midterm practice:

Question 1

Why are restriction enzymes so useful in molecular biology?

Answer 1

Restriction enzymes are extremely useful in molecular biology because they cut DNA at specific sequences, allowing scientists to precisely manipulate DNA. By generating either blunt or sticky ends, they make it possible to insert, remove, or replace genes in plasmids or other vectors for cloning, protein expression, or genetic engineering. Additionally, they enable DNA mapping, sequencing, and diagnostic applications, because their predictable cutting patterns let researchers analyze and assemble DNA fragments in a controlled way.

Question 2

If you had the privilege to discover the fourth restriction enzyme from the bacterial species Bifidobacterium longum, what would be its

Answer 2

BloIV

Question 3

Name the three types of DNA ends that can be generated by restriction enzymes and illustrate them indicating the position of the 5’ and 3’ ends for both DNA strands.

Answer 3

Blunt: Ends are flush = SmaI

Sticky 5′ overhang = 5′ single-stranded overhang EcoRI

Sticky 3′ overhang 3′ single-stranded overhang KpnI

5’ overhand means on top of 5’, theres something more

Question 4

After a cloning reaction (i.e. ligation of an insert into a plasmid), what are the two essential functional features needed for plasmid maintenance in colonies of transformed cells on solid medium?

Answer 4

A plasmid must have an origin of replication to replicate within the host and a selectable marker (e.g., antibiotic resistance) to allow identification of transformed cells. Without these features, the plasmid cannot be maintained or selected in colonies.

Question 5

what is lacIq, how does it work and why is it so useful?

Answer 5

lacIq is a high-expression version of the lacI repressor gene that produces more repressor protein, tightly blocking transcription from lac promoters until an inducer like IPTG is added. It is useful because it allows precise, controlled induction of gene expression, preventing unwanted or toxic protein production.

Question 6

Expression plasmids, like pGEX-3T, contain Multiple Cloning Sites (MCS). Where are MCS located on these expression plasmids (1 mark) and what is there function

Answer 6

The MCS is located downstream of the promoter, often within the fusion tag coding sequence, and provides multiple unique restriction sites to enable easy insertion of a gene of interest.

Question 7

Describe the different steps of a PCR reaction needed to amplify a 3 kb DNA fragment with the Taq DNA polymerase.

Answer 7

PCR involves repeated cycles of denaturation at 95°C to separate DNA strands, annealing at 50–°C for primers to bind the template, and extension at 72°C where Taq DNA polymerase synthesizes new DNA. For a 3 kb fragment, longer extension times (~1–2 minutes per kb) are required, and the cycles are typically repeated 25–35 times, followed by a final extension to complete synthesis of all products.

Question 8

1. If you amplify, using PCR, a DNA segment starting with a single copy of a longer template, how many copies you have after 6 cycles?
2. How many copies of that segment are of the correct size on both strands after 6 cycles?

Answer 8

1. 2^6=64
2. 2^6−2=2^4=16

ONLY AT 3RD CYCLE, THE COPIES ARE OF CORRECT SIZE

Question 9

1. When screening for positive clones after a cloning reaction, what would be the best location for your PCR primers to find out if your isolated plasmids contain (positive clone) or not (negative clone) your insert?
2. What relative size bands would you expect for your positive and negative clones

Answer 9

When screening clones by PCR, both primers should anneal to plasmid sequences flanking the DNA insert. In positive clones, the PCR amplifies the insert plus the flanking sequences, producing a larger DNA fragment that travels more slowly during gel electrophoresis. In negative clones, there is no insert between the primers, so only the short flanking sequences (essentially the primer-to-primer distance) are amplified, resulting in a smaller fragment that travels farther in the gel. This size difference allows easy distinction between positive and negative clones.

Question 10

Following plasmid transformation what are the three reasons we grow the bacterial cells at 37 C for 1h00 in SOC media?

Answer 10

1. Expression of antibiotic resistance gene
   * The cells need time to transcribe and translate the selectable marker (e.g., antibiotic resistance) before being exposed to the antibiotic.
2. Recovery of cell membranes
   * Transformation (especially chemical or electroporation) damages the cell membrane.
   * The recovery period allows cells to repair their membranes and resume normal growth.
3. Allow initial plasmid replication
   * Plasmids need to replicate inside the host to ensure sufficient copies per cell for selection.

Question 11

After transforming 10 nanogram of plasmid DNA you obtained the next morning 50 isolated colonies on one of your agar plates (10-7 dilution). Calculate the efficiency of transformation (# colonies / microgram of DNA) . What transformation method was used?

Answer 11

Transformation efficiency: 5 × 10¹⁰ CFU/µg DNA
Likely method: Electroporation

Question 12

If you are provided with a pGEX3-T expression plasmid that contains your favorite gene cloned downstream of the pTac promoter and fused to the GST tag, what are the five consecutive steps you need to take in order to purify the corresponding fusion protein?

Answer 12

1. Transform good clone (correct orientation) in E coli cells
2. Plate transformed E. coli on LB/Amp to get isolated colonies
3. Inoculate LB/Amp with a single colony
4. After sufficient bacterial growth induce protein expression with IPTG, which binds and represses the Tac repressor thus allowing translation of the gene by
   RNA polymerase.
5. Lyse cells and purify GST-fusion protein with glutathionine affinity column.