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## PowerPoint Slideshow about ' HOMOZYGOSITY MAPPING USING LOD SCORE METHOD' - evan-hopkins

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CONTENTS

- INTRODUCTION
- METHODS OF HOMOZYGOSITY MAPPING
- HOMOZYGOSITY MAPPER
- GENETIC LINKAGE
- LOD SCORE METHOD

HOMOZYGOSITY

- Containing two identical allelic forms
- Can be homozygous dominant
- Can be homozygous recessive
- PEA PLANT

HETEROZYGOUS

- BOTH ALLELES OF A GENE ARE DIFFERENT
- ONE GENE IS DOMINANT
- ONE GENE IS RECESSIVE

GENETIC MAPPING

- SETTING A LOCATION WITH RESPECT TO A MARKER
- PLOTTING DNA FRAGMENTS ON CHROMOSOMES
- HELPFUL IN PREDICTING A DISEASE

GENETIC MARKER

- A GENE OR A DNA SEQUENCE FOR A PARTICULAR TRAIT
- HAS A PARTICULAR LOCATION ON A CHROMOSOME
- DETECTION HELPFUL IN PREDICTING A DISEASE

RECOMBINATION

- BREAKING AND REJOINING OF DNA MOLECULES
- EXAMPLE IS CROSSING OVER
- EXCHANGE OF GENETIC MATERIAL TAKES PLACE
- RESULTING MOLECULES ARE CALLED RECOMBINANTS

RECOMBINATION FREQUENCY

- TOTAL NUMBER OF RECOMBINANTS/TOTAL NUMBER OF PROGENIES IN A TEST CROSS
- USED TO DETERMINE THE GENETIC DISTANCE
- CREATION OF GENETIC MAP
- CENTIMORGAN

HOMOZYGOSITY MAPPING

- METHOD USED TO DETECT THE DISEASE OF THE HOMOZYGOUS CONDITION
- HELPFUL FOR THE INHERITED DISORDERS

THREE MAIN METHODS

- SNP MICROARRAYS
- RFLP
- MICROSATELLITE MARKERS

1. SNP MICROARRAYS

- STANDS FOR SINGLE NUCLEOTIDE POLYMORPHISM
- INVOLVES A SINGLE PCR METHOD FOLLOWED BY GEL ELECTROPHORESIS
- TETRA-PRIMER ARMS PCR

APPLICATIONS

- HIGH DENSITY SNP ARRAYS FOR GENOTYPING
- MUTATION IDENTIFICATION BY POSITIONAL CLONING

BARDET BEIDL SYNDROME

- ABBREVIATED AS BBS
- CHARACTERIZED BY: OBESITY; PIGMENTARY RETINOPATHY; POLYDACTYLY; HYPOGONADISM
- RENAL AND CARDIAC ABNORMALITES
- COGNITIVE IMPAIRMENT

2. RFLP

- STANDS FOR RESTRICTION FRAGMENT LENGTH POLYMORPHISM
- USED TO FOLLOW THE PATH OF A SPECIFIC GENE
- VARIATIONS IN THE HOMOLOGOUS DNA SAMPLES

METHOD

- CUTTING DNA SAMPLES WITH RESTRICTION ENZYMES
- SEPERATION BY AGAROSE GEL ELECTROPHORESIS
- DETERMINING THE NUMBER OF FRAGMENTS AND SIZES

APPLICATIONS

- DNA FINGERPRINTING
- TRACING ANCESTORY
- STUDYING EVOLUTION AND MIGRATION
- DETECTION AND DIAGNOSIS
- GENETIC MAPPING

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:
- Restricted
- Public
- Integrated with GeneDistiller engine

Microsatellite Markers:

- SSTRs
- VSTMs
- Acting as markers
- Di, tri, tetra, penta nucleotides
- Present on non-coding sequences
- Amplified by locus specific primers with PCR

- Example:
- Presence of AC (n) in birds where n varies from 8 to 50.

- 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

Pedigree

- Lineage 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 Linkage

- Morgan’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

- Crossed:
- White eyed males (original) X F1 daughters…

- 129 red-eyed females
- 132 red-eyed males
- 88 white-eyed females
- 86 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 distance
- One 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

Constructing Linkage Map:

- Based on frequency of genetic markers passing together.

LOD Score Method

- Developed by Newton E. Morton
- LOD:Logarithm (base 10) Of Odds
- A statistical test for linkage analysis in
- Human
- Animal
- Plant populations
- It checks whether the two loci are:
- Indeed linked or
- They occur together by chance
- Usually done to check linkage of symptoms in syndromes

- The Method:
- Establish a pedigree
- Make a number of estimates of recombination frequency
- Calculate a LOD score for each estimate
- The estimate with the highest LOD score will be considered the best estimate

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 Result

- LOD score can be either positive or negative
- Positive LOD score means Linkage present
- Negative 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

Mapping Genes with LOD Score Method

- 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 genes
- 0.5 2 completely unlinked genes

Steps Involved

- Determine the expected frequencies of F2 phenotypes
- Determine the likelihood that the family data observed resulted form given R value
- Determine LOD ratio
- Add LOD scores from different families to achieve a high LOD score so a most likely R value can be assigned

EXAMPLE

- We are using two COMPLETELY DOMINANT GENES
- Heterozygote is indistinguishable from dominant homozygote
- Two genes are
- A: with A and a alleles
- B: with B and b alleles

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

- Determine the frequency of each gamete produced by F1 generation
- For example if R=0.20, then 20% of the gametes produced will be recombinants which in our example are Ab and aB.
- As there are 2 types of recombinant gametes, frequency of each type will be 0.10
- 80% gametes are parental, [AB and ab type] frequency of each of them is 0.40 or 40%

Determine the phenotype of each cell in Punnet square

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

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

- Done by determining the likelihood (L)
- Likelihood:
- the probability of the observed family
- determined using the multinomial theorem
- an extension of the binomial theorem.

First define the terms for the observed family

- a = number of A_ B_ offspring
- b = number of A_ bb offspring
- c = number of aaB_ offspring
- d = number of aabb offspring
- n = total offspring (= a+b+c+d)
- Define the terms for the expected family proportions
- p = expected proportion of A_B_ offspring
- q = expected proportion of A_ bb offspring
- r = expected proportion of aaB_ offspring
- s = expected proportion of aabb offspring

Multinomial Theorem

- Multinomial theorem describing actual family: paqbrcsdmultiplied by a coefficientn! /(a! b! c! d!)
- Thus the likelihood equation is

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

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

STEP III and IV

- 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)

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

= 4.331

- LOD score = Log10 4.331 (Log10 OR)

= 0.637

- It 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.

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