1Which form of DNA is the most common and biologically predominant right-handed double helix under physiological conditions?
Types and structure of DNA
Easy
A.C-DNA
B.A-DNA
C.B-DNA
D.Z-DNA
Correct Answer: B-DNA
Explanation:
B-DNA is the right-handed helix described by Watson and Crick and is the predominant form found in cells under normal physiological conditions.
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2In the DNA double helix, adenine (A) pairs with thymine (T) through how many hydrogen bonds?
Types and structure of DNA
Easy
A.Four hydrogen bonds
B.One hydrogen bond
C.Two hydrogen bonds
D.Three hydrogen bonds
Correct Answer: Two hydrogen bonds
Explanation:
A–T base pairs are held together by two hydrogen bonds, whereas G–C pairs are held by three hydrogen bonds.
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3Which unusual form of DNA has a left-handed helical structure with a zig-zag sugar-phosphate backbone?
Types and structure of DNA
Easy
A.Z-DNA
B.A-DNA
C.H-DNA
D.B-DNA
Correct Answer: Z-DNA
Explanation:
Z-DNA is a left-handed double helix in which the backbone follows a zig-zag path, giving it its name.
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4According to Chargaff's rule, in double-stranded DNA the amount of adenine equals the amount of which base?
Types and structure of DNA
Easy
A.Uracil
B.Cytosine
C.Guanine
D.Thymine
Correct Answer: Thymine
Explanation:
Chargaff's rule states that in double-stranded DNA, and because of complementary base pairing.
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5The temperature at which half of the DNA molecules become single-stranded during denaturation is called the:
denaturation and re-naturation of DNA
Easy
A.Freezing point
B.Melting temperature ()
C.Annealing point
D.Boiling point
Correct Answer: Melting temperature ()
Explanation:
The melting temperature () is defined as the temperature at which 50% of the DNA has denatured into single strands.
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6DNA with a higher G–C content will have a melting temperature () that is:
denaturation and re-naturation of DNA
Easy
A.Zero
B.Higher
C.Unchanged
D.Lower
Correct Answer: Higher
Explanation:
G–C pairs have three hydrogen bonds compared to two in A–T pairs, so higher G–C content requires more energy to separate, raising the .
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7The process by which separated complementary DNA strands come back together to reform a double helix is called:
denaturation and re-naturation of DNA
Easy
A.Renaturation
B.Denaturation
C.Transcription
D.Replication
Correct Answer: Renaturation
Explanation:
Renaturation (or reannealing) is the reassociation of complementary single strands to reform double-stranded DNA when conditions are favorable.
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8Which enzyme is responsible for relieving supercoiling tension in DNA?
supercoiling of DNA
Easy
A.Ligase
B.DNA polymerase
C.Topoisomerase
D.Primase
Correct Answer: Topoisomerase
Explanation:
Topoisomerases relax supercoiled DNA by cutting one or both strands, allowing the DNA to unwind, and then resealing them.
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9When DNA is coiled in the direction opposite to that of the right-handed helix, it is said to be:
supercoiling of DNA
Easy
A.Positively supercoiled
B.Negatively supercoiled
C.Linear
D.Relaxed
Correct Answer: Negatively supercoiled
Explanation:
Negative supercoiling represents underwinding of the DNA helix and is the common form found in most bacterial and cellular DNA.
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10Which type of RNA carries amino acids to the ribosome during protein synthesis?
types and structures of RNA
Easy
A.snRNA
B.mRNA
C.rRNA
D.tRNA
Correct Answer: tRNA
Explanation:
Transfer RNA (tRNA) delivers specific amino acids to the ribosome, matching its anticodon to the codon on the mRNA.
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11In RNA, which nitrogenous base replaces thymine found in DNA?
types and structures of RNA
Easy
A.Guanine
B.Cytosine
C.Adenine
D.Uracil
Correct Answer: Uracil
Explanation:
RNA contains uracil (U) in place of thymine; uracil pairs with adenine during transcription and translation.
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12Which type of RNA is the most abundant in the cell and forms a structural part of ribosomes?
types and structures of RNA
Easy
A.miRNA
B.tRNA
C.rRNA
D.mRNA
Correct Answer: rRNA
Explanation:
Ribosomal RNA (rRNA) is the most abundant RNA type and combines with proteins to form the ribosome.
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13Histone proteins are rich in which type of amino acids, giving them a positive charge?
Hierarchical Packaging of DNA: histone proteins
Easy
A.Aromatic amino acids
B.Acidic amino acids
C.Sulfur-containing amino acids
D.Basic amino acids
Correct Answer: Basic amino acids
Explanation:
Histones are rich in basic amino acids like lysine and arginine, giving them a positive charge that binds the negatively charged DNA.
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14Which histone protein acts as a linker histone rather than a core histone?
Hierarchical Packaging of DNA: histone proteins
Easy
A.H4
B.H3
C.H2A
D.H1
Correct Answer: H1
Explanation:
Histone H1 is the linker histone that binds to the DNA between nucleosomes, while H2A, H2B, H3, and H4 form the nucleosome core.
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15The core of a nucleosome is composed of a histone octamer containing how many histone molecules?
the nucleosome assembly
Easy
A.Six
B.Ten
C.Eight
D.Four
Correct Answer: Eight
Explanation:
The nucleosome core is a histone octamer made of two copies each of H2A, H2B, H3, and H4, totaling eight histone molecules.
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16Which enzyme unwinds the DNA double helix at the replication fork?
Enzymes and proteins in DNA replication
Easy
A.Helicase
B.Primase
C.Ligase
D.Topoisomerase
Correct Answer: Helicase
Explanation:
Helicase separates the two DNA strands by breaking the hydrogen bonds between base pairs at the replication fork.
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17Which enzyme joins Okazaki fragments together by forming phosphodiester bonds?
Enzymes and proteins in DNA replication
Easy
A.DNA ligase
B.Helicase
C.Primase
D.DNA polymerase
Correct Answer: DNA ligase
Explanation:
DNA ligase seals the nicks between adjacent Okazaki fragments on the lagging strand by forming phosphodiester bonds.
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18Which DNA polymerase is the main enzyme responsible for synthesizing new DNA strands in prokaryotes like E. coli?
DNA replication in prokaryotes
Easy
A.DNA polymerase IV
B.DNA polymerase III
C.DNA polymerase I
D.DNA polymerase II
Correct Answer: DNA polymerase III
Explanation:
In prokaryotes, DNA polymerase III is the primary enzyme carrying out the bulk of new DNA synthesis during replication.
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19In eukaryotic DNA replication, the ends of linear chromosomes are maintained by which enzyme?
DNA replication in eukaryotes
Easy
A.Telomerase
B.Helicase
C.Ligase
D.Primase
Correct Answer: Telomerase
Explanation:
Telomerase adds repetitive nucleotide sequences to the ends of linear eukaryotic chromosomes, preventing loss of genetic information.
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20The proofreading ability of DNA polymerase that removes incorrectly added nucleotides is due to its:
Fidelity of DNA replication
Easy
A.Helicase activity
B. exonuclease activity
C. exonuclease activity
D. polymerase activity
Correct Answer: exonuclease activity
Explanation:
The exonuclease activity of DNA polymerase allows it to remove mismatched nucleotides, improving replication fidelity.
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21A DNA sample is found to be a left-handed double helix with a zig-zag sugar-phosphate backbone and about 12 base pairs per turn. Which conformation does this sample most likely represent?
Types and structure of DNA
Medium
A.Z-DNA
B.A-DNA
C.C-DNA
D.B-DNA
Correct Answer: Z-DNA
Explanation:
Z-DNA is the only left-handed form, has a characteristic zig-zag backbone, and contains ~12 bp per helical turn, unlike the right-handed A- and B-forms.
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22In a segment of double-stranded DNA, adenine constitutes 30% of the total bases. Applying Chargaff's rules, what is the percentage of guanine?
Types and structure of DNA
Medium
A.40%
B.70%
C.20%
D.30%
Correct Answer: 20%
Explanation:
By Chargaff's rules, , so together they are 60%. The remaining 40% is split equally between G and C, giving .
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23Two DNA samples of equal length are heated. Sample X has a higher melting temperature () than sample Y. What can be concluded about sample X?
denaturation and re-naturation of DNA
Medium
A.It has more mismatched bases
B.It has a higher A-T content
C.It has a higher G-C content
D.It is single-stranded
Correct Answer: It has a higher G-C content
Explanation:
G-C pairs form three hydrogen bonds versus two for A-T, so a higher G-C content requires more energy to separate the strands, raising the .
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24During DNA renaturation experiments, the rate of reassociation is often expressed using the value. A DNA with highly repetitive sequences will show which behaviour?
denaturation and re-naturation of DNA
Medium
A.Reassociation at very high values
B.No reassociation at all
C.A constant absorbance regardless of
D.Reassociation at low values
Correct Answer: Reassociation at low values
Explanation:
Repetitive sequences find complementary partners quickly because many identical copies exist, so they reanneal at low values, while unique sequences require higher .
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25A circular DNA molecule has a linking number lower than that of its relaxed form. This molecule is best described as:
supercoiling of DNA
Medium
A.Negatively supercoiled
B.Positively supercoiled
C.Relaxed
D.Denatured
Correct Answer: Negatively supercoiled
Explanation:
When the linking number () is less than that of relaxed DNA (), the molecule is underwound and negatively supercoiled, which favours strand separation during replication and transcription.
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26Which enzyme relieves positive supercoils that build up ahead of a replication fork by introducing negative supercoils in bacteria?
supercoiling of DNA
Medium
A.Primase
B.DNA gyrase (topoisomerase II)
C.Topoisomerase I
D.Helicase
Correct Answer: DNA gyrase (topoisomerase II)
Explanation:
DNA gyrase, a type II topoisomerase, uses ATP to introduce negative supercoils and thereby cancels the positive supercoiling generated ahead of the fork.
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27A researcher isolates an RNA molecule containing an anticodon loop, a T\u03a8C loop, and a D-loop arranged in a cloverleaf secondary structure. Which type of RNA is this?
types and structures of RNA
Medium
A.Messenger RNA (mRNA)
B.Transfer RNA (tRNA)
C.Ribosomal RNA (rRNA)
D.Small nuclear RNA (snRNA)
Correct Answer: Transfer RNA (tRNA)
Explanation:
The cloverleaf pattern with an anticodon loop, T\u03a8C loop, and D-loop is the hallmark secondary structure of tRNA, which folds further into an L-shaped tertiary structure.
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28In eukaryotic cells, which RNA type is the most abundant by mass and forms the catalytic core of the ribosome?
types and structures of RNA
Medium
A.MicroRNA (miRNA)
B.Ribosomal RNA (rRNA)
C.Messenger RNA (mRNA)
D.Transfer RNA (tRNA)
Correct Answer: Ribosomal RNA (rRNA)
Explanation:
rRNA makes up roughly 80% of total cellular RNA and provides the peptidyl transferase (ribozyme) activity of the ribosome, making it both the most abundant and catalytic RNA.
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29Which property of histone proteins allows them to bind tightly to DNA?
Hierarchical Packaging of DNA: histone proteins
Medium
A.A high content of negatively charged aspartate residues
B.An abundance of phosphorylated serine residues
C.A high content of positively charged lysine and arginine residues
D.The presence of many hydrophobic aromatic residues
Correct Answer: A high content of positively charged lysine and arginine residues
Explanation:
Histones are rich in basic amino acids (lysine and arginine) that carry positive charges, enabling electrostatic attraction to the negatively charged phosphate backbone of DNA.
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30Which histone is NOT part of the nucleosome core octamer but instead binds to linker DNA?
Hierarchical Packaging of DNA: histone proteins
Medium
A.Histone H3
B.Histone H1
C.Histone H4
D.Histone H2A
Correct Answer: Histone H1
Explanation:
The core octamer consists of two copies each of H2A, H2B, H3, and H4. Histone H1 is the linker histone that binds DNA where it enters and exits the nucleosome, aiding higher-order folding.
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31Approximately how many base pairs of DNA are wrapped around a single histone octamer core to form the nucleosome core particle?
the nucleosome assembly
Medium
A.1000 bp
B.200 bp
C.10 bp
D.147 bp
Correct Answer: 147 bp
Explanation:
About 147 bp of DNA make roughly 1.65 turns around the histone octamer to form the nucleosome core particle; the additional linker DNA connects adjacent nucleosomes.
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32The 'beads-on-a-string' appearance of chromatin under the electron microscope corresponds to which level of packaging?
the nucleosome assembly
Medium
A.The metaphase chromosome
B.The 30 nm solenoid fibre
C.The 10 nm fibre of nucleosomes
D.Naked double-helical DNA
Correct Answer: The 10 nm fibre of nucleosomes
Explanation:
The extended 10 nm fibre, in which nucleosome cores are separated by linker DNA, gives the classic beads-on-a-string image; further coiling produces the compact 30 nm fibre.
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33Why is a primer required for DNA polymerase to begin synthesizing a new strand?
Enzymes and proteins in DNA replication
Medium
A.The primer supplies energy for polymerization
B.The primer stabilizes the replication origin
C.DNA polymerase can only add nucleotides to an existing free 3'-OH group
D.DNA polymerase can only add nucleotides to a free 5'-phosphate group
Correct Answer: DNA polymerase can only add nucleotides to an existing free 3'-OH group
Explanation:
DNA polymerases cannot initiate synthesis de novo; they extend an existing chain by adding nucleotides to a free 3'-OH, which the RNA primer provides.
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34What is the primary role of single-strand binding (SSB) proteins during replication?
Enzymes and proteins in DNA replication
Medium
A.Joining Okazaki fragments together
B.Unwinding the parental duplex
C.Preventing separated strands from reannealing
D.Synthesizing RNA primers
Correct Answer: Preventing separated strands from reannealing
Explanation:
SSB proteins coat the exposed single strands after helicase unwinds the duplex, keeping them apart and protecting them until the polymerase copies them.
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35In E. coli, which enzyme removes the RNA primers and replaces them with DNA during replication?
DNA replication in prokaryotes
Medium
A.Primase
B.DNA polymerase III
C.DNA ligase
D.DNA polymerase I
Correct Answer: DNA polymerase I
Explanation:
DNA polymerase I uses its exonuclease activity to excise RNA primers and its polymerase activity to fill the gaps with DNA; ligase then seals the nicks.
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36During replication of the leading and lagging strands, the lagging strand is synthesized discontinuously because:
DNA replication in prokaryotes
Medium
A.Helicase moves in the direction on it
B.Primase acts only on the lagging strand
C.The lagging strand template is more tightly supercoiled
D.DNA synthesis can only proceed in the direction
Correct Answer: DNA synthesis can only proceed in the direction
Explanation:
Since polymerases add nucleotides only in the direction, the strand running antiparallel to fork movement must be made in short Okazaki fragments.
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37Which feature distinguishes eukaryotic chromosomal replication from prokaryotic replication?
DNA replication in eukaryotes
Medium
A.Eukaryotes use multiple origins of replication per chromosome
B.Eukaryotes do not use Okazaki fragments
C.Eukaryotes use only a single origin per chromosome
D.Eukaryotic replication does not require primers
Correct Answer: Eukaryotes use multiple origins of replication per chromosome
Explanation:
Because eukaryotic chromosomes are very large, replication starts at many origins simultaneously to finish in a reasonable time, whereas prokaryotes typically use one origin.
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38The enzyme telomerase is required in eukaryotes to solve which specific problem of linear chromosome replication?
DNA replication in eukaryotes
Medium
A.Excessive supercoiling at replication origins
B.Failure of helicase to unwind telomeres
C.Loss of DNA at the 3' ends due to primer removal
D.Inability of ligase to seal terminal nicks
Correct Answer: Loss of DNA at the 3' ends due to primer removal
Explanation:
After the terminal RNA primer on the lagging strand is removed, no upstream 3'-OH exists to fill the gap, shortening the ends. Telomerase extends the template to counteract this end-replication problem.
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39Which activity of DNA polymerase III immediately corrects a mismatched nucleotide during synthesis?
Fidelity of DNA replication
Medium
A.Mismatch repair by MutS
B. exonuclease activity
C.Nucleotide excision repair
D. exonuclease proofreading
Correct Answer: exonuclease proofreading
Explanation:
The exonuclease (proofreading) activity of the polymerase removes an incorrectly paired nucleotide right after insertion, greatly improving replication accuracy before repair pathways act.
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40In rolling circle replication, what is the initial event that starts the process on the circular template?
Rolling circle replication
Medium
A.An endonuclease nicks one strand to create a free 3'-OH
B.Two replication forks form at a single origin
C.Primase synthesizes an RNA primer on the plus strand
D.Helicase fully separates both circular strands
Correct Answer: An endonuclease nicks one strand to create a free 3'-OH
Explanation:
Rolling circle replication begins when a specific endonuclease nicks one strand of the double-stranded circle, generating a free 3'-OH that the polymerase extends around the intact circular template.
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41A DNA sample is analyzed and found to have a helical repeat of 12 base pairs per turn, a left-handed helix, and a zig-zag sugar-phosphate backbone. Under high salt conditions, alternating purine-pyrimidine sequences favor this form. Which conformation and driving factor combination is correct?
Types and structure of DNA
Hard
A.C-DNA, stabilized by intermediate humidity conditions
B.A-DNA, stabilized by dehydration of the double helix
C.Z-DNA, stabilized by high ionic strength reducing backbone charge repulsion
D.B-DNA, stabilized by physiological hydration
Correct Answer: Z-DNA, stabilized by high ionic strength reducing backbone charge repulsion
Explanation:
Z-DNA is a left-handed helix with ~12 bp/turn and a zig-zag backbone, favored by alternating purine-pyrimidine (e.g., GC) sequences. High salt stabilizes it by shielding the electrostatic repulsion between closely apposed backbone phosphates.
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42Two DNA samples have identical length but sample X melts at and sample Y at under identical buffer conditions. What can be reliably concluded about sample Y relative to X?
denaturation and re-naturation of DNA
Hard
A.Sample Y has a higher G+C content
B.Sample Y has a higher A+T content
C.Sample Y is shorter in effective length
D.Sample Y contains more single-strand breaks
Correct Answer: Sample Y has a higher G+C content
Explanation:
increases with G+C content because G-C pairs form three hydrogen bonds and contribute stronger base-stacking than A-T pairs (two H-bonds). A higher under identical conditions indicates greater G+C content.
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43In a (renaturation kinetics) experiment on a eukaryotic genome, three distinct components reassociate at different values. The fraction reassociating at the lowest most likely represents:
denaturation and re-naturation of DNA
Hard
A.Moderately repetitive gene families
B.Unique single-copy genes
C.Denatured mitochondrial DNA only
D.Highly repetitive sequences
Correct Answer: Highly repetitive sequences
Explanation:
In analysis, reassociation rate is inversely proportional to sequence complexity. Highly repetitive sequences are present in many copies, so complementary strands find partners quickly and reassociate at the lowest values.
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44A relaxed circular DNA has . After the action of an enzyme, becomes 490 while the molecule is kept under conditions favoring -form. Assuming no change in twist (), what is the writhe and the type of supercoiling?
supercoiling of DNA
Hard
A., negatively supercoiled
B., negatively supercoiled
C., relaxed
D., positively supercoiled
Correct Answer: , negatively supercoiled
Explanation:
. With and , the change is absorbed entirely as writhe: . Negative writhe corresponds to negative (underwound) supercoiling, characteristic of DNA gyrase activity.
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45Which statement correctly distinguishes the ATP requirements and topological outcomes of bacterial topoisomerase I and DNA gyrase (topoisomerase II)?
supercoiling of DNA
Hard
A.Topo I relaxes negative supercoils without ATP; gyrase introduces negative supercoils using ATP
B.Topo I requires ATP to relax; gyrase acts without ATP to relax
C.Both require ATP and both introduce negative supercoils
D.Topo I introduces positive supercoils using ATP; gyrase relaxes them without ATP
Correct Answer: Topo I relaxes negative supercoils without ATP; gyrase introduces negative supercoils using ATP
Explanation:
Bacterial topoisomerase I relaxes negative supercoils by changing in steps of +1, requiring no ATP. DNA gyrase (topo II) introduces negative supercoils by changing in steps of −2, using energy from ATP hydrolysis.
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46A tRNA molecule loses its ability to be recognized by its cognate aminoacyl-tRNA synthetase after a point mutation, yet its cloverleaf secondary structure and anticodon are unchanged. The most likely explanation is a mutation affecting:
types and structures of RNA
Hard
A.An identity element such as the acceptor stem discriminator base
B.The loop pseudouridine modification
C.The variable loop length only
D.The 5' phosphate group of the tRNA
Correct Answer: An identity element such as the acceptor stem discriminator base
Explanation:
Aminoacyl-tRNA synthetases recognize specific identity elements, often located in the acceptor stem (including the discriminator base) and sometimes the anticodon. A mutation in an acceptor-stem identity element can abolish recognition without altering the cloverleaf fold or anticodon.
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47Ribosomal RNA is often described as the catalytic component of the ribosome. Which observation most directly supports rRNA (not ribosomal protein) as the peptidyl transferase catalyst?
types and structures of RNA
Hard
A.Ribosomal proteins cluster at the peptidyl transferase center
B.The peptidyl transferase center is composed entirely of 23S rRNA with no protein side chains at the active site
C.The catalytic activity requires the small subunit proteins exclusively
D.Removal of all rRNA still permits peptide bond formation
Correct Answer: The peptidyl transferase center is composed entirely of 23S rRNA with no protein side chains at the active site
Explanation:
Crystallographic studies of the large subunit show that the peptidyl transferase center is lined solely by 23S rRNA, with no protein within reach of the reacting groups. This demonstrates the ribosome is a ribozyme, with rRNA performing catalysis.
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48Histone H1 differs functionally from core histones in chromatin organization. Which statement best captures its distinct structural role?
Hierarchical Packaging of DNA: histone proteins
Hard
A.H1 is required for the initial deposition of the (H3-H4) tetramer
B.H1 binds linker DNA at the nucleosome dyad, promoting formation of the 30-nm fiber
C.H1 replaces H2A-H2B dimers during transcription activation
D.H1 forms the histone octamer core around which DNA wraps
Correct Answer: H1 binds linker DNA at the nucleosome dyad, promoting formation of the 30-nm fiber
Explanation:
Core histones (H2A, H2B, H3, H4) form the octamer. Linker histone H1 binds outside the core particle, at the DNA entry/exit point near the dyad, sealing DNA around the nucleosome and facilitating higher-order compaction into the 30-nm fiber.
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49During replication-coupled nucleosome assembly, histones are deposited in a defined order onto newly synthesized DNA. What is the correct sequence and its functional rationale?
the nucleosome assembly
Hard
A.All four core histones deposited simultaneously as a preformed octamer
B.H1 deposited first to organize linker DNA
C.(H3-H4) tetramer deposited first, followed by two H2A-H2B dimers
D.Two H2A-H2B dimers deposited first, then the (H3-H4) tetramer
Correct Answer: (H3-H4) tetramer deposited first, followed by two H2A-H2B dimers
Explanation:
Nucleosome assembly is ordered: chaperones (e.g., CAF-1) deposit the (H3-H4) tetramer onto DNA first, forming a subnucleosomal particle, after which two H2A-H2B dimers are added to complete the octamer.
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50In E. coli, a temperature-sensitive mutation inactivates the (tau) subunit of the DNA polymerase III holoenzyme at the restrictive temperature. What is the most direct consequence?
Enzymes and proteins in DNA replication
Hard
A.Loss of primase recruitment to the primosome
B.Failure of proofreading 3'→5' exonuclease activity
C.Loss of dimerization coupling the leading- and lagging-strand polymerases at the fork
D.Inability to load the sliding clamp onto DNA
Correct Answer: Loss of dimerization coupling the leading- and lagging-strand polymerases at the fork
Explanation:
The subunit dimerizes the two core polymerases and links them to the DnaB helicase, coordinating leading- and lagging-strand synthesis. Its loss uncouples the two polymerases, disrupting coordinated replication at the fork.
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51A researcher observes that in the absence of a particular protein, Okazaki fragments accumulate as short primed segments that are never joined, and RNA primers persist. Which combined defect best explains this?
Enzymes and proteins in DNA replication
Hard
A.Loss of primase and helicase activities
B.Loss of DNA polymerase I and DNA ligase activities
C.Loss of single-strand binding protein and topoisomerase
D.Loss of the clamp and clamp loader
Correct Answer: Loss of DNA polymerase I and DNA ligase activities
Explanation:
RNA primers persisting and unjoined fragments indicate failure of primer removal (DNA polymerase I 5'→3' exonuclease and fill-in) and failure of the final nick-sealing step (DNA ligase). Both are required to mature Okazaki fragments into a continuous strand.
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52In the E. coli origin oriC, replication initiation is tightly regulated. Which mechanism prevents immediate re-initiation at newly replicated origins?
DNA replication in prokaryotes
Hard
A.Sequestration of hemimethylated oriC by SeqA blocking Dam remethylation
B.Complete degradation of DnaA after each initiation event
C.Physical removal of the DnaB helicase from the cell
D.Permanent hypermethylation of both strands by Dam methylase
Correct Answer: Sequestration of hemimethylated oriC by SeqA blocking Dam remethylation
Explanation:
After replication, oriC is transiently hemimethylated (only the parental strand carries Dam methylation at GATC sites). SeqA binds hemimethylated oriC, sequestering it and delaying remethylation and DnaA binding, thereby preventing premature re-initiation.
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53Bidirectional replication from a single origin on a circular bacterial chromosome terminates in a specific region. What is the role of the Tus-Ter system?
DNA replication in prokaryotes
Hard
A.Tus binds Ter sites to arrest replication forks in a polar (direction-dependent) manner
B.Tus removes supercoils generated ahead of the fork at Ter
C.Tus recruits primase to reinitiate lagging-strand synthesis at Ter
D.Tus binds Ter sites to accelerate fork progression through the terminus
Correct Answer: Tus binds Ter sites to arrest replication forks in a polar (direction-dependent) manner
Explanation:
The Ter sites are arranged so that Tus-Ter complexes act as polar replication fork traps: they permit fork entry from one direction but block it from the other, ensuring forks meet and terminate within the terminus region.
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54Eukaryotic origins fire only once per cell cycle. Which regulatory logic involving the pre-replication complex (pre-RC) enforces this?
DNA replication in eukaryotes
Hard
A.Licensing (MCM loading) occurs only in G1 when CDK is low; high S-phase CDK blocks re-licensing
B.Licensing occurs in S phase when CDK is high, then CDK falls to allow firing
C.MCM helicase is loaded continuously throughout the cell cycle
D.ORC is degraded after G1 and resynthesized only in mitosis
Correct Answer: Licensing (MCM loading) occurs only in G1 when CDK is low; high S-phase CDK blocks re-licensing
Explanation:
Origin licensing—loading of the MCM2-7 helicase by ORC, Cdc6, and Cdt1—requires low CDK activity in G1. Once cells enter S phase, elevated CDK triggers firing while simultaneously inhibiting new MCM loading, preventing re-replication within one cycle.
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55The end-replication problem in linear eukaryotic chromosomes arises specifically because:
DNA replication in eukaryotes
Hard
A.DNA ligase cannot seal the final nick at internal Okazaki fragments
B.Removal of the terminal RNA primer on the lagging strand leaves a gap that cannot be filled
C.Topoisomerase cannot relax supercoils at the chromosome ends
D.Leading-strand synthesis cannot reach the extreme 3' end of the template
Correct Answer: Removal of the terminal RNA primer on the lagging strand leaves a gap that cannot be filled
Explanation:
On the lagging strand, once the terminal RNA primer is removed there is no upstream 3'-OH for a polymerase to extend from, leaving a short unreplicated gap at the 5' end of each new strand. This progressive shortening is addressed by telomerase.
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56Overall replication fidelity in E. coli approaches 1 error per – bp. If base selection gives , proofreading adds , what approximate contribution must mismatch repair provide to reach an overall error rate of ?
Fidelity of DNA replication
Hard
A.
B. (no contribution)
C.
D. to
Correct Answer: to
Explanation:
Fidelity steps multiply: base selection () × proofreading () = . To reach – overall, mismatch repair must contribute an additional factor of about to .
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57In E. coli methyl-directed mismatch repair, how does the system correctly identify and excise the newly synthesized (error-containing) strand rather than the template?
Fidelity of DNA replication
Hard
A.The leading strand is always designated for repair
B.The strand with more G+C content is always excised
C.The transiently unmethylated (hemimethylated) daughter strand at GATC sites is recognized as newer and targeted
D.The strand containing 5-methylcytosine is targeted for excision
Correct Answer: The transiently unmethylated (hemimethylated) daughter strand at GATC sites is recognized as newer and targeted
Explanation:
MutH nicks the unmethylated (newly synthesized) strand at hemimethylated GATC sites because Dam methylation lags behind replication. This strand discrimination ensures the error-containing daughter strand, not the correct template, is excised and resynthesized.
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58During rolling circle replication (e.g., of X174 or F plasmid transfer), what event initiates the process and generates the free 3'-OH used for leading-strand extension?
Rolling circle replication
Hard
A.Helicase unwinds both strands to create a bidirectional fork
B.Primase synthesizes an RNA primer at a fixed internal origin
C.Topoisomerase cleaves both strands simultaneously to open the circle
D.An initiator protein nicks one strand, becoming covalently attached to the 5' end and freeing a 3'-OH
Correct Answer: An initiator protein nicks one strand, becoming covalently attached to the 5' end and freeing a 3'-OH
Explanation:
Rolling circle replication begins when a sequence-specific initiator (Rep) protein nicks the plus strand, covalently linking to the 5'-phosphate end and generating a free 3'-OH. DNA polymerase extends from this 3'-OH, displacing the old strand as the template circle rolls.
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59A rolling circle intermediate produces a single-stranded linear tail that is many genome-lengths long (a concatemer). What downstream processing is required to yield unit-length genomes, as seen in some bacteriophages?
Rolling circle replication
Hard
A.Reverse transcription of the tail into DNA
B.Immediate ligation of the tail into a single giant circle
C.Site-specific cleavage of the concatemer at defined sequences followed by circularization or packaging
D.Random endonucleolytic fragmentation into any length
Correct Answer: Site-specific cleavage of the concatemer at defined sequences followed by circularization or packaging
Explanation:
The long single-stranded (or later double-stranded) concatemer is cut at specific recognition sequences by dedicated nucleases/terminases to release unit-length genomes, which are then circularized or packaged into phage heads (e.g., headful or cos-site packaging).
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60Chargaff's rules state and for double-stranded DNA. In a single strand of a given DNA, and . What are the percentages of T and C in the complementary strand?
Types and structure of DNA
Hard
A. and
B. and
C. and
D. and
Correct Answer: and
Explanation:
By complementary base pairing, every A in one strand pairs with a T in the other. Thus the complementary strand's T content equals this strand's A content (), and its C content equals this strand's G content ().
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