1. WELCOME IN MERIT LIFE SCIENCES e-TUTORIAL NEET/NET JRF CSIR EXAM /JAM/DBT DR SHEELENDRA Ph.D. FROM BHU 3-TIME NET-CSIR-JRF QUALIFIED GATE 98AUTHOR OF 2 BOOKS ANIMAL CELL CULTURE FROM OXFORD AND ENZYME TECHNOLOGY2 PATENTSBEST RRESEARCH AWARDEE ANDMORE THAN 50 INT PAPERNOW RUNNING MERIT E -TUTORIAL BRIEF INTRO ABOUT ME Get 20% discount On enrollment till Dec 2018

2. 3-point cross 19—NOV-2018 FACULTY: DR SHEELENDRA BHATT DR S M BHATT (PhD BHU) RESEARCH AWARDEE EX. DIRECTOR SBBS UNIVERSITY

3. Three Point Cross The easiest way to map genes is to compare them in groups of 3. This allows both the distances between them and their order to be determined. Further genes can be added to the map by using overlapping groups of 3. Mapping is usually done in test crosses. One parent is heterozygous for two versions of the chromosome being mapped, and the other parent is homozygous for the recessive mutants being mapped. Recombination in the heterozygous parent gives different combinations of alleles, which are counted. Note that recombination also occurs in the homozygous recessive parent, but it has no effect on the alleles in the offspring because it is homozygous.

4. 3 point cross A testcross discovered by Mendel generally involves crossing of phenotypically dominant individual with a phenotypically recessive individual to determine the recombinant frequency and zygosity of the inherited genes. In linkage analysis, a three point testcross refers to analyzing the inheritance pattern of 3 alleles by testcrossing a triple heterozygote with a triple recessive homozygote. It enables us to determine the distance between the 3 alleles and the order in which they are located on the chromosome. Example Three-point testcross can be best understood by considering an example. Let us consider 3 loci: A, B, and C. They can have 3 possible linear order arrangements: A-B-C; B-C-A; or A-C-B.

6. A cross between AABBCC with aabbcc is done to obtain F1 progeny that is testcrossed with aabbcc (homozygous recessive). A systematic approach after the crossover can help 1. Identify the order and distance between the genes. 2. Identify the parental classes (ABC and abc, with no crossover). 3. Map any two loci. 4. Map any other two loci and try to draw a rough map. 5. Map the remaining two loci. 6. Chose the right map with the help of distance value that fits better. 7. Draw the map with real order. 8. Analyze the single crossover classes. 9. Analyze the double crossover classes. 10. Refine the distance Important points that help in three-point testcrosses: Double crossover (DCO) gametes always have the lowest frequency. DCO shifts the middle allele from one sister chromatid to the other. A mathematical relationship can be used to determine the order the order of the genes. If recombination frequency between genes A and B is equal to X, between genes B and C is equal to Y, and between genes C and A is equal to Z, then it follows that: ○ If Z = X + Y, the order of the genes is A-B-C; ○ If Z = X - Y, the order of the genes is A-C-B; ○ If Z = Y - X, the order of the genes is B-A-C.

7. Table 04.01 Interpreting in a Three-Point Cross.

8. Figure 4.14 The order of genes in a three-point testcross

9. Figure 4.15 Result of single crossovers in a triple heterozygote

10. Figure 4.16 Result of double crossovers in a triple heterozygote

11. Table 4.2 Comparing Reciprocal Products in a Three-Point Cross

12. In corn, c gives a green plant body, while its wildtype allele c + gives a purple plant body. bz (bronze) gives brown seeds, while the wildtype allele bz + gives purple seeds. wx (waxy) gives waxy endosperm in the seeds; wx + gives starchy endosperm. The genes are arranged on the chromosome in the order c-bz- wx. The cross: c bz wx / + + + x c bz wx. Note the +’s are the dominant wildtype alleles of the corresponding gene. phenotypecount c bz wx318 + + +324 c bz +105 + + wx108 c + +18 + bz wx20 c + wx4 + bz +3 total900 Basic process: determine the percentage of offspring that had a crossover between each pair of genes. 1.Examine c and bz first. 2.Parental configuration was c bz and + +. 3. Therefore, the recombinant configurations are c + and + bz. Count recombinants, ignoring the other gene (wx): 18 c + +, 20 + bz wx, 4 c + wx, 3 + bz +. Total is 45 recombinants out of 900 total offspring. 45 / 900 = 0.05. Need to multiply by 100 to get percentage: 0.05 x 100 = 5.0 map units between c and bz. 2. Next examine bz and wx. Parentals are bz wx and + +, so recombinants are bz + and + wx. Ignoring c, the count of recombinants is: 105 c bz +, 108 + + wx, 4 c + wx, 3 + bz +. Total = 220 recombinants. 220 / 900 = 0.244. 0.244 x 100 = 24.4 map units between bz and wx. 3. Now do c and wx. Parentals are c wx and + +, so recombinant offspring are c + and + wx. Ignoring bz, the recombinants are 105 c bz +, 108 + + wx, 18 c + +, 20 + bz wx. Total = 251. 251/ 900 x 100 = 27.9 map units.

13. Notes on the Data Genes are arranged in reciprocal pairs: each pair has 1 copy of the mutant allele and the wildtype allele for each gene. The counts are roughly equal for reciprocal pairs, because they are both products of the same crossing over events. Parentals are the largest groups: c bz wx and + + +. Double crossovers, one between c and bz and another between bz and wx, are the smallest groups.

14. Map of c, bz, and wx All 3 genes are in the proper order, and all 3 distances between pairs of genes are shown. Note that distances don’t add up. This is due to double crossovers, which we will discuss next.

15. Double Crossovers and Mapping A double crossover is two crossovers both occurring between the two genes being examined. The first crossover changes the parental configuration of alleles to the recombinant configuration. The second crossover changes the recombinant configuration back to the parental. The net result is that the genes are in the parental configuration, same as if no crossovers had occurred. Thus, any even number of crossovers is the same as 0 crossovers, and any odd number is the same as 1 crossover. Since you only see the offspring and not the actual crossovers, it is very easy to undercount the number that occurred Consider the c bz wx cross. If you were just looking at c and wx, and hadn’t examined bz, the c + wx and + bz + offspring would be parental and count as 0 crossovers. The only way you know that 2 crossovers occurred is by examining bz. Perhaps other crossovers also occurred that we didn’t detect, since we didn’t examine any other genes in between bz and wx. This is the main reason why the sum of the c--bz and bz--wx distances didn’t add up: double crossovers were counted as parentals.

16. Mapping Function The further apart 2 genes are, the more likely it is that undetected double crossovers will occur between them. For this reason, gene maps are created using short intervals. One consequence: the maximum percentage of recombinant offspring is 50%, but many chromosomes are several hundred map units long. For instance, the human chromosome 13 is 125 map units long. Genes on opposite ends of this chromosome would have a 50% frequency of recombinant offspring. Mapping function: number of actual crossovers on x-axis, frequency of recombinant offspring on y-axis. In general, for less than 20% recombinants, the number of recombinants is roughly equal to the number of crossovers; for 20-50% recombinants, the number of recombinants is significantly less than the number of crossovers. for 50% recombinants, the number of crossovers is not determinable.

17. Interference There is a second issue with double crossovers: interference. Interference is the inability of 2 crossovers to occur very close to each other. Think of the chromosome as a thick rope: it is impossible to bend it too tightly. It is possible to measure the amount of interference, by comparing the actual number of double crossovers to the number that you would expect based on the number of single crossovers that occurred.

18. Measuring Interference Consider our c--bz--wx example. The c--bz interval was 5.0 map units, or 5% recombinants between those 2 genes. The probability of a crossover between c and bz is 0.05, which you get by dividing the map distance by 100 to put it on a 0-1 scale instead of a 0-100 scale. The bz--wx interval was 24.4 map units, or 24.4% recombinants, or a 0.244 probability. If a crossover between c and bz had no effect on the possibility of a crossover between bz and wx (i.e. no interference), then a chance of a c--bz crossover AND and bz--wx crossover would be the product of their individual probabilities. That is, the expected frequency of double crossovers would be 0.05 * 0.244 = 0.0122. Since there were 900 offspring, the expected number of double crossover offspring would be 10.98. The observed number of double crossovers was 7. The coefficient of coincidence is the ratio of observed double crossovers to expected: 7 / 10.98 = 0.638. The interference is 1 minus the coefficient of coincidence, = 1 - 0.638 = 0.362. This means that about 36% of the expected double crossovers are not occurring due to interference.

19. Interference Formulas

20. Gene Order Another problem that can arise is that when mapping genes you usually don’t know their order. You need to infer the order from the data. Also, not all the alleles are necessarily going to be in coupling. Sometimes some alleles are in repulsion. Getting the gene order is a matter of comparing the alleles present on the parental chromosomes to those on the double crossover chromosomes.

21. Example Note that the offspring counts are arranged in reciprocal pairs. Each member of the pair has the opposite alleles, and the counts for the two members of the pair are approximately equal. The parental class of offspring is the largest: a + d, and + b +. No crossovers have occurred here: the original parents had these combinations of alleles. The double crossovers class (2CO) is the smallest class: + b d and a + +. The other two pairs are the two single crossover classes. offspring phenotype count a + d510 + b +498 a + +3 + b d1 a b +61 + + d59 a b d35 + + +37 total1204

22. Getting Gene Order Since we don’t know (or care about) the orientation of these genes relative to the chromosome as a whole, there are only 3 possible orders. These are based on which gene is in the middle. The genes could be a--b--d, b--a--d, or a--d--b. Note that a--b--d is identical to d--b--a, etc. The order in which the genes are listed in the offspring counts has nothing to do with their order on the chromosome!!! The gene order is unknown at this point. To determine gene order, set up the parental chromosomes in the F1, then see what the resulting double crossover offspring would look like. If the observed 2CO’s match the expected, then you have the correct gene order. If not, try a different order.

23. Gene Order 1. Try a--b--d. The F1 chromosomes would be a + d/ + b +. A double crossover would give a b d and + + + 2CO offspring. This does not match the observed a + + and + b d. 2. Try b--a--d. The F1 chromosomes are + a d / b + +. A double crossover would give b a + and + + d 2CO offspring. This does not match the observed b + d and + a +. 3. Try a--d--b. The F1 chromosomes are a d + and + + b. A double crossover gives a + + and + d b, which matches the observed 2CO offspring. Therefore, a--d--b is the correct order; d is in the middle.

24. Gene Distances a--d. Parentals are a d and + +, so recombinants are a + and + d. There are 61 + 59 + 3 + 1 = 124 recombinant offspring. Total offspring = 1204. So map distance = 124/1204 * 100 = 10.3 map units. d--b. Parentals are d + and + b, so recombinants are d b and + +. There are 35 + 37 + 3 + 1 = 76 of them. 76 / 1204 * 100 = 6.3 map units. a--b. Parentals are a + and + b, so recombinants are a b and + +. There are 61 + 59 + 35 + 37 = 192 of them. 192 / 1204 * 100 = 15.9 map units.

25. Interference and Map There were 3 + 1 = 4 observed 2CO’s. Expected 2CO: = (10.3/100) * (6.3/100) * 1204 = 7.81 Coef. of Coincidence = obs 2CO / exp 2CO = 4 / 7.81 = 0.512 Interference = 1 - C. of C. = 1 - 0.512 = 0.487

26. Steps to Solving 3-Point Cross Problems 1. Organize the data into reciprocal pairs. For example, a + c and + b + are a reciprocal pair, and so are a b c and + + +. The number of offspring for each member of a pair will be similar. 2. Determine which pair is the parental class: it is the LARGEST class. 3. Determine which pair is the double crossover (2CO) class: the SMALLEST class. 4. Determine which gene is in the middle. If you compare the parentals with the 2COs, the gene which switched partners is in the middle. If you think two genes have switched sides, it is the other gene that is in the middle. 5. Determine map distances for all three pairs of genes. Count the number of offspring that have had a crossover between the genes of interest, then divide by the total offspring and multiply by 100. 6. Figure the expected double crossovers: (map distance for interval I / 100 ) * (map distance for interval II / 100 ) * (total offspring) = exp 2CO. 6. Figure the interference as: I = 1 - (obs 2CO / exp 2CO). The observed 2CO class comes from the data.

27. One More Example Here are some test cross data for a 3 point cross. Determine the order of the genes, make a map showing all map distances, and determine the interference value. pr = purple eyes bl = black body dp = dumpy bristles Answer will be given in class.

28. Data phenotypecount wild type756 purple59 dumpy414 black21 purple dumpy23 purple black417 dumpy black60 purple dumpy black750 total2500

29. More Realistic Data phenotypecount wild type60 purple756 dumpy23 black417 purple dumpy414 purple black21 dumpy black750 purple dumpy black59 total2500

30. Coupling and Repulsion Configuration of Linked Genes Coupling (cis configuration): Wild type alleles are found on one chromosome; mutant alleles are found on the other chromosome. p + b + p b Repulsion (trans configuration): Wild-type allele and mutant allele are found on the same chromosome. p + b p b +

32. Genetic Mapping Mapping function: the relation between genetic map distance and the frequency of recombination Chromosome interference: crossovers in one region decrease the probability of a second crossover close by Figure 04.17: Mapping functions.

33. Genetic Mapping Coefficient of coincidence = observed number of double recombinants divided by the expected number Interference = 1-coefficient of coincidence

34. Mapping Genes in Human Pedigrees Methods of recombinant DNA technology are used to map human chromosomes and locate genes Genes can then be cloned to determine structure and function Human pedigrees and DNA mapping are used to identify dominant and recessive disease genes Polymorphic DNA sequences are used in human genetic mapping.

35. Mapping Genes in Human Pedigrees Most genes that cause genetic diseases are rare, so they are observed in only a small number of families. Many mutant genes of interest in human genetics are recessive, so they are not detected in heterozygous genotypes. The number of offspring per human family is relatively small, so segregation cannot usually be detected in single sibships. The human geneticist cannot perform testcrosses or backcrosses, because human matings are not dictated by an experimenter.

36. Genetic Polymorphisms The presence in a population of two or more relatively common forms of a gene or a chromosome is called polymorphism A prevalent type of polymorphism is a single base pair difference, simple-nucleotide polymorphism (SNP) SNPs in restriction sites yield restriction fragment length polymorphism (RFLP) Polymorphism resulting from a tandemly repeated short DNA sequence is called a simple sequence repeat (SSR)

37. SNPs SNPs are abundant in the human genome. The density of SNPs in the human genome averages about one per 1300 bp Identifying the particular nucleotide present at each of a million SNPs is made possible through the use of DNA microarrays composed of about 20 million infinitesimal spots on a glass slide the size of a postage stamp.

38. Figure 04.18: SNP genotype of an individual.

39. RFLPs Restriction endonucleases are used to map genes as they produce a unique set of fragments for a gene There are more than 200 restriction endonucleases in use, and each recognizes a specific sequence of DNA bases EcoR1 cuts double-stranded DNA at the sequence 5'-GAATTC-3' wherever it occurs

40. Figure 04.19: The restriction enzyme EcoRI cleaves double-stranded DNA wherever the sequence 5-GAATTC-3 is present.

41. RFLPs Differences in DNA sequence generate different DNA cleavage sites for specific restriction enzymes Two different alleles will produce different fragment patterns when cut with the same restriction enzyme due to differences in DNA sequence Figure 04.20: A minor difference in the DNA sequence of two molecules can be detected if the difference eliminates a restriction site.

42. SSRs A third type of DNA polymorphism results from differences in the number of copies of a short DNA sequence that may be repeated many times in tandem at a particular site in a chromosome When a DNA molecule is cleaved with a restriction endonuclease that cleaves at sites flanking the tandem repeat, the size of the DNA fragment produced is determined by the number of repeats present in the molecule There is an average of one SSR per 2 kb of human DNA

43. Figure 04.21B: A type of genetic variation that is widespread in most natural populations of animals and plants.

44. Mapping Genes in Human Pedigrees Human pedigrees can be analyzed for the inheritance pattern of different alleles of a gene based on differences in SSRs and SNPs Restriction enzyme cleavage of polymorphic alleles that are different in RFLP pattern produces different size fragments by gel electrophoresis

45. Figure 04.22: Human pedigree showing segregation of SSR alleles.

46. Copy-number variations (CNVs) A substantial portion of the human genome can be duplicated or deleted in much larger but still submicroscopic chunks ranging from 1 kb to 1 Mb. This type of variation is known as copy-number variation (CNV). The extra or missing copies of the genome in CNVs can be detected by means of hybridization with oligonucleotides in DNA microarrays.

48. 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 Figure 04.24: Formation of an ascus containing all of the four products of a single meiosis.

49. 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

50. Figure 4.24 Life cycle of the yeast Saccharomyces cerevisiae

51. 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

52. Tetrad Analysis: Unordered Tetrads 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 Figure 04.26: Types of unordered asci produced with two genes in different chromosomes.

53. Tetrad Analysis: Unordered Tetrads 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

54. Tetrad Analysis 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

55. Tetrad Analysis: 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

56. Figure 04.28: The life cycle of Neurospora crassa.

57. 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

58. 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

59. Figure 04.29 (top): First- and second-division segregation in Neurospora.

60. Figure 04.28 (bottom): First- and second-division segregation in Neurospora.

61. Gene Conversion Most asci from heterozygous Aa diploids demonstrate normal Mendelian segregation and contain ratios of 2A : 2a in four-spored asci, or 4A : 4a in eight-spored asci. Occasionally, aberrant ratios are also found, such as 3A : 1a or 1A : 3a and 5A : 3a or 3A : 5a. The aberrant asci are said to result from gene conversion because it appears as if one allele has “converted” the other allele into a form like itself

62. Gene Conversion Gene conversion is frequently accompanied by recombination between genetic markers on either side of the conversion event, even when the flanking markers are tightly linked Gene conversion results from a normal DNA repair process in the cell known as mismatch repair Gene conversion suggests a molecular mechanism of recombination

63. Figure 4.29 Mismatch repair

64. Figure 04.31: Mismatch repair resulting in gene conversion.

65. Recombination Recombination is initiated by a double-stranded break in DNA The size of the gap is increased by nuclease digestion of the broken ends These gaps are repaired using the unbroken homologous DNA molecule as a template The repair process can result in crossovers that yield chiasmata between nonsister chromatids

66. Figure 04.32: Double-strand break in a duplex DNA molecule. Adapted from D. K. Bishop and D. Zickler, Cell 117 (2004): 9-15

67. Recombination: Holliday Model The nicked strands unwind, switch partners, forming a short heteroduplex region with one strand and a looped-out region of the other strand called a D loop The juxtaposed free ends are joined together, further unwinding and exchange of pairing partners increase the length of heteroduplex region—process of branch migration

68. Recombination: Holliday Model One of two ways to resolve the resulting structure, known as a Holliday junction, leads to recombination, the other does not The breakage and rejoining is an enzymatic function carried out by an enzyme called the Holliday junction- resolving enzyme

69. Figure 04.33: EM of a Holliday structure. Illustration modified from B. Alberts. Essential Cell Biology. Garland Science, 1997. Illustration reproduced with permission of Huntington Potter, Johnnie B. Byrd Sr., Alzheimer’s Center & Research Institute