1. Chapter 07 *Lecture Outline Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. *See separate FlexArt PowerPoint slides for all figures and tables pre-inserted into PowerPoint without notes.

2. INTRODUCTION Bacteria and viruses account for a quarter to a third of human deaths worldwide. Their impact on health is a major reason for studying them. Like eukaryotes, bacteria often possess allelic differences that affect their cellular traits –However, these allelic differences (such as different sensitivity to antibiotics) are between different strains of bacteria because Bacteria are usually haploid –This fact makes it easier to identify loss-of-function mutations in bacteria than in eukaryotes These usually recessive mutations are not masked by dominant genes in haploid species 7-2 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

3. INTRODUCTION Bacteria reproduce asexually –Therefore crosses are not used in the genetic analysis of bacterial species Rather, researchers rely on a similar phenomenon called genetic transfer –In this process, a segment of bacterial DNA is transferred from one bacterium to another 7-3 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

4. 7.1 GENETIC TRANSFER AND MAPPING IN BACTERIA Like sexual reproduction in eukaryotes, genetic transfer in bacteria enhances genetic diversity Transfer of genetic material from one bacterium to another can occur in three ways: –Conjugation Involves direct physical contact –Transduction Involves viruses –Transformation Involves uptake from the environment –See Table 7.1 7-4 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

5. 7-5

6. Conjugation Genetic transfer in bacteria was discovered in 1946 by Joshua Lederberg and Edward Tatum 7-6 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display They were studying strains of Escherichia coli that had different nutritional growth requirements Auxotrophs cannot synthesize a needed nutrient Prototrophs make all their nutrients from basic components One strain was designated met – bio – thr + leu + thi + It required one vitamin (biotin) and one amino acid (methionine) It could produce the amino acids phenylalanine, leucine and threonine Another strain was designated met + bio + thr – leu – thi – Had the opposite requirements for growth It required one vitamin (thiamine) and two amino acids, leucine and threonine Their experiment is described in Figure 7.1

7. Figure 7.1 7-7 10 8 cells No coloniesBacterial coloniesNo colonies Nutrient agar plates lacking amino acids, biotin, and thiamine met – bio – thr + leu + thi + met + bio + thr – leu – thi – Mixed together met – bio – thr + leu + thi + and met + bio + thr – leu – thi – Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

8. 7-8 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display The genotype of the bacterial cells that grew on the plates has to be met + bio + thr + leu + thi + Lederberg and Tatum reasoned that some genetic material was transferred between the two strains Either the met – bio – thr + leu + thi + strain got the ability to synthesize biotin and methionine (bio + met + ) Or the met + bio + thr – leu – thi – strain got the ability to synthesize threonine and leucine and thiamine (thr + leu + thi + ) The results of this experiment cannot distinguish between these two possibilities

9. 7-9 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Bernard Davis later showed that the bacterial strains must make physical contact for transfer of genetic material to occur He used an apparatus known as U-tube It contains a filter at the bottom which has pores that were Large enough to allow the passage of the genetic material But small enough to prevent the passage of bacterial cells Davis placed the two strains in question on opposite sides of the filter Application of pressure or suction promoted the movement of liquid through the filter Refer to Figure 7.2

10. 7-10 Figure 7.2 Nutrient agar plates lacking biotin, thiamine and amino acids No colonies Thus, without physical contact, the two bacterial strains did not transfer genetic material to one another StopperCotton ball met – bio – thr + leu + thi + met + bio + thr – leu – thi – Pressure/suction Filter Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

11. 7-11 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display The term conjugation now refers to the transfer of DNA from one bacterium to another following direct cell-to-cell contact Many, but not all, species of bacteria can conjugate Moreover, only certain strains of a bacterium can act as donor cells Those strains contain a small circular piece of DNA termed the F factor (for Fertility factor) Strains containing the F factor are designated F + Those lacking it are F – Plasmid is the general term used to describe extra- chromosomal DNA

12. 7-12 Plasmids, such as F factors, which are transmitted via conjugation are termed conjugative plasmids These plasmids carry genes required for conjugation Figure 7.3 These genes play a role in the transfer of DNA They are thus designated tra and trb followed by a capital letter F factor oriT Region of F factor that encodes genes that are necessary for conjugation Pilin protein Proteins that are components of the exporter Coupling protein Relaxase oriT traM traJ traYtraA traL traEtraKtraBtraP traRtraC traW traUtraN traF traQ traH traG traS traT traD tral traX traD trbG trbl trbC trbEtrbAtrbB trbJ trbF trbH traV Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

13. 7-13 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display The first step in conjugation is the contact between donor and recipient cells This is mediated by sex pili (or F pili) which are made only by F + strains These pili act as attachment sites for the F – bacteria Once contact is made, the pili shorten Donor and recipient cells are drawn closer together A conjugation bridge is formed between the two cells The successful contact stimulates the donor cell to begin the transfer process Refer to Figure 7.4a for the molecular details of conjugation

14. 7-14 Figure 7.4 F factor Bacterial chromosome Coupling factor Exporter Inner membrane Conjugation bridge Relaxosome Relaxase Origin of transfer Donor cell (F + ) Recipient cell (F – ) Relaxosome makes a cut at the origin of transfer and begins to separate the DNA strands. Most proteins of the relaxosome are released. The DNA/relaxase complex is recognized by the coupling factor and transferred to the exporter. Coupling factor Exporter Outer membrane T DNA Transferred DNA Protein complex encoded by the F factor Accessory proteins of the relaxosome are released One protein, relaxase, remains bound to the end of the T-DNA Together, these form the conjugation bridge a complex of 10-15 proteins encoded by the F factor that span both inner and outer membranes Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

15. F + cell The exporter pumps the DNA/relaxase complex into the recipient cell. In the donor cell, the F-factor DNA is replicated to become double-stranded. In the recipient cell, relaxase joins the ends of the single- stranded DNA. It is then replicated to become double-stranded. (a) Transfer of an F factor via conjugation 7-15 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Figure 7.4 The F - strain is now F + The result of conjugation is that the recipient cell has acquired an F factor Thus, it is converted from an F – to an F + cell The F + cell remains unchanged

16. Hfr Strains In the 1950s, Luca Cavalli-Sforza discovered a strain of E. coli that was very efficient at transferring chromosomal genes –He designated this strain as Hfr (for High frequency of recombination) Hfr strains are derived from F + strains 7-16 Figure 7.5a An episome is a segment of DNA that can exist as a plasmid and can integrate into the chromosome Bacterial chromosome Origin of transfer Origin of transfer (a) When an F factor integrates into the chromosome, it creates an Hfr cell. F factor Integration of F factor into chromosome by recombination Hfr cellF + cell lac + pro + lac + pro + Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

17. 7-17 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display In some cases, the integrated F factor is excised in an imprecise fashion may carry genes that were once found on the bacterial chromosome These types of F factors are called F’ factors F’ factors can be transferred through conjugation This may introduce new genes into the recipient cell and thereby alter its genotype

18. 7-18 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display William Hayes demonstrated that conjugation between an Hfr and an F – strain involves the transfer of a portion of the Hfr bacterial chromosome The origin of transfer of the integrated F factor determines the starting point and direction of the transfer process When the DNA is cut, or nicked, at this site it becomes the starting point for the transfer of the Hfr chromosome to the F – cell Then, a strand of bacterial DNA begins to enter the recipient cell in a linear manner

19. 7-19 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display It generally takes about 1.5-2 hours for the entire Hfr chromosome to be passed into the F – cell Most matings do not last that long Only a portion of the Hfr chromosome gets into the F – cell Since the nick is internal to the integrated F factor, only part of the plasmid is transferred and the F – cells does not become F + The F – cell does pick up chromosomal DNA from the Hfr cell This DNA can recombine with the homologous region on the chromosome of the recipient cell This may provide the recipient cell with a new combination of alleles Refer to Figure 7.6

20. Origin of transfer (toward lac + ) Transfer of Hfr chromosome Short time Longer time Hfr cell F – cell F – recipient cell lac + pro + lac + pro + lac + pro + lac – pro – lac – pro – lac + pro + lac + pro + pro – Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. lac + Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 7-20 Figure 7.6 lac +  Ability to metabolize lactose lac –  Inability pro +  Ability to synthesize proline pro –  Inability Therefore, the order of transfer is lac + – pro + F – cell received short segment of the Hfr chromosome It has become lac + but remains pro – F – cell received longer segment of the Hfr chromosome It has become lac + AND pro +

21. Experiment 7A Interrupted Mating Technique Developed by Elie Wollman and François Jacob in the 1950s The rationale behind this mapping strategy –The time it takes genes to enter the recipient cell is directly related to their order along the bacterial chromosome –The Hfr chromosome is transferred linearly to the F – recipient cell Therefore, interrupting matings at different times would lead to various lengths being transferred –The order of genes along the chromosome can be deduced by determining the genes transferred during short matings vs. those transferred during long matings 7-21 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

22. Wollman and Jacob started the experiment with two E. coli strains –The donor (Hfr) strain had the following genetic composition thr + : Able to synthesize the essential amino acid threonine leu + : Able to synthesize the essential amino acid leucine azi s : Sensitive to killing by azide (a toxic chemical) ton s : Sensitive to infection by T1 (a bacterial virus) lac + : Able to metabolize lactose and use it for growth gal + : Able to metabolize galactose and use it for growth str s : Sensitive to killing by streptomycin (an antibiotic) –The recipient (F – ) strain had the opposite genotype thr – leu – azi r ton r lac – gal – str r r = resistant 7-22 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

23. Wollman and Jacob already knew that –The thr + and leu + genes were transferred first, in that order –Both were transferred within 5-10 minutes of mating Therefore their main goal was to determine the times at which genes azi s, ton s, lac +, and gal + were transferred –The transfer of the str s was not examined Streptomycin was used to kill the donor (Hfr) cell following conjugation The recipient (F – cell) is streptomycin resistant 6-23 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

24. The Hypothesis –The chromosome of the donor strain in an Hfr mating is transferred in a linear manner to the recipient strain The order of genes along the chromosome can be deduced by determining the time various genes take to enter the recipient cell 7-24 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Testing the Hypothesis Refer to Figure 7.7

25. 7-25 Figure 7.7 Experimental level Conceptual level 1. Mix together a large number of Hfr donor and F – recipient cells. 2. After different periods of time, take a sample of cells and interrupt conjugation in a blender. 3. Plate the cells on growth media lacking threonine and leucine but containing streptomycin. Note: The general methods for growing bacteria in a laboratory are described in the Appendix. The conclusion is that the colony that was picked contained cells with a genotype of thr + leu + azi s ton s lac + gal – str r. 4. Pick each surviving colony, which would have to be thr + leu + str r, and test to see if It is sensitive to killing by azide, sensitive to infection by T1 bacteriophage, and able to metabolize lactose or galactose. Flask with bacteria Bacterial growth Surviving colonies No growth Plaques Overnight growth Solid growth medium and streptomycin +Azide+T1 phage Additional tests Cannot survive on plates with streptomycin Can survive on plates with streptomycin +Lactose+Galactose Sterile loop In this conceptual example, the cells have been incubated about 20 minutes. Separate by blending; donor DNA recombines with recipient cell chromosome. HfrF–F– Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. gal + str s thr + ton s azi s leu + lac + thr – lac – gal – str r ton r leu – azi r str s gal + lac + ton s azi s leu + thr + str r gal – lac – ton r leu – thr – azi s lac + ton s leu + thr + azi r str s gal + lac + ton s azi s leu + thr + str r gal + lac + ton s azi s leu + thr +

26. The Data 7-26 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Minutes that Bacterial Cells were Allowed to Mate Before Blender Treatment Percent of Surviving Bacterial Colonies with the Following Genotypes thr + leu + azi s ton s lac + gal + 5–– 10 10012300 15 100703100 20 1008871120 25 1009280280.6 30 1009075365 40 10090753820 50 10091784227 60 10091784227

27. Interpreting the Data 7-27 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Minutes that Bacterial Cells were Allowed to Mate Before Blender Treatment Percent of Surviving Bacterial Colonies with the Following Genotypes thr + leu + azi s ton s lac + gal + 5–– 10 10012300 15 100703100 20 1008871120 25 1009280280.6 30 1009075365 40 10090753820 50 10091784227 60 10091784227 After 10 minutes, the thr + leu + genotype was obtained The azi s gene is transferred first It is followed by the ton s gene The lac + gene enters between 15 and 20 minutes The gal + gene enters between 20 and 25 minutes There were no surviving colonies after 5 minutes of mating

28. 7-28 From these data, Wollman and Jacob constructed the following genetic map: They also identified various Hfr strains in which the origin of transfer had been integrated at different places in the chromosome Comparison of the order of genes among these strains, demonstrated that the E. coli chromosome is circular Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display thrleuazitonlacgal Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

29. The E. coli Chromosome Conjugation experiments have been used to map more than 1,000 genes on the E. coli chromosome The E. coli genetic map is 100 minutes long –Approximately the time it takes to transfer the complete chromosome in an Hfr mating –Refer to Figure 7.8 7-29 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

30. 7-30 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Figure 7.8 Arbitrarily assigned the starting point Units are minutes Refer to the relative time it takes for genes to first enter an F – recipient during a conjugation experiment polA oriC dnaA dnaB melA uvrA IacA,Y,Z proA,B galE hipA cheA uvrC purL recA mutS pyrG pyrB argR pabB xylA trpA,B,C,D,E thrA 0.0 gyrA 100/0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. argG

31. 7-31 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display The distance between genes is determined by comparing their times of entry during an interrupted mating experiment –The approximate time of entry is computed by extrapolating the time back to the x-axis Therefore these two genes are approximately 9 minutes apart along the E. coli chromosome Figure 7.9 % of F – recipient cells that have received the gene during conjugation Duration of mating (minutes) 30 20 10 0 0 162025304050 lacZ galE Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

32. Plasmids Plasmids occur naturally in many strains of bacteria and a few eukaryotic cells such as yeast –Range in size from a few thousand to 500,000 bp –Carry from one to hundreds of genes –Different plasmids will have one to 100 copies per cell Have their own origins of replication which are strong or weak –Plasmids can provide a growth advantage to the cell –Some plasmids can integrate into the chromosome Called episomes 7-32 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

33. Plasmids Plasmids fall into five different categories: –1. Fertility plasmids-allow bacteria to mate to each other –2. Resistance plasmids-confer resistance to antibiotics or toxins –3. Degradative plasmids-enable the digestion of unusual substances –4. Col-plasmids-encode colicines which are proteins that kill other bacteria –5. Virulence plasmids-turn a bacterium into a pathogenic strain 7-33 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

34. Transduction Transduction is the transfer of DNA from one bacterium to another via a bacteriophage Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display A bacteriophage is a virus that specifically attacks bacterial cells It is composed of genetic material surrounded by a protein coat Depending on the type of virus, it may follow one of two different cycles, or both Lytic Lysogenic Refer to Figure 7.10 7-34

35. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Phage DNA Bacterial chromosome Phage injects its DNA into cytoplasm. On rare occasions, a prophage may be excised from host chromosome. New phages can bind to bacterial cells. Cell lyses and releases the new phages. Phage DNA directs the synthesis of many new phages. Host DNA is degraded. Phage DNA integrates into host chromosome. Prophage DNA is copied when cells divide. Prophage Lytic cycleLysogenic cycle Bacteriophage Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Figure 7.10 Virulent phages only undergo a lytic cycle Temperate phages can follow both cycles 7-35 Prophage can exist in a dormant state for a long time It can switch to the lytic cycle

36. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Transduction Phages that can transfer bacterial DNA include P22, which infects Salmonella typhimurium P1, which infects Escherichia coli Both are temperate phages Figure 7.11 illustrates the process of transduction 7-36

37. 7-37 Figure 7.11 Any piece of bacterial DNA can be incorporated into the phage This type of transduction is termed generalized transduction Recombination Phage infects bacterial cell. Host DNA is hydrolyzed into pieces, and phage DNA and proteins are made. Phages assemble; occasionally a phage carries a piece of the host cell chromosome. Transducing phage injects its DNA into a new recipient cell. The transduced DNA is recombined into the chromosome of the recipient cell. The recombinant bacterium’s genotype has changed from his – lys – to his + lys –. Recombinant bacterium Transducing phage with host DNA Phage DNA his + lys + his – his + lys + lys – his + lys – lys + his + Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Recipient cell (his – lys – )

38. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Transduction was discovered in 1952 by Joshua Lederberg and Norton Zinder They used an experimental strategy similar to that of Figure 7.1 They used two strains of the bacterium Salmonella typhimurium One strain, designated LA-22, was phe – trp – met + his + Unable to synthesize phenylalanine or tryptophan Able to synthesize methionine and histidine The other strain, designated LA-2, was phe + trp + met – his – Able to synthesize phenylalanine and tryptophan Unable to synthesize methionine or histidine Their experiment is described next 7-38

39. ~ 1 cell in 100,000 was observed to grow 7-39 Nutrient agar plates lacking the four amino acids phe – trp – met + his + phe + trp + met – his – Genotypes of surviving bacteria must be phe + trp + met + his + Therefore, genetic material had been transferred between the two strains However, Lederberg and Zinder obtained novel results when repeating the experiment using the U-tube apparatus Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

40. 7-40 Nutrient agar plates lacking the four amino acids No colonies phe – trp – met + his + phe + trp + met – his – LA-22LA-2 Colonies Genotypes of surviving bacteria must be phe + trp + met + his + Transfer was directional-Only one strain was donating

41. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Therefore, some agent was being transferred from LA-2 to LA-22 through the filter Norton and Zinder conducted the same experiment with filters of different pore sizes They found out that the filterable agent was less then 0.1  m in diameter They correctly concluded that the filterable agent was a bacteriophage In this case, the LA-2 strain contained a prophage (such as P22) The prophage switched to the lytic cycle Packaged a segment of DNA containing the phe + and trp + genes Passed through the filter and injected the DNA into the LA-22 strain 7-41

42. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Cotransduction Mapping There is a maximum size to the DNA that can be packaged by bacteriophages during transduction P1 can pack up to 2-2.5% of the E. coli chromosome P22 can pack up to 1% of the S. typhimurium chromosome Cotransduction refers to the packaging and transfer of two closely-linked genes It is used to determine the order and distance between genes that lie fairly close together 7-42

43. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Cotransduction Mapping Researchers select for the transduction of one gene They then monitor whether a second gene is cotransduced Consider for example the following two E. coli strains The donor strain with genotype arg + met + str s The recipient strain with genotype arg – met – str r The steps in a cotransduction experiment are presented in Figure 7.12 7-43

44. 7-44 Figure 7.12 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Donor cell P1 Recipient cell Infection, production of new phages, and lysis arg + met + arg + met + arg – met + arg + met + arg – met + arg – met + str s arg – met – str r An occasional phage will contain a piece of the bacterial chromosome. Mix P1 lysate with recipient cells that are arg –, met –, str r. P1 lysate 50 colonies Growth of bacterial cells New flask containing millions of recipient cells Occasionally, a recipient cell will receive arg + and/or met + from a P1 phage. Plate on growth media with arginine and streptomycin but without methionine. Growth media without arginine Pick each of the 50 colonies and restreak. (Only the restreaking of 5 colonies is shown.) Genotype of cells in each colony Selected gene Nonselected gene Number of colonies that grew on Cotransduction frequency Results 50210.42 media + arginine media – arginine met + Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

45. Recipient cell arg + met + arg – met + arg + met + arg – met + arg – met + arg – met – str r 50 colonies Growth of bacterial cells New flask containing millions of recipient cells Occasionally, a recipient cell will receive arg + and/or met + from a P1 phage. Plate on growth media with arginine and streptomycin but without methionine. Growth media without arginine Pick each of the 50 colonies and restreak. (Only the restreaking of 5 colonies is shown.) Genotype of cells in each colony Selected gene Nonselected gene Number of colonies that grew on Cotransduction frequency Results 50210.42 media + arginine media – arginine met + 7-45 These colonies must be met + To determine whether they are also arg + streak onto plate that lacks both amino acids 21/50 Figure 7.12 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

46. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display T. T. Wu derived a relationship between cotransduction frequency and map distances obtained from conjugation experiments Cotransduction frequency = (1 – d/L) 3 where d = distance between two genes in minutes L = the size of the transduced DNA (in minutes) For P1 transduction, this size is ~ 2% of the E. coli chromosome, which equals about 2 minutes 7-46 Cotransduction Mapping

47. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Let’s use the equation in our example, 7-47 Transduction experiments can provide very accurate mapping data for genes that are fairly close together Conjugation experiments, on the other hand, are usually used for genes that are far apart on the chromosome Therefore, the distance between the met + and arg + genes is approximately 0.5 minutes

48. Transformation Transformation is the process by which a bacterium will take up extracellular DNA released by a dead bacterium It was discovered by Frederick Griffith in 1928 while working with strains of Streptococcus pneumoniae There are two types –Natural transformation DNA uptake occurs without outside help –Artificial transformation DNA uptake occurs with the help of special techniques Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 7-48

49. Transformation Natural transformation occurs in a wide variety of bacteria Bacterial cells able to take up DNA are termed competent cells –They carry genes that encode proteins called competence factors These proteins facilitate the binding, uptake and subsequent incorporation of the DNA into the bacterial chromosome The steps of bacterial transformation are presented in Figure 7.13 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 7-49

50. 7-50 Figure 7.13 A region of mismatch caused by sequence differences between the two alleles DNA fragment binds to a cell surface receptor of a competent bacterium. An extracellular endonuclease cuts the DNA into smaller fragments. One strand is degraded and a single strand is transported into the cell via an uptake system. Uptake system The DNA strand aligns itself with a homologous region on the bacterial chromosome. The DNA strand is incorporated into the bacterial chromosome via homologous recombination. The heteroduplex DNA is repaired in a way that changes the lys – strand to create a lys + gene. Transformed cell lys – lys + lys – lys + lys – lys + Receptor Heteroduplex Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. By DNA repair enzymes

51. Transformation Sometimes, the DNA that enters the cell is not homologous to any genes on the chromosome –It may be incorporated at a random site on the chromosome –This process is termed nonhomologous or illegitimate recombination Some bacteria preferentially take up DNA bacteria from the same or a related species –Streptococcus pneumoniae secretes a competence- stimulating peptide which leads to competence only when many cells of the same species are nearby –Other species use DNA uptake signal sequences 9 or 10 bp long repeated 1-2,000 times throughout genome DNA with this sequence is preferentially taken up Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 7-51

52. Transformation Can be used to map genes Method is similar to cotransduction –Cotransformation frequency is high for two genes that are close together –Will be very low or even zero for distant genes –Used only to map genes that are relatively close together Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 7-52

53. Horizontal Gene Transfer Vertical gene transfer is the transfer of genes from mother to daughter cell or from parents to offspring Horizontal gene transfer is the transfer of genes into an organism which is not the offspring of the donor –Can be members of the same species or different species A sizable fraction of bacterial genes are derived from horizontal gene transfer –Roughly 17% of E. coli and S. typhimurium genes during the past 100 million years Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 7-53

54. Horizontal Gene Transfer The types of genes acquired through horizontal gene transfer are quite varied and include –Genes that confer the ability to cause disease –Genes that confer antibiotic resistance –Genes that give the ability to degrade toxins Horizontal gene transfer has dramatically contributed to the phenomenon of acquired antibiotic resistance –Bacterial resistance to antibiotics is a serious problem worldwide In many countries, nearly 50% of Streptococcus pneumoniae strains are resistant to penicillin Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 7-54

55. 7.2 INTRAGENIC MAPPING IN BACTERIOPHAGES Viruses are not living –They rely on a host cell for existence and replication –However, they have unique biological structures and functions, and therefore have traits We will focus our attention on bacteriophage T4 –Its genetic material contains approximately 300 genes These genes encode a variety of proteins needed for the viral life cycle –Refer to Figure 7.14 for the T4 structure 7-55 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

56. Head Tail fiber Shaft Tail fiber composed of 5 different kinds of proteins Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Base plate 7-56 Figure 7.14 Contains the genetic material Used for attachment to the bacterial surface

57. In the 1950s, Seymour Benzer embarked on a ten-year study focusing on the function of the T4 genes –He conducted a detailed type of genetic mapping known as intragenic or fine structure mapping –The difference between intragenic and intergenic mapping is: 7-57 Intergenic mapping gene A gene B Distance between gene A and gene B DNA Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Intragenic mapping (also known as fine-structure mapping) Distance between 2 mutations in gene C gene C (allele 1) gene C (allele 2)

58. Plaques A plaque is a clear area on an otherwise opaque bacterial lawn on the agar surface of a petri dish It is caused by the lysis of bacterial cells as a result of the growth and reproduction of phages 7-58 Figure 7.15 Bacterial lawn on petri plate Each infected cell lyses and releases phages that infect nearby cells. Nearby cells lyse, infecting more cells. Process continues. Plaque Infected cell lysing and releasing new phages that infect nearby cells Plaque is a clear area where the bacterial lawn has been destroyed. Bacterial cells Phage DNA Infected cell Phages Phage capsid Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

59. Some mutations in the phage’s genetic material can alter the ability of the phage to produce plaques –Thus, plaque formation can be viewed as a trait of bacteriophages Plaques are visible with the naked eye –So mutations affecting them lend themselves to relatively easy genetic analysis Visualizing bacteriophages requires an electron microscope An example is a rapid-lysis mutant of bacteriophage T4, which forms unusually large plaques –Refer to Figure 7.16 –This mutant lyses bacterial cells more rapidly than do the wild-type phages Rapid-lysis mutant forms large, clearly defined plaques Wild-type phages produce smaller, fuzzy-edged plaques 7-59 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

60. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Plaque caused by a rapid-lysis mutantWild-type plaque (a) Plaques caused by wild-type bacteriophages (b) Plaques caused by rapid-lysis bacteriophage strains © Carolina Biological Supply/Phototake © 7-60 Figure 7.16

61. Benzer studied one category of T4 phage mutant, designated rII (r stands for rapid lysis) It behaved differently in three different strains of E. coli –In E. coli B rII phages produced unusually large plaques that had poor yields of bacteriophages –The bacterium lyses so quickly that it does not have time to produce many new phages –In E. coli K12S rII phages produced normal plaques that gave good yields of phages –In E. coli K12( ) (has phage lambda DNA integrated into its chromosome) rII phages were not able to produce plaques at all As expected, the wild-type phage could infect all three strains 7-61 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

62. Complementation Tests Benzer collected many rII mutant strains that can form large plaques in E. coli B and none in E. coli K12( ) But, are the mutations in the same gene or in different genes? To answer this question, he conducted complementation experiments Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 7-62

63. 7-63 Figure 7.17 shows the possible outcomes of complementation experiments involving coinfection of plaque formation mutants Figure 7.17 rII strain 1 (gene A is defective, gene B is normal) (a) Noncomplementation: The phage mutations are in the same gene.(b) Complementation: The phage mutations are in different genes. Coinfect E. coli K12(λ) Plate and observe if plaques are formed. No plaques gene A Coinfect E. coli K12(λ) Plate and observe if plaques are formed. Viral plaques No complementation occurs, because the coinfected cell is unable to make the normal product of gene A. The coinfected cell will not produce viral particles, thus no bacterial cell lysis and no plaque formation. Complementation occurs, because the coinfected cell is able to make normal products of gene A and gene B. The coinfected bacterial cell will produce viral particles that lyse the cell, resulting in the appearance of clear plaques. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. rII strain 2 (gene A is defective, gene B is normal) rII strain 3 (gene A is defective, gene B is normal) rII strain 4 (gene A is normal, gene B is defective) gene B gene A gene B gene A gene B gene A gene B

64. 7-64 Benzer carefully considered the pattern of complementation and noncomplementation –He determined that the rII mutations occurred in two different genes, which were termed rIIA and rIIB Benzer coined the term cistron to refer to the smallest genetic unit that gives a negative complementation test –So, if two mutations occur in the same cistron, they cannot complement each other A cistron is equivalent to a gene –However, this term is no longer commonly used Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

65. gene A Rare crossover Double mutantWild type rll mutation Coinfection Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. gene A rll mutation gene A At an extremely low rate, two noncomplementing strains of rII mutants can produce an occasional viral plaque if intragenic recombination has occurred 7-65 rII mutation Virus cannot form plaques in E. coli K12( ) Function of protein A will be restored Therefore new phages can be made in E. coli K12( ) Viral plaques will now be formed Figure 7.18 Virus cannot form plaques in E. coli K12( )

66. Figure 7.19 describes the general strategy for intragenic mapping of rII phage mutations 7-65 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. r103 gene A Isolate this new population of phages. It will primarily contain nonrecombinant phages, but it will occasionally contain intragenic recombinants of wild-type and double mutant phages (depicted in white and black, respectively). The phage preparation can contain several billion phages per milliliter. Phage E. coli B Isolate 2 different (noncomplementing) rII phage mutants, r103 and r104. Mix the 2 phages together. Coinfect E. coli B. A new population of phages will be made. The E. coli B cells will eventually lyse. r104 gene A

67. 66 plaques 11 plaques Plate the cells and observe the number of plaques. The number of plaques observed from the E. coli B infection provides a measure of the total number of phages in the population. The number of plaques observed from the E. coli K12(λ) infection provides a measure of the wild-type phage produced by intragenic recombination. Take some of the phage preparation, dilute it greatly (10 –8 ), and infect E. coli B. Also, take some of the phage preparation, dilute it somewhat (10 –6 ), and infect E. coli K12(λ). 10 –8 Double mutant phage Phage E. coli B Phage E. coli K12(λ) Wild-type phage Nonrecombinant phages 10 –6 7-67 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Total number of phages Number of wild-type phages produced by intragenic recombination r103 r104 Both rII mutants and wild-type phages can infect this strain rII mutants cannot infect this strain Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

68. 7-68 The data from Figure 7.19 can be used to estimate the distance between the two mutations in the same gene Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display The phage preparation used to infect E. coli B was diluted by 10 8 (1:100,000,000) 1 ml of this dilution was used and 66 plaques were produced Therefore, the total number of phages in the original preparation is 66 pfu/ml X 10 8 (d.f.) = 6.6 X 10 9 pfu/ml or 6.6 billion phages/ml The phage preparation used to infect E. coli k12( ) was diluted by 10 6 (1:1,000,000) 1 ml of this dilution was used and 11 plaques were produced Therefore, the total number of wild-type phages is 11 pfu/ml X 10 6 (d.f.) = 1.1 x 10 7 or 11 million phages/ml

69. 7-69 In this experiment, the intragenic recombination produces an equal number of recombinants –½ are wild-type phages –½ are double mutant phages However, only the wild-type phages are detected in the infection of E. coli k12( ) –Therefore, the total number of recombinants is the number of wild- type phages multiplied by two Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 2 [wild-type plaques obtained in E. coli k12( )] Frequency of recombinants = Total number of plaques obtained in E. coli B 2(1.1 X 10 7 ) 6.6 X 10 9 Frequency of recombinants = = 3.3 X 10 –3 = 0.0033 In this example, there were approximately 3.3 recombinants per 1,000 phages

70. 7-70 As in eukaryotic mapping, the frequency of recombinants can provide a measure of map distance along the bacteriophage chromosome –In this case the map distance is between two mutations in the same gene The frequency of intragenic recombinants is correlated with the distance between the two mutations –The farther apart they are the higher the frequency of recombinants Homoallelic mutations –Mutations that happen to be located at exactly the same site in a gene –They are not able to produce any wild-type recombinants So the map distance would be zero Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

71. Deletion Mapping Benzer used deletion mapping to localize many rII mutations to a fairly short region in gene A or gene B He utilized deletion strains of phage T4 –Each is missing a known segment of the rIIA and/or rIIB genes Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 7-71

72. Let’s suppose that the goal is to know the approximate location of an rII mutation, such as r103 –E. coli k12( ) is coinfected with r103 and a deletion strain If the deleted region includes the same region that contains the r103 mutation –No intragenic wild-type recombinants are produced »Therefore, plaques will not be formed If the deleted region does not overlap with the r103 mutation –Intragenic wild-type recombinants can be produced »And plaques will be formed Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 7-72

73. 7-73 Figure 7.20 Mutation must be in region contained in BP242 but not in PT1. This corresponds to region A4 in the rIIA gene Deletion strain Deleted region 1272 Deleted region Del. reg. 1241 J3 PTI PB242 A 105 638 Wild-type recombinants when coinfected with r103? No Yes gene rIIAgene rIIB A1A2A3A4A5A6B1–B10 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

74. As described in Figure 7.20, the first step in the deletion mapping strategy localized rII mutations to one of seven regions –Six in rIIA and one in rIIB Other deletion strains were used to eventually localize each rII mutation to one of 47 regions –36 in rIIA and 11 in rIIB At this point, pairwise coinfections were made between mutant strains that had been localized to the same region –This would precisely map their location relative to each other This resulted in a fine structure map depicting the locations of hundreds of different rII mutations –Refer to Figure 7.21 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 7-74

75. A1aA1b1A1b2A2aA2cA2eA2g A2h1 A2h2 A2h3A3a–dA3eA3fA3gA3hA3iA4aA4bA4cA4d A4e A4g A5aA5bA5c1A5c2A5dA6a1A6a2 A6b A6c A6dB1B2B3B4B5B6 B7 B8B9aB9b B10 A2bA2d Start of rIIB gene Start of rIIA gene End of rIIA gene End of rIIB gene Hot spots Hot spot Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A2f A4f Hot spots 7-75 Figure 7.21 Contain many mutations at exactly the same site within the gene

76. Intragenic mapping studies were a pivotal achievement in our early understanding of gene structure Some scientists had envisioned a gene as being a particle-like entity that could not be further subdivided However, intragenic mapping revealed convincingly that this is not the case –It showed that Mutations can occur at different sites within a single gene Intragenic crossing over can recombine these mutations, resulting in wild-type genes So a gene’s structure can be subdivided Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 7-76