Molecular Biology MCQ Set 1
1. During DNA cloning which of the following is not a crucial requirement?
a) DNA inserts
c) Protein expression
d) Molecular cutter
Answer: c [Reason:] DNA cloning typically involves a vector that carried the DNA insert into a host cell. Molecular cutters are important to incorporate the DNA insert into the vector thus giving rise to the chimera molecule. Hence cloning is complete if the organism replicates and the DNA insert replicates along with it.
2. Transformation does not involve ____________
Answer: d [Reason:] Transformation specifically involves the cutting of vector and gene of interest with specific cutters. These are then ligated and thus formed molecule is is the recombinant molecule. This molecule is then propagated into the host cell thereby ending the process of transformation of the host.
3. Which is the most common organism considered for genetic manipulations?
a) E. coli
b) Saccharomyces cerevisiae
d) Bacillus Subtilis
Answer: a [Reason:] The DNA fragment to be cloned needs to be inserted within the vector to be replicated within the host. By far the most common host used to propagate DNA and thus genetic manipulations is E. coli.
4. Which of the following is an essential feature for being a perfect vector?
a) Origin of replication
b) Selectable marker
c) Restriction site
d) Virulent gene
Answer: d [Reason:] A vector typically has three characteristics:
i. It must contain an origin of replication.
ii. It must contain a selectable marker.
iii. It must have one or more restriction sites for restriction endonucleases.
5. DNA ligase can ligate restriction site ends produced by EcoRI to the ends of DNA insert cut by the same enzyme _________
Answer: a [Reason:] EcoRI generates protruding 5’ ends (sticky ends) that are complementary to each other. Thus the ends are capable of reannealing with each other. Thus when both vector the same enzyme and DNA insert are cut using same cutter the strands anneal themselves but leaves two nicks in both the strands. Thus treatment with the enzyme ligase seals the nicks using ATP.
6. The vector and the DNA insert are cut by different enzymes for convenience __________
Answer: b [Reason:] A target DNA is cleaved with a restriction enzyme to generate potential DNA inserts with sticky ends. Vector DNA that has been cut with the same enzyme produces compatible over hangings thus making it convenient for the purpose of annealing and ligation.
7. What is the major difference between cloning vectors and primary vectors?
a) Selectable marker
b) DNA inserts
c) Presence of promoter
d) Presence of two Ori
Answer: c [Reason:] The major difference is the presence of suitable promoter before the DNA insert. In case of an expression vector the main motto is the production of protein thus a promoter in an essential criteria but in case of the cloning vector it is mainly used for amplification and/or production of library.
8. Which of the following is the primary use of an expression vector?
a) DNA library
b) DNA purification
c) Protein production
d) DNA cloning
Answer: c [Reason:] An expression vector is the type of vector that is designed for the expression of genes in cells. the can produce a large amount of proteins particularly useful if the gene product is toxic or a subunit vaccine.
9. Under the influence of which ion does E. coli takes up plasmid from the environment?
Answer: b [Reason:] The presence of copper (Cu2+) makes the cell wall of E. coli porous. This facilitates the uptake of external plasmid DNA by applying little agitation.
10. The process by which every type of transformant can be identified is __________
a) Replica plating
d) Insertional inactivation
Answer: d [Reason:] Insertional inactivation is the process by which a marker gene is inactivated by inserting the DNA insert within that gene. This leads to the visual identification if the gene inactivated gives a coloured product when grown in suitable medium. Also, the marker inactivated may be an antibiotic resistant gene where the cells are grown in suitable antibiotics following the process of replica plating.
Molecular Biology MCQ Set 2
1. Without using primers for RNA synthesis the polymerase produces small stretches of RNA of _____________ nucleotides.
Answer: b [Reason:] Without using primers for RNA synthesis the polymerase produces small stretches of RNA of 10 nucleotides. These transcripts are released from the polymerase which without dissociating from the template begins the synthesis of RNA again.
2. To facilitate binding of σ70 the melting of the template and non – template strands occurs between the positions ____________
a) -11 to +3
b) -9 to +5
c) -10 to +1
d) -10 to +2
Answer: a [Reason:] The transition from the closed to open complex involves structural changes in the enzyme and opening of the DNA duplex to reveal the template and the non-template strands. This melting occurs between positions -11 to +3, in relation to the transcription start site.
3. The isomerisation of σ70 of bacterial polymerase is an energy dependent process.
Answer: b [Reason:] The isomerisation of σ70 of bacterial polymerase is not an energy dependent process. Instead isomerisation is the result of a spontaneous conformational change in the DNA-enzyme complex to a more energetically favorable form.
4. With respect to the formation of complex between polymerase and DNA template which of the following is correct?
a) Open: reversible; Closed: irreversible
b) Open: irreversible; Closed: reversible
c) Open and closed: reversible
d) Open and closed: irreversible
Answer: b [Reason:] Isomerisation is essentially irreversible thus open complexes are irreversible and once forms transcription will be initiated. But the closed complex is readily reversible and the polymerase can easily dissociate from the promoter and is not bound to transcribe the DNA.
5. How many channels are found in the RNA polymerase clamp?
Answer: c [Reason:] The RNA polymerase clamp has 5 channels. They are NTP uptake channel, NT channel, T channel, RNA-exit channel and downstream DNA channel.
6. The channel that allows the exit of the coding strand is known as ____________
a) NTP channel
b) NT channel
c) T channel
d) Downstream DNA channel
Answer: b [Reason:] The channel that allows the exit of the coding strand is known as the NT channel. NT channel is the non-template exit channel. Non-template strand is also known as the coding strand.
7. Region 1.1 of the σ factor mimics the DNA and is negatively charged.
Answer: a [Reason:] Yes, the region 1.1 of the σ factor mimics the DNA and is negatively charged. The space in the active center cleft which is either occupied by the region 1.1 or by DNA, is highly positively charged.
8. RNA polymerase adds the nucleotides de novo. The first nucleotide added by the polymerase is ____________
Answer: a [Reason:] The first nucleotide added by the polymerase is always a purine. In this case adenine is more common than guanine thus ATP is the first nucleotide added.
9. The ternary complex is formed by the polymerase, DNA duplex and the nascent RNA.
Answer: b [Reason:] The ternary complex is formed by the polymerase, single stranded DNA template and the nascent RNA. This is formed after the polymerization of atleast 10 nucleotides which stabilizes the structure.
10. After the binding of polymerase to the DNA template the σ factor is ejected.
Answer: b [Reason:] No, the σ factor is not ejected right after the binding of polymerase to the DNA template. It is rather ejected after the addition of atleast 10 nucleotides. It has been seen to play some role in the addition of the first and second nucleotide and it cannot leave the complex during the transition of closed complex to open complex.
Molecular Biology MCQ Set 3
1. Where do the chimera molecules propagate?
Answer: c [Reason:] Chimera or recombinant molecules needs host for replication and propagation. Thus host selection is very essential for the propagation of a certain recombinant molecule.
2. Mutation sometimes assists the uptake of DNA material. It is a transformational regulator.
Answer: a [Reason:] DeoR is a mutation which helps in the uptake of the DNA. This process is one of the transformational regulators and it also has a DNA binding activity.
3. What is the DNA sequence for cleavage by the endonuclease that is coded by hsdR?
Answer: a [Reason:] hsdR encodes for a type I restriction endonuclease. This endonuclease is known to identify the methylated sequence AACGCNNTGC.
4. In which of the following methylation dependent restriction system is necessary?
a) Type III
b) Type II
c) Type I
Answer: d [Reason:] Methylation dependent restriction enzymes are the products of mcrA, mcrB and mrr loci. It can degrade the sites that contain the methylated cytosine or adenine residue.
5. The DAM protein methylates which residue in the sequence GATC?
Answer: a [Reason:] The DAM protein is responsible for the methylation of the newly formed DNA strands. It is known to methylate the adenine residue in the sequence GATC.
6. The quality and yield of plasmid DNA preparation both are enhanced by end A mutation.
Answer: a [Reason:] The gene for DNA specific endonuclease is inactivated by the end A mutations. This mutation is sometimes induced in the host molecules which is seen to enhance both quality and yield of plasmid DNA.
7. What is the main characteristic that is present in useful transformation method?
a) Property of plasmid
b) Ca2+ concentration
c) Efficiency of transformation
d) Availability of optimal physical conditions
Answer: c [Reason:] All transformation methods cannot be used equivalently if the participants are not efficient. Thus efficiency of the vector and its complementary host is the most important factor in case of transformation which determines the range of transformants produced.
8. Plasmids contribute to the survivality of its host cells.
Answer: a [Reason:] The genes contained in the plasmid are not of essential use for the bacterium for its day-to-day survival. Instead they help the bacterium to overcome stressed situations without which the bacterium may die.
9. Which of the following types of resistance is not provided by the plasmid for its host?
a) Antibiotic resistance
b) Heat resistance
c) Phage infection
d) Heavy metals
Answer: c [Reason:] Resistance to phage infection requires the production of nucleases which is provided by the bacterial genome itself and not the plasmid. The resistances such as antibiotic, temperature, metal, toxic etc are provided by the plasmid DNA.
10. The drug, quinolone, resistance is observed in which of the following?
a) Klebsiella pneumoniae
b) Bacillus thuringiensis
c) Streptococcus pneumoniae
d) Staphylococcus lentus
Answer: [Reason:] Quinolone resistance gene is are frequently found on the plasmids that contain the ESBL genes. They provide resistance to substances such as aminoglycose, Qnr proteins etc. These resistances are provided in the plasmid containing bacterial family Enterobacteriaceae such as, E. coli or Klebsiella pneumoniae.
11. Certain products produced by the plasmid genes can kill other microbes in its vicinity.
Answer: a [Reason:] Some plasmids produce substances that kill other type of bacteria in its vicinity and improve its own multiplication. An example of this type of survivality is shown by the bacterium Agrobacterium tumifacience which produces a substance called opine which promotes the growth and multiplication of the same and decreases the growth of other bacteria.
Molecular Biology MCQ Set 4
1. How many active sites are present in the DNA polymerase to catalyze the addition of the four dNTPs?
Answer: a [Reason:] Only 1 active site is dedicated for the addition of all the four types of dNTPs in the DNA polymerase. This is due to the identical geometry of all the four types of dNTPs.
2. How does the polymerase recognize the correct dNTP for addition?
a) Structure of dNTP
b) Molecular weight of dNTP
c) Purine and pyrimidine orientation
d) Ability of hydrogen bond formation
Answer: d [Reason:] Only 1 active site is dedicated for the addition of all the four types of dNTPs in the DNA polymerase. This is due to the identical geometry of all the four types of dNTPs. The polymerase recognizes the correct dNTP for addition is by the ability of hydrogen bond formation of the new bases with the existing bases on the template strand.
3. DNA polymerase can distinguish between dNTPs and rNTPS due to ___________
a) Structural differentiation
b) Steric exclusion
c) Steric hindrance
d) Enzyme substrate mismatch
Answer: b [Reason:] DNA polymerase shows an impressive ability of distinguishing between dNTPs and rNTPS. This is facilitated by the steric exclusion of the rNTPS from the DNA polymerase active site which is too small to accommodate the 2’-OH of rNTPs.
4. When we compare the structure of DNA polymerase to the structure of a body part, it resembles to _________________
a) Right hand
b) Left hand
c) Right foot
d) Left foot
Answer: a [Reason:] From the studies of the atomic structures of the various DNA polymerases bound to the primer:template junction it reveals that the structure resembles partially to that of a closed right hand. Based on the analogy to a hand the three domains of polymerase are called the thumb, fingers and palm.
5. With respect to the palm domain of the DNA polymerase which of the following is not its property?
a) Contains primary elements of the catalytic site
b) Binds to 2 divalent ions
c) Composed of α helix
d) Brings about the environmental changes around 3’-OH of dNTP
Answer: c [Reason:] The palm domain of the DNA polymerase is composed of β sheet and contains the primary elements of the catalytic site. In particular, this region of the polymerase binds to 2 divalent metal ions that bring about the chemical environmental changes around 3’-OH of dNTP for its polymerization.
6. The two divalent metal ions of the active sites of DNA polymerase are major catalytic elements that bring about changes required for the joining of the dNTPs.
Answer: a [Reason:] One of the metal ions reduces the affinity of the 3’-OH for its hydrogen. This generates a 3’-O- that is primed for nucleophilic attack of the α – phosphate of the incoming dNTP. The second metal ion coordinates the negative charges of the β – and γ – phosphates of the dNTPs and stabilizes the pyrophosphate produced by the joining the primer and the incoming nucleotide.
7. Mismatched DNA does not affect the rate of activity of DNA polymerase.
Answer: b [Reason:] Te accuracy of the base – pairing is monitored by the palm domain of the polymerase. The palm makes extensive hydrogen bond contacts with the base pairs in the minor groove of the newly synthesized strand. These contacts are not base pair specific but only bond if the base – pairing is correct. Thus mismatched DNA dramatically slows the rate of activity of DNA polymerase.
8. Which is the rate limiting step of DNA replication?
a) Formation of the RNA primer
b) Binding of primer to the DNA template
c) Binding of DNA polymerase to the primer:template junction
d) Binding of first dNTP to the primer
Answer: c [Reason:] The rate of DNA synthesis is dramatically increased by adding multiple nucleotides per binding site. It is the initial step of binding the DNA polymerase to the primer:template junction the slowest step.
Molecular Biology MCQ Set 5
1. Methylation at which of the following bases is possible within the following oligonucleotides?
Answer: b [Reason:] In eukaryotes, DNA methylation occurs predominantly at CpG dinucleotide sequence. In CpG, the intervening ‘P’ represents the phosphodiester bond linking cytosine- and guanine-containing nucleotides.
2. Which is known to be methylated in eukaryotic cells?
Answer: c [Reason:] In eukaryotes only the cytosine residue is methylated. In prokaryotes both adenine and cytosine residue is methylated.
3. Among the following which shows the maximum rate of cytosine methylation?
Answer: d [Reason:] Cytosine methylation is relatively rare in lower eukaryotes. But the cytosine residue in vertebrates is upto 10% of the total number of cytosine bases are methylated and in plants upto 30%.
4. Which of the following is the prototype of correctly methylated base?
Answer: a [Reason:] The enzyme DNA methyltransferase mediates the transfer of methyl group to cytosine generating 5-methyl-cytosine. Thus the reaction occurs as-
5. The region of maximum methylation in the genome is predominantly activated.
Answer: b [Reason:] The regions in the genomic DNA which has a high number of methylated cytosine are usually transcriptionally inactive. Silencing is thought to be either due to direct inhibition of transcription factors binding as a result of methylated cytosine or the binding of other proteins with methyl-binding domain of the DNA.
6. Prokaryotic methylation occurs at ____________
Answer: d [Reason:] Prokaryotic methylation occurs at both adenine and cytosine residue. The main function of DNA methylation in prokaryotes is to provide protection from its own endonucleases. It also plays an important role in mismatch repair.
7. How many types of methylation processes are known in eukaryotic cells?
Answer: b [Reason:] There are two types of methylation processes known in eukaryotic cells. They are maintenance methylation and de-novo methylation.
8. De-novo methylation is done in the newly synthesized DNA strand at opposite positions of the methylated sites on the parent strand. …
Answer: b [Reason:] Maintenance methylation is done in the newly synthesized DNA strand at opposite positions of the methylated sites on the parent strand. In de-novo methylation, addition of methyl groups occur at totally new positions.
9. Both de-novo methylation and maintenance methylation ensures the methylation pattern is inherited from parent to daughters.
Answer: b [Reason:] Maintenance methylation is done in the newly synthesized DNA strand at opposite positions of the methylated sites on the parent strand. The maintenance activity ensures that the two daughter DNA molecules retain the methylation pattern of the parent molecule.
10. Methylation of cytosine increases the transition mutation of cytosine to __________
d) Adenine and thymine
Answer: c [Reason:] The methylation of cytosine increases the transition mutation of cytosine to thymine. This is caused by the process of deamination of the 5-methyl-cytosine residue.
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