genetic linkage and gene mapping

genetic linkage, types, crossing over, types,
gene mapping for diploid and haploid organisms, types

1. WELCOME …… GENETIC LINKAGE AND GENE MAPPING

2. INTRODUCTION After detail study of meiosis we came to know that not all traits are independtly assorted,there are few traits which are responsible to stay gether generation to generation,this trait governed by traits is known as linked gene . Set of genes present in whole chromosome is known as Linkage group. And mapping of this genes is know as gene mapping .

3. HISTORY In 1906 Bateson and Punnet first evidently discovered effect of linkage in dihybrid cross in pea plant using traits like flower color and pollen shape but they did not interpret their result in terms of gene behaviour . In 1910 T.H.Morgan had successful in discovery of phenomenon of linkage using Drosophila traits like mutants yellow body with white eye and wild type grey body with red eye.

4. If two genes are on different chromosomes… ( Mendels theme)

5. Test cross.. Half look like they got a set of the parents chromosomes… And half look like they got a mix of both parents chromosomes…

6. 6 Now suppose both gene A and B were next to each other on the same chromosome . (linked genes) What happens to the ratios in this diagram?

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9. TYPES OF LINKAGE 1. LINKED GENES Genes on same chromosomes and tends to dtay to gether during the formation of gamets . doesn’t assort independently. 2.UNLINKED GENES Genes on different homologous chromosomes assort indepently .

10. COUPLING AND REPULSION HYPOTHESIS. This concept was given by Bateson and Punnet to explain the lack of independent assortment in the result of their experiment.but they couldn’t explain exact reason. later re-explained by T.H.Morgan .

11. 1.Coupling phase When two linked genes on each chromosomes are the same type. Is also known as cis phase . Parental type. i.e. both dominante AB or both recessive ab. 2.Repulsion phase When linked genes on each chromosome are different type Is also known as trans phase . Recombinant type. i.e. Ab or Ab. recombination of linked genes occurs in same frequency.

12. GENE MAPPING Linkage of genes in chromosome can be represented in form of genetics map or linkage or chromosomal map. First chromosomal map was succcesfully done by Alfred Sturtvent in Drosophilla using gene recombination frequency. recombination frequency= total no. recombinantes ×100 total number of progenies.

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14. FEATURES OF GENETIC RECOMBINATION Gene mapping determines the order of genes and the relative distances between them in map units 1 map unit = 1 cM ( centimorgan ) Gene mapping methods use recombination frequencies between alleles in order to determine the relative distances between them Recombination frequencies between genes are inversely proportional to their distance apart Distance measurement: 1 map unit = 1 percent recombination (true for short distances)

15. Genes with recombination frequencies less than 50 percent are on the same chromosome = linked ( recombinantes ) Two genes that undergo independent assortment have recombination frequency of 50 percent and are located on non homologous chromosomes or far apart on the same chromosome = unlinked (parental type)

16. 16 Gene Mapping: Crossing Over Exchange of genes between two homologous chromosomes is known as crossing over. Crossovers which occur outside the region between two genes will not alter their arrangement The result of double crossovers between two genes is indistinguishable from independent assortment of the genes Crossovers involving three pairs of alleles specify gene order = linear sequence of genes Fig. 4.13

17. 17 Fig. 4.12

18. GENETIC V/S PHYSICAL DISTANCE Map distances based on recombination frequencies are not a direct measurement of physical distance along a chromosome Recombination “hot spots” overestimate physical length Low rates in heterochromatin and centromeres underestimate actual physical length Physical mapping depends on base pairs(bps) ex:- in humans 1cM=1×10-6bps. In 1992 genetic mapping of s.cerevisiae chromosome 3, it determines that physical mapping and genetic mapping are not same.

19. Gene mapping for two point cross A cross involving two loci ussually refered as two point cross. STEPS:- 1. genes are linked are unlinked. 2.parental combination. 3.recombination frequency calculation. 4.genetic distance.

20. 21 Now cross ( AB ab ) F1 progeny with ( ab ab ) tester to look for recombination on these chromosomes. Suppose you Get…… AB ab 583 <parental > ab ab 597 < parental> Ab ab 134 < recombinant> aB ab 134 < recombinant> total= 1448 so…. 268 recombinants /1448 progeny = 0.185 recombinants/progeny= 18.5% recombinants= 18.5 cM. Starting with pure breeding lines, Cross Parent 1( AA BB ) with Parent 2( aa bb ) So Parental chromosomes in the F1 have to be AB and ab Mapping the distance between two genes

21. Gene mapping for three point cross A cross involving three loci ussually refered as three point cross. STEPS:- 1. genes are linked are unlinked. 2.parental combination. 3.gene ordering. 4.recombination frequency calculation. 5.genetic distance.

22. 23 Cross ( ABD abd ) F1 progeny with ( abd abd ) tester Suppose you Get…… ABD abd 580 <parental ABd abd 3 abD abd 5 <parental abd abd 592 AbD abd 45 <recombinant Abd abd 89 aBD abd 94 aBd abd 40 <recombinant total= 1448 Mapping (and ordering) three genes Starting with pure breeding lines, Cross Parent 1( AA BB DD ) with Parent 2( aa bb dd ) So you know the Parental chromosomes in the F1 have to be ABD and abc Ab + aB = (45+89)+(94+40) recom 268 recom/1448 total =0.185 A-B =18.5mu Bd + bD = (3+40)+(5+45) 93 recom/1448 total= 0.064 B-D =6.4mu Ad + aD = (3+89)+(5+94) 191 recom/1448 total= 0.132 A-D =13.2mu so the order must be A-----D---B -13.2--6.4- ----18.5----

23. 24 Cross ( ABD abd ) F1 progeny with ( abd abd ) tester Suppose you Get…… ABD abd 580 <parental ABd abd 3 abD abd 5 <parental abd abd 592 AbD abd 45 <recombinant Abd abd 89 aBD abd 94 aBd abd 40 <recombinant total= 1448 Ab + aB = (45+89)+(94+40) recom 268 recom /1448 total =0.185 A-B =18.5mu A-----D---B -13.2--6.4- ----18.5---- So How come 13.2 + 6.4 does not equal 18.5? We missed the double recombinants on the first pass… longer the distance, more potential to underestimate recomb freq. (45+89)+(94+40)+2(3+5) recom 284recom/1448 total = 0.196 A-B =19.6mu

24. INTERFERENCE AND COINCIDENCE Chromosome interference : crossovers in one region decrease the probability of a second crossover close by. Intereference can meassured by coefficient of coincidence. Coefficient of coincidence = observed number of double recombinants divided by the expected number. Expected double cross over = expected double cross over frequency×total number of progenies. Interference = 1-Coefficient of coincidence

25. If the coefficient of coincidence is, =0; then interference is complete and no double crossovers are observed. 0-1; partial interference. =1; there is no interference and expected double cross overs are observed. If the two crossovers were independent, we would expect that the probability of seeing two recombination events occur would be 0.132 between A-D AND 0.064 between D-B 0.132 X 0.064 = 0.008 For every 1448 progeny, this would be (1448x0.008)=12.23 double recombinants We actually observed only (5+3)= 8 double recombinants So the Coefficient of coincidence = observed / expected = 8/12.23 = 0.65 Interference = 1-Coefficient of coincidence = 1- 0.65 = 0.35

26. 27 Tetrad Analysis In some species of fungi, each meiotic tetrad is contained in a sac-like structure, called an ascus Each product of meiosis is an ascospore , and all of the ascospores formed from one meiotic cell remain together in the ascus . Formation of 4 cell after meiosis is known as TEDTRAD. Further mitosis leads to formation of 8 cell is known as OCTAD. Fig. 4.22

27. 28 Features Of Tetrad Analysis Several features of ascus -producing organisms are especially useful for genetic analysis: They are haploid, so the genotype is expressed directly in the phenotype They produce very large numbers of progeny Their life cycles tend to be short

28. 29 Ordered and Unordered Tetrads Organisms like Saccharomyces cerevisiae , produce unordered tetrads : the meiotic products are not arranged in any particular order in the ascus . Unordered tetrads have no relation to the geometry of meiosis. Bread molds of the genus Neurospora have the meiotic products arranged in a definite order directly related to the planes of the meiotic divisions — ordered tetrads. The geometry of meiosis is revealed in ordered tetrads.

29. 30 Tetrad Analysis In tetrads when two pairs of alleles are segregating, three patterns of segregation are possible parental ditype (PD) = two parental genotypes nonparental ditype (NPD) = only recombinant combinations tetratype (TT) = all four genotypes observed Fig. 4.24

30. 31 The existence of TT for linked genes demonstrates two important features of crossing-over. The exchange of segments between parental chromatids takes place in prophase I, after the chromosomes have duplicated. The exchange process consists of the breaking and rejoining of the two chromatids , resulting in the reciprocal exchange of equal and corresponding segments.

31. 32 Neurospora : Ordered Tetrads Ordered asci also can be classified as PD, NPD, or TT with respect to two pairs of alleles, which makes it possible to assess the degree of linkage between the genes The fact that the arrangement of meiotic products is ordered also makes it possible to determine the recombination frequency between any particular gene and its centromere

32. 33 Fig. 4.26

33. 34 Tetrad Analysis: Ordered Tetrads Homologous centromeres of parental chromosomes separate at the first meiotic division The centromeres of sister chromatids separate at the second meiotic division When there is no crossover between the gene and centromere, the alleles segregate in meiosis I A crossover between the gene and the centromere delays segregation alleles until meiosis II

34. 35 Tetrad Analysis: Ordered Tetrads The map distance between the gene and its centromere equals 1/2 x (Number of asci with second division segregation/ Total number of asci) x 100 This formula is valid when the gene is close enough to the centromere and there are no multiple crossovers

35. 36 Fig. 4.27 top

36. 37 Fig. 4.27 bottom

37. Tetrad analysis:- unordered type It contain spores that are randomly arranged. Estimation involves dihybrid cross. Mapping can done based on wether the genes are linked or not. Three types of segregation are possible. parental ditype (PD) = two parental genotypes nonparental ditype (NPD) = only recombinant combinations tetratype (TT) = all four genotypes observed

38. 39 Tetrad Analysis unordered types. When genes are unlinked, the parental ditype tetrads and the nonparental ditype tetrads are expected in equal frequencies: PD = NPD Linkage is indicated when nonparental ditype tetrads appear with a much lower frequency than parental ditype tetrads: PD » NPD Map distance between two genes that are sufficiently close that double and higher levels of crossing-over can be neglected, equals 1/2 x (Number TT / Total number of tetrads) x 100

39. Type 1:- If crossover doesnot occurs between two loci or if two strand double crossover between them. Resulted meiotic product will be of two kind Referred as parental ditype or PD .

40. Type 2:- Four strand double cross between two genes. Results in formation of two kinds which are parental combination. This is refered as non-parental ditype or NPD .

41. Type 3:- A two strand single cross or three strand double cross over between two gene. Results in formation of both parental and non-parental type combination This is refered as tetra types or TT. Formation of 50% parental type and 50% recombinant type .

42. REFERENCES:- Life science,fundamentals and practices-2,pranav kumar and usha mina,5 th edition,2016. Fumdamentals of genetics,B.D.singh,4 th edition,2001. Principles of genetics,Eldon Jhon Gardner,Michael J.simmon,D.Peter Snustad,8th edition,2015. Slideshare.com

43. THANK YOU…… Presented by, MAHAMMED FAIZAN MH2TAG0165 Jr.M.Sc (HORT) in GPB COLLEGE OF HORTICULTURE MUDIGERE.