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Dr. Chris Rongo rongo@waksman.rutgers.edu GENETICS 502 C. elegans Section, Lecture 2 10:00-11:25, Waksman Auditorium Tu

Dr. Chris Rongo rongo@waksman.rutgers.edu GENETICS 502 C. elegans Section, Lecture 2 10:00-11:25, Waksman Auditorium Tuesday, March 3, Spring 2009 Today’s Lecture: Complementation Linkage Analysis Two-factor Mapping Three-factor Mapping. Genetic dissection of biological.

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Dr. Chris Rongo rongo@waksman.rutgers.edu GENETICS 502 C. elegans Section, Lecture 2 10:00-11:25, Waksman Auditorium Tu

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  1. Dr. Chris Rongo rongo@waksman.rutgers.edu GENETICS 502 C. elegans Section, Lecture 2 10:00-11:25, Waksman Auditorium Tuesday, March 3, Spring 2009 Today’s Lecture: Complementation Linkage Analysis Two-factor Mapping Three-factor Mapping

  2. Genetic dissection of biological processes Wild type animal WT WT WT mutant WT WT WT WT 1) Identify a process to study GOAL: 2) Mutagenize To obtain mutants that are defective in a specific biological process. Mutant screens allow researchers to identify the molecular players that underlie a given biological process. By discovering these players, researchers obtain an experimental handle on the gene or process. 3) Screen for a phenotype 4) Define genes by complementation testing 5) Map gene 6) Clone gene

  3. Complementation tests Q: When do you use this? Note: requires recessive mutations mut-1 Complementation: the production of a wild-type phenotype when two different mutations are combined in trans. The two mutations are probably in different genes. WT phenotype mut-3 mut-1 Noncomplementation or failure to complement between two mutations usually indicates that the two mutations are in the same gene. Mut phenotype mut-2

  4. Is the new unc* mutation an allele of the gene unc-44? unc* + P0 X unc* + unc-44(e1111) unc* F1 X + unc-44(e1111) unc* + F2 unc* + unc-44(e1111) unc-44(e1111) unc-44(e1111) unc-44(e1111) Unc or WT? Unc or WT? WT WT 50% If unc* is an allele of unc-44 _____% of progeny are Unc. If unc* complements unc-44, it is probably not a mutation in the same gene. Q: What do you do next if unc* fails to complement unc-44? What molecular experiment could you perform to determine whether it is a new allele of unc-44?

  5. a Complex Complementation Tests Intragenic complementation: the production of a wild-type phenotype when two mutations in the same gene are combined in trans. Easy to explain for proteins that have more than one functional domain: Mutation 1 in Protein AB Alone, they lack either A or B activity . A B Mutation 2 in Protein AB Combined in trans, they have both A and A B B activity . Keep in mind that there are other ways that intragenic complementation can occur (polycistronic messages, alternative splicing, et cetera.)

  6. Complex Complementation Tests Non-allelic non-complementation: the production of a mutant phenotype when two mutations in two different genes are combined in trans. Two recessive mutations involved in embryogenesis: emb-1 & emb-4. emb-1 maps to chromosome I emb-4 maps to chromosome III GENOTYPE PHENOTYPE emb-1/emb-1 ; +/+ mutant emb-1/+ ; +/+ WT +/+ ; emb-4/emb-4 mutant +/+ ; emb-4/+ WT Seems normal for two recessives, but… emb-1/+ ; emb-4/+ mutant A wild-type copy of each gene is present, yet the animal still has a mutant phenotype.

  7. “Threshold Levels” model for Non-allelic Non-Complementation Perhaps EMB-1 and EMB-4 dimerize to form a functional protein, and the mutations make non-functional subunits. +/+ ; +/+ 100% Level of Heterodimer emb-1/+ ; +/+ 50% Level of Heterodimer +/+ ; emb-4/+ 50% Level of Heterodimer emb-1/+ ; emb-4/+ 25% Level of Heterodimer If the threshold limit for the level of the heterodimer in a cell is 50% before you see a phenotype, then any genotype combination that makes less than 50% will show a mutant phenotype. Heterozygosity for either gene alone gives 50%, which is enough. Heterozygosity for both genes gives 25%, which is not enough, so mutant phenotype appears.

  8. “Poison” model for Non-allelic Non-Complementation Perhaps EMB-1 and EMB-4 dimerize to form a functional protein, and the mutant proteins, when together, form a nonfunctional protein that can interfere with the function of wild-type dimers. +/+ ; +/+ no Poisoned Heterodimers emb-1/+ ; +/+ no Poisoned Heterodimers +/+ ; emb-4/+ no Poisoned Heterodimers emb-1/+ ; emb-4/+ 25% are Poisoned Heterodimers For example, perhaps EMB-1/EMB-4 heterodimers are subunits in a polymer chain. A poisoned heterodimer might be able to add itself to the end of the chain, but not allow additional subunits to be added (i.e., chain terminator).

  9. Linkage analysis in C. elegans Q: Why do geneticists map genes? How to begin mapping: • Cross your mutation with • known markers on the six • chromosomes. • Genes on C. elegans • chromosomes tend to be in • clusters, making the • assignment of a new • gene to a chromosome • easier.

  10. a+ a+ a- a- Mendel’s First Law: The Law of Segregation Alternative versions (alleles) of genes account for variation during heredity. For each inherited characteristic (gene), an organism inherits two alleles, one from each parent. If two alleles differ, then the dominant allele is fully expressed for the characteristic, whereas the recessive is not expressed. Two alleles for a given gene segregate during gamete production. 1. Alleles are on homologous chromosomes. 2. They line up at metaphase. 3. Each homolog goes into a different gamete. 4. You are drawing Metaphase I every time you write a genotype.

  11. a a a a If two genes are closely linked... unc-5 + + mut P0 X unc-5 + + mut unc-5 + F1 + mut unc-5 + unc-5 + + mut F2 + mut + mut unc-5 + The mutations repel or exclude each other in the F2 generation: 1/__ are Unc non Mut 1/__ are Mut non Unc 4 4 Question: Do you ever get worms that are both Unc and Mut? If perfectly linked, then never. If closely linked, then rarely (recombinants)

  12. Punnett Square For Closely Linked Genes F1: unc-5/mut Gametes From Sperm Gametes From Oocytes

  13. What if the genes are unlinked? More help from the 19th century: “During gamete formation the segregation of the alleles of one gene is independent of segregation of the alleles of another gene.” - Gregor Mendel, 2nd Law 20th century update: “This is true IF the genes are on separate chromosomes or are far apart on the same chromosome.” - Barbara McClintock, Thomas Hunt Morgan and others.

  14. If two genes are unlinked... + mut unc-5 + X ; ; P0 + mut unc-5 + unc-5 + F1 ; Self fertilize + mut 1 +/+ ; +/+ F2 2 +/+ ; +/mut 9 nonUnc nonMut 2 unc-5/+ ; +/+ 4 unc-5/+ ; +/mut 1 +/+ ; mut/mut 3 Mut nonUnc 2 unc-5/+ ; mut/mut 1 unc-5/unc-5 ; +/+ 3 Unc nonMut 2 unc-5/unc-5 ; +/mut 1 Unc Mut 1 unc-5/unc-5 ; mut/mut a a a

  15. Punnett Square For Unlinked Genes F1: unc-5/ + ; mut/ + Gametes From Sperm Gametes From Oocytes

  16. unc-5 + F1 ; Self fertilize + mut If you pick Unc F2s, what % are also Mut? 25% If you pick Mut F2s, what % are also Unc? 25% Why? Unlinked mutations sort independently. unc-5 F1 Self fertilize mut 0% If you pick Unc F2s, what % are also Mut? If you pick Mut F2s, what % are also Unc? 0% Why? Linked mutations repel each other.

  17. Mapping takes advantage of two kinds of information: Physical map Genetic map

  18. C. elegans genome is about 20,000 genes, 97Mb Molecular and genetic data is stored in public databases: http://www.wormbase.org/ http://elegans.swmed.edu/ The genome project generated large clones that cover the entire genome: Cosmid ~ 30KB insert (acts like a phage and plasmid) can contain 5-20 genes. YAC = Yeast artificial chromosome ~250KB insert contains hundreds+ of genes

  19. The genetic (left) and molecular (right) maps can be aligned.

  20. Connecting the genetic map to the molecular map: mut (+) mnDf10 (+) mnDf100 (-) mnDf101 (-) mnDf102 unc-3 genetic map molecular map SNP8 SNP9 YACs } Cosmids Genes

  21. Deficiency Mapping 1 2 3 4 1 2 3 4 Reannealing 1 2 3 4 1 2 4 Breakpoint Deficiency: A deletion of a portion of a chromosome Genotype:mnDf22/mnC1 dpy-10(e128) unc-52(e444)II. Description: Hets are WT and segregate WT, paralysed DpyUnc and dead eggs. Maintain by picking WT. Balancer chromosome: suppresses recombination (above, mnC1). Breakage X-rays or gamma rays 1 2 3 4 1 2 4

  22. + mDf10 Balancer unc-3(m22) lon-2(e43) The balancer is marked with two recessive mutations: unc-3 (which is within the deficiency and therefore “uncovered” ) lon-2 (which is outside of the deficiency, but closely linked so that recombination rarely occurs) Self fertilization results in the following progeny: mDf10/mDf10 lethal Q: Why are Dfs lethal? mDf10/unc-3 lon-2 Unc unc-3 lon-2/unc-3 lon-2 Unc Lon

  23. + mut P0 mut X + mnDf10 mut F1 X + unc-3 lon-2 mut + F2 mut + mnDf10 unc-3 lon-2 mnDf10 unc-3 lon-2 phenotypes: Mut or WT WT WT WT F3 phenotypes: ____ ____ ____ ____ 1/4 Mut 1/4 Dead 1/4 Mut 1/4 Unc Lon 1/4 Dead 1/4 Unc Lon

  24. MEIOSIS IN C. ELEGANS Special Feature: Holometabolous chromosomes (multiple centromeres) pairing DNA meiosis replication Division I recombination meiosis Division 2 Gametes = products of meiosis

  25. recombination distance = map distance Two Factor Mapping: • When do you use it?: • - How far is the gene from a known marker? • To confirm or establish which chromosome the gene maps to. • - When you want to generate more recombinants for mapping. dpy-9(-) mut(+) unc-34(-) ; mut(-) unc-34(+) dpy-9(+) 25% Pick Dpys: % Unc: _______ 0% if perfectly linked, >0% if partially linked. Pick Uncs: % Mut: _____ 25% Pick Dpys: % Mut: _____

  26. Two Factor Mapping: How far is a gene from a marker? b c + + P0 (BC) X (WT) b c + + b c F1 + + Parental Gametes Recombinant Gametes b c + + b + + c a

  27. MAPPING: where is the gene? Linkage MAPPING: what is the gene close to or linked to? dpy-14 unc-31 dpy-9 +6 -27 +12 ced-2 -19 How often does recombination occur between dpy-9 & ced-2? 8% How often does recombination occur between dpy-9 & dpy-14? 39% (not accurate) Which mutations will show linkage to dpy-9? ced-2

  28. Three Factor Mapping: Where Is A Mutation Compared To An Interval? unc-31 dpy-14 F1 Heterozygote: mut unc-31 dpy-14 Recombinant Gamete: mut Unc non-Dpy non-Unc Dpy mut unc-31 dpy-14 F2 Recombinant /Parental: unc-31 dpy-14 unc-31 dpy-14 1/2 Unc 1/4 Unc Dpy 1/4 Unc Mut 3/4 Dpy 1/4 Unc Dpy F3: Conclusion: All Unc non-Dpy F2 segregate Mut F3, and no non-Unc Dpy F2 segregate Mut F3; thus, the mut gene is to the right of the unc-31…dpy-14 interval.

  29. Three Factor Mapping: Where Is A Mutation Compared To An Interval? unc-31 dpy-14 F1 Heterozygote: mut unc-31 dpy-14 Recombinant Gamete: mut Unc non-Dpy non-Unc Dpy mut unc-31 dpy-14 F2 Recombinant /Parental: unc-31 dpy-14 unc-31 dpy-14 1/4 Unc Dpy 3/4 Unc 1/2 Dpy 1/4 Unc Dpy 1/4 Dpy Mut F3: Conclusion: No Unc non-Dpy F2 segregate Mut F3, and all non-Unc Dpy F2 segregate Mut F3; thus, the mut gene is to the left of the unc-31…dpy-14 interval.

  30. Three Factor Mapping: Where Is A Mutation Compared To An Interval? unc-31 dpy-14 F1 Heterozygote: mut unc-31 unc-31 dpy-14 dpy-14 Recombinant Gamete: mut mut Unc non-Dpy Unc non-Dpy non-Unc Dpy non-Unc Dpy F2 Recombinant /Parental: unc-31 mut unc-31 dpy-14 mut dpy-14 unc-31 dpy-14 unc-31 dpy-14 unc-31 dpy-14 unc-31 dpy-14 1/4 Unc Mut 1/2 Unc 1/4 Unc Dpy 3/4 Unc 1/4 Unc Dpy 3/4 Dpy 1/4 Unc Dpy 1/4 Dpy Mut 1/2 Dpy 1/4 Unc Dpy F3: Conclusion: Some Unc non-Dpy F2 segregate Mut F3, and some non-Unc Dpy F2 segregate Mut F3; thus, the mut gene is within the unc-31…dpy-14 interval.

  31. For example, you mate your marker strain, bc/bc, to males that are homozygous mutant for your gene of interest, a/a. The F1 cross progeny are wild-type in phenotype, and genotypically bc/a. You collect 19 recombinants from the F2 generation by isolating 10 B nonC and 9 C nonB. Of the 10 B nonC, 9 picked up a. The ratio 9/10 is large so a is far away from b. Of the 9 C nonB, 2 picked up a. The ration 2/9 is small so a is close to c. b c a 9 1 B nonC 7 2 C nonB 16 3 T otal a

  32. How can you use this information to figure out a map position? b a c ? -15 -20 # recombinants (B nonC): # recombinants ( C nonB): # recombinants Total: Total as % of Interval: 9 7 16 16/(16+3)=84% 1 2 3 3/(16+3)=16% Size of Interval (in cM) = (-20) - (-15) = 5 cM Gene a is to the right of gene b, at a distance that is 84% of the b-to-c interval. This distance is (0.84)(5) = 4.2 cM to the right of gene b. Thus, position of gene a = (-20) + 4.2 = -15.8

  33. What if a is so close to c that you don’t get any recombinants? Q: How would you figure out if a is close or to the right?

  34. Mapping with SNPs (Single Nucleotide Polymorphisms) A T Bristol, “WT” N2 T C Hawaiian HA Shotgun cloned about 5 Mbp (11,000 clones) of Hawaii 6222 potential polymorphisms 4670 SNPs 1552 small insertions/deletions 1 polymorphism/873 bp aligned sequence- only 5 Mpb out of 97 Mbp were aligned. Q: Are there other SNPs in the genome?

  35. SNIP/SNPs A GAATTC Bristol, “WT” N2 T GAACTC Hawaiian HA EcoRI Restriction Enz. sequence GAATTC Of the 6222 Hawaiian polymorphisms, 3457 produce RFLPs (Restriction Fragment Length Polymorphisms) SNPs that also create RFLPs are called snip-SNPs. So far, about 500 confirmed snip-SNPs 1 confirmed snip-SNP/220 Kbp (0.6 cM on genetic map)

  36. N2 HA N2 HA SNP Genomic DNA PCR product PCR with SNP-specific primers from Bristol (N2) strain PCR with SNP-specific primers from Hawaiian strain RsaI PCR Products Digested PCR products

  37. Clone F59H5 at position 23175 Variation & Probability: VarS=GT Psnp=0.9430 CB4856 read: vc87b04.s1@270,g,62 Verified: Yes 5' primer = TGCTCTTCCTCCTGAACTTC 3' primer = CAGGATGTTTTCGTCTGGAC enzyme used = ApoI N2 digest RAATTY ApoI, AcsI, FsiI, XapI HA digest none -- -- TCCTCCGACACTGGTACTCCAACATCAGCGATCATAGCTTGTCCTTCTCT CGAACGTCTTTCGACGTGATCGATAACTTTTCAGCTTGTCCTTCTAATCC CGAAAGTCTTTCGACGTGGTCGATAACTTCTCCTGGTCGCtgctcttcct cctgaacttcTCCTCCTTGCTGTTGACGAAGATCCAAAATCCAACGGACT CTGAATGTCCCGAAGCCAACTCCATCGTCCGACGTTGCTGCTTCTCTGAT GTACGACGTCTTCCGATCTCCGCTGCTGGTGATCCTCAGAGCGCACCGTC CTGATCCATCCCTTCGTGCGTACAGCACGACACTCCGGGGGAAAAAAGGC TCGAAATTCAAAGGAATTTTGAGTGGACCAAATTGTACAGAACGTTGCAA ATGTGCTCGGATCCACGACGACATCCGTATTTCTTGTCATCTTCGACTGT TCCTCTTCGTCTTCTGGAGTCATAACTTGTGCACGTTGAATCGGAAAATT [T/G]TTGTTCTTTTTACCGGAATTTGAACTTCTCGCAATGTGTAGACTT CTGCATTCGATGCCAATTCCAACGAAATTAACTCCATCTTCGGAATGGTA GATTCTGCTCCACTATTTGTCGGCTTTGTCCTCGCTTTTCGACATCACCA GTGGCATCCCGATAAAGCCACCCGGATATGAAAAATCCTCTTGATTCTTG TACAATTGAAGAATTTCTCCTTGTGACTCCGCCAACATCGTCACATTCAT CATGTAGAGATTCGTCTTGATTTGCTCCTTGATTTGATGCGGAAAATGTC GATTCCACGAAACGACTTCTATCCATTGAAAGCTGCTGTCTGATTGACAA ATCGTCCAGCACAATCACATCTTCAAGCCAAGGCGGGTGATAAATCCATC GATTTTTCGATATCATCCTTTGCCTGCCGAACATCCAACTTCCAACTACA TCTTTGAATTCCACGAAACGACACGACGAAATACTTTCCATCCAGCAAAA TTCCA

  38. Combining Three Factor Mapping With SNPs X unc-31 mut dpy-14 (from N2 background) P0 Hawaiian (HA) unc-31 mut dpy-14 HA F1 F2 Unc non-Mut non-Dpy Unc Mut non-Dpy non-Unc non-Mut Dpy non-Unc Mut Dpy 1 2 3 4 5 6 recombinant unc-31 unc-31 mut dpy-14 unc-31 mut unc-31 mut dpy-14 dpy-14 unc-31 mut dpy-14 mut dpy-14 unc-31 mut dpy-14 parental 1 2 3 6 5 4 unc-31 mut dpy-14 unc-31 mut dpy-14 SNP SNP

  39. Combining Three Factor Mapping With SNPs 1 2 3 6 5 4 N2 unc-31 mut dpy-14 unc-31 mut dpy-14 SNP SNP HA F2 Rec. Genotype recombinant (N2 & HA) unc-31 HA unc-31 mut N2 dpy-14 1 parental (N2) unc-31 mut HA unc-31 mut N2 dpy-14 2 unc-31 mut N2 unc-31 mut N2 dpy-14 3 PCR/Digestion of F2 Worms HA dpy-14 unc-31 mut N2 dpy-14 1 2 3 4 5 6 4 N2 HA N2 dpy-14 unc-31 mut N2 dpy-14 5 mut N2 dpy-14 unc-31 mut N2 dpy-14 6

  40. Using SNPs To Map A Mutation • Steps • Make a table that represents the recombinant chromosome for each of your recombinants. • Label each marker and SNP according to whether it came from the N2 or the HA chromosome. • Label whether the recombinant chromosome contains the mutation. F2 Rec. Genotype recombinant (N2 & HA) unc-31 HA unc-31 mut N2 dpy-14 1 parental (N2) unc-31 mut HA unc-31 mut N2 dpy-14 2 unc-31 mut N2 unc-31 mut N2 dpy-14 3 HA dpy-14 unc-31 mut N2 dpy-14 4 N2 dpy-14 unc-31 mut N2 dpy-14 5 mut N2 dpy-14 unc-31 mut N2 dpy-14 6

  41. Using SNPs To Map A Mutation Steps 2. Mark the crossover on the table.

  42. Using SNPs To Map A Mutation • Steps • Figure out on which side of the crossover the mutation lies: • If the recombinant chromosome contains the mutation (i.e., the recombinant animal has the mutant phenotype), then the mutation is on the N2 side of the recombination site. • If the recombinant chromosome does not contain the mutation (I.e, the recombinant animal does not have the mutant phenotype), then the mutation is on the HA side of the recombination site. The arrows point to where the mutation is relative to the recombination site

  43. Using SNPs To Map A Mutation • Steps • There will be at least two critical recombinants: one that defines the left hand border of where the mutation lies, and one that defines the right hand border of where the mutation lies. • For the left hand border, find the recombinant that has a rightward arrow that is the furthest to the right. • For the right hand border, find the recombinant that has a leftward arrow that is the furthest to the left. For the left hand border, recombinants 1 and 6 are the same: between unc-31 and the SNP. Thus, the left hand border is within this interval. For the right hand border, recombinants 2 and 5 are the same, and more to the left than the arrows in recombinants 3 and 4. The site for recombinant 2 and 5 falls between unc-31 and the SNP. Thus, the right hand border is within this interval. Conclusion: the mutation falls somewhere between unc-31 and the SNP.

  44. Which mapping method do you use? Some considerations… Deletion mapping: Also tells you if your mutation is recessive, etc. Falling out of fashion with the rise of powerful SNP mapping. Mapping with genetic markers: Generates marked strains that can be used for other types of mapping. No PCR problems. Mapping with molecular markers (SNPs): In theory, the single fastest method. Makes it possible to follow dominant alleles.

  45. Practice Problem, part a: You are mapping a new ced (cell death defective) mutation.

  46. Practice Problem, part b,c: b. You decide to use SNP mapping to narrow down where ced* is. You generate a dpy-24 ced* unc-75 triple mutant, and cross it to the Hawaiian strain. You collect Dpy nonUnc recombinants that are also Ced in phenotype. R e c o m b i n a n t s c. You tak e 5 of your Dpy D p y n o n U n c C e d nonUnc recombin a nts that N 2 H A 1 2 3 4 5 have ced* and make DNA from the Ced animals. Then you perf o rm PCR with S N P - S N P 1 specific primers, digesting the DNA with the correspond i ng enzyme. R e c o m b i n a n t s D p y n o n U n c C e d H A 1 2 3 4 5 N 2 S N P 2 W here is ced* relative to the two SNPs? Which recombina n ts show this? What would be your next mapping exper i ment?

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