0304 Mid Mark

BIO4320 Molecular Biology & Genetic Engineering Name: S.I.D.

Oct. 28, 2003

ANSWER ALL QUESTIONS Question 1 The figure on the right shows a diagrammatic representation of the lambda ZAP vector. During the excision process, XL1-Blue MRF’ and SOLR are employed as the first and second bacterial hosts, respectively. (a) What will happen if we wrongly use the SOLR strain as the first host? (1 mark) No λ or fl helper phage infection. (b) What will happen if we wrongly use the XL1-Blue MRF’ strain as the second host? (1 mark) fl helper phage continues to propagate and being secreted into the growth medium. (c) What will happen if the F’ plasmid is lost from the XL1-Blue MRF’ strain? (1 mark) No F pili and hence no fl helper phage infection. (d) What will happen if the ∆M15lacZ region on the F’ is deleted? (1 mark) Cannot perform blue-white screening even in the presence of X-gal and IPTG. (e) What will happen if the I (initiation site for fl phage packaging) reciprocally changes position with T (termination site for fl phage packaging)? ( 1 mark) The fl phage coat will pack another strand of the DNA flanked by I and T. After all subsequent processes, double-stranded plasmids will be formed but the clone will be in opposite orientation. (f) Will the lambda ZAP vector better be used to construct cDNA library or genomic library? Why? (1 mark) cDNA library is better because the plasmid portion occupied some of the packaging capacity. 1

Question 2 A student tried to amplify a highly G-C rich DNA template using PCR. (Note: ladders on the first and last lanes of the gels are molecular weight markers. The single bands on other lanes are the PCR products.) (a) Based on the results shown in the figure on the right, propose a possible function of DMSO (1 mark) DMSO may help to reduce the annealing at G-C rich regions and hence allows better PCR. (b) The student concluded that the FastStart Taq DNA Polymerase exhibited a higher enzymatic activity than the regular Taq DNA Polymerase, when G-C rich templates were used. Explain whether you agree with the student’s conclusion. (3 marks) The conclusion may not be correct (0.5), since PCR products cannot be formed without DMSO, despite of the type of Taq enzyme used (1). DMSO may have some inhibitory effects to the enzymes (1). The FastStart Taq enzyme may be more resistant to these inhibitory effects by DMSO (1). If the answer is “agreed”, check to see if valid reasons were given.

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Oct. 28, 2003

(in PCR reactions)

0% DMSO

No products

BIO4320 Molecular Biology & Genetic Engineering Name: S.I.D.

BIO4320 Molecular Biology & Genetic Engineering Name: S.I.D.

Oct. 28, 2003

Question 3 A patient is infected by a pathogenic bacterium. There are two subtypes of this bacterium: strain A and strain B. For proper treatments, the doctor needs to check whether the patient is infected by strain A or strain B. The genome of strain A is identical to that of strain B except the presence of a 40-bp region in strain A. The sequence of the 40-bp region is: 5’-ATCCGGTACATTAGCCGATTGGCGATTTTAACCGGCTTTC-3’ 3’-TAGGCCATGTAATCGGCTAACCGCTAAAATTGGCCGAAAG-5’ Suppose a laboratory is provided with the following materials: (i) The patient’s serum sample that contains only a trace amount (NOT enough for detection of the pathogenic bacterium by hybridization). (ii) Four radioactively labeled primers with the following sequences: Primer A: 5’-ATCCGGTACATTAGCCGATT-3’ Primer B: 5’-GGCGATTTTAACCGGCTTTC-3’ Primer C: 3’-TAGGCCATGTAATCGGCTAA-5’ Primer D: 3’-CCGCTAAAATTGGCCGAAAG-5’ (iii) A heat stable DNA ligase that can seal nicks on double-stranded DNA molecules. (Remark: NO heat stable DNA polymerase is provided). (iv) A thermal cycle machine. (v) Reagents for running a polyacrylamide gel (i.e. able to separate DNA molecules of small molecular weight). (vi) X-ray films and the film developing machine. Design a strategy to diagnose whether the patient is infected by strain A or strain B. Use diagrams whenever appropriate (7 marks)

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BIO4320 Molecular Biology & Genetic Engineering Name: S.I.D.

Oct. 28, 2003

(1) Prepare DNA sample (0.5); (2) Mix DNA sample (0.5), all four primers (0.5), and heat stable DNA ligase (0.5); (3) Heat to denature double-stranded DNA template (0.5) and then anneal the primers to the DNA templates (0.5); (4) The DNA ligase will seal the nick between two primers that hybridize side-by-side to the template DNA (1); (5) Primer A + Primer B and Primer C + Primer D will become Primer A–Primer B and Primer C–Primer D fragments, respectively (1); (6) Primer A–Primer B and Primer C–Primer D fragments will become templates for the ligation reaction in the next cycle (1); (7) Repeat the cycle of heating, annealing and ligating (1); (8) Run on polyacrylamide gel, Southern blot, detection by X-ray autoradiography (1); (9) If the sample gives a 40-bp band, it contains strain A (1). Question 4 The same sample from Question 3 was sent to a different laboratory. In this laboratory, only heat stable DNA polymerase but not heat stable DNA ligase is available. Also, this laboratory only has reagents for running an agarose gel (i.e. NOT able to separate DNA molecules of small molecular weight). Other reagents and facilities are the same as that in Question 1. You can synthesize more primers if needed. Design a strategy to perform the diagnosis in this laboratory. (3 marks) (1) Choose one primer from A, B, C, or D (1); (2) Synthesize another primer away from the 40-bp region (1) and this new primer should orient in an opposite direction to the primer chosen above (0.5); (3) Run on agarose gel and perform Southern blot to detect DNA band of expected size (1); (4) If the sample gives positive results, it contains strain A (0.5). Remarks: if use two primers flanking the 40-bp region, give 1 mark only (since the differences resulting from strain A and strain B cannot be seen using agarose gel.

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BIO4320 Molecular Biology & Genetic Engineering Name: S.I.D.

Oct. 28, 2003

Question 5 A new 5’-RACE method was invented to specifically target full-length mRNAs. In eukaryotes, each mRNA contains a poly A tail at the 3’ end as well as a G-cap (m7G-p-p-p) at the 5’end. A truncated mRNA does not contain a G-cap, but ended with a 5’ phosphate group (p). The following enzymes will be used: (i) (ii) (iii)

CIP (calf intestinal phosphatase): removes 5’ phosphate but not G-cap; TAP (pyrophosphatase): removes G-cap to expose the 5’ phosphate group; RNA ligase: ligate the 3’-OH group of a RNA molecule to the 5’ phosphate group of a RNA molecule.

You are provided with a RNA oligo with known sequence. (a) Draw a flow chart to illustrate the sequence of event leading to the addition of the RNA oligo to the 5’ end of only full-length but not truncated mRNA molecules). (6 marks) (1) Add CIP to remove 5’ phosphate groups from truncated mRNA (1); (2) Add TAP to remove G-cap (1) and expose 5’ phosphate groups from full-length mRNAs (1); (3) Add RNA oligo and RNA ligase (1); (4) RNA oligo will only ligate to full-length mRNAs but not truncated mRNAs (1) because truncated mRNAs do not 5’ phosphate groups (1); (b) Propose the subsequent steps for this enhanced 5’-RACE. (Do not forget that you now have an additional known sequence added to the 5’ end of mRNAs.) (4 marks) (1) Reverse transcription using GSP1 and reverse transcriptase (1); (2) RNaseH to remove mRNA (0.5); (3) Making second strand using a primer designed based on the sequence of the RNA oligo (0.5); (4) First round PCR using GSP2 and a primer designed based on the sequence of the RNA oligo (1); (5) If necessary, second round PCR using nested GSP and a primer designed based on the sequence of the RNA oligo (1).

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