EXAM Q

Question 1

Which technique could indicate whether a particular cloned DNA sequence was presented in a single copy in the genome

Answer 1

Southern blotting:combines transfer of electrophoresis-separated DNA fragments to a filter membrane and subsequent fragment detection by probe hybridization

Question 2

Transcription start points can be precisely identified by

Answer 2

Identification of TFIID binding sites

Question 3

Bacterial mRNA is UNLIKELY to contain

Answer 3

polyadenylation signal

Question 4

Eukaryotes 5’ cap and 3’ polyA tails are added to

Answer 4

Pol II transcripts

Question 5

RNA editing

Answer 5

Primary nucleotide sequence of RNA is altered post-transcriptionally

Question 6

Bacterial promoter likely to contain

Answer 6

Region which will be recognised by sigma factors

Question 7

Enhancers

Answer 7

Act in Cis
Short (50-1500 bp) region of DNA that can be bound with proteins (activators) to activate transcription of a gene or genes

Question 8

Histone deacetylase activities are often associated with

Answer 8

Repressors of gene expression
remove acetyl groups from an ε-N-acetyl lysine amino acid on a histone, allowing the histones to wrap the DNA more tightly

Question 9

Yeast 2-hybrid analysis

Answer 9

molecular biology technique used to discover protein–protein interactions

Transcription factor is split into two separate fragments, called the binding domain (BD) and activating domain (AD).

1. Gal4 transcription factor gene produces two domain protein (BD and AD), which is essential for transcription of the reporter gene (LacZ).
2. Two fusion proteins are prepared: Gal4BD+Bait and Gal4AD+Prey. None of them is usually sufficient to initiate the transcription (of the reporter gene) alone.
   D. When both fusion proteins are produced and Bait part of the first interact with Prey part of the second, transcription of the reporter gene occurs.

Question 10

Cyclins

Answer 10

proteins that are synthesized and degraded at specific times within each cell cycle and regulate protein kinases.

Question 11

Processed pseudogenes

Answer 11

Do not contain introns

Question 12

What would not be expected to contain intronic sequences

Answer 12

Plasmid cDNA library

Question 13

M female x P male –> progeny that are sterile but otherwise visibly normal DUE TO

Answer 13

INSERTION of P element in fertility genes

Question 14

C1 protein acts as

Answer 14

neg regulator of Prm and Pr

Question 15

C1

Answer 15

Encodes lambda repressor

Mutation: prevents PR and PL from being repressed so expression of lysis (late) genes cannot be prevented

Question 16

CII

Answer 16

Positive regulator of PRE (required to initiate CI) and PI (expression of int)
Mutation: CII prevents C1 repressor being established and prophage from integrating into E.coli genome

Question 17

CIII

Answer 17

Binds and stabilises C2 protein

Mutant: Increased degradation of C2 and decreased expression of C1 and Int

Question 18

Clear plaque

Answer 18

WT plaques are turbid - represent lysis (plaque)and lysogeny (turbidity)
Mutants with clear plaques cannot form lysogens

Question 19

Integration and excision

Answer 19

Formation of lysogen: High ratio of Int:Xis to allow integration of phage genome by Int
When entering lysogen PL (can express Xis and Int) is being repressed by CI repressor
At same time, PI activated by CII (with CIII). PI
is situated within the Xis gene and PI transcripts express only Int.

Question 20

Lac operon CAP AND adneyl cyclase (cya)

Answer 20

E.Coli uses glucose preferentially as a C source over 2o sources (lactose). If glucose is present, Lac operon is expressed at basal level even if lactose is present
CAP: positive regulator, cAMP = inducer
CAP and cAMP bind to upstream site and interact directly with RNAP –> increase transcription initiation
Presence of gluocse: cAmp is low, CAP binding and Lac expression are low
In CAP mutant: Lac expression always low, unable to metabolise lactose
Adenyl cyclase forms cAMP from ATP. Mutant lacks cAMP and CAP cannot bind and activate LAC expression

Question 21

TrpR regulatory mechanism

Answer 21

trpLD deletion mutation deletes attenuation mechanism (transcription terminator) resulting in increase in expression of trp operon (5 fold)
Mutation in trpR –> constitutive expression of trp operon (60 fold)
Make a strain containing both mutatations. If 2 mechanisms are acting independently, expect 300 fold increase in expression

Question 22

2 step gene replacement - HIS 4

Answer 22

Regulated by aa availability

Created set of deletions in HIS4 promoter