401. Two hypothetical nucleic acids are compared as hereditary molecules.
| Property | Molecule P | Molecule Q |
|---|
| Strandedness | Usually double-stranded | Usually single-stranded |
| Sugar | Deoxyribose | Ribose with \(2'\)-OH |
| Catalytic potential | Low in the standard description | Present in some molecules |
| Direct role as translation message | Requires RNA intermediate | Can act directly as message |
Which conclusion best distinguishes their biological suitability?
ⓐ. Q must be the preferred long-term hereditary store because greater chemical reactivity guarantees stability
ⓑ. P favours stable storage; Q provides greater versatility in expression and catalysis
ⓒ. P cannot mutate because double-stranded nucleic acids are chemically fixed
ⓓ. Q cannot act as genetic material because only DNA can undergo complementary copying
Correct Answer: P favours stable storage; Q provides greater versatility in expression and catalysis
Explanation: Molecule P has the properties associated with DNA, whereas Q resembles RNA. DNA is chemically more stable because deoxyribose lacks the reactive \(2'\)-OH characteristic of RNA, and its double-stranded structure also provides complementary information that supports stable maintenance of sequence. RNA is comparatively more reactive, yet that same chemical versatility allows RNA molecules to serve as messenger, adapter, structural component and, in some cases, catalyst. RNA can also function as genetic material in some viruses. The comparison separates two criteria that should not be confused: long-term storage stability favours DNA in most organisms, while functional versatility in information transfer and catalysis is especially characteristic of RNA.
402. A proposed early biological system uses one type of molecule both to store hereditary information and to catalyse reactions required for its own biological functioning. Later, a chemically more stable molecule becomes the principal hereditary archive. Which interpretation best matches the RNA-world concept?
ⓐ. Proteins must have been the first hereditary material because enzymes are usually proteins today
ⓑ. DNA must have evolved before RNA because DNA is chemically more stable
ⓒ. RNA could function only as a catalyst and therefore required DNA for information storage from the beginning
ⓓ. RNA could precede DNA by combining information storage and catalysis; stable DNA could evolve later
Correct Answer: RNA could precede DNA by combining information storage and catalysis; stable DNA could evolve later
Explanation: The RNA-world idea depends on RNA possessing two major capacities that are now often divided among different molecular classes. RNA can carry sequence information and, in some cases, perform catalytic functions. Modern biological processes retain evidence of this versatility, including the structural and catalytic participation of RNA in the translation machinery. DNA is chemically more stable and better suited to long-term information storage, but that advantage does not require DNA to have preceded RNA historically. The RNA-world hypothesis places RNA as an earlier genetic material, with more stable DNA emerging later. The reasoning is based on functional combination followed by increasing storage stability rather than on the assumption that today's dominant hereditary molecule must also have been the earliest.
403. Which process is essential for hereditary continuity but is not itself part of the standard expression sequence represented by the central dogma?
ⓐ. \(DNA\rightarrow DNA\) replication for hereditary continuity
ⓑ. \(DNA\rightarrow RNA\) transcription of a selected gene
ⓒ. \(RNA\rightarrow Protein\) translation of a coding message
ⓓ. \(RNA\rightarrow DNA\) information flow in the standard viral exception
Correct Answer: \(DNA\rightarrow DNA\) replication for hereditary continuity
Explanation: The standard central dogma describes expression of genetic information through \(DNA\rightarrow RNA\rightarrow Protein\). Replication is different: it copies DNA into DNA so that hereditary information can be transmitted when cells or organisms reproduce. Replication is indispensable to inheritance but is not one of the two standard expression steps represented by the central dogma. Some viral systems also show an exceptional reverse nucleic-acid information flow, \(RNA\rightarrow DNA\), in some viruses, but this remains an information-transfer relation rather than ordinary DNA replication. Keeping these processes distinct prevents a common conceptual error in which every DNA-directed molecular event is placed under the central dogma even though genome copying and gene expression answer different biological needs.
404. Two adjacent genes lie on the same double-stranded DNA molecule. Gene P is transcribed using the lower strand as template, while Gene Q is transcribed in the opposite direction using the upper strand as template. Which conclusion is correct?
ⓐ. The arrangement is impossible because only one physical DNA strand can ever be transcribed on a chromosome
ⓑ. Both genes must produce complementary RNAs that pair with each other
ⓒ. Each unit uses one template strand; different units can use opposite strands set by their promoters
ⓓ. One gene must be undergoing replication rather than transcription
Correct Answer: Each unit uses one template strand; different units can use opposite strands set by their promoters
Explanation: The one-strand rule applies to an individual transcription unit, not to an entire chromosome. For a particular gene, promoter orientation establishes the direction in which RNA polymerase moves and determines which DNA strand has the required \(3'\rightarrow5'\) template orientation. A neighbouring gene can have its promoter oriented in the opposite direction and can use the other physical DNA strand as its template. There is no requirement that every gene on one chromosome use the same strand. The arrangement also does not mean both strands of one transcription unit are being copied simultaneously. The correct interpretation separates local gene orientation from whole-chromosome strand identity and preserves the principle that one template strand is used for each transcriptional unit.
405. A selective inhibitor completely blocks eukaryotic RNA polymerase III but leaves RNA polymerases I and II functional. Which early molecular pattern is most likely?
ⓐ. hnRNA and major rRNAs disappear immediately, but tRNA remains normal
ⓑ. Only \(28\mathrm{S}\) rRNA synthesis declines, while all other RNAs remain normal
ⓒ. All mRNA synthesis stops because polymerase III transcribes protein-coding genes
ⓓ. tRNA and \(5\mathrm{S}\) rRNA fall while hnRNA and major Pol-I rRNAs initially persist
Correct Answer: tRNA and \(5\mathrm{S}\) rRNA fall while hnRNA and major Pol-I rRNAs initially persist
Explanation: Eukaryotic nuclear RNA polymerases have different major transcriptional responsibilities. RNA polymerase II produces hnRNA, which is processed into mRNA. RNA polymerase I produces the major \(28\mathrm{S}\), \(18\mathrm{S}\) and \(5.8\mathrm{S}\) rRNAs. RNA polymerase III is responsible for products that include tRNA and \(5\mathrm{S}\) rRNA. Selective inhibition of polymerase III directly reduces those RNA classes while initially sparing hnRNA and the major polymerase-I rRNAs. Translation may later become impaired because tRNAs and ribosomal components are essential for protein synthesis, but that is a downstream consequence. The immediate transcriptional pattern is determined by polymerase-product specificity rather than by the final effect on protein production.
406. A bacterial transcription unit has a normal promoter and structural gene, but its functional terminator sequence is deleted. RNA polymerase initiates normally and successfully transcribes the structural gene. What is the most direct predicted consequence?
ⓐ. RNA polymerase cannot recognise the promoter because terminators are required for initiation
ⓑ. Transcription may continue beyond the terminator because termination information is lost
ⓒ. The structural gene is converted into an intron
ⓓ. Translation must begin before RNA synthesis starts
Correct Answer: Transcription may continue beyond the terminator because termination information is lost
Explanation: The promoter and terminator define different boundaries and stages of a transcription unit. A functional promoter allows RNA polymerase to initiate transcription in the correct orientation. The terminator contributes to stopping transcription at the appropriate downstream position. Since the promoter is normal in this case, initiation and passage through the structural gene can occur. Loss of the terminator specifically disrupts the normal termination step, so RNA polymerase can continue synthesising RNA beyond the usual endpoint. This differs from a promoter deletion, which would affect initiation before the structural gene is transcribed. The changed condition tests stage specificity: successful initiation does not guarantee correct termination as the two events depend on distinct signals within the transcriptional unit.
407. The supplied genetic-code information is:
GAA = glutamate
GAG = glutamate
GUA = valine
UAA = stop
Starting from the mRNA codon GAA, three separate single-base substitutions produce GAG, GUA and UAA. Which classification is correct?
ⓐ. GAG is missense, GUA is synonymous and UAA is frameshift
ⓑ. GAG is synonymous, GUA is stop-producing and UAA is missense
ⓒ. GAG is synonymous, GUA is missense, and UAA produces nonsense termination
ⓓ. All three changes produce the same amino acid because each is a substitution
Correct Answer: GAG is synonymous, GUA is missense, and UAA produces nonsense termination
Explanation: GAA and GAG both specify glutamate, so the GAA-to-GAG substitution changes the nucleotide sequence without changing the encoded amino acid. It is synonymous at the protein level. GUA specifies valine, so changing GAA to GUA replaces glutamate with valine and produces a missense-type amino-acid substitution. UAA is a stop codon, so changing GAA to UAA converts a sense codon into a termination signal and can shorten the polypeptide if it occurs before the normal endpoint. None of these substitutions changes the number of nucleotides, so none is a frameshift. The biological consequence depends on the identity of the new codon rather than merely on the fact that one base changed.
408. An mRNA contains two AUG codons. The first AUG functions as the initiation codon, while the second AUG occurs later within the established reading frame and specifies an internal methionine. A mutation changes only the second AUG to a codon for valine. What is the most direct effect?
ⓐ. The first AUG still initiates; internal methionine may become valine
ⓑ. Translation cannot initiate because every AUG in the mRNA must remain unchanged
ⓒ. The first AUG automatically becomes a stop codon
ⓓ. The reading frame shifts from the second AUG onward
Correct Answer: The first AUG still initiates; internal methionine may become valine
Explanation: AUG has a dual contextual role. At the appropriate beginning of a translated region it serves as the initiation codon and also specifies methionine. The same triplet can occur internally after the reading frame has already been established, where it simply codes for methionine. In this case, the initiating AUG is unchanged, so the ribosome can still begin translation normally. The mutation affects only the later methionine codon and can replace that amino acid with valine. Since the mutation is a substitution rather than an insertion or deletion, it does not shift the reading frame. Codon identity must be distinguished from codon position: the same AUG triplet can participate in initiation at one location and ordinary elongation at another.
409. A tRNA retains its normal anticodon but is experimentally attached to the wrong amino acid. It enters translation and pairs correctly with its usual mRNA codon. What is the most direct consequence?
ⓐ. The codon is converted chemically into the amino acid attached to the tRNA
ⓑ. The wrong amino acid can be inserted at a position selected by the tRNA's unchanged anticodon
ⓒ. The ribosome automatically detects the amino-acid error and changes the codon sequence
ⓓ. The tRNA cannot enter the ribosome because anticodon pairing depends on the attached amino acid
Correct Answer: The wrong amino acid can be inserted at a position selected by the tRNA's unchanged anticodon
Explanation: The tRNA performs an adapter function by linking two different kinds of information. Its anticodon determines which mRNA codon it can recognise, while its acceptor end carries the amino acid that will be contributed to the growing polypeptide. If the anticodon remains normal, the tRNA can still pair with its usual codon. If the wrong amino acid has been attached experimentally, that incorrect amino acid can be delivered to the codon-selected position. The mRNA sequence itself has not changed. This altered condition shows why correct charging is essential: codon recognition alone cannot guarantee correct protein sequence unless the amino acid attached to the corresponding tRNA is also appropriate.
410. A tRNA normally carries amino acid X and has an anticodon complementary to codon P. Its anticodon mutates so that the tRNA now recognises codon Q, but it continues to carry amino acid X. If codon Q normally specifies amino acid Y, what can occur?
ⓐ. Codon Q becomes permanently converted into codon P
ⓑ. Amino acid Y is attached automatically because codon identity determines charging after the tRNA enters the ribosome
ⓒ. The tRNA can recognise neither P nor Q because any anticodon mutation destroys all pairing
ⓓ. Amino acid X can be delivered at a position where codon Q would normally direct amino acid Y
Correct Answer: Amino acid X can be delivered at a position where codon Q would normally direct amino acid Y
Explanation: The anticodon controls which mRNA codon the tRNA can recognise during translation. In the altered tRNA, that recognition property has changed from codon P to codon Q. The altered tRNA is also specified to continue carrying amino acid X. The result is a mismatch between codon recognition and normal amino-acid meaning: the tRNA can pair at Q but deliver X instead of the amino acid Y normally associated with that codon. This thought experiment demonstrates the two-part adapter role of tRNA. Accurate translation requires both correct codon-anticodon pairing and correct attachment of the corresponding amino acid; disrupting either relation can alter the resulting polypeptide.
411. A mutant release factor can recognise UAA and UAG normally but cannot recognise UGA. An mRNA contains UGA as its normal stop codon and a downstream in-frame UAA. What is the most likely outcome?
ⓐ. Translation can pass the UGA and continue until the downstream UAA is recognised
ⓑ. Translation must stop at UGA because stop codons terminate without any protein factor
ⓒ. UGA is translated normally as methionine
ⓓ. The reading frame shifts immediately when the ribosome reaches UGA
Correct Answer: Translation can pass the UGA and continue until the downstream UAA is recognised
Explanation: Stop codons do not terminate translation by supplying a normal aminoacyl-tRNA. Termination requires recognition of the stop signal by a release factor. In this altered condition, UGA remains present in the mRNA, but the relevant release-factor recognition has been lost. The normal termination event at that position can fail, allowing translation to continue into downstream sequence. Because the same mutant release factor can still recognise UAA, the later in-frame UAA can provide a termination point. No nucleotide has been inserted or deleted, so the reading frame itself is unchanged. The case distinguishes the nucleotide signal from the molecular machinery that interprets it: an intact stop codon is not sufficient for normal termination if its recognition mechanism is defective.
412. An mRNA contains a \(60\)-nucleotide \(5'\) UTR, followed by a translated region of \(900\) nucleotides from the first base of AUG through the final base of the stop codon, followed by a \(120\)-nucleotide \(3'\) UTR. How many amino acids are encoded by the translated region?
ⓐ. \(360\)
ⓑ. \(300\)
ⓒ. \(299\)
ⓓ. \(279\)
Correct Answer: \(299\)
Explanation: Only the translated region is used for this calculation; neither UTR contributes amino acids. The \(900\)-nucleotide translated segment contains \(900/3=300\) codons. One of those codons is the terminal stop codon, which functions as a termination signal and does not specify an amino acid. The number of amino acids in the completed polypeptide is \(300-1=299\). The \(5'\) and \(3'\) untranslated regions can influence efficient expression but are not themselves converted into the amino-acid sequence of the protein. This problem integrates UTR identification with triplet decoding and the special role of stop codons rather than treating total mRNA length as equivalent to protein-coding length. The total mRNA length is \(60+900+120=1080\) nucleotides, but using \(1080/3\) would be biologically invalid: the untranslated regions are not decoded into amino acids. The correct count is anchored specifically to the translated segment and then adjusted for the non-coding stop signal.
413. A tRNA has a perfectly complementary anticodon for its normal codon, but a chemical modification blocks attachment of any amino acid to its acceptor end. What is the most direct effect?
ⓐ. The tRNA becomes an mRNA because it can still base-pair
ⓑ. The codon is converted into a stop codon
ⓒ. The tRNA can complete peptide-bond formation because anticodon recognition is sufficient
ⓓ. Codon recognition remains possible, but the tRNA cannot deliver its amino acid
Correct Answer: Codon recognition remains possible, but the tRNA cannot deliver its amino acid
Explanation: The two important functional regions of tRNA perform different jobs. The anticodon provides sequence-specific recognition of the mRNA codon, whereas the amino-acid acceptor end carries the substrate that will be incorporated into the growing polypeptide. Blocking the acceptor end does not necessarily eliminate anticodon complementarity, so recognition of the codon can remain chemically possible. However, the uncharged tRNA lacks the amino acid required for productive protein synthesis at that position. Translation depends on more than RNA-RNA base pairing. The adapter must connect the codon to the correct amino acid physically. This changed condition highlights why tRNA charging is an essential preparatory step before the molecule can function effectively during elongation.
414. A long genomic sequence has been assembled correctly from overlapping reads, but investigators do not know which human chromosome contains it. The sequence includes a polymorphic marker whose chromosomal position has already been established on a genome map. What information is most useful for assigning the assembled sequence?
ⓐ. The number of amino acids encoded by every nearby gene
ⓑ. The mapped chromosomal position of the shared polymorphic marker
ⓒ. Whether the sequence was obtained using an EST method
ⓓ. The total number of chromosomes in a bacterial host used for cloning
Correct Answer: The mapped chromosomal position of the shared polymorphic marker
Explanation: Sequence assembly and chromosome assignment are related but distinct tasks. Overlap information can establish the order of neighbouring sequence fragments and create a long contiguous sequence, yet that contig may still lack a known chromosomal address. A mapped polymorphic marker can connect the assembled sequence to a previously established genomic location. If the contig contains the same marker, its known chromosome position provides positional evidence for assigning the sequence. Protein length is not required for that mapping decision, and EST methodology addresses expressed sequences rather than physical chromosome placement. This illustrates why genome projects combine sequence assembly with mapping information: one reconstructs continuity, while the other helps anchor reconstructed DNA to defined genomic locations.
415. Consider the following Human Genome Project observations.
I. Less than \(2\%\) of the human genome is protein-coding.
II. Functions were unknown for many identified genes in the reported data.
III. Repetitive DNA constitutes a substantial genomic fraction.
Which conclusion is most appropriate?
ⓐ. All non-protein-coding DNA must be repetitive DNA
ⓑ. Any sequence with unknown function was necessarily absent from the genome sequence
ⓒ. Sequence presence, coding status and known function are separate properties
ⓓ. Only protein-coding DNA can be detected by whole-genome sequencing
Correct Answer: Sequence presence, coding status and known function are separate properties
Explanation: The three observations concern different aspects of genome organisation. A sequence can be present in the genome without directly encoding protein, so non-protein-coding DNA cannot be equated automatically with repetitive DNA. Likewise, identification of a gene by sequence analysis does not guarantee that its biological function is already understood. Whole-genome sequencing includes coding and non-coding DNA and can record regions whose functions are uncertain. The correct interpretation separates at least three questions: whether a sequence exists in the genome, whether it contributes directly to protein coding and whether its biological role has been established. Conflating these categories leads to incorrect conclusions about the meaning of the Human Genome Project's sequence and annotation observations.
416. DNA from the same individual is analysed twice. In Experiment P, a labelled probe complementary to one VNTR family is used. In Experiment Q, a different labelled probe complementary to another VNTR family is used. The resulting band patterns differ. What is the best interpretation?
ⓐ. Different VNTR probes can reveal different polymorphic fragment subsets from the same individual's genome
ⓑ. One of the samples must belong to another person because a person's DNA can produce only one possible band pattern
ⓒ. VNTR probes determine the person's genotype by changing repeat copy number during hybridisation
ⓓ. Different probes should always reveal identical bands because all repetitive DNA has the same sequence
Correct Answer: Different VNTR probes can reveal different polymorphic fragment subsets from the same individual's genome
Explanation: Probe hybridisation is sequence-specific. A probe complementary to one repetitive sequence family will detect fragments containing that target, while a probe directed against another repetitive family can reveal a different set of genomic fragments. The underlying DNA sample can be identical even though the visible band pattern changes when the molecular target of detection changes. What should remain consistent is the result obtained when the same loci and same probe system are compared across suitable tissues from the same individual. Hybridisation does not create or alter VNTR copy number; it reveals DNA fragments already present. The observation reflects assay specificity rather than a change in inherited identity.
417. A restriction fragment containing a VNTR retains exactly the same length after a single-base mutation occurs within the sequence to which the labelled probe normally hybridises. If the mutation prevents effective probe binding, what can happen during classical fingerprinting?
ⓐ. Electrophoresis must move the fragment to a different position because every base substitution changes DNA length
ⓑ. The fragment can remain at the same physical size yet fail to appear as its usual probe-detected band
ⓒ. The mutation must create an additional tandem repeat
ⓓ. Autoradiography will reveal the fragment even if no labelled probe binds it
Correct Answer: The fragment can remain at the same physical size yet fail to appear as its usual probe-detected band
Explanation: Electrophoretic position depends principally on fragment size in the classical procedure. A single-base substitution does not necessarily change the length of the restriction fragment, so the fragment can migrate to the same physical position as before. Detection is a separate issue. The labelled probe must hybridise effectively to its complementary target sequence. If the mutation disrupts that complementarity sufficiently, the fragment may no longer become labelled and may fail to appear in the autoradiographic pattern. The case distinguishes physical separation from sequence-specific detection. A band is visible not merely because a DNA fragment exists at a certain size, but because the appropriate labelled probe recognises and marks that fragment after transfer to the membrane.
418. At one VNTR locus, an allele gains several additional copies of the same tandem repeat. The flanking restriction sites and the sequence recognised by the VNTR probe remain intact. What change is most likely in the classical fingerprint?
ⓐ. The locus disappears because longer fragments cannot hybridise with probes
ⓑ. The probe binds a different chromosome automatically
ⓒ. The fragment remains the same size because repeat copy number does not affect DNA length
ⓓ. The band moves to the larger-fragment position while still binding the probe
Correct Answer: The band moves to the larger-fragment position while still binding the probe
Explanation: Increasing tandem-repeat copy number adds DNA to the region between the unchanged restriction sites. The resulting restriction fragment is longer. During electrophoresis, its migration position changes according to the larger fragment size. The probe-recognition sequence is stated to remain intact, so the labelled VNTR probe can still hybridise to the fragment after transfer to the membrane. The expected fingerprint effect is a size shift of the detected band rather than complete disappearance. Two independent determinants shape the final pattern: repeat copy number influences fragment length, while sequence complementarity determines whether the labelled probe can detect the fragment. Both must be considered when interpreting a VNTR-based fingerprint.
419. Three germ-line mutations occur independently.
| Mutation | Location | Immediate molecular effect |
|---|
| P | Promoter of a protein-coding gene | Strongly reduces transcription |
| Q | Sense codon in the same gene | Changes one encoded amino acid |
| R | Non-coding VNTR locus | Changes tandem-repeat copy number |
Which comparison is correct?
ⓐ. P alters expression, Q alters coding information, and R can alter a fingerprint marker
ⓑ. P, Q and R must all produce the same altered polypeptide
ⓒ. Only R is heritable because promoters and coding regions are not part of DNA
ⓓ. R must reduce transcription of the protein-coding gene because every VNTR is a promoter
Correct Answer: P alters expression, Q alters coding information, and R can alter a fingerprint marker
Explanation: The three mutations affect different information layers. Mutation P leaves the protein-coding sequence intact but reduces transcription by altering a regulatory DNA region, so its primary effect is on how much RNA is produced. Mutation Q changes a sense codon and directly alters the amino-acid information carried through transcription and translation. Mutation R occurs at a non-coding VNTR locus and changes repeat copy number, which can alter the fragment-size pattern used in DNA fingerprinting without requiring a change in the protein encoded by the separate gene. Since all three are specified as germ-line mutations, each can be inherited. Their biological consequences differ because genomic location determines which molecular process is affected.
420. A germ-line substitution occurs in the promoter of a gene. The coding region, the gene's transcript-processing signals and all VNTR loci used in a routine fingerprinting panel remain unchanged. Which chain of consequences is most plausible if the promoter mutation greatly reduces transcription?
ⓐ. The VNTR fingerprint must change because every inherited mutation alters all marker loci
ⓑ. The coding DNA sequence changes automatically to compensate for reduced RNA production
ⓒ. RNA and protein output can fall while coding sequence and unrelated VNTRs stay unchanged
ⓓ. The mutation cannot be inherited because regulatory DNA is not part of the genome
Correct Answer: RNA and protein output can fall while coding sequence and unrelated VNTRs stay unchanged
Explanation: A promoter is part of genomic DNA and can carry an inherited mutation. Its principal role is regulatory rather than protein-coding. If promoter function is strongly reduced, transcription of the associated gene can fall, producing less mRNA and less protein even though the coding sequence itself has not changed. A routine DNA fingerprint examines selected polymorphic loci such as VNTR regions. If those marker loci are unrelated to the promoter mutation and remain unchanged, the person's fingerprint at those loci can remain the same. The case integrates regulation, inheritance and fingerprinting while keeping their information levels separate: an inherited genomic change can alter gene expression without altering protein-coding sequence or every other polymorphic marker in the genome.
