pGLO and pUC

Question 1

What is GFP? Where is it naturally found? In general, how is it used in cellular/molecular research?

Answer 1

GFP stands for Green Fluorescent Protein, it is gene and protein found naturally in Jellyfish. A copy of that gene has been inserted into a plasmid and is used for cellular and molecular research because it will make a gene product that glows green under UV light so it can be used as a protein tag.

Question 2

What is a plasmid? What is a recombinant plasmid?

Answer 2

Plasmids are small circular pieces of DNA. Recombinant plasmids are plasmids that have been rearranged to contain a chosen DNA fragment. They are also extra chromosomal, so they aren’t strictly necessary for survival, but often provide some sort of benefit to the host cell (such as antibiotic resistance)

Question 3

Explain the difference between pGLO, GFP gene, and GFP

Answer 3

pGLO- plasmid that contains the gene GFP.

GFP gene- the gene we want to insert into pUC18

GFP- controlled by the promoted araB. protein encoded by GFP gene

Question 4

Explain the purpose for including an antibiotic resistance gene to a vector

Answer 4

To determine if the bacteria acquired the gene through transformation. For example, the pUC18 contains a vector called the bla gene that degrades antibiotics. If the cell is able to grow on media, it means it acquired the gene through transformation

Question 5

Explain the roles of the following reagents as they apply to plasmid minipreps.

TE (Tris/EDTA)
SDS/NaOH
KOAc
Isopropanol

Answer 5

Tris- buffers the cell
SDS/NaOH- lyses the cell.
EDTA also binds up metal ions which are often co-factors for DNA degrading enzymes - so it helps keep the plasmid DNA intact. SDS dissolves the lipids in the plasma membrane and NaOH denatures proteins and DNA
KOAc- allows cells to renature
Isopropanol- used to precipitate the plasma DNA from the supernatant.

Question 6

Explain why plasmid DNA isn’t removed along with the chromosomal DNA.

Answer 6

Plasmid DNA. s smaller and intertwined. These can renature. Chromosomal DNA is a lot larger and is cut into larger fragments. These fragments are single-stranded and form clumps with other proteins and membrane fragments. They cannot renature.

Question 7

Explain the purpose of centrifuging in this procedure (mini prep)

Answer 7

The first centrifugation pellets all the cellular debris, membrane components, denatured proteins and Chromosomal DNA chunks. The plasmid DNA stays in the supernatant. The 2nd centrifugation helps to concentrate the plasmid DNA and get it away from all the other chemicals.

Question 8

Explain the role of each component of the mastermix.

Primers
dNTPs
Taq polymerase

Answer 8

Primers- used to specify which region of DNA to be amplified.

dNTP- build nucleic acid monomers.

Taq- specific primer that anneals to DNA strands

Question 9

Explain why we must include the araC gene along with the gene for GFP.

Answer 9

the araC gene product, the protein araC, is a RNA Pol repressor unless araC is in the presence of Arabinose – in which case it will bind to arabinose and dissociate from the DNA allowing RNA Pol to bind to the promoter and begin to transcribe the gene of interest..lkbv,

Question 10

Why couldn’t we simply cut pGLO with EcoRI? Why do we have to use PCR at this step?

Answer 10

GLO doesn’t have an EcoR1 site, we needed to add them at the ends of our PCR product.

Question 11

95oC
60oC
72oC

Answer 11

95oC- separating strands of DNA (denature)
60oC- rapidy cool to this temp to allow primers to anneal to the separated strands.
72oC- allow the primer Taq, to extend the the primers and make complete copies of the DNA strands

Question 12

Why did we need to “clean” our PCR product? Predict what would happen if we had skipped this step.

Answer 12

It removes the Taq polymerase,dNTPs, and primer dimers. We need the purified sample to set up in electrophoresis. If we did not remove those, we would not be able to read the bands in the electrophoresis.

Question 13

Why did we have to filter the PCR product instead of using heat to inactivate the unwanted molecules?

Answer 13

We can’t heat inactivate an enzyme that is stable at 100 C, so we have to filter it out.

Question 14

What are restriction endonucleases and what is their role in nature?

Answer 14

Restriction enzymes are responsible for cutting the phosphate sugar backbone of DNA. They recognize specific locations in DNA to cut. Because of this, bacteria cells use this as a defense mechanism against viruses

Question 15

EcoRI leaves “sticky ends” when it cuts. Explain this. Why is it preferred to restriction enzymes that would leave “blunt ends”?

Answer 15

It allows ligation to happen by covalently bonding the sticky ends

Question 16

Explain the reason for running a gel of the PCR product.

Answer 16

whether or not we have a good solid band at the correct size following PCR.

Question 17

Explain why DNA ligase is required to join DNA strands together. Why isn’t the hydrogen bonding between complementary “sticky ends” enough?

Answer 17

DNA ligase is responsible for joining DNA strands together. It specifically bonds the OH from 3’ with the PO4 in the 5’ from another nucleotide. Instead of a hydrogen bond, it forms a covalent phosphodiester bond, a much stronger bond.

Question 18

Why did we need to heat-inactivate EcoRI prior to ligation?

Answer 18

Ligation will not occur if the enzyme is not heated. We need to stop the enzyme from continuing its function (cleaving off the sugar-phosphate backbones of DNA). Ligase needs to but the DNA back together and cannot do that if the restriction enzyme is still cutting.

Question 19

What was the purpose of the 30-minute incubation following the “heat shock” step?

Answer 19

It’s more to start expressing the genes on the plasmid - especially the bla gene

Question 20

What is the bla gene. What is the purpose of including it in the vector? How would the results differ if it wasn’t part of our vector (pUC18)?

Answer 20

The bla gene is the ampicillin resistance gene.The bla gene is the ampicillin resistance gene. We need it to distinguish which bacteria have been transformed. We will not be able to see which cells were transformed without it.

Question 21

Why did we plate our transformed cells on LB plates? If we had not used these controls, how else could you explain the lack of colonies on LB+++ plates?

Answer 21

It demonstrates that our bacteria were alive at the time of plating.

Question 22

ampicillin, and X-gal in addition to the standard nutrient broth. Explain the role of each in this experiment.

Answer 22

Arabinose- Activates polymerase to transcribe.
Ampicillin- kills non-transformed bacteria
X-gal, color indicator to know if the LacZ gene is still functional or not.
Broth: allow cells to recover.

Question 23

Explain why it was important to have the multiple cloning site (MCS) within the lacZ gene.

Answer 23

Foreign DNA can be inserted into pUC18 to make a recombinant plasmid. It disrupts the lacZ gene and cannot use X-gal.

Question 24

Explain how this blue/white screening technique allows us to know which bacteria have been successfully transformed and, more importantly, transformed with a recombinant plasmid.

Answer 24

X-gal is metabolized and will turn blue colonies. When X-gal is not metabolized, it will give rise to white colonies. When white colonies appear, that means the cells were transformed by a recombinant plasmid

Question 25

Imagine you found a white colony growing on LB+++ that did not glow green. Propose 2 hypotheses for how this could have occurred.

Answer 25

we could have taken up some other “insert” into the lacZ gene, not our desired PCR product insert with the GFP gene.
Arabinose was not present in order to start the sequence/activation of the GFP gene

Question 26

Explain how a diagnostic restriction digest could be used to help test these hypotheses.

Answer 26

If a second pUC was inserted backwards into the MCS, the restriction digest would show one band only at the size of a cut pUC18 2. If the Taq POL made a mutant GFP gene, the restriction digest would show two bands, corresponding to one pUC18 and one insert