Molecular Biology MCQ Set 1
1. How many charged tRNA substrate is used for the polypeptide formation at a time?
Answer: b [Reason:] Both the growing chain and the incoming amino acid are attached to tRNAs; as a result, during peptide bond formation, the growing polypeptide is continuously attached to a tRNA. The actual substrates for each round of amino acid addition are two charged species of tRNAs – an aminoacyl-tRNA and a peptidyl tRNA.
2. The peptidyl transferase reaction is a result of ____________ consecutive steps.
Answer: a [Reason:] The peptidyl transferase reaction is a result of 2 consecutive steps. First, the N-terminus of the protein is synthesized followed by the transfer of the growing polypeptide from the peptidyl-tRNA to the aminoacyl tRNA.
3. The polypeptide bond formation is an ATP dependent process.
Answer: b [Reason:] Peptide bond formation takes place without the simultaneous hydrolysis of a nucleoside triphosphate. This is because peptide bond formation is driven by breaking the high energy acyl bond that joins the growing polypeptide chain to the tRNA.
4. During the addition of amino acid the codon on the mRNA is identified by the direct interaction of the anticodon loop of the tRNA.
Answer: b [Reason:] The 16S rRNA of the small subunit also plays a critical role in the addition of amino acid in the growing polypeptide chain. The anticodon loops of the charged tRNAs and the codons of the mRNA contact the 16S rRNA, not the ribosomal proteins of the small subunit.
5. The peptidyl transferase reaction center of the ribosome contains ___________ sites.
Answer: c [Reason:] The peptidyl transferase reaction center of the ribosome contains 3 sites. The A site for the addition of the aminoacyl-tRNA. The P site is the binding site for the peptidyl tRNA.
6. How many channels are observed in a ribosome molecule?
Answer: c [Reason:] Three channels are observed in a ribosome molecule. The first two are found in the small subunit that serves as the entry and exit channels for the mRNA. The third channel is found in the large subunit and provides an exit path for the newly synthesized polypeptide chain.
7. The channel for the entry of the mRNA into the decoding center is a crucial feature for the proper decoding of them RNA.
Answer: a [Reason:] This is because the entry channel is only wide enough for unpaired RNA to pass through. This feature ensures that the mRNA is in an extended form as it enters the decoding center by removing any intramolecular base-pairing interactions that may have formed in the mRNA.
8. The polypeptide elongation channel allows the formation of which of the following structures?
a) α – helix
b) β – sheet
Answer: a [Reason:] The channel found in the large subunit, provides an exit path for the newly synthesized polypeptide chain. The size of this channel limits the formation all secondary structures except the α – helix. For this reason, the final 3D form of the newly synthesized protein is not attained until after it is released from the ribosome.
Molecular Biology MCQ Set 2
1. The 3’ end of tRNA is __________
a) 3’ CCA 5’
b) 3’ ACC 5’
c) 3’ CCG 5’
d) 3’ GCC 5’
Answer: b [Reason:] All tRNA has a conserved sequence of 3’ ACC 5’. This sequence is conserved so as to facilitate cognate amino acid binding.
2. Thymine is present in tRNA _____________
Answer: a [Reason:] Some unusual bases are present in the tRNA for its improved functioning. One of such pos – transcriptional modification done enzymatically is the presence of thymine base in place of uracil in the primary structure.
3. With respect to tRNA which of the following is not its characteristic?
a) Complimentary region
b) Double helix molecule
c) Highly conserved pattern of fold
d) Variable loop
Answer: b [Reason:] The complementary region enables the tRNA to form limited stretches of double helix held together by base pairing. The other stretches of the tRNA are non – complementary thus remain single stranded as loops. Thus the tRNA is a mixture of single and double stranded regions represented by the clover leaf model.
4. How many loops are present in the clover leaf model of tRNA?
Answer: c [Reason:] There are 4 loops and an acceptor stem in the clover leaf model of tRNA. The names of the 4 loops are ΨU loop, D loop, anticodon loop and the variable loop.
5. Why the variable loop is named so?
a) Variable number of bases
b) Variable region
c) Multiple loops present
d) Variability of presence
Answer: a [Reason:] The variable loop sits between the anticodon loop and the ΨU loop and its region is fixed. The name is so given because they vary in size from 3 to 21 bases.
6. The distance between the amino acid binding site and the anticodon site is ______
a) 64 Å
b) 70 Å
c) 76 Å
d) 80 Å
Answer: b [Reason:] X – ray crystallography reveals the tRNA has an L – shaped tertiary structure. In this structure the amino acid binding site o acceptor site is approximately 70 Å away from the anticodon site residing on the opposite end of the structure.
7. What is the angle between the D loop and the anticodon loop?
Answer: b [Reason:] The X – ray crystallography shows that the D loop and the ΨU loop comes together making the anticodon loop an extended helix. These to helices align at right angle, that is, 90˚ to each other in their tertiary structure.
8. Which of the following does not contribute to the stability of tRNA?
a) Base and sugar – phosphate backbone interaction
b) Hydrogen bonding
c) Hydrophobic interactions
d) Base pairing
Answer: c [Reason:] Due to the negative charge of the tRNA backbone, it is unlikely for the molecule to have a hydrophobic nature. Thus, hydrophobic interactions do not play any role in the structural stability of the molecule. On the other hand, the other interactions such as the hydrogen bonding leads to base pairing which leads to its classic clover leaf model.
Molecular Biology MCQ Set 3
1. Topoisomerases I and II differ only on the number of steps required to change the linking number.
Answer: b [Reason:] Type II topoisomerase changes the linking number in two steps whereas type I in one step. Also for this process type II topoisomerase requires ATP but type I do not require ATP hydrolysis.
2. Which special enzyme introduces negative supercoils in DNA?
a) Type I topoisomerase
b) Type II topoisomerase
Answer: c [Reason:] A special type of type II topoisomerase found in prokaryotes, known as the DNA gyrase introduces negative supercoils rather than removing it. DNA gyrase is responsible for the negative supercoiling of DNA in prokaryotes.
3. Which of the following is not a property of topoisomerase?
a) Removal of supercoil
b) Helps in DNA replication
c) Cannot catenate and decatenate
d) Utilizes ATP to change linking number
Answer: c [Reason:] Topoisomerases promotes several other reactions important to maintaining the proper DNA structure within the cell. The enzymes use the same transient break and strand passage reaction for both catenation and decatenation of circular DNA molecules.
4. Topoisomerases introduces catenation to promote cell division.
Answer: b [Reason:] Topoisomerases play an essential role for unlinking DNA molecules and thus allowing them to separate into two daughter cells promoting cell division. The process of separation of these cccDNA is known as decatenation. This reaction is readily promoted by type II topoisomerase.
5. Type I topoisomerase can never decatenate catenated circular DNA?
Answer: b [Reason:] Type II topoisomerase can decatenate circular DNA in any situation. But type I topoisomerase can only decatenate catenated circular DNA when any one of the strand caries a nick or gap.
6. The promotion of DNA breaking and rejoining by topoisomerase requires______________
a) High – energy cofactor
c) DNase I
d) No external enzyme required
Answer: d [Reason:] Topoisomerases are able to promote both DNA cleavage and rejoining without the assistance of other proteins or high – energy cofactors. This is because they use a covalent – intermediate mechanism which stores the energy of phosphodiester bond cleavage to reseal the backbone later.
7. Attack of the topoisomerase gives rise to ____________
a) Hydroxy – valine linkage
b) Phospho – tyrosine linkage
c) Hydroxy – cystine linkage
d) Phospho – proline linkage
Answer: b [Reason:] DNA cleavage occurs when a tyrosine residue in the active site of the topoisomerase attacks a phosphodiester bond in the backbone of the target DNA. This attack generates a break in the DNA whereby the topoisomerase is covalently joined to one of the broken ends via a phospho – tyrosine linkage.
8. To reseal the DNA backbone where does topoisomerase gets the energy from?
a) ATP hydrolysis
b) High energy cofactor
c) Phosphodiester cleavage
d) Phospho – tyrosine cleavage
Answer: c [Reason:] The Phospho – tyrosine linkage conserves the energy in the phosphodiester bond that was cleaved. Therefore, the DNA can be resealed simply by reversing the original reaction: the OH group from one broken DNA end attack the Phospho – tyrosine bon reforming the DNA phosphodiester bond.
9. Which of the following does not occur during cell division in prokaryotes?
a) Multiple decatenations
b) DNA cleavage
c) DNA rejoining
d) Strand passage
Answer: a [Reason:] Between the steps of DNA cleavage and DNA rejoining, the topoisomerase promotes passage of a second segment of DNA through the break. Topoisomerase function thus involves DNA cleavage, strand passage and DNA rejoining in highly coordinated manner.
10. The active site of the topoisomerase contains a tyrosine residue.
Answer: a [Reason:] Yes the active site of the topoisomerase contains a tyrosine residue. This is a major requirement because it is used to produce the covalent DNA – tyrosine intermediate. This intermediate thus formed, stores the energy evolved during the phosphodiester bond breakage to reseal the backbone after the process of decatenation is complete.
Molecular Biology MCQ Set 4
1. Which of the following facilitates nucleosome positioning?
a) Nucleosome remodeling complex
b) Topoisomerase II
c) SMC protein
d) DNA – binding protein
Answer: d [Reason:] Nucleosome positioning can be directed by DNA – binding protein or particular DNA sequences. Just as many proteins cannot bind to DNA within the nucleosome, prior binding of a protein to a site on DNA can prevent association of the core histone with that stretch of DNA. If two such DNA – binding proteins are bound to sites closer than the minimal nucleosomal DNA requirement (≈150 bp) the DNA stretch between the two proteins will remain nucleosome free.
2. Which of the following regions promote histone – DNA association?
a) A, T
b) A, G
c) G, C
d) C, T
Answer: A [Reason:] A:T rich DNA has an intrinsic tendency to bend toward the minor grove. Thus A:T rich DNA is favored in positions in which the minor grove faces the histone octamer. G:C rich DNA has the opposite tendency thus, is favored when the major grove faces away from the histone octamer.
3. With respect to end terminal modification of histone tail which of the following options is true?
a) Lysine – phosphate modification
b) Serine – methylation
c) Acetylation – transcriptionally active
d) Modifications – involvement in gene expression
Answer: d [Reason:] Acetylation marks transcriptionally active region whereas methylation marks both transcriptionally active and repressed regions. Finally, phosphorylation histone H3 is commonly seen in highly condensed chromatin. Thus, these modifications result in a code that can be read by the proteins involved in gene regulation and expression.
4. Which of the following is a function of a modified histone tail?
a) Association with adjacent nucleosome
b) Interaction with DNA backbone
c) Interaction with chromodomains
d) Formation of 40-fold DNA structure
Answer: c [Reason:] Modification of histone tail generates binding sites for proteins. Specific protein domains called bromodomains and chromodomains mediate these interactions. Chromodomains containing proteins interact with methylated histone tails and are generally associated with tail specific methylating enzymes.
5. Bromodomains of proteins are associated with phosphorylated histone tails.
Answer: b [Reason:] Bromodomains of proteins interacts with the acetylated histone tails. These proteins are generally associated with histone tail – specific acetyl transferases. Such complexes facilitate the maintenance of acetylated chromatin by further modifying the already acetylated regions.
6. Which of the following enzymes of transcription commonly contains a bromodomain?
c) Pre – RC
d) DNA polymerase
Answer: a [Reason:] Bromodomain containing proteins are involved in regulating transcription or formation of heterochromatin. Only TFIID of the above options is involved in transcription of DNA whereas the others are involved in DNA replication.
7. Which of the following enzymes is not correctly paired with its function?
a) Acetyl transferase – adds acetyl group to lysine
b) Methyl transferase – adds methyl group to serine
c) Topoisomerase – associated with nuclear scaffold
d) Deacetylases – removes acetyl groups from serine
Answer: b [Reason:] Methylation is an N – terminal modification of histone tail induced by the enzyme methyl transferase. But methylation is induced in the amino acid lysine. Serine is generally involved with phosphorylation.
8. Which of the following is not promoted by histone tail modification?
a) Formation of repressive structures
b) Gene expression
c) Nucleosome remodeling
d) Nucleosome sliding
Answer: d [Reason:] Nucleosome sliding is a type of Nucleosome remodeling. It is facilitated by Nucleosome remodeling complex. It cannot be promoted alone by modifying histone tail unless a nucleosome remodeling complex comes along.
Molecular Biology MCQ Set 5
1. The β subunit of polymerase has a function of ____________
a) Promoter binding
b) Catalytic center
c) Template binding
d) Cation binding
Answer: b [Reason:] The β subunit of polymerase is thought to be its catalytic center. Studies suggested that the β subunit may contain two domains responsible for the transcription initiation and elongation.
2. Presence of the heparin promotes transcription.
Answer: b [Reason:] A poly heparin has been shown to bind to the β’ subunit. Heparin inhibits transcription invitro and also competes with DNA for binding to polymerase.
3. The β’ subunit of polymerase has a function of ________________
a) Promoter binding
c) Cation binding
Answer: c [Reason:] The β’ subunit of polymerase binds to two Zn2+ ions which are thought to participate in catalytic function of the polymerase. It is also responsible for the binding of the enzyme to the template DNA.
4. The RNA polymerase core enzyme converts into a holoenzyme on the addition of the _________ subunit.
Answer: d [Reason:] Binding of the σ factor converts the core RNA polymerase enzyme into the holoenzyme. The σ factor has a critical role in promoter recognition, but is not required for transcription elongation.
5. Four types of σ factors are known of them which one is heat stable?
Answer: a [Reason:] σ70 is the most common σ factor.
σ32 is highly heat stable.
σ54 is used during N2 deficiency.
σ28 is used for chemotaxis.
6. What is the function of the ω subunit of RNA polymerase?
a) Subunit association
b) Promoter binding
c) Initiation and elongation
d) Cation binding
Answer: a [Reason:] The ω subunit of RNA polymerase helps in the association of all the subunits. Promoter binding is promoted by α subunit. Initiation and elongation is promoted by the β subunit. And β’ subunit binds to the Mg2+ cations.
7. Which of the following is not the part of the RNA polymerase core enzyme?
Answer: d [Reason:] The σ subunit is not the part of the RNA polymerase core enzyme. The RNA polymerase core protein has the structural formula of α2ββ’ω. Addition of the σ subunit converts the RNA polymerase core protein into a holoenzyme.
8. The bacterial system has _____________ RNA polymerases.
Answer: [Reason:] The bacterial system has 3 RNA polymerases. They are RNA polymerase I, RNA polymerase II and RNA polymerase III. RNA polymerase I transcribes mRNAs, and RNA polymerases II and III transcribes the other specialized RNAs.
Molecular Biology MCQ Set 6
1. What is the central protein of homologous recombination?
Answer: c [Reason:] RecA is the central protein of homologous recombination. It is the founding member of a family of enzymes called the strand exchange protein.
2. What is the joint molecule?
a) RecB of the RecBCD pathway
b) B DNA
c) Z DNA
d) Rec A bound three stranded structure
Answer: d [Reason:] The RecA bound three stranded structure is known as the joint molecule. It usually contains several hundred base pairs of hybrid DNA.
3. The RuvAB complex specifically recognizes the Holliday junction in a ATP dependent manner.
Answer: a [Reason:] RuvA recognizes and binds to the Holliday junction and recruits RuvB protein to this site. RuvB is a hexameric ATPase thus making the process an ATP dependent one.
4. Which component cleaves the specific DNA strands at the Holliday junction?
Answer: [Reason:] RuvC recognizes the Holliday junction. It specifically nicks two of the homologous DNA strands that have the same polarity.
5. After cleaved by RuvC the resulting ligated recombination products will only be the splice product.
Answer: b [Reason:] Depending on which pair of strands is cleaved by RuvC the resulting recombinant product will depend. It either forms a spliced product or a patched product.
6. What is the cleavage site of RuvC?
Answer: a [Reason:] Cleavage by RuvC takes place at sites conforming to the consensus sequence 5’….A/TTTG/C….3’. cleavage occurs after the second T in this sequence.
7. What is the source of energy during the action of SPO11?
a) Cleavage of peptide bond
b) Cleavage of amide bond
c) Cleavage of glycoside bond
d) Cleavage of phosphodiester bond
Answer: d [Reason:] During the action of SPO11 the cleavage of DNA phosphodiester bond is stored in the bound protein-DNA linkage. By this process the DNA strands can be released.
8. What are the components of the MRX protein?
a) MreII, Rad50, Xrs2
b) Rad51, Rad52, Rad59
c) RecA, RuvC, RuvAB
d) Rad51C, WRN, BLH
Answer: a [Reason:] MRX enzyme complex is used for the event of DNA processing. It is composed of protein subunits, MreII, Rad50 and Xrs2. The first letters of the subunits gives the complex its name.
9. Processing of the DNA at the break site occurs at the 3’ end.
Answer: b [Reason:] Processing of the DNA at the break site occurs exclusively at the 5’ terminus. The strand ic covalently attached to the Spo11 protein.
10. How many cycles are required for the configuration of chromosomes?
Answer: b [Reason:] Chromosomes are configured during two division cycles. This is because meiosis involves two rounds of nuclear divisons.
11. What is non-disjunction?
a) Proper alignment
b) Sister chromatid formation
c) Failed alignment
d) Chromosome segregation
Answer: c [Reason:] The absence of recombination leads to the failed alignment of the chromosomes in the 1st meiotic division and as a result there is a high incidence of chromosome loss. This improper segregation of chromosomes is called nondisjunction.
12. For which biological phenomenon homologous recombination is critical?
Answer: b [Reason:] Homologous recombination is critical for meiosis during the pachytene stage of meiosis I. This is important for the introduction of genetic variations in every generations.
13. What are homologs?
a) Part of DNA
b) Pair of nucleus
c) Pair of chromosomes
d) Part of a cell
Answer: c [Reason:] Before cell division the cell has two copies of each chromosome, one of each was inherited from either parent. These are known as homologs.
14. What is the function of the Spo11 in meiotic recombination?
a) Introduction of mutation
b) Introduction of nick
c) Introduction of DSB
d) Introduction of lesion
Answer: c [Reason:] Spo11 gene encodes a protein that introduces DSBs in chromosomal DNA to initiate meiotic recombination. The Spo11 protein cuts the DNA at many chromosomal locations with little sequence selectivity but at a very specific time during meiosis.
15. How many subunits of Spo11 are involved in the process of cleavage of DNA?
Answer: b [Reason:] The two subunits of Spo11 cleave the DNA two nucleotides apart on the two DNA strands. It makes staggered double strand break by cleavage.
Molecular Biology MCQ Set 7
1. What is the condition required to observe B form of DNA?
a) Low humidity
b) High humidity
c) Low temperature
d) High temperature
Answer: b [Reason:] X – ray diffraction studies of DNA shows two different kinds of structures of DNA, A and B forms. Of the two, B form of DNA is generally observable at high humidity and closely resembles to the structure of the DNA in physiological condition.
2. What is the condition required to observe A form of DNA?
a) Low humidity
b) High humidity
c) Low temperature
d) High temperature
Answer: a [Reason:] A form of DNA is generally observed at low humidity. The conformation is usually taken by DNA – RNA and RNA – RNA complexes. They have a much more compact structure than any another form of DNA.
3. How many bases are there in per turn of the B form of DNA?
a) 8 bp
b) 9 bp
c) 10 bp
d) 11 bp
Answer: c [Reason:] B form of DNA has 10 base pairs per turn. The height of one complete turn, that is, the pitch is 2.46 nm.
4. Which of the following is not a character of A DNA?
a) Short and broad
b) Left handed helix
c) Narrow minor groove
d) Broad and shallow major groove
Answer: b [Reason:] A form of DNA is known to have a right handed helical structure. The left handed helical conformation is the characteristic form of the Z form of DNA.
5. How many bases are there in per turn of the A form of DNA?
a) 11 bp
b) 10 bp
c) 12 bp
d) 8 bp
Answer: a [Reason:] A form of DNA has 11 base pairs per turn. The height of one complete turn, that is, the pitch is 3.32 nm.
6. Which of the following is not a character of B form of DNA?
a) Longer and thinner
b) Right – handed helix
c) Narrow minor groove
d) Flat major groove
Answer: d [Reason:] B form of DNA has a wide and intermediate major groove. Flattened major groove is normally found Z form of DNA.
7. In a right handed DNA the glycosidic bond is in syn configuration ____________
Answer: b [Reason:] Glycosidic bond connects the base to the 1’ position of 2’-deoxyribose. This bond can be in one of the two conformations called syn and anti. In right handed DNA, the glycosidic bond is always in the anti configuration.
8. In the Z form of DNA the fundamental repeating unit is __________
a) Purine – pyrimidine nucleotide
b) Purine – pyrimidine dinucleotide
c) Purine – pyrimidine tri-nucleotide
d) Purine – pyrimidine tetra-nucleotide
Answer: d [Reason:] As we know Z form of DNA is the left–handed DNA. Thus it has the anti–conformation at the pyrimidine and the syn–conformation at the purine and is usually a dinucleotide unit.
9. What is the factor responsible for the left – handed helical conformation in Z form of DNA?
a) Syn conformation at pyrimidine
b) Anti conformation at pyrimidine
c) Syn conformation at purine
d) Anti conformation at purine
Answer: c [Reason:] It is the syn conformation at the purine nucleotides that is responsible for the left – handed helical conformation. The change to the syn position in the purine residues to alternating anti – syn conformation gives the backbone of the left – handed DNA a zig – zag look.
10. In higher salt concentration the DNA helix assumes the right – handed conformation _______
Answer: b [Reason:] In higher salt concentration the DNA helix assumes the left – handed conformation. This is because the higher concentration of the positively charged ions is required to shield the negatively charged phosphate groups to maintain the stability of the conformation.
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