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NEUR2001 > Workshop 1 > Flashcards

Workshop 1 Flashcards

1
Q

What is a replicon ?

A

A replicon = a self replicating genetic unit.

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2
Q

What molecular organization do CCC plasmids show under native coniditions ?

A

Covalently closed circular (CCC) plasmids under native conditions show negative supercoiling

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3
Q

How long would the E. Coli genome be if it was linearized instead of -vely supercoiled ?

A

Linearized E. coli genome would be ~700x longer than E. coli (see BIOC1001)
DNA is further organized, supercoiled, to fit.

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4
Q

Is DNA a right-handed or left-handed double helix ?

A

DNA is a right-handed double helix. The removal of turns in a CCC plasmid results in -ve and the addition of turns in +ve supercoiling.

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5
Q

What is the role of DNA topoisomerase I ?

A

DNA topoisomerase I = primarily responsible for relaxing positively supercoiled (over-wound) and/or negatively supercoiled (under-wound) DNA.
This involves rotating the broken strand around the intact strand to relax (unwind) the strain on the DNA helix, followed by resealing the ends of the broken strand.

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6
Q

What is the role of DNA gyrase (topoisomerase II) ?

A

DNA gyrase causes negative supercoiling of the DNA or relaxes positive supercoils.

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7
Q

What is the difference between a stringent and a relaxed plasmid ?

A

Copy number: stringent = low Vs relaxed = high

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8
Q

How do stringent plasmids segregate plasmids to different daughter cells ?
Why is this useful ?

A

Stringent plasmids have a mechanism using par genes to segregate plasmids to different daughter cells, thereby preventing plasmid loss.

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9
Q

What is plasmid incompatibility ?

How does this affect the was different plamids can co-exist ?

A

Incompatibility refers to the inability of two plasmids with similar replication or segregation mechanism to coexist in the same bacterium. A number of different incompatibility (Inc) groups exist and plasmids belonging to different Inc groups can co-exist. Plasmid incompatibility is basic requirement to enable “cloning”.

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10
Q

ColEI and pMBI are 2 Inc groups. What are the -ve control elements and how do they work ?

A

-ve control elements : RNAI (small inhibitory RNA), works by controlling the processing of pre-RNAII into primer

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11
Q

IncFII and pT181 are two Inc groups. What are the -ve control elements and how do they work ?

A

-ve control elements : RNA, works by controlling the synthesis of the RepA protein

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12
Q

PI, F, R6K, pSC101 and p15A are Inc groups. What are the -ve control elements and how do they work ?

A

-ve control elements : iterons, work by sequestering the RepA protein

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13
Q

How is replication of ColE1-derived plasmids regulated ?

A

RNAII must be processed by RNAase before it can prime replication. Most of the time, RNAI binds to RNAII and inhibits processing, thereby regulating the copy number.
The Rop protein dimer enhances the initial pairing of RNAI and RNAII.

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14
Q

What are the (usual) requirements for a good plasmid in gene technology ?
Are these requirements fulfilled naturally ?

A
  • Replicates autonomously
  • At least one, but potentially more markers (resistance to antibiotic)
  • A region with a number of unique restriction endonuclease cleavages sites & some unique sites in marker
  • Neither transmissible or mobilisable - biological safety
  • Small molecular weight, because efficiency of transformation is inversely proportional the plasmid size
  • Copy number under relaxed control
    Naturally occurring plasmids usually do not fulfill all the requirements listed above. Artificial plasmid vectors combining several elements of various naturally occurring plasmids were therefore devised early in the development of cloning technique.
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15
Q

What is pBR322 ?

What are its characteristics ?

A
  • Three plasmids contribute to pBR322 (Bolivar, F., Rodriguez, R. et al., 1977)
  • Replicon - pMB1
  • Major advantages: combination of ampR and tetR, increase replication rate, “reduced” conjugal transfer, size (4361 bp)
  • rop gene: gene product stabilises RNAI–RNAII
  • Copy number: ~15-20/cell
  • Resistance: tet gene (R6-5), bla/amp gene (Tn3, isolated in London in 1963)
  • Identification of cloning: insertional inactivation
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16
Q

What are the three plasmids that contribute to pBR322 ?

A

pBR322 originally made of R1, R6-5 and pMB1.

  • R1 plasmid : Ampicillin resistance (ampR) is encoded by transposon Tn3 and its transposability was exploited in that is was allowed to “jump” unspecifically between DNA molecules
  • Plasmid R6-5 : Tetracycline resistance (tetR) is encoded by tet C in R6-5DNA rearrangements can be accomplished effectively by random or direct reassociation of mixtures of certain restriction fragments pBR313 to pBR322 (reduction in size and loss of 2 recognition sites for Pst I, reduction in size)
  • copy number ~ 15-20 but can be amplified by protein synthesis inhibitor chloramphenicol
  • In pBR327 another 1089 bp were eliminated
  • pBR327 copy number is 30-45 per E. coli but more importantly the deletion destroyed residual conjugative properties of pBR322 - important for biological containment
17
Q

When protein inhibitor chloramphemicol is added in a mdeium with bacteria, chromosomal DNA completes its current round of replication and ceases while plasmid DNA (ColE1) continues to replicate and increases its rate.
What does this mean ?

A

Protein synthesis is not required for plasmid replication.

18
Q

Whn chloramphemicol is added, as much as 3000 copies per cell of ColE1 can accumulate, representing a 125-fold increase over the normal 24 copies per cell.
What is the consequence of this significant proliferation ?

A

Plasmids might start taking up ribonucleatides instead of deoxyribonucleaotides.

19
Q

What is rafampicin ? How does it work ?
What effect does it have on plasmid replication ?
What is the implication of this ?

A

Rifampicin inhibits bacterial DNA-dependent RNA synthesis by inhibiting bacterial DNA-dependent RNA polymerase. Rifampicin binds to the pocket of the RNA polymerase β subunit within the DNA/RNA channel, but away from the active site. The inhibitor prevents RNA synthesis by physically blocking elongation, and thus preventing synthesis of host bacterial proteins. By this “steric-occlusion” mechanism, rifampicin blocks synthesis of the second or third phosphodiester bond between the nucleotides in the RNA backbone, preventing elongation of the 5’ end of the RNA transcript past more than 2 or 3 nucleotides.
Rifampicin stops replication of Col E1.
Thus, RNA synthesis is required for plasmid replication.

20
Q

What is ethidum bromide (EtBr) ?

A

EtBr = an intercalating agent commonly used as a fluorescent tag (nucleic acid stain) in molecular biology laboratories for techniques such as agarose gel electrophoresis

21
Q

Why is ethidum bromide a mutagen ?

A

Because ethidium bromide can bind with DNA, it is highly toxic as a mutagen. It may potentially cause carcinogenic or teratogenic effects, although no scientific evidence showing either health effect has been found. Exposure routes of ethidium bromide are inhalation, ingestion, and skin absorption.

22
Q

What is the effect of EtBr on plasmid structure ?

What is the maximum limit of EtBr binding to DNA ?

A

EtBr either relaxes -ve supercoiling or causes +ve supercoiling.
EtBr can bind DNA up to 1 EtBr/2 bases.

23
Q

How does ampicilin (Amp) work ?

How can a bacterium defend itself against ampicilin ?

A

Ampicillin inhibits bacterial cell wall synthesis by disrupting peptidoglycan cross-linking. ß-lactamase is secreted, which hydroylzes the ß-lactam ring of ampicillin.

24
Q

How does tetracycline (Tet) work ?

How can a bacterium defend itself against ampicilin ?

A

Binds to 30S ribosomal subunit and prevents association of the aminoacyl-t-RNA to the ribosomal acceptor A site inhibiting bacterial protein synthesis. The tetracyclin efflux proteins are membrane-associated proteins that recognise and export tetracyclin from the cell. Tetracyclin binding is reversible.

25
Q

How do Kanamycin (Kan), Neomycin (Neo), Streptomycin (Str) and Hygromycin (HyB) work ?
How can a bacterium defend itself against these ?

A

Aminoglycosides: Kanamycin binds irreversibly to 70S ribosomal subunit and interferes with translocation of aminoacly-t-RNA to peptidyl site inhibiting protein synthesis. The other aminglycosides interfere also with protein synthesis at the 30S ribosomal subunit streptomycin (tRNA selection), Hygromycin (translocation). They are all inactivated by aminophosphotransferase I or II transferring a gamma-phosphate from ATP.

26
Q

How does Zeocin (Zeo) work ?

How can a bacterium defend itself against Zeo ?

A

Belonging to the bleomycing family induces DNA strand scission (degradation) in the presence of Cu2+. It is inactivated by the Sh ble gene that binds the antibiotic in a 1:1 ratio.

27
Q

What is the basis of cloning by insertional activation ?

A
  1. Bacteria that contain no plasmid are sensitive to both ampicillin and tetracyclin.
  2. Desired colonies are those that do NOT grow on the tetracycline plates.
28
Q

What are the characteristics of pUC19 ?

A

Length: 2686 bp
Replicon: mutated pMB1
rop gene: deleted (RNAI-RNAII pairing not as stable as in BpBR322) –> easier for RNAII to prime replication
Copy number: ~500-700/cell (result of mutated pMB1 and deleted rob gene).
Resistance: bla gene (Tn3 –> amp) & deletion of tet gene
MCS: multiple cloning site introduced (removal of endonuclease recognition sites at other locations).
lacZ: regulatory sequence and coding information for the first 146aa of ß-galactosidase
Identification of clones: alpha-complementation (Blue/white screen) but there are requirement for E.coli host.

29
Q

What were the 7 first steps of the cloning experiment we carried out ?

A

Step 1: Selecting and isolating DNA from an organism of interest
Step 2: DNA is cleaved with restriction enodnucleases to generate inserts of suitable size for the cloning vector chosen.
Step 3: Fragments are inserted (DNA ligation) into the
vector of choice.
Step 4: Introduce (transform) the recombinant DNA (vector with insert) into bacteria. => library
Step 5: Grow bacteria on appropriated selection medium.
Step 6: Select colonies, grow bacteria.
Step 7: Isolated a plasmid DNA and analyse.

30
Q

Your are given the following scenario:
1. Lambda-phage grows well in E. coli C (EOP=1)
2. Lambda-phage (C) does not grow well in E. coli K
3. Survivor lambda-phage (K) now grow well on E. coli K
4. However that ability is lost after one grow cycle in E. coli C
Explain what has happened.

A 
Restricting host (E. coli K) reduced biological activity of 
lamdba-phage (C) caused by cleavages of infecting DNA by endonucleases. This process is called restriction => Restriction Endonucleases
Restricting host (E. coli K) modifies some of the infecting DNA, which allows them to grow well on E. coli (K), which is caused by DNA methylation and is lost if grown again in E. coli (C).
31
Q

What are the 4 main types of restriction endonucleases ?

What are their characteristics ?

A

Type I Subtype 1A-1D : Multisubunit proteins that function as single complex for recognition, cleavage, and methylation. Require ATP, Mg2+ and S-adenosylmethionine. Interact with two asymmetrical bi-partite recognition sites, and cut half-way between cutting site not clearly defined.

Type II 11 subtypes : Endonucleases and methylases are separate enzymes. The endonucleases recognize symmetric sequences, palindrom, and cleave at constant positions at or close to that sequence to produce 5’-phosphate and 3’-OH. Most of them require only Mg 2+ ions as a cofactor.

Type III : Multisubunit proteins that require ATP & Mg2+. Enzyme must interact with two copies of non palindromic recognition sequence and the sites must be in inverse orientation.

Type VI : The enzymes cleave only modified DNA, recognition site not well defined

32
Q

The recognition sequences of restriction enzymes are palindromic.
What does this mean ?

A

This means that the sequence reads the same on both strands starting from the 5’ end.

33
Q

How are restriction enzymes named ?

Give two examples.

A
  1. 1st letter of the genus
  2. 1st two letters of the specific epithet.
  3. Followed by a particular strain (if used).
  4. When host strain has several different restriction and modification systems, these are identified by a roman number
    E.g : Escherichia coli: EcoRI
    Haemophilus influenzae: HindIII
34
Q

What are the cuts produced by EcoRI, MluI and HindIII ?

A

EcoRI : 5’- G/AATTC - 3’
MluI : A/CGCGT
HindIII : A/AGCT

35
Q

What do sticky ends facilitate ?

A

Sticky ends make the joining of DNA molecules more efficient

NEUR2001 (14 decks)

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