0001 Mid Mark

BIO4320 Mid-Term Examination Name:

Oct. 24, 2000 Student ID:

Part II (Dr. Lam’s Portion; total 32 marks)

Answer all questions.

Question 1 A special lambda phage vector (λYES) was constructed as shown in the following diagram: λ Arm

λ Arm

Promoter GAL1 lox sequence

ori

ARS1

Promoter lac

Apr

lox sequence

URA3

Single Cloning Site: XhoI ori: plasmid replication origin (ColE1 type) that functions in E. coli ARS1: replication origin that works in yeast Apr: ampicillin (an antibiotic) resistance gene Promoter GAL1: yeast promoter inducible by galactose Promoter lac: E. coli promoter inducible by IPTG URA3: yeast URA3 gene required for uracil biosynthesis

lox sequence: Cre-lox system is a site specific recombination mechanism found in P1 phage (P1 phage is not a filamentous phage); two lox sequences will recombine in the presence of the Cre protein

lox

lox

+ Cre Protein

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lox

BIO4320 Mid-Term Examination Name:

Oct. 24, 2000 Student ID:

A modified λYES vector containing an amber mutation in a coat protein gene was used to generate a cDNA library of mammalian cells. (a) What are the two major properties of the E. coli strain used to amplify the library. (2 mark) It should be a suppressor strain (1) and free of Cre protein activities (1). (b) A putative mammalian cell cycle gene was identified in the above cDNA library. You are provided with a clone of the Cre gene. What strategies will you use to convert the λ phage clone into a plasmid clone AND propagate it in E. coli? (5 marks) Choose a non-suppressor strain (1) and transform it with a clone containing the Cre gene (1). Infect the resulting strain with the λ phage clone (1). The Cre protein will release the circular (0.5) plasmid DNA between two lox sites (0.5) on the λ phage DNA (1). The λ phage cannot lyse the cell because the amber mutation (0.5) it carries will prevent protein coat synthesis (0.5) in a non-suppressor strain. Selection of the bacteria containing the desire plasmid clone can be done using the ampillicin resistant marker (1).

(c)

Draw a diagram representing the plasmid clone resulting from (b). (2 marks) Putative Promoter mammalian GAL1 cell cycle gene Ap

Marking Note Promoter lac

r

XhoI

XhoI

Reduce 0.25 mark for each missing item; reduce 0.5 mark for 2 lox sites Wrong arrangement: no mark; Only lox site on a circular plasmid: 0.5 mark

ARS1

Ura3 ori

lox

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BIO4320 Mid-Term Examination Name:

Oct. 24, 2000 Student ID:

The mammalian cell cycle gene in the plasmid clone obtained in (c) was sequenced and found to locate at the right orientation relative to the lac promoter (i.e. sense RNA will be generated using the lac promoter). With only the information given above, describe how you will reconstruct the clone so that the mammalian cell cycle gene is now located at the right orientation relative to the yeast GAL1 promoter on the plasmid. (4 marks) Cut out the mammalian cell cycle gene from the plasmid clone using XhoI (1). Purified the vector and cDNA by gel electrophoresis (0.5). Remove the 5’ phosphate group from the cut vector by phosphatases (0.5) to prevent self-ligation of vector (0.5). Ligate the cDNA fragment with the cut vector (1). Since it is a non-directional cloning (1), the cDNA fragment can be inserted in both orientation (1) and some will at the right orientation as that of the promoter GAL1 (1). OR Cut out the mammalian cell cycle gene from the plasmid clone using XhoI (1). Generate probes using the target genes (1) and used the probes to re-screen the λ phage library mentioned above (1). λ phage clones with positive signals will be purified (0.5) and plasmid clones will be obtained as described in (b) (1). Clones with desired the orientation will be identified by DNA sequencing (0.5) or restriction mapping (0.5).

(d) Since the function of the mammalian cell cycle gene cannot be tested in E. coli, the new plasmid clone constructed in (d) was transformed into yeast for further experiments. Describe how you can select for the successful transformation event. (3 marks) Transform the plasmid into a URA3- mutant (1). Since the plasmid carries a normal URA3 gene (1), successful transformants can be selected as survivors (0.5) on uracil free medium (1; 0.5 for minimal medium).

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BIO4320 Mid-Term Examination Name:

Oct. 24, 2000 Student ID:

Question 2 A cDNA fragment encoding a glutelin protein from wheat was cloned. No EcoRI site was found in this cDNA fragment. The wheat genomic DNA was isolated and cut with EcoRI before separated on an agarose gel. Southern blot experiments were performed using high stringency conditions that do not allow mismatch. (a) If no DNA sequence information is available, which method you will use to generate probes based on this cDNA fragment. (1 mark) Random primed labeling (1) or nick translation (1).

(b) Two (instead of one) bands were found to hybridize to the probes generated in (a). Explain this observation. (1 mark) Presence of an intron (0.5) with a EcoRI site (0.5) on the genomic region spanned by the cDNA fragment (0.5). OR More than one copy of the gene existed in the genome (1). (c) With only the information mentioned above, describe the key experimental conditions that you will use to check if similar glutelin genes exist in rice. (3 marks) Extract genomic DNA from rice (0.5) and perform Southern blot (0.5; accept description of procedure without stating the term Southern blot). Using the wheat glutelin gene as probe (0.5), low stringency hybridization and washes (1) will be performed. Low stringency conditions can be achieved by lowering the hybridization temperature (0.5), increase the NaCl concentration (0.5), or reducing the amount of formamide (0.5).

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BIO4320 Mid-Term Examination Name:

Oct. 24, 2000 Student ID:

Question 3 RNA samples were prepared from normal cells and cancer cells respectively. After reverse transcription, the cDNAs were subject to real time PCR measurement. Two pairs of primers, using for the amplification of an unknown gene “X” and a housekeeping gene “Y” respectively, were tested. The following results were obtained. Standard Curve Sample Normal Cancer Normal Cancer

Number of Cycle 16 12 18 16

Number of PCR Cycle

Gene X X Y Y

20 18 16 14 12 10

1.0 0.1 0.01 0.001 Relative Relative Amount Amount of ofPCR PCRProducts Templates

(a) Describe the relative expression of Gene “X” in normal cells and cancer cells. (4 mark) In normal cells, expression of Gene X relative to Gene Y = 0.1/0.01 = 10 (1). In cancer cells, expression of Gene X relative to Gene Y = 10/0.1 = 100 (1). Using Gene Y as the reference (0.5), expression of Gene X in normal cells relative to cancer cells = 10/100 = 0.1 (1). Therefore, Gene X is induced in cancer cells as compared to normal cells (1).

(b) Explain why it is important to measure the expression of Gene “Y”. (2 marks) Gene Y is a housekeeping gene that will express at relatively constant level (1) in different cell types. Gene Y thus provides a reference point for comparison (0.5) and will eliminate experimental artifacts during tissue collection (0.5), RNA preparation (0.5), and reverse transcription (0.5).

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BIO4320 Mid-Term Examination Name:

Oct. 24, 2000 Student ID:

Question 4 An AFLP (amplified fragment length polymorphism) experiment was performed using the enzyme EcoRI and MseI. EcoRI

MseI

BamHI

5’GAATTC3’ 3’CTTAAG5’

5’TTAA3’ 3’AATT5’

5’GGATTC3’ 3’CCTAAG5’

(a) If BamHI was used instead of MseI in the first step of the experiment, what problems will occur? Explain. (2 marks) BamHI is a 6 bp cutter and MseI is a 4 bp cutter (0.5) and hence restriction by MseI will be more random (0.5) and more fragments will be generated (0.5) compared to the restriction by BamHI. Therefore, the number of bands in the final gel electrophoresis would be decreased if BamHI was used (1).

(b) What subsequent step can be altered to minimize the problem caused in (a)? Explain. (3 marks) One way to increase the number of bands in the final gel electrophoresis is to use a less selective primer (1) during the PCR steps involved. For example, the Primer (+2) can be used instead of Primer (+3) (1) during the selective amplification step (1).

- End of Part II -

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