Molecular Biology MCQ Set 1
1. Which of the following statements are true about aminoacyl tRNA
i. Recognition and attachment of correct amino acid depends on aminoacyl tRNA synthetase.
ii. Transfer of amino acid to the polypeptide chain.
iii. Recognition of specific codon
iv. Recognition of specific anticodon
Chose the correct option
a) i and ii
b) iii and iv
c) i, ii and iii
d) i, ii, iii and iv
Answer: d [Reason:] Recognition and attachment of correct amino acid is the primary role of aminoacyl tRNA synthetase and requires the identification of anticodon on the tRNA to attach the correct amino acid. This aminoacyl tRNA now transfers the amino acid attached to it to the elongating polypeptide chain by recognizing the codon on the mRNA during translation.
2. For the charging of tRNA molecules the acyl linkage occurs between the carboxyl group of the amino acid to the ____________
a) 2’ hydroxyl group of A
b) 3’ hydroxyl group of T
c) 2’ hydroxyl group of G
d) 3’ hydroxyl group of C
Answer: a [Reason:] The tRNA molecules to which amino acids are attached are said to be charged. Charging requires an acyl linkage between the carboxyl group of the amino acid and 2’ or 3’ hydroxyl group of the adenosine nucleotide that protrude from the acceptor stem.
3. The formation of the acyl linkage is significant for protein synthesis even though it is a high energy bond.
Answer: a [Reason:] Acyl linkage is formed due to hydrolysis which results in the change in free energy. This is significant for protein synthesis: the energy released when the bond is broken helps drive the formation of the peptide bonds that links amino acids to each other in polypeptide chain.
4. The joining of the amino acid to the tRNA requires ___________ steps.
Answer: b [Reason:] All aminoacyl tRNA synthetases attach amino acids to the tRNA in two enzymatic steps. Step 1: adenylation in which the amino acid reacts with ATP to become adenylated. Step 2 is tRNA charging in which the adenylated amino acid reacts with the tRNA while bound to the tRNA synthetases.
5. The principle driving force for adenylation reaction during the formation of the aminoacyl tRNA is carried out by ___________
Answer: c [Reason:] The principle driving force for adenylation reaction is the subsequent hydrolysis of pyrophosphate by pyrophosphatases. As a result of adenylation, the amino acid is attached to adenylic acid via a high energy ester bond in which in which the carbonyl group of the amino acid is joined to the phosphoryl group of the AMP.
6. How many types of tRNA synthetases are found?
Answer: c [Reason:] There are two classes of tRNA synthetases. Class I enzymes attach the amino acid to the 2’ – OH of the tRNA and are generally monomeric. Class II enzymes attach the amino acid to the 3’ – OH of the tRNA and are generally dimeric or tetrameric.
7. How many tRNA synthetases are found in a cell?
Answer: d [Reason:] Each 20 amino acids are attached to the appropriate tRNA by a single, dedicated tRNA synthetase. Because most amino acid are identified by more than one codon, it is not uncommon for one synthetase to identify and charge more than one tRNA. Thus, only one tRNA synthetase is responsible for charging all tRNAs for a particular amino acid.
8. The quaternary structure of which of the aminoacyl tRNA synthetase is the odd one among the following?
Answer: a [Reason:] The quaternary structure of the aminoacyl tRNA synthetase of Alanine, Proline and Serine are monomeric. Glycine has a dimeric aminoacyl tRNA synthetase molecule.
9. Which of the following pair is an example for using only one type of tRNA synthetase in bacteria?
a) Glutamine and cystine
b) Glutamic acid and asparagine
c) Cystine and Valine
d) Glutamine and glutamic acid
Answer: d [Reason:] Some bacteria lack a synthetase for charging the tRNA for glutamine (tRNAGln) with its cognate amino acid. Instead, a single species of aminoacyl tRNA synthetases charges tRNAGln (glutamic acid) as well as tRNAGlu.
10. Which of the following parts of the mRNA determines the specificity of the amino acid attached?
a) Acceptor stem
b) Variable loop
c) ΨU loop
d) D loop
Answer: a [Reason:] Genetic, biochemical and X-ray crystallographic evidence indicate that the specificity determinants are clustered at the two distant sites on the molecule. They are the acceptor stem and the anticodon loop.
11. Any mutation in the sequence of the acceptor stem does not pose a threat to the amino acid incorporation.
Answer: b [Reason:] The acceptor stem is an especially important determinant for the specifically of tRNA synthetase recognition. In some cases changing a single base pair in the acceptor stem is sufficient to convert the recognition specificity of a tRNA from one synthetase to another.
12. For the addition of amino acids to the tRNA molecules the tRNA synthetases rely on the guidance provided by the anticodon sequence.
Answer: b [Reason:] For the addition of amino acids to the tRNA molecules by the tRNA synthetases, guidance provided by the anticodon sequence is not enough for the specificity as amino acids are usually specified by more than one codon. Thus to recognize the specific tRNAs, the synthetases for that amino acid must rely on determinants that lie outside of the anticodon.
13. The set of tRNA determinants that enable synthetases to discriminate among tRNAs are called ___________
a) Primary genetic code
b) First genetic code
c) Secondary genetic code
d) Second genetic code
Answer: d [Reason:] The set of tRNA determinants that enable synthetases to discriminate among tRNAs are called second genetic code because of its central importance in information flow. This genetic code is significantly more complex than the first genetic code and cannot be readily tabulated. Without such a code, synthetases could not distinguish one tRNA from another, and the translation machinery would not produce polypeptides with a reproducible sequence.
Molecular Biology MCQ Set 2
1. With respect to DNA polymerase holoenzyme pick the odd one out.
a) 2 copies of DNA polymerase III
b) 5 protein γ complex
c) Presence of ATP
d) Presence of DNA
Answer: d [Reason:] The sliding clamp loader or the γ complex are five protein subunit protein complexes whose activity is controlled by ATP binding and hydrolysis. To catalyze the sliding clamp opening, clamp loader must be bound to ATP. Once bound to ATP the clamp loader binds to the clamp and opens the ring at one of the sub unit interface. The resulting complex now binds to DNA.
2. A complete DNA polymerase III holoenzyme contains ____________sub units.
Answer: c [Reason:] The complete DNA polymerase III holoenzyme contains 6 subunits. They are two copies of DNA polymerase III core proteins, one sliding clamp connected to a clamp loader. This clamp loader is further connected to the two DNA polymerase III core proteins by a pair of τ proteins.
3. How many polymerases are present in a replication fork?
Answer: b [Reason:] Three different polymerases are present in a replication fork. They are DNA polymerase α or primase, DNA polymerase δ and DNA polymerase ε. The DNA polymerase α or primase initiate new strand synthesis and DNA polymerase δ and DNA polymerase ε extend these strands.
4. With respect to eukaryotic replication which of the following enzyme is not correctly paired with its function?
a) Primase – initiation of new strand synthesis
b) RF-C – recruitment, maintenance and coordination of DNA polymerases
c) DNA polymerase ε – elongation of DNA strand
d) τ protein – binds to DNA polymerase
Answer: b [Reason:] In case of prokaryotic replication the clamp loader complex is responsible for the recruitment, maintenance and coordination of DNA polymerases. But its analog, the RF-C complex found in the eukaryotes does not perform this function.
5. The formation of the phosphodiester bond is an example of_________reaction.
Answer: b [Reason:] The phosphodiester bond is formed in an SN2 reaction in which the hydroxyl group at the 3’ end of the primer strand attacks the α-phosphoryl group of the incoming nucleoside triphosphate. The leaving group for the reaction is pyrophosphate, which arises from the release of the β- and γ-phosphates of the nucleotide substrate.
6. DNA synthesis is known as a coupled process.
Answer: a [Reason:] The addition of a nucleotide to a growing polynucleotide chain of length n is indicated by the reaction: NTP + (NMP)n → (NMP)n+1 + P – P.
But the free energy for this reaction is rather small. Additional free energy is provided by the rapid hydrolysis of the pyrophosphates into two phosphate groups (P – P → 2Pi). The net result of nucleotide and pyrophosphate hydrolysis is the breaking of two high – energy phosphate bonds. Therefore, the DNA synthesis is known as a coupled process.
7. DNA helicase travels along _________
a) Leading strand template in 3’→5’ direction
b) Leading strand template in 5’→3’ direction
c) Lagging strand template in 3’→5’ direction
d) Lagging strand template in 5’→3’ direction
Answer: d [Reason:] The DNA helicase travels along the lagging strand template in 5’→3’ direction in the replication fork. The DNA primer associates with this DNA helicase to synthesize new primers on the lagging strand template.
8. DNA polymerase III holoenzyme does not need to interact with the helicase during replication.
Answer: b [Reason:] The DNA polymerase III holoenzyme interacts with the DNA helicase through the τ subunit, which also binds to the DNA polymerases. One of these polymerases replicates the leading strand and the other replicates the lagging strand.
9. Binding of polymerase III and helicase ensures helicase efficiency.
Answer: a [Reason:] The association of helicase with polymerase III stimulates the activity of the helicase by increasing the rate of its movement tenfold. Thus the DNA helix slows down if it is separated from the polymerase III. This binding also controls and coordinates the function of both the helicase and the DNA polymerase holoenzyme.
10. Association of primase with DNA template is brought about by the interaction of primase with _________________
a) Dna polymerase III
b) Clamp-loader complex
c) Helicase and SSB protein
d) τ protein
Answer: c [Reason:] Association of primase with DNA template is brought about by the interaction of primase with the helicase and SSB protein. Although this interaction is relatively weak, it strongly stimulates primase function.
11. How is the strength of interaction between primase and helicase which regulates the length of Okazaki fragments?
a) Weak association short fragment
b) Strong association short fragment
c) Strong association long fragment
d) Does not depend on strength of association
Answer: b [Reason:] The relatively weak interaction between the primase and DNA helicase is important for regulating the length of the Okazaki fragments. A tighter association would ensure more frequent primer synthesis on the lagging strand and, therefore, shorter Okazaki fragments. Similarly, a weaker interaction would result in longer Okazaki fragments.
12. Which of the following enzyme is not required for the synthesis of lagging strand?
Answer: b [Reason:] The functions of the above enzymes are
i) Primase – synthesis of primer for replication
ii) Helicase – unwinds the DNA during replication
iii) Exonuclease – hydrolysis of the rNTP of the primer that is directly bound to the dNTP of the newly synthesized strand
iv) Gyrase – releases stress by removing supercoils within the DNA helix. Thus gyrase does not take part in the synthesis of lagging strand.
13. Length of Okazaki fragments in eukaryotes ranges between ____________ nucleotides.
a) 100 – 400
b) 400 – 800
c) 800 – 1200
d) 1200 – 1600
Answer: a [Reason:] The length of Okazaki fragment varies in between 100 – 400 nucleotides in eukaryotes. The length of Okazaki fragments in bacteria is about 1000 – 2000 nucleotides. They are the transient intermediates in lagging strand DNA replication.
14. Polymerization of dNTP by DNA polymerase is in 3’ → 5’ direction.
Answer: b [Reason:] DNA polymerase needs a free 3’ – OH to polymerize dNTPs. This free 3’ – OH end is provided only when the DNA strand is elongated in 5’ → 3’ direction. Thus, polymerization of dNTP by DNA polymerase is in 5’ → 3’ direction.
Molecular Biology MCQ Set 3
1. RecA protein of E. coli is involved in the process of ____________
a) Pairing of homologous DNA
b) Introduction of DSB
c) Assembly of strand exchange
d) Resolution of Holiday junction
Answer: a [Reason:] In E. coli, RecA protein is involved in pairing homologous DNA and strand invasion. In eukaryotes, Rad 51 is involved in this same process.
2. RecBCD and RecFOR in E. coli is involved in the process ____________
a) Assembly of strand exchange
b) Holiday junction recognition
c) Introduction of DSB
d) Resolution process
Answer: a [Reason:] RecBCD and RecFOR in E. coli are involved in the assembly of strand exchange proteins. In eukaryotes Rad52 and Rad59 are involved in this same process.
3. In bacteria specific proteins are involved in the RecBCD pathway.
Answer: b [Reason:] In bacteria, no specific proteins have been found that performs the RecBCD pathway. Breaks result in the damage of DNA, missteps in DNA repair or collapse of a replication fork.
4. What is holiday junction resolving enzyme?
Answer: b [Reason:] RuvC in E. coli is the holiday junction resolving enzyme. In eukaryotes this function is performed by the enzyme Rad51C-XRCC3 complex.
5. RecBCD enzyme has polymerase and transcriptase activity.
Answer: b [Reason:] RecBCD enzyme has both DNA helicase and nuclease activity. It processes broken DNA molecules of DNA by the process of recombination.
6. RecBCD pathway is an ATP dependent process.
Answer: a [Reason:] The RecBCD complex binds to DNA molecules at the site of DSB and tracks along DNA using the energy of ATP hydrolysis as a result the DNA is unwound.
7. Chi is the control site of the RecBCD pathway.
Answer: a [Reason:] The activities of RecBCD are controlled by specific DNA sequence elements known as Chi sites. Chi sites were discovered because they stimulate the frequency of homologous recombination.
8. What are the helicase components of RecBCD pathway?
a) RecB, RecC
b) RecC, RecD
c) RecB, RecD
d) RecA, RecB
Answer: c [Reason:] RecB and RecD subunits are both DNA helicases. These enzymes use ATP hydrolysis to melt and unwind the base pairing of DNA double strand.
9. What is the function of Rec component?
a) 5’→3’ helicase activity
b) 3’→5’ helicase activity
c) Recognize Chi sites
d) Nuclease activity
Answer: c [Reason:] Rec functions recognize Chi sites. Chi sites are the controlled sites of RecBCD pathway.
10. What are the two components of RecBCD pathway involved as motor parts?
a) RecB, RecD
b) RecB, RecC
c) RecB, RecA
d) RecC, RecD
Answer: a [Reason:] RecB and RecD helicase acts as motors help to move independently along opposite strands of the DNA duplex and at different speeds. RecB and RecD are helicase parts of the RecBCD pathway.
11. What is the speed of the driving the RecBCD along the DNA?
a) More than 2000 bp per second
b) More than 1000 bp per second
c) More than 200 bp per second
d) More than 100 bp per second
Answer: b [Reason:] RecB and RecD are capable together of driving the RecBCD complex along the DNA at rates of greater than 1000 bp per second. Chi sites within DNA act as a sort of molecular throttle to regulate the activities of the helicases and therefore the speed of DNA translocation.
12. How many times the frequency is increased by the Chi sites?
a) ≈10 fold
b) ≈20 fold
c) ≈30 fold
d) ≈40 fold
Answer: a [Reason:] Chi sites increase the frequency of recombination by ≈10 fold. This stimulation is most pronounced directly adjacent to the Chi sites.
13. The nuclease activity of the RecBCD is prone to digest the bacterial cell wall from foreign DNA.
Answer: b [Reason:] The ability of Chi sites to control the nuclease activity of RecBCD also helps bacterial cell protect them from foreign DNA. This foreign DNA may enter via phage infection or conjugation.
14. What is the sequence of the Chi site that is highly found in the E. coli genome?
Answer: B [Reason:] The 8 nucleotide Chi site is GCTGGTGGC. It is found in high concentration in the E. coli genome.
15. If the foreign DNA lacks the Chi site what will happen?
a) Chi site of DNA will destroy
b) Foreign DNA is not affected
c) Foreign DNA easily enters the cell
d) DNA sequence will be intact as previous
Answer: c [Reason:] RecBCD sometimes function simply to destroy DNA as it does when foreign DNA lacking frequent Chi sites enters cells. RecBCD can protect cells from the potentially deleterious consequences of taking up foreign sequences.
Molecular Biology MCQ Set 4
1. Which of the following is not a part of a gene?
Answer: a [Reason:] Ori is the site for origin of replication. This does not take part in any kind of gene expression and thus is not a part of any gene. Its function is to govern genome duplication.
2. What is the correct order for increasing gene density?
a) Bacteria, Virus, Fruit fly, Human
b) Fruit fly, Bacteria, Virus, Human
c) Human, Fruit fly, Bacteria, Virus
d) Virus, Bacteria, Fruit fly, Human
Answer: c [Reason:] As the complexity of the organism increases the gene density decreases. Thus, gene density is inversely proportional to organism complexity. Therefore, viruses being the simplest organisms have the highest gene density and humans the lowest.
3. Which of these factors contribute in decrease in gene density?
b) Intergenic sequence
c) Exon sequence
d) Cell size
Answer: b [Reason:] As the genome size increases the DNA between two consecutive sequences increase, known as the Intergenic sequences. Thus, with the increase in genome complexity, regulatory sequences are required for the proper functioning of the different mechanisms. These regulatory sequences are stored in these Intergenic sequences.
4. What do you think is the requirement of Intergenic DNA in higher organisms?
a) Just genetic load
b) To avoid viable mutations
c) Helps in regulation of transcription
d) Helps in genome organization
Answer: b [Reason:] With the increase in genome size the gene density decreases thus the rate of mutation at a viable point also decreases. This is why viable mutations in bacteria and viruses are more frequent than in humans.
5. Which of the following is not a character of unique Intergenic DNA?
a) Mutant genes
d) Nonfunctional gene fragments
Answer: c [Reason:] Palindromes are specialized sequences identified by different enzymes and proteins. They are unique structures but are not found in unique Intergenic regions of DNA. These are generally found in the highly functioning DNA.
6. Which of the following is responsible for the formation of Pseudogenes?
b) Reverse transcriptase
c) Klenow fragment
d) Wrong primer
Answer: b [Reason:] Pseudogenes arises from the action of enzyme reverse transcriptase. This enzyme copies RNA into double stranded DNA and are found in certain RNA viruses. Thus, when infected by these specific viruses the reverse transcriptase might copy the host cellular mRNA into a double stranded copy DNA. This copy DNA might get incorporated into the host genome giving rise to Pseudogenes.
7. With respect to microsatellite DNA which of the following is correct?
a) Tandem repeats
b) Dinucleotide repeats
c) 100 bp units
d) Inaccurate duplicating
Answer: c [Reason:] Microsatellites are dinucleotide repeats such as “ATATATATATAT”. Thus, they are of a small stretch of DNA and are very short and less than 13 base pairs.
8. Genome – wide repeat unit are generally transposable elements ____________
Answer: a [Reason:] Genome – wide repeat unit are greater than 100 bp in length. They are found in either as single copies dispersed throughout the genome or in clusters. They are of numerous classes but their most common feature is that all are forms of transposable elements.
9. Repeated DNA can be referred to as junk DNA .
Answer: b [Reason:] Although it is tempting to refer repeated DNA as junk DNA, they are stably maintained over the hundreds and thousands of years. This suggests that Intergenic DNA confers a positive value to the host organisms.
10. The simpler the organism the more likely they are to have transposable elements .
Answer: b [Reason:] Even in the compact genomes of simpler organisms transposable elements are found. But the rate of success of occupying the genome is very less for these elements. This lack of success is likely a combination of inefficient duplication and/or more efficient elimination.
Molecular Biology MCQ Set 5
1. Replication of chromosome occurs during the____________ phase.
Answer: b [Reason:] Replication of chromosome occurs during the S – phase of the cell cycle. During this time all the DNA is duplicated exactly once. Incomplete replication of any part causes inappropriate links between daughter chromosomes.
2. It is important to replicate the DNA correctly and only once.
Answer: a [Reason:] During S – phase the DNA is duplicated exactly once. Incomplete replication of any part causes inappropriate links between daughter chromosomes. Segregation of this linked chromosome causes chromosome breaks or loss. Rereplication of DNA can also have severe consequences.
3. Selection of replicator occurs in ___________phase.
Answer: a [Reason:] Replicator selection is the process of identifying sequences that will direct the initiation of replication and occurs in the G1 phase. This process leads to the assembly of a multiprotein complex at each replicator in the genome.
4. Activation of the origin of replication occurs in ____________ phase.
Answer: b [Reason:] Origin activation only occurs after the cell enters the S phase. This triggers the replicator – associated protein complex to initiate DNA unwinding and DNA polymerase recruitment.
5. The replicator selection is mediated by the formation of pre – replicative complex. It comprises of _________________
a) ORC and the initiator protein
b) ORC and helicase
c) A helicase loader and a helicase
d) An ORC, two helicase loaders and a helicase
Answer: d [Reason:] The replicator selection is mediated by the formation of pre – replicative complex. It comprises of four separate proteins assembled together. The four parts are the eukaryotic initiator, ORC which recognizes and binds to the replicator sequence of the DNA. This further facilitates the binding of 2 helicase loading proteins and finally a helicase.
6. The pre – RCs are auto – activated as soon as the cell enters the S – phase .
Answer: b [Reason:] The pre – RCs are activated by two protein kinases. Protein kinases are those enzymes which covalently attaches phosphate groups to target enzymes to activate them.
7. Which of the following is wrong about the cyclin dependent kinases?
a) Phosphorylates enzymes
b) Activates pre – RCs
c) Promotes formation of new pre – RCs
d) Activates in the S – phase
Answer: c [Reason:] Cyclin dependent kinases play 2 strongly contradictory roles in regulating pre – RC function. First they activate the pre – RC complex to initiate DNA replication. Cdk activity also inhibits formation of new pre – RCs.
8. Active cyclin dependent kinase is absent in ____________phase.
Answer: a [Reason:] Active Cdk is absent during G1, where as elevated levels of Cdk are present during the remainder of the cell cycle. Thus, during each cycle there is only one opportunity for pre – RCs to form and to be activated.
Synopsis and Project Report
You can buy synopsis and project from distpub.com. Just visit https://distpub.com/product-category/projects/ and buy your university/institute project from distpub.com