Block 1 DNA-RNA Mini 1
.JPG "Block 1 DNA-RNA Mini 1 - Quiz")
Questions and Answers
- 1.
A 3 y.o. patient with delayed mental acuity is shown to have an enzyme deficiency of Histone Acetyltransferase (HAT) activity. What is the most likely outcome of this patient's deficiency, at the level of gene expression?
- A.
Ubiquitination of histones H3 and H4 and increased overall transcription
- B.
Fragmentation of poorly protected chromosomal DNA and reduced transcription
- C.
Activation of histone H3 and H4 gene expression
- D.
Reduced transcription rates because nucleosomes occupy promoters
- E.
Uncoupling of transcription and translation in neurons
Correct Answer
D. Reduced transcription rates because nucleosomes occupy promotersExplanation
The most likely outcome of the patient's deficiency in Histone Acetyltransferase (HAT) activity is reduced transcription rates because nucleosomes occupy promoters. Histone acetylation is a process that loosens the structure of chromatin, allowing for easier access to DNA by transcription factors and RNA polymerase. Without HAT activity, histones remain tightly bound to DNA, forming nucleosomes that obstruct the binding of transcription factors and RNA polymerase to gene promoters. This obstruction leads to reduced transcription rates and subsequently delayed mental acuity in the patient.Rate this question:
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- 2.
DNA contains which one of the following components?
- A.
Nitrogenous bases joined by phosphodiester bonds
- B.
Negatively charged phosphate groups in the interior of the molecule
- C.
Base pairs stacked along the exterior of the molecule
- D.
Two strands that run in the opposite direction
- E.
The sugar ribose
Correct Answer
D. Two strands that run in the opposite directionExplanation
DNA is a double-stranded molecule, where the two strands run in opposite directions. This is known as antiparallel orientation. One strand runs in the 5' to 3' direction, while the other runs in the 3' to 5' direction. This arrangement is important for DNA replication and transcription processes. The other options listed in the question, such as nitrogenous bases joined by phosphodiester bonds and base pairs stacked along the exterior, are also components of DNA, but they do not specifically describe the opposite directionality of the strands.Rate this question:
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- 3.
When synthesis of segment C begins, which other segment is also being synthesized?
- A.
A
- B.
B
- C.
C
- D.
D
- E.
E
Correct Answer
E. EExplanation
When synthesis of segment C begins, segment E is also being synthesized.Rate this question:
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- 4.
If adenine is the first base on thee template strand corresponding to the initiation point for segment E, which precursor molecule would serve as the substrate that formed the first nucleotide in segment E?
- A.
DUTP
- B.
UTP
- C.
DTTP
- D.
TTP
- E.
DTMP
Correct Answer
B. UTPExplanation
If adenine is the first base on the template strand corresponding to the initiation point for segment E, the precursor molecule that would serve as the substrate to form the first nucleotide in segment E is UTP. This is because adenine pairs with uracil in RNA, and UTP is the precursor molecule that contains uracil.Rate this question:
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- 5.
Which of the following phrases describes nucleosomes?
- A.
Single ribosomes attached to mRNA
- B.
Complexes of DNA and all the histones except H4
- C.
Subunits of chromatin
- D.
Structures that contain DNA in the core with histones wrapped around the surface
- E.
Complexes of protein and the 45S rRNA precursors found in the nucleolus
Correct Answer
C. Subunits of chromatinExplanation
Nucleosomes are subunits of chromatin. Chromatin is the material that makes up chromosomes, and it consists of DNA wrapped around histone proteins. Nucleosomes are the basic building blocks of chromatin, with DNA coiled around a core of histones. This structure helps to compact and organize the DNA in the nucleus of a cell.Rate this question:
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- 6.
Which one of the following point mutations would NOT produce a change in the protein translated from an mRNA?
- A.
UCA → UAA
- B.
UCA → CCA
- C.
UCA → UCU
- D.
UCA → ACA
- E.
UCA → GCA
Correct Answer
C. UCA → UCUExplanation
The UCA → UCU mutation would not produce a change in the protein translated from an mRNA because both codons code for the same amino acid, serine. This is known as a silent mutation, where the change in the DNA sequence does not result in a change in the corresponding amino acid sequence of the protein.Rate this question:
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- 7.
Tetracycline, streptomycin, and erythromycin are effective antibiotics because they inhibit
- A.
RNA synthesis in prokaryotes
- B.
RNA synthesis in eukaryotes
- C.
Protein synthesis in prokaryotes
- D.
Protein synthesis on cytoplasmic ribosomes of eukaryotes
- E.
Protein synthesis on mitochondrial ribosomes of prokaryotes
Correct Answer
C. Protein synthesis in prokaryotesExplanation
Tetracycline, streptomycin, and erythromycin are effective antibiotics because they inhibit protein synthesis in prokaryotes. Prokaryotes, such as bacteria, rely on protein synthesis for various cellular processes, including growth and replication. These antibiotics specifically target the ribosomes, which are responsible for protein synthesis, in prokaryotic cells. By inhibiting protein synthesis, these antibiotics disrupt the ability of the bacteria to produce essential proteins, ultimately leading to their death.Rate this question:
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- 8.
When this in vitro protein-synthesizing system was treated with the new antibiotic, the only product was methionyl-phenylalanyltRNA. What process in protein synthesis is inhibited by the antibiotic?
- A.
Binding of an aminoacyl-tRNA to the A site of the ribosome
- B.
Initiation
- C.
Translocation
- D.
Peptidyl transferase catalyzed formation of a peptide bond
Correct Answer
C. TranslocationExplanation
The correct answer is "Translocation". Translocation is the process in protein synthesis where the ribosome moves along the mRNA strand, shifting the tRNA from the A site to the P site and the growing polypeptide chain from the P site to the A site. This movement allows for the addition of the next amino acid to the growing chain. In this case, the antibiotic inhibits translocation, resulting in the only product being the methionyl-phenylalanyltRNA.Rate this question:
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- 9.
A double-stranded RNA genome isolated from a virus in the stool of a child with gastroenteritis was found to contain 15 % uracil. What is the percentage of guanine in this genome?
- A.
15
- B.
25
- C.
35
- D.
75
- E.
85
Correct Answer
C. 35Explanation
The percentage of guanine in the double-stranded RNA genome can be determined by subtracting the percentage of uracil from 100%. Since the genome contains 15% uracil, the percentage of guanine would be 100% - 15% = 85%. Therefore, the correct answer is 85%.Rate this question:
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- 10.
It is now believed that a substantial proportion of the single nucleotide substitutions causing human genetic disease are due to misincorporation of bases during DNA replication. Which proofreading activity is critical in determining the accuracy of nuclear DNA replication and thus the base substitution mutation rate in human chromosomes?
- A.
3’ to 5’ polymerase activity of DNA polymerase δ
- B.
3’ to 5’ exonuclease activity of DNA polymerase γ
- C.
Primase activity of DNA polymerase α
- D.
5’ to 3’ polymerase activity of DNA polymerase δ
- E.
3’ to 5’ exonuclease activity of DNA polymerase δ
Correct Answer
E. 3’ to 5’ exonuclease activity of DNA polymerase δExplanation
The 3' to 5' exonuclease activity of DNA polymerase δ is critical in determining the accuracy of nuclear DNA replication and the base substitution mutation rate in human chromosomes. This activity allows DNA polymerase δ to proofread its own work by removing incorrect nucleotides that have been incorporated during replication. By removing these errors, DNA polymerase δ ensures that the newly synthesized DNA strand is an accurate copy of the original template strand, reducing the likelihood of base substitution mutations.Rate this question:
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- 11.
The proliferation of cytotoxic T-cells is markedly impaired upon infection with a newly discovered human immunodeficiency virus, designated HIV-V. The defect has been traced to the expression of a viral-encoded enzyme that inactivates a host-cell nuclear protein required for DNA replication. Which protein is a potential substrate for the viral enzyme?
- A.
TATA-box binding protein (TBP)
- B.
Cap binding protein (CBP)
- C.
Catabolite activator protein (CAP)
- D.
Acyl-carrier protein (ACP)
- E.
Single-strand binding protein (SBP)
Correct Answer
E. Single-strand binding protein (SBP)Explanation
The correct answer is Single-strand binding protein (SBP). The passage states that the viral-encoded enzyme inactivates a host-cell nuclear protein required for DNA replication. Single-strand binding protein (SBP) is involved in DNA replication, specifically in stabilizing single-stranded DNA during replication. Therefore, it is a potential substrate for the viral enzyme.Rate this question:
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- 12.
The anti-Pseudomonas action of norfloxacin is related to its ability to inhibit chromosome duplication in rapidly dividing cells. Which of the following enzymes participates in bacterial DNA replication and is directly inhibited by this antibiotic?
- A.
DNA polymerase I
- B.
DNA polymerase II
- C.
Topoisomerase I
- D.
Topoisomerase II
- E.
DNA ligase
Correct Answer
D. Topoisomerase IIExplanation
Norfloxacin inhibits chromosome duplication in rapidly dividing cells by directly inhibiting the enzyme topoisomerase II. Topoisomerases are enzymes involved in the winding and unwinding of DNA during replication, transcription, and recombination. By inhibiting topoisomerase II, norfloxacin disrupts the normal DNA replication process, leading to the inhibition of bacterial cell division and ultimately the anti-Pseudomonas action of the antibiotic.Rate this question:
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- 13.
Enhancers are transcriptional regulatory sequences that function by enhancing the activity of
- A.
General transcriptional factors
- B.
RNA polymerase to enable the enzyme to transcribe through the terminating region of a gene
- C.
Transcription factors that bind to the promoter but not to RNA polymerase
- D.
RNA polymerase at a single promoter site
- E.
Spliceosomes
Correct Answer
D. RNA polymerase at a single promoter siteExplanation
Enhancers are transcriptional regulatory sequences that function by enhancing the activity of RNA polymerase at a single promoter site. They do not bind to the promoter or RNA polymerase directly, but instead interact with other transcription factors to increase the efficiency of transcription initiation. This allows RNA polymerase to transcribe through the terminating region of a gene, leading to increased gene expression.Rate this question:
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- 14.
A gene encodes a protein with 150 amino acids. There is one intron of 1,000 bp, a 5’-untranslated region of 100 bp, and a 3’-untranslated region of 200 bp. In the final processed mRNA, how many bases lie between the start AUG codon (include the start codon) and the final termination codon?
- A.
1,750
- B.
750
- C.
650
- D.
450
- E.
150
Correct Answer
D. 450Explanation
The total number of bases between the start AUG codon and the final termination codon can be calculated by adding the lengths of the intron, the 5'-untranslated region, and the 3'-untranslated region. In this case, the intron is 1,000 bp, the 5'-untranslated region is 100 bp, and the 3'-untranslated region is 200 bp. Adding these lengths together gives a total of 1,300 bp. However, the question specifies that the start AUG codon should be included, so we need to add 3 more bases to the total. Therefore, the correct answer is 1,300 + 3 = 1,303, which is not one of the given options. Since none of the options match the correct answer, the question is incomplete or not readable.Rate this question:
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- 15.
A messenger RNA is 336 nucleotides long, including the initiator and termination codons. The number of amino acids in the protein translated from this mRNA is:
- A.
999
- B.
630
- C.
330
- D.
111
- E.
110
Correct Answer
D. 111Explanation
The number of amino acids in a protein is determined by the number of codons in the mRNA sequence. Each codon codes for one amino acid. In this case, the mRNA is 336 nucleotides long, including the initiator and termination codons. Since there are 3 nucleotides in each codon, we can divide the total number of nucleotides by 3 to get the number of codons. 336 divided by 3 is 112. However, we need to subtract 1 from this number because the termination codon does not code for an amino acid. Therefore, the number of amino acids in the protein translated from this mRNA is 111.Rate this question:
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- 16.
You have obtained a sample of DNA, and you transcribe mRNA from this DNA and purify it. You then separate the two strands of the DNA and analyze the base composition of each strand and of the mRNA. You obtain the data shown in the table to the right. Which strand of the DNA is the coding strand, serving as a template for mRNA synthesis?
- A.
Strand 1
- B.
Strand 2
- C.
Both strands 1 and 2
- D.
Neither strand 1 nor 2
- E.
Too little information to tell
Correct Answer
B. Strand 2Explanation
Based on the given data, the base composition of Strand 2 is complementary to the mRNA sequence. This suggests that Strand 2 is the coding strand, serving as a template for mRNA synthesis.Rate this question:
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- 17.
Given that the LD50 (the dose of which 50% of the recipients die) of amanitin is 0.1 mg per kg body weight, and that the average mushroom contains 7 mg amanitin, how many mushrooms must be consumed by someone who is 100 kg to be above the LD50?
- A.
1
- B.
2
- C.
3
- D.
4
- E.
5
Correct Answer
B. 2Explanation
The LD50 of amanitin is 0.1 mg/kg, and the average mushroom contains 7 mg of amanitin. To calculate the number of mushrooms needed to be above the LD50, we can divide the LD50 dose by the dose per mushroom. So, 0.1 mg/kg divided by 7 mg per mushroom gives us approximately 0.014 mushrooms. Since we cannot consume a fraction of a mushroom, we would need to consume at least 1 whole mushroom, which means the answer is 2 mushrooms.Rate this question:
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- 18.
If a 1,000-kilobase fragment of DNA has 10 evenly spaced and symmetric replication origins and DNA polymerase moves at 1 kilobase per second, how many seconds will it take to produce two daughter molecules (ignore potential problems at the ends of this linear piece of DNA)? Assume that the 10 origins are evenly spaced from each other, but not from the ends of the chromosome.
- A.
20
- B.
30
- C.
40
- D.
50
- E.
100
Correct Answer
D. 50Explanation
The 1,000-kilobase fragment of DNA has 10 evenly spaced and symmetric replication origins. This means that each replication origin is 100 kilobases apart from each other. Since DNA polymerase moves at a rate of 1 kilobase per second, it will take 100 seconds to replicate the entire length between two replication origins. Therefore, it will take 100 seconds to replicate one daughter molecule. Since we need to produce two daughter molecules, it will take a total of 100 seconds * 2 = 200 seconds. However, the question states to ignore potential problems at the ends of the DNA fragment, so we can assume that the time taken to replicate the ends is negligible. Therefore, the correct answer is 50 seconds.Rate this question:
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- 19.
In eukaryotes, histone acetylation is typically associated with:
- A.
Gene activation
- B.
Gene inactivation
- C.
Transposition
- D.
Translation
- E.
Protein secretion
Correct Answer
A. Gene activationExplanation
Histone acetylation is a process in which acetyl groups are added to histone proteins, which are involved in packaging DNA into a compact structure called chromatin. This modification loosens the chromatin structure, allowing for easier access of transcription factors and other regulatory proteins to the DNA. This increased accessibility promotes gene activation, as it facilitates the binding of transcription factors to the DNA and the initiation of transcription. Therefore, histone acetylation is typically associated with gene activation.Rate this question:
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- 20.
The following piece of DNA is being replicated. The replication fork moves from left to right along this DNA. What is the sequence of an Okazaki fragment on this piece of DNA whose 5’ and 3’ ends are located right at the borders of the shown sequence?
- A.
5’ AGGTGCATGATAACATCC 3’
- B.
5’ CCTACAATAGTACGTGGA 3’
- C.
5’ TCCACGTACTATTGTAGG 3’
- D.
5’ GGATGTTATCATGCACCT 3’
- E.
5’ GGATGTTATTATTGTAGG 3’
Correct Answer
D. 5’ GGATGTTATCATGCACCT 3’Explanation
The sequence of an Okazaki fragment is determined by the lagging strand synthesis during DNA replication. In this case, the 5' end of the Okazaki fragment is located at the border of the shown sequence, which is "GGATGTTATCATGCACCT". The 3' end of the Okazaki fragment is not shown, but it would be complementary to the 5' end and would have the sequence "CCTACAATAGTACGTGGA". This is the correct answer because it matches the criteria given in the question.Rate this question:
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- 21.
The average size of a human gene is
- A.
1,000 bp
- B.
40,000 bp
- C.
2 x 106 bp
- D.
1.5 x 108 bp
- E.
3 x 109 bp
Correct Answer
B. 40,000 bpExplanation
The correct answer is 40,000 bp because the average size of a human gene is typically around 40,000 base pairs. This is based on studies and research that have been conducted on the human genome, which have found that most genes range in size from a few hundred to several thousand base pairs. While there can be variation in gene size, 40,000 bp is a commonly accepted average.Rate this question:
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- 22.
A child presents with severe growth failure, accelerated aging that causes adult complications such as diabetes and coronary artery disease, and microcephaly (small head) due to increased nerve cell death. In vitro assay of labeled thymidine incorporation reveals decreased levels of DNA synthesis compared to controls, but normal-sized labeled DNA fragments. The addition of protein extract from normal cells, gently heated to inactivate DNA polymerase, restores DNA synthesis in the child’s cell extracts to normal. Which of the enzymes used in DNA replication is likely to be defective in this child?
- A.
DNA-directed DNA polymerase
- B.
Unwinding proteins
- C.
DNA polymerase I
- D.
DNA-directed RNA polymerase
- E.
DNA ligase
Correct Answer
B. Unwinding proteinsExplanation
The child in this case presents with severe growth failure, accelerated aging, microcephaly, and decreased levels of DNA synthesis. The fact that the addition of protein extract from normal cells, heated to inactivate DNA polymerase, restores DNA synthesis suggests that the issue lies with a different enzyme involved in DNA replication. Unwinding proteins are responsible for separating the DNA strands during replication, allowing DNA polymerase to synthesize new DNA strands. Therefore, a defect in unwinding proteins would explain the decreased DNA synthesis observed in the child's cells.Rate this question:
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- 23.
If a completely radioactive double-stranded DNA molecule undergoes two rounds of replication in a solution free of radioactive label, what is the radioactivity status of the resulting four double-stranded DNA molecules?
- A.
Half should contain no radioactivity
- B.
All should contain radioactivity
- C.
Half should contain radioactivity in both strands
- D.
One should contain radioactivity in both strands
- E.
None should contain radioactivity
Correct Answer
A. Half should contain no radioactivityExplanation
When a completely radioactive double-stranded DNA molecule undergoes replication in a solution free of radioactive label, the resulting four double-stranded DNA molecules will have half of them containing no radioactivity. This is because during replication, each original DNA strand acts as a template for the synthesis of a new complementary strand. As a result, one of the newly synthesized strands in each of the four molecules will be radioactive, while the other strand will be non-radioactive. Therefore, half of the resulting molecules will contain no radioactivity.Rate this question:
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- 24.
The hypothetical “stimulin” gene contains two exons that encode a protein of 100 amino acids. They are separated by an intron of 100 bp beginning after the codon for amino acid 10. Stimulin messenger RNA (mRNA) has 5’ and 3’ untranslated regions of 70 and 30 nucleotides, respectively. A complementary DNA (cDNA) made from mature stimulin RNA would have which of the following sizes?
- A.
500 bp
- B.
400 bp
- C.
300 bp
- D.
100 bp
- E.
70 bp
Correct Answer
B. 400 bpExplanation
The cDNA made from mature stimulin RNA would have a size of 400 bp. This is because the cDNA is made from the mature mRNA, which includes only the exons and excludes the intron and untranslated regions. The two exons that encode the protein are 100 amino acids long, which corresponds to 300 nucleotides. Therefore, the cDNA would have a size of 300 bp. However, since the mRNA also includes 70 nucleotides of 5' untranslated region and 30 nucleotides of 3' untranslated region, the total size of the cDNA would be 400 bp.Rate this question:
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- 25.
An immigrant from Eastern Europe is rushed into the emergency room with nausea, vomiting, diarrhea, and abdominal pain. His family indicates he has eaten wild mushrooms. They have brought a bag of fresh, uncooked mushrooms from a batch he had not yet prepared. You note the presence of Amanita phalloides, the death-cap mushroom. Care is supportive. A major toxin of the death-cap mushroom is the hepatotoxic octapeptide a-amanitin, which inhibits
- A.
DNA primase
- B.
RNA nuclease
- C.
DNA ligase
- D.
RNA polymerase
- E.
RNA/DNA endonuclease
Correct Answer
D. RNA polymeraseExplanation
The correct answer is RNA polymerase because a-amanitin, a major toxin of the death-cap mushroom, inhibits RNA polymerase. This inhibition prevents the synthesis of RNA, leading to disruption of protein synthesis and ultimately causing liver damage. This is consistent with the symptoms of nausea, vomiting, diarrhea, and abdominal pain observed in the patient. Supportive care is the recommended treatment in such cases.Rate this question:
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- 26.
The DNA sequence M, shown below, is the sense strand from a coding region known to be mutational “hot spot” for a gene. It encodes amino acids 21-25. Given the genetic and amino acid codes CCC = proline (P), GCC = alanine (A), TTC = phenylalanine (F), and TAG = stop codon, which of the following sequences is a frame-shift mutation that causes termination of the encoded protein? M 5’-CCC-CCT-AGG-TTC-AGG-3’
- A.
-CCA-CCT-AGG-TTC-AGG-
- B.
-GCC-CCT-AGG-TTC-AGG-
- C.
-CCA-CCC-TAG-GTT-CAG-
- D.
-CCC-CTA-GGT-TCA-GG- -
- E.
-CCC-CCT-AGG-AGG- - -
Correct Answer
C. -CCA-CCC-TAG-GTT-CAG-Explanation
The correct answer is "-CCA-CCC-TAG-GTT-CAG-". This sequence is a frame-shift mutation because it shifts the reading frame by one nucleotide, causing a change in the codons. As a result, the codons and their corresponding amino acids are altered. In this mutated sequence, the codon CCC (proline) is replaced by CCA (proline), which causes a frameshift. Additionally, the TAG codon (stop codon) is introduced earlier in the sequence, leading to premature termination of the encoded protein.Rate this question:
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- 27.
Erythromycin is the antibiotic of choice when treating respiratory tract infections in legionnaire’s disease, whooping cough, and Mycoplasma-based pneumonia because of its ability to inhibit protein synthesis in certain bacteria by
- A.
Inhibiting translocation by binding to 50S ribosomal subunits
- B.
Acting as an analogue of mRNA
- C.
Causing premature chain termination
- D.
Inhibiting initiation
- E.
Mimicking mRNA binding
Correct Answer
A. Inhibiting translocation by binding to 50S ribosomal subunitsExplanation
Erythromycin is the antibiotic of choice for treating respiratory tract infections in legionnaire's disease, whooping cough, and Mycoplasma-based pneumonia because it inhibits protein synthesis in certain bacteria. It does this by binding to the 50S ribosomal subunits, specifically inhibiting translocation. This prevents the ribosome from moving along the mRNA strand during protein synthesis, effectively stopping bacterial growth. Other mechanisms listed, such as acting as an analogue of mRNA, causing premature chain termination, inhibiting initiation, and mimicking mRNA binding, are not the primary mode of action for erythromycin.Rate this question:
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.JPG "If adenine is the first base on thee template strand corresponding to the initiation point for segment E, which precursor molecule would serve as the substrate that formed the first nucleotide in segment E?")
.JPG "You have obtained a sample of DNA, and you transcribe mRNA from this DNA and purify it. You then separate the two strands of the DNA and analyze the base composition of each strand and of the mRNA. You obtain the data shown in the table to the right. Which strand of the DNA is the coding strand, serving as a template for mRNA synthesis?")
.JPG "The following piece of DNA is being replicated. The replication fork moves from left to right along this DNA.
What is the sequence of an Okazaki fragment on this piece of DNA whose 5’ and 3’ ends are located right at the borders of the shown sequence?")
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Mar 22, 2023Quiz Edited by
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Dec 09, 2011Quiz Created by
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