101. Consider the following statements about DNA-dependent DNA polymerase.
I. It uses a DNA strand as a template.
II. Deoxyribonucleoside triphosphates serve as substrates and also provide energy for polymerisation.
III. New DNA is synthesised only in the \(3'\rightarrow5'\) direction.
IV. Template complementarity helps determine the sequence of the newly formed strand.
ⓐ. I and III only
ⓑ. I, II and IV only
ⓒ. II, III and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: DNA-dependent DNA polymerase synthesises a new DNA strand by using an existing DNA strand as the template. The incoming deoxyribonucleoside triphosphates have a dual role: they supply the nucleotide units incorporated into the growing chain, and their high-energy terminal phosphate bonds provide energy associated with the polymerisation process. Complementary base pairing ensures that the sequence of the template directs the sequence of the daughter strand. The directional statement in III is reversed. New DNA is synthesised only in the \(5'\rightarrow3'\) direction, which means the relevant template is read in the opposite \(3'\rightarrow5'\) direction. This directional restriction later becomes the key reason that the two daughter strands at a replication fork cannot both be synthesised in exactly the same continuous manner.
102. A replicating cell has intact template DNA and functional DNA polymerase, but its available deoxyribonucleoside triphosphates become severely depleted. What is the most direct consequence?
ⓐ. Replication continues normally but produces only RNA strands
ⓑ. Existing parental DNA loses its complementary base pairing
ⓒ. DNA synthesis reverses automatically to the \(3'\rightarrow5'\) direction
ⓓ. Elongation slows or stops when nucleotide substrates become limiting
Correct Answer: Elongation slows or stops when nucleotide substrates become limiting
Explanation: DNA polymerase cannot extend a daughter strand without a supply of deoxyribonucleoside triphosphates. Each incoming triphosphate contributes the appropriate deoxyribonucleotide to the growing DNA chain and also provides high-energy phosphate bonds associated with driving polymerisation. Depleting these molecules removes both the immediate building blocks and an energy source required for continued chain growth. The template and enzyme may remain structurally intact, but that alone is insufficient for replication. The process does not switch to RNA synthesis or reverse its normal direction in response to substrate shortage. The prediction follows from the dual role of the triphosphates: availability of template, polymerase and substrates must coincide for efficient daughter-strand elongation.
103. A DNA template is written as \(3'\text{-TACGGA-}5'\). If DNA polymerase copies it in the permitted direction, the newly synthesised complementary strand is:
ⓐ. \(5'\text{-ATGCCT-}3'\)
ⓑ. \(3'\text{-ATGCCT-}5'\)
ⓒ. \(5'\text{-TACGGA-}3'\)
ⓓ. \(3'\text{-TGGCAT-}5'\)
Correct Answer: \(5'\text{-ATGCCT-}3'\)
Explanation: The template runs \(3'\rightarrow5'\) from left to right, so DNA polymerase can synthesise the daughter strand in its permitted \(5'\rightarrow3'\) direction over the same interval. Complementary pairing gives T with A, A with T, C with G, G with C, G with C and A with T. The daughter sequence is \(5'\text{-ATGCCT-}3'\). Two decisions are required: the bases must be complementary to the template, and the newly synthesised strand must be antiparallel to it. Writing the complementary bases with a \(3'\) end on the left would represent an incorrect synthesis direction. DNA replication problems involving sequence cannot be solved from complementarity alone; strand polarity must be applied at the same time.
104. Assertion: DNA polymerase synthesises a new DNA strand only in the \(5'\rightarrow3'\) direction.
Reason: During synthesis, DNA polymerase moves along the template in the \(5'\rightarrow3'\) direction.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not correctly explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Assertion is true, but Reason is false
Explanation: The assertion gives the correct direction of daughter-strand synthesis. DNA polymerase extends the new strand in the \(5'\rightarrow3'\) direction. Since the two strands involved are antiparallel, the template must be read in the opposite \(3'\rightarrow5'\) direction relative to the advancing polymerase. The reason incorrectly assigns the same \(5'\rightarrow3'\) direction to movement along the template. Polymerisation direction becomes especially important at a replication fork, where the two parental templates have opposite polarities. Polymerase retains the same direction of new-strand synthesis on both sides, but the antiparallel templates force one daughter strand to be produced continuously and the other discontinuously. Correct directional reasoning must distinguish synthesis direction from template orientation rather than treating them as identical.
105. Escherichia coli contains about \(4.6\times10^6\,\text{bp}\) of DNA and completes replication in approximately \(18\,\text{min}\). The effective whole-genome duplication rate required over that interval is closest to:
ⓐ. \(4.3\times10^3\,\text{bp/s}\)
ⓑ. \(2.0\times10^3\,\text{bp/s}\)
ⓒ. \(4.3\times10^2\,\text{bp/s}\)
ⓓ. \(2.6\times10^5\,\text{bp/s}\)
Correct Answer: \(4.3\times10^3\,\text{bp/s}\)
Explanation: The known genome size is \(4.6\times10^6\,\text{bp}\), and the stated replication time is \(18\,\text{min}\). First convert time to seconds: \(18\times60=1080\,\text{s}\). The effective rate needed to duplicate the complete genome over that interval is \((4.6\times10^6\,\text{bp})/(1080\,\text{s})\), which is approximately \(4.26\times10^3\,\text{bp/s}\). Rounded appropriately, this is \(4.3\times10^3\,\text{bp/s}\). This is an aggregate genome-level rate, not necessarily the speed of one individual polymerising site. Replication of a long chromosome involves simultaneous synthesis at replication forks, so total chromosome completion cannot automatically be equated with the rate of a single DNA-polymerase trajectory. The calculation connects genome scale with the need for rapid, coordinated DNA synthesis.
106. The following values are given for Escherichia coli.
| Feature | Value |
|---|
| Genome size | \(4.6\times10^6\,\text{bp}\) |
| Replication time | about \(18\,\text{min}\) |
| Average polymerisation rate cited for a synthesis site | about \(2000\,\text{bp/s}\) |
Which interpretation best reconciles the chromosome-scale completion time with the local polymerisation rate?
ⓐ. The chromosome must be synthesised by converting DNA temporarily into protein
ⓑ. Concurrent forks are needed to finish chromosome replication in that time
ⓒ. The genome-size value must represent nucleotides rather than base pairs
ⓓ. DNA polymerase must switch repeatedly to \(3'\rightarrow5'\) synthesis
Correct Answer: Concurrent forks are needed to finish chromosome replication in that time
Explanation: At \(2000\,\text{bp/s}\), one isolated polymerisation front would require \((4.6\times10^6)/2000=2300\,\text{s}\), or about \(38.3\,\text{min}\), to account for the entire stated genome length. The reported chromosome completion time is shorter. The biologically meaningful interpretation is that chromosome replication is organised through replication forks so that synthesis can proceed concurrently over the DNA rather than as one polymerase copying the whole chromosome in a single linear pass. Nothing in the data requires reversal of the permitted synthesis direction or reinterpretation of base pairs as single nucleotides. The comparison distinguishes a local polymerisation rate from an aggregate chromosome-duplication rate. Rapid genome replication depends on coordinated parallel synthesis while preserving the same molecular rules at each active site.
107. A mutant DNA polymerase operates at approximately the normal replication speed but incorporates incorrect nucleotides much more frequently. If the errors persist in the DNA, the most important consequence is:
ⓐ. chromosome length must decrease after every replication cycle
ⓑ. replication becomes transcription because sequence accuracy is reduced
ⓒ. daughter DNA can no longer contain phosphodiester linkages
ⓓ. the frequency of persistent heritable sequence changes can increase
Correct Answer: the frequency of persistent heritable sequence changes can increase
Explanation: Replication speed and replication fidelity are different performance requirements. A polymerase may copy DNA rapidly yet still be biologically defective if it frequently incorporates incorrect nucleotides. An error that remains in the DNA changes the nucleotide sequence and can become a mutation that is transmitted when the altered molecule is copied in later cell generations. High fidelity is essential for preserving hereditary information even when replication is fast. Reduced accuracy does not by itself convert DNA synthesis into transcription or remove the covalent backbone of the daughter strand. The important consequence is informational: a copying mistake can alter the stored genetic sequence. Replication must balance rapid chromosome duplication with sufficiently accurate template copying so that mutation remains occasional rather than becoming an unavoidable consequence of every synthesis event.
108. Two bacterial strains duplicate chromosomes of the same size in equal time. Strain P produces very few persistent copying errors, whereas strain Q produces many. What property most clearly distinguishes their replication performance?
ⓐ. Q has a higher genome-copying speed than P
ⓑ. P uses RNA rather than DNA as a template
ⓒ. P has higher fidelity at a similar replication rate
ⓓ. Q must synthesise DNA in the \(3'\rightarrow5'\) direction
Correct Answer: P has higher fidelity at a similar replication rate
Explanation: Chromosome size and completion time are held constant, so the two strains do not differ in the overall speed of genome duplication. Their difference lies in the number of persistent copying errors. Strain P reproduces the template sequence more accurately and has higher replication fidelity. Strain Q can complete replication just as quickly but produces more sequence alterations that may become mutations. Replication quality cannot be judged from speed alone. Hereditary continuity requires both efficient synthesis and accurate copying of the parental information. The observed difference does not imply a change in template type or synthesis direction. Fidelity specifically describes how reliably nucleotide sequence information is reproduced, making it the property that separates the two replication systems in this controlled comparison.
109. A replication fork moves from left to right. The upper parental template runs \(3'\rightarrow5'\) from left to right, while the lower template runs \(5'\rightarrow3'\) in the same direction. Which synthesis pattern is expected?
ⓐ. Both daughter strands are synthesised discontinuously
ⓑ. The upper-template daughter is continuous and the lower-template daughter is discontinuous
ⓒ. The upper-template daughter is discontinuous and the lower-template daughter is continuous
ⓓ. Both daughter strands are synthesised continuously
Correct Answer: The upper-template daughter is continuous and the lower-template daughter is discontinuous
Explanation: DNA polymerase can synthesise a new strand only in the \(5'\rightarrow3'\) direction. At the stated fork, the upper template is encountered in the \(3'\rightarrow5'\) direction as the fork moves rightward. Polymerase can follow the opening fork and extend the complementary daughter continuously in its permitted \(5'\rightarrow3'\) direction. The lower template has the opposite polarity, \(5'\rightarrow3'\) toward the moving fork. A daughter strand copied from this template still has to grow \(5'\rightarrow3'\), forcing synthesis to occur in separate fragments rather than continuously in the same direction as fork opening. The continuous-versus-discontinuous distinction is determined from template polarity and polymerase direction, not from whether a strand happens to be drawn above or below the fork.
110. A replication fork moves from right to left. The upper parental strand is labelled \(5'\) at the left end and \(3'\) at the right end. Which statement follows from this orientation?
ⓐ. The upper strand runs \(3'\rightarrow5'\) toward the moving fork, allowing continuous synthesis
ⓑ. The upper strand is encountered \(5'\rightarrow3'\) and must support discontinuous synthesis
ⓒ. The upper strand cannot act as a template because its \(3'\) end is on the right
ⓓ. The upper strand must first reverse its chemical polarity before replication can begin
Correct Answer: The upper strand runs \(3'\rightarrow5'\) toward the moving fork, allowing continuous synthesis
Explanation: The important step is to read the parental strand in the direction the fork is actually moving. The strand is \(5'\rightarrow3'\) from left to right, so when followed from right to left it is encountered as \(3'\rightarrow5'\). That orientation is compatible with continuous synthesis of a complementary daughter strand in the \(5'\rightarrow3'\) direction as the fork progresses leftward. The conclusion would be reversed if the fork direction were reversed while the parental labels stayed fixed. This is why leading-versus-discontinuous behaviour should be solved from polarity rather than memorised from a standard drawing. A DNA strand never has to reverse its own chemical polarity; the same strand can support a different synthesis mode depending on how its orientation relates to the direction of fork movement.
111. Assume a hypothetical DNA polymerase could synthesise DNA efficiently in both \(5'\rightarrow3'\) and \(3'\rightarrow5'\) directions while all other replication requirements remained satisfied. What consequence would most directly follow at a replication fork?
ⓐ. Complementary base pairing would no longer be required
ⓑ. The parental DNA would no longer need to separate locally
ⓒ. Replication would necessarily become conservative
ⓓ. One strand might no longer require discontinuous synthesis
Correct Answer: One strand might no longer require discontinuous synthesis
Explanation: In normal DNA replication, the two parental strands are antiparallel while DNA polymerase can synthesise only \(5'\rightarrow3'\). This combination creates the polarity problem at the fork: one template can be copied continuously in the direction of fork movement, whereas the opposite template must be copied discontinuously in fragments. In the hypothetical condition, polymerisation would be possible in either chemical direction. That specific directional restriction would no longer force one daughter strand to be assembled discontinuously. Complementary pairing and local strand separation would still be required to expose templates and specify nucleotide sequence, and the semiconservative distribution of parental strands need not change. The prediction isolates the causal link between polymerase directionality and discontinuous synthesis rather than treating DNA fragmentation as an arbitrary feature of replication.
112. Consider the following statements about a replication fork.
I. A long DNA molecule opens locally rather than becoming completely separated at once.
II. Continuous synthesis occurs on the template oriented \(3'\rightarrow5'\) in the direction of fork movement.
III. Discontinuous synthesis occurs on the oppositely oriented template because new DNA still must grow \(5'\rightarrow3'\).
IV. Discontinuous synthesis occurs only because DNA ligase initially prevents continuous polymerisation.
ⓐ. I and IV only
ⓑ. II, III and IV only
ⓒ. I, II and III only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: Replication proceeds through local opening of the DNA molecule at replication forks. DNA polymerase retains one directional rule on both templates: new DNA is synthesised \(5'\rightarrow3'\). A parental template running \(3'\rightarrow5'\) toward the advancing fork can be copied continuously. Its antiparallel partner is oriented oppositely and must be copied discontinuously while still obeying the same synthesis direction. DNA ligase does not create this asymmetry. Its role comes later, when separately synthesised DNA fragments on the discontinuous daughter strand are joined into a continuous covalent chain. The origin of discontinuity is the combination of antiparallel parental strands and unidirectional polymerisation. Keeping cause and repair of the discontinuous pattern separate is essential for understanding the replication fork mechanism.
113. Three events involved in discontinuous daughter-strand formation are listed.
P. A replication fork exposes the parental template locally.
Q. DNA polymerase synthesises separate daughter-DNA fragments in the permitted \(5'\rightarrow3'\) direction.
R. DNA ligase joins the fragments into a continuous strand.
The correct order is:
ⓐ. P, Q, R
ⓑ. Q, P, R
ⓒ. P, R, Q
ⓓ. R, Q, P
Correct Answer: P, Q, R
Explanation: Template DNA must first become locally accessible at the replication fork. Once the relevant parental region is exposed, DNA polymerase can use it to synthesise complementary daughter DNA while maintaining the required \(5'\rightarrow3'\) direction. On the strand that cannot be copied continuously with fork movement, this produces separate DNA fragments. Those fragments must already exist before DNA ligase can perform its role. Ligase then joins neighbouring fragments to establish continuity along the daughter strand. The order follows the dependency among the events: exposure permits synthesis, and synthesis produces the substrates that ligase subsequently joins. Placing ligase before fragment formation would assign it a function it cannot perform. The sequence also distinguishes the cause of discontinuous synthesis from the later step that completes the discontinuously produced strand.
114. DNA polymerase remains functional on both templates at a replication fork, but DNA ligase activity is selectively lost. Which result is most directly expected?
ⓐ. Neither daughter strand can incorporate any deoxyribonucleotides
ⓑ. Both daughter strands become continuous but lose complementarity
ⓒ. Replication changes from semiconservative to conservative
ⓓ. Unjoined fragments accumulate on the discontinuous strand
Correct Answer: Unjoined fragments accumulate on the discontinuous strand
Explanation: DNA ligase acts after separate DNA fragments have been synthesised on the discontinuous daughter strand. If polymerase remains functional, complementary DNA can still be produced from the exposed templates, including formation of those fragments in the required \(5'\rightarrow3'\) direction. The defect appears when the fragments must be connected into a continuous covalent strand. Without functional ligase, unjoined pieces accumulate rather than being converted efficiently into one continuous daughter chain. The continuous daughter strand is not expected to acquire the same joining problem simply from loss of ligase, and the basic semiconservative template relationship remains unchanged. This prediction separates polymerisation from fragment joining: DNA polymerase creates new DNA, whereas DNA ligase restores continuity between independently synthesised portions of the discontinuous strand.
115. Assertion: DNA ligase is important for completion of the discontinuously synthesised daughter strand.
Reason: Deoxyribonucleoside triphosphates serve as substrates and energy sources for DNA polymerisation.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not correctly explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, but Reason does not correctly explain Assertion
Explanation: The assertion is correct because discontinuous synthesis produces separate DNA fragments that must subsequently be joined, and DNA ligase performs that joining function. The reason is also biologically correct: deoxyribonucleoside triphosphates provide the nucleotide substrates and high-energy phosphate bonds associated with DNA polymerisation. However, this property of the triphosphates explains how DNA polymerase extends new DNA, not why ligase is required to connect preformed fragments. The need for ligase arises from the discontinuous pattern created by antiparallel template orientation together with the \(5'\rightarrow3'\) restriction on DNA synthesis. The assertion and reason concern different mechanistic steps. Both statements are valid, but the reason does not account for the specific completion function assigned to ligase.
116. Consider the following statements about initiation of DNA replication.
I. Escherichia coli has a definite origin from which chromosome replication begins.
II. A promoter and an origin of replication perform the same molecular function.
III. A DNA molecule intended to be propagated through a vector requires an origin that permits replication.
IV. An origin of replication is primarily the signal that begins translation.
ⓐ. I and II only
ⓑ. II and IV only
ⓒ. I and III only
ⓓ. I, III and IV only
Correct Answer: I and III only
Explanation: Chromosomal replication is initiated at defined origin regions rather than beginning randomly at any position. Escherichia coli has a specific origin associated with initiation of chromosome replication. The same functional principle matters when a DNA segment is propagated using a vector: the DNA construct must be associated with an origin that permits it to be copied in the host. An origin should not be confused with a promoter. A promoter is related to initiation and orientation of transcription, not duplication of the entire DNA molecule. Likewise, translation begins from signals on an RNA template rather than from a DNA replication origin. Distinguishing these control points prevents three different processes, replication, transcription and translation, from being assigned interchangeable initiation signals.
117. A circular DNA construct carrying an inserted DNA segment is introduced into bacteria. The construct enters the cells successfully but lacks any functional origin of replication. What problem is most directly expected during continued bacterial growth?
ⓐ. The insert must immediately be translated into a protein
ⓑ. The construct will replicate only after every promoter is removed
ⓒ. The DNA will automatically acquire a bacterial chromosome origin
ⓓ. Replication cannot initiate, so the DNA is not stably propagated
Correct Answer: Replication cannot initiate, so the DNA is not stably propagated
Explanation: Stable propagation of an introduced DNA molecule requires the molecule to be copied as host cells grow and divide. Replication does not normally begin from arbitrary positions; an appropriate origin provides the site from which the replication process can be initiated. If the circular construct lacks a functional origin, merely entering the bacterial cell does not ensure that it will be duplicated and inherited by subsequent cells. The defect is specifically one of DNA replication rather than transcription or translation. A promoter could allow transcription of an adjacent sequence but would not replace the function of an origin. The example illustrates why vectors used to maintain inserted DNA must supply an origin of replication while avoiding the need to invoke more detailed biotechnology-vector design.
118. A DNA construct can be transcribed from its promoter after entering a bacterial cell, but it is progressively lost from the population because it is not copied as cells divide. Adding a functional [blank] would most directly correct the replication defect.
ⓐ. origin of replication
ⓑ. transcription terminator
ⓒ. operator region
ⓓ. translation stop codon
Correct Answer: origin of replication
Explanation: The observation separates expression from propagation. The construct can already be transcribed, so the promoter is functioning and the main defect is not initiation of RNA synthesis. Its progressive loss during cell multiplication indicates that the DNA itself is not being reliably duplicated. Replication requires a functional origin from which copying can begin. Supplying such an origin addresses the specific failure to maintain the construct across successive bacterial generations. A transcription terminator regulates the endpoint of RNA synthesis, while an operator participates in regulation of transcription and a stop codon acts during translation. None substitutes for the initiation site required for DNA replication. The missing relation is identified from the biological failure described in the case rather than from isolated recall of the term origin.
119. Four eukaryotic cell populations are followed through one cell-cycle interval.
| Population | DNA replication during S phase | Subsequent cell division |
|---|
| P | Completed | Completed |
| Q | Completed | Failed |
| R | Failed | Failed |
| S | Failed | Completed normally |
Which population most directly shows the condition that can lead to polyploidy through uncoupling of replication and division?
ⓐ. P
ⓑ. R
ⓒ. Q
ⓓ. S
Correct Answer: Q
Explanation: Polyploidy can arise when genome duplication occurs but the subsequent cellular division needed to distribute the replicated genetic material fails. Population Q has exactly that combination: DNA replication is completed during S phase, but cell division does not occur. The replicated chromosome complements can remain within one cellular lineage rather than being separated into two daughter cells. Population P shows normal coordination of replication and division, while R does not complete the genome-duplication step needed for this particular route to increased chromosome-set number. Population S does not represent a biologically normal sequence under the stated conditions because successful division should not proceed as if genome replication were complete. The table tests coordination between two stages rather than asking only where DNA synthesis occurs.
120. A eukaryotic cell completes DNA replication during S phase, but the subsequent division process is blocked and the replicated chromosome complements remain within the same cellular lineage. Which outcome is most consistent with repeated uncoupling of these events?
ⓐ. permanent conversion of DNA into RNA
ⓑ. an increase in chromosome-set number leading toward polyploidy
ⓒ. loss of all replicated DNA before the next cell cycle
ⓓ. replacement of semiconservative replication by conservative replication
Correct Answer: an increase in chromosome-set number leading toward polyploidy
Explanation: S phase duplicates the cellular DNA so that replicated genetic material is available for distribution during the subsequent division process. If replication is completed but division repeatedly fails, replicated chromosome complements can remain together rather than being partitioned into separate daughter cells. Continued cycles of genome duplication without corresponding cell division can increase the number of chromosome sets in the resulting cell, producing polyploidy. The phenomenon illustrates why DNA replication and cell division must be coordinated. Failure of division does not change the molecular mechanism of DNA replication from semiconservative to conservative, nor does it convert DNA into RNA. The key consequence lies at the chromosome-set level: genome duplication has occurred, but the cellular event that normally separates the duplicated complements has not kept pace with it.
