1. PowerPoint Presentation Materials to accompany Genetics: Analysis and Principles Robert J. Brooker Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display CHAPTER 6 GENETIC TRANSFER AND MAPPING IN BACTERIA AND BACTERIOPHAGES

2. INTRODUCTION Bacteria are usually haploid This fact makes it easier to identify loss-of-function mutations in bacteria than in eukaryotes Bacteria reproduce asexually Therefore crosses are not used in the genetic analysis of bacterial species Bacterial genetics is based on genetic transfer A segment of bacterial DNA is transferred from one bacterium to another 6-3 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

3. 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 6-4 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 6.1 GENETIC TRANSFER AND MAPPING IN BACTERIA

4. Genetic transfer in bacteria was discovered in 1946 by Joshua Lederberg and Edward Tatum 6-5 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Two strains of Escherichia coli with different nutritional growth requirements: One strain was designated bio – met – phe + thr + It required one vitamin (biotin) and one amino acid (methionine) It could produce the amino acids phenylalanine and threonine The other strain was designated bio + met + phe – thr – Conjugation

5. Figure 6.1 6-6 ~ 100

6. 6-7 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 bio + met + phe + thr + Lederberg and Tatum reasoned that some genetic material was transferred between the two strains The “direction” of gene transfer was unclear

7. 6-8 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 to occur He used an apparatus known as U-tube It contains at the bottom a filter 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

8. 6-9 Figure 6.2 Nutrient agar plates lacking biotin, methionine, phenylalanine and threonine No colonies Thus, without physical contact, the two bacterial strains did not transfer genetic material to one another

9. 6-10 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 Only certain strains of a bacterium can act as donor cells Those strains contains 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 – Conjugation 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

10. 6-11 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Plasmids, such as F factors, which are transmitted via conjugation are termed conjugative plasmids These plasmids carry genes required for conjugation Figure 6.3 These genes play a role in the transfer of DNA They are thus designated tra and trb followed by a capital letter

12. Fig. 6.4b

13. 6-15 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 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 In some cases, the F factor may carry genes that were once found on the bacterial chromosome These types of F factors are called F’ factors F’ factors (and donor genes) can be transferred through conjugation

15. 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 6-16 Hfr Strains Figure 6.5a An episome is a segment of DNA that can exist as a plasmid and integrate into the chromosome

16. 6-17 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 The cut, or nicked site is the starting point that will enter the F – cell

17. Fig. 6.5 (TE Art) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. (b) An Hfr donor cell can pass a portion of its chromosome to an F – recipient cell. Origin of transfer (toward lac + ) Transfer of Hfr chromo- some Short time Longer time Hfr cell F – cell F – recipient cell lac + pro + pro – lac + pro + lac + pro – pro + lac + lac – lac + lac – pro – pro + lac + (a) When an F factor integrates into the chromosome, it creates an Hfr cell. Bacterial chromosome Origin of transfer Origin of transfer F factor Integration of F factor into chromosome Hfr cellF + cell lac + pro + lac + pro +

18. 6-18 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 The F – cells does not become F + The transferred DNA can recombine with the homologous region on the chromosome of the recipient cell

19. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 6-19 Figure 6.5b 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 +

20. The time it takes genes to enter the recipient cell is directly related to their order along the bacterial chromosome Interruptions of mating at different times 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 6-20 Interrupted Mating Technique

21. 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 6-21 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

22. 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 6-22 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

23. 6-24 Figure 6.6

24. Interpreting the Data 6-26 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 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

26. 6-27 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

27. Figure 6-9 Copyright © 2006 Pearson Prentice Hall, Inc.

28. Conjugation experiments have been used to map 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 6-28 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display The E. coli Chromosome

29. 6-30 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 Therefore these two genes are approximately 9 minutes apart along the E. coli chromosome Figure 6.7

30. Figure 6-8 Copyright © 2006 Pearson Prentice Hall, Inc.

31. Fig. 6.7(TE Art) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. polA oriC dnaA dnaB melA uvrA lacA,Y,Z proA,B galE hipA cheA uvrC purL recA mutS pyrG pyrB argG 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

32. Transfer of F’ produces merozygotes

33. 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 Transduction A bacteriophage is a virus that specifically attacks bacterial cells two types of cycles Lytic Lysogenic 6-31

34. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Figure 6.9 Virulent phages only undergo a lytic cycle Temperate phages can follow both cycles 6-32 Prophage can exist in a dormant state for a long time It will undergo the lytic cycle

35. 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 6-33

36. 6-34 Figure 6.10 Any piece of bacterial DNA can be incorporated into the phage This type of transduction is termed generalized transduction

37. 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 two strains of the bacterium Salmonella typhimurium One strain, designated LA-22, was phe – trp – met + his + The other strain, designated LA-2, was phe + trp + met – his – 6-35

38. ~ 1 cell in 100,000 was observed to grow 6-36 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

39. 6-37 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 +

40. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display The filterable agent was less then 0.1  m in diameter Conclusion: the filterable agent was a bacteriophage In this case, the LA-2 strain contained a prophage 6-38

41. 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 6-39

42. 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 6-40

43. 6-41 Figure 6.11 Mix P1 lysate with recipient cells that are arg – met – str r

44. 6-42 Plate on minimal plates with arginine and streptomycin but without methionine These colonies must be met + To determine whether they are also arg + streak onto plate that lacks both amino acids 21/50

45. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 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 6-43 Cotransduction Mapping

46. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Let’s use the equation in our example, 6-44 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

47. Transformation is the process by which a bacterium will take up extracellular DNA Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Transformation 6-45

48. Horizontal gene transfer is the transfer of genes between two 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 Genes that confer the ability to cause disease Genes that confer antibiotic resistance Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 6-49 Horizontal Gene Transfer

50. Viruses are not living However, they have unique biological structures and functions, and therefore have traits We will focus our attention on bacteriophage T4 Its genetic material contains several dozen genes These genes encode a variety of proteins needed for the viral cycle 6-51 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 6.2 INTRAGENIC MAPPING IN BACTERIOPHAGES

51. 6-52 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Figure 6.13 Contains the genetic material Used for attachment to the bacterial surface

52. 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: 6-53

53. 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 6-54 Plaques Figure 6.14

54. A rapid-lysis mutant of bacteriophage T4 forms unusually large plaques This mutant lyses bacterial cells more rapidly than do the wild-type phages 6-55

55. 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 6-56 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

56. 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 6-57 Complementation Tests

57. Fig. 6.16a(TE Art) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Noncomplementation: The phage mutations are in the same gene. Coinfect E. coli K12 ( ) Plate and observe if plaques are formed. No plaques No complementation occurs since 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. rII strain 1 (gene A is defective, gene B is normal) gene Agene B rII strain 2 (gene A is defective, gene B is normal) gene Agene B

58. Fig. 6.16b(TE Art) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Complementation: The phage mutations are in different genes. Plate and observe if plaques are formed. Viral plaques Complementation occurs since 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. Coinfect E. coli K12 ( ) rII strain 3 (gene A is defective, gene B is normal) rII strain 4 (gene A is normal, gene B is defective) gene Agene Bgene Agene B

59. 6-59 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 A cistron is equivalent to a gene However, it is not as commonly used Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

60. At an extremely low rate, two noncomplementing strains of viruses can produce an occasional viral plaque, if intragenic recombination has occurred Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 6-60 Coinfection (E. coli B) rII mutations Viruses 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 6.17

62. General strategy for intragenic mapping of rII phage mutations 6-61

63. 6-62 r103 r104 Take some of the phage preparation, dilute it greatly (10 -8 ) and infect E. coli B Take some of the phage preparation, dilute it somewhat (10 -6 ) and infect E. coli K12( ) 66 plaques 11 plaques Total number of phages Number of wild-type phages produced by intragenic recombination Both rII mutants and wild-type phages can infect this strain rII mutants cannot infect this strain

64. 6-63 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 X 10 8 = 6.6 X 10 9 or 6.6 billion phages per milliliter 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 X 10 6 or 11 million phages per milliliter

65. 6-64 In this experiment, the intragenic recombination produces an equal number of recombinants Wild-type phages and 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(11 X 10 6 ) 6.6 X 10 9 Frequency of recombinants = = 3.3 X 10 –3 = 0.0033 In this example, there was approximately 3.3 recombinants per 1,000 phages

66. 6-65 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

67. 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 6-66 Deletion Mapping

68. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 6-68 Figure 6.19

69. 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 6-67

70. The first step in the deletion mapping strategy localized rII mutations to seven regions Six in rIIA and one in rIIB Other 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 with depicting the locations of hundreds of different rII mutations Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 6-69

71. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 6-70 Figure 6.20 Contain many mutations at exactly the same site within the gene

72. 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 parts within a single gene Intragenic crossing over can recombine these mutations, resulting in wild-type genes Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 6-71