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

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.

d d D D

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2

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One locus: penetrance probabilitiesPedigree 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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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 1=0.1 and 2=0.3 is the product:

D d D d

D dd d D D

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2

(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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Tutorial #2by Ma’ayan 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 << 50% between 2 genes shows that they are linked.

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Two Loci Inheritance

Recombinant

2 1A AB B

a ab b

A aB b 3 4

a ab b

A ab b 5 6

A aB b

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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.

The farther apart U & V are the greater the chance that a crossing over would occur between them the greater the chance of recombination between them.

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Recombination Fraction

Linkage) No(5.0)ionRecombinat(0)Linkage( P

• 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.

tsrecombinan double of # expected

tsrecombinan double of # observerd1ce(I)Interferen

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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 wasn’t 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: It’s 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 geta more reliable estimate of the recombination fraction.

1. 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.

2. Calculate the Lod score = log(b/a).

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