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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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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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3. MICROSATELLITE MARKERSSIMPLE SEQUENCE TANDEM REPEATS
[SSTRs]HIGHLY POLYMORPHICSLIPPAGE REPLICATION: MUTATION
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ADVANTAGESLOCUS SPECIFICCODOMINANTSPCR-BASEDUSEFUL AT A WIDE RANGE OF SCALE
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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.