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Molecular Biology MCQ Set 1

1. Nonstop mediated decay is observed mostly in ____________
a) Bacteria
b) Fungi
c) Protozoa
d) Animal

View Answer

Answer: d [Reason:] Nonstop mediated decay rescues ribosomes that translate mRNAs that lack a stop codon. Eukaryotic mRNAs, due to tailing, terminate with a poly-A tail. When mRNA lacking a stop codon is translated it translates through the poly-A tail. This results in the addition of multiple lysines to the end of the protein thus stalling the ribosome at the end of the mRNA and thus triggering nonstop mediated decay.

2. Ski7 is known to recruit ____________ enzyme for mRNA degradation.
a) Endonuclease
b) 3’→5’ exonuclease
c) 5’→3’ exonuclease
d) Nuclease

View Answer

Answer: b [Reason:] The protein Ski7 is related to the class II release factor eRF3, and stimulates the dissociation of the ribosome bound to the mRNA without a stop codon. Thus, Ski7 recruits the 3’→5’ exonuclease to the damaged mRNA for mRNA degradation.

3. Translation of an mRNA is the only way to proofread mRNAs.
a) True
b) False

View Answer

Answer: a [Reason:] In the absence of translation, the damaged mRNA is not rapidly degraded and has normal stability. Thus eukaryotic cells rely on the translation mechanism to proofread their mRNA.

4. Introns makes up about ____________ % of the total mRNA.
a) 50
b) 70
c) 90
d) 80

View Answer

Answer: c [Reason:] Over about 90% of the total pre-mRNA sequence are introns. These intronic sequences are degraded within the nucleus after they have been excised from the mRNA.

5. The enzyme that degrades the mRNA recognizes ____________ bond of mRNA.
a) 2’ – 5’
b) 2’ – 3’
c) 2’ – 5’
d) 3’ – 5’

View Answer

Answer: a [Reason:] The enzyme that degrades the mRNA recognizes three specific sites in the intronic mRNA. These are the unique 2’ – 5’ bond formed at the branching point as well as the 3’ and 5’ ends of the RNA molecule.

6. The introns are degraded due to ____________
a) Specific enzymatic sequence
b) Unprotected ends
c) Lariat formation
d) Endonuclease activity

View Answer

Answer: b [Reason:] The ends of the processed mRNA are protected from degradation by capping and tailing. In case of the bare ends of the introns it is easily identified by 3’ or 5’ exonucleases and is thus degraded.

7. How many ways of RNA degradation are found in eukaryotic cell?
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] Two ways of RNA degradation are found in eukaryotic cell. They are nonsense mediated mRNA decay where mRNA with premature stop codon and/or incomplete open reading frame is degraded. And secondly, nonstop mediated decay rescues ribosomes that translate mRNAs that lack a stop codon.

8. Nonsense mediated mRNA decay is proceeded due to the ____________
a) Absence of tailing
b) Absence of capping
c) Absence of stop codon
d) Absence of ORF

View Answer

Answer: d [Reason:] Nonsense mediated mRNA decay is proceeded due to two triggering factors as a quality control system. These two factors are: i. Absence of open-reading frame (ORF) ii. Presence of premature termination codon.

9. Presence of premature stop codon triggers nonstop mediated decay.
a) True
b) False

View Answer

Answer: b [Reason:] Nonsense mediated decay is triggered when ribosomes encounter premature termination/stop codon. These premature stop codons are identified when they occurs more than 50 nucleotides upstream of an exon-exon junction in the spliced mRNA.

10. Which of the following types of RNA is the most stable?
a) mRNA
b) hnRNA
c) tRNA
d) snRNA

View Answer

Answer: c [Reason:] Only two types of RNA are most stable in the cell and are rRNA and tRNA. This ability largely accounts for the high levels of these RNAs in both prokaryotic and eukaryotic cells.

11. Which of the following is not a step to degrade mRNA?
a) Deadenylation
b) 5’→3’ exonuclease activity
c) 3’→5’ exonuclease activity
d) Site-specific degradation

View Answer

Answer: d [Reason:] The steps to degrade mRNA are as follows: i. Deadenylation that is, shortening of the poly-A tail, for the degradation from the 3’ end by 3’→5’ exonuclease ii. Removal of 5’ cap for degradation from the 5’ end by 5’→3’ exonuclease.

12. The shorter half life of mRNAs is essential for the cells environmental adaptation in prokaryotes.
a) True
b) False

View Answer

Answer: a [Reason:] Bacterial mRNAs are rapidly degraded and usually have half-lives of only 2 to 3 minutes. This rapid turnover of bacterial mRNAs allows the cell to respond quickly to alterations in its environment, such as changes in the availability of nutrients require for growth.

13. The half life of mRNA of both prokaryotic and eukaryotic cells is same.
a) True
b) False

View Answer

Answer: b [Reason:] Bacterial mRNAs are rapidly degraded and usually have half-lives of only 2 to 3 minutes. Instead the mRNAs of eukaryotic cells have a half life ranging from 30 minutes to approximately 20 hours.

14. mRNA for which of the following has a longer half life?
a) SSB protein
b) Polymerase
c) G protein
d) Nucleases

View Answer

Answer: c [Reason:] The unstable mRNAs generally code for regulatory proteins, such as transcription factors, whose level within the cell vary rapidly in response to the environmental stimuli. In contrast, mRNAs encoding structural proteins, such as G protein, or central metabolic enzymes, such as kinases, generally have long half lives.

Molecular Biology MCQ Set 2

1. How many types of protein degradation pathways are seen in a eukaryotic cell?
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] Proteins are rapidly degraded in response to specific signals, providing another mechanism for the regulation of intracellular enzyme activity. In addition, faulty or damaged proteins are recognized and rapidly degraded within cells, thereby eliminating the consequences of mistakes made during protein synthesis. In eukaryotic cells, two major pathways – the Ubiquitin-proteasome pathway and lysosomal proteolysis – mediate protein degradation.

2. Ubiquitin degradation mechanism recognizes ____________
a) Cytosolic protein
b) Nuclear protein
c) Structural protein
d) Both cytosolic and nuclear proteins

View Answer

Answer: d [Reason:] Ubiquitin degradation mechanism recognizes both cytosolic and nuclear proteins. The major pathway of selective protein degradation in eukaryotic cells uses Ubiquitin as a marker that targets both cytosolic and nuclear proteins for rapid proteolysis.

3. Ubiquitin has ___________ of amino acids.
a) 70
b) 75
c) 76
d) 72

View Answer

Answer: c [Reason:] Ubiquitin is a 76 amino acid polypeptide. This is highly conserved in all eukaryotes (yeasts, animals and plants).

4. Ubiquitin binds to the ___________ amino acid residue for degradation.
a) Proline
b) Lysine
c) Serine
d) Valine

View Answer

Answer: b [Reason:] Proteins are marked for degradation by the attachment of Ubiquitin to the amino group of the side chain of a lysine residue. Additional ubiquitins are then added to form a multiubiquitin chain.

5. The Ubiquitin-proteasome degradation pathway is an energy dependent process.
a) True
b) False

View Answer

Answer: a [Reason:] Ubiquitin-proteasome degradation pathway is an energy independent process. Thus, both the attachment of ubiquitin and the degradation of marked proteins require energy in form of ATP.

6. Which of the following is an Ubiquitin activating enzyme?
a) E1
b) E2
c) E3
d) E4

View Answer

Answer: a [Reason:] Ubiquitinization is a multistep process. In this process the first step is the activation of ubiquitin. This is done by the attachment of Ubiquitin activating enzyme, E1 to the ubiquitin.

7. The Ubiquitin-proteasome pathway is an active target for anticancer drugs.
a) True
b) False

View Answer

Answer: a [Reason:] The ubiquitin-proteasome pathway is responsible for the degradation of several important regulatory proteins, including proteins that control cell proliferation and cell survival. Since the growth of cancer cells depends on the degradation of these regulatory proteins, the proteasome has emerged as a target for anti-cancer drugs.

8. Which of the following is the substrate specific enzyme of the following?
a) E1
b) E2
c) E3
d) E4

View Answer

Answer: c [Reason:] Most cells contain a single E1 but have several E2s and al large number of E3 enzymes. Different E3s recognize different substrate proteins, and the specificity of these enzymes is what selectively targets cellular proteins for degradation by the ubiquitin-proteasome pathway.

9. The Cdk1 activates which of the following type of Ubiquitin enzymes?
a) E1
b) E2
c) E3
d) E4

View Answer

Answer: c [Reason:] Cdk1 activates an ubiquitin ligase that targets cyclin B for degradation towards the end of mitosis. The E3 enzymes are known as the ubiquitin ligases that proceeds this step.

10. Ubiquitinization always leads to the degradation of protein.
a) True
b) False

View Answer

Answer: b [Reason:] Although polyubiquitin chains usually target proteins for degradation, the addition of ubiquitin to some proteins serves other functions. For example, the addition of single ubiquitin molecules to some proteins is involved in regulation of DNA repair, transcription and endocytosis.

11. Lysosomes does not have which one of the following properties?
a) Membrane bound organelle
b) Digestive enzymes
c) Cell metabolism
d) Carrier vesicles

View Answer

Answer: d [Reason:] Lysosomes are membrane-enclosed organelles that contain an array of digestive enzymes, including several proteases. They have several roles in cell metabolism, including the digestion of extracellular proteins taken up by endocytosis as well as the turnover of cytoplasmic organelles and cytosolic proteins.

12. For the process of autophagy the formation of vesicles occurs. The membrane of these vesicles are derived from ___________
a) Nuclear membrane
b) Cell membrane
c) Mitochondria
d) Golgi membrane

View Answer

Answer: c [Reason:] The principle pathway of uptake of cellular proteins is known as autophagy. This process involves the formation of vesicles in which small areas of cytoplasm or cytoplasmic organelles are enclosed in membranes derived from the ER or mitochondria.

Molecular Biology MCQ Set 3

1. Semiconservative nature of replication of eukaryotic chromosome was first demonstrated by _______________
a) Walter Flemming on root tip cells of Vicia faba
B) J. Herbert Taylor, Philip Wood and Walter Hughes on root tip cells of Vicia faba
c) Walter Flemming on root tip cells of Phaseolus vulgaris
d) J. Herbert Taylor, Philip Wood and Walter Hughes on root tip cells of Phaseolus vulgaris

View Answer

Answer: b [Reason:] Walter Flemming discovered cell division not semi conservative nature of replication. Semiconservative nature of replication was first demonstrated by J. Herbert Taylor, Philip Wood and Walter Hughes on root tip cells of broad bean which is the common name for Vicia faba.

2. Pick the correct pair with respect to primers used in DNA replication.
a) RNA primer- for prokaryotes only
b) DNA primer-for eukaryotes only
c) DNA primer- for both prokaryotes and eukaryotes
d) RNA primer- for both prokaryotes and eukaryotes

View Answer

Answer: d [Reason:] Short oligonucleotides of RNA are required by DNA polymerase for the synthesis of both leading and lagging strands of DNA due to the requirement of free 3’ end for DNA synthesis. As formation of oligonucleotides of DNA by DNA polymerase also requires a free 3’ end thus, DNA primers are not applicable for the synthesis of new strands of DNA during replication.

3. Which of the following is correctly matched with its subsequent role?
a) Topoisomerase II- can remove both positive and negative supercoil in the DNA duplex
b) Polymerase I- larger fragment responsible for exonuclease activity
c) DnaA protein- responsible for “melting” of the DNA double helix during replication
d) DnaB protein- attaches to the newly unwounded single strand of DNA to prevent folding of the strand

View Answer

Answer: c [Reason:] The reasons are i) Topoisomerase II- converts a positive supercoil to a negative supercoil, also known as gyrase ii) Polymerase I- larger (klenow) fragment is responsible for polymerase activity iii) DnaB protein- also known as helicase helps in unwinding of DNA duplex to form the open complex.

4. Replication fork is the junction between the two ___________
a) Unreplicated DNA
b) Newly synthesized DNA
c) Newly separated DNA strands and newly synthesized DNA strands
d) Newly separated DNA strands and the unreplicated DNA

View Answer

Answer: d [Reason:] As both the strands of DNA occur simultaneously the two template strands undergo separation. The junction between the newly separated DNA strands and the unreplicated DNA is known as the replication fork.

5. Which of the following does not affect DNA replication?
a) Antiparallel nature of DNA
b) End specificity of polymerase
c) SSB protein
d) Helicase

View Answer

Answer: c [Reason:] The antiparallel nature of DNA and end specificity of polymerase of polymerase leads to two types of strand synthesis, leading and lagging strands. Helicase helps in the unwinding of the DNA helix. SSB protein binds to the single stranded DNA during replication to stabilize it but does not take any part in new strand synthesis.

6. Who was the first person to analyse the process of replication and on which organism?
a) Arthur Kornberg: E. coli
b) John Cairns: E. coli
c) Arthur Kornberg: Bacillus subtilis
d) John Cairns: Bacillus subtilis

View Answer

Answer: b [Reason:] The person who first analyzed the process of replication was John Cairns in which E. coli were grown in the presence of radioactive Thymidine. This allowed subsequent visualization of newly replicated DNA by autoradiography.

7. In the case of a circular DNA synthesis how many replication forks are observed?
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] In case of replication of a complete circular DNA molecule 2 replication forks are observable. This represents the regions of active DNA synthesis and is represented in form of a bubble known as the replication bubble.

8. Primer synthesis short stretches of DNA for replication.
a) True
b) False

View Answer

Answer: b [Reason:] Primer synthesizes short stretches of RNA for replication. These RNA serves as primers for the synthesis of both the leading and lagging strand during replication.

9. Which of the following is not used for degrading RNA from RNA:DNA hybrid in replication?
a) RNase A
b) RNase H
c) Polymerase I
d) Exonuclease

View Answer

Answer: a [Reason:] RNase H dissolves RNA in eukaryotes and the last rNTP connected directly to the dNTP is digested by exonuclease. Polymerase I hydrolyses RNA in the 5’ – 3’ direction in prokaryotes. RNase A is not required in replicative purpose.

10. The gaps formed by the hydrolyzing the RNA from RNA:DNA hybrid is filled by DNA polymerase I and the nicks are ligated by T4 ligase.
a) True
b) False

View Answer

Answer: b [Reason:] The gaps formed by the hydrolyzing the RNA from RNA:DNA hybrid is filled by DNA polymerase δ. The nicks are then filled in by T4 ligase at the cost of an ATP yielding an intact lagging strand.

11. Which of the following is correctly matched with its subsequent role?
a) Topoisomerase II- can remove both positive and negative supercoil in the DNA duplex
b) Polymerase I- larger fragment responsible for exonuclease activity
c) DnaA protein- responsible for “melting” of the DNA double helix during replication
d) DnaB protein- attaches to the newly unwounded single strand of DNA to prevent folding of the strand

View Answer

Answer: c [Reason:] The reasons are i) Topoisomerase II- converts a positive supercoil to a negative supercoil, also known as gyrase. ii) Polymerase I- larger (klenow) fragment is responsible for polymerase activity. iii) DnaB protein- also known as helicase helps in unwinding of DNA duplex to form the open complex.

Molecular Biology MCQ Set 4

1. The two strands of cccDNA can be easily separated without any breakage of bonds.
a) True
b) False

View Answer

Answer: b [Reason:] There are no interruptions in either of the polynucleotide strands. The two strands of cccDNA cannot be separated from each other without the breaking of a covalent bond.

2. Passing of one strand through the other in order to get separated is known as _______
a) Twist
b) Linking number
c) Writhing number
d) Toroid

View Answer

Answer: b [Reason:] The two circular strands can be separated without permanently breaking any bonds in the sugar – phosphate backbone by passing one strand through the other strand repeatedly. The number of times one strand would have to be passed through the other strand in order for the two strands to be entirely separate from each other is called the linking number.

3. Linking number is always a ___________
a) Whole number
b) Prime number
c) Integer
d) Even number

View Answer

Answer: c [Reason:] The linking number is always an integer which is an invariant topological property of cccDNA. It does not depend on the shape of the DNA molecule.

4. Linking number is the sum of which geometric components?
a) Twist and writhe
b) Toroid and writhe
c) Twist and wrap
d) Twist and crossover

View Answer

Answer: a [Reason:] Linking number is the sum of two geometric components twist and writhe. Twist is simply the number of helical turns of one strand about the other. cccDNA is usually torsionally stressed such that the long axis of the double helix crosses over itself called writhe.

5. 3 – dimension structure of cccDNA is ___________
a) Nicked
b) Supercoiled
c) Flat
d) Torsionally stressed

View Answer

Answer: d [Reason:] 3 – dimension structure of cccDNA is torsionally stressed. cccDNA is usually torsionally stressed such that the long axis of the double helix crosses over itself called writhe.

6. Plectonomic writhe is __________
a) Cylindrical axis winding
b) Long axis twisted around itself
c) Helical turns of one strand about the other
d) Number of encounters of the two strands of a helix

View Answer

Answer: b [Reason:] Writhe can take two forms. One form is the interwound or plectonomic writhe, in which the long axis is twisted around itself. The other forms of writhe are toroid or spiral.

7. Interwound writhe and spiral writhe are topologically different.
a) True
b) False

View Answer

Answer: b [Reason:] Interwound writhe and spiral are topologically equivalent to each other. They have readily interconvertible geometric properties of cccDNA.

8. Twist and writhe are interconvertible.
a) True
b) False

View Answer

Answer: a [Reason:] Twist and writhe are interconvertible. A molecule of cccDNA can readily undergo distortions that convert some of its twist to writhe to twist without the breakage of any covalent bonds.

9. What is the relation between linking number (Lk), writing number (Wr) and twist number (Tw)?
a) Lk = Tw – Wr
b) Lk = Wr – Tw
c) Tw = Lk – Wr
d) Tw = Lk + Wr

View Answer

Answer: c [Reason:] The sum of the twist number (Tw) and the writhing number (Wr) must remain equal to the linking number (Lk). This constraint is described by the equation: Lk = Tw + Wr =>Tw = Lk – Wr.

10. How can we remove supercoils from cccDNA?
a) Treat with DNase
b) Treat with detergent
c) Treat with endonuclease
d) Application of heat

View Answer

Answer: a [Reason:] One procedure is to treat the DNA mildly with the enzyme DNase I, so as to break on average one phosphodiester bond in each DNA molecule. Once the DNA has been “nicked” in this manner, it is no longer topologically constrained and the strands can rotate freely.

Molecular Biology MCQ Set 5

1. The first X-ray diffraction patterns of DNA were taken in 1938 by ___________
a) William Asbury
b) Rosalind Franklin
c) Francis H. Crick
d) Linus Pauling

View Answer

Answer: a [Reason:] The first X–ray diffraction patterns of DNA of low quality taken in 1983 by William Asbury. He used using DNA supplied by Ola Hammarsten and Torbjorn Caspersson.

2. In early 1950s high quality X-ray diffraction photographs of DNA suggesting the DNA being double helix and composed of two nucleotide strands. Who took those photographs?
a) Rosalind Franklin
b) William Asbury
c) Francis H. Crick and James D. Watson
d) Rosalind Franklin and Maurice Wilkins

View Answer

Answer: d [Reason:] The high quality X-ray diffraction photographs were taken by Rosalind Franklin and Maurice Wilkins. It suggested that DNA structure was helical and might be composed of more than one nucleotide strand in early 1950s.

3. In 1952, an unambiguous established work in the laboratory of Alexander Todd led to the discovery of _____________
a) Chemical nature of DNA
b) X-ray diffraction structure of DNA
c) 3’-5’ phosphodiester linkage regularly links the nucleotides of DNA
d) Nucleic acid strands are held together by hydrogen bonds

View Answer

Answer: c [Reason:] Chemical nature of DNA was discovered by Watson and Crick in 1952. They also found that nucleic acid strands are held together by hydrogen bonds. Again the first X-ray diffraction structure was discovered by William Asbury in 1938.

4. What should be the complementary strand of 3’….ATGGCTTGA….5’?
a) 3’….TACCGAACT….5’
b) 5’….TACCGAACT….3’
c) 3’….TAGGCAAGT….5’
d) 5’….TAGGCAAGT….3’

View Answer

Answer: b [Reason:] The complementary of A is T and for G is C, therefore the complementary sequence should be ….TACCGAACT…. . And the two strands are antiparallel as the given strand is in 3’ to 5’ direction the complementary strand should be in 5’ to 3’ direction.

5. The hydrogen bonds formed during A and T bonding occurs between C6NH2 of A and _____
a) C4O of T
b) PO3
c) C6O of A
d) C1O of T

View Answer

Answer: a [Reason:] As the sugar molecule is connected to N1 in T the C4O and C1O are free. Also the C1O is farthest from C6NH2 thus the bond forms between C6NH2 and C4O. This stabilizes the model because of being in preferred tautomeric state.

6. In a DNA double helix the bases are held together by hydrogen bonds. These hydrogen bonds are _________
a) Covalent bonds
b) Non-covalent bonds
c) Ionic bonds
d) Van der Waals forces

View Answer

Answer: b [Reason:] Hydrogen bond is a weak non covalent bond. This occurs between two molecules resulting from an electrostatic attraction between a proton and an electronegative atom of the other molecule. No share of electron takes place which is the criteria for non-covalent bonding.

7. It is easy to break the bond between A and T than in between G and C.
a) True
b) False

View Answer

Answer: a [Reason:] It is easy to break the bond between A and T than in between G and C. This is because A and T has a double bond whereas G and C has triple bond connecting them. Thus, the energy required to break a double bond is lower compared to a triple bond.

8. With respect to the importance of hydrogen bonding and DNA double helix stability, which of the following statements is false?
a) Favorable tautomeric form of nucleotide bases
b) Contributes to the thermodynamic stability
c) Decreases the entropy
d) Specificity of base pairing

View Answer

Answer: c [Reason:] When the two complementary strands of DNA comes together in an aqueous solution the water molecules attached to the polynucleotide strands are displaced from the bases. This creates disorder and this increases the entropy. Therefore, it finally increases the stability of the DNA duplex.

9. In the late 1970s non double helical form of DNA was also suggested but was discarded on the basis of certain factors. Which of the following is not a factor responsible?
a) X-ray crystallography
b) Nuclein
c) Topoisomerase
d) Nucleosome core particle

View Answer

Answer: b [Reason:] Nuclein, later named as nucleic acid and eventually DNA was first discovered by Swiss physiological chemist Friedrich Miescher inside the nuclei of human WBCs in 1869. Thus nuclein is not a term related to the structural orientation and chemical nature of DNA.

10. Levene investigated and found that the nucleic acid is composed of poly-nucleotides and each nucleotide is composed of one base, a sugar molecule and a phosphate. This was performed on the genome of________________
a) Bacteria
b) Fungi
c) WBCs
d) Yeast

View Answer

Answer: d [Reason:] Levene placed his initial proposal of nucleic acids being composed of nucleotides and nucleotides being composed of one of the four bases, a phosphate and a sugar molecule in 1919. He gathered all his evidences regarding the structure were from the repeated experiments of hydrolysis and analysis of the yeast nucleic acid.

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