Exam 2 Notes
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Date Created: 10/25/15
BIO 198 Exam 2 Study Guide Chapter 5 Notes 51 Syntenic genes genes located on the same chromosome 0 when two are too close to each other cannot assort independently gt linked genesquot genetic linkage produce distinct pattern of gamete genotypes gt can be used to map locations of genes on chromosomes 0 can still assort independently if they are far apart enough on a chromosome recombinant chromosomes produced from the reshuffling of syntenic genes through crossing over bw homologous chromosomes parental chromosomesnonrecombinant chromosomes ones that do not undergo crossing over to reshuffle alleles 0 genetic linkage mapping plots positions of genes on chromosomes 0 connection that causes alleles of linked genes to segregate together during meiosis can be broken by crossing over Observations and conclusions about genetic linkage o Linked genes are ALWAYS syntenic and are ALWAYS located near one another on a chromosome lf far apart enough and can independently assort they are NOT linked 0 Gene linkage gt produces greater number of gametes with parental chromosomes than would be expected and smaller number of gametes with recombinant chromosomes than expected o Crossing over gt less likely to occur bw linked genes that are close to each other than bw genes that are farther apart frequency of cross over distance bw genes quot proportional to 0 Complete genetic linkage gt no recombination at all occurs between linked genes 0 Incomplete genetic linkage gt more common for linked genes 0 produces a mixture of parental and non parental gametes 0 generally same frequency bw parentals and same frequency bw recombinants Recombination frequency r the rate of recombination for a given pair of linked genes r number of recombinants total number of progeny o varies bw different pairs of syntenic genes 0 depends roughly on distance separating the genes on chromosome Discovery of genetic linkage 0 William Bateson and Reginald Punnett Studied the traits of flower color and the shape of pollen grains in sweet peas First as independent traits in different plants follow the law of independent assortment Next gt together in the same plant F2 produced more parental and less recombination than expected 0 Morgan confirmed this coupling vs repulsion effect that alleles exhibited crossed female pure breeding white eyesminiature wings with hemizygous WT males red eyefull wings F1 gt expected ratio in progeny produced F2 gt unexpected ratio in progeny produced 0 Two point test cross analysis 0 dihybrid F1 fly X pure breeding mate with recessive phenotypes recessive will only give recessive but dihybrid fly can contribute recessive OR dominant allele 0 used to fully interpret the linkage of autosomal genes 0 Results of Morgan s experimental observations and explanations of genetic linkage 0 genetic linkage gt physical relationship bw genes located near each other on a chromosome 0 recombination bw linked genes on homologous chromosomes occur in less than 50 of meiotic divisions and parental combinations occur in more than 50 of gametes produced from meiotic divisions o recombo frequency distance bw genes on a chromosome 52 0 Examine construction of genetic maps from the recombination data for two linked genes 0 Smaller recombination frequencies 2 genes residing closer to each other on the chromosome 0 Larger recombination frequencies 2 greater distances between genes on the chromosome 0 First genetic map made by Alfred Sturtevant 0 Map Units Table 52 Sturtevant s Recombination Data for Five X Linked Genes in Drosophila Gene Pairs Recombination Frequency Yellow y and white w 21421736 0010 Yellow y and vermilion v 14644551 0322 Vermilion v and white w 4711584 0297 Vermilion v and miniature m 17573 0030 Miniature m and white w 20626116 0337 White w and rudimentary r 406898 0452 Rudimentary r and vermilion v 109405 0269 yw vm r Centromere Sturtevant39s map 0010 307 337 576 yw vm r Centromere Contemporary map 0015 330 361 545 677 smallest recombination frequency closest in proximity In this case gene producing white eye w and gene carrying yellow y body 0 recombination frequencies bw genes on a chromosome gt units of physical distance gt map unit mulcentiMorgan CM 0 1 recombination 1 mu or 1 GM of distance between linked genes 0 Chisquare analysis 2 observed ex eeted X2 2 p expected 0 Used to tell whether the observed data shows evidence of genetic linkage rather than just a chance variation from expected values 0 The larger the chisquare number the greater the deviation is from the expected results 0 For a trihybrid cross expected genotypic ratio is 279993331 and the expected phenotypic ratio is 11 1 1 1 1 1 1 53 0 Three Point Test Cross Analysis used to effectively map three linked genes simultaneously 0 Constructing a ThreePoint Recombination Map 0 Started with Rollins Emerson on his 1935 study on maize Tested three genes green seedling 0 vs yellow seedling vv rough leaf Gl vs glossy leaf gl gl gene fro normal fertility Va and variable fertility va va crossed 0 GI Va V GI Va x v gl va v gl va gt produced heterozygous dominant F1 F1 crossed with pure breeding recessive and the results are as follows Parental cross VGI VaVGI Va gtlt vgl vavgl va Green rough normal yellow glossy variable Test cross V GI Vavgl va gtlt v Q vav gl va Green rough normal yellow glossy variable Testcross progeny Number Number Genotype Phenotype Observed Expected Q gamete0quot gamete lYeow rough normal 60 9075 vGI Vav gl va 2Yellow glossy normal 48 9075 vgl Vav gl va 3Yeow rough variable 4 9075 vGI vav gl va 4 Yellow glossy variable 270 9075 vgl vavgl va 5 Green rough normal 235 9075 VG Vav g va 6 Green glossy normal 7 9075 Vgl Vavglva 7 Green roughvariabe 40 9075 VGvavgl va 8 Green glossy variable 62 9075 Vglvavglva 726 726 o In a trihybrid cross two parental chromosomes and six recombinant chromosomes 0 Of the recombinant four gt single crossovers two gt double crossovers these will be the least frequent because in order for double recombinants to be produces both crossovers must occur at the same time 0 Two ways to determine gene order 0 list each gene order possible for the parental chromosomes draw corresponding double crossover chromosomes and determine whether double crossover gametes produced by this match the predicted double crossover progeny 0 compare parental and doublecrossover chromosomes gt alleles of outside genes appear to remain the same while the middle allele switches when compared we will notice that two alleles match and one does not Example parental gt abclabc vs recombinant abclabc look at the doublecrossover chromosomes and find the allele that seems to have been switched This will help you determine the parental chromosome 0 Recombination Frequency of the Gene pairs 0 recombination frequency sum of all single and double recombinants for this gene pair total number of progeny 0 Reduction in the observed number of double crossovers is caused by the effect interference I gt limits the number of crossovers that can occur in a short length of chromosome quantified by comparing the number or freq of observed doublecrossover events to number of frequency expected from independent assortment Ex in the data above double crossovers 11726 0015 15 if independently assorted then this would be the product of two single crossover frequencies 01830136 0025 25 The expected number of doublecrossover progeny would therefore be 0025726 182 It is also I 1c c coefficient of coincidence o Coefficient of coincidence c observed double recombinantsexpected double recombinants uhxcrx 2d double rcwmbintmts expected double l39CL39UlIlblllLlIltN l l182 2 using numbers 2 00150025 usingfrequencies 54 0 Three patterns of double crossover between two genes 0 twostrand double crossover gt no recombinants o threestrand double crossover gt involves 3 sister chromatids and can happen in two ways produce the same outcome two parental and 2 recombinants o fourstrand double crossover gt all four chromatids gt four chromosomes are recombinant a Twostrand double crossover three equivalent ways one position held constant Gametes Recombinants A B A B E w A B a b O a b A B 4 a a b a b by any two strand dou ble crossover b Three strand double crossover one position held constant Gametes Recombinants A B A B E A B A b 2 a b a B 4 a a b a b A B A B E A B A b 2 a b a b 4 a b a B l Half ofall gametes are recombinant c Four strand double crossover one position held constant Gametes Recombinants A B A b A B A b zi a b a B 4 a b a B All gametes are recombinant l 16 Recombination limit is 50 o Haplotype haploid genotype the specific array of alleles in a set of linked genes on a single chromsomes o alleles belong to linked genes so they tend to be passed together during meiosis 56 Chapter 6 Notes 61 o Bacteria propagate through binary fission gt bacterial chromosome replicates and a copy is distributed to each of the progeny cells through cell division each cell produced is genetically identical 0 Reasons why geneticists use bacteria 0 Genomic simplicity Most bacterial genomes contain fewer genes and fewer base pairs in their haploid genomes than do other organisms o Uncomplicated genotypes The haploid genomes of most bacteria allow all mutations to be observed directly without interference from dominance interactions between alleles 0 Short generation times Bacteria reproduce rapidly their generation times can be measured in minutes 0 Large numbers of progeny Enormous numbers of clonal progeny can be examined increasing the likelihood that statistically rare events will be observed 0 Ease of propagation Microbes may be grown either in liquid culture or on culture plates The cultures are easy and inexpensive to maintain and they require little laboratory space 0 Numerous heritable differences Mutants are easily created identified isolated and manipulated for examination 0 Transferring of genetic material occur in three ways conjugation donor bact to recipient bact transformation gt uptake of DNA from environment transduction gt use of viral vector to transfer from donor bact to recipient bact o Transferred DNA either extrachromosomal plasmid or a portion of the donor bacterial chromosome 0 homologous parts from donor and recipient can undergo recombination gt change in genotype of recipient cell 0 Each AA are examples of lateral gene transfer 0 Plasmids we will talk about 0 Fertility F plasmids genes that promote its own transfer from a donor bacterium to a recipient 0 Resistance R plasmids carries antibiotic genes that can be transferred from donors to recipients o Allelic Identification 0 complete medium with everything that is needed for the bacteria to grow gt minimal medium bacteria needs to produce what it needs for metabolism DNA replication etc gt 1 Minimal alanine 2 Minimal proline 3 Minimal alanine and proline THIS IS AN EXAMPLE 0 prototrophs gt wild types gt can synthesize all the products required for growth on a minimal medium 0 auxotrophs gt can grow in supplemented minimal medium medium with the compounds that an auxotroph cannot produce on its own 0 Lederberg and Tatum s experiment 0 mixed two auxotrophs that had genes that could produce what each other was missing gt grew them in complete medium gt formed prototrophic bacteria that could grow on minimal medium 0 showed conjugation between the two bacteria types 0 genetic information conveyed between donor and recipient bacterial cell through a hollow tube known as conjugation pilusconjugation tube 0 Transfer of the F Factor 0 William Hayes proposed that the ability to act as a donor was hereditary and determined by a fertility factor F Factor and was transferable from donors to recipients Donors gt FF cells Recipients gt F F cells 0 Conjugation can only occur between a donor cell and a recipient Cannot occur between two donors 0 exconjugant cell gt produced by conjugation it is a recipient cell that has had its genetic content modified by receiving DNA from the donor cell 0 After contact is est by the pilus gene expression from the F factor leads to a protein complex called the relaxosome gt this binds to oriT origin of transfer gt relaxosome allows for the cleavage of one phosphodiester bond on DNA strand l39 strand to show that this is the transferred strand to the recipient cell c episome a bacterial plasmid that is able to both replicate autonomously and integrate into the host genome o Luigi Luca Cavalli Sforza discovered highfrequency recombination Hfr strains of bacteria gt donor strains that transferred donor bacterial genes to recipients at extraordinarily high rates 0 IS element gt allows the F factor plasmid to be inserted into the bacterial chromosome making Hfr chromosome 0 This process is called the rolling circle replication process Bacterial chromosome F factor ce F H orIT element Recombination of bacterial chromosome and F factor at an IS element Hfr cell o Hfr x F conjugation Mix in coniugation culture l Hfr donor thr leu strS F recipient thr leu strquot 39 39 39 R plasmid Bacterial chromosome ONT X str Conjugation and partiali strand transfer due to interrupted mating Crossover sites F factor seg me nt s Homologous R j recombination 5quot Donor Chromosomal fragment Enzymatic cf degradatioN J str One kind of exconjugant cell thr leu str str Minimal medium plus streptomycin l Gui g Gav W C v 9 9i9 49 6 9 I Only 1hrquot lau sir exconjugants grow o F F prime donor gt contains a functional but altered F factor derived from imperfect excision of the F factor out of the Hfr chromosome a Hfr chromosome s x orrT l2 Bacterial A F factor chromosome loci I I l Normal excusmn Aberrant excusmn l vl g A segment of the bacterial l ilk DNA loops out 1 during excision l FormationJVof F factor Formationiof F39factor lg lac T 1T3 l v o r l r f u r A l lac llQ on A worry I 1 QVLI 1 V c were Bacterial F plasmid Bacterial F39 plasmid chromosome chromosome The F factor contains the donor lac in addition to a full set of F factor genes b F cell F39 cell 6 lac r H i r V M Jr rf orrT x I x Z Bacterial F factor Bacterial chromosome chromosome Grows on a lactose medium l Unable to grow on a lactose mediu m Frcequot Conjuigation Ircequot lac I39ll in r kLA 3 ll ll l ll l l V A H I 42 V V Transfer com lete F39Cequot i p F39exconjugant Allquot if lac 4 33 lac W K I 1 R it f I I l l l l V ll l r I l l l quvx 1le 4 The exconjugant is a lachac partial diploid and has acquired the ability to grow on a lactose medium Because F plasmid transfer was complete the exconjugant can act as an F donor 0 partial diploid gt an exconjugant bacterium that acquires a second copy of one or more genes by conjugation with an F donor cell 64 Bacterial Transformation Produces Genetic Recombination 0 Transformation gt recipient cell takes up a fragment of donor cell DNA from the surrounding growth medium 0 DNA fragment passes through wall and membrane of recipient and gets incorporated into the recipient cell chromosome by homologous recombination 0 Four step process lysis breakage of a donor cell and the release of fragmented DNA from the donor chromosome transforming DNA is double stranded Recipient Doublestranded chromosome donor DNA Ech al ReceptorgO site 0 Donor DNA binds at the receptor site One strand 0 is degraded as it enters the recipient cell at Transforming strand Degraded DNA binding V nucleotides co m p ex at rece ptor Recipient Cytoplasmic cell wall membrane a The transforming strand pairs with the homologous region ofthe recipient chromosome Transforming strand Heteroduplex l DNA 9 The transforming strand displaces a recipient strand forming complementary heteroduplex DNA a aquot The excess strand degrades DNA replication and cell division Nontransformant Transformant 0 DNA replication and cell division produce one transformant and one nontransformant 0 After the subsequent DNA replication and cell division cycle one daughter cell is a transformed ce Cl39ransformant and one daughter cell retains the recipient s chromosome and is not genetically altered 65 o cotransformation simultaneous transformation of two or more genes carried on a donor DNA fragment into a recipient o for it to occur crossover events must incorporate closely linked genes on a single fragment of transforming DNA T4 phage Base plate fibers A phage Head DNA t o Sheath xx x s 2 xvi be 7 5L 0 Lytic cycle causes the lysis of the host cell OOOOOO Lytic cycle Infection gt Lysogenic cycle e A phage injects DNA a through hollow tail Lysogenic cycle a ProgenyA phages are released by lysis from host bacteria I 7777gt Phage l Phage chromosome circularizes to protect it from degradation t 39 A prophage DNA 9 Integration of Q N L into the host 1 chromosome 6 DNA and proteins are I I assemblF d into Progeny A Multiple divisionsand quot P h phages Lytic cycle many generations may mp age I occur in this state A V K 7 rophage DNA is copied N S Q a Replication of phage p g q chromosome occurs host When cell deeS39 7 DNA breaks down 0 Underthe direction of phage A ea Q 9 The Iytic cycle ex genes transcription and 7 3 3 resumes o Excision of translation produce new 39 g lt prophage from the h rt V p age pa IC es 39i host chromosome I attachment of phage particle to the host cell injection of the phage chromosome into the host cell replication of phage DNA Transcription and translation of phage genes Packaging of phage chromosomes into phage heads lysis of the host cell c Temperate phages phages capable of a temporary alternative life cycle that leads to the integration of the phage chromosome into the bacterial host chromosome gt lysogeny O O O O O 0 ex lambda phages attachment of the phage particle to the host cell injection 0 phage chromosome into the host cell integration of the phage chromosome into the host chromosome phage DNA gt prophage excision of the prophage resumption of the Iytic cycle 0 Generalized transduction phages formed when a random piece of donor bacterial DNA of the appropriate length is mistakenly packed into the phage head instead of similarly sized length of phage DNA 0 occurs because packaging mechanism discriminated DNA by length rather than sequence 0 transductant bacterium that has acquired one or more donor genes through transduction o cotransduction simultaneous transduction of two or more genes o cotransduction frequency depends on how close two genes are to one another on the donor chromosome 0 Finding cotransductants in an experiment 0 identify cells transducer with one donor allele and then screen those transductants for the acquisition of additional donor alleles selected marker screen identify transductants for one of the donor alleles of interest unselected marker screen happens to transductants that have gone through the selected marker screen test and are to be screened for a second donor allele goal is to determine the percentage of transductants for the selected marker that are also transducer for the unselected marker while reducing unnecessary colony genotyping o Specialized transduction phages Transduction from a donor cell to a recipient cell of a few selected genes located near the site of bacteriophage integration A chromosome 0 a tt P gaIK 0mg bioA Bacterial chromosome lnteg rationl l Normal induction galK Prophage bioA Aberrant induction I l l bioA QLIK bI39Ol l l N A Specialized lt 3 Adb 39 M GP phaSQO transducmg phage Phgge gaK A DNA A DNA bioA Bacterial chromosome 0 Lateral gene transfer horizontal gene transfer gt transfer of genetic material bw individual bacteria or archaea and other organisms Chapter 7 71 o antigenic variation 2 when a bacterium or virus changes its surface proteins in order to evade the host immune response 0 Frederick Griffith s experiment 0 identify a transformation factor responsible for heredity o Sstrain bacteria killed the mice but that heatkilled S strain bacteria did not kill o R strain did not produce illness and kill but heat killed SStrain Rstrain induced illness and killed the mice 0 proof for transformation 0 Hershey and Chase s experiment 0 reconfirmed that DNA was the material responsible for transformation 0 they grew phages in 2 different mediums 1 was radioactive 835 gt to label protein the other was radioactive P32 gt to label DNA 0 Used them to infect unlabeled host bacterial cells in parallel experiments 72 0 DNA nucleotides have 3 parts a deoxyribose sugar one of four nitrogenous bases and three phosphate groups 0 purines A and G amp pyrimidines C and T 0 base stacking the tight packing of DNA bases in the duplex leads to the offsetting of adjacent base pairs so that their planes are parallel and imparts a twist to the double helix 0 major groove vs minor groove a Ribbon diagram b Spacefilling diagram c Ball and stickdiagram St d1 co plementaw Complementary Complementary ran ase palr Strand 1 base pair Strand 1 base Pair St d 2 l d ran 3 Strgn 2 3 Str alndZ 3r I 5 u G C Phosphate Axisof T 1 RKEPhosphate AXis of helical grOUps helical quot V A groups symmetry SymmEtry Sugar Cm rings xi rings 39 Minor Minor groove groove V J One helical I 34393 quot3394 A turn 2 34A 2 34 nm 34 A 105 base pairs Major Major groove groove 20A2 nm 20A 0 Three forms of DNA 0 most common and stable gt Bform DNA gt righthand twisting of the sugar phosphate backbone o Aform DNA gt more compact than Bform o Z form DNA gt left handed twist that gives the sugar phosphate a zig zag appearance tends to happen in the presence of a high concentration of positively charged ions 73 0 Three models of replication 0 semiconservative DNA replication each daughter duplex contains one original strand and one complementary strand 0 conservative DNA replication one daughter duplex contains both strands of parental molecule and the other contains two newly synthesized daughter strands o dispersive DNA replication each daughter duplex is a composite of interspersed parental duplex segments and daughter duplex segments 0 Meselson Stahl Experiment 0 suggested the three modes of replication of DNA 0 grew Ecoli in medium containing N15 all DNA duplexes contained only the heavy nitrogen isotope because they were placed under these growth conditions 0 A DNA duplex with normal isotope of Nitrogen N14 0 Watched the DNA replication cycle happen and saw that the daughter strands had N15N14 meaning that it was semiconservative Generation 0 Cycle 1 Cycle 2 Cycle 3 E coli cultures Transfer cells to 14N growth medium E 15N growth l 14Ngrowth It 14N growth l 14N growth medium medium medium medium 9 DNA samples A 95 9Q 9 mm z c x W f W immom Heavy Light Light Hybrid Hybrid moooo DNA analysis Hybrid Densitometric DNA bands bands Light 14N14N 15N14N l 15N15N Heavy Results All heavy DNA L lzi light to hybrid DNA 31 light to hybrid DNA 0 Origin and Directionality of Replication in Bacterial DNA 0 bidirectional gt happens in both directions 0 Only one origin of replication gt John Cairns was the first to report evidence of a single origin of replication in Ecoli forms a replication bubble Replication forks Replication bubbm DNA Replication oHelicase breaks hydrogen bondsTopoisomerase 9 DNA polymerase III elongates the leading strand relaxes supercoiling continuously and the lagging strand discontinuously 3 Hencase Leading strand Topoisomerase IN 5 Origin of replication Okazak39 fragment 1 fragment 2 95ingle stranded binding 558 protein prevents reannealing 553 6 DNA polymerase I removes and replaces nucleotides ofthe RNA primer Helicase Topoisomerase e DnaG synthesizes RNA primers RNA primer 0 DNA polymerase III synthesizes daughter strand Leading strand DNA polymerase III pol lll L TW Dkazaki Okazaki fragment 1 fragment 2 0 DNA ligase joins Okazaki fragments Okazaki Okazaki kazaki fragment 1 fragment 3 fragment 2 3 5 W 1M 3 Lagging strand ll l llll 5 539 l Okazaki fragment DNA Helicase Protein topoisomerase DnaB SSB Primase DNA pol l DNA poll DNA ligase 0 a a m Role Relaxes Unwinds the Prevents Synthesizes Synthesizes Removes Joins DNA supercoiling double helix reannealing RNA primers DNA and replaces segments of separated RNA primer strands with DNA 0 In Ecoli three replicationinitiating enzymes DnaA DnaB and DnaC that need to locate and bind to the consensus sequences in oriC TATA box 0 DnaA bind to 9mer components of oriC gt bends DNA and breaks hydrolyzes Hbonds in the 13 mer component gt creates a replication bubble gt DnaB helicase carried by DnaC attaches to both strands in the open complex and continues to unwind the DNA by breaking more Hbonds gt singlestranded binding proteins 888 prevents the reannealing of the separated strands o In order to begin DNA strand synthesis DNA polymerase requires a primer sequence provides an OH end to which a new nucleotide can be added by DNA polymerase Primase begins DNA replication by synthesizing in short RNA primers 0 To avoid random breakage in the molecule topoisomerases catalyze a controlled cleavage and rejoining of DNA to allow overwound DNA strands to unwind gt relief from this supercoiling is accomplished by topoisomerase cutting either one or both strands of DNA allowing DNA to unwind and reseal the strands 0 DNA pol Ill synthesizes new daughter strand by bringing in new nucleotides to a daughter strand aSEJaUJAIOd 310 m Od weon 1 dwep uuous Japeol dump aSEJawqod 9103 III IOd aauanbas snsuasuog Vi VVV CJi ViVVV Vi UV 111 BV iViii iV S 393LJVLLLLVIV9V V VEJViCLLLJV VVVVBBOVO V V V VVV VVVVBiVViVVO31VLLL3LLV9OViiiLLOVOOVVViim E E OVVlVVVVlGlDlVVVV V VVV DlVSVVVV V VV VVVV lllOOBlBl V VV VllliOVilVLLBBVlVVVBVVLOBlVVVVVOiBOiiiVVw 5 quHJ Lia L18 L984 dq 96 l LSHV aauanbas Bugleaudaj snowouolneabjsmaja 39s q Aeue made wapuel Jaw6 Rene medal LuapueJ JawEL mmavwv aauanbas vvvwvwvsvm aluanbas VOVDDLVLL Jaw6 iiiViiiViOiVE Jaw2L I JBLU 6 l LJauJs l LJSUJ 6 l LJaw5 J I JauJg L I JauJg 1 I Jawg L J 1 h 39 dq svz 39 BLUOSOUJOJHD 102 393 31101103 393 e 75 DNA pol III has tau protein clamp loader and sliding clamp has editing properties along with DNA pol l gt 3 to 5 exonuclease activity errors gt once every billion nucleotides telomeres gt ends for lagging strand at the end of replication gt synthesized by telomerase a ddCTP FeaCtiO quot0 lane Incorporation of dCTP allows the chain to continue growing but incorporation of ddCTP terminates chain elongation TTACCCATCES 3 7 5 Length of synthesized fragment 23 18 mer AATG 25 5 28 5 31 E MAATGCGCTGCAT 36 S MAATGCGCTGCATCGTAG Partial replication products terminate at each cytosine of the chain due to the incorporation of ddCTP b ddGTP reaction quot6 lane Length of Partial synthesized replication fragment products 22 5 24 539 27 18 mer G 32 s39mAATGCGCTGCATc 35 S MAATGCGCTGCATCGTA c ddTTP reaction T lane Length of Partial synthesized replication fragment products 21 539 26 18 mer T 30 S MAATGCGCTGCA 33 S MAATGCGCTGCATCG 38 S MAATGCGCTGCATCGTAGC d ddATP reaction Aquot lane Length of Partial synthesized replication fragment products 19 539 20 539 29 S MAATGCGCTGC 34 S MAATGCGCTGCATCGT 38 S MAATGCGCTGCATCGTAGCT
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