Quiz Review Biologi Molekuler 2009/2010 (I)
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Quiz Review Biologi Molekuler 2009/2010 (() Program Studi Biologi (S1) Fakultas Biologi UNSOED ini mencakup materi pendahuluan sampai dengan transkripsi. Pada hari Rabu, 28 April 2010, Jam 10.00 - 22.00 WIB, Quiz ini dapat diakses tanpa password.
Questions and Answers
- 1.
The nucleic acids DNA and RNA are polymers of ___________
- A.
Amino acids
- B.
Nucleotides
- C.
Nucleic acids
- D.
Glucose
- E.
Nucleosides
Correct Answer
B. NucleotidesExplanation
Nucleic acids such as DNA and RNA are composed of repeating units called nucleotides. These nucleotides consist of a sugar molecule (ribose in RNA and deoxyribose in DNA), a phosphate group, and a nitrogenous base (adenine, cytosine, guanine, or thymine/uracil). These nucleotides link together through phosphodiester bonds to form the polymer chains of DNA and RNA. Therefore, the correct answer is nucleotides.Rate this question:
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- 2.
Polysaccharides are polymers of simple sugars covalently linked by ___________
- A.
Peptide bonds
- B.
Hydrogen bonds
- C.
Glycosidic bonds
- D.
Phosphodiester bond
Correct Answer
C. Glycosidic bondsExplanation
Polysaccharides are polymers of simple sugars, meaning that they are made up of many sugar molecules linked together. The correct answer is glycosidic bonds because these are the covalent bonds that join the sugar molecules in a polysaccharide. Peptide bonds are found in proteins, hydrogen bonds are weak attractions between molecules, and phosphodiester bonds are found in nucleic acids.Rate this question:
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- 3.
In nucleic acids, the bond between the bases and the sugars is _________
- A.
Peptide bonds
- B.
Glycosylic (or glycosidic) bond
- C.
Phosphodiester bond
- D.
Hydrogen bonds
Correct Answer
B. Glycosylic (or glycosidic) bondExplanation
The bond between the bases and the sugars in nucleic acids is called a glycosylic (or glycosidic) bond. This bond forms between the sugar molecule and the nitrogenous base, connecting them together to form the backbone of the nucleic acid. The glycosylic bond is a type of covalent bond that is formed through a condensation reaction, where a water molecule is released. This bond is important for the stability and structure of nucleic acids, allowing them to carry and store genetic information.Rate this question:
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- 4.
In a DNA or RNA molecule, deoxyribonucleotides or ribonucleotides respectively are joined into a polymer by the covalent linkage of a phosphate group between the 5-hydroxyl of one ribose and the 3-hydroxyl of the next.This kind of bond or linkage is called ____________
- A.
Peptide bonds
- B.
Glycosylic (or glycosidic) bond
- C.
Phosphodiester bond
- D.
Hydrogen bonds
Correct Answer
C. Phosphodiester bondExplanation
In a DNA or RNA molecule, deoxyribonucleotides or ribonucleotides respectively are joined together through a covalent bond called a phosphodiester bond. This bond is formed between the 5-hydroxyl group of one ribose and the 3-hydroxyl group of the next ribose, creating a chain-like structure. Peptide bonds are found in proteins, glycosylic (or glycosidic) bonds are found in carbohydrates, and hydrogen bonds are weak bonds that help stabilize the structure of DNA and RNA, but they are not involved in joining nucleotides together.Rate this question:
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- 5.
There is a convention to write the sequences of DNA or RNA with the 5-end at the right.
- A.
TRUE
- B.
FALSE
Correct Answer
B. FALSEExplanation
The given statement is false. The convention for writing the sequences of DNA or RNA is to have the 5-end at the left. This convention is based on the direction in which the nucleotides are added during DNA or RNA synthesis. The 5-end refers to the end of the nucleic acid molecule that has a phosphate group attached to the 5' carbon of the sugar molecule, while the 3-end refers to the end that has a hydroxyl group attached to the 3' carbon of the sugar molecule. Therefore, the correct answer is false.Rate this question:
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- 6.
In strong acid and at elevated temperatures, for example perchloric acid (HClO4) at more than 100°C, nucleic acids are hydrolyzed completely to their constituent
- A.
TRUE
- B.
FALSE
Correct Answer
A. TRUEExplanation
In strong acid and at elevated temperatures, nucleic acids undergo complete hydrolysis, breaking down into their constituent parts. This process is particularly evident with perchloric acid (HClO4) at temperatures exceeding 100°C. Therefore, the statement "In strong acid and at elevated temperatures, nucleic acids are hydrolyzed completely to their constituent parts" is true.Rate this question:
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- 7.
Chemical agents can cause the denaturation of DNA or neutral pH.
- A.
Urea and ehidium bromide
- B.
Urea and formamide
- C.
Formamide and perchloric acid
- D.
Ehidium bromide and perchloric acid
- E.
Ehidium bromide and cesium chloride
Correct Answer
B. Urea and formamideExplanation
Urea and formamide are both chemical agents that can cause the denaturation of DNA. Denaturation refers to the process of unraveling the double helix structure of DNA, leading to the separation of the DNA strands. Urea and formamide disrupt the hydrogen bonds between the bases of DNA, which are responsible for holding the two strands together. This disruption causes the DNA to lose its native structure and become denatured.Rate this question:
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- 8.
The absorbance at 260 nm is used to determine the __________ of nucleic acids.
- A.
Concentration
- B.
Purity
- C.
Double-stranded DNA
- D.
Single-stranded DNA
Correct Answer
A. ConcentrationExplanation
The absorbance at 260 nm is used to determine the concentration of nucleic acids because nucleic acids have a characteristic absorption peak at this wavelength. By measuring the absorbance at 260 nm, one can quantify the amount of nucleic acids present in a sample, providing information about their concentration.Rate this question:
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- 9.
The approximate purity of dsDNA preparations may be estimated by determination of the ratio of absorbance at 260 and 280 nm (A260/A280).
- A.
TRUE
- B.
FALSE
Correct Answer
A. TRUEExplanation
The statement is true because the ratio of absorbance at 260 nm and 280 nm is commonly used to estimate the purity of dsDNA preparations. A260/A280 ratio is typically around 1.8-2.0 for pure DNA, indicating minimal contamination by proteins or other impurities. A lower ratio suggests protein contamination, while a higher ratio suggests RNA contamination. Therefore, by determining the A260/A280 ratio, one can get an approximate estimate of the purity of dsDNA preparations.Rate this question:
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- 10.
The pure dsDNA has an A260/A280 of ____
- A.
2.0
- B.
0.18
- C.
1.8
- D.
18
Correct Answer
C. 1.8Explanation
The A260/A280 ratio is a measure of the purity of DNA samples. A ratio of 1.8 indicates that the DNA sample is relatively pure, with minimal contamination from proteins or other impurities. This is the ideal ratio for a pure dsDNA sample, as it suggests that there is no significant contamination that could interfere with downstream applications such as sequencing or PCR.Rate this question:
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- 11.
The pure RNA has an A260/A280 of ______
- A.
2.0
- B.
0.18
- C.
1.8
- D.
18
Correct Answer
A. 2.0Explanation
The A260/A280 ratio is commonly used to assess the purity of RNA samples. A ratio of 2.0 indicates that the RNA is pure and free from protein or other contaminants. This ratio is obtained by measuring the absorbance of the RNA sample at wavelengths of 260 nm and 280 nm. A higher ratio indicates that the sample contains more protein contamination, while a lower ratio suggests the presence of other contaminants. Therefore, a pure RNA sample should ideally have an A260/A280 ratio of 2.0.Rate this question:
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- 12.
The DNA in E. coli cells consists of a single closed-circular DNA molecule of length 4.6 million base pairs. The DNA is packaged into a region of the cell known as _________
- A.
The nucleoid
- B.
The nucleus
- C.
The nucleolus
- D.
Tjhe nicleosome
- E.
The nucleoplasm
Correct Answer
A. The nucleoidExplanation
The DNA in E. coli cells is packaged into a region of the cell known as the nucleoid. This is a distinct region within the cell where the circular DNA molecule is organized and compacted. Unlike eukaryotic cells, E. coli cells do not have a nucleus or nucleolus, which are found in higher organisms. The nucleosome refers to the basic structural unit of DNA packaging in eukaryotic cells, and the nucleoplasm is the fluid inside the nucleus. Therefore, the correct answer is the nucleoid.Rate this question:
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- 13.
The major protein components of chromatin are the histones; small, basic (positively charged) proteins which bind tightly to DNA. There are four families of core histone:______________
- A.
H2A, H2B, H3, H4
- B.
H2A, H2B, H3, H1
- C.
H2A, H2B, H1, H4
- D.
H1, H2B, H3, H4
Correct Answer
A. H2A, H2B, H3, H4Explanation
The major protein components of chromatin are the histones, which are small, basic proteins that tightly bind to DNA. The four families of core histones are H2A, H2B, H3, and H4. These histones play a crucial role in packaging DNA into a compact and organized structure, known as a nucleosome, which is the basic unit of chromatin. Each nucleosome consists of DNA wrapped around a histone octamer composed of two copies each of H2A, H2B, H3, and H4. Therefore, the correct answer is H2A, H2B, H3, H4.Rate this question:
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- 14.
Chromatin is organized into a larger structure, known as the ____________
- A.
Nucleosome
- B.
Chromatosome
- C.
Solenoid
- D.
Histone
Correct Answer
C. SolenoidExplanation
Chromatin, which consists of DNA and proteins, is organized into a larger structure called the solenoid. The solenoid is a tightly coiled and compacted structure formed by the wrapping of DNA around histone proteins. This coiling helps to condense the DNA and allows for efficient packaging within the nucleus of a cell. The solenoid structure plays a crucial role in regulating gene expression and maintaining the stability of the genome.Rate this question:
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- 15.
A nucleosome core plus H1 is known as a _________________
- A.
Chromatosome
- B.
Chromatin
- C.
Chromosome
- D.
Loops
- E.
Solenoid
Correct Answer
A. ChromatosomeExplanation
A nucleosome core plus H1 is known as a chromatosome. This structure consists of DNA wrapped around a histone octamer, with an additional linker histone H1. The H1 histone helps to stabilize the nucleosome and regulate access to the DNA. The chromatosome plays a crucial role in compacting the DNA and organizing it into higher-order structures, such as chromatin fibers and chromosomes.Rate this question:
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- 16.
Chromosomes greatly alter their level of compactness as cells progress through the cell cycle, varying between highly condensed chromosomes at _________ and very much more diffuse structures in _______
- A.
Anaphase, interphase
- B.
Metaphase, prophase
- C.
Prophase, interphase
- D.
Metaphase, interphase
Correct Answer
D. Metaphase, interphaseExplanation
During the cell cycle, chromosomes undergo changes in their level of compactness. In metaphase, chromosomes are highly condensed and tightly packed, which allows them to be easily separated and distributed to daughter cells during cell division. On the other hand, during interphase, chromosomes become much more diffuse and less compact. This is because interphase is the resting phase of the cell cycle, where the cell prepares for the next round of division by carrying out normal cellular functions such as growth and DNA replication.Rate this question:
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- 17.
The __________ is the constricted region where the two sister chromatids are joined in the metaphase chromosome.
- A.
Centromere
- B.
Telomere
- C.
Kinetochore
- D.
Microtubules
- E.
Heterochromatin
Correct Answer
A. CentromereExplanation
The centromere is the correct answer because it is the region where the two sister chromatids are joined in the metaphase chromosome. This region plays a crucial role in the separation of the chromatids during cell division. It serves as an attachment point for the spindle fibers, which are responsible for pulling the chromatids apart. Without a functional centromere, proper chromosome segregation would not occur, leading to genetic abnormalities and potential cell death.Rate this question:
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- 18.
Specialized DNA sequences that form the ends of the linear DNA molecules of the eukaryotic chromosomes
- A.
Kinetochore
- B.
Microtubules
- C.
Centromere
- D.
Telomeres
Correct Answer
D. TelomeresExplanation
Telomeres are specialized DNA sequences that form the ends of linear DNA molecules in eukaryotic chromosomes. They consist of repetitive nucleotide sequences that protect the chromosomes from degradation and ensure the stability of the genetic material during cell division. Telomeres also play a crucial role in preventing the loss of genetic information and the fusion of chromosomes. Therefore, telomeres are essential for maintaining the integrity and stability of the genome.Rate this question:
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- 19.
This enzyme has been used to map the regions of transcriptionally active chromatin in cells.
- A.
The nuclease deoxyribonuclease I (DNase I)
- B.
DNA polymerase III
- C.
DNA polymerase I
- D.
DNA helicase
- E.
DNA primase
Correct Answer
A. The nuclease deoxyribonuclease I (DNase I)Explanation
DNase I is an enzyme that specifically cleaves DNA at phosphodiester bonds. It is commonly used in molecular biology experiments to map regions of transcriptionally active chromatin in cells. DNase I can digest the exposed DNA in regions where transcription factors and other proteins are bound, indicating the presence of active genes. By analyzing the resulting DNA fragments, researchers can identify and study the regulatory elements and protein-DNA interactions involved in gene expression. DNA polymerase III, DNA polymerase I, DNA helicase, and DNA primase are not involved in mapping transcriptionally active chromatin and do not have the same specificity as DNase I.Rate this question:
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- 20.
Enzymes which use the energy of ATP hydrolysis to move into and melt double-stranded DNA (or RNA).
- A.
DNA primase
- B.
DNA polymerase III
- C.
DNA helicase
- D.
DNA polymerase I
- E.
DNA ligase
Correct Answer
C. DNA helicaseExplanation
DNA helicase is the correct answer because it is an enzyme that uses the energy from ATP hydrolysis to separate the two strands of a double-stranded DNA molecule. This process is known as DNA melting or DNA unwinding. DNA helicase plays a crucial role in DNA replication, transcription, and repair by unwinding the DNA strands and creating a replication fork where other enzymes, such as DNA polymerase, can bind and synthesize new DNA strands.Rate this question:
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- 21.
The single-stranded bubble created by DNA helicase is coated with _______________ to protect it from breakage and to prevent the DNA renaturing.
- A.
DNA helicase
- B.
RNA primer
- C.
DNA polymerase I
- D.
DNA ligase
- E.
Singlestranded binding protein (Ssb)_
Correct Answer
E. Singlestranded binding protein (Ssb)_Explanation
The single-stranded bubble created by DNA helicase is coated with single-stranded binding protein (Ssb) to protect it from breakage and to prevent the DNA renaturing. Ssb proteins bind tightly to the single-stranded DNA, stabilizing it and preventing it from forming secondary structures or reannealing with its complementary strand. This allows the DNA replication machinery to access the single-stranded template for DNA synthesis. DNA polymerase, DNA ligase, and RNA primer are not involved in coating the single-stranded bubble.Rate this question:
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- 22.
The enzyme ____________ attaches to the DNA and synthesizes a short RNA primer to initiate synthesis of the leading strand of the first replication fork.
- A.
DNA polymerase I
- B.
DNA primase
- C.
DNA ligase
- D.
DNA polymerase III
- E.
Singlestranded binding protein (Ssb)_
Correct Answer
B. DNA primaseExplanation
DNA primase is the correct answer because it is the enzyme that attaches to the DNA and synthesizes a short RNA primer to initiate the synthesis of the leading strand of the first replication fork. This primer is necessary for DNA polymerase III to start adding nucleotides and continue the replication process. DNA polymerase I is involved in removing RNA primers and replacing them with DNA, DNA ligase is responsible for joining the Okazaki fragments, and single-stranded binding protein (Ssb) helps stabilize the single-stranded DNA during replication.Rate this question:
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- 23.
Inhibitors of , such as novobiocin and oxolinic acid, are effective inhibitors of bacterial replication and have antibiotic activity.
- A.
TRUE
- B.
FALSE
Correct Answer
A. TRUEExplanation
Novobiocin and oxolinic acid are mentioned as inhibitors of bacterial replication and have antibiotic activity. This implies that they are effective in stopping the growth and replication of bacteria. Therefore, the statement is true.Rate this question:
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- 24.
DNA primase synthesizes RNA primers every 1000–2000 nt on the lagging strand.
- A.
TRUE
- B.
FALSE
Correct Answer
A. TRUEExplanation
DNA primase is an enzyme responsible for synthesizing short RNA primers during DNA replication. These primers are necessary for DNA polymerase to initiate replication on the lagging strand. The RNA primers are synthesized every 1000-2000 nucleotides on the lagging strand to provide a starting point for DNA synthesis. Therefore, the statement that DNA primase synthesizes RNA primers every 1000-2000 nt on the lagging strand is true.Rate this question:
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- 25.
Both leading and lagging strand primers are elongated by ______
- A.
DNA polymerase III
- B.
DNA primase
- C.
DNA gyrase
- D.
DNA polymerase I
- E.
DNA ligase
Correct Answer
A. DNA polymerase IIIExplanation
DNA polymerase III is responsible for elongating both the leading and lagging strand primers during DNA replication. It is the main enzyme involved in DNA synthesis and is responsible for adding nucleotides to the growing DNA strand. DNA primase is responsible for synthesizing the RNA primers that are needed to initiate DNA replication, but it is not involved in the elongation of the primers. DNA gyrase is responsible for relieving the torsional strain that builds up ahead of the replication fork. DNA polymerase I is involved in removing the RNA primers and replacing them with DNA, but it is not responsible for elongating the primers. DNA ligase is responsible for joining the Okazaki fragments on the lagging strand, but it is not involved in the elongation of the primers.Rate this question:
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- 26.
Once the lagging strand primers have been elongated, the primers are removed and the gaps filled by ___________
- A.
DNA polymerase III
- B.
DNA primase
- C.
DNA gyrase
- D.
DNA ligase
- E.
DNA polymerase I
Correct Answer
E. DNA polymerase IExplanation
After the lagging strand primers have been elongated, DNA polymerase I is responsible for removing the primers and filling in the gaps. It has exonuclease activity that allows it to remove the RNA primers, and then it synthesizes DNA to fill in the gaps left behind. DNA polymerase III is involved in the elongation of the DNA strands, while DNA primase is responsible for synthesizing the RNA primers. DNA gyrase is involved in the supercoiling of DNA. DNA ligase is responsible for joining the Okazaki fragments on the lagging strand.Rate this question:
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- 27.
This enzyme has 5'→3' polymerase, 5'→3' exonuclease and 3'→5' proofreading exonuclease activities on a single polypeptide chain.
- A.
DNA primase
- B.
DNA gyrase
- C.
DNA polymerase I
- D.
DNA ligase.
- E.
DNA helicase
Correct Answer
C. DNA polymerase IExplanation
DNA polymerase I is the correct answer because it is an enzyme that possesses all three activities mentioned in the question - 5'→3' polymerase, 5'→3' exonuclease, and 3'→5' proofreading exonuclease. This enzyme is involved in DNA replication and repair processes, where it synthesizes new DNA strands, removes RNA primers, and corrects any errors in the DNA sequence.Rate this question:
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- 28.
The final phosphodiester bond between the Okazaki fragments is made by ___________
- A.
DNA ligase.
- B.
DNA helicase
- C.
DNA gyrase
- D.
DNA primase
- E.
DNA polymerase I
Correct Answer
A. DNA ligase.Explanation
DNA ligase is responsible for joining the Okazaki fragments during DNA replication. Okazaki fragments are short segments of DNA that are synthesized on the lagging strand during DNA replication. DNA polymerase synthesizes these fragments in the 5' to 3' direction, creating a gap between the fragments. DNA ligase then seals the gaps by catalyzing the formation of a phosphodiester bond between the adjacent fragments. This process is essential for completing the replication of both strands of DNA. DNA helicase unwinds the DNA double helix, DNA gyrase relieves the tension caused by unwinding, DNA primase synthesizes RNA primers, and DNA polymerase I removes the RNA primers and fills the gaps with DNA nucleotides.Rate this question:
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- 29.
The E. coli RNA polymerase is one of the largest enzymes in the cell. The holoenzyme of RNA polymerase enzyme consists of the alpha, beta, beta prime, and omega subunits.
- A.
TRUE
- B.
FALSE
Correct Answer
B. FALSEExplanation
The given statement is false. The holoenzyme of E. coli RNA polymerase consists of the alpha, beta, and beta prime subunits, but not the omega subunit.Rate this question:
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- 30.
In E. coli promoters, this sequence is centered at around the –10 position with respect to the transcription start site. This is sometimes referred to as the Pribnow box.
- A.
TRUE
- B.
FALSE
Correct Answer
A. TRUEExplanation
The statement is true because in E. coli promoters, the sequence mentioned is indeed centered around the -10 position with respect to the transcription start site. This sequence, also known as the Pribnow box, plays a crucial role in the initiation of transcription in E. coli.Rate this question:
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- 31.
Many eukaryotic promoters contain a sequence called the TATA box around 25–35 bp upstream from the start site of transcription
- A.
TRUE
- B.
FALSE
Correct Answer
A. TRUEExplanation
The TATA box is a sequence found in many eukaryotic promoters, located around 25-35 base pairs upstream from the start site of transcription. It plays a crucial role in initiating transcription by binding to transcription factors and recruiting RNA polymerase to the promoter region. Therefore, the statement that many eukaryotic promoters contain the TATA box is true.Rate this question:
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- 32.
RNA polymerase binds to specific initiation sites downstream from transcribed sequences, these are called promoters.
- A.
TRUE
- B.
F:ASE
Correct Answer
B. F:ASEExplanation
RNA polymerase does not bind to specific initiation sites downstream from transcribed sequences; instead, it binds to specific initiation sites upstream from transcribed sequences. These initiation sites are called promoters. Therefore, the given statement is false.Rate this question:
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- 33.
Transcription is the synthesis of a single-stranded RNA from a doublestranded DNA template. RNA synthesis occurs in the 5'→3' direction and its sequence corresponds to that of the DNA strand which is known as the sense strand.
- A.
TRUE
- B.
FALSE
Correct Answer
A. TRUEExplanation
Transcription is the process in which a single-stranded RNA molecule is synthesized from a double-stranded DNA template. During transcription, the RNA synthesis occurs in the 5'→3' direction, meaning that the RNA molecule is built in the 5' to 3' direction. Additionally, the sequence of the RNA molecule corresponds to that of the DNA strand known as the sense strand. Therefore, the statement that transcription is the synthesis of a single-stranded RNA from a double-stranded DNA template, occurring in the 5'→3' direction and with a sequence corresponding to the sense strand, is true.Rate this question:
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- 34.
The E. coli RNA polymerase can synthesize RNA at a rate of around 40 nt per sec at 37°C and requires Mg2+ for its activity.
- A.
TRUE
- B.
FALSE
Correct Answer
A. TRUEExplanation
The given statement is true because it states that the E. coli RNA polymerase can synthesize RNA at a rate of around 40 nt per sec at 37°C and requires Mg2+ for its activity. This implies that the E. coli RNA polymerase is capable of RNA synthesis and that its activity is dependent on the presence of Mg2+.Rate this question:
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- 35.
The alfa subunit of E. coli RNA polymerase is encoded by the rpoA gene.
- A.
TRUE
- B.
FALSE
Correct Answer
A. TRUEExplanation
The statement is true because the alfa subunit of E. coli RNA polymerase is indeed encoded by the rpoA gene. This gene provides the instructions for the synthesis of the alfa subunit, which is an essential component of the RNA polymerase enzyme in E. coli.Rate this question:
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- 36.
RNA polymerase II (RNA Pol II) transcribes all protein-coding genes and some small nuclear RNA (snRNA) genes. It is located in the nucleoplasm and is very sensitive to alpha-amanitin.
- A.
TRUE
- B.
FALSE
Correct Answer
A. TRUEExplanation
RNA polymerase II (RNA Pol II) is responsible for transcribing all protein-coding genes and some small nuclear RNA (snRNA) genes. It is located in the nucleoplasm, the fluid inside the nucleus of a cell. Additionally, RNA Pol II is known to be highly sensitive to alpha-amanitin, a toxin produced by certain mushrooms. Therefore, the statement that RNA Pol II transcribes all protein-coding genes and some snRNA genes and is sensitive to alpha-amanitin is true.Rate this question:
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- 37.
RNA polymerase I (RNA Pol I) transcribes most rRNA genes. It is located in the nucleoli and is insensitive to alpha-amanitin
- A.
TRUE
- B.
FALSE
Correct Answer
A. TRUEExplanation
RNA polymerase I is responsible for transcribing most ribosomal RNA (rRNA) genes. It is primarily located in the nucleoli, which are specialized regions within the nucleus where rRNA synthesis occurs. Additionally, RNA polymerase I is known to be insensitive to alpha-amanitin, a toxin that inhibits the activity of other RNA polymerases. Therefore, the statement that RNA polymerase I transcribes most rRNA genes and is insensitive to alpha-amanitin is true.Rate this question:
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- 38.
RNA polymerase III (RNA Pol III) transcribes the genes for tRNA, 5S rRNA, U6 snRNA and certain other small RNAs. It is located in the nucleoplasm and is moderately sensitive to alpha-amanitin.
- A.
TRUE
- B.
FALSE
Correct Answer
A. TRUEExplanation
RNA polymerase III is responsible for transcribing specific genes, including those for tRNA, 5S rRNA, U6 snRNA, and other small RNAs. It is located in the nucleoplasm and is moderately sensitive to alpha-amanitin. Therefore, the statement that RNA polymerase III transcribes these genes and is moderately sensitive to alpha-amanitin is true.Rate this question:
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- 39.
The strands are joined noncovalently by hydrogen bonding between the bases on opposite strands, to form base pairs.There are around ______ base pairs per turn in the DNA double helix
- A.
19
- B.
10
- C.
29
- D.
20
- E.
16
Correct Answer
B. 10Explanation
The correct answer is 10 because there are 10 base pairs per turn in the DNA double helix. The strands of DNA are joined together by hydrogen bonding between the bases on opposite strands, specifically adenine (A) with thymine (T) and cytosine (C) with guanine (G). This base pairing allows the two strands to twist around each other in a helical structure, with each complete turn consisting of 10 base pairs.Rate this question:
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- 40.
The strands are joined noncovalently by hydrogen bonding between the bases on opposite strands, to form base pairs.There are around ______ base pairs per turn in the DNA double helix
- A.
10
- B.
15
- C.
20
- D.
25
- E.
30
Correct Answer
A. 10Explanation
The DNA double helix is formed by two strands that are joined together by hydrogen bonding between their bases. Each base pairs with a complementary base on the opposite strand (adenine with thymine, and cytosine with guanine). The number of base pairs per turn in the DNA double helix is 10. This means that for every complete turn of the helix, there are 10 base pairs.Rate this question:
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