Molecular Biology MCQ Set 1
1. According to pairing concept of wobble hypothesis base “I” in the anticodon does not pair with?
Answer: c [Reason:] The wobble hypothesis does not permit any single tRNA molecule to recognize four different codons. Three codons, A, U and C can be recognized only when inosine occupies the first (5’) position of the anticodon.
2. Which of the following is not a chain termination codon?
Answer: d [Reason:] The universal chain termination codons are UAA, UAG and UGA. UGG codes for tryptophan according to the universal genetic code.
3. The first amino acid added by the tRNA is added to the anticodon ____________
Answer: b [Reason:] The first amino acid added is methionine which is coded by the codon AUG. Thus, the anticodon for this codon in the tRNA is UAC.
4. The modification of which base gives rise to inosine?
Answer: a [Reason:] Inosine arises through enzymatic modification of a base present in an otherwise completed tRNA chain. The base from which it is derived is adenine, whose carbon 6 is deaminated to give the 6 – keto group of inosine.
5. The stop codon on the mRNA is read by __________
Answer: c [Reason:] Chain termination codons UAA, UAG and UGA, are read not by special tRNAs but by specific proteins known as release factors. RF1 and RF2 are found in bacteria and eRF1 in eukaryotes.
6. With respect to the genetic code reading frame which of the following is wrong?
a) 5’ → 3’ direction reading frame
b) The code is non – overlapping
c) No gaps present
d) Flexible reading frame
Answer: d [Reason:] The message from the mRNA is translated in a fixed reading frame, which is set by the initiation codon. Translation starts at an initiation codon located at the 5’ end of protein encoding sequence. Because codons are non – overlapping and consists of the three consecutive nucleotides, a stretch of nucleotides could be translated in principle in any of the three reading frames.
7. The mitochondrial mRNA code for methionine is _____________
Answer: c [Reason:] The universal mRNA code for methionine is AUG. The mitochondrial mRNA code for methionine is AUA.
8. The mitochondrial genome has _____________ stop codons.
Answer: d [Reason:] The mitochondrial genome has 4 stop codons. They are UAA, UAG, AGA and AGG. But the universal code has 3 stop codons that are UAA, UAG and UGA.
9. Fruitfly mitochondrial codes AGA and AGG codes for _____________
Answer: c [Reason:] Fruitfly mitochondrial codes AGA and AGG codes for serine. In case of the universal codon AGA and AGG codes for arginine.
10. With respect to the universal genetic code pick the odd one out.
a) UAA – Glutamine
b) UGG – Tryptophan
c) AGA – Stop
d) CUG – Serine
Answer: b [Reason:] UAA, AGA and CUG are variations found in the universal codon where as UGG is not a variation.
UAA – Glutamine codon in some unicellular protozoa whereas codes for chain termination in the universal codon.
AGA – Stop codon in mammalian mitochondria but is the universal code for arginine.
CUG – Serine codon in the yeast (Candida) but is the universal code for leucine.
Molecular Biology MCQ Set 2
1. How many nucleotides of the growing RNA chain remain base paired with the template DNA?
Answer: a [Reason:] Only 8 or 9 nucleotides of the growing RNA chain remain base paired with the template DNA. The remainder of mRNA chain is peeled off and directed out of the enzyme through the RNA – exit channel.
2. The transcription bubble formed for facilitating RNA synthesis is about _____________ nucleotides long.
Answer: b [Reason:] The transcription bubble formed for facilitating RNA synthesis is about 17 nucleotides long. This is the region of unwound DNA where the two strands are separated. The bubble seems to move along with the RNA polymerase as the RNA grows in length.
3. The nascent RNA forms a double helix with the non – coding strand.
Answer: a [Reason:] The nascent RNA forms a double stranded hybrid helix with the antisense strand which is also known as the non-coding strand. The helix formed by the RNA and DNA is about 12 base pairs long.
4. How many proof-reading activities does RNA polymerase have?
Answer: b [Reason:] The RNA polymerase has 2 proof-reading activities. They are pyrophospholytic editing and hydrolytic editing.
5. The RNA polymerase can remove only the incorrect bases from the nascent RNA.
Answer: b [Reason:] The RNA polymerase is able to remove both correct and incorrect bases from the nascent RNA. As it spends longer hovering over the mismatched bases than the matched ones thus it removes the mismatched bases more frequently.
6. The Gre factor enhances the hydrolytic editing function of the polymerase.
Answer: a [Reason:] The hydrolytic editing is enhanced by the Gre factor and also serves as the elongation stimulating factor. That is, they ensure that the polymerase elongates efficiently and helps overcome “arrest” at sequences that are difficult to transcribe.
7. E. coli polymerase adds upto _____________ nucleotides per second.
Answer: b [Reason:] The E. coli polymerase adds upto an average of 40 nucleotides per second. This rate varies depending upon the local DNA sequence.
8. The Gre factor is homologous to the eukaryotic factor _____________
Answer: d [Reason:] The Gre factor is homologous to the eukaryotic factor TFIIS. The TFIIS factor stimulates the RNA polymerase II during transcription in eukaryotes by increasing its efficiency and proof-reading activity.
Molecular Biology MCQ Set 3
1. What is the correct definition of excision repair?
a) Repair of a single damaged nucleotide
b) Repair of a damaged oligonucleotide
c) Removal of a single damaged nucleotide
d) Removal of a damaged oligonucleotide
Answer: c [Reason:] In the excision repair damaged nucleotide is not repaired but removed from the DNA. The other undamaged strand serves as the template for reincorporation of the correct nucleotide by DNA polymerase.
2. How many types of excision repair systems are known?
Answer: b [Reason:] There are two types of excision repair systems. One involves the removal of only the damaged nucleotide. The other involves the removal of a short stretch of single stranded DNA containing the lesion.
3. Why recombinational repair system is called double strand break repair?
a) Both strands are broken
b) One strand is broken
c) No strand is broken
d) Both strand act as template
Answer: a [Reason:] Recombinational repair system is used when both strands are damaged such as when the DNA is broken. In such situation one strand cannot serve as template for the repair of the other.
4. Which enzyme is activated during double stranded break?
a) DNA polymerase
b) Translesional polymerase
c) RNA polymerase
d) Klenow fragment
Answer: b [Reason:] In double strand break repair when damaged bases block progression of a replicating DNA polymerase a special translesional polymerase copies the bases across the damaged site. This process does not depend on base pairing between the template and newly synthesized DNA strand.
5. What is the main enzyme that plays the major role in formation of thymine dimer?
a) DNA glycosylase
b) DNA photolyase
c) DNA gyrase
d) DNA ligase
Answer: b [Reason:] In photoreactivation DNA photolyase captures energy from light and uses it to break the covalent bond linking the adjacent pyrimidines. In this process the damaged bases are mended directly.
6. Which two Uvr component molecules scan the DNA during nucleotide excision repair?
a) UvrC, UvrA
b) UvrA, UvrB
c) UvrB, UvrC
d) UvrD, UvrA
Answer: b [Reason:] In nucleotide excision repair complex of two UvrA and UvrB molecules scan the DNA with two UvrA subunits being responsible for detecting the distortions to the helix. Upon encountering a distortion, UvrA exits the complex and create a single stranded bubble with UvrB.
7. After creation of the bubble which Uvr component are recruited?
a) Uvr B
b) Uvr A
c) Uvr C
d) Uvr D
Answer: c [Reason:] After creation of the single stranded bubble Uvr B recruits Uvr C. Uvr C creates two incisions in the double stranded DNA.
8. Which complex in E. coli performs in translesion synthesis?
a) DNA polymerase IV, DNA polymerase V
b) DNA polymerase II, DNA polymerase III
c) DNA polymerase III, DNA polymerase IV
d) DNA polymerase V, RNA polymerase I
Answer: a [Reason:] DNA polymerase IV (Din B) and DNA polymerase V (Umv C) performs translesional synthesis. Din B and Umv C are members of a distinct family of DNA polymerases found in many organisms known as the family of DNA polymerases.
9. In human which DNA polymerase is involved?
a) DNA polymerase ƞ
b) DNA polymerase α
c) DNA polymerase β
d) DNA polymerase ν
Answer: a [Reason:] DNA polymerase ƞ correctly inserts two A residues opposite a thymine dimer. The active site of DNA polymerase ƞ is better at accommodating a thymine dimer than is the active site of another translesional DNA polymerase.
10. The genes encoding the translesional polymerases are expressed as part of a pathway known as the SOS response.
Answer: a [Reason:] The gene encoding the translesional polymerases are expressed as a part of a repair mechanism known as the SOS pathway. Damage leads to the proteolytic destruction of a transcriptional repressor.
11. In eukaryotes the other name of PCNA is __________
a) Sliding clamp
b) Sliding clamp loader
c) Replicative DNA polymerase
d) Translesional DNA polymerase
Answer: a [Reason:] Sliding clamp which is known as PCNA in eukaryotes, anchors the replicative polymerase to the DNA template. The chemical modification is the covalent attachment to the sliding clamp of a peptide known as ubiquitin in a process known as ubiquitinisation.
12. In SOS repair system cleavage of LexA and UmuD is mediated by ___________
Answer: b [Reason:] LexA and UmuD complex is cleaved by a protein called RecA. This process is activated by single stranded DNA resulting from DNA damage.
13. What are the two fundamental components of NHEJ?
a) KU70, KU80
b) KU50, KU70
c) KU40, KU50
d) KU30, KU40
Answer: a [Reason:] KU70 and KU80 are the most fundamental components of NHEJ. They constitute a heterodimers that binds to the DNA ends.
14. NHEJ is ubiquitin independent in eukaryotic organisms.
Answer: b [Reason:] NHEJ mechanism is ubiquitin dependent in eukaryotic organisms. The occurrence of NHEJ in prokaryotic cells is less frequent and is predominantly found in eukaryotic cells.
15. In the following compounds which are involved in nucleotide excision repair?
a) UvrA, UvrB, UvrC
b) UvrC, UvrD
c) Uvr C, XPA, XPD
d) XPA, XPC, XPD, XPG
Answer: d [Reason:] In the nucleoside excision repair, damage occurs in pyrimidine dimer. In humans XPA, XPC, XPD and XPG are activated to repair the damage.
Molecular Biology MCQ Set 4
1. Which of the following is the most energetically costly process among the following?
c) Post transcriptional processing
Answer: d [Reason:] The process of translation is the most conserved of all other cellular processes mentioned above. It is also the most energetically costly process costing over 80% of the cells energy in rapidly growing cells.
2. The accurate ordering of amino acids in a polypeptide is determined by the direct interactions between the mRNA template and the amino acid.
Answer: b [Reason:] It is unlikely that the accurate ordering of amino acids in a polypeptide is determined by the direct interactions between the mRNA template and the amino acid. This is because the side chains of amino acids have little or no affinity for nucleotides found in the RNAs.
3. Who was the first to propose indirect interaction between mRNA template and amino acids to produce accurate ordering of amino acid in a polypeptide chain?
a) Paul C. Zamecnic
b) James Watson
c) Francis H. Crick
d) Mahlon B. Hoagland
Answer: c [Reason:] Francis H. Crick was the first to propose indirect interaction between mRNA template and amino acids to produce accurate ordering of amino acid in a polypeptide chain in 1955. He suggested that certain adaptor molecules were used to interact with the codons of mRNA and respective amino acids itself acting as a bridging element. This was later demonstrated by Paul C. Zamecnic and Mahlon B. Hoagland that a special class of RNA, the tRNA, was the adaptor molecules as suggested by Crick and thus named transfer RNA.
4. How many major components are used for the process of translation?
Answer: d [Reason:] Four major components are used for the process of translation. They are mRNAs, tRNAs, aminoacyl tRNA synthetases and ribosome.
5. Which of the following is not a property of open reading frame?
b) Non overlapping
c) Encodes a single protein
d) Starts and ends at either end of the mRNA.
Answer: d [Reason:] The protein coding region of each mRNA is composed of a contiguous, non-overlapping string of codons called open reading frame. Each ORF specifies a single protein and starts and ends at internal sites within the mRNA and thus have distinct start and stop codons used for translation.
6. Both Prokaryote and Eukaryote has one start codon 5’-AUG-3’.
Answer: b [Reason:] Eukaryotes have one start codon that is 5’-AUG-3’. Prokaryotes, instead of one, have three start codons: 5’-AUG-3’, 5’-GUG-3’ and 5’-UUG-3’.
7. The start codon has a vital role to play in incorporating the specific amino acids in the peptide chains.
Answer: a [Reason:] The start codon plays two important roles in the synthesis of a protein. First, it specifies the first amino acid to be incorporated in the growing polypeptide chain. Second, it defines the reading frame for all the subsequent codons.
8. How many reading frames are applicable in case of translation?
Answer: c [Reason:] As the codons are immediately adjacent to each other and because the codons are three nucleotides long, any stretch of mRNA can be translated into three different reading frames. However once the translation starts the reading frame is fixed as the codons are subsequent to each other.
9. Both prokaryotes and Eukaryotes have one ORF.
Answer: b [Reason:] Eukaryotes have only one ORF and are thus known as monocistronic mRNAs. Prokaryotes may have more than one ORF and are thus known as polycistronic mRNAs.
10. Shine – Dalgarno sequence is also known as the __________
c) Stop codon
d) Start codon
Answer: b [Reason:] In prokaryotic mRNA a short sequence is found on the 5’ side where the ribosome binds to the mRNA. This sequence is known as the ribosome binding site (RBS) or the Shine – Dalgarno sequence after the scientist who discovered it.
11. Which component of the rRNA binds to the mRNA?
Answer: a [Reason:] The ribosome binding site is typically located 3 – 9 bp on the 5’ side of the start codon. This sequence is complementary to the sequence located near the 3’ end of the 16S rRNA. The core of the mRNA binding site of the 16S rRNA has a sequence of 5’ – CCUCCU – 3’ and not surprisingly prokaryotic RBS is often the subset of 5’ – AGGAGG – 3’.
12. The phenomenon of translating two a protein by two ribosomes is known as translational coupling.
Answer: b [Reason:] Some prokaryotic ORFs internal to a polycistronic message lack a strong RBS but are nonetheless translated. In such cases the start codon often overlaps with the 3’ end of the adjacent ORF. Thus, a ribosome that has just completed translating the upstream ORF is appropriately positioned to begin translating from the start codon for the downstream ORF. This process of translating a protein is known as translational coupling.
Molecular Biology MCQ Set 5
1. Double – stranded structure of nucleic acid is the basic requirement for palindromic sequences.
Answer: a [Reason:] Palindromic sequence is a nucleic acid sequence present in double – stranded nucleic acid. This is because 5’ to 3’ reading frame on one strand matches the 5’ to 3’ reading frame of its complementary strand, which is the requisite for any sequence to be called a palindrome.
2. Which of the following words represents a palindrome?
Answer: d [Reason:] A palindrome is a word which can be read as the same from both forward and backward. Thus, MADAM is the same when spelt in either way.
3. Which of the following will form a palindromic sequence?
Answer: a [Reason:] The complementary sequence of ATTGCAAT is TAACGTTA. Thus, when the first is read from left to right and the later read from right to left the sequence of the bases is exactly the same. This is the criteria for a sequence to be palindromic.
4. Which of the following is affected by palindromic sequences?
a) Acetylation site
b) Phosphorylation site
c) Methylation site
d) Promoter site
Answer: c [Reason:] Palindromic sites are frequently methylated in many organisms. These are the sites recognized by methylases where a methyl group is attached to inactivate a gene or mark a restriction site for an endonuclease.
5. Palindromic sequences play a very important role in gene manipulation.
Answer: a [Reason:] Many restriction endonucleases recognize and cut the palindromic sequence, for example, EcoR1. This is important for the genetic manipulation to meet the purpose of insertion and excision of gene or genome fragment from certain genetic material (eg. plasmid).
6. Which of the following palindromes is not a restriction site?
Answer: b [Reason:] GAATTC serves as the restriction site for the EcoR1 endonuclease. CCTAGG serves as the restriction site for the endonuclease BamH1. Again AGCT acts as the restriction site for the endonuclease Alu1. Only TACGTA does not serve as a restriction site or else it is not yet known.
7. Molecular cutters do not recognize palindromic sequence.
Answer: b [Reason:] Molecular cutters or endonucleases always recognize a palindromic sequence. Type I endonucleases generally cuts the DNA 1000 bp away from the 5’ end of its recognition sequence. Similarly Type II cuts within the sequence and Type III cuts 25-27 bp away.
8. Which of the following is an example of an inverted repeat?
Answer: b [Reason:] Inverted repeat is a single stranded sequence of nucleotides followed downstream by its reverse complement separated by a few or more nucleotides. When the intervening length iin between the repeats is zero, the composite sequence is a palindromic sequence.
9. Inverted repeat have a number of biological functions. Which of the following is a biological function of an inverted repeat?
b) Central dogma
c) Cellular metabolism
d) Genetic stability
Answer: a [Reason:] The Inverted repeat defines the region of self complementation thus can be identified by many proteins, enzymes and transposons. They also serve as sites for many mutations. Thus all of these factors may lead to alteration in the basic genome sequence. If any such alteration occurs in any vital gene it could cause genetic disorders and diseases. Example of certain disease is osteogenesis imperfecta.
10. Most commonly known hairpin structures are found in _____________
Answer: c [Reason:] Hairpin structure is a type of stem – loop structure involving intra molecular base pairing. This is generally found in single – stranded DNA and RNA. Most commonly known hairpin structures are observed in the cloverleaf model of tRNA namely the anticodon loop, D loop and the ΨU loop.
11. Which of the following does not promote stability in stem – loop structure?
a) Length of the stem
b) Size of the loop
c) A : U base pairing
d) π orbital of aromatic ring
Answer: c [Reason:] A : U base pairing has only two hydrogen bonds to hold them together whereas G : C has three hydrogen bonds. Thus a G : C rich stem promotes stem – loop structure more by providing more stability to the stem.
12. What is the minimum number of bases required for loop stability?
Answer: b [Reason:] Loops less than 3 bases long are sterically impossible to form. The optimal loop length is 4 – 8 bases which perfectly stabilize the loop of the stem – loop structure.
13. Which of the following does not contain a stem – loop structure?
Answer: d [Reason:] tRNA contains three stem loops to meet the cloverleaf model. The pseudoknot is the structure with two nested loops. Again the ribozyme also contains a stem – loop structure, best example of which is the hammerhead ribozyme containing three stem – loops.
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