DNA contains deoxyribose sugar, while RNA contains ribose sugar.
Incorrect! Try again.
2Which type of RNA carries genetic information from DNA to the ribosome?
Types of DNA and RNA
Easy
A.Messenger RNA
B.Small nuclear RNA
C.Ribosomal RNA
D.Transfer RNA
Correct Answer: Messenger RNA
Explanation:
Messenger RNA (mRNA) carries the genetic code copied from DNA to the ribosome.
Incorrect! Try again.
3What are the three basic components of a DNA nucleotide?
Nature, structure, and replication of genetic material
Easy
A.Sugar, phosphate, and base
B.Sugar, protein, and lipid
C.Sugar, lipid, and base
D.Protein, phosphate, and base
Correct Answer: Sugar, phosphate, and base
Explanation:
A DNA nucleotide consists of deoxyribose sugar, a phosphate group, and a nitrogenous base.
Incorrect! Try again.
4How are the two strands of a DNA double helix oriented?
Nature, structure, and replication of genetic material
Easy
A.Parallel
B.Perpendicular
C.Antiparallel
D.Circular
Correct Answer: Antiparallel
Explanation:
The two DNA strands run in opposite directions and are therefore described as antiparallel.
Incorrect! Try again.
5Which base pairs with adenine in double-stranded DNA?
Nature, structure, and replication of genetic material
Easy
A.Thymine
B.Cytosine
C.Guanine
D.Uracil
Correct Answer: Thymine
Explanation:
In DNA, adenine pairs with thymine through hydrogen bonds.
Incorrect! Try again.
6DNA replication is described as semiconservative because each new DNA molecule contains:
Nature, structure, and replication of genetic material
Easy
A.One DNA strand and one RNA strand
B.Two old strands joined together
C.One old strand and one new strand
D.Two entirely new DNA strands
Correct Answer: One old strand and one new strand
Explanation:
Each daughter DNA molecule contains one parental strand and one newly synthesized strand.
Incorrect! Try again.
7Which enzyme unwinds the DNA double helix during replication?
Nature, structure, and replication of genetic material
Easy
A.Amylase
B.Peptidase
C.Ligase
D.Helicase
Correct Answer: Helicase
Explanation:
Helicase separates the two DNA strands by breaking the hydrogen bonds between paired bases.
Incorrect! Try again.
8Where does protein synthesis occur in a cell?
Protein synthesis
Easy
A.Ribosomes
B.Centrioles
C.Peroxisomes
D.Lysosomes
Correct Answer: Ribosomes
Explanation:
Ribosomes read mRNA codons and join amino acids to form a protein.
Incorrect! Try again.
9Which type of RNA brings amino acids to the ribosome?
Protein synthesis
Easy
A.Ribosomal RNA
B.Messenger RNA
C.Small nuclear RNA
D.Transfer RNA
Correct Answer: Transfer RNA
Explanation:
Transfer RNA (tRNA) carries specific amino acids to the ribosome during protein synthesis.
Incorrect! Try again.
10What is transcription?
Transcriptional and translational mechanisms of genetic material
Easy
A.Synthesis of lipids from RNA
B.Synthesis of protein from DNA
C.Synthesis of DNA from protein
D.Synthesis of RNA from DNA
Correct Answer: Synthesis of RNA from DNA
Explanation:
During transcription, the information in a DNA template is copied into an RNA molecule.
Incorrect! Try again.
11Which enzyme synthesizes RNA during transcription?
Transcriptional and translational mechanisms of genetic material
Easy
A.RNA polymerase
B.DNA polymerase
C.Aminoacyl synthetase
D.DNA ligase
Correct Answer: RNA polymerase
Explanation:
RNA polymerase reads the DNA template and joins RNA nucleotides together.
Incorrect! Try again.
12Which codon usually begins translation?
Transcriptional and translational mechanisms of genetic material
Easy
A.UAA
B.AUG
C.UGA
D.UAG
Correct Answer: AUG
Explanation:
AUG is the usual start codon and also codes for methionine.
Incorrect! Try again.
13What is a codon?
Transcriptional and translational mechanisms of genetic material
Easy
A.A region joining two chromosomes
B.A sequence of three bases on mRNA
C.A chain of three amino acids
D.A sequence of two bases on DNA
Correct Answer: A sequence of three bases on mRNA
Explanation:
A codon is a group of three mRNA bases that specifies an amino acid or a stop signal.
Incorrect! Try again.
14Which statement best defines a gene?
Gene concept
Easy
A.A protein that copies DNA
B.A complete set of chromosomes
C.A basic functional unit of heredity
D.A membrane surrounding the nucleus
Correct Answer: A basic functional unit of heredity
Explanation:
A gene is a unit of heredity that contains information for producing a functional product.
Incorrect! Try again.
15Which region of a gene commonly provides a binding site for RNA polymerase?
Gene structure
Easy
A.Telomere
B.Promoter
C.Exon
D.Intron
Correct Answer: Promoter
Explanation:
The promoter is a DNA region where RNA polymerase binds to begin transcription.
Incorrect! Try again.
16Which regions generally remain in mature eukaryotic mRNA after splicing?
Gene structure
Easy
A.Introns
B.Exons
C.Promoters
D.Operators
Correct Answer: Exons
Explanation:
Introns are removed during RNA splicing, while exons are joined to form mature mRNA.
Incorrect! Try again.
17What is the main function of a gene?
Gene function
Easy
A.Digesting nutrients inside the cell
B.Moving chromosomes during cell division
C.Storing energy for cellular respiration
D.Directing synthesis of a functional product
Correct Answer: Directing synthesis of a functional product
Explanation:
A gene contains instructions for producing a functional RNA or protein.
Incorrect! Try again.
18According to the central dogma, genetic information usually flows in which direction?
Gene function
Easy
A.Protein to DNA to RNA
B.DNA to protein to RNA
C.DNA to RNA to protein
D.RNA to protein to DNA
Correct Answer: DNA to RNA to protein
Explanation:
Genetic information is transcribed from DNA into RNA and then translated into protein.
Incorrect! Try again.
19What is an operon?
Gene regulation
Easy
A.A chromosome with identical chromatids
B.A protein that repairs damaged DNA
C.A structure that produces ribosomes
D.A group of genes regulated together
Correct Answer: A group of genes regulated together
Explanation:
An operon is a group of genes controlled together by shared regulatory DNA sequences.
Incorrect! Try again.
20In a typical operon, where does a repressor protein bind to block transcription?
Gene regulation
Easy
A.Operator
B.Exon
C.Telomere
D.Centromere
Correct Answer: Operator
Explanation:
A repressor can bind to the operator and prevent RNA polymerase from transcribing the genes.
Incorrect! Try again.
21A DNA segment containing alternating guanine and cytosine bases is placed under high-salt conditions. Which DNA conformation is this segment most likely to adopt?
Types of DNA and RNA
Medium
A.Triple-helical H-DNA
B.Left-handed Z-DNA
C.Right-handed B-DNA
D.Right-handed A-DNA
Correct Answer: Left-handed Z-DNA
Explanation:
Alternating purine-pyrimidine sequences, especially GC repeats, can form left-handed Z-DNA under high-salt or negative-supercoiling conditions.
Incorrect! Try again.
22A mutation prevents a small nuclear RNA from binding to pre-mRNA. Which process is most directly affected?
Types of DNA and RNA
Medium
A.Removal of introns
B.Replication of telomeres
C.Transport of proteins
D.Addition of amino acids
Correct Answer: Removal of introns
Explanation:
Small nuclear RNAs are components of the spliceosome, which recognizes splice sites and removes introns from pre-mRNA.
Incorrect! Try again.
23Bacteria containing DNA labeled with heavy nitrogen are transferred to light-nitrogen medium. After two rounds of semiconservative replication, what DNA molecules are expected?
Nature, structure, and replication of genetic material
Medium
A.Only heavy DNA molecules
B.Equal heavy and hybrid DNA molecules
C.Equal hybrid and light DNA molecules
D.Only hybrid DNA molecules
Correct Answer: Equal hybrid and light DNA molecules
Explanation:
After one round, all molecules are hybrid. After the second round, half remain hybrid and half contain only light nitrogen.
Incorrect! Try again.
24A drug inhibits DNA ligase during bacterial DNA replication. Which molecular consequence is most likely?
Nature, structure, and replication of genetic material
Medium
A.Replication origins cannot be recognized
B.Okazaki fragments remain unjoined
C.DNA strands cannot be unwound
D.RNA primers cannot be synthesized
Correct Answer: Okazaki fragments remain unjoined
Explanation:
DNA ligase seals breaks in the sugar-phosphate backbone, including the gaps between adjacent Okazaki fragments on the lagging strand.
Incorrect! Try again.
25A human cell line lacks functional telomerase but continues dividing. What change is most likely after many cell divisions?
Nature, structure, and replication of genetic material
Medium
A.Progressive expansion of centromeric DNA
B.Immediate duplication of every chromosome
C.Progressive shortening of chromosome ends
D.Immediate loss of all replication origins
Correct Answer: Progressive shortening of chromosome ends
Explanation:
Without telomerase, conventional DNA polymerases cannot fully maintain chromosome ends, so telomeres become progressively shorter.
Incorrect! Try again.
26A coding sequence mutation changes the mRNA codon 5′-UAU-3′ to 5′-UAA-3′. What is the most likely effect on the protein?
UAU encodes tyrosine, whereas UAA is a stop codon. The substitution therefore creates a nonsense mutation.
Incorrect! Try again.
27An aminoacyl-tRNA synthetase incorrectly attaches valine to a tRNA whose anticodon recognizes an alanine codon. What will occur during translation?
Protein synthesis
Medium
A.Translation will stop at the alanine codon
B.Valine will be inserted at an alanine codon
C.Alanine will be inserted at a valine codon
D.The ribosome will remove the incorrect tRNA
Correct Answer: Valine will be inserted at an alanine codon
Explanation:
The ribosome checks codon-anticodon pairing but does not verify the attached amino acid. The mischarged tRNA therefore inserts valine.
Incorrect! Try again.
28An antibiotic specifically inhibits the peptidyl transferase center of the large ribosomal subunit. Which event is directly blocked?
Protein synthesis
Medium
A.Charging of transfer RNAs
B.Binding of mRNA to DNA
C.Recognition of start codons
D.Formation of peptide bonds
Correct Answer: Formation of peptide bonds
Explanation:
The peptidyl transferase center catalyzes peptide-bond formation between amino acids during polypeptide elongation.
Incorrect! Try again.
29A DNA template strand has the sequence 3′-TAC GGA TTT-5′. Which RNA sequence will be produced?
Transcriptional and translational mechanisms of genetic material
Medium
A.5′-AAA UCC GUA-3′
B.5′-ATG CCT AAA-3′
C.5′-AUG CCU AAA-3′
D.5′-UAC GGA UUU-3′
Correct Answer: 5′-AUG CCU AAA-3′
Explanation:
RNA is synthesized antiparallel and complementary to the DNA template, with uracil replacing thymine.
Incorrect! Try again.
30Why can translation begin before transcription is complete in bacteria but not normally in eukaryotic cells?
Transcriptional and translational mechanisms of genetic material
Medium
A.Bacteria lack a nuclear membrane
B.Eukaryotes lack messenger RNA
C.Bacteria use only one RNA polymerase
D.Eukaryotes have smaller ribosomes
Correct Answer: Bacteria lack a nuclear membrane
Explanation:
Because bacteria have no nucleus, ribosomes can bind and translate an mRNA while RNA polymerase is still transcribing it.
Incorrect! Try again.
31A mutation weakens the promoter of a gene but leaves its coding and termination sequences intact. Which result is most likely?
Transcriptional and translational mechanisms of genetic material
Medium
A.Longer proteins are produced from each transcript
B.Translation continues beyond the stop codon
C.Fewer full-length transcripts are produced
D.Introns are converted directly into exons
Correct Answer: Fewer full-length transcripts are produced
Explanation:
A weaker promoter reduces transcription initiation, decreasing transcript abundance without necessarily changing transcript length.
Incorrect! Try again.
32Two homozygous recessive mutants have the same visible phenotype. Their offspring have the wild-type phenotype. What does this complementation result indicate?
Gene concept
Medium
A.One mutation reverted during fertilization
B.Both mutations are dominant alleles
C.The mutations are in the same codon
D.The mutations are in different genes
Correct Answer: The mutations are in different genes
Explanation:
Wild-type offspring indicate that each parent supplies a functional allele of the gene mutated in the other parent.
Incorrect! Try again.
33A single eukaryotic gene produces one protein in liver cells and a structurally different protein in muscle cells. Which mechanism best explains this observation?
Gene concept
Medium
A.Semiconservative DNA replication
B.Alternative RNA splicing
C.Independent chromosome assortment
D.Random nucleotide substitution
Correct Answer: Alternative RNA splicing
Explanation:
Alternative splicing combines different sets of exons from the same pre-mRNA, allowing one gene to produce multiple protein isoforms.
Incorrect! Try again.
34A mutation destroys the 5′ splice site of an intron, and no alternative splice site is available. What is the most likely effect on the mature mRNA?
Gene structure
Medium
A.The poly(A) tail is removed
B.The intron is retained
C.The promoter is duplicated
D.The start codon is restored
Correct Answer: The intron is retained
Explanation:
Without a functional 5′ splice site, the spliceosome cannot correctly remove the intron, so it is likely to remain in the transcript.
Incorrect! Try again.
35A mutation deletes a tissue-specific enhancer located several thousand base pairs upstream of a gene. Which outcome is most likely?
Gene structure
Medium
A.Replacement of introns with coding exons
B.Constitutive expression in all cell types
C.Reduced expression in the relevant tissue
D.Loss of DNA replication in every tissue
Correct Answer: Reduced expression in the relevant tissue
Explanation:
Enhancers bind regulatory proteins that increase transcription in particular cells. Their deletion can reduce tissue-specific gene expression.
Incorrect! Try again.
36A metabolic pathway is arranged as precursor A → intermediate B → product C. A mutation in the enzyme converting B to C would most likely cause which pattern?
Gene function
Medium
A.B decreases and C increases
B.A and B both disappear
C.A decreases and C accumulates
D.B accumulates and C decreases
Correct Answer: B accumulates and C decreases
Explanation:
Blocking the second reaction prevents B from being converted into C, leading to accumulation of B and reduced production of C.
Incorrect! Try again.
37Under which conditions is transcription of the bacterial lac operon expected to be highest?
Gene regulation
Medium
A.Lactose absent and glucose absent
B.Lactose present and glucose present
C.Lactose absent and glucose present
D.Lactose present and glucose absent
Correct Answer: Lactose present and glucose absent
Explanation:
Lactose inactivates the repressor, while low glucose increases cAMP-CAP activation. Together these conditions produce maximal transcription.
Incorrect! Try again.
38When tryptophan is abundant, how does attenuation regulate the bacterial trp operon?
Gene regulation
Medium
A.The repressor is permanently degraded
B.A terminator structure forms in the leader RNA
C.RNA polymerase binds more strongly to the promoter
D.The ribosome stalls at tryptophan codons
Correct Answer: A terminator structure forms in the leader RNA
Explanation:
At high tryptophan levels, the ribosome rapidly translates the leader peptide, allowing formation of the terminator hairpin and early transcription termination.
Incorrect! Try again.
39A histone acetyltransferase is recruited to the promoter of a normally inactive eukaryotic gene. What is the most likely result?
Gene regulation
Medium
A.Chromatin condenses and transcription decreases
B.DNA replication stops at the promoter
C.Chromatin opens and transcription increases
D.Messenger RNA is translated in the nucleus
Correct Answer: Chromatin opens and transcription increases
Explanation:
Histone acetylation weakens histone-DNA interactions, creating more accessible chromatin and generally promoting transcription.
Incorrect! Try again.
40A partial diploid bacterium has the genotype lacOᶜ lacZ⁺ / lacO⁺ lacZ⁻ and contains functional lacI. What occurs when lactose is absent?
Gene regulation
Medium
A.Both operons are transcribed because lacOᶜ acts in trans
B.Neither operon is transcribed because lacI is functional
C.Beta-galactosidase is produced from the lacOᶜ-linked gene
D.Beta-galactosidase is produced from the lacO⁺-linked gene
Correct Answer: Beta-galactosidase is produced from the lacOᶜ-linked gene
Explanation:
The constitutive operator mutation acts only in cis. It prevents repressor binding and permits expression of the linked functional lacZ gene.
Incorrect! Try again.
41A covalently closed 4200-bp DNA circle contains a 40-bp segment that changes from B-DNA, with 10.5 bp per right-handed turn, to Z-DNA, with 12 bp per left-handed turn. No strand break occurs. Approximately how must writhe change?
Types of DNA and RNA
Hard
A.Writhe must increase by about
B.Writhe must decrease by about
C.Writhe must remain approximately unchanged
D.Writhe must increase by about
Correct Answer: Writhe must increase by about
Explanation:
The segment changes from to , so . Because remains fixed, .
Incorrect! Try again.
42Purified bacterial RNase P RNA cleaves pre-tRNA slowly at high concentration, whereas addition of its protein subunit permits rapid cleavage under physiological ionic conditions. Which conclusion is best supported?
Types of DNA and RNA
Hard
A.The RNA supplies energy, while the protein hydrolyzes the phosphodiester bond
B.The protein and RNA are independently active enzymes with identical specificity
C.The RNA is catalytic, while the protein improves activity under cellular conditions
D.The protein catalyzes cleavage, while RNA only recognizes pre-tRNA
Correct Answer: The RNA is catalytic, while the protein improves activity under cellular conditions
Explanation:
Catalysis by purified RNA establishes RNase P RNA as a ribozyme. The protein enhances substrate binding, stability, or catalysis under physiological conditions.
Incorrect! Try again.
43Cells with fully -labeled DNA are transferred to medium for two generations and then returned to medium for one generation. Assuming semiconservative replication, what DNA classes occur after the final generation?
Nature, structure, and replication of genetic material
Hard
A. heavy and hybrid
B. light and hybrid
C. heavy and hybrid
D. heavy, hybrid, and light
Correct Answer: heavy and hybrid
Explanation:
After two generations in light medium, half the molecules are hybrid and half are light. One generation in heavy medium produces two heavy and six hybrid molecules out of eight total.
Incorrect! Try again.
44A bacterial mutant replicates its circular chromosome to completion, but the two daughter chromosomes remain topologically interlinked and cannot segregate. Which enzyme is most directly defective?
Nature, structure, and replication of genetic material
Hard
A.Topoisomerase IV, which decatenates replicated daughter chromosomes
B.DNA gyrase, which removes positive supercoils before the fork
C.DNA ligase, which seals nicks between adjacent Okazaki fragments
D.DNA polymerase I, which removes primers from lagging strands
Correct Answer: Topoisomerase IV, which decatenates replicated daughter chromosomes
Explanation:
Completed circular daughter chromosomes form catenanes. Topoisomerase IV performs double-strand passage reactions that separate these interlinked chromosomes.
Incorrect! Try again.
45An engineered telomerase uses the RNA template segment 3′-CAAUCC-5′ for repeat synthesis. Ignoring flanking alignment nucleotides, which DNA sequence will be added to the chromosome's 3′ end?
Nature, structure, and replication of genetic material
Hard
A.5′-GTTAGG-3′
B.5′-CCTAAC-3′
C.5′-GGATTA-3′
D.5′-CAATCC-3′
Correct Answer: 5′-GTTAGG-3′
Explanation:
Telomerase synthesizes DNA antiparallel and complementary to its RNA template. The complement of 3′-CAAUCC-5′ is 5′-GTTAGG-3′.
Incorrect! Try again.
46In E. coli, a replication error lies near a hemimethylated GATC site. Which change would most directly compromise methyl-directed mismatch repair by obscuring identification of the newly synthesized strand?
Nature, structure, and replication of genetic material
Hard
A.Failure to proofread the terminal nucleotide of each primer
B.Reduced negative supercoiling behind the replication fork
C.Premature methylation of the newly synthesized GATC strand
D.Failure to remove RNA primers from Okazaki fragments
Correct Answer: Premature methylation of the newly synthesized GATC strand
Explanation:
MutH-directed repair uses transient hemimethylation to identify the unmethylated daughter strand. Premature Dam methylation removes this strand-discrimination signal.
Incorrect! Try again.
47A mutant aminoacyl-tRNA synthetase attaches alanine to a tRNA whose anticodon normally recognizes phenylalanine codons. If the mischarged tRNA enters the ribosome, what is the most likely outcome?
Protein synthesis
Hard
A.Translation terminates because the esterified amino acid mismatches the anticodon
B.Alanine is inserted because codon recognition does not verify the attached amino acid
C.The codon is skipped because EF-Tu rejects every mischarged tRNA
D.Phenylalanine is inserted because the ribosome checks the tRNA identity
Correct Answer: Alanine is inserted because codon recognition does not verify the attached amino acid
Explanation:
The ribosome monitors codon–anticodon pairing but generally cannot determine whether the tRNA carries the correct amino acid. Thus alanine is incorporated at a phenylalanine codon.
Incorrect! Try again.
48After acute addition of a translation inhibitor, ribosomes already on mRNAs complete elongation and dissociate, while few new ribosomes load. Polysome profiles progressively shift from heavy polysomes toward monosomes. Which process is primarily inhibited?
Protein synthesis
Hard
A.Peptide-bond formation during elongation
B.Release-factor recognition at stop codons
C.Translation initiation on new coding regions
D.Ribosome recycling after peptide release
Correct Answer: Translation initiation on new coding regions
Explanation:
Blocking initiation allows existing ribosomes to run off transcripts without replacement, causing progressive loss of heavy polysomes. Elongation inhibition would instead retain ribosomes on mRNAs.
Incorrect! Try again.
49Rifampicin is added to bacteria actively transcribing a long operon. Full-length transcripts continue to appear briefly, but production then stops. Which mechanism explains this pattern?
Transcriptional and translational mechanisms of genetic material
Hard
A.Rifampicin blocks initiation but permits engaged RNA polymerases to finish
B.Rifampicin blocks elongation only after RNA polymerase reaches a terminator
C.Rifampicin prevents translation and thereby immediately degrades RNA polymerase
D.Rifampicin destroys completed transcripts but leaves new initiation unaffected
Correct Answer: Rifampicin blocks initiation but permits engaged RNA polymerases to finish
Explanation:
Rifampicin primarily prevents formation of productive new transcription complexes. Polymerases that had already escaped the promoter can transiently complete their transcripts.
Incorrect! Try again.
50During spliceosomal pre-mRNA splicing, which nucleophile performs the first transesterification reaction?
Transcriptional and translational mechanisms of genetic material
Hard
A.The 2′-OH of branch-point adenosine attacks the 5′ splice site
B.The 3′-OH of the upstream exon attacks the branch-point adenosine
C.The 3′-OH of the downstream exon attacks the 5′ splice site
D.The 2′-OH of the upstream exon attacks the 3′ splice site
Correct Answer: The 2′-OH of branch-point adenosine attacks the 5′ splice site
Explanation:
The branch-point adenosine's 2′-OH attacks the 5′ splice site, forming the 2′–5′ linkage of the intron lariat and releasing the upstream exon.
Incorrect! Try again.
51A suppressor tRNA is engineered to recognize the amber stop codon 5′-UAG-3′. When the anticodon is written 5′ to 3′, which sequence is required for direct Watson–Crick pairing?
Transcriptional and translational mechanisms of genetic material
Hard
A.5′-GAU-3′
B.5′-CUA-3′
C.5′-UAG-3′
D.5′-AUC-3′
Correct Answer: 5′-CUA-3′
Explanation:
The antiparallel complement of 5′-UAG-3′ is 3′-AUC-5′. Written in the requested 5′-to-3′ direction, the anticodon is 5′-CUA-3′.
Incorrect! Try again.
52In a trpR deletion strain, the two consecutive tryptophan codons in the trp leader peptide are replaced by alanine codons. Alanine remains abundant. What happens to operon expression during tryptophan starvation?
Gene regulation
Hard
A.Expression becomes high because tryptophan starvation prevents leader transcription
B.Expression oscillates because the repressor is required for attenuation to occur
C.Expression remains low because the ribosome no longer stalls at the former Trp codons
D.Expression becomes high because uncharged tRNA directly disrupts the terminator
Correct Answer: Expression remains low because the ribosome no longer stalls at the former Trp codons
Explanation:
Without Trp codons, low charged tRNA no longer causes leader-ribosome stalling. The ribosome permits formation of the 3–4 terminator, so attenuation continues despite starvation.
Incorrect! Try again.
53Two recessive mutations produce the same phenotype. A diploid carrying one mutation on each homolog remains mutant, yet rare wild-type meiotic recombinants are recovered. What is the strongest interpretation?
Gene concept
Hard
A.The mutations occupy different genes whose products form a required complex
B.The mutations occupy different sites within the same functional cistron
C.The mutations are identical alleles repaired by meiotic gene conversion
D.The mutations are dominant alleles located in separate regulatory pathways
Correct Answer: The mutations occupy different sites within the same functional cistron
Explanation:
Failure to complement indicates disruption of the same functional unit, while wild-type recombinants show that the mutations are at separable sites within that gene.
Incorrect! Try again.
54Two recessive missense mutations mapped to the same gene unexpectedly restore near-normal function when present in trans, although neither allele functions alone. Which explanation best fits this intragenic complementation?
Gene concept
Hard
A.Each mutant subunit supplies a different intact function within a multimeric protein
B.Both alleles independently activate transcription of an unrelated compensating gene
C.One mutation converts the other allele back to wild type through DNA repair
D.The two mutations must actually lie in separate genes that cannot recombine
Correct Answer: Each mutant subunit supplies a different intact function within a multimeric protein
Explanation:
Different mutant polypeptides can complement within a multimer if each retains a domain or activity missing from the other. This is complementation between alleles of one gene.
Incorrect! Try again.
55A candidate regulatory sequence is integrated with the same minimal promoter and reporter at one defined chromosomal locus. Which result most specifically supports classification of the sequence as an enhancer rather than a core promoter element?
Gene structure
Hard
A.It stimulates expression at a distance in either orientation relative to the promoter
B.It recruits RNA polymerase only when the minimal promoter has been deleted
C.It determines the exact nucleotide used to initiate every reporter transcript
D.It stimulates expression only when placed directly over the transcription start site
Correct Answer: It stimulates expression at a distance in either orientation relative to the promoter
Explanation:
Enhancers characteristically activate compatible promoters over substantial distances and often retain activity after inversion. Core promoter elements are position- and orientation-dependent.
Incorrect! Try again.
56A genomic substitution changes the invariant +1 guanine of an intron's 5′ splice site. The resulting mature mRNA lacks the entire upstream exon, while transcription initiation is unchanged. Which molecular interaction was directly disrupted?
Gene structure
Hard
A.CPSF recognition of the polyadenylation signal
B.U1 snRNP recognition of the 5′ splice site
C.TFIID recognition of the core promoter sequence
D.U2 snRNP recognition of the branch-point sequence
Correct Answer: U1 snRNP recognition of the 5′ splice site
Explanation:
The conserved GU at the 5′ splice site is recognized initially through base pairing with U1 snRNA. Its disruption can force alternative splice-site use or exon skipping.
Incorrect! Try again.
57A signaling pathway has the order ligand → receptor R → inhibition of repressor X → expression of gene G. An R null mutant cannot express G, an X null mutant expresses G constitutively, and the R X double null also expresses G constitutively. What does the double-mutant phenotype establish?
Gene function
Hard
A.R acts downstream of X and directly activates gene G
B.X acts downstream of R as a negative regulator of gene G
C.R and X act in independent pathways that converge on gene G
D.X is required to produce the ligand that activates receptor R
Correct Answer: X acts downstream of R as a negative regulator of gene G
Explanation:
The X null phenotype masks the R null phenotype, making X epistatic to R. This supports X acting downstream as the repressor normally inhibited by R.
Incorrect! Try again.
58Consider the partial diploid E. coli genotype . What pattern of -galactosidase production is expected in the absence and presence of lactose?
Gene regulation
Hard
A.Low both without and with lactose
B.High both without and with lactose
C.Low without lactose and high with lactose
D.High without lactose and low with lactose
Correct Answer: High both without and with lactose
Explanation:
The functional allele is linked in cis to , which cannot bind repressor. The trans-acting product therefore cannot repress that structural gene.
Incorrect! Try again.
59A microRNA reduces expression of a reporter carrying a candidate 3′-UTR target site. A point mutation in the site's seed-complementary region abolishes repression. Which result would provide the strongest evidence for direct base pairing?
Gene regulation
Hard
A.A compensatory microRNA mutation restores repression of the mutant reporter
B.Moving the target site into the promoter eliminates reporter transcription
C.Deleting Dicer increases expression of several unrelated cellular reporters
D.Overexpressing the original microRNA further represses the wild-type reporter
Correct Answer: A compensatory microRNA mutation restores repression of the mutant reporter
Explanation:
Restoration by reciprocal, compensatory mutations demonstrates sequence-specific pairing between the microRNA seed and the target site rather than an indirect regulatory effect.
Incorrect! Try again.
60A CTCF-bound boundary lies between an enhancer and a normally insulated promoter. Acute CTCF degradation causes new enhancer–promoter contacts and ectopic transcription; restoring CTCF reverses both effects. Which inference is best supported?
Gene regulation
Hard
A.CTCF normally serves as the basal transcription factor for the insulated promoter
B.CTCF normally constrains regulatory contacts across a chromatin boundary
C.CTCF normally methylates the enhancer to prevent nucleosome displacement
D.CTCF normally blocks transcription by degrading enhancer-derived RNAs
Correct Answer: CTCF normally constrains regulatory contacts across a chromatin boundary
Explanation:
The reversible gain of cross-boundary contact and transcription indicates that CTCF-dependent insulation limits inappropriate enhancer access to promoters in neighboring domains.
Incorrect! Try again.
Did this save you a night before the exam?
LPU Notes is free, and it stays free. Ads cover part of the server bill.
The rest comes out of a student's own pocket: the domain, the storage,
and keeping the site up through the weeks everyone needs it at once.
The payment button didn't load. An ad blocker or a filtered network is the usual reason.
to try again.
Nothing here is ever locked, and nothing unlocks. Chip in only if it was worth it.
What it pays for →