Molecular Biology MCQ Set 1
1. Which of the following is the genetically engineered insulin?
Answer: a [Reason:] The first genetically engineered insulin is humulin. It was manufactured by the company Lily in USA.
2. Genetically manufactured GH is not effective for?
Answer: c [Reason:] New pharmaceutical products such as genetically manufactured GH are created by cloning certain genes. The genetically manufactured GH is extremely effective for the treatment of skin burns, bone fractures and ulcers in digestive tract and other ailments but not against infections. Vaccines are effective against infections.
3. Active insulin consists of how many polypeptide chains?
Answer: b [Reason:] Active insulin consists of two polypeptide chains, the A chain and the B chain. The A chain consists of 21 amino acids and the B chain consists of 30 amino acids. The chains are joined together by two disulphide linkages.
4. The first successful active insulin was synthesized in __________
Answer: c [Reason:] K. Itakura and collaborators in 1977 chemically synthesized DNA sequence for the two chains A and B. these were incorporated in two separate pBR322 vectors and were synthesized then combined in vitro to produce active insulin.
5. The Insulin A vector does not contain __________
c) Lac promoter
d) B chain
Answer: d [Reason:] The Insulin A vector does not contain B-chain it contains the A chain. A separate pBR322 vector is needed for the production of the B-chain known as insulin B vector.
6. Pick the odd one out.
Answer: b [Reason:] Somatostatin, somatotropin and β-endorphin all are genetically engineered growth hormones. Insulin is a genetically engineered hormone but it is not a growth hormone.
7. Genetically engineered somatotropin is formed by introducing DNA fragment from the genome to the plasmid.
Answer: b [Reason:] Genetically engineered somatotropin is formed by introducing cDNA of the gene to the plasmid. The cDNA is produced by the reverse transcription of the mRNA produced from the gene encoding somatotropin.
8. Which one of the following was first expressed as a fusion protein?
d) HBsAg vaccine
Answer: c [Reason:] Somatostatin is a 14 residue polypeptide hormone synthesized in the hypothalamus. It was the first polypeptide which was expressed in E. coli as a part of a fusion protein which inhibited the secretion of growth hormone, glucagon and insulin.
9. How many classes of interferons are found in humans?
Answer: c [Reason:] The humans have 3 classes of interferons. They are alpha (leukocyte) interferon, beta (fibroblast) interferon and gamma (immune) interferon.
10. Synthesis of hGH is an inducer dependent process.
Answer: a [Reason:] Synthesis of hGH is induced by an inducer of lac operon, that is isopropylthiogalactoside. The hGH thus produces is subsequently purified for commercial use.
11. The major enzyme required for the production of a chimera molecule is __________
b) Reverse transcriptase
d) Restriction endonuclease
Answer: c [Reason:] The major enzyme required for the production of a chimera molecule is restriction endonuclease. It is the enzyme responsible for the isolation of desired fragment from the genome and the specified end producing cleavages that are essential for the ligation of the DNA insert in correct position.
12. The subunit vaccine for hepatitis B is created against __________
a) Surface protein
b) Core protein
d) Whole virus
Answer: a [Reason:] The subunit vaccine for hepatitis B is created against the viral surface protein. It shows an antigenic property in the body but is not harmful and cannot infect the cells of the body as it does not contain any genetic material thus fulfilling the sole purpose of producing subunit vaccine.
13. Pick the odd one out.
Answer: b [Reason:] Antibodies are proteins that are produced by the hybridoma technology. The others, vitamins, antibiotics and alcohols are organic chemicals that are produced by using recombinant DNA technology for commercial purposes.
14. Monoclonal antibodies are produced by __________
a) Recombinant DNA technology
c) Hybridoma technology
Answer: c [Reason:] Hybridoma technology is a process for the production of monoclonal or identical antibodies. The production of monoclonal antibodies by the hybridoma technology was invented by Cesar Milstein and Georges J. F. Kohler in 1975.
Molecular Biology MCQ Set 2
1. The binding of U2 snRNP is aided by BBP and thus promotes 5’ site splicing.
Answer: b [Reason:] U2 snRNP binds to the branch point site by the help of U2AF by displacing BBP. The base pairing between the U2 snRNA and the branch point site is such that the branch site A is extruded from the resulting stretch of the double helical RNA as a single nucleotide bulge. This A residue is thus unpaired and available to react with the 5’ splice site.
2. Which three snRNPs together make the tri-snRNP particle?
a) U1, U2, U4
b) U2, U4, U5
c) U1, U5, U6
d) U4, U5, U6
Answer: d [Reason:] After the formation of the A complex, it rearranges itself in such a fashion that all the 3 splice sites come close together. This is achieved when U4, U5 and U6 snRNPs join the A complex. Together these three snRNPs are called the tri-snRNP particle.
3. Which of the following pairs is held together by the complementary base pairing in the B complex of splicosome?
a) U1, U2
b) U2, U4
c) U4, U6
d) U2, U5
Answer: c [Reason:] U4 and U6 snRNPs are held together by the complementary base pairing between their RNA components in the tri-snRNP particle. The U5 snRNP is more loosely associated through protein:protein interactions. The entry of this tri-snRNP into the A complex converts it into the B complex.
4. Which of the following complexes produces the active site of the splicosome?
a) A complex
b) A complex rearrangement
c) B complex
d) C complex
Answer: d [Reason:] U4 is released from the complex, allowing U6 to interact with U2. This arrangement, called the C complex, produces the active site.
5. Formation of the active site only in the C complex is a necessity for avoiding wrong splicing.
Answer: a [Reason:] Formation of the active site only in the C complex is a necessity for avoiding the chance of any aberrant splicing. Linking the formation of the active site to the successful completion of earlier steps in the splicosome assembly makes it highly likely that the active site is available only at legitimate splice sites.
6. Rearrange the order of the steps in the splicing mechanism.
i) Transesterification of 3’ splice site
ii) Formation of lariat complex
iii) Formation of active site
iv) Release of snRNPs
a) iii, i, ii, iv
b) iii, ii, i, iv
c) ii, iii, i, iv
d) ii, iv, iii, i
Answer: a [Reason:] Formation of the active site juxtaposes the 5’ splice site and the branch point site facilitating the first transesterification reaction. The second reaction, between the 5’ and 3’ splice sites, is aided by U5 snRNP, which helps bring together the two exons. This thus facilitates the complete formation of the lariat complex. The final step involves the release of the lariat complex and the snRNPs from the complex.
7. Group II introns are spliced by the splicosome complex.
Answer: b [Reason:] Group II introns are spliced by the RNA enzyme encoded by the intron. Nuclear pre-mRNA is the only mRNA that utilizes the splicosome complex for its splicing mechanism.
8. The branch point site conserved sequence for group II intron is ___________
Answer: c [Reason:] The branch point site conserved sequence for group II intron is G. they are very rare and are found in the nuclear rRNA of some eukaryotes along with some organelle gene and a few prokaryotic gene.
9. Which of the following groups can perform self splicing?
a) mRNA, rRNA
b) rRNA, mtRNA
c) tRNA, mRNA
d) rRNA, tRNA
Answer: b [Reason:] Self splicing is observed in two groups of very rare intron that are, group I and group II introns. The attain splicing by ribozyme encoded by the intron and are found in some eukaryotic genes along with some organelle gene and a few prokaryotic gene.
10. Self-splicing is achieved by the formation of specific structures within the introns.
Answer: d [Reason:] A typical self-splicing intron is between 400 to 1000 nucleotides long, and much of the sequence is critical for the splicing reaction. This sequence requirement holds because the intron must fold into a precise structure to perform the reaction chemistry.
Molecular Biology MCQ Set 3
1. The transcription starts with a purine that ends with a __________
Answer: d [Reason:] Transcription starts with a nucleoside triphosphate (A/G) and a 5’ – triphosphate groups is retained at the first position. The transcript thus ends with the triphosphate of the nucleotide and is represented as 5’….pppApNpNp…….3’.
2. The enzyme used for capping mRNA is a __________
Answer: c [Reason:] To the 5’ end is added a terminal G with the help of guanyl transferase. It is represented as follows:
5’Gppp + 5’….pppApNpNp….3’ → GpppApNpNp….3’
3. The guanyl residue added during capping is added in the same direction as the other nucleotides
Answer: b [Reason:] The guanyl transferase adds a guanine cap to the 5’ end of the transcript. The guanosine triphosphate thus added is in the opposite direction that is the triphosphate pointing inward rather than outward. The guanine cap also undergoes methylation.
4. How many methylated nucleotides are observed during capping?
Answer: b [Reason:] The eukaryotic mRNAs are capped with a Guanosine triphosphate residue which is methylated and is denoted as Cap 0. But most of the mRNAs undergo a second methylation at the penultimate base and is termed as Cap 1.
5. During capping the methylated penultimate base is known as the __________
a) Cap 1
b) Cap 2
c) Cap G
d) Cap 0
Answer: a [Reason:] The cap with a single methyl group is added at the 5’ terminal which is found in eukaryotic cells and is describes as Cap 0. But most eukaryotes methyl group may also be present on the penultimate base at 2’ position of the sugar moiety, so that with methyl groups at two terminal nucleotides, it is now described as Cap 1.
6. The specific sequence present in the DNA for the triggering of polyadenylation is __________
Answer: d [Reason:] The polyadenylation reaction requires a specific sequence in the DNA strand and thus in the pre – mRNA transcript as well. This sequence is known as the polyadenylation signal in the mRNA transcript and is a consensus sequence of 5’….AAUAAA….3’. Thus the DNA complement of this sequence will be 3’….TTATTT….5’.
7. The poly A adding site is 5’….GA….3’.
Answer: b [Reason:] The polyadenylation site is 5’….NA….3’which is followed by a GU rich region and preceded by the polyadenylation signal. The “N” in the polyadenylation site is always a pyrimidine, C or T.
8. The requirements of polyadenylation site does not include __________
a) 5’….AATAAA….3’ sequence
b) GU rich region downstream
c) Cleavage site of “C/TA”
d) Polyadenylation after GU rich region at cleavage site
Answer: d [Reason:] The requirements of polyadenylation site include a 5’….AATAAA….3’ polyadenylation signal sequence, a cleavage site of “C/TA” and a GU rich region downstream. The polyadenylation occurs at the cleavage site after the adenine residue and the polyadenylation site is in between the polyadenylation signal sequence and the GU rich region.
Molecular Biology MCQ Set 4
1. How is the extent of supercoiling measured?
a) Lk + Lk0
b) Lk0 – Lk
c) Lk – Lk0
d) Lk + Lk1
Answer: c [Reason:] The extent of supercoiling is measured by the difference between Lk and Lk0. This is known as the linking difference and is expressed as ΔLk = Lk – Lk0.
2. If Lk < Lk0 then DNA is ____________
b) Negatively coiled
c) Positively coiled
d) ΔLk > 0
Answer: b [Reason:] If ΔLk of a cccDNA is significantly different from zero, then the DNA is torsionally strained and hence it is supercoiled. If LK < ΔLk and ΔLk < 0, then DNA is said to be negatively supercoiled.
3. What is ΔLk and Lk0 dependent upon?
a) Histone concentration
b) G : A ratio
c) Composition of DNA
d) Length of DNA
Answer: d [Reason:] ΔLk and Lk0 are dependent upon the length of the DNA molecule thus it is more convenient to refer to a normalized measure of supercoiling. This is known as the superhelical density and is assigned the symbol σ.
4. Strand separation is easily accomplished in positively supercoiled DNA.
Answer: b [Reason:] Regions of negatively supercoiled DNA has a tendency to partially unwind. Thus strand separation can be accomplished more easily in negatively supercoiled DNA than in relaxed DNA.
5. Which of the following organism have been found to have a positively supercoiled DNA?
a) Saurida tumbil
b) Monarda didyma
c) Thermus aquaticus
Answer: c [Reason:] The only organisms that are found to have positively supercoiled DNA are certain thermophiles, microorganisms living under conditions of extremely high temperature. Example of one such bacteria found in hot springs are Thermus aquaticus.
6. Which of the following is not true about positive supercoils?
a) Helps to avoid denaturing
b) Over wounded DNA
c) Strand separations requires low energy
d) Rare form of DNA supercoiling
Answer: c [Reason:] The positive supercoils are thought to store free energy that helps keep the DNA from denaturing at levels of elevated temperature. Positive supercoils can be converted into more twists and thus requires a very high amount of energy to separate its DNA strands than any other type of negatively supercoiled DNA.
7. Packaging of DNA around nucleosome is known as ____________
a) Toroid writhes
c) Plectonemic writhes
Answer: a [Reason:] In the nucleosomes the double helix is wrapped almost two times around the outside circumference of a protein core. This wrap is recognized as a toroid or spiral form of writhe.
8. The nucleosome is seen to have negative supercoils.
Answer: a [Reason:] The spiral writhe found in the nucleosome is seen to have a left – handed spiral that is equivalent to negative supercoils. Thus, the packaging of DNA into nucleosomes introduces negative superhelical density.
9. Which of the following is known to relax supercoiled DNA?
b) DNase I
Answer: d [Reason:] The linking number is an invariant property of DNA that is topologically constrained. It can only be changed by introducing interruptions into the sugar – phosphate backbone. Topoisomerases are able to do the same by introducing transient single or double stranded breaks.
10. With respect to topoisomerases which of the following is true?
a) Only of one type
b) Never changes linking number
c) Requires ATP
d) Introduces knots
Answer: c [Reason:] Topoisomerases are of two general types. Type II topoisomerase changes the linking number in two steps. They make transient double stranded breaks in the DNA through which they pass a segment of uncut duplex DNA before resealing the break. They require ATP hydrolysis for this action.
Molecular Biology MCQ Set 5
1. Histone H1 binds two DNA helices ____________
Answer: a [Reason:] Upon interacting with a nucleosome, H1 binds to the linker DNA at one end of the nucleosome and the central DNA helix of the nucleosome bound DNA. This protects the ≈20 base pair linker DNA from nuclease digestion.
2. The distinct zig – zag appearance of the chromatin fibre is due to ____________
b) Histone H1
c) Histone core
d) Linker DNA
Answer: b [Reason:] Histone H1 binding produces a more defined angle of DNA entry and exit from the nucleosome. This effect results in the nucleosomal DNA taking on a distinct zig – zag appearance.
3. With respect to Histone H1 DNA packaging which of the following is false?
a) H1 induce 10 nm fibre formation
b) H1 binding gives nucleosome a defined angle
c) H1 stabilizes higher order chromatin structures
d) H1 binds to two DNA helices at once
Answer: a [Reason:] As the salt concentration increases in a test tube, addition of H1 results in the nucleosomal DNA forming a 30 nm fibre. This structure formed in vivo is known to be the next level of DNA compaction.
4. With respect to the solenoid model of nucleosome, which of the following is true?
a) Helical model
b) 8 nucleosomes per turn
c) Structure supported by X – ray diffraction
d) Helical pitch of approximately 13 nm
Answer: c [Reason:] The solenoid model of nucleosomal DNA packaging is supported by both EM and X – ray diffraction studies. This study indicated that the 30 nm fibre is composed of nucleosome stacked on edge in the form of a helix.
5. What marks the difference between the solenoid and the zig – zag models of 30 nm fibres?
a) Linker histone molecule
b) Linker DNA
c) Nucleosome structure
d) 10 nm fibre
Answer: b [Reason:] Unlike the solenoid model, the zig – zag conformation requires the linker DNA to pass through the central axis of the fibre in a relatively straight form. Thus, longer linker DNA favors this conformation. Because the average linker DNA varies between species, the form of the 30 nm fibre may not always be the same.
6. The histone tails stabilizes the 30 nm fibre by interacting with adjacent nucleosomes __________
Answer: a [Reason:] Core histones lacking their N – terminal tails are incapable of forming the 30 nm fibre. The tails is to stabilize the 30 nm fibre by interacting with adjacent nucleosomes. This model is supported by the three dimensional structure of the nucleosome., which shows that the amino terminal tails of H2A, H3 and H4 each interact with adjacent nucleosomes in the crystal lattice.
7. Which of the following is not a characteristic of nuclear scaffold?
a) Associated with loops of 40 – 90 kb
b) Topoisomerase I
c) SMC protein
d) Proteinacious in nature
Answer: b [Reason:] Two classes of protein contributing to nuclear scaffold have been identified, that are, topoisomerase II and SMC protein. Presence of Topo II as a protein associated with the structure can be proved when the cells are treated with drugs which results in DNA breaks at the sites of Topo II DNA bindings. The treatment generates DNA fragments of about 50 kb size.
8. Which histone molecule produces novel binding sites for protein components of the kinetochore?
a) CENP – A
Answer: a [Reason:] A histone variant CENP – A, is associated with the nucleosome that include centromeric DNA. In this region CENP – A replaces H3 subunit in the nucleosomes. These nucleosomes are incorporated into the kinetochore which mediates attachment of the chromosome to the mitotic spindle. The extended tail of CENP – A may generate these novel binding sites for other binding protein components of the kinetochore.
9. With respect to remodeling complex of nucleosome, which of the following is not true?
a) Does not require ATP
b) Favors sliding of histone
c) Favors transfer of histone
d) Favors remodeling of histone
Answer: a [Reason:] The stability of the DNA-histone core interaction is influenced by large protein complexes referred to as nucleosome remodeling complexes. These multi – protein complexes facilitate changes in the nucleosome location or interaction with the DNA using the energy of ATP hydrolysis.
10. Positioning of nucleosome is beneficial ___________
Answer: a [Reason:] Due to dynamic interactions with DNA, most nucleosomes are not fixed in their locations. But there are occasions when restricting nucleosome location, called positioning, can be beneficial. Positioning allows DNA binding site to remain as accessible linker DNA region for regulatory proteins.
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