Molecular Biology MCQ Set 1
1. RNA genomes were first discovered in ___________ viruses.
Answer: a [Reason:] RNA genomes were first discovered in the plant Tobacco mosaic virus. The first direct proof that RNA acts as the genetic material was obtained in 1950s by experiments demonstrating that RNA purified from TMV could infect new host cells giving rise to progeny viruses.
2. The initial hypothesis of RNA provirus was proposed by __________ in 1960s.
a) David Baltimore
b) Howard Temin
c) Robert Gallo
d) David J. Griffiths
Answer: b [Reason:] The initial hypothesis of RNA provirus was proposed by Howard Temin in 1960s. In his experiments he found that the RNA tumor viruses replicate via synthesis of DNA intermediates called DNA proviruses, thus leading to his hypothesis.
3. Reverse transcription is an important phenomenon for eukaryotic cells.
Answer: a [Reason:] Reverse transcription is not restricted to retroviruses it also occurs in cells. For example, reverse transcription is essential for the replication of telomeres of eukaryotic chromosomes. Also reverse transcription is responsible for the transposition of sequences from one position to the other.
4. The virus on which the first experiment performed by Howard Temin was __________
a) Human Immunodeficiency Virus
b) T lymphatic Virus
c) Rous Sarcoma Virus
d) Human papiloma virus
Answer: c [Reason:] Rous sarcoma virus (RSV) the first cancer causing virus to be described was of considerable interest as an experimental system for studying the molecular biology of cancer. Howard Temin began his research in this area when, as a graduate student in 1958, he developed the first assay for the transformation of normal cell to cancer cells in culture following infection with RSV.
5. The viral genome contains which of the following characteristic sequences?
Answer: d [Reason:] The viral genome contains the characteristic sequence of long terminal repeats (LTRs), which are direct repeats of several hundred base pairs. Viral genes including genes for reverse transcriptase, integrase and structural proteins of the virus particle, are located between the LTRs.
6. The integration of viral DNA into the host DNA is promoted by the host integrase enzyme.
Answer: b [Reason:] The viral DNA integrates into the host cell chromosome by a process that resembles the integration of DNA transposable elements. This integration is catalyzed by a viral integrase and occurs at many different target sequences of the host chromosome.
7. Which of the following is not a function of reverse transcriptase?
a) RNA dependent DNA polymerase
b) DNA dependent DNA polymerase
c) RNase H
Answer: d [Reason:] Reverse transcription has a high error rate due to no proofreading activity. Thus the reverse transcriptase that facilitates reverse transcription has no exonuclease activity.
8. Reverse transcription has a high error rate.
Answer: a [Reason:] As the reverse transcriptase has no exonuclease activity thus, proofreading of the synthesized cDNA is not possible. This accounts to the high error rate of 1.7*10-4 of reverse transcription.
9. What will be the transcription product of 3’….AUCCGAGCUAAC….5’ by reverse transcriptase?
Answer: a [Reason:] The reverse transcriptase facilitates reverse transcription from RNA to DNA. The given sequence in the question is of RNA as it consists of “U” instead of “T” thus the resulting cDNA sequence should be 3’….AUCCGAGCUAAC….5’5’….TAGGCTCGATTG….3’.
10. Reverse transcription is a very important tool in modern molecular biology techniques.
Answer: a [Reason:] Reverse transcription is a vital role in modern molecular biology techniques. It is the mere foundation of building the cDNA libraries. The enzyme reverse transcriptase also plays an important role in the PCR technique known as RT – PCR. This process is also used in cloning, sequencing and other molecular analyses.
Molecular Biology MCQ Set 2
1. siRNA are produced by __________
b) DNA damage
c) RNA damage
d) Enzymatic action
Answer: d [Reason:] siRNA or short interfering RNAs are enzymes involved in the process of RNA interference. They are generally generated by the action of certain enzymatic activities such as dicer.
2. Which of the following is not the handiwork of siRNA?
a) Translational inhibition
b) mRNA destruction
c) Base dimerization
d) Promoter silencing
Answer: c [Reason:] siRNA causes gene silencing in three pathways. They operate through translational inhibition, mRNA degradation and promoter silencing. Base dimerization is caused by DNA exposure to UV rays causing errors during replication.
3. Introduction of ds DNA can repress gene expression.
Answer: a [Reason:] The discovery that simply introducing a double stranded RNA into a cell can repress the gene containing sequencings identical to that dsRNA. This phenomenon was first reported in 1998 found by the experiment in the worm C. elegans.
4. Dicer is __________
a) DNAse-like enzyme
b) RNAse-like enzyme
c) An endonuclease
d) An exonuclease
Answer: b [Reason:] Dicer is an RNAse III-like enzyme that recognizes and digests long double stranded RNAs. The products of this are short double-stranded fragments about 23 nucleotides long.
5. Which of the following pathways involve the Dicer for gene silencing?
a) Chromatin modification
b) Structural gene triple helix formation
c) DNA polymerase binding
d) RNA polymerase binding
Answer: a [Reason:] Dicer is involved in the production of siRNA that inhibit the expression of homologous genes in three ways. They are translational inhibition, mRNA degradation and promoter silencing by remodeling of chromatin.
6. The RNAi mediated gene silencing includes the binding of siRNA to form a complex with the target gene called RISC. .
Answer: b [Reason:] RISC or RNA induced silencing complex is machinery that includes the siRNA themselves, various proteins including members of the Argonaut family. The RNAi mediated gene silencing includes the binding of siRNA to form a complex with proteins to form the RISC.
7. The RNAi mediated gene silencing is an ATP-dependent procedure.
Answer: a [Reason:] Once the RISC complex is formed with the double-stranded siRNA it is denatured in an ATP-dependent manner. The appearance of the single stranded RNA activates the RISC complex.
8. If the siRNA is a complete homolog of the target mRNA sequences what is the net result?
a) Double helix formation
b) Translation inhibition
c) Chromatin modification
d) mRNA degradation
Answer: d [Reason:] If the siRNA is a complete homolog of the target mRNA sequence the latter is seen to be degraded. In case is there is not a complete complementation the inhibition of translation takes place. A nuclease is responsible for the degradation in case of complete complementation.
9. Binding of siRNA to the DNA does not lead to __________
a) Chromatin remodeling
b) Promoter unavailability
c) Transcriptional inhibition
d) Triple helix formation
Answer: d [Reason:] The RISC complex can also be directed by a siRNA into the nucleus where it associates with the complementary region of the DNA. Once there the complex recruits the other proteins that modify the structure around the promoter of the gene thus inhibiting transcription.
10. Which of the following RNAs can induce gene silencing?
Answer: a [Reason:] Apart from siRNA, miRNA and piRNA causes gene silencing. The miRNAs or microRNAs are naturally occurring RNAs that direct the repression of genes, but mostly in plants and worms.
11. The miRNAs are transcribed from non-protein encoding genes and are typically __________ nucleotides long.
a) 10 – 15
b) 12 – 15
c) 18 – 20
d) 20 – 25
Answer: d [Reason:] miRNA or microRNAs are small, non-coding RNA molecules encoded in the genome of plants, animals and their viruses. These are highly conserved sequences ranging about 20 – 25 nucleotides that appear to regulate post-transcriptional gene regulation.
12. The first miRNA was identified in C. elegans named __________
Answer: c [Reason:] The first miRNA was first identified by Victor Ambros and colleagues in C. elegans named Lin-4. The Lin-4 gene was unusual as it did not encoded a protein but rather a small RNA that imperfectly base-pairs with the complementary sequences on the target mRNAs in order to block gene expression. The lin-4 gene controls the expression of the lin-14 gene.
13. Pick the odd one out.
Answer: b [Reason:] siRNA, piRNA and miRNA can all mediate regulation of expression of the gene by the process of gene silencing and RNA interference. snoRNA does not mediate gene regulation by RNA interference.
14. Inactive miRNA undergoes how many cleavages before incorporation into the RISC complex?
Answer: c [Reason:] The process of formation of the active RNA complex needs the cleavage miRNA. The miRNA gene is transcribed into pri-miRNA that folds into a hairpin structure. Nuclear enzyme Drosha cleaves pri-miRNA into pre-miRNA which is exported out from the nucleus. In the cytosol dicer cleaves the pre-miRNA into miRNA. This miRNA is now incorporated into RISC.
15. piRNAs are synthesized in a Dicer dependent mechanism.
Answer: b [Reason:] piRNAs or Piwi-interacting RNAs are most recently discovered class of longer small RNAs of about 25-30 nucleotides. This piRNA binds to the Piwi clade of argonaute proteins. These are thought to be derived from ssRNA precursors and are produced without the dicer cleavage step and thus are dicer-independent.
16. Stem loop precursors are generally seen in which of the following?
d) Both siRNA and miRNA
Answer: b [Reason:] miRNAs are processed from stem-loop precursors with incomplete double-stranded character. siRNAs are derived from a long, fully complementary double stranded RNAs. piRNAs are derived from long ssRNA precursors and does not contain stem-loop structures.
Molecular Biology MCQ Set 3
1. Which of the following cannot be used for the separation of nucleic acids?
a) SDS – PAGE
c) Northern blotting
Answer: a [Reason:] Sodium dodecyl sulphate is a detergent, often used in biochemical preparations, binds to proteins and causes them to form a rod like structure. Most proteins bind SDS in the same ratio (1.4g per g of protein). Thus, the electrophoresis of proteins in an SDS – containing polyacrylamide gel separates them in order of their molecular masses. It is not known to have the similar effect on nucleic acids.
2. The polymerization of the gel used in PAGE occurs between polyacrylamide and ____________
a) N, N – acrylamide
c) N – methyleneacrylamide
d) N, N – methylene bisacrylamide
Answer: d [Reason:] In PAGE, polyacrylamide gel electrophoresis, gel is made by free radical by polymerizing the monomers together. The two major monomers between which the polymerization occurs are polyacrylamide and N, N – methylene bisacrylamide in the buffer of choice.
3. If DNA is digested by endonucleases in four sites giving rise to fragments of which two are equal in length how many bands would be seen after electrophoresis?
Answer: b [Reason:] Digestion at four sites gives rise to five fragments. Two fragments are of same size thus they will form a single band. Therefore only four bands is observed.
4. The fluorescent dye such Ethidium is used for visualizing DNA. How do ethidium binds to DNA?
a) Stacked between histone molecules
b) Binds to the nucleotide base
c) Intercalated between the stacked bases
d) Binds to the phosphodiester backbone
Answer: c [Reason:] Ethidium is intercalated between the stacked bases. This increases the spacing of successive base pairs, distorts the regular sugar phosphate backbone and decreases the twist of the helix.
5. Pulse field gel electrophoresis separates DNA molecules of size ___________
a) 10 – 20 bp
b) 20 – 30 Kb
c) 30 – 50 Kb
d) 40 – 50 bp
Answer: c [Reason:] DNA molecules ranging from 30 – 50 kb migrate in a snake like manner thus, cannot be readily be resolved. These long DNAs can be separated by changing the orientation of the electric field. This process of separation is known as PFGE.
6. Which of the following will migrate faster? The condition is the molecular weight of the following is equal.
a) Supercoiled circular DNA
b) Nicked circular DNA
c) Single stranded DNA
d) Double stranded DNA
Answer: a [Reason:] Supercoiled circular DNA has a less effective volume than the others. Thus, it migrates more rapidly when subjected to electrophoresis due to its compact structure.
7. Agarose can be extracted from which of the following?
a) Gracilaria esculentus
b) Lycazusican esculentum
c) Ficum benghalensis
d) Agrostis stolonifera
Answer: a [Reason:] As we know agarose is extracted from seaweeds. Precisely two types of species are used for the extraction; they are the Gracilaria esculentus and Gelidium nudifrons.
8. Electrophoresis cannot be used to separate _______________
c) Amino acid
Answer: c [Reason:] DNA, RNA and protein can be separated by the by the method of electrophoresis as it can be separate charged molecules. Amino acids are generally separated by the process of chromatography.
9. Which of the following is not a character of polyacrylamide gel?
b) Ionic strength
c) Stable over a wide range of pH
d) Separate upto a few 100 bp of DNA
Answer: d [Reason:] The pore size of polyacrylamide gel is very small due to high resolution power. Thus, it can separate DNA fragments upto a few 100 base pairs only.
10. Pulse field gel electrophoresis was developed by ____________
a) Collins and John
b) Kary Mullis
c) Patrick O’ Farrell
d) Schwartz and Cantor
Answer: d [Reason:] PFGE was developed by Schwartz and Cantor in 1984. This mechanism was developed to separate molecules as long as 10 Mb in an agarose gel.
Molecular Biology MCQ Set 4
1. Which of the following properties is improved by site directed mutagenesis?
a) Physical property
b) Chemical property
c) Kinetic property
Answer: c [Reason:] Site directed mutagenesis is a process used to achieve protein engineering. Protein engineering improves the kinetic property of the protein by altering the amino acid structure and sequence.
2. What is the formula for annealing temperature for polymerase chain reaction primer?
a) Tm = 81.5 + 0.41 (%AT) – (675/N)
b) Tm = 81.5 + 0.41 (%GC) – (675/N)
c) Tm = 81.2 + 0.41 (%GC) – (672/N)
d) Tm = 81.5 + 0.42 (%AT) – (675/N)
Answer: b [Reason:] For calculating the annealing temperature for a PCR primer is Tm = 81.5 + 0.41 (%GC) – (675/N). Here the % of GC is the total amount of G and C present in the oligonucleotide primer and the target of interested and N is the length of the oligonucleotide primer.
3. What is the function of Dpm I endonuclease in Tm method of site directed mutagenesis?
a) Joining of blunt ends
b) Addition of dNTPs
c) Sensitive to cleavage of methylated DNA
d) Breaking of DNA strand
Answer: c [Reason:] In Tm method the template DNA was deprived from an E. coli cell with an intact restriction modification system. This strand is sensitive to restriction by the Dpn I endonuclease.
4. Which phage is used in oligonucleotide directed mutagenesis?
d) λ – phage
Answer: a [Reason:] M13 phage is single stranded phage. It is used for cloning a specific DNA sequence which will be mutated.
5. Write down the name of scientist who has discovered the method of site directed mutagenesis?
a) Bostein Shortle
d) Joller Smith
Answer: d [Reason:] In 1985 Joller Smith discovered site directed mutagenesis technique. By this technique the nucleotide sequence of a cloned DNA fragment may be changed by site directed mutagenesis using synthetic oligonucleotide.
6. Which two genes are absent in the E. coli strain CJ236?
Answer: a [Reason:] CJ236 which lacks functional dut-phase and uracil N-glycosylase. These are used for generating uracilated single stranded DNA.
7. Which polymerase is used in PCR based mutagenesis?
a) Deep vent R polymerase
b) pfu polymerase
c) Taq polymerase
d) DNA polymerase
Answer: b [Reason:] In PCR based mutagenesis pfu polymerase is used. After 12 – 15 cycles of amplification new strands will be generated.
8. Homologous recombination in germ cells occur in the ____________ phase.
Answer: c [Reason:] Homologous recombination in germ cells occurs in the pachytene phase of meiosis I. This exchange is classically termed as crossing over and occurs via the formation of a synaptonemal complex between the two homologous chromosomes.
9. The frequency of crossing over between two genes on a same chromosome is independent of their position.
Answer: b [Reason:] The frequency of crossing over between two genes on a same chromosome is dependent of their locus. The more distance between the two genes the more will be the probability of recombination.
10. When two genes on a chromosome is inherited without any record of recombination the genes are said to be ____________
Answer: d [Reason:] When two genes on a chromosome is inherited without any record of recombination the genes are said to be linked. This is because they are placed so close together that no recombination is able to take place. The phenomenon is known as linkage.
11. Homologous recombination does not provide ____________
a) Genetic variation
b) Sequence retrieval
c) Restart of stalled replication
d) Random base incorporation
Answer: d [Reason:] The homologous recombination provides three functions:
i) Provides genetic variation
ii) Allows the retrieval of sequences lost through DNA damage
iii) Provides mechanism for restarting stalled or damaged replication forks.
12. Recombination regulates the function of genes.
Answer: a [Reason:] Special types of recombination regulate the expression of some genes. For example by switching specific segments within chromosomes, cells can put otherwise dormant genes into sites where they are expressed.
13. What is strand invasion?
a) Fork collapse
b) Nick in template strand
c) Formation of lesion
d) Complementary strand pairing
Answer: d [Reason:] The pairing between short regions occurs when a single stranded region of DNA originating from one molecule with its complementary strand in the homologous DNA duplex. This event is known as the strand invasion process.
14. What is holiday junction?
a) The site of strand break
b) The site of heteroduplex DNA formation
c) Formation of a crossing over complex
d) The site of strand invasion
Answer: c [Reason:] After strand invasion, the two DNA molecules become connected by crossing over of DNA strands. It forms a structure called the Holiday junction.
Molecular Biology MCQ Set 5
1. How many classes of sequences are carried by recombination site?
Answer: b [Reason:] Recombination sites carry two classes of sequence elements. Sequences specifically bound by the recombinants and sequences where DNA cleavage and rejoining occur.
2. CSSSR can generate ____________ Different types of DNA rearrangement.
Answer: c [Reason:] CSSSR can generate three different types of DNA rearrangement.
i) Insertion of a segment of DNA into a specific site
ii) Deletion of a DNA segment
iii) Inversion of a DNA segment.
3. In recombinase recognition sequences what is the short sequence?
a) Crossover region
b) Cleavage region
c) Restriction region
d) Attachment region
Answer: a [Reason:] In the recombinase recognition sequences, the recognition sequence flank a central short asymmetric sequence known as the crossover region. In this site DNA cleavage and rejoining occur.
4. What are the two families of conservative site specific recombinases?
a) Alanine and phenylalanine
b) Threonine and valine
c) Arginine and lysine
d) Serine and tyrosine
Answer: d [Reason:] There are two families of conservative site specific recombinases, the serine and tyrosine recombinases. This mechanism is used when they cleave the DNA, a covalent protein-DNA intermediate is generated.
5. Site specific recombination is ATP dependent process.
Answer: a [Reason:] This process is required ATP hydrolysis. This process is needed for DNA cleavage and joining by the proteins.
6. What are the two proteins which creates site specific recombination?
a) DNA topoisomerases and Spo11
b) DNA gyrase
c) DNA ligase
d) DNA helicase
Answer: a [Reason:] Both DNA topoisomerases and Spo11 the protein that introduces double strand break into DNA. this initiates homologous recombination during meiosis.
7. How many single strands are generated during recombination?
Answer: d [Reason:] During recombination four single stranded DNA must be cleaved. They rejoined after the process with a different partner to generate the rearranged DNA.
8. Which recombinases cleaves the four strands before strand exchange occur?
Answer: a [Reason:] The serine recombinase cleaves all four strands before strand exchange occurs. One molecule of the recombinase protein promotes each of these cleavage reactions.
9. What is the correct pair of the four segments (R1, R2, R3, and R4)?
a) R2 and R3, R1 and R4
b) R2 and R1, R3 and R4
c) R1 and R2, R3 and R4
d) R1 and R3, R2 and R4
Answer: a [Reason:] For recombination to occur, the R2 segment of the Trp DNA molecule must recombine with R3 segment of the bottom DNA molecule. R1 segment of the top molecule must recombine with R4 segment of the bottom DNA molecule.
10. What is the nature of the DNA complex formed by the recombination process?
d) Non polar
Answer: b [Reason:] The structure of the DNA complex is hydrophobic. It provides little barrier to impede rotation of the top and bottom halves of the complex around each other.
11. How many mechanisms are involved in recombination process?
Answer: c [Reason:] Three mechanisms support the model of recombination:
i) DNA cleavage to form the covalent enzyme-DNA intermediate
ii) An 180˚ rotation of the dimmers in the protein-DNA complex
iii) Attack of the 3’-OH DNA ends on the resolvase DNA linkages.
12. How does tyrosine recombinase acts?
a) First join then cleave
b) First cleave then join
c) Both process occur simultaneously
d) Cleaves and rejoin two DNA pairs one after another
Answer: d [Reason:] The tyrosine recombinase cleaves and rejoins two DNA strands first and only then cleaves and rejoins the other two strands. The two DNA molecules with their recombination sites then align.
13. What is the enzyme involved in the tyrosine recombination process?
a) Cre recombinase
b) DNA polymerase
d) DNA recombinase
Answer: a [Reason:] In tyrosine recombinase, Cre recombinase is an example of the enzyme involved in the mechanism. This enzyme is encoded by phage P1 and acts by binding to to two different configurations of the recombining DNA.
14. What is the name of the site where Cre enzyme acts?
Answer: b [Reason:] The recombination site on the DNA where Cre acts are called the LOX site. Cre-LOX is a simple example of recombination by the tyrosine recombinase family.
15. What is the function of Salmonella Hix invertase?
a) Inversion of chromosomal region
b) Promotion of DNA deletion
c) Promotion of DNA integration
d) Promotion of base inversion
Answer: a [Reason:] It inverts a chromosomal region to flip a gene promoter by recognizing Hix sites. It allows expression of two distinct surface antigens.
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