Molecular Biology Quiz2

Molecular Biology Quiz 2-2007
Total: 120’
1. What are the components required for a cloning vector? (10’)
A vector should have:
(1) An origin of replication
(2) A selectable marker
(3) Multiple cloning sites
Total 10 points: minus 3’ for omitting one key point
2. How to clone a gene, express it in E. coli and purify the expressed protein from
E. coli? (25’)
⑴ How to clone a gene:
1 Cut the cloning vector and target DNA with the same restriction enzyme(s) (insert
preparation) (2.5’)
2 Use DNA ligase to ligate the vector and target DNA to form recombinant DNA (ligation)
(2.5’)
3 Transform the ligation product into the E.coli competent cells (transformation) (2.5’)
4 Grow the transformed cells on the corresponding selective medium to identify the
recombinant plasmid. (selection) (2.5’)
Additional points: (1’ bonus for each)
1 Obtain target DNA by PCR
2 Restriction enzyme digestion to confirm the presence of target DNA in the plasmid
3 Using double restriction endonucleases to digest the vector DNA and the target DNA to
allow directional cloning
⑵ How to express the gene:
1 Subclone the gene into an expression vector (4’)
2 An expression vector should contain a promoter and a terminator for transcription (2’)
3 The inserted DNA fragment should have a start codon and a stop codon for translation.
(1’)
⑶ How to purify the expressed protein:
A sequence encoding a His-tag should be inserted at the N- or C-termini of the target ORF,
which allows purification of the fusion protein by binding to the Ni2+ column (Affinity
Chromatography). (8’)
Alternative: for each protein purification technique, plus 2’; the total points ≦5’
3. Please describe a method that you can use to detect the mRNA level of your
interested gene in human cells. (10’)
I. RT-PCR (10’)
In the first step (RT step), the cDNA is copied from the mRNA template by using dNTPs,
the reverse transcriptase and the oligo(dT) primer. (5’)
In the second step, the synthesized cDNA is amplified by PCR to an easily detectable level,
using the DNA polymerase and the upstream and downstream DNA primers specific for the gene.
(5’)

Ⅱ . Northern blot (10’)
(1) The mRNA is extracted from the cells;
(2) The mRNA is loaded onto a denatured agarose gel for electrophoresis;
(3) Transfer mRNA onto a nylon membrane;
(4) A labeled probe specific for the RNA fragment is produced and incubated with the membrane,
allowing hybridization with the target mRNA;
(5) The membrane is washed and exposed to an X-ray film or a phosphor screen to visualize the
signal.
4. How the replication errors are repaired in E. coli? (10’)
Using mismatch repair:
(1) MutS scans the DNA, recognizing the mismatch. (3’)
(2) The complex of MutS and the mismatch-containing DNA recruits MutL and MutH. MutH
makes a nick on the newly synthesized strand of the DNA. (3’)
(3) A helicase and an exonuclease then unwind and digest this mismatch-containing DNA segment.
(2’)
(4) The gap is then filled by DNA polymerase Ⅲ and sealed by DNA ligase. (2’)
5. What kind of DNA damage can be caused by UV light irradiation? Please
describe two mechanisms that bacteria can use to repair such damage. (25’)
UV radiation can cause two adjacent pyrimidines to form a pyrimidine dimer (typically, a thymine
dimer). (5’)
Two mechanisms that can repair this kind of damage:
(1) Direct reversal by photoreactivation. (4’) DNA photolyase uses energy from light to break the
covalent bonds linking the adjacent pyrimidines. (6’)
(2) Nucleotide excision repair. (4’) UvrA and UvrB complex scans the DNA to find the distortion
caused by the pyrimidine dimer (UvrA responsible for detection), and UvrB unwinds the DNA
around while UvrA leaves the complex. UvrC is recruited and creates nicks at the both sides
of the lesion. UvrD (DNA helicase) releases the single stranded fragment and DNA Pol I seals
the gap with the DNA ligase. (6’)
Note: Base excision repair is not appropriate here. And if you mention the translesion synthesis,
you should point out that this damage has accumulated a lot in E. coli and has been a threat to
DNA replication, because under normal circumstances E. coli does not resort to this SOS
response.
6. What are homologous recombination models? Please describe the difference
between these two models. (12’)
The Holliday model, (3’) the double-strand break repair model. (3’)
The main difference of these two models:
In Holliday model, recombination is initiated by the introduction of a nick in each of the two
aligned DNA molecules at an identical location. But in the DSB repair model, the initiating event
is the introduction of a double-stranded break (DSB) in one DNA molecule. (4’)
What’s more, there is one Holliday junction in Holliday model but two in the DSB repair model.

(2’)
7. Please describe the bacterial and eukaryotic proteins that involve in
homologous recombination. (18’)
I. In bacteria:
(1) RecBCD helicase/nuclease processes broken DNA ends to generate a 3’ ssDNA tail for
recombination ( 答 出 具 体 χ site 的 功 能 再 加 1’). (3’)
(2) RecA protein binds to the 3’ ssDNA tail and promotes strand invasion. (3’)
(3) RuvAB complex promotes branch migration. (2’)
(4) RuvC resolves the Holliday junction. (2’)
II. In eukaryotes:
(1) The Spo11 protein introduces DSBs in chromosomal DNA at many locations to initiate meiotic
recombination. (3’)
(2) MRX protein processes the cleaved DNA ends for assembly of the RecA-like strand–exchange
proteins. (3’)
(3) Dmc1 is a RecA-like protein that specifically functions in meiotic recombination. Rad51 has a
similar function. ( 答 出 一 个 即 2’)
8. Describe the mating type switching mechanism in yeast. (10’)
Mating type switching: (synthesis-dependent strand annealing, SDSA)
(1) Alignment of MATαlocus with HMR locus, or MATa locus with HML locus by homologous
pairing. (3’)
(2) The HO endonuclease makes a double strand break at MAT locus. (3’)
(3) MRX protein processes the DSB to generate a 3’ ssDNA tail, followed by Rab51-dependent
strand invasion. (1’)
(4) Synthesis of two new DNA strands from the HMR or HML locus, resulting in the switch
of a orαinformation at MAT locus to the opposite one. (3’)
Score Distribution Chart
25
21
Average: 88.6’
20
18
18
15
10
5
6
10
6 6
0
110-120 100- 90-99.5 80-89.5 70-79.5 60-69.5 0-59.5
109.5