Molecular Biology MCQ Set 1
1. Reverse transcriptase produces DNA from RNA _______
Answer: b [Reason:] Production of DNA from RNA is the reverse process of transcription. Such a DNA produce will not be exactly same as its parent gene as it will lack the introns as it is converted from the RNA. This reaction is facilitated by the enzyme reverse transcriptase and the DNA thus formed is known as the complementary DNA or the cDNA.
2. Which of the following serves as the first primer in RT – PCR for eukaryotic RNA?
a) Oligo A
b) Oligo T
c) Oligo G
d) Oligo C
Answer: b [Reason:] All RNAs have a poly – A tail added to its 3’ end as the post transcriptional modification. Thus, during reverse transcription this poly – A tail is used as the template strand and a primer of oligo T that is used for the synthesis of the cDNA strand.
3. In RT – PCR the enzyme deoxynucleotydil transferase adds poly – G residues in the __________
a) 5’ end of RNA
b) 3’ end of RNA
c) 3’ end of cDNA
d) 5’ end of cDNA
Answer: c [Reason:] During the synthesis of the complementary strand for the cDNA molecule, its 3’ end is unknown. Thus no primer can be formed which can synthesize its complementary strand. Here the enzyme deoxynucleotydil transferase adds poly – G residues in the 3’ end of cDNA so that oligo C primer can be used for the synthesis of its complementary strand.
4. The digestion of mRNA during RT – PCR is carried out by the enzyme ____________
b) RNase H
c) Bal 31
Answer: b [Reason:] Exonuclease and endonuclease both restricts DNA fragments and Bal 31 is a type of exonuclease. RNase H brings about the mRNA digestion during RT – PCR.
5. How many primers are used in the process of reverse transcriptase amplification?
Answer: b [Reason:] The first primer of oligo – T is used during the reverse transcription of the mRNA to cDNA. The cDNA produced is a single stranded molecule thus synthesis of its complementary strand is necessary. A poly G/C tail is added to its 3’ end and thus the second primer is used against this poly G/C tail for the synthesis of the cDNA duplex.
6. From which organism is the enzyme reverse transcriptase isolated?
Answer: c [Reason:] Certain viruses have RNA as their genetic material. These viruses carry an extra protein for the synthesis of cDNA so that it can be incorporated efficiently into the host chromosome. This enzyme used by the viruses is reverse transcriptase. Examples of such viruses are retrovirus, human papiloma virus, etc.
7. The cDNA library provides several advantages over RT – PCR _________
Answer: b [Reason:] The use of 5’ specific primer eliminates the risk of amplification of partial cDNAs. Thus the procedure is so sensitive that the total cellular RNA can be used for RT – PCR, and mRNA separation step is unnecessary. Thus, RT – PCR offers several advantages over cDNA library for the isolation of a specific gene provided enough information is available.
8. ________________ is a modification of RT – PCR.
a) Overlap extension PCR
b) Inverse PCR
c) Thermal cycle sequencing PCR
Answer: d [Reason:] A modification of RT – PCR is rapid amplification of cDNA ends or RACE. This process is designed to obtain full length copies of the mRNAs. In RACE, 5’ or 3’ ends of the mRNA are copied into cDNA and are based on one primer specific to an internal site of the mRNA.
Molecular Biology MCQ Set 2
1. How many polypeptides are required for the structural assembly of ribosome?
b) Less than 10
d) More than 50
Answer: d [Reason:] Single polypeptides can perform DNA or RNA synthesis. In contrast, the machinery for polymerizing amino acids is composed of atleast three RNA molecules and more than 50 different proteins.
2. Which of the following is the slowest process among the following?
Answer: b [Reason:] The speed of DNA replication that is, 200 to 1000 nucleotides per second. Translation takes place at a rate of only 2 to 20 amino acids per second.
3. The ribosome is designed in such a way so as to keep up with the speed of the RNA polymerase.
Answer: a [Reason:] The typical prokaryotic rate of translation of 20 amino acids per second corresponding to the translation of 20 amino acids per second corresponding to the translation of 60 nucleotides (20 codons) of mRNA per second. This is similar to the rate of 50 to 100 nucleotides per second synthesized by RNA polymerase.
4. Transcription and translation occurs in the same compartment in the cases of both prokaryotes and eukaryotes.
Answer: b [Reason:] In contrast to the situation in prokaryotes, translation in the eukaryotes is completely separate from transcription. Indeed, these events occur in separate compartments of the cell: transcription occurs in the nucleus, whereas translation occurs in the cytoplasm.
5. Transcriptional coupling ensures fast rate f translation.
Answer: a [Reason:] Due to transcriptional coupling prokaryotic translation occurs at a rate of 2 -20 amino acid per second. As eukaryotes lack transcriptional coupling translation proceeds at a more leisurely speed of 2 – 4 amino acid per second.
6. Ribosome has two subunits with 4 rRNA molecules. Which of the four rRNAs is found in the decoding center of the ribosome?
Answer: d [Reason:] The ribosome is composed of two subunits the large subunit and the small subunit. The small subunit contains the decoding center and is composed of 16S rRNA and other proteinaceous components.
7. The large ribosomal subunit in prokaryotes has the sedimentation velocity of ___________
Answer: c [Reason:] In bacteria the large subunit has a sedimentation velocity of 50 Svedberg units and is accordingly known as the 50S subunit whereas the small subunit is called the 30S subunit. The intact prokaryotic ribosome is referred to as the 70S subunit.
8. Svedberg unit used to distinguish among the rRNAs and proteins by measuring it’s ___________
a) moecular weight
d) sedimentation velocity
Answer: d [Reason:] The ribosome is made up of two subunits comprising of both RNA and protein. Svedberg unit are once again used to distinguish among ribosomal RNAs and is done by measuring the sedimentation velocity during centrifugation.
9. Correct the order of occurrence of the following steps.
i. Binding of the large and small subunit
ii. Binding of mettRNA
iii. Binding of small subunit
iv. mRNA scanning
a) ii, iii, i, iv
b) iii, i, ii, iv
c) iv, iii, i, ii
d) iii, iv, ii, i
Answer: a [Reason:] The correct order of occurrence is as follows:
i. Binding of mettRNA
ii. Binding of small subunit
iii. Binding of the large and small subunit
iv. mRNA scanning.
10. Multiple proteins can be synthesized from an mRNA at a time.
Answer: a [Reason:] Although a ribosome can synthesize only one polypeptide at a time, each mRNA can be translated simultaneously by multiple ribosomes. An mRNA bearing multiple ribosomes is known as a polyribosome or a polysome.
Molecular Biology MCQ Set 3
1. With respect to polymerization of amino acids which of the following statements is wrong?
a) The polymerization machinery consists of fifty proteins and two RNA molecules
b) Overall molecular mass 2.5 megadalton
c) Occurs at a slow pace of only 2 – 20 amino acids per second
d) In eukaryotes the process occurs in the cytoplasm
Answer: a [Reason:] Though the polymerizing machinery consists of 50 proteins, it atleast should consist of 3 RNA molecules. The molecules that take part in translation are mRNA, tRNA and rRNA. Messenger or mRNA is the transcript, transfer or tRNA is the carrier of amino acid and the ribosome contains rRNA or ribosomal RNA which decodes the transcript codon.
2. Ribosome begins its translation at the 3’ end of the mRNA in prokaryotes.
Answer: b [Reason:] In prokaryotes the transcription and translation occurs simultaneously. Thus, the end of mRNA which emerges from transcription is translated first. As the RNA polymerase reads the frame in 5’ to 3’ direction thus the 5’ end of mRNA is synthesized first thus, it is translated first.
3. Check which one of the following mechanisms is wrongly paired with its location of occurrence.
a) Transcription – nucleus
b) Post transcriptional mechanism – ER
c) Translation – cytoplasm
d) Post translational mechanism – ER
Answer: b [Reason:] Post transcriptional mechanisms are splicing, capping and tailing. These three mechanisms occurs in the nucleus as if the nascent mRNA (hnRNA) leaves the nucleus without the modification ribosome would not be able to read it and also it would not be able to read it and also it would be hydrolysed by the enzyme RNase present in the cell cytoplasm.
4. With respect to the subunits of ribosome which of the following is wrongly paired?
a) Ribosome = rRNA + protein
b) Large subunit = decoding center
c) Small subunit = decoding center
d) Subunit sedimentation unit = Svedberg
Answer: b [Reason:] The large subunit contains the three sites A, P and E. These are the centers where the amino acyl tRNA brings the amino acid peptide linkage is formed and the tRNA exits respectively. Thus the subunit is known as peptidyl transferase center.
5. An mRNA bearing multiple ribosomes is known as______________
a) Small subunit-mRNA-initiator tRNA complex
b) mRNA ribosome complex
c) Polyamine-ribosome complex
Answer: d [Reason:] A single ribosome contacts around so nucleotides on mRNA whereas the due to the large density of ribosome it allows one ribosome per 80 nucleotides. Thus a typical mRNA could be very long (example: 1000 nucleotides). Therefore more than one ribosome can attach to the mRNA forming a structure known as polyribosome or polysome.
6. What is the actual substrate for each round of amino acid addition?
b) Amino acid
Answer: d [Reason:] The actual substrate for the addition of amino acid to the growing polypeptide chain in each round are two charged species of t RNAs. They are an aminoacyl tRNA which is the carrier of amino acid to the ribosome-mRNA complex and the peptidyl tRNA which forms the peptide linkages between the amino acids.
7. The reaction to form a new peptide bond is known as the peptidyl transferase reaction
Answer: a [Reason:] The mechanism for the formation of new peptide bond first requires the synthesis of amino terminus before carboxyl terminus. The second step of the mechanism is the transfer of growing polypeptide chain from the peptidyl tRNA to the amino acyl tRNA. Thus reaction is called peptidyl transferase reaction.
8. Where does ribosome gets the energy for the formation of peptide bond?
a) Breaking of the keto bond
b) Breaking of acyl bond
c) Simultaneous hydrolysis of ATP
d) Energy released during peptidyl transferase reaction
Answer: b [Reason:] The energy provided for the formation of peptide bond is provided by the breaking of high energy acyl bonds. This is because the charging reaction of the tRNA involves the hydrolysis of a molecule of ATP. Thus, the energy of peptide bond formation originates from the molecule of ATP that was hydrolysed during tRNA charging reaction but not by simultaneous hydrolysis.
9. With respect to the mRNA – ribosome complex for translation mechanism which of the following pair is wrong?
a) Small subunit – decoding center
b) A site – aminoacylated tRNA binding site
c) P site – peptidyl tRNA binding site
d) E site – polypeptide exit site
Answer: d [Reason:] E site is the binding site for the tRNA that is released after the growing polypeptide chain has been transferred to the aminoacyl tRNA. Thus E site is the peptidyl tRNA exit site.
10. Only a single stranded RNA can pass through the ribosome for decoding.
Answer: a [Reason:] The mRNA enters the decoding center through two narrow channels in the small subunit. The entry channel is only wide enough for unpaired RNA to pass through. This features ensures that mRNA is in a single stranded from as it enters the decoding center by removing any intermolecular base pairing interactions that might have been formed in the mRNA.
11. How many channels are present in the ribosome?
Answer: b [Reason:] There are three channels present in the smaller subunit of ribosome which facilitates the entering and exit of the mRNA. The third channel is for the exit of the growing polypeptide chain.
Molecular Biology MCQ Set 4
1. The most common signal for termination of transcription is self complementation.
Answer: a [Reason:] Yes, the most common signal for termination of transcription is self complementation. This self – complementation leads to the formation of a hairpin loop. It is also known as the Rho independent termination.
2. The stem of the hairpin loop of RNA consists mostly of _____________
a) A, T
b) G, C
c) A, G
d) C, T
Answer: b [Reason:] The stem of the hairpin loop of RNA consists mostly of G, C. This makes the structure more stable and thus facilitating proper termination.
3. The hairpin structure generated in the RNA is followed by a stretch of oligonucleotide complementary to the base ____________
Answer: a [Reason:] The RNA hairpin is followed by a sequence of many “U” residues. Thus the complementary stretch of nucleotides in the DNA strand is adenine or “A”.
4. A stretch of which nucleotide in the DNA acts as the termination signal?
a) Poly G
b) Poly C
c) Poly T
d) Poly A
Answer: d [Reason:] A hairpin loop followed by a stretch of oligo U nucleotides acts as the termination sequence. This subsequent stretch of the “U” residue pairs weakly to the “A” residues of the DNA thus favoring easy dissociation.
5. The rho protein has ____________ subunit.
Answer: b [Reason:] The rho protein is a hexameric protein containing 6 subunits. This protein is known to mediate transcription and is known as the Rho dependent transcription.
6. The rho proteins are ATP independent proteins.
Answer: b [Reason:] The rho protein is an ATP dependent protein. It uses the hydrolysis of ATP to terminate the transcription in the presence of a single stranded RNA.
7. The rho protein binds to the RNA recognizes special structure. It consists of ____________ nucleotides.
Answer: c [Reason:] The rho protein binds to the RNA recognizes special structure. It consists of 40 nucleotides. Rho moves along the nascent RNA towards the transcription complex and helps terminate the RNA synthesis.
8. The rut sites are rich in ____________ residues.
Answer: d [Reason:] The rut sites are special sequences used by the Rho protein to bind to terminate the RNA synthesis. These stretches of RNA do not fold into any secondary structure and is rich in “C” residues.
9. The rho protein can even bind within operons to terminate transcription.
Answer: b [Reason:] The Rho protein fails to bind to any transcript of RNA that is being translated. Thus rho protein cannot bind within operons to terminate transcription and binds beyond the operon or gene to terminate transcription.
Molecular Biology MCQ Set 5
1. mRNA of which of the following organism does not undergo processing?
Answer: c [Reason:] Most newly synthesized RNAs must be modified in various ways to the converted to their functional forms. Bacterial mRNAs are an exception; they are used immediately as templates for protein synthesis while still being transcribed.
2. rRNA and tRNA of prokaryotes does not undergo any post transcriptional processing.
Answer: b [Reason:] rRNA and tRNA of prokaryotes do undergo post transcriptional processing such as splicing and structural modification. Only bacterial mRNAs are an exception; they are used immediately as templates for protein synthesis while still being transcribed.
3. Splicing of mRNA occurs after they arrive the cytoplasm.
Answer: b [Reason:] Primary transcript of eukaryotic mRNAs undergo extensive modifications, including the removal of introns by splicing, before they are transported from the nucleus to the cytoplasm to serve as templates for protein synthesis. Regulation of these processing steps provides an additional level of control of gene expression, as does regulation of the rates at which different mRNAs are subsequently degraded within the cell.
4. Which of the following is not a type of RNA processing?
a) Polyadenylation at the 3’ end
b) Capping of 5’ end
c) Removal of exons
Answer: c [Reason:] The processing events include the following:
i) Capping of the 5’ end of RNA.
ii) Splicing for removing the introns.
iii) Polyadenylation of the 3’ end of the RNA.
5. Elongation of transcription and RNA processing are all interconnected.
Answer: b [Reason:] One elongation factor (hSPT5) also recruits and stimulates the 5’ capping enzyme. In another case, elongation factor TAT – SF1 recruits components of the splicing machinery. Thus it seems that elongation, termination of transcription, and RNA processing are interconnected to ensure their proper coordination.
6. The first RNA processing event is __________
Answer: a [Reason:] The first RNA processing event is capping. This involves the addition of the modified guanine base to the 5’ end of the RNA.
7. Capping is done by the addition of __________
a) Methylated A
b) Methylated T
c) Methylated G
d) Methylated C
Answer: c [Reason:] The addition of methylated guanine to the RNA transcript by an unusual 5’ – 5’ linkage involving three phosphates. The 5’ cap is created in three enzymatic steps. In the first step, a phosphate group is moved from the 5’ of the transcript. Then, the GTP is added. And in the final step, that nucleotide is modified by the addition of a methyl group.
8. Capping occurs after the mRNA synthesis is complete.
Answer: b [Reason:] The RNA is capped when it is still only some 20 – 40 nucleotides long – when the transcription cycle has progressed only to the transition between the initiation and elongation phases. After capping, dephosphorylation of Ser5 within the tail repeats leads to dissociation of the capping machinery and further phosphorylation causes recruitment of the machinery needed for RNA splicing.
9. As the polymerase reaches the end of RNA which of the following event does not occur as a response?
a) Transfer of Polyadenylation enzyme
b) Cleavage of the RNA
c) Addition of poly A at the 3’ end
d) Termination of transcription
Answer: c [Reason:] Once polymerase reaches the end of a gene, it encounters specific sequences that, after being transcribed into RNA, trigger the transfer of the polyadenylation enzymes to that RNA. This leads to three events:
i) Cleavage of the RNA
ii) Addition of many adenine residues at the 5’ end
iii) Subsequent termination of transcription by polymerase.
10. About how many “A” are added to the nascent RNA in the 5’ end during Polyadenylation?
Answer: d [Reason:] Polyadenylation is mediated by an enzyme called poly – A polymerase, which adds about 200 adenines to the RNA’s 3’ end product by the cleavage. This enzyme uses ATP as a precursor and adds the nucleotides using the same process as RNA polymerase, but does not require a template.
11. The long tail of As is unique to the RNA synthesized by the RNA polymerase __________
Answer: b [Reason:] It is noteworthy that the long tail of As is unique to eukaryotic mRNAs. Thus the long poly A tail is added to the RNA transcript produced by RNA polymerase II. This feature allows the experimental isolation of protein coding mRNAs by affinity chromatography.
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