HOMOZYGOSITY MAPPING USING LOD

SCORE METHOD

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BBS- 6

CONTENTSINTRODUCTIONMETHODS OF HOMOZYGOSITY MAPPINGHOMOZYGOSITY MAPPERGENETIC LINKAGELOD SCORE METHOD

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LET US START WITH THE BASICS!

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KEY TERMS

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HOMOZYGOSITY• CONTAINING TWO IDENTICAL ALLELIC

FORMS• CAN BE HOMOZYGOUS DOMINANT• CAN BE HOMOZYGOUS RECESSIVE• PEA PLANT

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HETEROZYGOUSBOTH ALLELES OF A GENE ARE

DIFFERENTONE GENE IS DOMINANTONE GENE IS RECESSIVE

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GENETIC MAPPINGSETTING A LOCATION WITH RESPECT TO A

MARKERPLOTTING DNA FRAGMENTS ON

CHROMOSOMESHELPFUL IN PREDICTING A DISEASE

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EXAMPLE

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GENETIC MARKERA GENE OR A DNA SEQUENCE FOR A

PARTICULAR TRAITHAS A PARTICULAR LOCATION ON A

CHROMOSOMEDETECTION HELPFUL IN PREDICTING A

DISEASE

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RECOMBINATIONBREAKING AND REJOINING OF DNA

MOLECULESEXAMPLE IS CROSSING OVEREXCHANGE OF GENETIC MATERIAL TAKES

PLACERESULTING MOLECULES ARE CALLED

RECOMBINANTS

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RECOMBINATION FREQUENCYTOTAL NUMBER OF

RECOMBINANTS/TOTAL NUMBER OF PROGENIES IN A TEST CROSS

USED TO DETERMINE THE GENETIC DISTANCE

CREATION OF GENETIC MAPCENTIMORGAN

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HOMOZYGOSITY MAPPINGMETHOD USED TO DETECT THE DISEASE

OF THE HOMOZYGOUS CONDITIONHELPFUL FOR THE INHERITED

DISORDERS

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METHODS OF HOMOZYGOSITY

MAPPING

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THREE MAIN METHODSSNP MICROARRAYSRFLPMICROSATELLITE MARKERS

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1. SNP MICROARRAYSSTANDS FOR SINGLE NUCLEOTIDE

POLYMORPHISMINVOLVES A SINGLE PCR METHOD

FOLLOWED BY GEL ELECTROPHORESISTETRA-PRIMER ARMS PCR

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PROCEDURE

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APPLICATIONSHIGH DENSITY SNP ARRAYS FOR

GENOTYPINGMUTATION IDENTIFICATION BY

POSITIONAL CLONING

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BARDET BEIDL SYNDROMEABBREVIATED AS BBSCHARACTERIZED BY: OBESITY;

PIGMENTARY RETINOPATHY; POLYDACTYLY; HYPOGONADISM

RENAL AND CARDIAC ABNORMALITESCOGNITIVE IMPAIRMENT

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2. RFLPSTANDS FOR RESTRICTION FRAGMENT

LENGTH POLYMORPHISMUSED TO FOLLOW THE PATH OF A

SPECIFIC GENEVARIATIONS IN THE HOMOLOGOUS DNA

SAMPLES

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METHODCUTTING DNA SAMPLES WITH

RESTRICTION ENZYMESSEPERATION BY AGAROSE GEL

ELECTROPHORESISDETERMINING THE NUMBER OF

FRAGMENTS AND SIZES

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APPLICATIONSDNA FINGERPRINTINGTRACING ANCESTORYSTUDYING EVOLUTION AND MIGRATIONDETECTION AND DIAGNOSISGENETIC MAPPING

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HOMOZYGOSITYMAPPER

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LAYOUT

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HOMOZYGOSITY MAPPER:Web based approach

for homozygosity mapping.

Stores markers data in its database…users can upload their SNP files there.

Data analysis is quick, detects homozygous alleles, and represents graphically.

Zooming in and out of a chromosome.

Access:RestrictedPublic

Integrated with GeneDistiller engine

Microsatellite Markers:SSTRsVSTMsActing as markersDi, tri, tetra, penta

nucleotidesPresent on non-

coding sequencesAmplified by locus

specific primers with PCR

Example: Presence of AC (n) in

birds where n varies from 8 to 50.

Uses:

Important most tool in mapping genome

Serve in biomedical diagnosis as markers for certain disease conditions

Primary marker for DNA testing in forensics for high specificity.

Markers for parentage analysis

address questions concerning degree of relatedness of individuals or groups

PedigreeLineage or

Genealogical study of family lines.

Gives list or family tree of ancestors.

Used for studies of certain inheritance pattern.

Genetic Linkage:Staying together of

physically close loci.

Offspring acquires more parental combinations.

Discovery: An Exception to

“Mendel's Law of independent Assortment”

Thomas Morgan : Linked genes are physical objects, linked in close proximity

Genetic LinkageMorgan’s

Experiments:

1st Cross:

F1 Progeny: Heterozygous red eyed males

and females

2nd Cross:

F2 Progeny:

2,459 red-eyed females 1,011 red-eyed males 782 white-eyed males

Sex limited trait…evidence

Crossed:White eyed males

(original) X F1 daughters…

129 red-eyed females132 red-eyed males88 white-eyed females86 white-eyed males

Conclusions:Eye color is Sex Linked….Physically closer genes do not assort independently

LINKAGE MAP

Genetic Map for location determination of genes and genetic markers.

Based on markers recombination frequency during cross over.

Predicts the relative position, not the physical distance between genes.

separated

Lesser the distance, more tightly they are bound, more often inherited together.

Centi Morgan: unit to calculate linkage distanceOne centimorgan

corresponds to about 1 million base pairs in humans.

Two markers on a chromosome are one centimorgan apart if they have a 1% chance of being

Based on frequency of genetic markers passing together.

Constructing Linkage Map:

Developed by Newton E. Morton

LOD: Logarithm (base 10) Of Odds

A statistical test for linkage analysis in HumanAnimalPlant populations

It checks whether the two loci are:Indeed linked orThey occur together

by chance

Usually done to check linkage of symptoms in syndromes

LOD SCORE METHOD

The Method:Establish a pedigreeMake a number of estimates of recombination

frequencyCalculate a LOD score for each estimateThe estimate with the highest LOD score will

be considered the best estimate

LOD SCORE METHOD

Calculations:

Where:

•NR denotes the number of non-recombinant offspring •R denotes the number of recombinant offspring•Theta is the recombinant fraction, it is equal to R / (NR + R)•0.5 in the denominator means that alleles that are completely unlinked have a 50% chance of recombination

LOD score can be either positive or negativePositive LOD score means Linkage presentNegative LOD score means No Linkage

>3 Evidence for linkage

+3 1000 to 1 odds that the linkage did not occur by chance

<-2 Evidence to exclude linkage

LOD score Result

Determines R (Recombination Fraction, fraction of gametes that are recombinant) using data from small families

R value varies from 0 – 0.5

0 2 completely linked genes0.5 2 completely unlinked genes

Mapping Genes with LOD Score Method

1. Determine the expected frequencies of F2 phenotypes

2. Determine the likelihood that the family data observed resulted form given R value

3. Determine LOD ratio

4. Add LOD scores from different families to achieve a high LOD score so a most likely R value can be assigned

Steps Involved

We are using two COMPLETELY DOMINANT GENES

Heterozygote is indistinguishable from dominant homozygote

Two genes areA: with A and a allelesB: with B and b alleles

EXAMPLE

P1: AABB X aabbGametes

F1 AaBb

Parental Combinations Recombinants

abaBAb

AB

AB

ab

1. Determine the frequency of each gamete produced by F1 generation

2. For example if R=0.20, then 20% of the gametes produced will be recombinants which in our example are Ab and aB.

3. As there are 2 types of recombinant gametes, frequency of each type will be 0.10

4. 80% gametes are parental, [AB and ab type] frequency of each of them is 0.40 or 40%

STEP I: Calculate the expected frequency of offspring for values of R from 0-0.5

AB0.40

Ab0.10

aB0.10

ab0.40

AB0.40

AABB0.16

AABb0.04

AaBB0.04

AaBb0.16

Ab0.10

AABb0.04

Aabb0.01

AaBb0.01

Aabb0.04

aB0.10

AaBB0.04

AaBb0.01

aaBB0.01

aaBb0.04

ab0.40

AaBb0.16

Aabb0.04

aaBb0.04

Aabb0.16

5. Draw Punnet Square to determine offspring

6. Determine the phenotype of each cell in Punnet square

7. Add up the frequencies to get the total frequency of each offspring phenotype

F2 Phenotype Cell Sums Expected Frequency

A_B_ 0.16+0.04+0.04+0.16+0.04+0.01+0.04+0.01+0.16

0.66

A_bb 0.01+0.04+0.04

0.09

aaB_ 0.01+0.04+0.04

0.09

Aabb 0.16 0.16

Done by determining the likelihood (L)Likelihood:

the probability of the observed familydetermined using the multinomial theorem

an extension of the binomial theorem.

STEP II: Examine the observed Family Data in light of expected distribution of offspring for each R value

First define the terms for the observed familya = number of A_ B_ offspringb = number of A_ bb offspringc = number of aaB_ offspringd = number of aabb offspringn = total offspring (= a+b+c+d)

Define the terms for the expected family proportions p = expected proportion of A_B_ offspringq = expected proportion of A_ bb offspringr = expected proportion of aaB_ offsprings = expected proportion of aabb offspring

Multinomial theorem describing actual family: pa qb rc sd multiplied by a coefficient n! /(a! b! c! d!)

Thus the likelihood equation is

Multinomial Theorem

We have calculated phenotypic proportions for R = 0.20 (20 map units between A and B)

A family of 5 children has - 2 children with A_B_ phenotype- 1 with aaB_ - And 2 with aabb

Hence Likelihood is:

Likelihood needs to be calculated between each value of R i.e. 0.01 – 0.5.

Data from several families are added and compared to get a good estimate of R

Standardization of L value which means calculation of Odds Ratio (OR)

Then Logarithm of OR is taken which is LOD score

LOD scores from various families are added (this is like AND rule for two events i.e. Family 1 AND family 2 ---- Both occurring)

STEP III and IV

A total LOD score for some R value of 3 is considered proof of linkage of two genes

In our example, Odds Ratio = L0.20 /

L0.50

= 0.0301 / 0.00695

= 4.331LOD score = Log10

4.331 (Log10 OR)

= 0.637It is evident from this

score that data from several families of this size is needed to reach a lod score of 3.0 as a proof of linkage.