Vectors

Question 1

Cosmids

Answer 1

Plasmids with med/high copy number origin + lambda packaging sequence (Cos ends)

• 8kp
• introduced into E.coli by transduction

Question 2

An example of a cosmid

Answer 2

SupercosI -> selection marker for eukaryotic cells with a SV40 origin

Question 3

What are the benefits of using a cosmid?

Answer 3

Injects and circularises like phage DNA but replicates as a normal plasmid without phage functions

Question 4

BACS

Answer 4

Plasmid with origin of F1 - plasmid with copy number 1

• contains genome inserts e.g. sacB replaced lacZ or insertion of transposon
• includes lambda cos N + PI loxP sites + restriction enzyme sites
• can be transferred efficiently without packaging

Question 5

YACS

Answer 5

Plasmid with additional autonomously replicating sequence

-> generates 2 arms that ligate to genomic DNA -> linear construct

Question 6

Benefit of YACs

Answer 6

No hard size limit, introduced by electroporation

Question 7

YAC vs BAC

Answer 7

YAC’s are less stable and more labour-intensive so slower than BACs. Both can be manipulated by homoogous recombination but easier in YAC’s.

Question 8

Construct design for high-level eukaryotic expression:

Answer 8

1. Strong + constituive promoter
2. Intron included
3. PolyA signal -> defined 3’ end to mRNA + expor mRNA
4. Remove unnecessary untranslated sequences (au reduce stability)
5. Optimisation of transgene for translational efficiency e.g. AUG in kozak

Question 9

Benefits of using yeast:

Answer 9

• easy to handle in eukaryotic system
• overproduction of proteins -> research and commercial value
• able to clone large pieces of DNA

Question 10

cos ends:

Answer 10

12 nucleotide 5’ overhangs of linear lambda DNA cut from terminase -> long 5’ overhang means stickier than sticky ends

Question 11

Ligase:

Answer 11

joins 5’ phosphate to 3’OH (phosphodiester bond) within double-stranded DNA using mg2+ ATP/NAD+

Question 12

E.coli ligase uses

Answer 12

NAD+

Question 13

T4 DNA ligase uses

Answer 13

ATP -> joins more end types

Question 14

Steps of ligation

Answer 14

1. DNA ends collide by chance (higher chance at lower temp)
2. ATP/ NAD+ used to NH-AMP bond to produce PPi
3. Adenyl group is transferred to 5’ phosphorylated donor
4. nucleophilic attack of 3’OH on 5’ donor forms phosphodiester bond

Question 15

In bactoinfection

Answer 15

DNA packaged inside a bacterium -> usually transient -> DNA degrades if not integrated into nuclear genome but doubling rate fast

Question 16

Regulon

Answer 16

> 1 operon is under control of a single regulatory protein

Question 17

LacZ consists of

Answer 17

4 identical sub-units, 2 loop at 5’end

Question 18

If lactose is not hyrdolysed it’s ______

Answer 18

acetylated by galactoside transacetylase coded by lacA

Question 19

LacZ lies next to ___ then ____

Answer 19

Lac Y then Lac A

Question 20

Lac Y functions

Answer 20

lactose permease

Question 21

Lac I codes for _____

Answer 21

lac repressor -> ihibitory -> negative control

Question 22

Characteristics of lac repressor

Answer 22

-tetramer with a sigma 70 promoter

Question 23

How does lac repressor bind DNA

Answer 23

binds to O2 +O3/ O1 + O3 non-specifically -> binding prevents RNAp binding and open conformation

Question 24

How is the gene activated by CRP-CAP

Answer 24

cAMP - CRP forms in absence of glucose and binds C site near promoter -> more binding sites for CRP-cAMP which activate promoters (signalling) so RNAp binds at promoter with higher affinity

Question 25

When lactose is present a small number of lactose does what to aid operon

Answer 25

converts to allolactose which binds tac repressor to decrease affinity

Question 26

Each sub-unit of lac repressor contains an

Answer 26

alpha helix-turn-helix only when DNA present at ends of from hinge region

Question 27

When bound repressor shape is :

Answer 27

1 alpha helix lies in major groove where amino acid side chains interact with bp of operator sequence