Molecular Biology MCQ Set 1
1. PCR is a DNA amplifying in vivo method ______
Answer: b [Reason:] A powerful method for amplifying particular DNA segment is the polymerase chain reaction. This procedure is carried out entirely biochemically, that is invitro.
2. Which of the following is a mismatch?
a) Polymerase – Taq polymerase
b) Template – double stranded DNA
c) Primer – oligonucleotide
d) Synthesis – 5’ to 3’ direction
Answer: b [Reason:] PCR uses the enzyme DNA polymerase that directs the synthesis of DNA from deoxynucleotide substrates on one stranded DNA template. DNA polymerase synthesizes DNA in a 5’→3’ direction and can add nucleotides in the 3’ end of a custom – designed oligonucleotide.
3. Annealing of primer is facilitated by complementary region ___________
Answer: a [Reason:] A synthetic oligonucleotide is annealed to a single stranded DNA template that contains a complementary region. DNA polymerase thus uses this annealed primer to elongate the new strand.
4. Primer used for the process of polymerase chain reaction are ______________
a) Single stranded DNA oligonucleotide
b) Double stranded DNA oligonucleotide
c) Single stranded RNA oligonucleotide
d) Double stranded RNA oligonucleotide
Answer: a [Reason:] Two single stranded, oligonucleotide is synthesized and used as a primer. The primers thus produces binds to the denatured DNA and polymerase starts its synthesis in the 5’ to 3’ direction.
5. Polymerase used for PCR is extracted from _____________
a) Escherichia coli
b) Homo sapiens
c) Thermus aquaticus
d) Saccharomyces cerevisiae
Answer: c [Reason:] Amplification is usually carried out by the DNA polymerase I enzyme from Thermus aquaticus. As this organism lives in hot springs, Taq polymerases are thermostable thus they are resistant to high temperature.
6. How many DNA duplex is obtained from one DNA duplex after 4 cycles of PCR?
Answer: c [Reason:] After each cycle the number of duplexes doubles itself thus after the first cycle there are 2 DNA duplexes. A duplex has 2 DNA strands thus after second cycle there will be 4 Duplexes, after third cycle there will be 8 DNA duplex. Lastly, after the 4th cycle there will be 16 duplexes.
7. At what temperature do denaturation of DNA double helix takes place?
Answer: d [Reason:] The polymerase chain reaction is started by heating the mixture at 94˚C. At this temperature the hydrogen bonding between the nucleotide bases melts thus separating the two DNA strands.
8. At what temperature do annealing of DNA and primer takes place?
Answer: d [Reason:] After the denaturation of the two strands the temperature is decreased to 50 – 60˚C. At this temperature the primers anneal to their complementary segments. Thus as 54˚ lies within this range option b is the correct option.
9. The temperature at which DNA synthesis takes place is 74˚C _______________
Answer: a [Reason:] The DNA synthesis in polymerase chain reaction takes place at 74˚C. This is usually set at 74˚C because this is just below the optimum temperature of functioning of the Taq polymerase.
Molecular Biology MCQ Set 2
1. Regulation of enzyme activity is accomplished in ___________ steps.
Answer: b [Reason:] Regulation of enzyme activity plays the key role in governing the cell behavior. Thus its regulation comes in two specific steps. The first step is accomplished in part at the level of gene expression, which determines the amount of any protein synthesized by the cell. A further level of control is then obtained by regulation of protein function, which allows the cell to regulate not only the amount but also the activities of its proteins.
2. Binding of small molecules to enzymes leads to feedback inhibition.
Answer: a [Reason:] Conformational changes in the enzymes result from the binding of small molecules, such as, amino acids or nucleotides that regulate enzyme activity. This type of regulation commonly is responsible for controlling metabolic pathways by the process of feedback inhibition.
3. The regulatory molecules bind to the catalytic site of enzyme to regulate its activity.
Answer: b [Reason:] Feedback inhibition is an example of allosteric regulation, in which a regulatory molecule binds to a site on an enzyme that is distinct from the catalytic site. The binding of such a regulatory molecules alters the conformation of the protein, thereby changing the shape of the catalytic site and affecting catalytic activity.
4. Which of the following is incorrect about the inactive state of the Ras protein?
a) GDP bound form
b) Cell proliferation activation
c) Mutation leads to cancer
d) Changes configuration when bound to regulatory molecules
Answer: b [Reason:] Ras protein when bound to GDP, that is, in its inactive form, cannot activate cell proliferation. It can only activate cell proliferation when it is in the active GTP bound state.
5. The regulatory proteins bind to the enzymes with covalent linkages.
Answer: b [Reason:] The regulatory proteins bind to the enzymes with non-covalent linkages. Since no covalent bonds are formed, the binding of these regulatory molecules to the protein is readily reversible, allowing the cell to respond rapidly to environmental changes.
6. The activity of which of the following proteins is not regulated by covalent modifications?
a) Digestive enzymes
c) Cell signaling
d) Blood clotting
Answer: c [Reason:] The activity of many proteins is regulated by covalent modifications. One example of this type of regulation is the activation of some enzymes by proteolytic cleavage of inactive precursors. Digestive enzymes and proteins involved in blood clotting and apoptosis are regulated in this mechanism.
7. Protein kinases transfer phosphate groups to the side chains of which of the following?
Answer: a [Reason:] Protein phosphorylation is catalyzed by protein kinases, most of which transfer phosphate groups from ATP to the hydroxyl groups of the side chains of serine, threonine or tyrosine residues. The protein kinases are one of the largest protein families in eukaryotes, accounting for approximately 2% of eukaryotic genes.
8. Like kinases all phosphatases are also species specific in nature.
Answer: b [Reason:] Like protein kinases, most protein phosphatases are specific either for serine and threonine or for tyrosine residues. But some protein phosphatases can recognize all the three phosphoamino acids.
9. Which of the following is false about amino acid modification and activity?
a) Acetylation of lysine residue
b) Methylation of lysine and arginine residue
c) Glycosylation of serine and threonine residue
d) Nitrosylation of arginine residue
Answer: d [Reason:] Nitrosylation is the addition of NO groups to the side chains of amino acids. Nitrosylation generally occurs to the cystine residues of protein polypeptide chains and not arginine.
10. The cAMP-dependent protein kinase has how many subunits?
Answer: d [Reason:] The cAMP-dependent protein kinase is composed of two regulatory subunits and two catalytic subunits. In this state the enzyme is inactive. The enzyme is activated by cAMP, which binds to the regulatory subunits and induces a conformational change leading to dissociation of the complex; the free catalytic subunits are then enzymatically active protein kinase.
Molecular Biology MCQ Set 3
1. Which of the following is not required for DNA sequencing?
a) Restriction digestion
d) Polymerase chain reaction
Answer: c [Reason:] DNA sequencing is the process of determining the precise order of nucleotides within a DNA molecule. It includes the methods and technologies of restriction endonuclease, electrophoretic techniques and Polymerase Chain Reaction.
2. End labeled DNA sequencing is known as dideoxy method of sequencing
Answer: b [Reason:] The end labeled DNA sequencing is known as Maxem and Gilbert method. This procedure involves either 3’ or 5’ end labeling thus is also known as end labeled DNA sequencing.
3. The 32P is added at the 3’ end by polynucleotidyl kinase
Answer: b [Reason:] The 32P dNTP is added at the 5’ end by polynucleotidyl kinase. The 32P dNTP is added at the 3’ end by deoxynucleotydil transferase. The end labeling is done in either one of the two ends.
4. The end labeled fragment is cleaved in how many pieces?
Answer: a [Reason:] The end labeled fragment is digested with a restriction endonuclease which cleaves it into two unequal lengths of fragments. As a result, only one end of each of the two fragments thus produced will be labeled.
5. Restriction digestion is the only process to achieve sequencing by the Maxem and Gilbert method.
Answer: b [Reason:] Restriction digestion is the one of the two processes to achieve sequencing by the Maxem and Gilbert method. The alternative method includes denaturation of its two complementary strands separated by gel electrophoresis.
6. The denatured strands cannot be separated.
Answer: b [Reason:] The two complementary strands of DNA generally show different mobility during electrophoresis. This is because of their different molecular weight as one of the strands has a higher number of purines and the other pyrimidines.
7. What are the basic base – specific cleavage sites used in Maxem and Gilbert method?
a) A, T, G, C
b) C, T, A+G, T+C
c) A, G, A+T, G+C
d) G, C, A+G, C+T
Answer: d [Reason:] The single end labeled double or single stranded DNA samples produced is subjected to base – specific cleavage. The bases – specific cutters are used to cut at one of the four sites G, C, A+G and C+T.
8. Which step is not involved in base – specific cleavage of DNA fragment?
a) Modification of concerned base
b) Removal of modified base from DNA strand
c) Induction of random strand break
d) End labeled DNA fragments of variable lengths produced
Answer: c [Reason:] The strand breaks is induced in a specific position. This type of break is induced in the position from which a modified base is removed.
9. The process of electrophoresis is the key to sequencing.
Answer: a [Reason:] The digests from the four reaction mixtures, that is, G, C, A+G, C+T, are separately subjected to electrophoresis. This separates the fragments according to their sizes. The base sequence is thus determined by the sequential reading of the bands developed in the four lanes of the gel through electrophoresis.
10. The dideoxy method is also known as ……………………………
a) Maxem and Gilbert method
c) Sanger’s enzymatic sequencing
Answer: c [Reason:] The dideoxy method is also known as Sanger’s enzymatic sequencing. It was developed by Fred Sanger and his co-workers in 1070s. It is also known as the chain termination sequencing.
11. Which enzyme is used for the replication in case of Sanger’s method of sequencing?
a) Polymerase I
b) Smaller subunit polymerase I
c) Polymerase III
d) Larger subunit polymerase I
Answer: d [Reason:] One of the two complementary strands is used as a template for DNA replication in the dideoxy method of sequencing. DNA replication is catalyzed by the larger subunit known as the klenow fragment.
12. With respect to Sanger’s enzymatic method of sequencing pick the odd one out.
a) Radioactive dideoxyribonucleotides
c) Klenow fragment
d) Restriction digestion
Answer: d [Reason:] In the reaction system for DNA replication, at least one of the four deoxyribonucleotides is radioactive in order to permit the autoradiographic development of bands after gel electrophoresis. A small primer sequence with a free 3’ – OH group must be provided with the template strand of DNA replication to proceed, since a free 3’ – OH is absolutely essential for the larger subunit of DNA polymerase I to catalyze DNA replication.
13. The four reaction mixtures that were prepared, contained ……………………………
a) dCTP, dTTP, dGTP, dATP
b) dCTP, ddTTP, dGTP, ddATP
c) ddCTP, dTTP, ddGTP, dATP
d) dCTP, ddTTP, dGTP, dATP
Answer: d [Reason:] Four different reaction mixtures are prepared for the replication of each DNA strand to be sequences. One of the systems contains 2’, 3’ – dideoxycytidine triphosphate in a concentration of about 1/100th of the normal amount of 2’ – dideoxycytidine triphosphate present in the mixture. In each one of the other 3 reaction mixtures using the same DNA an template, 2’, 3’ – dideoxythymidine triphosphate (),2’, 3’ – dideoxyadinosine triphosphate () and 2’, 3’ – dideoxyguanosine triphosphate () is used.
14. ddNTP acts as the chain radioactive markers.
Answer: b [Reason:] ddNTP acts as a terminator of the polynucleotide chain being newly synthesized on the template strand. Chain termination by ddNTP is achieved due to the fact that it does not have a 3’ – OH group, as a result of which further nucleotides cannot be added to the new chain.
15. Strand separation for DNA sequencing is done to ……………………………
a) sort the fragments according to their size
b) sort the fragments according to their molecular weight
c) obtain consecutive sequence as per termination
d) obtain specific lanes of sequencing
Answer: c [Reason:] The four singe stranded sample reaction mixtures are subjected to electrophoresis. This separates the strands according to their size. The smallest is the fastest moving strand that migrates furthest from the negative terminal, and vice versa. Therefore, by comparing the bands of the four gels thus obtained, nucleotide sequence of the DNA fragment can be determined.
16. The position of band during electrophoresis indicates the dNTP used as chain terminator in that mixture.
Answer: b [Reason:] The portion of a band during electrophoresis indicates the ddNTP used as chain terminator in that mixture. ddNTP is used as the chain terminator, dNTP is the normal nucleotide base.
Molecular Biology MCQ Set 4
1. Plasmid replication is dependent on the host cell.
Answer: b [Reason:] The plasmid is an autonomous replicating genetic material. It has its own origin of replication and complete replicating machinery thus can replicate freely and is thus independent of the replication of the host genome.
2. How the plasmid clones can be screened?
a) By selectable markers
b) By bacterial resistance gene
c) For restriction site
d) By ARS sequence
Answer: a [Reason:] Plasmid contains a selectable marker that allows cells that contain the vector to be easily identified. Thus selectable markers are used to screen clones.
3. How many restriction sites are contained by a plasmid?
d) More than 1
Answer: d [Reason:] Plasmid has one or more than one site for one or more restriction enzymes. Artificial plasmids contain single restriction site for one or more restriction enzymes. This allows DNA fragments to be inserted at a definite position.
4. Who were the scientists who discovered the plasmid pBR322?
a) Rodriguez and Bolivar
b) Joller smith
c) Herbert Boyer
d) Stanley Cohen and Joller smith
Answer: a [Reason:] Plasmids are most commonly used as vector DNA. pBR322 is a plasmid vector discovered by Rodriguez and Bolivar in 1977.
5. Under relaxed condition how many copies of plasmid are present in the cell?
a) 10 – 100 copies
b) 100 – 500 copies
c) 1 – 300 copies
d) 10 – 700 copies
Answer: d [Reason:] Most plasmids used in molecular cloning are generally present under relaxed condition. They are normally present in 10 to as many as 700 copies per cell.
6. What is length of the polylinker segment of plasmid?
a) Less than 100 base pair
b) Less than 10 base pair
c) Less than 70 base pair
d) Less than 50 base pair
Answer: a [Reason:] Plasmid vector contains a strategically located short, less than 100 base pair long segments of DNA. This segment of DNA is known as polylinker segment.
7. The full for of pUC is polylinker university cloning.
Answer: b [Reason:] The E. coli derived plasmid.
8. The host bacterium takes up a plasmid in presence of ______________
a) Monovalent cations
b) Monovalent anions
c) Divalent cations
d) Divalent anions
Answer: c [Reason:] The host bacteria can take up a plasmid from its surrounding. This process is greatly enhanced by the presence of divalent cation such as Ca2+.
9. What is the temperature at which bacteria can takes up the plasmid?
Answer: b [Reason:] Bacteria efficiently take up the plasmid DNA at -42˚C. This increases cell membrane permeability to DNA.
10. Which gene in the pUC18 vector confers antibiotic resistance to the transformed cells?
Answer: c [Reason:] pUC18 is a well known cloning vector. It contains the AmpR gene that confers resistance to the antibiotic ampicillin which is used as a selectable marker.
11. What is the characteristic of lacZ gene of pUC18 vector among the following?
a) Encodes for antibiotic resistance
b) Encodes for β-galactosidase enzyme
c) Encodes for β-lactamase enzyme
d) Encodes for β-galactoside transferase enzyme.
Answer: b [Reason:] The lacZ gene of the pUC18 vector is a gene of the lac operon encoding for the β-galactosidase enzyme. This enzyme cleaves galactoside present in the medium as a carbon source and liberates a blue coloured product.
12. Which one of the following is the first engineered plasmid vector?
Answer: d [Reason:] Although pBR322 is the most commonly used vector in scientific research. pSC101 is the first engineered plasmid vector. The pUC family vectors are the derivative of the pBR family of vectors.
13. What is the expanded form of pBR in pBR322?
a) Plasmid Boliver and Rodriguez
b) Plasmid Baltimore and Rodriguez
c) Plasmid bacterial recombination
d) Plasmid bacterial replication
Answer: a [Reason:] pBR322 was the first artificial cloning vector developed in 1977 by Boliver and Rodriguez. Thus pBR represents its creators leading to the expansion- Plasmid Boliver and Rodriguez.
14. What is incorrect about plasmid?
a) Helps in reproduction
b) Contains stress resistant genes
c) Serves as the transformation vehicle
d) They are the genetic material of the bacteria
Answer: d [Reason:] Plasmids are the extra genetic materials that are found in the bacterial cell along with the genetic component. They are autonomously replicating cyclic double strand DNA molecules used as vectors for gene transfer and also for replication.
15. The repressor for the β-galactosidase gene is encoded by ___________
c) Lac I
Answer: c [Reason:] In the pUC family vector the gene Lac I encodes for the repressor protein. This protein binds to the promoter of the lacZ thus inhibiting the expression of the gene.
Molecular Biology MCQ Set 5
1. How is the genetic material expressed?
a) By replication and transcription
b) By transcription and translation
c) By translation and modification
d) By mutation and transposition
Answer: b [Reason:] Expression of the genetic material is the series of processes how the sequence of bases in the DNA directs the production of the RNAs and proteins that perform cellular functions and define cellular identity. The basic processes responsible for gene expression are transcription and RNA processing followed by translation.
2. RNA polymerase requires a DNA primer for RNA synthesis.
Answer: b [Reason:] RNA polymerase does not require any primer for RNA synthesis. It can start transcription of the DNA template de novo.
3. The transcription process carried out by the RNA polymerase is very accurate but less accurate than replication.
Answer: a [Reason:] Transcription, though accurate, is less accurate than replication. One mistake occurs in every 10,000 nucleotides added in transcription, compared to 10,000,000 nucleotides for replication. This difference reflects the lack of extensive proofreading mechanism for transcription, although two forms of proofreading for RNA synthesis do exist.
4. Which of the following RNA polymerases are responsible for the production of 5S rRNA?
a) RNA polymerase I
b) RNA polymerase II
c) RNA polymerase III
d) RNA polymerase IV
Answer: c [Reason:] Polymerase I is responsible for the transcription of the different types of rRNA except the 5S rRNA. 5S rRNA is transcribed by polymerase III along with some small nuclear RNA genes and the tRNAs.
5. Which RNA polymerase deals with the production of mRNA?
a) RNA polymerase I
b) RNA polymerase II
c) RNA polymerase III
d) RNA polymerase IV
Answer: a [Reason:] Polymerase I is responsible for the transcription of the different types of rRNA except the 5S rRNA. 5S rRNA is transcribed by polymerase III along with some small nuclear RNA genes and the tRNAs. Polymerase II deals with the transcription of the mRNAs.
6. The eukaryotic subunit of RNA polymerase homologous to the ω subunit of bacterial RNA polymerase is ___________
Answer: d [Reason:] The eukaryotic subunit of RNA polymerase homologous to the ω subunit of bacterial RNA polymerase is RPB6. RPA1, RPB2 and RPC5 subunits of eukaryotic RNA polymerase are homologous to the bacterial RNA polymerase subunits β’, β and αI respectively.
7. The E. coli RNA polymerase adds __________nucleotides per second.
Answer: c [Reason:] The E. coli RNA polymerase can synthesize RNA at a rate of 40 nucleotides per second. For proper functioning of the RNA polymerase enzyme Mg2+ ion and an optimum temperature of 37˚C is required.
8. The RNA polymerase holoenzyme has the structural formula of ___________
Answer: a [Reason:] In a complete RNA polymerase, called the holoenzyme there are 5 sub units. Of which two are α and one of the each of the other 4 subunits namely β, β’, ω and σ.
9. Which of the following statements are true about the roles of RNA polymerase subunits?
i. α subunit is encoded by “rpo A” gene, required for core protein assembly.
ii. β subunit is encoded by “rpo C” gene is the catalytic center of RNA polymerase.
iii. ω subunit is encoded by “rpo Z” gene and helps in proper association of all the subunits.
iv. σ factor is encoded by “rpo D” gene and contributes in promoter recognition
Choose the correct option
a) i, ii, iii
b) ii, iii, iv
c) i, iii, iv
d) i, ii, iii, iv
Answer: c [Reason:] The β subunit of the RNA polymerase is coded by “rpo B” gene. The β’ subunit is known to be encoded by “rpo C gene.
10. The α subunits of polymerase has a function of ____________
a) Promoter binding
Answer: a [Reason:] The α subunits of polymerase is required for the core protein assembly, but has no clear role in transcription assigned to it. However, this subunit plays an important role in promoter binding.
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