Molecular Biology Practice Test
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Do you wish to practice molecular biology? Take this test and see how well you understand molecular biology or if you need to learn more about it. We have got a question about molecular biology that will help you recall whatever you have learned and prepare for any upcoming exams. This exam cover chapters 14,15,17, 18 and 24. It will be an easy quiz for you, as your concepts of molecular biology are already clear. All the best! Even if you miss out on some question, we are here to help you with the correct answer.
Questions and Answers
- 1.
Which of the following experiments was useful in determining the existence of introns in the adenovirus genome?
- A.
DNA/DNA hybridization
- B.
RNA/RNA hybridization
- C.
R-looping experiments
- D.
RT-PCR
- E.
CDNA cloning experiments
Correct Answer
C. R-looping experimentsExplanation
R-looping experiments were useful in determining the existence of introns in the adenovirus genome. R-looping is a technique that involves the formation of RNA-DNA hybrids, where the RNA molecule hybridizes with the DNA template. This technique allows for the identification and mapping of introns, as the RNA molecule will form loops where the introns are present. Therefore, R-looping experiments provide valuable evidence for the existence of introns in the adenovirus genome.Rate this question:
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- 2.
Which enzyme is used in the unwinding of DNA?
- A.
Exonuclease
- B.
Helicase
- C.
Topoisomerase
- D.
Ligase
Correct Answer
B. HelicaseExplanation
Helicase is the enzyme used in the unwinding of DNA. It plays a crucial role in DNA replication and transcription by breaking the hydrogen bonds between the DNA strands, allowing the double helix to separate and unwind. This unwinding is necessary for the DNA to be accessible for replication or transcription processes. Helicase acts as a molecular motor, moving along the DNA strands and separating them in a process called DNA unwinding. Once the helicase unwinds the DNA, other enzymes can then bind to the separated strands and carry out their respective functions.Rate this question:
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- 3.
You have isolated a new species of mRNA using an oligo dT column. You then perform an R-looping experiment to determine if the gene encoding for your mRNA contains introns. Below is a representation of the electron micrograph results. Which of the following can you conclude from your experiment?
- A.
The gene contains no introns.
- B.
The gene contains one intron.
- C.
The gene contains two introns.
- D.
The gene contains three intron.
Correct Answer
C. The gene contains two introns.Explanation
From the electron micrograph results, it can be concluded that the gene contains two introns. The R-looping experiment is used to identify the presence of introns in a gene. In this experiment, the mRNA is hybridized with the DNA template, forming loops where the introns are present. The number of loops observed in the electron micrograph indicates the number of introns in the gene. In this case, the micrograph shows two loops, suggesting that the gene contains two introns.Rate this question:
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- 4.
The function of EF-Tu is to
- A.
Properly fold proteins
- B.
Ensure proper ribosome assembly
- C.
Protect rRNA from the degradation of RNase
- D.
Escort aminoacyl-tRNA to the ribosome
- E.
Stabilize mRNA within the ribosome
Correct Answer
D. Escort aminoacyl-tRNA to the ribosomeExplanation
EF-Tu is a protein that plays a crucial role in protein synthesis. It acts as a molecular escort, facilitating the delivery of aminoacyl-tRNA molecules to the ribosome during translation. Aminoacyl-tRNA molecules carry the correct amino acids that correspond to the codons on the mRNA template. EF-Tu binds to aminoacyl-tRNA and helps to bring it to the ribosome, ensuring accurate and efficient protein synthesis. Therefore, the function of EF-Tu is to escort aminoacyl-tRNA to the ribosome.Rate this question:
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- 5.
You have isolated a new species of mRNA using an oligo-dT column. You then perform an R-looping experiment to determine if the gene encoding for your mRNA contains introns. Below is a representation of the electron micrograph results. Which of the following can you conclude from your experiment?
- A.
The gene contains no introns
- B.
The gene contains one intron
- C.
The gene contains two introns
- D.
The gene contains three introns
- E.
The results are inconclusive
Correct Answer
C. The gene contains two intronsExplanation
Based on the electron micrograph results from the R-looping experiment, it can be concluded that the gene contains two introns. The presence of two loops in the electron micrograph indicates that there are two regions of single-stranded DNA where the mRNA has hybridized with the DNA template. These regions correspond to the introns in the gene. Therefore, the presence of two loops suggests that the gene contains two introns.Rate this question:
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- 6.
The first two bases and the last two bases in the splicing signal consensus sequence are
- A.
GT-AG
- B.
GU-AG
- C.
CU-AG
- D.
GU-AC
- E.
GT-TG
Correct Answer
B. GU-AGExplanation
The correct answer is GU-AG because it matches the consensus sequence for splicing signals. In splicing, the GU dinucleotide is found at the 5' end of the intron, while the AG dinucleotide is found at the 3' end. These dinucleotides are important for the recognition and cleavage of the intron during RNA splicing. Therefore, GU-AG is the correct answer as it follows the pattern of the splicing signal consensus sequence.Rate this question:
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- 7.
Which of the following is the name for the YEAST 40s particle where mRNA splicing takes place?
- A.
Ribosome
- B.
Nucleolus
- C.
Spliceosome
- D.
HnRNA
- E.
R-loop
Correct Answer
C. SpliceosomeExplanation
The correct answer is spliceosome. The spliceosome is a complex of proteins and small nuclear ribonucleoproteins (snRNPs) that is responsible for removing introns and joining exons in pre-mRNA during the process of mRNA splicing. It plays a crucial role in the maturation of mRNA molecules before they can be translated into proteins. The other options, ribosome, nucleolus, hnRNA, and R-loop, are not directly involved in mRNA splicing.Rate this question:
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- 8.
"Snurps" or snRNPs are composed of
- A.
RNA
- B.
Protein
- C.
DNA
- D.
RNA and protein
- E.
DNA and RNA
Correct Answer
D. RNA and proteinExplanation
Snurps, also known as snRNPs, are a type of small nuclear ribonucleoproteins that play a crucial role in the processing of pre-mRNA molecules. These molecules are composed of both RNA and protein components. The RNA component of snurps is responsible for binding to specific sequences in the pre-mRNA, while the protein component helps in stabilizing the structure and facilitating the splicing process. Therefore, the correct answer is RNA and protein.Rate this question:
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- 9.
The catalytic center of the spliceosome appears to include
- A.
Mg2+
- B.
U2 and U6 snRNP
- C.
The branch point region of the intron
- D.
Mg2+ and the branch point region of the intron
- E.
Mg2+, U2, and U6 snRNP, and the branch point region of the intron
Correct Answer
E. Mg2+, U2, and U6 snRNP, and the branch point region of the intronExplanation
The catalytic center of the spliceosome is a complex composed of Mg2+, U2 and U6 snRNP, and the branch point region of the intron. These components work together to catalyze the splicing reaction, which involves removing the intron and joining the exons together. The Mg2+ ions facilitate the chemical reactions by stabilizing the transition state, while the U2 and U6 snRNPs play a role in recognizing the splice sites and positioning the intron for cleavage. The branch point region of the intron is also essential for proper splicing, as it helps to define the splice sites and guide the splicing machinery to the correct location.Rate this question:
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- 10.
Which of the following is the first snRNP to bind during the assembly stage of the spliceosome cycle?
- A.
U1
- B.
U2
- C.
U4
- D.
U5
- E.
U6
Correct Answer
A. U1Explanation
U1 is the correct answer because it is the first snRNP (small nuclear ribonucleoprotein particle) to bind during the assembly stage of the spliceosome cycle. The spliceosome is responsible for removing introns from pre-mRNA, and U1 snRNP recognizes the 5' splice site at the beginning of the intron. It forms a base pairing interaction with the 5' splice site, marking the start of the splicing process.Rate this question:
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- 11.
Which of the following snRNP is mismatched with its function?
- A.
U1: base pairs with 5' splice site of mRNA
- B.
U2: base pairs with the conserved sequence at splicing branch point
- C.
U4: base pairs with 3' splice site of mRNA
- D.
U5: associates with the last nucleotide in one exon and the first nucleotide in the next exon.
- E.
U6: base pairs with 5' end of the intron
Correct Answer
C. U4: base pairs with 3' splice site of mRNAExplanation
U4: base pairs with 3' splice site of mRNA is the mismatched snRNP and its function. The correct function of U4 snRNP is to base pair with U6 snRNA to form a complex that prevents U6 from pairing with the 5' splice site. This complex is essential for the regulation of splicing and the removal of introns from mRNA.Rate this question:
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- 12.
The binding of which of the following snRNPs to spliceosome requires ATP?
- A.
U1
- B.
U2
- C.
U4
- D.
U5
- E.
U6
Correct Answer
B. U2Explanation
The U2 snRNP binds to the spliceosome and requires ATP.Rate this question:
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- 13.
Spliceosomal lariat is
- A.
Cowboy's lasso
- B.
Texan lariat
- C.
Hairpin RNA
- D.
Stem and loop DNA
- E.
Splicing intermediate
Correct Answer
E. Splicing intermediateExplanation
The correct answer is "Splicing intermediate." Spliceosomal lariat refers to the structure formed during the splicing process of pre-mRNA molecules. It is a loop-like structure that is formed when the intron is removed and the exons are joined together. This intermediate structure plays a crucial role in removing the non-coding regions (introns) and joining the coding regions (exons) to produce the mature mRNA molecule. Therefore, the term "splicing intermediate" accurately describes the spliceosomal lariat.Rate this question:
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- 14.
Please put the following steps of Cap synthesis in the correct order. (1) N7 of the capping guanine is methylated (2) The terminal phosphate is removed from the pre-mRNA(3) A capping GMP is added to the pre-mRNA(4) The 2'-O-methyl group of the penultimate nucleotide is methylated
- A.
1,2,3,4
- B.
1,4,3,2
- C.
2,4,1,3
- D.
2,3,1,4
- E.
4,3,2,1
Correct Answer
D. 2,3,1,4Explanation
In the process of Cap synthesis, the correct order of steps is as follows: (2) The terminal phosphate is removed from the pre-mRNA, (3) A capping GMP is added to the pre-mRNA, (1) N7 of the capping guanine is methylated, and finally (4) The 2'-O-methyl group of the penultimate nucleotide is methylated. This order ensures the proper modification and protection of the mRNA molecule for its stability and efficient translation.Rate this question:
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- 15.
Which of the following is NOT a function of the mRNA Cap?
- A.
Protects the mRNA from degradation
- B.
Enhances translatability of the mRNA
- C.
Enhances the transport of the mRNA to the cytoplasm
- D.
Enhances splicing of the mRNA
- E.
Helps regulate the expression of the mRNA from the promoter
Correct Answer
E. Helps regulate the expression of the mRNA from the promoterExplanation
The mRNA cap has several functions, including protecting the mRNA from degradation, enhancing translatability, enhancing transport to the cytoplasm, and enhancing splicing. However, it does not directly regulate the expression of the mRNA from the promoter. This function is typically carried out by transcription factors and other regulatory elements.Rate this question:
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- 16.
Although snRNA U1 functions in the nucleus, it receives a 5' Cap to enhance its
- A.
Binding to snRNPs
- B.
Export to the cytoplasm
- C.
Export to the ER lumen
- D.
Export out of the cell
- E.
Binding to the mature mRNA
Correct Answer
B. Export to the cytoplasmExplanation
The snRNA U1 functions in the nucleus, but it receives a 5' Cap to enhance its export to the cytoplasm. This suggests that the snRNA U1 is needed in the cytoplasm for some cellular processes or interactions. The 5' Cap modification helps facilitate its transport from the nucleus to the cytoplasm, where it can carry out its functions.Rate this question:
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- 17.
Which of the following is the mRNA polyadenylation signal motif found in mammalian cells?
- A.
AAGAAA
- B.
AAUAAA
- C.
AAUGGG
- D.
UUAUUU
- E.
GGAUUU
Correct Answer
B. AAUAAAExplanation
The correct answer is AAUAAA. This sequence, known as the polyadenylation signal motif, is commonly found in the mRNA of mammalian cells. It plays a crucial role in the process of polyadenylation, where a string of adenine nucleotides (poly-A tail) is added to the mRNA molecule. This poly-A tail helps in stabilizing the mRNA and facilitating its export from the nucleus to the cytoplasm for translation into proteins.Rate this question:
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- 18.
When does the capping of the pre-mRNA occur?
- A.
It is the first nucleotide added by RNA polymerase.
- B.
It occurs before the mRNA reaches a chain length of 30 nt.
- C.
It occurs after the poly (A) tail has been added.
- D.
It occurs after splicing has occurred.
- E.
It occurs after the export of the mRNA into the cytoplasm.
Correct Answer
B. It occurs before the mRNA reaches a chain length of 30 nt.Explanation
The correct answer is that the capping of the pre-mRNA occurs before the mRNA reaches a chain length of 30 nt. This means that the capping process, which involves the addition of a modified guanine nucleotide to the 5' end of the mRNA, takes place early in the transcription process before the mRNA molecule has grown to a certain length. This capping is important for the stability and efficient processing of the mRNA molecule.Rate this question:
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- 19.
Which of the following is not required for mammalian cell polyadenylation of pre-mRNA?
- A.
PAP
- B.
CPSF
- C.
PABII
- D.
DNA polymerase I
- E.
RNA polymerase II
Correct Answer
D. DNA polymerase IExplanation
DNA polymerase I is not required for mammalian cell polyadenylation of pre-mRNA. Polyadenylation is the process of adding a poly(A) tail to the 3' end of the mRNA molecule. PAP (Poly(A) Polymerase), CPSF (Cleavage and Polyadenylation Specificity Factor), and PABII (Poly(A)-Binding Protein II) are all involved in this process. DNA polymerase I, on the other hand, is primarily involved in DNA replication and repair, and not in mRNA processing. Therefore, it is not required for mammalian cell polyadenylation.Rate this question:
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- 20.
Uncapped mRNA is much more stable than capped mRNA.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
The statement is false because capped mRNA is actually more stable than uncapped mRNA. The cap structure at the 5' end of capped mRNA provides protection against degradation by exonucleases. This cap structure also plays a crucial role in the initiation of translation by facilitating the binding of the mRNA to the ribosome. Therefore, capped mRNA is generally more stable and has a longer half-life compared to uncapped mRNA.Rate this question:
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- 21.
The 5' cap of mRNA has no effect on the translatability of the mRNA.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
The 5' cap of mRNA plays a crucial role in the translatability of the mRNA. It helps in the recognition and binding of the mRNA to the ribosome during translation. Additionally, the 5' cap protects the mRNA from degradation and enhances its stability. Therefore, the presence of the 5' cap positively affects the translatability of the mRNA.Rate this question:
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- 22.
Poly (A) tails are added to the 5' end of the mRNA molecule.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
Poly(A) tails are actually added to the 3' end of the mRNA molecule. This modification, known as polyadenylation, involves the addition of a string of adenine nucleotides to the mRNA molecule. The poly(A) tail helps protect the mRNA from degradation and is also involved in the export of the mRNA from the nucleus to the cytoplasm.Rate this question:
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- 23.
The addition of poly (A) to mRNA seems to be important for its translatability.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
The addition of poly (A) to mRNA is indeed important for its translatability. Poly (A) is a tail of adenine nucleotides that is added to the 3' end of mRNA molecules. This poly (A) tail helps in stabilizing the mRNA molecule, protecting it from degradation. It also plays a crucial role in the initiation of translation, as it interacts with various proteins involved in translation initiation. Additionally, the poly (A) tail enhances the efficiency of mRNA export from the nucleus to the cytoplasm. Therefore, the addition of poly (A) to mRNA is essential for its stability and efficient translation, making the statement true.Rate this question:
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- 24.
RNA polymerase II transcription of a gene usually stops at the polyadenylation site.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
RNA polymerase II transcription of a gene does not usually stop at the polyadenylation site. Instead, it continues beyond the polyadenylation site and transcribes a stretch of DNA known as the poly(A) tail. This poly(A) tail is important for mRNA stability and is added to the mRNA molecule during post-transcriptional processing. Therefore, the correct answer is False.Rate this question:
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- 25.
Pre-mRNA must be cleaved before it is polyadenylated.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
Pre-mRNA, which is the initial transcript of a gene, undergoes several modifications before it can become mature mRNA. One of these modifications is cleavage, where specific sections of the pre-mRNA molecule are removed. This cleavage is necessary to remove non-coding regions and join the coding regions together. After cleavage, the pre-mRNA is polyadenylated, which involves the addition of a poly-A tail at the 3' end. Therefore, the statement that pre-mRNA must be cleaved before it is polyadenylated is true.Rate this question:
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- 26.
Once an mRNA loses its poly (A) tail, it is imported back into the nucleus.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
mRNA is transcribed in the nucleus and then transported to the cytoplasm for translation. Once in the cytoplasm, the poly(A) tail of the mRNA is gradually shortened by enzymatic degradation. When the poly(A) tail is completely lost, the mRNA is targeted for degradation by exonucleases. Therefore, an mRNA that loses its poly(A) tail is not imported back into the nucleus. Hence, the correct answer is False.Rate this question:
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- 27.
Polyadenylation is not required for the efficient transport of mRNA out of the nucleus.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
Polyadenylation is a crucial step in mRNA processing that involves the addition of a poly(A) tail to the 3' end of the mRNA molecule. This poly(A) tail plays a vital role in the efficient transport of mRNA out of the nucleus. It helps to protect the mRNA from degradation, aids in the binding of mRNA to transport proteins, and is involved in the recognition and export of mRNA through nuclear pores. Therefore, polyadenylation is indeed required for the efficient transport of mRNA out of the nucleus, making the given statement false.Rate this question:
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- 28.
Once the poly (A) tail is added to the mRNA, it is very difficult to remove.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
The statement is false because the poly(A) tail can be removed from mRNA through enzymatic processes. One such process is called deadenylation, which involves the enzymatic removal of the poly(A) tail. Additionally, the poly(A) tail can also be shortened through other mechanisms such as mRNA decay pathways. Therefore, it is not accurate to say that the poly(A) tail is very difficult to remove once it is added to the mRNA.Rate this question:
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- 29.
Pre-mRNA processing begins during transcription.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
During transcription, the pre-mRNA molecule is synthesized by RNA polymerase. However, the pre-mRNA molecule is not immediately ready for translation into a protein. It undergoes a series of modifications known as pre-mRNA processing. These modifications include the addition of a 5' cap, the addition of a poly-A tail, and the removal of introns through a process called splicing. These modifications are necessary for the stability and functionality of the mRNA molecule. Therefore, it can be concluded that pre-mRNA processing begins during transcription.Rate this question:
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- 30.
Experiments using yeast as a model have shown that cleavage of the poly (A) tail is necessary for transcription termination.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
Experiments conducted with yeast as a model organism have provided evidence that supports the idea that cleavage of the poly (A) tail is indeed necessary for transcription termination. This means that the process of cutting off the poly (A) tail is crucial in order to properly end the transcription of genetic information. Therefore, the statement "Experiments using yeast as a model have shown that cleavage of the poly (A) tail is necessary for transcription termination" is true.Rate this question:
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- 31.
The Shine-Dalgarno sequence can be found in
- A.
MRNA
- B.
TRNA
- C.
5S rRNA
- D.
16S rRNA
- E.
30S ribosome
Correct Answer
A. MRNAExplanation
The Shine-Dalgarno sequence is a short nucleotide sequence found in mRNA molecules. It is involved in the initiation of protein translation in prokaryotes. This sequence interacts with the complementary sequence in the 16S rRNA of the 30S ribosome, allowing the ribosome to bind to the mRNA and start protein synthesis. Therefore, the Shine-Dalgarno sequence can be found in mRNA.Rate this question:
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- 32.
The Shine-Dalgarno sequence is
- A.
AGGAGGU
- B.
UCCUCCA
- C.
AUGAUGU
- D.
AGGAGGT
- E.
AGGUGGU
Correct Answer
A. AGGAGGUExplanation
The Shine-Dalgarno sequence is a conserved nucleotide sequence found in the mRNA of prokaryotes, typically located upstream of the start codon. It plays a crucial role in initiating translation by allowing the ribosome to bind to the mRNA and position itself correctly for protein synthesis. Among the given sequences, AGGAGGU is the only one that matches the Shine-Dalgarno sequence.Rate this question:
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- 33.
Put the following steps on prokaryotic translation initiation in the correct order. (1) Binding of IF1, IF2, and GTP to the 30 S subunit (2) Binding of IF3 to the 30S subunit. (3) Binding of the 50S subunit and loss of IF1 and IF3(4) Dissociation of the 70S ribosome. (5) Formation of the 70S initiation complex by dissociation of IF2 and GTP hydrolysis. (6) Formation of the 30S initiation complex
- A.
4,2,1,6,3,5
- B.
2,1,6,3,5,4
- C.
5,6,3,2,1,4
- D.
1,2,3,4,5,6
- E.
2,4,1,6,3,5
Correct Answer
A. 4,2,1,6,3,5Explanation
The correct order for prokaryotic translation initiation is as follows:
1) Binding of IF1, IF2, and GTP to the 30 S subunit.
2) Binding of IF3 to the 30S subunit.
3) Formation of the 30S initiation complex.
4) Binding of the 50S subunit and loss of IF1 and IF3.
5) Formation of the 70S initiation complex by dissociation of IF2 and GTP hydrolysis.
6) Dissociation of the 70S ribosome.
This order ensures that the correct components are bound to the ribosome and that the initiation complex is formed before the 50S subunit binds and the 70S ribosome is fully assembled.Rate this question:
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- 34.
Which of the following initiation factors prevents reassociation of the 70s ribosome by binding to the free 30s subunit?
- A.
IF1
- B.
IF2
- C.
IF3
- D.
IF4
- E.
IF5
Correct Answer
C. IF3Explanation
IF3 is the correct answer because it prevents reassociation of the 70s ribosome by binding to the free 30s subunit. This initiation factor acts as a dissociation factor, preventing the 30s subunit from reattaching to the 50s subunit after translation is complete. By binding to the free 30s subunit, IF3 helps to ensure that the ribosome remains in a dissociated state and ready for the next round of translation.Rate this question:
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- 35.
Which of the following is not part of the 30s initiation complex?
- A.
IF1, IF2, and IF3
- B.
5S rRNA
- C.
16s rRNA
- D.
Amino-acyl tRNA
- E.
MRNA
Correct Answer
B. 5S rRNAExplanation
The 30s initiation complex is a complex formed during the initiation of protein synthesis in bacteria. It consists of mRNA, 16s rRNA, IF1, IF2, and IF3, as well as the initiator amino-acyl tRNA. 5S rRNA is not part of the 30s initiation complex.Rate this question:
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- 36.
Match the following eukaryotic translation initiation factors with their correct function. A. eIF2 B. eIF1 C. eIF3 D. eIF4F E. eIF6 This initiation factor is involved in binding tRNA to the ribosome.
- A.
A
- B.
B
- C.
C
- D.
D
- E.
E
Correct Answer
A. AExplanation
eIF2 is involved in binding tRNA to the ribosome.Rate this question:
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- 37.
Match the following eukaryotic translation initiation factors with their correct function. A. eIF2 B. eIF1 C. eIF3 D. eIF4F E. eIF6This initiation factor binds to the 40s subunit and inhibits re-association of the 40s and 60s subunits.
- A.
A
- B.
B
- C.
C
- D.
D
- E.
E
Correct Answer
C. CExplanation
Eukaryotic translation initiation factor 6 (eIF6) binds to the 40s subunit and inhibits re-association of the 40s and 60s subunits.Rate this question:
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- 38.
Match the following eukaryotic translation initiation factors with their correct function. A. eIF2 B. eIF1 C. eIF3 D. eIF4F E. eIF6This initiation factor is a Cap-binding protein.
- A.
A
- B.
B
- C.
C
- D.
D
- E.
E
Correct Answer
D. DExplanation
eIF4F is the correct answer. It is a Cap-binding protein, which means it binds to the 5' cap structure of mRNA during the initiation of translation in eukaryotic cells. This binding helps in recruiting the ribosome to the mRNA and initiating protein synthesis.Rate this question:
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- 39.
Match the following eukaryotic translation initiation factors with their correct function. A. eIF2 B. eIF1 C. eIF3 D. eIF4F E. eIF6This initiation factor binds to the 60S subunit and inhibits re-association of the 40S and 60S subunits.
- A.
A
- B.
B
- C.
C
- D.
D
- E.
E
Correct Answer
E. EExplanation
Eukaryotic translation initiation factor eIF6 binds to the 60S subunit and inhibits the re-association of the 40S and 60S subunits.Rate this question:
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- 40.
Match the following eukaryotic translation initiation factors with their correct function. A. eIF2 B. eIF1 C. eIF3 D. eIF4F E. eIF6This initiation factor aids in ribosome scanning to locate the initiation codon.
- A.
A
- B.
B
- C.
C
- D.
D
- E.
E
Correct Answer
B. BExplanation
eIF1 aids in ribosome scanning to locate the initiation codon.Rate this question:
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- 41.
Which of the following is the best definition of translation?
- A.
A process by which pre-mRNA is transformed into mature mRNA
- B.
The process by which pre-mRNA is translated y ribozymes to produce proteins
- C.
A process by which a gene is converted into mRNA
- D.
The process by which ribosomes read the genetic message to produce a protein
- E.
A process by which a tRNA is converted into a protein
Correct Answer
D. The process by which ribosomes read the genetic message to produce a proteinExplanation
Translation is the process by which ribosomes read the genetic message carried by mRNA and use it to synthesize proteins. This process involves the decoding of the nucleotide sequence of mRNA into a specific amino acid sequence, which then forms a protein. This definition accurately describes the process of translation and distinguishes it from other cellular processes such as transcription or post-translational modifications.Rate this question:
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- 42.
Which is false?
- A.
In prokaryotes the first amino acid in all mature proteins is an N-formyl-methionine.
- B.
Similar to prokaryotes, eukaryotic translation incorporates N-formyl-methionine as the first amino acid.
- C.
Most eukaryotic mRNAs have 5' caps that function in translation initiation instead of Shine-Dalgarno sequences.
- D.
Translation initiation in eukaryotes involves canning of the mRNA to locate a favorable AUG.
- E.
Phosphorylation of eukaryotic initiation factors can play both an inhibitory and stimulatory role in the translational control of gene expression.
Correct Answer
B. Similar to prokaryotes, eukaryotic translation incorporates N-formyl-methionine as the first amino acid.Explanation
The given statement is false. In prokaryotes, the first amino acid in all mature proteins is indeed N-formyl-methionine, but in eukaryotes, the first amino acid is methionine without the formyl group. Therefore, the statement that eukaryotic translation incorporates N-formyl-methionine as the first amino acid is incorrect.Rate this question:
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- 43.
Which of the following enzymes is used to search for CpG island?
- A.
Uracil-N-glycosylase
- B.
Lysozyme
- C.
The restriction enzymes that cut both methylated and unmethylated CCGG
- D.
The restriction enzymes that cut methylated CCGG
- E.
The restriction enzymes that cut only unmethylated CCGG
Correct Answer
A. Uracil-N-glycosylaseExplanation
Uracil-N-glycosylase is an enzyme that is used to search for CpG islands. CpG islands are regions of DNA that contain a high frequency of CpG sites, which are cytosine nucleotides followed by guanine nucleotides. Uracil-N-glycosylase specifically recognizes and removes uracil bases from DNA molecules. This enzyme is commonly used in molecular biology techniques, such as bisulfite sequencing, to identify and study CpG islands. By removing uracil from the DNA, uracil-N-glycosylase allows for the detection and analysis of CpG sites in the DNA sequence.Rate this question:
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- 44.
Which of the following statements is not true concerning the genetic code?
- A.
It is an overlapping code.
- B.
It is a "comma-less" code.
- C.
It is an almost universal code.
- D.
It is capable of wobble base pairing.
- E.
It is a triplet code.
Correct Answer
A. It is an overlapping code.Explanation
The genetic code is not an overlapping code. In an overlapping code, a single nucleotide can be a part of multiple codons, leading to ambiguity in decoding the genetic information. However, in the genetic code, each nucleotide is read in a continuous, non-overlapping manner to form codons, which are then translated into specific amino acids. Therefore, the statement that the genetic code is an overlapping code is not true.Rate this question:
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- 45.
Genetic code has all EXCEPT
- A.
Non-overlapping codons
- B.
No gaps
- C.
Absolutely universal code
- D.
Triplets
- E.
Wobbles
Correct Answer
C. Absolutely universal codeExplanation
The genetic code is the set of rules by which information encoded within DNA or mRNA sequences is translated into proteins. It is characterized by non-overlapping codons, meaning that each codon consists of three nucleotides and is read in a continuous sequence. There are no gaps between codons, as they are read one after another. The genetic code also exhibits wobbles, which refers to the flexibility in base pairing at the third position of the codon, allowing some variation in the amino acid that is encoded. However, the genetic code is not absolutely universal, as there are some variations and exceptions found in certain organisms or organelles.Rate this question:
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- 46.
Direction of translation
- A.
Start from C-terminal
- B.
MRNA is read in the 3' to 5' direction
- C.
Protein is made in the C-to-N direction.
- D.
Start from both C and N terminal.
- E.
Protein is made in the N to C direction, starting from N- terminal.
Correct Answer
E. Protein is made in the N to C direction, starting from N- terminal.Explanation
The direction of translation refers to the process by which a protein is synthesized from an mRNA template. The mRNA is read in the 3' to 5' direction, while the protein is made in the N to C direction. This means that the amino acids are added to the growing polypeptide chain starting from the N-terminus and progressing towards the C-terminus. Therefore, the correct answer is that protein is made in the N to C direction, starting from the N-terminus.Rate this question:
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- 47.
EF-Tu is an important protein in cells whose function is to
- A.
Properly fold proteins
- B.
Ensure proper ribosomes assembly
- C.
Escort aminoacyl-tRNA to the ribosome
- D.
Protect rRNA from degradation by RNases
- E.
Stabilize mRNA within the ribosome
Correct Answer
C. Escort aminoacyl-tRNA to the ribosomeExplanation
EF-Tu is a protein that plays a crucial role in protein synthesis. It acts as a molecular escort, binding to aminoacyl-tRNA and delivering it to the ribosome during translation. This process ensures that the correct amino acid is added to the growing protein chain. Without EF-Tu, the accuracy and efficiency of protein synthesis would be compromised. Therefore, the correct answer is "escort aminoacyl-tRNA to the ribosome".Rate this question:
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- 48.
Please place the steps of translation elongation in the correct order. (1) Peptidyl transferase forms a peptide bond between the peptide in the P site and the newly arrived aminoacyl-tRNA in the A site. (2) EF-G, with GTP, translocates peptidyl-tRNA to the P site (3) Ef-Tu, with GTP, binds an aminoacyl-tRNA to the ribosomal A site
- A.
3,1,2
- B.
3,2,1
- C.
1,2,3
- D.
1,3,2
- E.
2,3,1
Correct Answer
A. 3,1,2 -
- 49.
Termination codons include three nonsense codons.
- A.
UAG, UAA, UGA
- B.
UUU, UAU, GAU
- C.
TAG, TAA, TGA
- D.
GAU, AAU, AGU
- E.
GAU, UAA, UGA
Correct Answer
A. UAG, UAA, UGAExplanation
Termination codons are specific sequences of nucleotides in mRNA that signal the termination of protein synthesis. In this case, the termination codons are UAG, UAA, and UGA. These codons do not code for any amino acids and instead signal the ribosome to stop translation. Therefore, when one of these codons is encountered during protein synthesis, the ribosome releases the newly synthesized protein and the process is terminated.Rate this question:
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- 50.
Which of the following are the three naturally occurring stop codons?
- A.
UAG, UAA, UGG
- B.
UGA, UGG, UAG
- C.
UAA, AUU, GUU
- D.
UAG, UAA, UGA
- E.
UAA, UUA, GGA
Correct Answer
D. UAG, UAA, UGAExplanation
The three naturally occurring stop codons are UAG, UAA, and UGA. These codons signal the end of protein synthesis during translation. When a ribosome encounters one of these codons, it releases the newly synthesized protein and disassembles. UAG, UAA, and UGA are recognized by release factors, which cause the termination of translation. These stop codons do not code for any amino acids and are essential for the accurate and efficient synthesis of proteins.Rate this question:
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Oct 06, 2023Quiz Edited by
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Apr 22, 2010Quiz Created by
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