QUESTION 1: You are inserting a gene into an MCS found
within the LacZ gene. Using blue/white colony selection,
why could you assume that white colonies have
modified plasmids?
a. A blue colony means the LacZ reading-frame was
disrupted
b. A blue colony means your gene has
mutations
c. A white colony means the LacZ reading-frame is
intact
d. A white colony means the LacZ reading-frame was
disrupted
QUESTION 2: You are performing a PCR using primers
with a sequence perfectly complementary to a template with an
expected melting temperature of 52°C. What do you risk if you use
an annealing temperature of 72°C?
a. Degrading the template DNA at higher
temperatures
b. This reaction is expected to work
perfectly
c. Failure to amplify because primers can’t
anneal
d. Nonspecific amplification
QUESTION 3: Binding between a DNA template and a
primer with an imperfect complementary sequence:
a. Is stabilized at lower temperatures
b. Is stabilized at higher temperatures
c. Has never been observed
d. Is not a major concern when optimizing PCR
QUESTION 4: Restriction enzyme sites, clamps, and
other engineered sequences can safely be put at the 5′ end of
primers because:
a. only the 3′ end of a primer must match the
template
b. This is false, you cannot modify the 5′
end
c. the 5′ end must bind to the template DNA
d. Taq polymerase will correct any
sequence errors
QUESTION 5: Gibson assembly will connect two
pieces of DNA by:
a. Blunt-end ligations
b. sticky ends produced by Type IIS restriction
enzymes
c. 20 nucleotides of identical sequence shared on the
ends of both fragments
d. sticky ends produced by palindromic restriction enzymes like EcoRI, SalI, and XbaI
QUESTION 6: Which enzyme used in Gibson
assembly is responsible for exposing single-stranded DNA?
a. Phusion polymerase
b. Taq DNA ligase
c. T5 exonuclease
d. BsaI
QUESTION 7 What are the benefits of calcium phosphate
precipitation for transfecting mammalian cells? Choose TWO
answers
a. Calcium phosphate precipitation is the most
efficient method for getting DNA into any cell
b. Less preparation is needed than for viral
vectors
c. Calcium phosphate precipitation requires packaging
cells
d. Calcium phosphate precipitation can be used to
transform intact tissues, like human embryos
e. Calcium and phosphate salts are
inexpensive
QUESTION 8 What is the difference between a
constitutive promoter and an inducible promoter?
a. Constitutive promoters are “always on”
b. Constitutive promoters are always
stronger
c. Inducible promoters are always stronger
d. Inducible promoters are never used in protein
expression
QUESTION 9 What would happen if you accidentally
encoded a TAA stop codon between a CDS and a C-terminal affinity
tag?
a. The protein would be translated without the
tag
b. The protein would not be translated at
all
c. The protein would be translated with the
tag
d. The tag would be translated into protein, but frame
shifted
QUESTION 10 In site-saturation mutagenesis using
degenerate primers, why do the mutated sites only appear at the
ends of PCR amplified fragments?
a. Taq polymerase only causes
mutations at the ends of DNA
b. Without a clamp, DNA is prone to
mutagenesis
c. Degenerate sequences are needed for any primer to
bind template DNA
d. The degenerate bases can only be introduced on the
primers
QUESTION 11 How does Prime editing insert
mutations into genomic DNA?
a. By directly modifying nitrogen bases
b. By providing an RNA template that is reverse
transcribed
c. By inducing non-homologous end joining
d. By inducing homology directed repair
QUESTION 12 During PCR mutagenesis, why do we use
an enzyme like DpnI, which degrades methylated DNA?
a. To destroy PCR-amplified DNA
b. To destroy template DNA that originated
from E. coli
c. DpnI is needed to mutate DNA sequences
d. To destroy excess, unused PCR primers
QUESTION 13 Why can scFvs be made
efficiently in E. coli?
a. They are a single polypeptide chain that lacks
posttranslational modification
b. They must be heavily glycosylated in the
endoplasmic reticulum
c. Any IgG can be efficiently made in E.
coli
d. They are made of multiple polypeptide chains
QUESTION 14 Which part of an antibody binds to an
antigen?
a. Complimentary determining region
b. Fc
c. Fab
d. Variable domains
e. Both (a) and (c)
f. (a), (c) and (d)
QUESTION 15:What is most likely to happen if you
forget the blocking step during an immunoblot?
a. Antibodies will bind directly to the
membrane
b. You will not observe a signal during
detection
c. Antibodies will no longer bind to
epitopes
d. Secondary antibodies will be unable to bind primary
antibodies
1. Blue white screening is based on the activity of B-galactasidase which cleaves lactose into glocose and galactose. If lactose is present in the medium, the lac operon gets activated amd produces functional B-galacyosidase dur to a-complementation. MCS present in lac Z site is nicked by restriction endonucleases and gene of interest is added to it leads to inactivation of a-complementation. As a result beta-galactosidase is not produced.
For screening the recimbinant chromogenic X-gal is added to the agar plate. If beta-galactosidase is present, X-gal will be hydrolyzed and will produce insoluble blue pigment. Means the colonies are not recombinant. In the absence of beta-galacyosidase, the plate will be white because X-gal will not be hydrolyzed.
So, option D is correct because we can insert the gene only after cutting the MCS present in lac Z frame. Means the recombinant colonies will be white in color with disrupted lac Z reading frame.
2. Option B is correct. Because for PCR the template gets separated at arounf 96°C and the melting temperature is good if it is kept in the range of 52 - 58°C. Anealing is done at around 72°C. So, the reaction condition is good.
Non specific products form if the annealing temperature is kept low.
3. Imperfect primer do bind with template DNA because the restriction sites are added if we have to do cloning or to insert new gene in the MCS. And it can be done with little lesser annealing temperature bht it shoukd not be too less to make undesirable products. So option A is correct.
4. Option A is correct because 3' end corresponds to the template DNA in order to proceed to elongation.
Clamps, RE sites etc are addeded to 5' end so that the enzyme can cleave it efficiently.
5. Option B is correct because gibson assembly do not need specific restriction enzymes. It can combine any DNA fragment at once having sticky ends created by T5 exonucleases.
6. Gibson assembly uses T5 exonucleases to create sticky ends to join them with the help of ligases.
7. Option A and D are correct.
8. Constitutive promoter means it is always expressed irrespective of the condition provided. And indicible promoter means it will be expressed once a suitable conditiin is provided, eg, in presence of certain metabolite like lactose.
So, option A is correct.
QUESTION 1: You are inserting a gene into an MCS found within the LacZ gene. Using...
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