Exam HT20 Flashcards
Does mammalian cells fit the following descirption?
PTM: Nearly none
Speed (from plasmid –> expression): weeks
General costs including investments: low
No! The answer is suiting for e.coli.
Does e.coli fit the following description?
PTM: Nearly none
Speed (from plasmid –> expression): weeks
General costs including investments: low
Yes.
No post-translational modifications. Quick expression from plasmids, and cheap.
Does the following description fit BEVS?
Speed (from plasmid to expression): Months
General costs including investments: medium
PTM: some
Yes!
It’s a jack of all trades.
Does the following descrption fit BEVS?
Speed (from plasmid to expression): Weeks or months.
General costs including investments: high
PTM: Nearly all
No. This is more fitting of mammalian cells than BEVS. BEVS TPM are somewhat occurring and the speed of expression is slow.
What’s a VLP?
VLP = Virus-like particles.
VLPs are non-infectious because they don’t carry any viral genetic material. VLPs can be used as carriers for other genetic material.
What’s a continuous cell line?
A continuous cell line has been immortalized and can thus grow forever. The immortalization can occur via hybridisation with a cancer cell line.
Which of the following methods utilizes the 3’ –> 5’ activity of T4 polymerase?
Ligation independent cloning
Bac-to-Bac
FlashBac
Topo-TA
The gateway system
LIC.
The T4 exonuclease activity generates large 3’ overhangs which are he main driving force behind the ligation.
Why do we add tags to recombinant proteins? Two answers are correct.
A. To allow affinity purification of the expressed protein
B. To increase the isoelectric point of the expressed protein
C. To increase the mass of the expressed protein
D. To improve solubility of the expressed protein
E. To decrease the pH of the expressed protein
F. To improve formation of inclusion bodies by the expressed protein
G. To allow the proteins to be analyzed by SDS-PAGE
A: To allow affinity purification of the expressed proteins (just like we did in the lab, using IMAC)
D. To improve the solubility of the expressed protein.
What’s the elongated wording of “CHO cell line”. What are the benefits to the cell line?
Chinese ovary hamster cell line.
Benefits:
- Post-translational modifications occur in a huge extent.
What’s methotrexate? Explain how DHFR and recombinant proteins are assocaited to it.
Methotrexate is a cytostatic drug which is similar to folic acid. In the CHO cell line, DHFR catalyses the formation of purines from folic acid.
In order for DHFR/methotrexate selection to work, you need to things:
1. DHFR defficicent CHO cell line.
2. Plasmid with gene of interest + DHFR
When introducing recombinant DNA to the CHO cell line, you conjugate the gene for DHFR to it. Then you cultivate the cell line in methotrexate. The dosage of methotrexate will increase over time, only the cells with proper incorporation of the recombinant plasmid will survive due to its surplus of DHFR. The cells which are not transfected do not have any DHFR, especially not a surplus which is required for mtx not to inhibit its function any way. This leads to them dying.
Excess primers and unincorporated nucleotides can be enzymatically removed from an amplified PCR product. What enzymes are needed?
a. DNA polymerase combined with DNA kinase
b. Exonuclease I combined with alkaline phosphatase
c. Ribonuclease H combined with alkaline phosphatase
d. DNA ligase combined with topoisomeras I
e. Exonuclease I combined with DNA ligase
b. Exonuclease I combined with alkaline phosphatase
Reasoning:
The exonuclease 1 will cleave all remaining primers into individual nucleotides. The alkaline phosphatase dephsosphorylises the 5’ phosphate of the nucleotides, making them effectively useless.
How does a bacterial cell protect its own DNA from restriction enzymes?
a. adding methyl groups to adenines and cytosines
b. using DNA ligase to seal the bacterial DNA into a closed circle
c. adding histones to protect the double-stranded DNA
d. forming “sticky ends” of bacterial DNA to prevent the enzyme from attaching
e. reinforcing the bacterial DNA structure with covalent phosphodiester bonds
a. adding methyl groups to adenines and cytosines
The Gibson assembly reaction mixture does not include:
a. all four deoxynucleoside triphosphates.
b. dsDNA fragments with overlapping ends.
c. DNA ligase.
d. 5’ Exonuclease.
e. Oligonucleotide primers.
e. Oligonucleotide primers.
You don’t need primers, nucleotides are needed because the plasmid needs to be repaired in the organism which it’s being transformed to.
Im a little usnure.
FRET is useful for nanoscale proximity detection in microscopy. This due to its inherent
a. linear distance dependence
b. inverse linear distance dependence
c. inverse-sixth-power distance dependence
d. non-radiative energy transfer
e. good dynamic range
c. inverse-sixth-power distance dependence
It is what it is.
You would like to generate a DNA fragment with a single stranded overhang as shown below. Which of the following reagents do you need in your reaction mixture.
5´TACTTCCAATCCATG 3´
3´————–C 5´
a. E. coli DNA polymerase + dNTP
b. T4 DNA polymerase + dNTP
c. E. coli DNA polymerase +dGTP, dTTP, dATP
d. T4 DNA polymerase + dGTP, dTTP, dATP
e. T4 DNA polymerase + dCTP
e. T4 DNA polymerase + dCTP
T4 DNA polymerase has 3’ –> 5’ exonucleic activity. The activity is only available when there are no nucleiotides to be incorporated. Whenvever the exonuclease finds a base which it can build upon, the polymerase activity will start up.
A small molecule that is not antigenic by itself, but is eliciting an immune response when attached to a large carrier molecule is called a
a. Nanobody
b. Hapten
c. Idiotope
d. Adjuvant
e. Epitope
b. Hapten
During agarose gel electrophoresis most of the power is dissipated as heat. Heating of the electrophoretic medium has the several effects. Which alternative is not correct?
a. An increased rate of diffusion of sample and buffer ions, leading to broadening of the separated samples.
b. The formation of convection currents, which leads to mixing of separated samples.
c. A decrease of buffer viscosity, and hence a reduction in the resistance of the medium.
d. An increase of buffer viscosity, and hence a reduction in the resistance of the medium.
e. Using a constant power eliminates fluctuations in heating
d. An increase of buffer viscosity, and hence a reduction in the resistance of the medium.
When you heat up a viscuous liquid, it’s very rare that the viscosity increases.