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2. INTRODUCTION METHODS OF HOMOZYGOSITY MAPPING HOMOZYGOSITY MAPPER GENETIC LINKAGE LOD SCORE METHOD 2

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5. CONTAINING TWO IDENTICAL ALLELIC FORMS CAN BE HOMOZYGOUS DOMINANT CAN BE HOMOZYGOUS RECESSIVE PEA PLANT 5

6. BOTH ALLELES OF A GENE ARE DIFFERENT ONE GENE IS DOMINANT ONE GENE IS RECESSIVE 6

7. SETTING A LOCATION WITH RESPECT TO A MARKER PLOTTING DNA FRAGMENTS ON CHROMOSOMES HELPFUL IN PREDICTING A DISEASE 7

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9. A GENE OR A DNA SEQUENCE FOR A PARTICULAR TRAIT HAS A PARTICULAR LOCATION ON A CHROMOSOME DETECTION HELPFUL IN PREDICTING A DISEASE 9

10. BREAKING AND REJOINING OF DNA MOLECULES EXAMPLE IS CROSSING OVER EXCHANGE OF GENETIC MATERIAL TAKES PLACE RESULTING MOLECULES ARE CALLED RECOMBINANTS 10

11. TOTAL NUMBER OF RECOMBINANTS/TOTAL NUMBER OF PROGENIES IN A TEST CROSS USED TO DETERMINE THE GENETIC DISTANCE CREATION OF GENETIC MAP CENTIMORGAN 11

12. METHOD USED TO DETECT THE DISEASE OF THE HOMOZYGOUS CONDITION HELPFUL FOR THE INHERITED DISORDERS 12

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14. SNP MICROARRAYS RFLP MICROSATELLITE MARKERS 14

15. STANDS FOR SINGLE NUCLEOTIDE POLYMORPHISM INVOLVES A SINGLE PCR METHOD FOLLOWED BY GEL ELECTROPHORESIS TETRA-PRIMER ARMS PCR 15

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17. HIGH DENSITY SNP ARRAYS FOR GENOTYPING MUTATION IDENTIFICATION BY POSITIONAL CLONING 17

18. ABBREVIATED AS BBS CHARACTERIZED BY: OBESITY; PIGMENTARY RETINOPATHY; POLYDACTYLY; HYPOGONADISM RENAL AND CARDIAC ABNORMALITES COGNITIVE IMPAIRMENT 18

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20. STANDS FOR RESTRICTION FRAGMENT LENGTH POLYMORPHISM USED TO FOLLOW THE PATH OF A SPECIFIC GENE VARIATIONS IN THE HOMOLOGOUS DNA SAMPLES 20

21. CUTTING DNA SAMPLES WITH RESTRICTION ENZYMES SEPERATION BY AGAROSE GEL ELECTROPHORESIS DETERMINING THE NUMBER OF FRAGMENTS AND SIZES 21

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23. DNA FINGERPRINTING TRACING ANCESTORY STUDYING EVOLUTION AND MIGRATION DETECTION AND DIAGNOSIS GENETIC MAPPING 23

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26. 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

27. 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.

28. 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

29. Lineage or Genealogical study of family lines. Gives list or family tree of ancestors. Used for studies of certain inheritance pattern.

30. 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

31. 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

32. 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

33. 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

34. Based on frequency of genetic markers passing together.

35. 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

36. 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

37. 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

38. 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

39. 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

40. 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

41. 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

42. P1:AABB Xaabb Gametes F1AaBb Parental Combinations Recombinants ab aBAb AB ab

43. 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%

44. 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

45. 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

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

47. 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

48. 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

49. 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

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

51. 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)

52. 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.