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
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.
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
Lineage or Genealogical study of family lines. Gives list or family tree of ancestors. Used for studies of certain inheritance pattern.
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
Morgan’s Experiments: 1 st Cross: F1 Progeny: Heterozygous red eyed males and females 2 nd 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 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
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
Based on frequency of genetic markers passing together.
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
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 negative Positive LOD score meansLinkage present Negative LOD scoremeansNo Linkage >3 Evidence for linkage +3 1000 to 1 odds that the linkage did not occur by chance <-2Evidence to exclude linkage
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 0.52 completely unlinked genes
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
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
P1:AABB Xaabb Gametes F1AaBb Parental Combinations Recombinants ab aBAb 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%
AB 0.40 Ab 0.10 aB 0.10 ab 0.40 AB 0.40 AABB 0.16 AABb 0.04 AaBB 0.04 AaBb 0.16 Ab 0.10 AABb 0.04 Aabb 0.01 AaBb 0.01 Aabb 0.04 aB 0.10 AaBB 0.04 AaBb 0.01 aaBB 0.01 aaBb 0.04 ab 0.40 AaBb 0.16 Aabb 0.04 aaBb 0.04 Aabb 0.16 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 PhenotypeCell SumsExpected Frequency A_B_0.16+0.04+0.04+0. 16+0.04+0.01+0.04 +0.01+0.16 0.66 A_bb0.01+0.04+0.040.09 aaB_0.01+0.04+0.040.09 Aabb0.16
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 describing actual family: p a q b r c s d multiplied by a coefficient n! /(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
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)
A total LOD score for some R value of 3 is considered proof of linkage of two genes In our example, Odds Ratio = L 0.20 / L 0.50 = 0.0301 / 0.00695 = 4.331 LOD score = Log 10 4.331 (Log 10 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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