Biology Questions

Answer all in red.

Question 1

1) Two strains of E. coli, B and K, can be infected by wild-type (WT) and rapid lysis mutant (RL) T4 bacteriophages.

 

a) A lawn of bacteria on a petri dish is infected by a mixture of WT and RL bacteriophages. Assuming a high bacteria strain B/bacteriophage mixture ratio is used for the experiment, what is (are) the size (s) of plaques (large or small) produced? In your explanation includes why RL bacteriophages produce large plaques.

 

b) A lawn of bacteria on a petri dish is infected by a mixture of WT and RL bacteriophages. Assuming a high bacteria strain K/bacteriophage mixture ratio is used for the experiment, what is (are) the size (s) of plaques (large or small) produced? Briefly explain your answer.

 

 

2) You use two rapid lysis mutants RL1 and RL2 in a recombination assay with E. coli strain B. Describe the first step of this assay. Include in your description the genotype of the progeny

 

 

3) At the end of the first step, the total concentration of recovered bacteriophages is 5 x 106 bacteriophages/ml. In the second step, 0.01 mL of the recovered bacteriophages are added to 9.99 mL of SM buffer, and 0.01 mL of this diluted bacteriophage solution is plated on a lawn of E. coli, strain B. We will assume that all the bacteriophages recovered in step 1 are infectious.

 

a) Calculate the total number of plaques (large and small) you expect on the plate of E. coli B. Show your calculation. 5 pts

 

b) If the frequency of recombination between the 2 mutants is 5%, how many plaques would you expect when 1 mL of the diluted bacteriophage solution (see above) is plated on E. coli strain K. Show your calculations. 5 pts

 

 

4) Pairwise complementation assays are done between 4 mutants RL1, 2, 3, and 4. Each mutant carry a single mutation

 

a) The following table indicates the number of plaques (large or small) are obtained after plating, on E. coli strain B, the bacteriophages produced during the first step of the complementation assay.

 

	RL2	RL3	RL4
RL1	>100	> 100	<10

 

 

For each of the 3 complementation assays, how many plaques (few or many) do you expect if the plating had been done, in the same conditions, on a lawn of E. coli strain K? Briefly explain 4 pts

 

b) Based on these 3 complementation assays what is the minimum number of complementation groups for this mutation? Briefly explain. 4 pts

 

c) Do mutants RL2 and RL3 belong to the same complementation group? Explain your answer 4 pts

 

Question 2

Figure 1 (see below) shows the plasmid pUC18. This plasmid contains a LacZ gene interrupted by a multiple cloning site (MCS), a gene conferring resistance to ampicillin (AmpR) and an origin of replication (Ori).

Figure 1: Plasmid pUC 18 map.

1) Colonies derived from bacteria transformed by this plasmid accumulate a blue product in the presence of Xgal. Xgal is a colorless substrate of -galactosidase (an enzyme coded by the LacZ gene) into a blue product. Untransformed bacteria would form white colonies in the presence of Xgal.

a) Does the insertion of the multiple cloning site affect the production of active -galactosidase? Explain 3 pts

 

b) What is the likely color of colonies transformed by a pUC18 plasmid that contain a piece of DNA cloned at the MCS? Explain 2 pts

 

c) What is the role of the AmpR gene in the plasmid? 4 pts

 

d) What is the role of the origin of replication in the plasmid? 4 pts

 

2) Cd, a circular molecule of DNA, is digested by EcoRI, or XbaI or Hind III and then is added, without additional purification, to a solution containing pUC18 linearized with the matching restriction enzyme (EcoRI, or XbaI or Hind III, all contained in the MCS) and some DNA ligase. The ligation mixture is then used to transformed bacteria and the bacteria are plated on solid media containing ampicillin and Xgal. After an overnight incubation at 37°C, you observe blue and white colonies on the plates. You isolate plasmids from both type of colonies and you digest the isolated plasmids with the restriction enzyme used initially to cleave Cd (EcoRI, or XbaI or Hind III). The table below shows the length of the DNA fragments generated by restriction digest. The circular molecule, Cd, by itself cannot be propagates in bacteria. The sites for the 3 restriction enzymes are contained in pUC18 multiple cloning site (MCS).

 

Restriction enzyme a	Type of colonies b	Length c
EcoRI	Blue	2690 bp
	White	2690 bp &950 bp
XbaI	Blue	2690 bp
Hind III	Blue	2690 bp
	White #1	2690 bp &230 bp
	White #2	2690 bp &120 bp
	White #3	2690 bp &600 bp

 

a: Restriction enzyme used for the cloning and to digest the plasmids purified from each colony

b: Different white colonies (#1,2, or 3) generate plasmids that give rise to restriction fragments of different sizes when digested with HindIII

c: Length of linear DNA fragments generated by restriction digest of the purified plasmid using the enzyme listed in the first column

 

a) What is the length of pUC18? Explain 2 pts

 

b) How many XbaI restriction sites are found in the pUC18 and in Cd? Explain? 2pts

 

c) Based on the EcoRI digestion of plasmids purified from white colonies, how many EcoRI site were contained into Cd, and pUC18? What is the length of Cd? Explain. 6 pts

 

d) When the plasmid purified from white colonies #1, 2, and 3 are digested by HindIII, restriction fragments of different sizes are produced (230 bp, 120 bp, and 600 bp). Explain why 3 types of colonies are found. 6 pts

 

Question 3

You have extracted DNA containing the same gene from two different individuals (A1 and A2) and then sequenced it using the dideoxy method. The position and orientation of the template DNA and the radiolabeled sequencing primer are indicated below.

 

The autoradiograms below show a partial sequence of the gene obtained for individuals A1 and A2, with the dideoxy nucleotide added to each reaction indicated at the top:

Top

Bottom

 

 

1) What are the sequences of the template DNA for individuals A1 and A2 (read from the bottom to the top of the autoradiogram). Indicate the 5’ and 3’ end of the sequences (8 pts).

 

2) By mistake you erased the identifying labels on the tubes containing the DNA from individuals A1 and A2. You rush to the freezer and find vials containing three different restriction enzymes: PvuI that cleaves the sequence 5′-CGATCG-3′, EcoRI that cleaves the sequence 5’-GAATTC-3’ and Tsp5091 that cleaves the sequence 5′-AATT-3′. Explain how you would re-identify the tubes (A1 and A2) without performing another sequencing reaction (10 pts).

 

3) In class, you have learned about one method of DNA sequencing (Sanger or dideoxy-termination method). More recently, a new method has been developed to sequence short sequences of single-stranded DNA. This method is called pyrosequencing. Here is a description of the 6-steps process.

 

Step 1:

Combine the following reagent:

Single stranded DNA template, primer, adenosine 5’ Phosphosulfate (APS), luciferin, DNA polymerase, ATP sulfurylase, luciferase and, apyrase

 

Luciferase is an enzyme that reacts with its substrate (luciferin) and ATP to produce light that can be easily detected and quantified.

 

 

Step 2:

Add to the mixture a single dNTP (dATP, dTTP, dCTP, or dGTP)

 

A reaction of elongation of the primer is possible if the correct dNTP has been added. This reaction catalyzed by the DNA polymerase is the following:

 

Primer (DNA)n + dNTP  Primer (DNA)n+1 + PPi

 

Step 3:

The pyrophosphate (PPi) reacts with APS to produce ATP. This reaction is catalyzed by the ATP sulfurylase. APS + PPi  ATP + SO4-

Luciferin can now convert its substrate since ATP is present:

 

ATP + Luciferin + O2 AMP + Oxyluciferin + CO2 + PPi + Light

 

Step 4: A detector measures the amount of light produced by the reaction.

 

 

Step 5:

 

The enzyme Apyrase catalyzes two reactions:

 

dNTP  dNMP + 2 Pi ATP  AMP + 2 Pi

Step 6: Back to step 2 for another reaction cycle (steps 2 to 6) using a different dNTP.

 

a) The following graph (see below) represents the intensity of the light produced at each reaction cycle after the addition of a specific dNTP. An intensity of 1 on the graph corresponds to the addition of one nucleotide to the elongating primer.

 

What is the sequence of the sequenced template DNA? Label the 5’ and 3’ ends. 8 pts

 

 

b) Explain the importance of step 5. 4 pts

 

 

You study two strains of bacteria, S and R.

 

 

Question 4

 

Strain S contains a gene coding for an enzyme responsible for the synthesis of AHL, a small molecule that can freely diffuses across bacteria cell wall and cell membrane. This gene is always transcribed because it is under the control of a constitutive promoter. This constitutive promoter is a weak promoter leading to the synthesis of AHL at a moderate rate.

Strain R contains a gene coding for GFP (Green Fluorescent Protein). This gene is under the control of a promoter activated by a dimer comprised of 2 Proteins R1 and R2 (R1 and R2 do not activate GFP expression on their own). The genes coding for R1 and R2 are both under the control of a weak constitutive promoter. Note that proteins R1 and R2 are not produced by strain S.

 

a) When strain R is grown in culture by itself, R1 and R2 are synthesized but GFP is not produced. In the following experiment, you incubate a preparation of cytoplasm from strain R with an antibody directed against R1 attached to agarose beads (assume that you have an excess of beads and that all the protein R1 contained in the cytoplasm is now attached to beads). After incubation, you centrifuge the reaction mixture to separate beads (in the pellet) from unbound cytoplasmic proteins (supernatant) and then you analyze the distribution of proteins R1 and R2 between pellet and supernatant fractions. The result is shown in the table below.

 

Proteins	Pellet	Supernatant
R1	Yes No	Yes No
R2	No	Yes

 

 

For protein R1, what are the correct answers (Yes or No) for the pellet and supernatant fractions. (2 pts)

 

b) Based on the previous experiment explain why the GFP protein is not produced when the strain R is grown by itself. (3 pts)

 

c) S and R Strains are grown in the same culture and both the AHL and GFP expressions are monitored over time.

Based on the graph shown above explain how AHL affect GFP expression (4 pts)

 

d) S and R strains are co-cultured as described in the previous question. After 1 hour of growth, 5 mL of the liquid culture are centrifuged and the bacterial pellet is then resuspended in 1 mL of liquid culture. You observe that GFP starts to be expressed in this new culture after only 10 minutes. Briefly explain this observation. (4 pts)

 

e) After 3 hours of co-culture, bacteria are harvested, and the bacterial cytoplasm is isolated. An immuno- purification with anti-R1 antibodies is performed as described in part 3a. This time R2 is found in the pellet fraction but not in the cytoplasmic fraction. Based on this observation and the previous experiments, explain the role of AHL in GFP expression. Remember from your past problem sets that an immuno-purification uses antibodies bound to beads (4 pts)

 

Question 5

1) Aliquots of a 7.8 kb linear piece of DNA are digested with the restriction enzymes PvuII, HincII, ClaI, and BanII, alone and in pairs. The digestion products are separated by gel electrophoresis, and the size of each fragment is determined by comparison to size standards. The fragment sizes obtained, in kb, are given below.

a) Digestion with PvuII: 1.3 kb and 6.5 kb

b) Digestion with HincII: 0.5 kb, 3.3 kb, and 4 kb

c) Digestion with ClaI: 1 kb and 6.8 kb

d) Digestion with BanII 1.3 kb and 6.5 kb

e) Digestion with PvuII and ClaI: 1 kb, 5.5 kb, and 1.3 kb

f) Digestion with PvuII and HincII: 0.5 kb, 0.8 kb, 2.5 kb, and 4 kb

g) Digestion with HincII and ClaI: 0.5 kb, 1 kb, 3 kb, and 3.3 kb

h) Digestion with HincII and BanII: 0.5 kb, 0.8 kb, 2.5 kb, 4 kb

i) Digestion with ClaI and BanII: 1kb, 1.3 kb, 5.5 kb

 

The diagram below represents the restriction map of the DNA fragment. Each vertical bar represents the site recognized by one of the restriction enzymes.

 

Match each letter (A, B, C, D) with one of the restriction enzymes and calculate the length of each restriction fragment (L1, L2, L3, L4 and L5). (10 pts)

Note: One site might be recognized by more than one enzyme and an enzyme might cut more than once along the 7.8 kb DNA

 

 

2) A 3.3 kb fragment created after the HincII digestion (see part 1) is ligated into a unique HincII site of the plasmid pUC18. After bacterial transformation, colonies are isolated and the recombinant plasmids for each colony are purified and digested with the restriction enzyme PvuII. Surprisingly, the plasmids do not all give the same pattern of restriction fragments on a gel.

The plasmids fall into two distinct groups. Both groups yield the same number of bands in the gel, the aggregate size of which is the same, but two of the fragment sizes are different.

a) How many PvuII sites are found in the original plasmid pUC18? Briefly explain your answer (3 pts)

b) Explain the reason for this difference between the two types of recombinant plasmids. (3 pts)

 

 

Question 6

Your goal is to amplify by PCR the following sequence: 5’…GAATGTAAACTTCGGCGAATCAG GATGAAGTTCAAATCATCGACGGC…3’

1) The first and last nucleotides (bold, red) correspond to nucleotides 747 and 1134 and define the first (747) and last (1134) nucleotide of the amplified sequence.

 

a) What are the sequences of the 18-nucleotides forward (or sense) and reverse (or antisense) primers that you plan to use to amplify the DNA sequence located between nucleotides 747 and 1134? Indicate 5’ and 3’ 6 pts

 

b) What is the length of the PCR product (expressed in base pairs or bp)? Show your calculation. 6 pts

 

2) The PCR mixture contains the template DNA, the dNTP mixture, the reaction buffer, both primers (forward and reverse) and the Taq polymerase.

a) You have stock solutions of your two primers at both a concentration of 20 mol/L. Your supervisor asked you to add in the PCR reaction 300ng of each primer. Determine the volume (expressed in l) of primers you need to add in your 100l reaction, knowing that the two primers have the same molecular weight (7500 g/mol). Show your calculation. 8 pts

Reminder: 1g =103 mg = 106g =109 ng. The same applies for mol, mmol, mol and nmol

 

b) You know that 3 l of the Taq polymerase catalyze the incorporation of 0.03 mol of dNTP in 15 min at 70°C. Your advisor asked you to use 2 units of the Taq polymerase. Calculate the volume of Taq polymerase you need to add in your 100 l PCR reaction? To help you, your advisor gives you the information that came with the Taq polymerase solution and states that 1 unit of this polymerase catalyzes the incorporation of 10 nmoles of dNTP in 30 min at 70°C. Show your calculation. 8 pts