Question

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 scF_vs 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. F_c

c. F_ab

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

Answer 1

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.