Block 1 DNA Technology
.JPG "Block 1 DNA Technology - 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. HAT is responsible for adding acetyl groups to histones, which helps in loosening the chromatin structure and allowing transcription factors to access the DNA and initiate gene expression. Without HAT activity, the chromatin remains tightly packed around the DNA, making it difficult for transcription factors to bind and initiate transcription. As a result, transcription rates are reduced.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 consisting of two strands that run in opposite directions. This is known as the antiparallel nature of DNA. The two strands are held together by hydrogen bonds between complementary nitrogenous bases, forming base pairs. The phosphate groups are located on the exterior of the DNA molecule, while the nitrogenous bases are stacked along the interior. The sugar component in DNA is deoxyribose, not ribose. Therefore, the correct answer is "Two strands that run in the opposite direction."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 for segment E, then the corresponding nucleotide on the newly synthesized strand would be uracil, as adenine pairs with uracil in RNA. Therefore, the precursor molecule that would serve as the substrate for the formation of the first nucleotide in segment E is UTP (uridine triphosphate).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 is composed of DNA tightly wrapped around histone proteins. Nucleosomes are formed when DNA is wound around a core of eight histone proteins. Therefore, the phrase "Subunits of chromatin" accurately describes nucleosomes.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 UCA and UCU code for the amino acid serine. This mutation is a silent mutation, meaning it does not result in a change in the 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, have different cellular machinery for protein synthesis compared to eukaryotes. These antibiotics specifically target the ribosomes in prokaryotic cells, preventing them from synthesizing proteins effectively. This disruption in protein synthesis weakens the bacteria and inhibits their growth and reproduction, making these antibiotics effective in treating bacterial infections.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 given correct answer for the question is "Translocation". Translocation is the process in protein synthesis where the ribosome moves along the mRNA strand, shifting the tRNA molecules from the A site to the P site and then to the E site. This movement allows for the synthesis of the polypeptide chain. In this case, the antibiotic inhibits the translocation process, resulting in the only product being 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 genome can be determined by subtracting the percentage of uracil (15%) from 100%. Since uracil and guanine pair up in RNA, if there is 15% uracil, there must be 15% guanine as well. Therefore, the percentage of guanine in this genome is 35%.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 and correct errors that may occur during DNA replication. By removing misincorporated bases, DNA polymerase δ ensures that the newly synthesized DNA strand is accurate and free from mutations. This helps maintain the integrity of the genetic information and reduces the occurrence of genetic diseases caused by single nucleotide substitutions.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 question states that a viral-encoded enzyme inactivates a host-cell nuclear protein required for DNA replication. Among the options given, SBP is the only protein that is directly involved in DNA replication. It binds to single-stranded DNA and stabilizes it, allowing for DNA replication to occur. Therefore, it is a potential substrate for the viral enzyme that inactivates the host-cell nuclear protein.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 the action of Topoisomerase II, an enzyme that plays a crucial role in bacterial DNA replication. Topoisomerase II is responsible for untangling and unknotting the DNA strands during replication, making it an essential enzyme for chromosome duplication in rapidly dividing cells. By inhibiting this enzyme, norfloxacin disrupts the replication process and prevents the bacteria from replicating their DNA effectively.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 they interact with other transcription factors to increase the efficiency of transcription initiation. By facilitating the binding of RNA polymerase to the promoter and stabilizing the transcription initiation complex, enhancers play a crucial role in regulating 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 start AUG codon is located at the beginning of the coding sequence, which is preceded by a 5'-untranslated region of 100 bp. The final termination codon is located at the end of the coding sequence, which is followed by a 3'-untranslated region of 200 bp. Therefore, the total number of bases between the start AUG codon and the final termination codon is 100 (5'-untranslated region) + 150 (coding sequence) + 200 (3'-untranslated region) = 450.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 a specific amino acid. The mRNA in this question is 336 nucleotides long, which includes the initiator and termination codons. Since there are 3 nucleotides per codon, we can divide 336 by 3 to get the number of codons. 336 divided by 3 equals 112 codons. However, we need to subtract 1 for the termination codon, so the final answer is 111 amino acids.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 information, the base composition of Strand 2 matches the base composition of the transcribed mRNA. 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 the dose at which 50% of the recipients die. In this case, the LD50 is 0.1 mg per kg body weight. The average mushroom contains 7 mg of amanitin. To calculate how many mushrooms must be consumed by someone who is 100 kg to be above the LD50, we need to divide the LD50 dose by the amount of amanitin in one mushroom. 0.1 mg/kg divided by 7 mg/mushroom equals approximately 0.014 mushrooms. Since you can't consume a fraction of a mushroom, the person would need to consume at least 2 mushrooms to be above the LD50.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 1,000-kilobase fragment of DNA. However, the question asks for the time it takes to produce two daughter molecules. Since there are 10 replication origins, each origin will produce one daughter molecule. Therefore, it will take 50 seconds to produce two daughter molecules.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 that involves the addition of acetyl groups to histone proteins, which are involved in packaging DNA into a compact structure called chromatin. This modification is typically associated with gene activation because it loosens the interaction between DNA and histones, allowing for easier access of transcription factors and other regulatory proteins to the DNA. This facilitates the transcription of genes, leading to increased gene expression and ultimately, 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 direction of DNA replication and the location of the 5' and 3' ends. In this case, the replication fork is moving from left to right, so the new strand being synthesized will be in the 5' to 3' direction. The 5' and 3' ends of the Okazaki fragment are located right at the borders of the shown sequence, which means that the new strand will start with the same nucleotide as the last nucleotide of the shown sequence and continue in the 5' to 3' direction. Therefore, the correct answer is 5’ GGATGTTATCATGCACCT 3’.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 average size of a human gene is 40,000 bp. This is because genes in humans can vary in size, ranging from a few hundred base pairs to over a million base pairs. However, when considering the average size, it is found to be around 40,000 base pairs. This is determined by analyzing the sizes of all known human genes and calculating the average length.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 scenario 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 restores DNA synthesis suggests that there is a defect in an enzyme involved in DNA replication. Unwinding proteins play a crucial role in DNA replication as they help in unwinding the double helix structure of DNA, allowing for DNA synthesis to occur. Therefore, the likely enzyme that is defective in this child is the unwinding proteins.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 radioactive double-stranded DNA molecule undergoes replication in a solution free of radioactive label, the resulting DNA molecules will have one radioactive strand and one non-radioactive strand. This is because during replication, the two strands of the original DNA molecule separate and each strand serves as a template for the synthesis of a new complementary strand. Since the solution is free of radioactive label, the newly synthesized strands will not be radioactive. Therefore, half of the resulting four double-stranded DNA 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 synthesized from the mature mRNA, which does not contain the intron. The two exons that encode the protein of 100 amino acids are separated by the intron of 100 bp, which is not present in the mature mRNA. Therefore, the cDNA will only include the two exons and the untranslated regions, resulting in a size of 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 the hepatotoxic octapeptide a-amanitin, which is present in the death-cap mushroom, inhibits RNA polymerase. This inhibition of RNA polymerase disrupts the process of transcription, leading to the dysfunction of protein synthesis and ultimately causing the symptoms of nausea, vomiting, diarrhea, and abdominal pain. Supportive care is provided as there is no specific antidote for the toxin.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 deleting one base pair (C) from the original sequence. This causes a shift in the codons and results in a premature stop codon (TAG) being introduced earlier in the sequence. As a result, the encoded protein is terminated prematurely.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 effective in treating respiratory tract infections in legionnaire's disease, whooping cough, and Mycoplasma-based pneumonia because it inhibits translocation by binding to 50S ribosomal subunits. This means that it prevents the movement of the ribosome along the mRNA strand during protein synthesis, effectively stopping the production of proteins in certain bacteria. By inhibiting this essential step in bacterial protein synthesis, erythromycin can effectively kill the bacteria and treat the respiratory tract infections.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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