Module 5: Bacterial Genetics (Basics) Flashcards

1
Q

Advantages of using bacteria in genetic research

A

1) Each cell is a complete organism
2) Usually grow rapidly
3)Simple genomic organization

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1
Q

Bacterial growth is measured by…

A

NUMBER of cells (NOT by the change in size of cells)

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2
Q

In what way do bacteria have a simple genomic organization?

A

Usually only contain ONE copy of their genome

–> Each gene exists as a single copy within the larger genome

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3
Q

How does the organization of bacterial genome make it ideal for genetic research?

A

No mutation masking

–> A mutation in a gene will often produce a readily observable phenotype (because there’s only one copy of the gene)

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4
Q

Why are bacteria “better” than eukarya for genetic research?

A

Bacteria = 1 copy of genome

Eukarya = 2 copies of genome

–> A mutation in one copy of a gene can get “masked” by the second copy being the WT variant == Harder to observe an induced mutation!

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5
Q

Why were microbial genetics limitedly studied in history?

A

1) It was believed that bacteria (and other microbes) were “too small/simple” to have a system of genetic exchange

2) Nobody had observed inheritance of phenotype in bacteria

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6
Q

What was the major objective of Joshua Lederberg’s work with nutritional mutants?

A

To examine if gene transfer could occur between bacteria

–> Did so by studying nutritional mutants of E. coli

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7
Q

Nutritional Mutant

A

AKA. Auxotroph

Organisms with altered metabolic requirements

–> typically have mutations which disrupt specific metabolic pathways

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8
Q

What auxotrophs did Joshua Lederberg study?

A

2 auxotrophs of E. Coli:

1) Met+ Pro- (could produce methionine but not proline)

2) Met- Pro+ (could produce proline but not methionine)

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9
Q

Growth conditions for: Met+ Pro-

A

Media supplemented with proline

(Can’t produce it)

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10
Q

Growth conditions for: Met- Pro+

A

Media supplemented with methionine

(Can’t produce it)

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11
Q

Phototroph

A

= PARENTAL STRAIN (Wild type; WT)

–> Grows on simple defined growth medium (w/ proper sources of C, N, P)

–> No supplementation required for growth!

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12
Q

Auxotrophic Mutations typically…

A

Make organisms unable to synthesize AAs and vitamins (that are produced within + required for parental strain growth)

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13
Q

Auxotrophs will only grow if…

A

The AA or vitamin they cannot produce is supplemented into the growth medium they are grown in

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14
Q

Met- auxotrophs only grow in…

A

Medium WITH methionine

(Because they can’t produce it)

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15
Q

Pro- auxotrophs only grow in…

A

Medium WITH proline

(because they can’t produce it)

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16
Q

WT, Met-, Pro- Growth In:

Media with Met + Pro

A

ALL GROW

–> WT, MET- PRO-

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17
Q

WT, Met-, Pro- Growth In:

Defined media

A

WT GROWS

–> Met- and Pro- do not grow

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18
Q

WT, Met-, Pro- Growth In:

Media with Pro

A

WT + PRO- grows!

MET- can’t grow

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19
Q

WT, Met-, Pro- Growth In:

Media with Met

A

WT and MET- grows

Pro- can’t grow

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20
Q

What was Lederberg’s hypothesis/prediction (for auxotroph experiment)?

A

If TWO auxotrophs were cultured together and produced progeny that could thrive on media WITHOUT nutrient supplement, the transfer of genetic material between auxotrophs will have occurred

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21
Q

What unexpected outcome did Lederberg observe with his control experiment?

(What was the control experiment?)

A

Experiment:
–> Individually plated WT, met-, pro- cultures onto plates with regular medium

Expected Results:
–> ONLY WT plate would display growth

Actual Results:
–> WT, met-, AND pro-cultures all grew!

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22
Q

What was the cause of Lederberg’s unexpected control experiment results?

A

REVERSION

–> The reversion rate was too high

(spontaneous mutation)

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23
Q

Reversion

A

Spontaneous recovery of a mutant to perform an action/function

–> Essentially, a spontaneous mutation that allows a once-mutant to REGAIN a specific ability

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24
Q

Why did Lederberg observe growth of Met- and Pro- in regular growth medium?

A

Due to reversion in Met- and Pro- cells!

– > Spontaneous mutation resulted in Met- and Pro- cells regaining their ability to produce their missing AAs! (= can grow in normal conditions!)

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25
Q

Rate of reversion =

A

Frequency of reversion (how many cells reverted per total # of cells)

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26
Q

Rate of reversion varies with…

A

The # of nutritional mutations in a given strain

–> As # of mutations increases, the rate of reversion decreases

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27
Q

Single vs Double vs Triple Mutant Strains

A

1) Single = ONE distinct nutritional mutation

2) Double = TWO distinct nutritional mutations

3) Triple = THREE distinct nutritional mutations

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28
Q

Rate of reversion for single mutant strain

A

10-6 to 10-7 (per 108 cells plated)

= 10-100 colonies (in reg. medium)

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29
Q

Rate of reversion for a double mutant strain

A

10-12 to 10-14 (per 108 cells plated)

= 0.0001-0.00001 colonies grow (in reg. medium)

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30
Q

Rate of reversion for a triple mutant strain

A

10-18 to 10-21

= Basically no colonies

31
Q

How did Lederberg redo his control experiments?

Why did he make this change?

A

He redid the experiments but with DOUBLE and TRIPLE mutants

–> The probability of two or three mutations simultaneously reverting is very low so using these mutants controlled for reversion

32
Q

What did Lederberg observe in his repeated control experiments?

A

He observed no growth of the double and triple mutants in regular medium

33
Q

Why did Lederberg need to control for reversion within his test experiments (not the controls)?

A

Because any reversion occurring would mask any gene transfer that might occur between cells

(because then you couldn’t be sure if what was growing in regular medium had undergone reversion or had done gene transfer)

34
Q

What was Lederberg’s Test experiment?

A

Co-cultured 2 triple mutants and plated them on defined (reg./unsupp.) medium

35
Q

What were the results of Lederberg’s test experiment?

Conclusions from this?

A

Results:
The co-cultured triple mutants were able to grow on regular medium (they weren’t able to on their own)

Conclusion:
The genes required for synthesis of the missing compounds (due to mutation) were transferred from one cell to another = both mutants regained ability to grow without supplement

–> **Bacterial gene exchange (mating) happened

36
Q

What did Lederberg have to do even after he got his results?

A

Confirm them.

–> Had to rule out possibility of exogenous DNA uptake accounting for the genetic changes rather than active cell-cell exchange

37
Q

How did Lederberg confirm that gene exchange had occurred (and it wasn’t something else)?

Results?

A

1) Prepared DNA extracts from one of the co-cultured mutants

2) Added this DNA extract to medium with the opposite strain of the co-cultured mutants (during growth)

== Observed NO GROWTH!

(no genetic change that “undid” the mutation which means DNA uptake DID NOT account for the genetic changes he observed in the test experiment, they WERE gene transfer!)

38
Q

What did Lederberg’s confirmation experiment show?

A

Showed that DNA ALONE is not sufficient for the changing of the nutritional mutants!

–> Active live cells are needed for such changes = GENE TRANSFER

39
Q

What were some larger applications of Lederberg’s work with nutritional mutants? (3)

A

1) E. coli became the most studied and well characterized organism

2) Led to the genetic mapping of E. coli

3) Was a turning point in the development of bacterial genetics as a science discipline (paved way for genomics revolution)

40
Q

Genome

A

ALL of the hereditary material within a cell

(For bacteria = Chromosomal AND plasmid DNA

41
Q

Characteristics of the COMMON bacterial genome (4)

A

1) SMALL (-er than eukarya)

2) COMPACT (most of genome is CODING DNA)

3) SINGLE/CIRCULAR CHROMOSOME

4) PLASMIDS (present in some)

42
Q

What is a difference between the compaction of bacterial and eukaryal genomes?

A

–> The amount of CODING vs NONCODING DNA

Bacteria:
Most of genome = coding DNA (encodes for functional proteins)

Eukarya:
Most of genome = NONcoding DNA (does NOT directly encode for proteins)

43
Q

Plasmid

A

A small, circular, EXTRACHROMOSOMAL DNA molecule

(Replicates independently of the chromosome/s)

44
Q

Difference between plasmid and chromosomal genes:

A

Plasmid Genes = Mainly for accessory functions; NOT always needed for survival

Chromosomal Genes = Generally REQUIRED for basic metabolic processes (and thus survival)

45
Q

What are some example of plasmid genes? (3)

A

1) Antibiotic resistance

2) Degradation of toxic substances

3) Mechanisms of bacterial infection (can be good or bad)
–> Good = establish symbiotic relationships
–> Bad = Disease and pathogenicity

46
Q

What are the NAMES of some plasmid examples: (6)

A

1) RI
2) pSym
3) pTi
4) pTol
5) pR773
6) PWR100

47
Q

What is the source and function of:

R1 plasmid

A

R1:

Salmonella paratyphi

–> Antibiotic resistance

Think “R” = “Resistance”

48
Q

What is the source and function of:

pSym

A

pSym:

Rhizobium

–> formation of nitrogen-fixing nodule on legume plant roots

Think “sym” = “symbiosis” and rhizosphere is a symbiosis of nitro-fixing bacteria w/ roots

49
Q

What is the source and function of:

pTi

A

pTi:

Agrobacterium

–> Tumor formation on plants

50
Q

What is the source and function of:

pTol

A

pTol:

Pseudomonas putida

–> Toluene degradation

Think “tol” = “toluene”

51
Q

What is the source and function of:

pR773

A

pR773:

Escherichia coli

–> Arsenic resistance

52
Q

What is the source and function of:

pWR100

A

pWR100:

Shigella flexneri

–> Entry into host cell

53
Q

What are two main features found to exhibit SOME variation in bacterial genome?

A

1) Chromosome shape –> Not all chromosomes are circular!

2) Number of chromosomes –> Some bacteria have > 1 chromosome!

54
Q

Example of bacteria with non-circular chromosome

A

1) Streptomyces (antibiotic producing soil bacteria)

2) Borrelia burgdorferi (Lyme’s Disease bacteria)

(both have linear chromosomes)

55
Q

Unique Genome Organization of:

Vibrio cholerae

A

TWO circular chromosomes

56
Q

Unique Genome Organization of:

Agrobacterium tumefaciens

A

TWO chromosomes

–> One circular, ONE LINEAR

+ 2 plasmids

57
Q

Unique Genome Organization of:

Burkholderia xenovarans

A

THREE circular chromosomes

58
Q

Unique Genome Organization of:

Borrelia burgdorferi

A

One circular chromosome

21 Plasmids!

–> NINE circular plasmids
–> TWELVE linear plasmids

59
Q

Unique Genome Organization of:

Bacillus Anthracis

A

1 circular chromosome

TWO circular plasmids

60
Q

Unique Genome Organization of:

Streptomyces coelicolor

A

1 LINEAR chromosome

2 plasmids: one circular, one LINEAR

61
Q

Replicon

A

DNA molecules that replicate from a SINGLE origin of replication

62
Q

Bacterial chromosome and plasmids are…

A

Replicons!

(Only have ONE origin of replication)

63
Q

What occurs at the origin of replication?

A

DNA polymerase initiates DNA synthesis

64
Q

Plasmid Incompatibility

A

–> Plasmids that CANNOT coexist stably within a population of cells

When 2 different plasmids use the SAME control mechanisms for replication initiation

–> Essentially, plasmids with the same origin of replication

65
Q

What causes plasmid incompatibility?

WHY?

A

Shared identical origin of replication between plasmids!

= Cell replication machinery treats the two distinct plasmids as copies of ONE plasmid

–> Therefore, will only initiate replication of ONE of them

66
Q

What is the product of cell division in cell that has 2 incompatible plasmids?

A

1) During plasmid replication: ONE of the two gets RANDOMLY replicated

= Cell has TWO copies of one and ONE copy of the other

2) During cell division, both daughter cells get a copy of the replicated plasmid and ONE of the daughter cells gets the unreplicated plasmid

= UNEQUAL daughter cells! One of the cells has LOST a plasmid!

67
Q

What is the greater implication of plasmid incompatibility?

A

Over multiple cell divisions, a plasmid can be lost entirely from a population

68
Q

In plasmid incompatibility, the plasmid that gets replicated is ______________. This is NOT a ___________ process

A

1) RANDOMLY chosen

2) Selective

69
Q

Copy Number

A

The number of plasmids in a cell

–> Highly controlled and is characteristic of different plasmids (each has a specific copy number)

70
Q

__________ controls copy number.

____________ does NOT control copy number.

A

1) DNA replication initiation

2) DNA replication rate
(Because DNA polymerase always works at a consistent rate)

71
Q

Incompatibility Group

A

Group of plasmids that have the same origin of replication

(cannot stably coexist in a cell)

72
Q

For cells with > 1 plasmid, the plasmids MUST be…

A

part of DIFFERENT incompatibility groups

73
Q

Other than plasmids and chromosomes, what other components may make up bacterial genome?

A

Bacteriophage DNA

–> Viral genetic material can exist within bacterial hosts for extended periods of time

74
Q

Bacteriophage DNA can be maintained within bacterial hosts as…

A

1) Extrachromosomal circular DNA molecules (like plasmids)

OR

2) Integrated into the bacterial host genome

75
Q

What genomic components contribute MOST to bacterial evolution?

Why?

A

Plasmids and Bacteriophage DNA!

–> Provide mechanisms for movement of bacterial genome segments from one cell to another