Plasmids and their uses

Question 1

What is the detection limit for dsDNA in gel electrophoresis ?

Answer 1

A single band at 10ng.

Question 2

What are the consequences of DNA overloading ?

Answer 2

Railing and smearing, more severe with longer fragments

Question 3

What are the three main purposes of the gel loading buffer ?

Answer 3

• Increase density (sucrose, glycerol, Ficoll) of sample (=>evenly distributed in well)
• Aid loading by adding colour
• Add mobility dye(s) Bromphenol Blue/Xylene cyanol FF/Orange Green to monitor electrophoretic process

Question 4

Why are the smaller fragments harder to identify ?

Answer 4

Smaller fragment = smaller region to which the DNA-binding dye attaches to –> less intense signal under UV light

Question 5

What is the structure and function of alkaline phospohatase Calf Intestinal (CIP) ?

Answer 5

AP = = enzyme that dephosphorylates 5’ and 3’ end of DNA, RNA, dNTPs and proteins –> prevents self-ligation of plasmids
Dimeric metalloenzyme of two identical subunits, each subunit contains two zinc ions and one magnesium ion

Question 6

What is the optimal pH of AP ?

Answer 6

An alkaline solution is optimal, with a pH optimum at 8 - 10 but can be used at reduced activity at (pH = 7.5-9).

Question 7

1 unit of CIP will dephosphorylate ~ 1 pmole of 5’ phosphorylated termini at 37 ̊C within 30 minutes.
You have 1 μg of a 3 kb plasmid and the CIP has a concentration of 0.2 U/μl.
How much enzyme do need to add?

Answer 7

Average MW of 1bp ~ 650 Daltons = 650 g/mol
MW of our 3 kb fragment = 3,000 * 650 = 1.95e6
Number of moles : (1e-6) / (1.95e6) = 5.13e-13 moles = 0.513 pmol
Number of units needed: 0.513 (1 U for 1 pmol)
0.513 / 0.2 = 2.565 uL of enzyme
Because we have 2 DNA strands, we need 2.565 * 2 = 5.13 uL of enzyme

Question 8

What is Southern blotting ?
Northern blotting ?
Western blotting ?

Answer 8

Southern blotting (named after Sir Edwin Southern) = identification of DNA
Northern blotting = identification of RNA
Western blotting = identification of proteins

Question 9

What are the three main gel blotting techniques ?

What are the advantages and limits of each ?

Answer 9

Capillary blotting:
- transfer time = 2-16hrs
- membrane = nitrocellulose or nylon
- equipment = standard
- resolution = excellent (short transfer) 
Vacuum blotting:
- transfer time = 0.5-1h
- membrane = nylon
- equipment = special
- resolution = excellent (optimal vacuum)
Electrophoretic blotting:
- transfer time = 2-16h
- membrane = nylon
- equipment = special
- resolution = good (external cooling)

Question 10

What is the size limit of inserts accepted by plasmids ?

How does this affect their use ?

Answer 10

<10kb (nat occurring multi-copy plasmids)

Used for sub-cloning and downstream manipulation, cDNA cloning and expression assays

Question 11

What is the size limit of insert accepted by phages ?

How does this affect their use ?

Answer 11

5-20kb (e.g. bacteriophage lambda)

Used for genomic DNA cloning, cDNA cloning and expression libraries

Question 12

What is the size limit of insert accepted by cosmids ?

How does this affect their use ?

Answer 12

35-45kb (e.g. plasmid containing a bacteriophage lambda cos site)
Used for genomic library constructions.

Question 13

What is the size limit of insert accepted by BACs ?

How does this affect their use ?

Answer 13

75-300kb (e.g. E. Coli F factor plasmid)

Used for the analysis of large genomes.

Question 14

What is the size limit of insert accepted by YACs ?

How does this affect their use ?

Answer 14

100-1000kb (e.g. S. Cerevisiae centromere, telomere, and autonomously replicating sequence)
Used for analysis of large genomes, construction of YAC transgenic mice.

Question 15

What is the size limit of insert accepted by MACs ?

How does this affect their use ?

Answer 15

100kb to > 1 MB (e.g. mammalian centromere, telomere and origin of replication)
Under development for use in animal biotechnology and human gene therapy.

Question 16

What are the cos ends of the lambda genome ?
How are they produced ?
What are their properties ?

Answer 16

cos ends = 12 nucleotide 5’ overhangs of linear DNA are the result of a cut by the enzyme terminase, which is encoded by itself and acts like a restriction enzyme during the replication of the phage DNA. The terminase is an endonuclease specific for the cos-site in multimeric phage DNA. The ends of the resulting monomeric DNA are similar to cohesive ends produced by common restriction enzymes. Because they have long 5’ overhangs cos ends are much more sticky than the cohesive ends generated by restriction enzymes.
cos end : 5’ - G/GGGCGGCGACCTC - 3’

Question 17

What is the function of DNA ligase ?

Answer 17

Joins 5’-phosphate to 3’-OH (phosphodiester bond) within

double stranded DNA using Mg2+ and ATP or NAD+.

Question 18

What is the difference between E. Coli DNA ligase and T4 DNA ligase ?

Answer 18

E. Coli DNA ligase : uses NAD+ as a cofactor, joins only cohesive ends
T4 DNA ligase : uses ATP as a cofactor, joins cohesive and blunt ends of DNA, oligonucleotides, RNA and RNA-DNA hybrids

Question 19

What are the four steps of the ligation reaction ?

Answer 19

Step 1 : DNA ends have to collide by chance and stay together long enough for the ligase to join them => easier at low T ̊C.
Step 2 : Self-adenylylation of an active site lysine using ATP or NAD+ (depending on the ligase) release of pyrophosphate (PPi)
Step 3 : Transfer of the adenylyl group to the 5’-phosphorylated terminus (donor) forming apyrophosphate bond.
Step 4 : Nucleophilic attack of 3-hydroxyl group (acceptor) on 5’- phosphate group (donor), loosing the AMP, forms
phosphodiester bond.

Question 20

Are sticky ends produced by the same restriction enzyme compatible.
What about ends formed by different restriction enzymes ?

Answer 20

Same enzyme = frequently yes

Different enzyme = sometimes, e.g. BamHI and Sau3A, PstI and Nsil