Science · AP Biology ★★★ Hard UNIT 6 OF 0

AP Biology Unit 6: Gene Expression and Regulation — Free Review Games.

This unit covers DNA replication, transcription, translation and gene regulation — essential concepts for AP Biology. Use our interactive study games to test your understanding, or review questions in traditional format below.

📋 200 questions ⏱ ~30 min 📊 12-16% of exam
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All 200 questions below, each with the worked answer and a written explanation. Click any question to expand it.

Q1. During DNA replication, which enzyme unwinds the double helix?
A DNA polymerase
B RNA polymerase
C Helicase
D Ligase

Helicase breaks the hydrogen bonds between complementary base pairs, separating the two strands to create replication forks.

Q2. Transcription produces which molecule from a DNA template?
A DNA
B mRNA
C Protein
D tRNA only

Transcription uses RNA polymerase to synthesize a messenger RNA (mRNA) strand complementary to the DNA template strand.

Q3. A codon on mRNA is read by a complementary sequence on tRNA called the:
A Promoter
B Anticodon
C Exon
D Operator

The anticodon is a three-nucleotide sequence on tRNA that base-pairs with the complementary codon on mRNA during translation.

Q4. DNA replication is described as semiconservative because:
A Both strands are entirely new
B Each new DNA molecule contains one original and one new strand
C Only one strand is replicated
D The original DNA is destroyed

In semiconservative replication, each daughter DNA molecule consists of one parental strand and one newly synthesized strand.

Q5. Translation occurs at which cellular structure?
A Nucleus
B Ribosome
C Golgi apparatus
D Lysosome

Ribosomes are the molecular machines that read mRNA codons and catalyze the formation of peptide bonds between amino acids during translation.

Q6. In the lac operon, the repressor protein binds to the operator to:
A Activate transcription of structural genes
B Block RNA polymerase from transcribing structural genes
C Enhance translation of the mRNA
D Degrade the mRNA produced

When lactose is absent, the repressor binds the operator and physically blocks RNA polymerase from transcribing the lac operon genes.

Q7. What is the role of RNA splicing in eukaryotic gene expression?
A It adds a poly-A tail to mRNA
B It removes introns and joins exons to form mature mRNA
C It translates mRNA into protein
D It degrades faulty mRNA molecules

RNA splicing removes non-coding introns from pre-mRNA and ligates the remaining exons, producing a mature mRNA ready for translation.

Q8. A point mutation that changes one amino acid in a protein is called a:
A Silent mutation
B Missense mutation
C Nonsense mutation
D Frameshift mutation

A missense mutation changes one codon to code for a different amino acid, which may or may not affect protein function.

Q9. Epigenetic modifications such as DNA methylation typically:
A Change the DNA base sequence
B Silence gene expression without altering the DNA sequence
C Only occur during DNA replication
D Increase mutation rates

DNA methylation adds methyl groups to cytosine bases, typically silencing gene expression by preventing transcription factor binding, without changing the DNA sequence.

Q10. The enzyme that synthesizes the leading strand during DNA replication continuously is:
A Primase
B Helicase
C DNA polymerase III
D Topoisomerase

DNA polymerase III synthesizes the leading strand continuously in the 5' to 3' direction as the replication fork opens.

Q11. A deletion of one nucleotide in the coding region of a gene would most likely cause:
A A silent mutation
B A missense mutation affecting one amino acid
C A frameshift mutation altering all downstream amino acids
D No change in the protein

Deleting one nucleotide shifts the reading frame of all subsequent codons, changing the amino acid sequence from that point onward and usually producing a nonfunctional protein.

Q12. Alternative splicing allows a single gene to code for multiple proteins by:
A Mutating the gene at different times
B Including different combinations of exons in the mature mRNA
C Using different start codons
D Replicating the gene multiple times

Alternative splicing selectively includes or excludes different exons from the pre-mRNA, producing different mRNA variants that encode distinct protein isoforms from one gene.

Q13. A researcher discovers that a gene is being actively transcribed but very little protein is produced. Which post-transcriptional mechanism could explain this?
A Histone acetylation increasing transcription
B microRNA (miRNA) binding to the mRNA and inhibiting translation
C Promoter methylation
D Enhanced ribosome production

miRNAs bind to complementary sequences on mRNA, blocking translation or targeting the mRNA for degradation, reducing protein output despite active transcription.

Q14. In eukaryotes, enhancer sequences can regulate transcription of genes located thousands of base pairs away. How do they accomplish this?
A Enhancers move along the DNA to the promoter
B DNA looping brings enhancer-bound activators into contact with the transcription complex at the promoter
C Enhancers directly bind to ribosomes
D Enhancers change the DNA sequence at the promoter

DNA looping allows transcription factors bound to distant enhancers to physically interact with the mediator complex and RNA polymerase at the promoter.

Q15. CRISPR-Cas9 genome editing works by:
A Inserting random transposons into the genome
B Using a guide RNA to direct Cas9 to cut a specific DNA sequence
C Methylating target genes to silence them
D Enhancing natural DNA polymerase proofreading

CRISPR-Cas9 uses a guide RNA complementary to the target sequence to direct the Cas9 nuclease, which creates a double-strand break at the precise genomic location.

Q16. Short, discontinuous DNA segments synthesized on the lagging strand during DNA replication are called:
A Okazaki fragments
B RNA primers
C Leading strand segments
D Telomeric repeats

Because DNA polymerase can only synthesize DNA in the 5' to 3' direction, the lagging strand must be built in short segments away from the replication fork. These are Okazaki fragments, each initiated by its own RNA primer. RNA primers (choice B) are what initiate each fragment, not the fragments themselves.

Q17. To initiate transcription, RNA polymerase binds to a specific DNA sequence known as the:
A Promoter
B Operator
C Terminator
D Enhancer

The promoter is the DNA sequence upstream of a gene where RNA polymerase binds to initiate transcription. The operator (choice B) is a regulatory sequence in prokaryotes where repressor proteins bind, not RNA polymerase. The terminator signals the end of transcription, and enhancers increase transcription but are not the binding site for RNA polymerase itself.

Q18. In the standard genetic code, which codon signals the beginning of translation and also codes for the amino acid methionine?
A AUG
B UAA
C GUG
D ACG

AUG is the universal start codon that codes for methionine. UAA (choice B) is one of the three stop codons and does not code for any amino acid. GUG can occasionally serve as a start codon in prokaryotes under special conditions, but AUG is the standard and universal answer expected at the AP level.

Q19. During transcription, RNA polymerase reads the template strand of DNA in which direction?
A 3' to 5'
B 5' to 3'
C Both directions simultaneously
D Either direction depending on the gene

RNA polymerase reads the template (antisense) strand from 3' to 5', which allows it to synthesize the new mRNA strand in the 5' to 3' direction. Choice B is incorrect because it describes the direction of mRNA synthesis, not template reading. These two directions are antiparallel, a fundamental rule of nucleic acid chemistry.

Q20. Which of the following is the primary function of the 5' methylguanosine cap and poly-A tail added to eukaryotic mRNA?
A Protecting mRNA from degradation and facilitating ribosome binding and nuclear export
B Marking introns so the spliceosome can remove them
C Signaling which reading frame ribosomes should use during translation
D Attaching the mRNA permanently to the endoplasmic reticulum

The 5' cap protects mRNA from exonucleases and helps recruit the small ribosomal subunit during translation initiation. The poly-A tail similarly protects the 3' end and aids in nuclear export. Choice B is incorrect because introns are identified by splice site sequences, not the cap or tail.

Q21. Which molecule is responsible for carrying specific amino acids to the ribosome during translation?
A Transfer RNA (tRNA)
B Messenger RNA (mRNA)
C Ribosomal RNA (rRNA)
D Small nuclear RNA (snRNA)

tRNA molecules are charged with specific amino acids by aminoacyl-tRNA synthetase enzymes and deliver those amino acids to the ribosome. The anticodon on the tRNA base-pairs with the complementary codon on the mRNA. rRNA (choice C) is a structural and catalytic component of the ribosome itself, not a carrier of amino acids.

Q22. What is the role of DNA ligase during DNA replication?
A It seals the nicks between Okazaki fragments on the lagging strand by forming phosphodiester bonds
B It unwinds the double helix at the replication fork
C It synthesizes short RNA primers needed to start replication
D It removes incorrectly paired nucleotides via proofreading

After RNA primers are replaced with DNA on the lagging strand, DNA ligase covalently joins the adjacent DNA fragments by forming phosphodiester bonds, producing a continuous strand. Helicase (not ligase) unwinds the double helix (choice B), and primase (not ligase) synthesizes RNA primers (choice C).

Q23. DNA polymerase requires a free 3'-hydroxyl group to begin adding nucleotides. Which enzyme supplies this requirement during DNA replication?
A Primase, by synthesizing a short RNA primer complementary to the template strand
B Helicase, by exposing single-stranded DNA at the replication fork
C Topoisomerase, by cutting and rejoining DNA strands to relieve supercoiling
D Ligase, by joining pre-existing DNA fragments

Primase is an RNA polymerase that synthesizes a short RNA primer using the DNA template. This primer provides the 3'-OH group that DNA polymerase III needs to begin extending the new strand. Helicase (choice B) unwinds DNA but does not provide the 3'-OH starting point.

Q24. In the trp operon of E. coli, transcription of tryptophan biosynthesis genes is repressed when tryptophan is abundant. This is because tryptophan acts as a:
A Corepressor that binds to the repressor protein, enabling it to bind the operator
B Direct inhibitor of RNA polymerase binding to the trp promoter
C Substrate that promotes methylation of the operon's DNA
D Signal that degrades the trp mRNA already being produced

The trp operon is a repressible operon. The repressor protein is inactive on its own but becomes active when tryptophan (the corepressor) binds to it. The activated repressor then binds the operator, blocking transcription. This is different from the lac operon, where an inducer inactivates the repressor. Choice B is incorrect because tryptophan does not directly contact RNA polymerase.

Q25. Histone acetylation is generally associated with increased gene transcription. The most direct reason for this effect is:
A Acetyl groups neutralize the positive charge on histone tails, weakening their attraction to negatively charged DNA and making chromatin more accessible
B Acetylation directly activates RNA polymerase by binding to its active site
C Acetyl groups recruit transcriptional repressors that paradoxically increase transcription
D Acetylation compacts DNA more tightly around nucleosomes, protecting it from degradation

Histones are positively charged proteins that interact tightly with negatively charged DNA phosphate groups. Acetylation adds negatively charged acetyl groups to lysine residues on histone tails, reducing this electrostatic attraction and relaxing chromatin into a more open euchromatin state that RNA polymerase can access. Choice D is the opposite of the correct effect.

Q26. A single base substitution in a gene changes a codon from UAC (tyrosine) to UAA (a stop codon). This type of mutation is classified as a:
A Nonsense mutation
B Missense mutation
C Silent mutation
D Frameshift mutation

A nonsense mutation converts a sense codon (one encoding an amino acid) into a stop codon, causing premature termination of translation and typically producing a truncated, nonfunctional protein. A missense mutation (choice B) changes one amino acid to a different amino acid. A silent mutation changes the codon but not the amino acid due to degeneracy of the genetic code.

Q27. The wobble hypothesis, proposed to explain the degeneracy of the genetic code, states that:
A Flexible base pairing at the third position of a codon allows a single tRNA to recognize multiple synonymous codons
B Ribosomes can shift reading frames when they encounter rare codons
C The same amino acid can be carried by structurally identical tRNA molecules from different genes
D Non-Watson-Crick base pairing at the first codon position allows misreading of the start codon

The wobble hypothesis explains that strict Watson-Crick base pairing is required at the first two codon positions, but the third position (3' end of the codon) allows non-standard base pairing. This means one tRNA can decode multiple codons that differ only at the third base, explaining why fewer than 61 tRNA types are needed despite 61 sense codons. Choice B describes a real phenomenon called frameshifting but is unrelated to wobble.

Q28. When glucose is absent from the environment of E. coli, lac operon transcription increases dramatically even when lactose is present. The molecular explanation is that:
A Low glucose causes cAMP levels to rise, which allows catabolite activator protein (CAP) to bind the lac promoter and enhance RNA polymerase recruitment
B Absence of glucose directly removes a glucose-mediated block on the lac repressor
C Low glucose increases the stability of lac mRNA, leading to more protein production
D Glucose normally competes with lactose for the inducer binding site on the repressor

This phenomenon is catabolite repression. When glucose is absent, adenylyl cyclase produces more cAMP. cAMP binds to CAP (also called CRP), causing a conformational change that allows CAP to bind near the lac promoter and recruit RNA polymerase more effectively. This is a positive regulatory mechanism layered on top of the repressor-based negative regulation. Choice B is incorrect because glucose regulation operates through a completely separate pathway involving cAMP and CAP.

Q29. The spliceosome is a large ribonucleoprotein complex whose primary function is to:
A Remove introns from pre-mRNA and join the remaining exons to produce mature mRNA
B Add the 5' methylguanosine cap to newly transcribed pre-mRNA
C Facilitate export of mature mRNA through nuclear pores into the cytoplasm
D Recruit RNA polymerase II to the promoter region of protein-coding genes

The spliceosome recognizes conserved splice site sequences at intron-exon boundaries and catalyzes two transesterification reactions that remove introns (as a lariat structure) and ligate adjacent exons. This RNA processing step is unique to eukaryotes. Choice B describes capping enzymes, and choice C describes nuclear export factors, both of which are separate processes.

Q30. An mRNA molecule has the sequence 5'-AUGCCGAAAUAG-3'. How many amino acids will be incorporated into the polypeptide during translation of this message?
A 3
B 4
C 2
D 5

Reading the codons: AUG (methionine, start), CCG (proline), AAA (lysine), UAG (stop — not decoded as an amino acid). Translation stops at UAG without incorporating another amino acid, so the polypeptide contains 3 amino acids. Choice B is incorrect because UAG is a stop codon, not a sense codon, so no fourth amino acid is added.

Q31. During DNA replication, DNA polymerase occasionally inserts the wrong nucleotide. The primary mechanism that corrects most of these errors immediately is:
A The 3' to 5' exonuclease proofreading activity built into DNA polymerase itself
B Post-replication mismatch repair enzymes that scan newly synthesized DNA
C The RNA primer, which marks newly synthesized DNA for error checking
D Topoisomerase, which removes mismatched bases when relieving supercoiling

DNA polymerase has an intrinsic 3' to 5' exonuclease domain that acts as a proofreader. When a mismatched base is incorporated, the polymerase detects the distortion, reverses direction, excises the incorrect nucleotide, and resynthesizes. This immediate proofreading reduces errors by roughly 100-fold. Choice B describes a real but separate backup system that corrects errors missed by proofreading.

Q32. A polyribosome (polysome) forms when multiple ribosomes attach to a single mRNA molecule. The biological advantage of this arrangement is:
A Many copies of the same protein can be produced simultaneously from one mRNA molecule, maximizing translational efficiency
B Multiple different proteins can be made from a single mRNA by using different reading frames
C The mRNA is protected from degradation when ribosomes coat its entire length
D Polysomes allow translation to begin before transcription is complete, but only in eukaryotes

Since each ribosome independently translates the mRNA from 5' to 3', a single mRNA with 10 ribosomes attached can produce 10 protein copies nearly simultaneously. This is highly efficient. Choice B is incorrect because all ribosomes on a polysome read the same reading frame and produce identical proteins. Choice D reverses reality: coupled transcription-translation occurs in prokaryotes, not eukaryotes.

Q33. A single base substitution in the promoter region of a gene reduces RNA polymerase binding affinity by approximately 90%. Which of the following best predicts the impact on cellular protein levels from this gene?
A Protein levels will decrease substantially because reduced transcription initiation limits mRNA abundance available for translation
B Protein levels will remain normal because ribosomes will compensate by translating the remaining mRNA more rapidly
C Protein levels will increase because RNA polymerase freed from this promoter will transcribe other genes and release regulatory factors
D Protein levels will decrease because the mutant mRNA will be unstable and rapidly degraded by the nonsense-mediated decay pathway

Gene expression follows the central dogma: promoter strength determines transcription rate, which determines mRNA abundance, which determines translation output. A 90% reduction in RNA polymerase binding means far fewer transcription initiation events, so far less mRNA is produced and thus far less protein. Choice B is incorrect because ribosomes cannot exceed their inherent elongation rate to compensate for reduced mRNA. Choice D is incorrect because a promoter mutation does not create a premature stop codon, so nonsense-mediated decay is not triggered.

Q34. Linear eukaryotic chromosomes shorten slightly with each round of DNA replication. The molecular basis of this end-replication problem is that:
A After the RNA primer at the 5' end of the lagging strand is removed, DNA polymerase cannot fill the resulting gap because there is no upstream 3'-OH to extend from
B Helicase is unable to unwind the highly repetitive sequences at chromosome ends called telomeres
C Topoisomerase cannot relieve the torsional stress that accumulates at the ends of linear chromosomes
D The leading strand cannot be replicated at chromosome ends because the template runs out before synthesis is complete

Each Okazaki fragment on the lagging strand requires an RNA primer. The terminal primer at the very 5' end of the lagging strand is eventually removed, but unlike interior gaps, there is no adjacent upstream 3'-OH group that DNA polymerase can use to fill in the gap. This leaves a short single-stranded 3' overhang and results in net shortening. Choice D is incorrect because the leading strand, being continuous, can be fully replicated to the end of the template.

Q35. RNA interference (RNAi) is a cellular mechanism triggered by double-stranded RNA. When a researcher introduces dsRNA matching a specific mRNA into a eukaryotic cell, gene silencing most likely occurs through:
A Processing of dsRNA into small interfering RNAs (siRNAs) that guide the RISC complex to bind and cleave complementary mRNA molecules
B Direct binding of the dsRNA to the gene's promoter, physically blocking RNA polymerase
C Reverse transcription of the dsRNA into cDNA that inserts into the genome and disrupts the target gene
D Activation of histone deacetylases that compact chromatin at the target gene locus

The Dicer enzyme cleaves long dsRNA into 21-23 nucleotide siRNA duplexes. One strand of each siRNA is loaded into the RNA-induced silencing complex (RISC), which uses it as a guide to find and cleave perfectly complementary mRNA. This post-transcriptional silencing degrades the target mRNA before it can be translated. Choice B is incorrect because dsRNA acts post-transcriptionally, not at the DNA level.

Q36. A mutation destroys the allolactose (inducer) binding site of the lac repressor protein while leaving the operator-binding domain fully functional. What effect would this mutation most likely have on lac operon expression?
A The operon would be constitutively repressed even when lactose is present, because the repressor can never be inactivated by allolactose
B The operon would be constitutively expressed regardless of lactose levels, because a non-functional repressor cannot bind the operator
C The operon would respond normally to lactose but would no longer respond to changes in glucose concentration
D Transcription would increase because a repressor unable to bind allolactose would also lose its affinity for the operator

Normally, allolactose binds to the repressor and causes a conformational change that releases it from the operator, allowing transcription. If the allolactose-binding site is destroyed, the repressor can never be inactivated, so it remains permanently bound to the operator blocking RNA polymerase access. This is a constitutively repressed (super-repressed) phenotype. Choice B describes a mutation in the operator-binding domain, not the inducer-binding domain.

Q37. Biochemical studies have shown that the peptidyl transferase activity responsible for forming peptide bonds during translation is catalyzed by ribosomal RNA rather than by a ribosomal protein. This finding means the ribosome is best classified as:
A A ribozyme, because catalysis is performed by RNA rather than by a protein enzyme
B An allosteric enzyme, because the ribosome changes shape during each elongation cycle
C A holoenzyme, because it requires protein cofactors in addition to its RNA core to function
D A chaperone, because it facilitates folding of the nascent polypeptide as it emerges

A ribozyme is an RNA molecule with catalytic activity. Because the 23S rRNA (prokaryotes) or 28S rRNA (eukaryotes) in the large subunit performs the peptidyl transferase reaction, the ribosome is a ribozyme. This was a landmark discovery supporting the RNA world hypothesis. Choice C is incorrect because the term holoenzyme applies to protein enzymes with their cofactors, not to RNA-based catalysts.

Q38. A eukaryotic gene has 5 exons. Exon 1 always appears in the final mRNA because it contains the start codon, but exons 2, 3, 4, and 5 can each be independently included or excluded through alternative splicing. How many distinct mRNA isoforms can theoretically be produced from this gene?
A 16
B 32
C 10
D 5

Since exon 1 is fixed, each of the 4 remaining exons has two possible states: included or excluded. By the multiplication principle, the number of combinations is 2 x 2 x 2 x 2 = 2^4 = 16. Choice B (32 = 2^5) would be correct only if exon 1 could also be excluded. Choice C (10 = 5 choose 2) incorrectly applies combinations rather than independent binary choices for each exon.

Q39. A mutation inactivates the nuclear pore complexes of a eukaryotic cell, preventing transport of macromolecules between the nucleus and cytoplasm. Which of the following is the most direct consequence for cytoplasmic protein synthesis?
A Translation of nuclear-encoded proteins would cease because mature mRNA could not be exported to cytoplasmic ribosomes
B Transcription would immediately stop because RNA polymerase proteins synthesized in the cytoplasm could not re-enter the nucleus
C Protein synthesis rates would be initially unchanged because ribosomes already in the cytoplasm retain all mRNA needed
D Only secretory proteins would be affected because they are translated at the endoplasmic reticulum rather than on free ribosomes

In eukaryotes, transcription and RNA processing occur in the nucleus, but ribosomes that translate mRNA are located in the cytoplasm. Mature mRNA must be exported through nuclear pores to reach ribosomes. If nuclear pores are non-functional, newly processed mRNA cannot exit the nucleus, so cytoplasmic ribosomes would exhaust existing mRNA and translation would stop for nuclear-encoded genes. Choice B describes a secondary effect that is slower and indirect.

Q40. A researcher compares two versions of a human gene: the wild type, which is expressed in all tissues, and a mutant version in which a single CpG site in the promoter region is methylated. In which tissue would the mutant gene most likely show reduced expression compared to the wild type?
A In a differentiated somatic cell where the methylated CpG recruits methyl-binding proteins and histone deacetylases that compact chromatin
B In a rapidly dividing stem cell where DNA methylation patterns are actively erased during each cell division
C In a cell where the promoter CpG is within an intron and therefore cannot influence RNA polymerase binding directly
D In a germ cell, because methylation of somatic promoters only affects gamete production

DNA methylation at CpG sites in promoter regions is a well-characterized epigenetic silencing mechanism. Methyl-CpG binding domain proteins recognize the methylated cytosine and recruit histone deacetylase complexes, which remove acetyl groups from histones, causing chromatin compaction and transcriptional repression. This effect is stable in differentiated somatic cells. Choice B is incorrect because stem cells actively maintain or remodel methylation patterns rather than erasing all methylation at each division.

Q41. What is the primary function of DNA ligase during DNA replication?
A It synthesizes RNA primers to initiate strand synthesis
B It joins Okazaki fragments by forming phosphodiester bonds on the lagging strand
C It unwinds the double helix ahead of the replication fork
D It removes mismatched nucleotides through proofreading

DNA ligase seals the nicks between Okazaki fragments on the lagging strand by catalyzing the formation of phosphodiester bonds, producing a continuous strand. Primase (not ligase) makes RNA primers, helicase unwinds the helix, and proofreading is carried out by the 3' to 5' exonuclease activity of DNA polymerase.

Q42. Which type of RNA molecule carries amino acids to the ribosome during translation?
A mRNA
B rRNA
C tRNA
D snRNA

Transfer RNA (tRNA) carries specific amino acids to the ribosome, matching each amino acid to the correct codon on the mRNA via its complementary anticodon. mRNA carries the coding sequence, rRNA forms the structural and catalytic core of ribosomes, and snRNA is involved in pre-mRNA splicing.

Q43. Which term describes the process by which the nucleotide sequence of an mRNA is used to assemble a chain of amino acids?
A Transcription
B Replication
C Translation
D Transduction

Translation is the process by which ribosomes decode the sequence of codons in an mRNA to produce a polypeptide chain. Transcription produces mRNA from a DNA template, replication copies DNA, and transduction refers to signal transduction or viral gene transfer.

Q44. In prokaryotic cells, transcription and translation are coupled, meaning they occur:
A In the nucleus at the same ribosome
B Sequentially, with translation only starting after transcription is fully complete
C Simultaneously in the cytoplasm, with ribosomes binding the mRNA while it is still being transcribed
D In separate compartments, separated by the nuclear envelope

Because prokaryotes lack a nucleus, ribosomes can begin translating the 5' end of an mRNA while RNA polymerase is still elongating the 3' end in the cytoplasm. This simultaneous coupling is impossible in eukaryotes, where transcription occurs in the nucleus and the mRNA must be processed and exported before translation begins in the cytoplasm.

Q45. Which codon on an mRNA molecule signals the start of translation and codes for the amino acid methionine?
A UAA
B UAG
C AUG
D UGA

AUG is the universal start codon that initiates translation and encodes methionine. UAA, UAG, and UGA are all stop codons that signal termination of translation rather than addition of an amino acid.

Q46. Which of the following correctly describes the molecular composition of a single nucleotide?
A A five-carbon sugar, a phosphate group, and a nitrogenous base
B Two five-carbon sugars joined by a glycosidic bond
C A nitrogenous base and a phosphate group, with no sugar component
D A five-carbon sugar and a nitrogenous base only, without phosphate

A nucleotide consists of three components: a five-carbon (pentose) sugar, one or more phosphate groups, and a nitrogenous base. In DNA the sugar is deoxyribose; in RNA it is ribose. The phosphate group and sugar together form the backbone of the nucleic acid strand, while the base carries the genetic information.

Q47. RNA polymerase reads the template strand of DNA in which direction as it synthesizes an mRNA molecule?
A 5' to 3', producing an mRNA in the 3' to 5' direction
B 3' to 5', producing an mRNA in the 5' to 3' direction
C Either direction, depending on the promoter orientation
D From the centromere outward toward the telomere

RNA polymerase moves along the template (antisense) strand of DNA in the 3' to 5' direction and synthesizes the new mRNA strand in the 5' to 3' direction, following the same polarity rule as DNA polymerase. The resulting mRNA sequence matches the non-template (sense) strand, with U replacing T.

Q48. During translation, a stop codon in the A site of the ribosome causes:
A The ribosome to shift to a new reading frame and continue synthesis
B Release factors to bind the ribosome, resulting in hydrolysis and release of the completed polypeptide
C The 5' cap to be removed from the mRNA
D The small ribosomal subunit to dissociate first, followed by continued elongation

When a stop codon (UAA, UAG, or UGA) enters the A site, release factors bind instead of a tRNA. These proteins stimulate peptidyl transferase to hydrolyze the bond between the polypeptide and the P-site tRNA, releasing the finished protein. The ribosomal subunits then dissociate from the mRNA.

Q49. A promoter sequence in a gene primarily functions as:
A The coding region that specifies the last amino acid of the protein
B The binding site for RNA polymerase and associated transcription factors to initiate transcription
C The sequence that is translated into a signal peptide targeting the protein to the endoplasmic reticulum
D The region that marks where the mRNA transcript will be cleaved and polyadenylated

The promoter is a DNA sequence located upstream of the transcription start site where RNA polymerase and transcription factors bind to initiate transcription. The promoter is not translated into protein. Polyadenylation signals are separate downstream sequences, and signal peptides are encoded within the coding region.

Q50. During translation elongation, the tRNA carrying the growing polypeptide chain occupies which ribosomal site?
A The E (exit) site
B The A (aminoacyl) site
C The P (peptidyl) site
D The decoding site on the small subunit only

The P (peptidyl) site holds the tRNA attached to the growing polypeptide chain. The A site accepts the incoming aminoacyl-tRNA, and peptide bond formation transfers the polypeptide from the P-site tRNA to the A-site amino acid. After translocation, the now-empty tRNA moves to the E site and exits.

Q51. A missense mutation differs from a nonsense mutation in that a missense mutation:
A Involves deletion of nucleotides, while a nonsense mutation involves substitution
B Results in incorporation of a different amino acid, while a nonsense mutation produces a premature stop codon
C Always abolishes protein function, while a nonsense mutation rarely affects function
D Occurs exclusively in introns, while a nonsense mutation occurs in exons

A missense mutation is a base substitution that changes one codon to a different amino acid codon, potentially altering protein function. A nonsense mutation changes a coding codon to a stop codon (UAA, UAG, or UGA), prematurely terminating translation and usually producing a nonfunctional truncated protein. Neither type involves insertions or deletions.

Q52. In the trp operon of E. coli, the biosynthetic genes are switched off when tryptophan is abundant because:
A Excess tryptophan binds directly to RNA polymerase and blocks its movement
B Tryptophan acts as a corepressor by binding to the repressor protein, enabling the complex to bind the operator and block transcription
C The CAP-cAMP activator complex dissociates from the promoter in the presence of tryptophan
D High tryptophan levels cause the ribosome to read through the operator sequence

The trp operon encodes a repressible system. The repressor protein is inactive (aporepressor) by itself and cannot bind the operator. When tryptophan is plentiful, it binds the aporepressor as a corepressor, causing a conformational change that allows the repressor-tryptophan complex to bind the operator and block RNA polymerase, shutting off transcription of the tryptophan biosynthesis genes.

Q53. Which set of processing events is required to convert a eukaryotic pre-mRNA into a mature mRNA ready for export from the nucleus?
A Addition of introns and enzymatic removal of exons
B Addition of a 5' methylguanosine cap, addition of a poly-A tail at the 3' end, and removal of introns by splicing
C Translation of the signal sequence within the nucleus before export
D Phosphorylation of the 3' end and acetylation of the 5' end

Eukaryotic pre-mRNA processing includes three main steps: (1) a 7-methylguanosine cap is added to the 5' end, protecting the mRNA from degradation and aiding ribosome binding; (2) a poly-A tail of ~200 adenine nucleotides is added to the 3' end; and (3) introns are removed by spliceosomes and exons are joined. Exons, not introns, are retained in the mature mRNA.

Q54. The wobble hypothesis in translation refers to the ability of:
A Ribosomes to slide along mRNA without decoding every codon
B A single tRNA to recognize multiple codons by allowing non-Watson-Crick base pairing at the third codon position
C One tRNA to carry more than one type of amino acid depending on cellular conditions
D Peptidyl transferase to form peptide bonds in both directions

The wobble hypothesis, proposed by Crick, explains why the number of tRNA species is less than the 61 sense codons. The third base (3' end) of a codon can form non-standard base pairs with the first base of the anticodon, allowing one tRNA to decode multiple synonymous codons. This is possible because the geometry at the third codon position is more flexible (wobbles) than positions 1 and 2.

Q55. Telomerase solves a specific problem in eukaryotic DNA replication. What is that problem?
A RNA primers on the leading strand cannot be removed without creating a gap
B The lagging strand template cannot be read because of its 3' to 5' orientation
C The very ends of linear chromosomes shorten with each replication cycle because the primer at the 5' end of newly synthesized DNA cannot be replaced with DNA
D Topoisomerases cannot relieve supercoiling at telomeric regions

Because DNA polymerase requires a primer and can only extend in the 5' to 3' direction, the RNA primer at the very 5' end of a new strand leaves a gap when removed that cannot be filled, shortening the chromosome with each round of replication (the end-replication problem). Telomerase is a reverse transcriptase that uses its own RNA template to extend the 3' overhang of the parental strand, providing material for the lagging strand machinery to fill in and preventing critical sequence loss.

Q56. A silent (synonymous) mutation in the coding sequence of a gene is one that:
A Prevents RNA polymerase from transcribing the gene entirely
B Changes an amino acid to one with very similar properties, preserving protein structure
C Alters a nucleotide but still encodes the same amino acid due to the degeneracy of the genetic code
D Removes an entire exon from the final mRNA

The genetic code is degenerate: most amino acids are specified by more than one codon (synonymous codons). A silent mutation changes one nucleotide, typically at the third codon position (wobble position), but the new codon still encodes the same amino acid, so the protein sequence is unchanged. A conservative substitution (choice B) changes the amino acid but preserves similar chemistry and is a missense, not silent, mutation.

Q57. Which statement correctly distinguishes the leading strand from the lagging strand during DNA replication?
A The leading strand is synthesized 3' to 5'; the lagging strand is synthesized 5' to 3'
B The leading strand requires multiple RNA primers; the lagging strand requires only one
C The leading strand is synthesized continuously toward the replication fork; the lagging strand is synthesized discontinuously in Okazaki fragments directed away from the fork
D The leading strand uses DNA polymerase I; the lagging strand uses DNA polymerase III

Both strands are synthesized 5' to 3', but only one template strand runs in a direction that allows continuous synthesis toward the fork (leading strand). The other template runs antiparallel, so its complementary strand must be synthesized in short Okazaki fragments moving away from the fork (lagging strand), each requiring its own RNA primer. Both strands are synthesized primarily by DNA polymerase III in prokaryotes.

Q58. Which of the following statements about ribosomes is accurate?
A In eukaryotes, ribosomes are found exclusively inside the nucleus
B The large ribosomal subunit contains peptidyl transferase activity that catalyzes peptide bond formation, and this activity resides in the ribosomal RNA itself
C Eukaryotic ribosomes are smaller than prokaryotic ribosomes
D Each ribosome contains exactly one rRNA molecule and one mRNA molecule

Peptidyl transferase activity in the large ribosomal subunit is a ribozyme activity — it is carried out by the 23S rRNA (in prokaryotes) or 28S rRNA (in eukaryotes), not by ribosomal proteins. This was a landmark discovery showing RNA can function as an enzyme. Eukaryotic ribosomes are 80S (larger than the prokaryotic 70S), and ribosomes contain multiple rRNA molecules along with many proteins.

Q59. A mutation eliminates the operator sequence of the lac operon so that the lac repressor can no longer bind to it. Which outcome would most likely result?
A The structural genes of the lac operon would never be transcribed, even when lactose is present
B The lac structural genes would be transcribed constitutively regardless of whether lactose is present or absent
C The cell would upregulate repressor production from the lacI gene to compensate
D The CAP-cAMP complex would be unable to bind the promoter, preventing all transcription

The operator is the DNA sequence where the repressor binds to block transcription. If the operator is mutated so the repressor cannot bind, the repressor can never block RNA polymerase, so transcription of the structural genes (lacZ, lacY, lacA) proceeds at full rates at all times — a constitutively active (operator-constitutive) mutation. This is distinct from a lacI mutation that eliminates repressor protein; the effect is the same but the mechanism differs.

Q60. An enhancer sequence that regulates a gene is located 15,000 base pairs upstream of the gene's promoter. Which mechanism best explains how an activator protein bound to this enhancer can stimulate transcription initiation at the promoter?
A The activator protein slides processively along the DNA backbone from the enhancer to the promoter
B DNA looping brings the enhancer and promoter into close physical proximity, allowing the activator to contact the transcription initiation complex directly
C The activator releases a diffusible chemical signal that travels through the nucleus to the promoter
D The activator methylates the intervening DNA, causing it to condense and bring the two sites together

Enhancers can function at great distances from the promoter because the intervening DNA forms a loop, bringing the enhancer-bound activator protein into direct physical contact with the mediator complex and transcription factors assembled at the promoter. This looping has been directly visualized by chromatin conformation capture (3C) experiments. Diffusible signals and sliding models do not account for the specificity and kinetics of enhancer action.

Q61. If the 3' to 5' exonuclease proofreading activity of DNA polymerase were completely abolished by a mutation, the most immediate consequence would be:
A Okazaki fragments would remain unjoined, leaving permanent gaps in the lagging strand
B RNA primers could not be removed from newly synthesized strands
C The rate of incorporation of mismatched nucleotides would increase significantly, raising the overall mutation rate
D Replication forks would collapse because helicase depends on polymerase for recruitment

The 3' to 5' exonuclease activity of DNA polymerase detects and excises incorrectly paired nucleotides immediately after incorporation, reducing the error rate by approximately 100-fold. Eliminating this activity does not prevent primer synthesis, fragment joining (which is done by ligase), or helicase activity — those processes involve distinct enzymes. The direct result is a dramatic increase in the mutation rate of the newly replicated DNA.

Q62. A single eukaryotic gene produces three distinct protein isoforms in three different cell types. The gene is transcribed at similar rates in all three cell types, and all three proteins have the same N-terminus but differ in their internal and C-terminal sequences. Which mechanism most directly explains this observation?
A Three distinct promoters drive transcription with different start sites in each cell type
B Cell-type-specific post-translational modifications cleave the protein at different sites
C Cell-type-specific splicing factors direct alternative splicing of the pre-mRNA, joining different combinations of exons
D Different ribosome populations in each cell type skip different codons during translation

Alternative splicing allows different exon combinations from the same pre-mRNA to be joined, producing multiple mRNAs and thus multiple protein isoforms from a single gene. Because all three proteins share the same N-terminus, they likely share the first exon(s), with divergence arising from different internal exon combinations — a hallmark of alternative splicing. This is a widespread regulatory mechanism in eukaryotes; the human genome encodes far more proteins than genes largely due to this process.

Q63. Histone acetylation by histone acetyltransferase enzymes is generally associated with increased transcription because:
A Acetylation neutralizes the positive charge on lysine residues in histones, reducing electrostatic attraction to the negatively charged DNA and relaxing chromatin structure to allow transcription factor access
B Acetyl groups bind directly to RNA polymerase and increase its processivity along the template
C Acetylation removes phosphate groups from the DNA backbone, making it more flexible
D Acetylation converts constitutive heterochromatin into mitochondrial DNA accessible to the nucleus

Histones are rich in positively charged lysine and arginine residues that interact tightly with the negatively charged DNA backbone, compacting chromatin. Acetylation of lysine residues by histone acetyltransferases adds an acetyl group that neutralizes the positive charge. This weakens histone-DNA interaction, loosens nucleosome packing (euchromatin), and allows transcription factors and RNA polymerase to access the DNA. Histone deacetylases reverse this process, compacting chromatin and repressing transcription.

Q64. A researcher treats eukaryotic cells with a drug that specifically blocks addition of the 7-methylguanosine cap to newly synthesized mRNA. Which of the following would be the most likely consequence for the affected transcripts?
A Pre-mRNA splicing would be dramatically enhanced because the 5' cap normally inhibits spliceosome assembly
B The uncapped mRNAs would be rapidly degraded in the nucleus and cytoplasm and would not be efficiently translated
C Translation would proceed normally because the poly-A tail alone is sufficient to recruit ribosomes
D The mRNAs would accumulate indefinitely in the nucleus because only capped mRNAs can be exported

The 5' methylguanosine cap protects the mRNA from 5' to 3' exonuclease degradation and is recognized by cap-binding proteins that facilitate nuclear export, ribosome recruitment (via eIF4E), and translation initiation. Without the cap, mRNAs are rapidly degraded by cytoplasmic and nuclear exoribonucleases. While the poly-A tail also contributes to stability and translation, it cannot fully substitute for cap-dependent initiation; uncapped mRNAs are translated very inefficiently and are quickly turned over.

Q65. A nonsense mutation near the 5' end of an open reading frame is generally more deleterious than one near the 3' end. Which explanation best accounts for this difference in severity?
A Mutations near the 5' end are more likely to disrupt the Shine-Dalgarno ribosome binding sequence in eukaryotes
B A premature stop codon early in the sequence truncates the vast majority of the protein, almost certainly eliminating its folded structure and function, whereas a late stop codon produces a nearly full-length protein that may retain significant activity
C The 5' region of an mRNA is translated more slowly, giving the cell less time to degrade the aberrant mRNA before a toxic protein accumulates
D Ribosomes translate from 3' to 5', so stop codons near the 5' end are encountered last and affect the most recently assembled portion of the protein

A nonsense mutation introduces a premature stop codon that terminates translation early. A stop codon near the 5' end of the coding sequence produces a very short, truncated polypeptide that almost certainly cannot fold correctly or carry out the protein's function, because most of the sequence — including catalytic sites, binding domains, and structural elements — is never synthesized. A stop codon near the 3' end produces a near-complete protein missing only a few C-terminal residues, which may preserve the majority of function. Additionally, early premature stop codons often trigger nonsense-mediated mRNA decay (NMD), degrading the transcript entirely. The Shine-Dalgarno sequence is a prokaryotic feature, and ribosomes translate 5' to 3'.

Q66. What is the primary function of DNA ligase during DNA replication?
A Unwind the double helix ahead of the replication fork
B Synthesize short RNA primers to initiate new strands
C Join Okazaki fragments by forming phosphodiester bonds between them
D Proofread newly synthesized DNA and remove mismatched nucleotides

DNA ligase seals the nicks between Okazaki fragments on the lagging strand by forming phosphodiester bonds, completing the continuous sugar-phosphate backbone. Helicase (not ligase) unwinds the helix, and primase synthesizes RNA primers — a common distractor confusion.

Q67. During transcription, RNA polymerase reads the DNA template strand in which direction?
A 5' to 3', producing mRNA in the 3' to 5' direction
B 3' to 5', producing mRNA in the 5' to 3' direction
C 5' to 3', producing mRNA in the 5' to 3' direction
D 3' to 5', producing mRNA in the 3' to 5' direction

RNA polymerase moves along the template strand in the 3' to 5' direction, synthesizing the mRNA strand in the 5' to 3' direction — the same rule that applies to all nucleic acid synthesis. The mRNA sequence matches the non-template (coding) strand, with U replacing T.

Q68. Which mRNA codon universally signals the initiation of translation?
A UAA
B UGA
C AUG
D UAG

AUG is the start codon recognized by the initiator tRNA (carrying methionine) and marks the beginning of the open reading frame. UAA, UGA, and UAG are all stop codons that signal termination — not initiation.

Q69. What is the primary function of the poly-A tail added to eukaryotic mRNA during processing?
A It facilitates ribosome attachment at the 5' end of the mRNA
B It codes for a string of lysine residues at the protein's C-terminus
C It protects the mRNA from enzymatic degradation and aids in nuclear export
D It signals the spliceosome to begin removing introns

The poly-A tail (a string of adenine nucleotides added to the 3' end) protects mRNA from exonuclease degradation and assists in export from the nucleus to the cytoplasm. Ribosome attachment is facilitated by the 5' cap, not the poly-A tail — a frequent point of confusion.

Q70. Okazaki fragments are produced during DNA replication on which strand, and why?
A The leading strand, because DNA polymerase must pause frequently near the fork
B The lagging strand, because it is oriented so that synthesis must proceed away from the replication fork in short segments
C The template strand, because helicase disrupts hydrogen bonds in short bursts
D Both strands equally, because DNA polymerase works in both directions simultaneously

DNA polymerase can only add nucleotides in the 5' to 3' direction. On the lagging strand, this direction points away from the moving replication fork, so synthesis must restart repeatedly, producing short Okazaki fragments. The leading strand is synthesized continuously toward the fork.

Q71. Which RNA molecule is responsible for carrying a specific amino acid to the ribosome during translation?
A mRNA
B rRNA
C tRNA
D snRNA

Transfer RNA (tRNA) is charged with a specific amino acid by aminoacyl-tRNA synthetase and delivers it to the ribosome. rRNA forms the structural and catalytic core of the ribosome, while mRNA carries the coding instructions. snRNA is involved in pre-mRNA splicing.

Q72. The 5' cap added to eukaryotic pre-mRNA during processing consists of:
A A string of adenine nucleotides added by poly-A polymerase
B A modified 7-methylguanosine nucleotide attached in an unusual 5'-to-5' linkage
C A phosphorylated serine residue that protects the mRNA end
D A short sequence of uracil nucleotides that recruits translation factors

The 5' cap is a 7-methylguanosine added to the 5' end of pre-mRNA via an unusual 5'-to-5' triphosphate linkage. It protects mRNA from degradation and is recognized by translation initiation factors. The poly-A tail (adenine nucleotides) is a common distractor since both are mRNA processing events.

Q73. A mutation in the promoter region of a gene that reduces RNA polymerase binding affinity would most likely result in:
A Production of a truncated, nonfunctional protein due to early stop codons
B Decreased transcription initiation and lower levels of mRNA produced
C Incorrect splicing of introns, producing an aberrant protein product
D Increased translation efficiency because fewer ribosomes compete for the mRNA

The promoter is the binding site for RNA polymerase and associated transcription factors. Reduced polymerase affinity means transcription is initiated less frequently, yielding lower mRNA levels. Splicing errors would require a mutation in splice site sequences, not the promoter — a plausible distractor since both affect gene expression.

Q74. In the trp operon of E. coli, transcription is repressed when:
A Tryptophan is absent and binds directly to the operator sequence
B Tryptophan is abundant and acts as a corepressor by binding and activating the repressor protein
C The repressor protein binds to the ribosome and blocks translation of trp genes
D High glucose levels cause catabolite repression of the operator

The trp operon is a repressible operon. When tryptophan is plentiful, it serves as a corepressor by binding the inactive repressor protein, activating it so it can bind the operator and block transcription. This is the opposite logic from the lac operon — a common point of confusion in AP Biology.

Q75. Which histone modification would most likely increase transcription of a nearby gene?
A Methylation of histone H3K27, a modification associated with chromatin compaction
B Deacetylation of histone tails, which tightens histone-DNA electrostatic interactions
C Acetylation of histone tails, which neutralizes positive charges and loosens chromatin
D Addition of methyl groups to cytosine residues in the DNA sequence itself

Acetylation of histone tails neutralizes their positive charge, reducing affinity for the negatively charged DNA backbone. This loosens chromatin structure (euchromatin), making genes accessible to transcription machinery. Deacetylation has the opposite effect, and H3K27 methylation is associated with gene silencing.

Q76. During translation, the formation of peptide bonds between adjacent amino acids is catalyzed by:
A A protein enzyme located in the A site of the small ribosomal subunit
B The rRNA component of the large ribosomal subunit, functioning as a ribozyme
C Aminoacyl-tRNA synthetase in the cytoplasm prior to ribosome entry
D The mRNA template strand at the P site of the ribosome

Peptide bond formation is catalyzed by peptidyl transferase activity, which resides in the 23S (prokaryotic) or 28S (eukaryotic) rRNA of the large ribosomal subunit — making it a ribozyme (catalytic RNA). This is a key example used in AP Biology to illustrate RNA's catalytic versatility.

Q77. The wobble hypothesis, proposed to explain features of the genetic code, states that:
A Silent mutations at the third codon position explain why the code appears degenerate but is actually not
B A single tRNA can recognize multiple codons because the 5' anticodon base can form non-standard base pairs
C The genetic code evolved independently in different lineages, producing slight variations across species
D Stop codons have looser base-pairing requirements, allowing any release factor to recognize them

The wobble hypothesis explains that the base at the 5' end of the anticodon (wobble position) can pair with more than one base at the 3' position of the codon. This allows a single tRNA to recognize synonymous codons, which is why fewer than 61 tRNA types are needed for all sense codons.

Q78. If a DNA template strand reads 3'-ATGCGT-5', what is the correct sequence of the mRNA produced during transcription?
A 3'-ATGCGT-5'
B 5'-TACGCA-3'
C 5'-UACGCA-3'
D 3'-UACGCA-5'

RNA polymerase reads the template strand 3' to 5' and synthesizes mRNA 5' to 3', using complementary base pairing with U substituting for T. Template 3'-ATGCGT-5' pairs to produce mRNA 5'-UACGCA-3'. Choice B is incorrect because it uses T instead of U, a common error when students forget that mRNA uses uracil.

Q79. A nonsense mutation differs from a missense mutation in that a nonsense mutation:
A Changes one codon to a synonymous codon, preserving the amino acid sequence
B Converts an amino acid codon to a stop codon, prematurely terminating the polypeptide
C Inserts or deletes a nucleotide, altering the reading frame of the entire downstream sequence
D Alters the promoter sequence, reducing the rate of transcription initiation

A nonsense mutation changes an amino acid codon to a stop codon (UAA, UAG, or UGA), causing premature termination and a truncated, usually nonfunctional protein. A missense mutation substitutes one amino acid for another. Frameshift mutations (insertions/deletions) are a separate category and are often more severe.

Q80. Which of the following is an example of post-transcriptional gene regulation in eukaryotes?
A Histone acetylation near the promoter increasing RNA polymerase access
B Transcription factor binding to an enhancer element 10,000 base pairs upstream
C Variation in mRNA stability controlling how long transcripts persist in the cytoplasm
D DNA methylation at CpG islands silencing a gene in differentiated cells

Post-transcriptional regulation occurs after mRNA is made. mRNA stability is controlled by factors such as the poly-A tail length, 5' cap status, RNA-binding proteins, and miRNA. The other three choices all regulate transcription itself — they act before mRNA is produced, making them transcriptional (or epigenetic) regulation.

Q81. Why is primase required during DNA replication if DNA polymerase is the main replicating enzyme?
A DNA polymerase degrades RNA and requires primase to protect the template strand
B DNA polymerase can only extend an existing 3'-OH group and cannot begin a new strand de novo
C Primase stabilizes the replication fork by holding the two template strands apart
D DNA polymerase works only on the lagging strand and primase covers the leading strand

DNA polymerase requires a free 3'-OH group to add nucleotides. It cannot initiate a new strand on its own. Primase synthesizes a short complementary RNA primer that provides the 3'-OH needed for DNA polymerase to begin elongation. These primers are later removed and replaced with DNA.

Q82. A researcher treats cells with a drug that completely inhibits histone deacetylase (HDAC) enzymes. What is the most likely effect on transcription?
A Widespread decrease in transcription because histones cannot be recycled after each round of synthesis
B Widespread increase in transcription because histone tails remain acetylated, keeping chromatin in an open state
C No change in transcription because HDAC activity is redundant with other chromatin remodeling factors
D Selective silencing of housekeeping genes because they depend on histone deacetylation for baseline expression

HDAC enzymes remove acetyl groups from histone tails, restoring positive charges and compacting chromatin to repress transcription. Inhibiting HDACs allows acetyl groups to accumulate, maintaining an open chromatin state that promotes transcription. HDAC inhibitors are actually studied as anti-cancer agents for this reason.

Q83. Which of the following correctly describes the role of the signal recognition particle (SRP) in eukaryotic protein synthesis?
A It delivers aminoacyl-tRNA molecules to the ribosomal A site during elongation
B It recognizes a signal peptide on a growing polypeptide and directs the ribosome to the rough ER membrane
C It cleaves the N-terminal methionine after the polypeptide is fully synthesized
D It stabilizes the mRNA-ribosome complex at the AUG start codon during initiation

The SRP recognizes a hydrophobic signal peptide at the N-terminus of proteins destined for secretion or membrane insertion. It pauses translation and docks the ribosome on the rough ER, allowing the growing polypeptide to be threaded into the ER lumen. Proteins without a signal peptide are synthesized by free ribosomes in the cytosol.

Q84. A transcription factor's DNA-binding domain is mutated such that it can no longer bind to its specific promoter elements, but it can still dimerize with its normal binding partner. What is the most likely consequence?
A The mutant factor is rapidly degraded because unbound transcription factors are always unstable
B Target genes show decreased transcription because the mutant factor sequesters the binding partner without activating transcription
C RNA polymerase binds more efficiently to target promoters because the mutant factor no longer competes with it
D Translation of the binding partner's mRNA is blocked by the non-functional mutant factor

This describes a dominant negative mechanism. The mutant factor dimerizes with the functional partner (as normal), but the complex cannot bind DNA, effectively titrating out the working partner and reducing transcription below even basal levels. This is more disruptive than a simple loss-of-function because it also inactivates the wild-type copy.

Q85. In a somatic cell lineage where telomerase is completely absent, what is the predicted long-term consequence of repeated cell divisions?
A Uncontrolled cell proliferation because shortened telomeres activate proto-oncogenes adjacent to chromosome ends
B Progressive telomere shortening with each division, eventually triggering replicative senescence or apoptosis
C Increased genome-wide mutation rate because telomere loss destroys mismatch repair enzyme binding sites
D Immediate arrest of the cell cycle because DNA polymerase cannot replicate any chromosome without telomerase

DNA polymerase cannot fully replicate the extreme 3' end of the lagging strand (the end-replication problem), so telomeres shorten with each division when telomerase is absent. When telomeres become critically short, p53-mediated checkpoints trigger senescence or apoptosis, preventing genomic instability. Telomerase is active in germ cells and many cancer cells but normally inactive in somatic cells.

Q86. A loss-of-function mutation destroys a silencer element located 2,000 base pairs upstream of a eukaryotic gene. Which outcome is most likely?
A The gene is permanently silenced because the silencer is required for RNA polymerase recruitment to the promoter
B Transcription of the gene increases because repressor proteins can no longer be recruited to suppress the gene
C The promoter becomes methylated as a compensatory epigenetic response to restore silencing
D RNA polymerase loses its ability to locate the transcription start site without the silencer as a reference point

Silencer elements recruit repressor proteins or chromatin-modifying complexes that reduce transcription. Destroying the silencer removes this repressive signal, resulting in constitutively elevated transcription. This parallels the logic of operator mutations in bacterial operons — loss of a repressor binding site leads to constitutive expression.

Q87. How do microRNA (miRNA) molecules regulate gene expression in eukaryotic cells?
A By binding to promoter regions and physically blocking RNA polymerase from initiating transcription
B By methylating cytosine residues within the coding sequence of target genes, preventing elongation
C By base-pairing with complementary sequences in target mRNAs, leading to mRNA degradation or translational repression
D By competing with aminoacyl-tRNA for the ribosomal A site, globally stalling translation of all mRNAs

miRNA molecules are processed from hairpin RNA precursors and loaded into the RNA-induced silencing complex (RISC). The miRNA guides RISC to complementary sequences in the 3' UTR of target mRNAs. Depending on complementarity, this causes mRNA cleavage or translational repression — a form of post-transcriptional regulation. This is distinct from siRNA (which requires near-perfect complementarity) and from transcriptional regulation.

Q88. A scientist observes that a particular protein maintains constant cellular levels even when its gene is artificially induced to transcribe at 10 times the normal rate. Which explanation best accounts for this observation?
A The excess mRNA transcripts lack functional 5' caps and are degraded before ribosomes can translate them
B Post-translational regulation via increased protein degradation through the ubiquitin-proteasome pathway compensates for the elevated synthesis rate
C The additional mRNA molecules are mis-spliced due to spliceosome overload and produce no functional protein
D A ribozyme embedded in the 3' UTR cleaves excess mRNA whenever cellular protein concentration exceeds baseline

Gene expression can be regulated at multiple levels. If mRNA increases but protein stays constant, the bottleneck is post-translational. Accelerated degradation via the ubiquitin-proteasome system is a well-documented homeostatic mechanism. This illustrates why mRNA levels alone are poor predictors of protein abundance — a key AP Biology and proteomics concept.

Q89. In a eukaryotic cell with a ribosome mutation that prevents recognition of all three stop codons, what would most likely occur during translation?
A Proteins would be shorter than normal because the ribosome would terminate prematurely at sense codons
B Translation would stall permanently at the start codon because AUG recognition depends on functional stop codon machinery
C Ribosomes would translate through the 3' UTR and potentially continue until the mRNA ends or the ribosome falls off
D The cell would compensate by reassigning rare sense codons as alternative stop signals within one cell cycle

Stop codons are recognized by release factors (not tRNAs) that trigger polypeptide release. Without functional stop codon recognition, ribosomes would continue translating past the normal stop codon, reading through the 3' UTR and producing an abnormally long polypeptide. This runaway translation would produce aberrant proteins and likely trigger mRNA surveillance pathways like nonstop decay.

Q90. A prokaryotic gene is transcribed at the same high rate whether or not its inducer molecule is present. The inducer normally prevents a repressor protein from binding the operator. Given this phenotype, which mutation most likely explains constitutive expression?
A A gain-of-function mutation in the structural gene that makes the protein hyperactive
B A mutation in the operator sequence that prevents repressor binding regardless of inducer presence
C A mutation in the promoter that dramatically increases RNA polymerase binding affinity
D A loss-of-function mutation in the gene encoding the inducer molecule itself

If the gene is expressed at the same high rate with or without inducer, the repressor can never shut transcription off — this is constitutive expression. A mutation in the operator that prevents repressor binding achieves this regardless of inducer status. A promoter mutation strengthening polymerase binding could increase basal transcription but would not make the system inducer-independent in the same mechanistic way. This is analogous to lac operator constitutive mutants classic in molecular genetics.

Q91. What is the primary function of DNA ligase during DNA replication?
A Unwinds the double helix at the replication fork
B Joins adjacent Okazaki fragments on the lagging strand
C Adds RNA nucleotides to form a primer
D Removes mismatched nucleotides during proofreading

DNA ligase seals the nicks between Okazaki fragments on the lagging strand by forming phosphodiester bonds, producing a continuous strand. Helicase unwinds the double helix, primase synthesizes RNA primers, and proofreading is carried out by the 3' to 5' exonuclease activity of DNA polymerase III.

Q92. Which type of RNA molecule carries individual amino acids to the ribosome during translation?
A Messenger RNA (mRNA)
B Ribosomal RNA (rRNA)
C Transfer RNA (tRNA)
D Small nuclear RNA (snRNA)

Transfer RNA (tRNA) has a specific anticodon that base-pairs with the mRNA codon and carries the corresponding amino acid to the ribosome. mRNA carries the genetic message, rRNA is a structural and catalytic component of the ribosome, and snRNA participates in RNA splicing.

Q93. The central dogma of molecular biology describes the typical flow of genetic information as:
A RNA to DNA to Protein
B DNA to RNA to Protein
C Protein to RNA to DNA
D DNA to Protein to RNA

The central dogma states that genetic information flows from DNA (via transcription) to RNA, and then from RNA (via translation) to protein. While reverse transcriptase in retroviruses can copy RNA back to DNA, the standard directional flow is DNA to RNA to Protein.

Q94. During transcription, which nitrogenous base in RNA is paired with adenine in the DNA template strand?
A Thymine
B Guanine
C Cytosine
D Uracil

RNA contains uracil instead of thymine. When RNA polymerase reads an adenine on the DNA template strand, it incorporates uracil into the growing RNA strand. Thymine is found only in DNA and would be the complement of adenine in a DNA-DNA pairing.

Q95. A three-nucleotide sequence on mRNA that specifies a particular amino acid during translation is called a:
A Codon
B Anticodon
C Exon
D Promoter

A codon is the three-nucleotide unit on mRNA that is recognized during translation. The anticodon is the complementary three-nucleotide sequence on tRNA that base-pairs with the mRNA codon. Exons are protein-coding regions of a gene, and a promoter is a regulatory DNA sequence for transcription initiation.

Q96. The promoter region of a gene is best described as:
A The nucleotide sequence that directly encodes the amino acid sequence of a protein
B A DNA sequence upstream of a gene where RNA polymerase binds to initiate transcription
C A non-coding region excised from pre-mRNA before translation occurs
D A sequence that signals termination of the polypeptide chain at the ribosome

The promoter is a regulatory DNA sequence typically located upstream (5') of the transcription start site where RNA polymerase and associated transcription factors bind to initiate transcription. The coding sequence encodes the protein, introns are excised from pre-mRNA, and stop codons signal termination of translation.

Q97. The two strands of a DNA double helix are described as antiparallel. This means:
A One strand contains only purines while the other contains only pyrimidines
B The two strands run in opposite 5' to 3' orientations relative to each other
C Only one strand serves as a template during transcription
D The two strands are chemically identical but differ in nucleotide sequence

Antiparallel means that the two DNA strands are oriented in opposite directions: one runs 5' to 3' while its complement runs 3' to 5'. This arrangement is essential because DNA polymerase can only synthesize DNA in the 5' to 3' direction, which is why replication of the two strands proceeds differently.

Q98. When a ribosome encounters the stop codon UAA on an mRNA strand, the result is:
A The ribosome shifts to a new reading frame and continues translation
B A release factor binds the A site and the completed polypeptide is released from the ribosome
C The poly-A tail is added to the mRNA to signal successful translation
D The ribosome returns to the start codon and repeats translation

Stop codons (UAA, UAG, UGA) are not recognized by tRNAs but instead by release factor proteins that bind the A site of the ribosome. This triggers hydrolysis of the bond between the polypeptide and the tRNA in the P site, releasing the completed protein. The poly-A tail is added in the nucleus during pre-mRNA processing, not during translation.

Q99. A DNA template strand has the sequence 3'-TACGCCATT-5'. What would be the sequence of the mRNA transcribed from this template?
A 5'-AUGCGGUAA-3'
B 3'-AUGCGGUAA-5'
C 5'-UACGCCAUU-3'
D 3'-ATGCGGTAA-5'

RNA polymerase reads the DNA template strand 3' to 5' and synthesizes mRNA 5' to 3' using complementary base pairing (A-U, T-A, G-C, C-G). Reading 3'-TACGCCATT-5' produces 5'-AUGCGGUAA-3'. Choice B has the wrong directionality, choice C is simply a copy of the template with U substituted for T, and choice D would be a DNA sequence.

Q100. The genetic code is described as 'degenerate.' Which of the following best explains this property?
A Some codons can code for more than one amino acid depending on cellular conditions
B Multiple different codons can specify the same amino acid
C The reading frame can shift during translation to produce different proteins
D Some codons have lost function through evolutionary mutation

Degeneracy means that multiple codons can encode the same amino acid. For example, six different codons all specify leucine. This redundancy provides a buffer against the effects of certain mutations. The genetic code is unambiguous — each codon codes for only one amino acid — so choice A is incorrect. Reading frame shifts and evolutionary loss of function describe different phenomena.

Q101. In eukaryotic cells, the 5' methylguanosine cap and the 3' poly-A tail added to pre-mRNA primarily serve to:
A Signal intron boundaries so the spliceosome can remove them accurately
B Protect mRNA from degradation by nucleases and facilitate ribosome recognition
C Terminate transcription by causing RNA polymerase to dissociate from the template
D Prevent mature mRNA from being exported out of the nucleus prematurely

The 5' cap and 3' poly-A tail are protective modifications that stabilize the mRNA against exonuclease degradation and increase its half-life in the cytoplasm. The 5' cap also assists in ribosome binding during translation initiation. Intron boundaries are marked by specific splice site sequences recognized by snRNAs, not by the cap or tail.

Q102. The lagging strand during DNA replication is synthesized in short, discontinuous segments called Okazaki fragments because:
A DNA polymerase can only add new nucleotides in the 5' to 3' direction, which is away from the replication fork on the lagging strand template
B The lagging strand template contains more repetitive sequences that pause the polymerase
C Helicase moves away from the lagging strand, leaving less template exposed
D RNA primase cannot bind the lagging strand until both strands are fully unwound

DNA polymerase synthesizes DNA only in the 5' to 3' direction. On the lagging strand template (which runs 5' to 3' toward the fork), synthesis must proceed away from the fork. As helicase unwinds more DNA, new RNA primers must be laid down and new Okazaki fragments started, creating a discontinuous pattern. This is a direct consequence of polymerase directionality, not template sequence or helicase movement.

Q103. A mutation in the operator region of the E. coli lac operon prevents the repressor protein from binding to the DNA. Which of the following outcomes would result?
A Constitutive transcription of lac genes, even in the absence of lactose
B Complete and permanent silencing of the lac operon
C The operon becomes more sensitive to catabolite repression by glucose
D Overproduction of the repressor protein to compensate for its inability to bind

Normally the lac repressor binds the operator to block RNA polymerase and silence the operon when lactose is absent. If the operator is mutated so the repressor cannot bind, the block is permanently removed and the structural genes (lacZ, lacY, lacA) are transcribed constitutively regardless of lactose availability. This is a classic constitutive mutant. The mutation does not affect repressor synthesis, only its binding.

Q104. Post-translational modifications such as phosphorylation affect protein function primarily by:
A Permanently altering the nucleotide sequence of the gene that encodes the protein
B Changing the protein's three-dimensional shape and its capacity to interact with other molecules
C Directing the mRNA encoding the protein to specific ribosomes in the endoplasmic reticulum
D Preventing the protein from entering the proteasome for degradation

Phosphorylation adds a charged phosphate group to specific amino acid residues (often serine, threonine, or tyrosine), which can cause conformational changes in the protein. These shape changes can activate or inactivate enzymes, alter binding affinities, and regulate protein-protein interactions. Post-translational modifications act on the protein, not the gene sequence, and do not inherently block all degradation.

Q105. Transcription factors in eukaryotic cells are best described as:
A Enzymes that catalyze the synthesis of RNA using a DNA template
B Proteins that bind specific DNA sequences and regulate the rate of transcription initiation
C Small RNA molecules that base-pair with mRNA to inhibit translation
D Structural proteins that are integral components of the large ribosomal subunit

Transcription factors are regulatory proteins that recognize and bind specific DNA sequences (such as enhancers or promoter elements) and interact with RNA polymerase or the general transcription machinery to increase or decrease transcription rates. RNA polymerase itself catalyzes transcription. Small interfering RNAs and miRNAs block translation post-transcriptionally. Ribosomal proteins are structural components of ribosomes.

Q106. The Shine-Dalgarno sequence in prokaryotes and the Kozak sequence in eukaryotes share which functional similarity?
A Both signal RNA polymerase to terminate transcription at the correct position
B Both help position the ribosome on the mRNA strand at the correct start codon to initiate translation
C Both mark the boundaries between introns and exons during RNA splicing
D Both recruit DNA polymerase to the origin of replication during cell division

The Shine-Dalgarno sequence in prokaryotes base-pairs with the 16S rRNA of the small ribosomal subunit, and the Kozak sequence in eukaryotes is recognized by the 43S ribosomal pre-initiation complex. Both function to correctly position the ribosome at the AUG start codon to initiate translation. They are regulatory sequences in mRNA, not involved in transcription termination, splicing, or DNA replication.

Q107. A gene encoding a liver enzyme is expressed at high levels in liver cells but is completely silent in muscle cells, despite both cell types containing identical genomic DNA. This phenomenon is best explained by:
A Liver cells accumulate somatic mutations that activate the gene in each generation
B Cell-type-specific transcription factors and chromatin states differentially regulate which genes are expressed
C The gene undergoes alternative splicing only in liver cells, producing a functional transcript
D Horizontal gene transfer introduces additional copies of the gene exclusively into liver cells

Differential gene expression is the fundamental mechanism by which genetically identical cells develop distinct identities. Liver cells express transcription factors that activate liver-specific genes by binding enhancers and remodeling local chromatin, while muscle cells lack these factors and maintain the gene in a repressed, inaccessible chromatin state. Somatic mutations, alternative splicing, and horizontal gene transfer do not account for the systematic tissue-specific patterns seen across development.

Q108. A researcher introduces a point mutation that destroys the TATA box in the core promoter of a eukaryotic gene. Which outcome is most likely?
A Transcription rate increases because the repressive TATA element can no longer block RNA polymerase
B Transcription initiation is severely reduced because TFIID cannot properly position RNA polymerase II at the transcription start site
C The gene produces a truncated mRNA because RNA polymerase terminates prematurely
D The mRNA is produced normally but cannot be exported from the nucleus due to faulty capping

The TATA box is bound by TBP (TATA-binding protein), a component of the general transcription factor TFIID. TFIID binding is a critical step in assembling the pre-initiation complex that positions RNA polymerase II at the correct transcription start site. Mutating the TATA box prevents this assembly, severely reducing or abolishing transcription initiation. The TATA box is not a repressor element, and its mutation would not directly affect RNA polymerase processivity, termination, or mRNA capping.

Q109. Cells are treated with a drug that inhibits histone deacetylases (HDACs). Which effect on gene expression is most likely?
A Decreased transcription because histones remain acetylated and wrap DNA more tightly around nucleosomes
B Increased transcription of many genes because accumulation of acetyl groups on histones loosens chromatin and increases DNA accessibility
C No measurable change in transcription because histone modifications do not affect RNA polymerase II directly
D Decreased transcription because excess acetyl groups directly inhibit RNA polymerase II elongation

HDACs remove acetyl groups from histone tails. Acetylated histones carry less positive charge, reducing their attraction to the negatively charged DNA backbone and creating a more open (euchromatin) state accessible to transcription factors. When HDACs are inhibited, histones remain hyperacetylated, chromatin stays open, and transcription of many genes increases. Choice A incorrectly reverses the relationship between acetylation and chromatin compaction.

Q110. A gene encodes a 200-amino-acid protein. A point mutation at nucleotide position 417 in the coding sequence changes codon 139 from GGU to GGC. What is the most likely effect on the protein?
A A premature stop codon is introduced, producing a truncated 139-amino-acid protein
B No change in the amino acid sequence, because both GGU and GGC encode glycine
C An amino acid substitution at position 139 alters the protein's tertiary structure and function
D A frameshift occurs at position 139, producing a completely different amino acid sequence downstream

GGU and GGC are synonymous codons — both encode glycine. This type of nucleotide change is called a silent (or synonymous) mutation because it does not alter the amino acid sequence. No truncation occurs because no stop codon is created, and no frameshift occurs because a single base substitution (not an insertion or deletion) does not alter the reading frame. This example illustrates the protective effect of codon degeneracy.

Q111. MicroRNAs (miRNAs) regulate gene expression post-transcriptionally by:
A Methylating CpG dinucleotides in the promoter regions of target genes to silence transcription
B Base-pairing with complementary sequences in target mRNAs, leading to mRNA degradation or translational repression
C Catalyzing the removal of introns from pre-mRNA as part of the spliceosome complex
D Acetylating histone proteins to alter chromatin accessibility around target gene loci

miRNAs are short, single-stranded RNA molecules that associate with the RISC (RNA-induced silencing complex). The miRNA guides RISC to complementary sequences typically in the 3' UTR of target mRNAs. Depending on the degree of complementarity, this leads to mRNA cleavage and degradation or stalling of ribosomes (translational repression), reducing protein output. DNA methylation and histone acetylation are epigenetic mechanisms that operate at the chromatin level, not at the level of mRNA.

Q112. In E. coli, transcription of the lac operon is dramatically higher when both lactose is present AND glucose is absent, compared to when lactose alone is present. This synergistic increase is best explained by:
A The lac repressor is more efficiently removed from the operator when glucose is depleted from the medium
B Absence of glucose raises intracellular cAMP levels, allowing the CAP-cAMP complex to bind the lac promoter and strongly stimulate RNA polymerase binding
C RNA polymerase has inherently higher affinity for the lac promoter sequence when glucose transporters are inactive
D The operator region undergoes a structural conformation change in response to falling glucose concentration

The lac operon is under dual control: negative regulation by the lac repressor (relieved by allolactose) and positive regulation by catabolite activator protein (CAP). When glucose is scarce, adenylyl cyclase is more active, raising cAMP. cAMP binds CAP, and the CAP-cAMP complex binds an upstream activator site, recruiting RNA polymerase to the promoter much more effectively. Both controls must be permissive (repressor off, CAP on) for maximal transcription. The repressor-operator interaction is not directly affected by glucose concentration.

Q113. A frameshift mutation caused by a single nucleotide insertion near the 3' end of a coding sequence would generally have a less severe effect on protein function than a frameshift near the 5' end. The best explanation is:
A DNA repair mechanisms preferentially correct frameshift mutations located closer to the 3' end of genes
B A 3' frameshift alters the amino acid sequence of only a short stretch near the C-terminus, while most of the protein's original structure is preserved
C The 3' region of a coding sequence encodes the N-terminal domain, which contributes less to protein function than the C-terminal domain
D Ribosomes translate mRNA from 3' to 5', so a 3' insertion is encountered last and produces the least disruption

A frameshift mutation scrambles all codons from the point of insertion to the stop codon. If the insertion is near the 3' end of the coding sequence, only the last few amino acids (near the C-terminus) are altered before a stop codon is reached, preserving the vast majority of the protein's sequence and structure. A frameshift near the 5' end disrupts nearly the entire protein. Choice C incorrectly reverses the N/C terminus relationship — the 5' end of the coding sequence encodes the N-terminus.

Q114. A eukaryotic pre-mRNA contains 5 exons separated by 4 introns. Exon 3 is 60 nucleotides long. If alternative splicing can either include or exclude exon 3, which of the following predictions is most accurate?
A Two mRNA isoforms are produced: the protein including exon 3 has 20 additional amino acids compared to the isoform lacking exon 3
B Exclusion of exon 3 shifts the reading frame in all downstream exons, producing non-functional protein variants
C Including or excluding exon 3 changes the reading frame by 60 nucleotides, generating completely different downstream amino acid sequences
D The protein produced without exon 3 is always non-functional because every exon is required for correct protein folding

Since 60 nucleotides is divisible by 3, skipping exon 3 removes exactly 20 codons without altering the reading frame of the downstream exons. The result is two functional isoforms that differ by a 20-amino-acid segment. If the exon length were not divisible by 3, its exclusion would cause a frameshift. Choice B is incorrect because 60/3 = 20 with no remainder, so no frameshift occurs. Many proteins remain functional when internal segments are removed.

Q115. A researcher uses chromatin immunoprecipitation (ChIP) and finds high levels of the histone modification H3K27me3 at a specific gene locus in terminally differentiated cells but negligible levels at the same locus in embryonic stem cells. This finding most directly supports which conclusion?
A The gene accumulates somatic mutations at higher frequency in differentiated cells than in stem cells
B The gene is epigenetically silenced in differentiated cells through Polycomb repressive complex-mediated chromatin compaction
C The gene locus is more accessible to transcription factors in differentiated cells, driving higher expression
D Differentiated cells produce greater total amounts of histone H3 protein compared to stem cells

H3K27me3 (trimethylation of lysine 27 on histone H3) is a repressive chromatin mark deposited by the Polycomb Repressive Complex 2 (PRC2). High H3K27me3 at a locus is associated with compacted, inaccessible chromatin and transcriptional silencing. Finding this mark specifically in differentiated cells but not stem cells suggests the gene was active in the pluripotent state and has been epigenetically silenced upon differentiation. Choice C contradicts what H3K27me3 signifies — it is a repressive, not activating, mark.

Q116. Which of the following best describes the function of DNA polymerase III during replication?
A It unwinds the double helix at the replication fork
B It adds new nucleotides to the 3' end of a growing strand
C It removes RNA primers and replaces them with DNA
D It joins Okazaki fragments on the lagging strand

DNA polymerase III is the main replicative polymerase in prokaryotes and adds nucleotides to the 3' hydroxyl end of a growing strand, always synthesizing in the 5' to 3' direction. Unwinding is performed by helicase. Primer removal is done by DNA polymerase I. Joining of Okazaki fragments is performed by DNA ligase.

Q117. Which of the following RNA molecules carries amino acids to the ribosome during translation?
A mRNA
B rRNA
C tRNA
D snRNA

Transfer RNA (tRNA) acts as the adaptor molecule that carries specific amino acids to the ribosome. Each tRNA has an anticodon that base-pairs with a complementary mRNA codon. mRNA carries the genetic message, rRNA forms the structural and catalytic core of the ribosome, and snRNA is involved in RNA splicing.

Q118. The process by which genetic information flows from DNA to RNA to protein is known as:
A Replication
B Transduction
C The central dogma
D Transcription

The central dogma of molecular biology describes the flow of genetic information: DNA is transcribed into RNA, which is then translated into protein. Replication is the copying of DNA. Transduction refers to viral transfer of DNA between bacteria. Transcription is only one step of the overall process described by the central dogma.

Q119. Which strand of DNA serves as the template during transcription?
A The coding strand, read 5' to 3'
B The template strand, read 3' to 5'
C Both strands simultaneously
D Whichever strand has more adenine bases

RNA polymerase reads the template strand (also called the antisense strand) in the 3' to 5' direction, producing an mRNA that is complementary to the template and identical in sequence to the coding strand (except with uracil replacing thymine). The coding strand itself is not directly read during transcription.

Q120. What is the role of a promoter sequence in transcription?
A It codes for the first amino acid in the protein
B It signals where translation should begin on the mRNA
C It is the DNA region where RNA polymerase binds to initiate transcription
D It marks the site where the mRNA is cleaved and polyadenylated

A promoter is a DNA sequence upstream of a gene where RNA polymerase (and associated transcription factors) binds to initiate transcription. It does not code for amino acids or signal translation start — that role belongs to the start codon and ribosome binding site. Polyadenylation signals are separate sequences near the 3' end of the gene.

Q121. In translation, the ribosome moves along the mRNA in which direction?
A 3' to 5'
B 5' to 3'
C Either direction depending on codon sequence
D From the poly-A tail toward the 5' cap

Ribosomes read mRNA in the 5' to 3' direction, moving codon by codon from the start codon (AUG) toward the stop codon. Moving 3' to 5' would produce a nonsensical sequence. The poly-A tail is at the 3' end, so moving from the tail toward the 5' cap would be backwards.

Q122. Which of the following nitrogenous bases is found in RNA but not in DNA?
A Adenine
B Guanine
C Thymine
D Uracil

Uracil is found only in RNA, where it pairs with adenine. DNA uses thymine in place of uracil. Adenine and guanine are purines found in both DNA and RNA. Thymine is exclusive to DNA, making uracil the correct answer here.

Q123. During eukaryotic transcription, the 5' cap added to pre-mRNA serves to:
A Signal the ribosome where translation should terminate
B Protect the mRNA from degradation and assist in ribosome binding
C Provide the site where the poly-A tail is added
D Recruit spliceosomes to remove introns

The 5' cap (a modified guanine nucleotide) protects the mRNA from exonuclease degradation from the 5' end and is recognized by translation initiation factors that help the ribosome bind. It does not signal termination — stop codons do that. The poly-A tail is added at the 3' end. Spliceosome recruitment is directed by splice-site sequences within the pre-mRNA.

Q124. A gene contains the following DNA template strand sequence: 3'-TACGGTCAA-5'. What would be the corresponding mRNA sequence?
A 3'-AUGCCAGUU-5'
B 5'-AUGCCAGUU-3'
C 5'-TACGGTCAA-3'
D 3'-ATGCCAGTT-5'

RNA polymerase reads the template strand 3' to 5' and synthesizes mRNA 5' to 3' using complementary base pairing (A-U, T-A, G-C, C-G). Reading 3'-TACGGTCAA-5' produces 5'-AUGCCAGUU-3'. The other choices either have incorrect directionality, use DNA bases (T instead of U), or represent the wrong complementary strand.

Q125. In the trp operon of E. coli, transcription is repressed when:
A Tryptophan binds to RNA polymerase, blocking elongation
B Tryptophan acts as a corepressor by binding to the inactive repressor protein
C The operator is methylated and physically blocks the ribosome
D Glucose is absent from the environment

The trp operon is a repressible system. The repressor protein is normally inactive and cannot bind the operator. When tryptophan (the corepressor) is present in excess, it binds to the repressor, activating it so it can bind the operator and block transcription. This is distinct from the lac operon, which responds to glucose and lactose availability. Methylation and ribosome blocking are not the mechanism here.

Q126. Which of the following post-transcriptional modifications is required for the export of mature mRNA from the nucleus in eukaryotes?
A Addition of multiple AUG codons to the 5' end
B Removal of exons via the spliceosome
C Addition of a 5' cap and poly-A tail
D Conversion of uracil to thymine

Eukaryotic pre-mRNA must be processed — including addition of the 5' methylguanosine cap and the 3' poly-A tail, as well as splicing of introns — before it is exported from the nucleus as mature mRNA. Exons are retained, not removed. Uracil is never converted to thymine — that distinction is inherent to RNA vs. DNA. AUG codons are translated, not added as modifications.

Q127. A ribosome has three sites: A, P, and E. During elongation, the A site functions to:
A Hold the polypeptide chain before it exits the ribosome
B Catalyze peptide bond formation between amino acids
C Accept the incoming aminoacyl-tRNA
D Release the completed polypeptide

The A (aminoacyl) site accepts the next incoming aminoacyl-tRNA that matches the mRNA codon. The P (peptidyl) site holds the tRNA carrying the growing polypeptide, and peptide bond formation occurs between the amino acid in the A site and the chain in the P site. The E (exit) site is where the empty tRNA leaves the ribosome.

Q128. A mutation converts a codon from UAC to UAG. What is the most likely effect on the protein?
A A conservative amino acid substitution occurs, preserving protein function
B Translation terminates prematurely, producing a truncated protein
C The mutation is silent and the same amino acid is incorporated
D A different amino acid is inserted, potentially altering function

UAG is one of the three stop codons (UAA, UAG, UGA). When a sense codon mutates into a stop codon, it is called a nonsense mutation, causing premature termination of translation and producing a truncated, typically nonfunctional protein. This is distinct from a missense mutation (different amino acid) or a silent mutation (same amino acid due to degeneracy).

Q129. Which of the following correctly describes the relationship between gene expression and histone acetylation?
A Histone acetylation adds positive charges, tightening DNA wrapping and silencing genes
B Histone acetylation neutralizes positive charges on histones, loosening chromatin and promoting transcription
C Histone acetylation directly binds transcription factors to the TATA box
D Histone acetylation methylates cytosine residues in the promoter region

Histones are positively charged proteins that bind tightly to negatively charged DNA. Acetylation of lysine residues on histone tails neutralizes their positive charge, weakening DNA-histone interactions and opening chromatin (euchromatin). This makes DNA more accessible to transcription machinery, promoting gene expression. Methylation of histones or DNA is a separate epigenetic modification, and acetylation does not directly recruit transcription factors to the TATA box.

Q130. Wobble base pairing in translation refers to:
A Mismatches in the Watson-Crick base pairs within the mRNA double helix
B Flexible base pairing at the third position of the codon, allowing one tRNA to recognize multiple codons
C The ability of ribosomes to skip over stop codons under certain conditions
D Non-standard pairing between the 5' cap and the ribosome

Wobble base pairing occurs at the third (3') position of the mRNA codon, which corresponds to the first (5') position of the tRNA anticodon. This position tolerates non-Watson-Crick base pairs (e.g., G-U pairing), allowing a single tRNA to recognize multiple synonymous codons. This explains why there are fewer tRNA molecules than codons. It has nothing to do with skipping stop codons or ribosome-cap interactions.

Q131. In prokaryotes, the Shine-Dalgarno sequence on mRNA is important because it:
A Codes for the initiator methionine amino acid
B Base-pairs with the 16S rRNA of the small ribosomal subunit to position the ribosome at the start codon
C Recruits sigma factor to initiate transcription
D Signals the end of the open reading frame

The Shine-Dalgarno sequence is a purine-rich region on prokaryotic mRNA located 5-10 nucleotides upstream of the AUG start codon. It base-pairs with a complementary sequence on the 16S rRNA of the 30S ribosomal subunit, positioning the ribosome correctly over the start codon to initiate translation. Sigma factor is involved in transcription initiation, not translation. Stop codons signal the end of the open reading frame.

Q132. Which of the following statements about the genetic code is correct?
A Each codon can specify more than one amino acid
B The code is non-overlapping: each nucleotide belongs to only one codon
C The code differs significantly between prokaryotes and eukaryotes
D Each amino acid is encoded by exactly one unique codon

The genetic code is read in non-overlapping triplets — each nucleotide is part of only one codon. The code is also degenerate (redundant), meaning multiple codons can specify the same amino acid, so most amino acids have more than one codon. The genetic code is nearly universal, with very minor variations between organisms. A single codon specifies only one amino acid (unambiguous), but one amino acid may be coded by several codons.

Q133. A scientist treats cells with a drug that prevents the phosphorylation of eIF-4E, a translation initiation factor. What is the most likely consequence?
A mRNA is degraded before it can be translated
B Transcription is globally increased to compensate
C Cap-dependent translation initiation is reduced, decreasing protein synthesis
D Ribosomes shift to reading from internal ribosome entry sites exclusively

eIF-4E is the cap-binding protein that recognizes the 5' methylguanosine cap of eukaryotic mRNA. Its phosphorylation promotes translation initiation by facilitating assembly of the translation initiation complex. Inhibiting its phosphorylation impairs cap-dependent translation, reducing overall protein synthesis. While some mRNAs can use IRES-dependent translation, this is not the default pathway for most cellular mRNAs, so the dominant effect is reduced translation.

Q134. During DNA replication, Okazaki fragments are necessary because:
A DNA polymerase can only synthesize short fragments before falling off the template
B Helicase unwinds DNA in short segments, allowing only brief windows of synthesis
C DNA polymerase can only add nucleotides in the 5' to 3' direction, but the lagging strand template runs 3' to 5' relative to fork movement
D The lagging strand template is single-stranded and must be stabilized before synthesis

Both strands of DNA are synthesized 5' to 3' by DNA polymerase. The leading strand is synthesized continuously because its template runs 3' to 5' in the direction of fork movement. However, the lagging strand template runs 5' to 3' in the direction of fork movement, so DNA polymerase must repeatedly reinitiate synthesis in short fragments (Okazaki fragments) moving away from the fork. This is a fundamental constraint of polymerase directionality, not processivity or helicase activity.

Q135. A gene is found to produce the same protein in both liver and muscle cells despite having different histone modification patterns at its locus in each cell type. Which conclusion is best supported?
A Epigenetic modifications always silence gene expression regardless of cell type
B The protein must be essential to cell survival, overriding epigenetic regulation
C Gene expression can be maintained by multiple regulatory mechanisms; histone modifications alone do not determine transcriptional output
D The gene must lack a promoter and is regulated entirely by enhancers

This scenario illustrates that gene expression is governed by a combinatorial system of regulatory inputs — transcription factor binding, chromatin remodeling, enhancer activity, and histone modifications. The same transcriptional outcome can be achieved through different chromatin states if other activating signals are present. It would be incorrect to conclude that epigenetic modifications always silence genes, or that a single mechanism (like promoters vs. enhancers) fully controls expression.

Q136. Which of the following best explains why a nonsense mutation early in a coding sequence is generally more deleterious than one near the 3' end of the same gene?
A Early nonsense mutations affect more codons because the genetic code is read from the 3' end
B A premature stop codon near the 5' end truncates most of the protein, likely eliminating its functional domain, while a late stop codon leaves most of the protein intact
C Nonsense mutations near the 5' end are more likely to be corrected by DNA repair mechanisms
D mRNA with early stop codons is more efficiently translated, producing more truncated protein

A nonsense mutation introduces a premature stop codon. If it occurs early in the coding sequence, only a small N-terminal fragment is made — most functional domains, binding sites, and structural regions are absent. A late nonsense mutation produces a protein missing only a short C-terminal segment, which may still fold and function near-normally. Nonsense-mediated decay (NMD) may degrade both transcripts, but the protein truncation effect is the key conceptual answer here.

Q137. A cell line is found to have a mutation in the gene encoding a subunit of the mediator complex. Which of the following outcomes is most expected?
A Only housekeeping genes will be affected because mediator is non-specific
B Broad disruption of transcription of many genes, since mediator transmits signals from enhancer-bound activators to RNA polymerase II
C Increased transcription of genes with strong promoters, because mediator normally represses basal transcription
D Loss of splicing efficiency because mediator coordinates pre-mRNA processing

The mediator complex is a large multi-subunit co-activator that bridges enhancer-bound transcriptional activators with the RNA polymerase II pre-initiation complex at promoters. A loss-of-function mutation in mediator would impair signal transduction from activators to the transcription machinery, broadly reducing inducible gene expression across many loci. Mediator promotes — not represses — transcription in response to activators, and it is not directly involved in splicing.

Q138. MicroRNAs (miRNAs) regulate gene expression post-transcriptionally primarily by:
A Inserting into the genome near target genes and blocking RNA polymerase binding
B Base-pairing with complementary sequences in the 3' UTR of target mRNAs, leading to mRNA degradation or translational repression
C Methylating histone H3 at lysine 9 to silence chromatin at target loci
D Competing with ribosomes for binding to the 5' cap of mRNAs

MicroRNAs are short (~22 nt) non-coding RNAs that, as part of the RISC complex, base-pair with complementary sequences typically in the 3' untranslated region (3' UTR) of target mRNAs. Depending on the degree of complementarity, this leads to mRNA cleavage and degradation or translational inhibition without degradation. miRNAs do not insert into the genome, do not directly methylate histones, and do not compete for the 5' cap.

Q139. Which of the following scenarios would most likely result in constitutive (always-on) expression of a normally inducible gene?
A A point mutation in the coding sequence that changes one amino acid in the active site
B A deletion of the operator sequence in a prokaryotic repressible operon
C A loss-of-function mutation in the gene encoding the activator protein for the operon
D Methylation of cytosines in the promoter CpG island

In a repressible operon, the repressor protein normally binds to the operator to block transcription when the repressor is active. If the operator is deleted, the repressor has no binding site and cannot inhibit transcription — the gene is expressed constitutively regardless of repressor status. A loss-of-function mutation in an activator would reduce expression, not cause constitutive expression. Promoter methylation typically silences expression. A coding-sequence mutation alters protein function, not regulation.

Q140. During replication, telomeres at chromosome ends cannot be fully replicated by DNA polymerase alone. Which of the following best explains the molecular basis of this limitation?
A DNA polymerase lacks the processivity to copy repetitive sequences
B DNA polymerase requires a 3' OH group to extend a primer, so the lagging strand's outermost RNA primer cannot be replaced once removed, leaving a gap at the 5' end of the daughter strand
C Helicase cannot unwind the G-quadruplex structures at telomeres without accessory proteins
D The high GC content of telomeric repeats causes polymerase stalling

This is the end-replication problem. DNA polymerase requires a pre-existing 3' OH (on an RNA primer) to begin synthesis. On the lagging strand, the terminal RNA primer at the very end of the chromosome is removed but cannot be replaced with DNA because there is no upstream 3' OH to extend. This leaves a progressively shorter 5' end on each daughter strand, causing telomere shortening with each division. Telomerase addresses this using its internal RNA template.

Q141. Which enzyme is directly responsible for synthesizing a new RNA strand during transcription?
A DNA polymerase
B RNA polymerase
C Primase
D Helicase

RNA polymerase reads the DNA template strand and synthesizes a complementary RNA strand. DNA polymerase synthesizes DNA, not RNA. Primase is a type of RNA polymerase but its role is specifically to synthesize short RNA primers during DNA replication, not to carry out transcription.

Q142. RNA polymerase reads the DNA template strand in which direction during transcription?
A 5' to 3'
B 3' to 5'
C Either direction, depending on the promoter location
D Both directions simultaneously to increase efficiency

RNA polymerase reads the template strand in the 3' to 5' direction, which allows it to synthesize the new RNA strand in the 5' to 3' direction. This is the same directionality rule followed by DNA polymerase. Option A describes the direction of RNA synthesis, not template reading.

Q143. Which mRNA codon signals the start of translation and codes for the amino acid methionine?
A UAA
B AUG
C UAG
D UGA

AUG is the universal start codon that initiates translation and codes for methionine. UAA, UAG, and UGA are all stop codons that signal the termination of translation rather than the beginning.

Q144. What is the primary function of transfer RNA (tRNA) during translation?
A Carrying the genetic message from the nucleus to the ribosome
B Catalyzing peptide bond formation between adjacent amino acids
C Delivering specific amino acids to the ribosome by matching anticodons to mRNA codons
D Unwinding the mRNA secondary structure to expose codons for reading

tRNA molecules each carry a specific amino acid and contain an anticodon that base-pairs with a complementary mRNA codon, ensuring the correct amino acid is added at each position. Carrying the genetic message from the nucleus is the role of mRNA. Peptide bond formation is catalyzed by ribosomal RNA (rRNA) acting as a ribozyme.

Q145. In eukaryotic pre-mRNA processing, the sequences removed during RNA splicing are called:
A Exons
B Introns
C Codons
D Promoters

Introns are intervening sequences that are removed from the pre-mRNA by the spliceosome and are not present in the mature mRNA. Exons are the expressed sequences that are retained and joined together. Promoters are DNA sequences upstream of the gene that control transcription initiation.

Q146. A promoter sequence in DNA functions primarily to:
A Encode the first amino acid of the resulting protein
B Signal the end of transcription
C Provide a binding site for RNA polymerase to initiate transcription
D Code for the ribosomal RNA used during translation

The promoter is a regulatory DNA sequence located upstream of a gene where RNA polymerase (and associated transcription factors in eukaryotes) binds to initiate transcription. Termination signals are found at the 3' end of the gene. The promoter itself is not transcribed into protein.

Q147. DNA polymerase can only synthesize new DNA strands in which direction?
A 3' to 5'
B Either direction, depending on which strand is being copied
C 5' to 3'
D From the centromere outward toward the telomeres

DNA polymerase adds new nucleotides only to the 3' hydroxyl end of a growing strand, meaning synthesis always proceeds in the 5' to 3' direction. This constraint is why the lagging strand must be synthesized discontinuously as Okazaki fragments. Option A describes the direction in which the template is read, not the direction of synthesis.

Q148. Which nitrogenous base is found in RNA but NOT in DNA?
A Adenine
B Guanine
C Thymine
D Uracil

Uracil replaces thymine in RNA. Both adenine and guanine are purines found in both DNA and RNA. Thymine is found in DNA but not in RNA — in RNA, the methyl group present on thymine is absent, giving uracil. This structural difference also affects base-pairing: uracil pairs with adenine just as thymine does.

Q149. Okazaki fragments are synthesized during DNA replication because:
A DNA helicase can only unwind the double helix in the 5' to 3' direction
B DNA polymerase can only add nucleotides in the 5' to 3' direction, requiring the lagging strand to be synthesized discontinuously
C RNA primase cannot synthesize primers on the leading strand template
D Multiple origins of replication fire simultaneously, creating gaps between replicons

Because DNA polymerase can only synthesize in the 5' to 3' direction, it can extend the leading strand continuously as the replication fork opens. However, the lagging strand runs antiparallel, so it must be synthesized in short fragments (Okazaki fragments), each requiring a new RNA primer, in the direction away from the fork. These fragments are later joined by DNA ligase.

Q150. The 5' methylguanosine cap added to eukaryotic mRNA primarily functions to:
A Signal the ribosome where translation should terminate
B Protect the mRNA from degradation and facilitate ribosomal recognition during translation initiation
C Allow the mRNA to be transported back into the nucleus for storage
D Serve as the binding site for the spliceosome during intron removal

The 5' cap protects the mRNA from degradation by exonucleases and is recognized by initiation factors that recruit the small ribosomal subunit to begin scanning for the start codon. The poly-A tail at the 3' end also aids in stability. Neither the cap nor the poly-A tail is involved in splicing, which is directed by specific sequences at intron-exon boundaries.

Q151. Histone acetylation by acetyltransferase enzymes generally leads to increased transcription because:
A Acetyl groups attract DNA-condensing proteins that open chromatin
B Acetyl groups neutralize the positive charges on histone tails, reducing their attraction to negatively charged DNA and loosening chromatin structure
C Acetylation directly activates RNA polymerase by binding to its active site
D Acetyl groups cross-link adjacent nucleosomes, stabilizing the open chromatin state permanently

Histone proteins carry positively charged lysine residues in their tails that tightly associate with the negatively charged DNA backbone. Acetylation of these lysines neutralizes the positive charge, weakening histone-DNA interactions and allowing chromatin to adopt a more open conformation accessible to transcription machinery. This is an example of epigenetic regulation without changing the DNA sequence.

Q152. A nonsense mutation differs from a missense mutation in that a nonsense mutation:
A Changes one amino acid to a chemically similar amino acid, preserving protein function
B Converts an amino acid-coding codon into a stop codon, causing premature termination of translation
C Deletes a single nucleotide, shifting the downstream reading frame
D Occurs only in the non-coding intron regions of a gene

A nonsense mutation changes a codon that specifies an amino acid into one of the three stop codons (UAA, UAG, or UGA), causing the ribosome to terminate translation prematurely and produce a truncated, typically non-functional protein. A missense mutation changes one amino acid to a different one. A frameshift results from insertions or deletions of nucleotides not in multiples of three.

Q153. The genetic code is described as 'degenerate.' This means that:
A Multiple codons can encode the same amino acid
B Some codons do not specify any amino acid or stop signal
C The genetic code varies significantly among different species
D A single codon can encode multiple different amino acids depending on context

Degeneracy means that most amino acids are encoded by more than one codon. For example, leucine is encoded by six different codons. The third position of a codon often varies without changing the amino acid (called the wobble position). This degeneracy provides a buffer against point mutations. The genetic code itself is nearly universal across all organisms, which is strong evidence for a common ancestor.

Q154. In prokaryotes, the sigma factor associated with RNA polymerase is required for which step of transcription?
A Elongation of the RNA strand along the template
B Termination of transcription at rho-dependent termination sites
C Promoter recognition and transcription initiation
D Covalent assembly of RNA polymerase core subunits

The sigma factor allows RNA polymerase holoenzyme to recognize and bind specifically to promoter sequences (such as the -10 and -35 elements in prokaryotes). Once transcription initiates and the polymerase clears the promoter, the sigma factor dissociates, and the core enzyme continues elongation on its own. Different sigma factors allow recognition of different promoter classes under varying environmental conditions.

Q155. In the lac operon, catabolite activator protein (CAP) bound to cAMP acts as a positive regulator. When glucose is absent and cAMP levels are high, CAP-cAMP:
A Directly inactivates the lac repressor, allowing constitutive operon expression
B Binds to a site upstream of the promoter and enhances RNA polymerase binding, increasing transcription of lac genes
C Directly removes the repressor from the operator by competitive binding
D Methylates the operator sequence, preventing repressor binding

CAP-cAMP is a positive regulator that binds to the CAP site upstream of the lac promoter and makes contact with RNA polymerase, stabilizing its binding and dramatically increasing transcription. This mechanism ensures the lac genes are maximally expressed only when glucose (preferred carbon source) is absent AND lactose is present. It is separate from and additive with repressor regulation.

Q156. Which of the following is an example of post-translational modification that directly alters protein function?
A Addition of a 7-methylguanosine cap to the 5' end of mRNA
B Removal of introns from pre-mRNA by the spliceosome
C Phosphorylation of specific serine residues in a signaling protein
D Addition of a poly-A tail to the 3' end of mRNA

Post-translational modifications occur after translation on the completed protein. Phosphorylation of serine (or threonine or tyrosine) residues by kinases can activate or inhibit protein function, alter protein localization, or mark proteins for degradation. The 5' cap, splicing, and poly-A tail are all mRNA processing events that occur before or during translation, not after.

Q157. During translation elongation, GTP hydrolysis by elongation factors is required for:
A Peptide bond formation between amino acids in the A and P sites of the ribosome
B Delivery of aminoacyl-tRNA to the A site and translocation of the ribosome along the mRNA
C Release of the finished polypeptide chain from the ribosome at a stop codon
D Initial assembly of the large and small ribosomal subunits on the mRNA

EF-Tu (in prokaryotes) or eEF1A (in eukaryotes) delivers aminoacyl-tRNA to the A site in a GTP-dependent manner. After proofreading and accommodation, peptide bond formation occurs (catalyzed by rRNA). EF-G or eEF2 then uses GTP hydrolysis to drive translocation, moving the ribosome three nucleotides along the mRNA. Peptide bond formation itself does not require GTP — it is driven by the ribosome's ribozyme activity.

Q158. DNA polymerase's proofreading activity reduces the mutation rate during replication primarily by:
A Scanning newly synthesized DNA for UV-induced thymine dimers and excising them
B Using 3' to 5' exonuclease activity to remove a mismatched nucleotide from the 3' end of the growing strand
C Resynthesizing both strands of the template if a mismatched base pair is detected
D Immediately recruiting mismatch repair proteins after each nucleotide addition event

DNA polymerase has an intrinsic 3' to 5' exonuclease activity that allows it to remove the most recently added nucleotide if it is mismatched. The polymerase then reinserts the correct nucleotide before continuing. This proofreading reduces the error rate by approximately 100-fold. UV damage repair is handled by nucleotide excision repair, a separate pathway. Mismatch repair is a post-replication process distinct from proofreading.

Q159. Attenuation in the trp operon is a regulatory mechanism in which transcription terminates prematurely when tryptophan is abundant. This occurs because:
A High tryptophan activates the trp repressor, which physically blocks RNA polymerase mid-elongation
B Ribosomes translating a leader peptide move rapidly through two Trp codons when charged tRNA-Trp is available, allowing a termination hairpin to form in the nascent RNA
C High tryptophan concentrations promote methylation of the DNA template, stalling RNA polymerase
D Excess tryptophan activates RNase enzymes that degrade the nascent mRNA before elongation completes

Attenuation couples translation to transcription in prokaryotes. The trp operon leader sequence contains two adjacent tryptophan codons. When tryptophan is abundant, charged tRNA-Trp is plentiful, ribosomes translate rapidly and stall at a position that allows a termination hairpin to form, aborting transcription. When tryptophan is scarce, ribosomes stall at the Trp codons, allowing an antitermination structure to form instead, permitting continued transcription. This is distinct from repressor-operator control, which is a separate regulatory layer.

Q160. RNA interference (RNAi) silences gene expression primarily by:
A Blocking RNA polymerase from binding to the promoter of the target gene in the nucleus
B Incorporating small RNA sequences into the RISC complex, which then cleaves or represses translation of complementary mRNA targets
C Methylating cytosines in the promoter region of the target gene to prevent transcription factor binding
D Physically preventing ribosomal subunit assembly on the 5' cap of the target mRNA

In RNAi, double-stranded RNA is processed by Dicer into small interfering RNAs (siRNAs) or microRNAs (miRNAs) are loaded into the RNA-Induced Silencing Complex (RISC). The single-stranded guide RNA directs RISC to complementary mRNA sequences, leading to mRNA cleavage (siRNA, perfect complementarity) or translational repression (miRNA, imperfect complementarity). RNAi operates post-transcriptionally in the cytoplasm, not by blocking transcription itself.

Q161. Telomerase solves the 'end replication problem' in eukaryotes by:
A Recruiting DNA ligase to join Okazaki fragments at the very ends of chromosomes
B Using an internal RNA template to extend the 3' overhang of the parental strand, thereby providing a template for lagging strand synthesis at chromosome ends
C Synthesizing a continuous leading strand that extends all the way to the chromosome terminus without a primer
D Preventing exonuclease degradation of the 5' recessed end of newly synthesized strands

Without telomerase, the outermost RNA primer on the lagging strand cannot be replaced, so chromosomes shorten with each replication cycle. Telomerase is a reverse transcriptase that carries its own RNA template (AAUCCC in humans) and uses it to add repetitive TTAGGG sequences to the 3' overhang of the chromosome end. This extended 3' template then allows primase and DNA polymerase to synthesize additional lagging strand sequence, maintaining chromosome length. Telomerase is highly active in germ cells and many cancer cells.

Q162. In prokaryotes, ribosomes can begin translating an mRNA while it is still being transcribed. This coupled transcription-translation is impossible in eukaryotes primarily because:
A Eukaryotic ribosomes are too large to associate with mRNA before transcription is complete
B Eukaryotic transcription occurs in the nucleus while translation occurs in the cytoplasm, and mRNA must be processed before export
C Eukaryotic mRNA is inherently unstable and would be degraded before translation could begin
D Eukaryotic RNA polymerase synthesizes mRNA too slowly for ribosomes to associate

Prokaryotes lack a nuclear envelope, so ribosomes in the cytoplasm can directly access nascent mRNA emerging from RNA polymerase. In eukaryotes, transcription occurs in the nucleus and the pre-mRNA must undergo processing (5' capping, splicing, and 3' polyadenylation) before being exported through nuclear pores to the cytoplasm for translation. This spatial and temporal separation prevents coupled transcription-translation.

Q163. Female mammals heterozygous for an X-linked coat color gene often display a mosaic (patchy) phenotype because:
A Each cell randomly and permanently inactivates one X chromosome early in development, and all descendant cells maintain the same inactive X, creating patches of cells expressing each allele
B Both X chromosomes are partially active in every cell, each contributing approximately 50% of the gene product
C The inactive X chromosome is gradually degraded in roughly half of somatic cells over the organism's lifetime
D X-inactivation occurs only after differentiation, so different tissue types independently choose which X to silence

X-inactivation (lyonization) occurs early in embryonic development. Each cell independently and randomly inactivates one X chromosome, which condenses into a Barr body. This inactivated state is epigenetically maintained through all subsequent cell divisions. Because patches of tissue derive from single progenitor cells, a mosaic pattern of expression results — each patch expresses only the allele on the active X chromosome of its founding cell. Calico cat coloration is a classic example.

Q164. A riboswitch is a regulatory element found in the 5' untranslated region of certain bacterial mRNAs. Riboswitches control gene expression by:
A Binding transcription factors that then recruit RNA polymerase to the upstream promoter
B Directly binding specific small metabolite molecules, causing RNA conformational changes that affect transcription termination or translation initiation
C Encoding short regulatory peptides that inhibit ribosome assembly on the downstream coding sequence
D Acting as decoy sequences that sequester repressor proteins away from the operator

Riboswitches are structured RNA elements that can directly sense small molecules (such as amino acids, vitamins, or nucleotides) without protein intermediaries. When the metabolite is abundant, it binds the aptamer domain of the riboswitch, inducing a conformational change in the expression platform that can either form a transcription termination hairpin or sequester the Shine-Dalgarno sequence to block ribosome binding. This is a form of direct metabolite-responsive gene regulation at the RNA level.

Q165. A transcriptional coactivator protein is knocked out, and expression of dozens of unrelated genes across different tissues is abolished. The coactivator does not bind DNA directly. The most likely explanation is that the coactivator:
A Methylates histones at gene promoters, a modification that is essential for RNA polymerase II recruitment at all active genes
B Bridges gene-specific DNA-bound activators and the general transcription machinery, facilitating assembly of the preinitiation complex at many different promoters
C Is a structural subunit of RNA polymerase II whose loss prevents the enzyme from translocating along the template
D Degrades transcriptional repressor proteins, freeing operator sequences for constitutive transcription across the genome

Coactivators such as the Mediator complex do not bind DNA directly but instead physically interact with sequence-specific activator proteins bound at enhancers and with general transcription factors at the promoter (including TFIID and RNA polymerase II). This bridging function allows diverse upstream signals to converge on the transcription machinery. Because many different genes rely on this shared coactivator, its loss would broadly abolish transcription. Option A describes a histone methyltransferase, which is a distinct type of chromatin-modifying enzyme.

Q166. Which of the following best describes the function of primase during DNA replication?
A It synthesizes a short RNA sequence to provide a free 3' end for DNA polymerase
B It unwinds the double helix at the replication fork
C It seals the gaps between Okazaki fragments on the lagging strand
D It proofreads newly synthesized DNA for errors

Primase synthesizes a short RNA primer that provides the free 3'-OH group required by DNA polymerase III to begin adding nucleotides. DNA polymerase cannot initiate a new strand on its own — it can only extend an existing one. Helicase unwinds the helix, DNA ligase seals gaps, and the proofreading 3' to 5' exonuclease activity belongs to DNA polymerase itself.

Q167. Which strand of DNA serves as the template during transcription?
A The template strand, read 3' to 5' by RNA polymerase
B The coding strand, read 5' to 3' by RNA polymerase
C Both strands simultaneously to maximize RNA output
D The strand with the highest adenine content

RNA polymerase reads the template strand in the 3' to 5' direction and synthesizes mRNA in the 5' to 3' direction. The coding strand (also called the sense strand or non-template strand) has the same sequence as the mRNA (with U replacing T). Only one strand is used as the template for any given gene, though different genes on the same chromosome may use different strands.

Q168. The 5' cap added to eukaryotic pre-mRNA is composed of:
A A modified guanine nucleotide
B A string of adenine nucleotides
C A short sequence of uracil nucleotides
D A methyl group attached to the first coding exon

The 5' cap is a 7-methylguanosine (m7G) cap added in an unusual 5'-to-5' triphosphate linkage. It protects mRNA from degradation, assists in ribosome binding during translation initiation, and helps export the mRNA from the nucleus. The poly-A tail (not cap) consists of adenine nucleotides and is added to the 3' end.

Q169. During translation, what determines which amino acid a tRNA carries?
A The anticodon sequence on the tRNA
B The size and shape of the tRNA molecule
C The ribosome subunit the tRNA binds to
D The order of introns in the original gene

Aminoacyl-tRNA synthetases recognize both the anticodon of a tRNA and charge it with the appropriate amino acid. The anticodon is complementary to a specific mRNA codon, ensuring the correct amino acid is incorporated. The ribosome provides the site for translation but does not determine amino acid identity; introns are removed from pre-mRNA before translation.

Q170. In eukaryotic cells, where does transcription occur?
A In the nucleus
B On the ribosome in the cytoplasm
C In the mitochondria only
D At the nuclear pore complex

Transcription in eukaryotes occurs in the nucleus, where DNA is housed. The resulting pre-mRNA is processed (capping, polyadenylation, splicing) before being exported to the cytoplasm for translation. Mitochondria do have their own transcription machinery, but nuclear genes are transcribed in the nucleus.

Q171. Which of the following correctly describes a stop codon?
A It signals the ribosome to release the polypeptide chain and does not code for an amino acid
B It codes for the amino acid methionine to begin the polypeptide chain
C It is recognized by a tRNA carrying no amino acid on the acceptor stem
D It is found only in the 5' untranslated region of mRNA

Stop codons (UAA, UAG, UGA) are recognized by release factors, not tRNAs. Release factors trigger hydrolysis of the bond between the polypeptide and the last tRNA, releasing the completed protein. AUG (the start codon) codes for methionine. Stop codons are found in the coding region of mRNA, not the 5' UTR.

Q172. What is the role of telomerase in eukaryotic cells?
A It extends the ends of linear chromosomes to prevent loss of genetic information during replication
B It repairs double-strand breaks in the middle of chromosomes
C It removes RNA primers from Okazaki fragments on the lagging strand
D It methylates DNA at repetitive sequences to silence telomere genes

Because DNA polymerase cannot replicate the very end of a linear chromosome (the end-replication problem), telomeres shorten with each cell division. Telomerase uses its own RNA template to extend the 3' end of telomeres, preserving chromosome integrity. RNase H and DNA polymerase I remove RNA primers; telomerase does not have a repair or methylation function.

Q173. In the lac operon system, which condition leads to the highest level of transcription of the structural genes?
A Lactose present and glucose absent
B Lactose absent and glucose present
C Both lactose and glucose present
D Both lactose and glucose absent

Maximum transcription requires two conditions: the repressor must be inactivated (lactose must be present so allolactose binds the repressor and removes it from the operator) and cAMP-CAP must activate transcription (which requires low glucose so adenylyl cyclase produces cAMP). When glucose is present, cAMP levels are low and CAP cannot activate the operon even if lactose is also present.

Q174. A mutation in a gene's promoter region that reduces its affinity for RNA polymerase would most likely result in:
A Decreased transcription of that gene
B Increased translation efficiency of the mRNA
C A frameshift in the resulting protein
D Altered splicing of the pre-mRNA

The promoter is the region where RNA polymerase binds to initiate transcription. Reduced binding affinity means RNA polymerase binds less frequently, producing fewer mRNA transcripts and therefore less protein. This mutation affects transcription initiation, not translation or mRNA processing. A frameshift would result from an insertion or deletion in the coding sequence.

Q175. Histone acetylation generally leads to increased gene expression because:
A Acetyl groups neutralize positive charges on histones, loosening their grip on negatively charged DNA
B Acetyl groups directly recruit RNA polymerase to the promoter
C Acetylation causes histones to be degraded, freeing DNA
D Acetylation promotes methylation of CpG islands near the gene

Histones are positively charged proteins that wrap negatively charged DNA tightly. Acetylation of lysine residues on histones neutralizes their positive charge, reducing the electrostatic attraction to DNA and relaxing chromatin structure (euchromatin). This makes the DNA more accessible to transcription factors and RNA polymerase. Acetylation does not degrade histones or directly recruit polymerase.

Q176. Which of the following is an example of post-translational regulation of gene expression?
A Phosphorylation of a transcription factor that prevents it from entering the nucleus
B Methylation of cytosine residues in the promoter region of a gene
C Binding of a miRNA to the 3' UTR of an mRNA to block translation
D Alternative splicing of pre-mRNA to produce different protein isoforms

Post-translational regulation occurs after a protein has been synthesized and modifies the protein directly. Phosphorylation of a transcription factor that changes its localization is a post-translational event. DNA methylation is an epigenetic/transcriptional regulation. miRNA acts at the mRNA level (post-transcriptional). Alternative splicing is a pre-translational (RNA processing) event.

Q177. During DNA replication, which of the following correctly describes synthesis on the lagging strand?
A Multiple Okazaki fragments are synthesized in the 5' to 3' direction, away from the replication fork
B One continuous strand is synthesized toward the replication fork in the 3' to 5' direction
C Multiple Okazaki fragments are synthesized in the 3' to 5' direction, toward the replication fork
D A single primer initiates continuous synthesis in the 5' to 3' direction

DNA polymerase can only synthesize in the 5' to 3' direction. Because the lagging strand template runs 5' to 3' toward the fork, synthesis must proceed away from the fork in short discontinuous segments called Okazaki fragments. Each fragment requires its own RNA primer. DNA ligase later joins the fragments into a continuous strand. Continuous synthesis toward the fork describes the leading strand.

Q178. A nonsense mutation differs from a missense mutation in that a nonsense mutation:
A Changes a codon that specifies an amino acid into a stop codon, terminating translation early
B Changes one amino acid to a chemically different amino acid in the polypeptide
C Inserts or deletes a nucleotide, altering the reading frame
D Has no effect on the amino acid sequence of the resulting protein

A nonsense mutation converts an amino acid codon to one of the three stop codons (UAA, UAG, UGA), causing premature termination of translation and typically producing a truncated, nonfunctional protein. A missense mutation changes an amino acid codon to one coding for a different amino acid. A frameshift involves insertions or deletions. A silent (synonymous) mutation does not change the amino acid.

Q179. MicroRNAs (miRNAs) regulate gene expression primarily by:
A Base-pairing with complementary mRNA sequences to block translation or promote mRNA degradation
B Binding to the promoter of a gene to recruit transcription factors
C Altering histone acetylation patterns at specific chromosomal loci
D Serving as templates for reverse transcription into regulatory DNA sequences

miRNAs are small non-coding RNAs (~22 nucleotides) that associate with the RISC complex and bind to complementary sequences in the 3' UTR of target mRNAs. Depending on complementarity, they either silence translation or trigger mRNA cleavage and degradation. This is post-transcriptional regulation. miRNAs do not bind promoters, alter chromatin directly, or serve as reverse transcription templates.

Q180. Which of the following describes the role of the Shine-Dalgarno sequence in prokaryotic translation?
A It is a ribosome-binding site on the mRNA that helps position the start codon in the ribosome
B It is the sequence at the 5' end of tRNA that pairs with the mRNA start codon
C It is the promoter consensus sequence recognized by sigma factor in prokaryotes
D It codes for the signal peptide that directs proteins to the endoplasmic reticulum

The Shine-Dalgarno sequence is a purine-rich sequence on prokaryotic mRNA located 5-10 nucleotides upstream of the AUG start codon. It base-pairs with the 16S rRNA of the 30S ribosomal subunit, properly positioning the mRNA so that translation begins at the correct AUG. The -10 and -35 boxes are the promoter elements recognized by sigma factor. Signal peptides are encoded within the protein coding sequence.

Q181. In eukaryotic cells, the spliceosome removes introns from pre-mRNA. Which molecules form the core of the spliceosome?
A Small nuclear RNAs (snRNAs) complexed with proteins to form snRNPs
B Ribozymes made entirely of double-stranded DNA
C Aminoacyl-tRNA synthetases that recognize splice-site sequences
D Methyl-CpG binding proteins that mark intron boundaries

The spliceosome is a large ribonucleoprotein complex composed of five snRNPs (U1, U2, U4, U5, U6), each containing snRNA and associated proteins. The snRNAs recognize the splice site sequences at intron boundaries through complementary base pairing and catalyze the two-step transesterification reactions that excise the intron and join the exons. Aminoacyl-tRNA synthetases function in translation, not splicing.

Q182. Which of the following correctly describes the relationship between a gene's template strand and the mRNA produced from it?
A The mRNA sequence is complementary and antiparallel to the template strand, with U replacing T
B The mRNA sequence is identical to the template strand, with U replacing T
C The mRNA sequence is complementary and parallel to the template strand
D The mRNA has the same polarity as the template strand but uses ribonucleotides

RNA polymerase reads the template strand in the 3' to 5' direction and synthesizes mRNA in the 5' to 3' direction, making the mRNA complementary and antiparallel to the template strand. Uracil is incorporated instead of thymine. The mRNA sequence is identical to the coding (non-template) strand, not the template strand — except that T is replaced by U.

Q183. A eukaryotic gene contains five exons. Through alternative splicing, how many different protein isoforms could potentially be generated if each exon can independently be included or excluded (excluding the possibility of no exons being included)?
A 31
B 25
C 10
D 120

If each of 5 exons can independently be included or excluded, there are 2^5 = 32 possible combinations. Subtracting the one combination where no exons are included gives 31 possible isoforms. This illustrates how alternative splicing dramatically expands the proteome. In reality, constraints such as required exons and reading frame maintenance reduce the actual number, but the mathematical maximum is 31.

Q184. A researcher introduces a mutation that prevents the assembly of the replisome at the origin of replication. Which of the following would be the most likely outcome?
A DNA replication would not initiate, and the cell would fail to divide
B Replication would proceed but only on the lagging strand
C Transcription would be unaffected, but translation would halt
D The cell would replicate its DNA in the 3' to 5' direction instead

The replisome is the multi-protein complex that assembles at the origin of replication and includes helicase, primase, and DNA polymerase. Without replisome assembly, replication cannot initiate at origins. Since DNA must be replicated before cell division (S phase), the cell would be unable to divide. The lagging strand still requires replisome components, so it cannot replicate independently. Transcription uses RNA polymerase and is a separate process.

Q185. In an experiment, researchers find that blocking RNA polymerase II specifically prevents the production of a particular set of mRNAs but not rRNA or tRNA. This result is consistent with which of the following conclusions?
A RNA polymerase II is specifically responsible for transcribing protein-coding genes in eukaryotes
B RNA polymerase II transcribes all classes of RNA in eukaryotic cells
C Inhibiting RNA polymerase II would halt translation but not transcription
D rRNA and tRNA are synthesized from DNA without using an RNA polymerase

Eukaryotes have three RNA polymerases with distinct roles: RNA pol I transcribes rRNA precursors, RNA pol II transcribes pre-mRNA and some small nuclear RNAs, and RNA pol III transcribes tRNA, 5S rRNA, and other small RNAs. The selective loss of mRNA production when RNA pol II is blocked directly supports its role in transcribing protein-coding genes, while rRNA and tRNA synthesis by pol I and pol III continues unaffected.

Q186. A gene in a bacterium has the following mutation: the operator sequence is altered so that the repressor can no longer bind. For a gene regulated by a repressible operon, what would be the phenotypic consequence of this mutation?
A Constitutive expression of the operon regardless of whether the corepressor is present
B Permanent repression of the operon regardless of substrate availability
C Increased sensitivity to the inducer molecule, lowering the threshold for activation
D No change in expression because repressors are not used in repressible operons

In a repressible operon (such as the trp operon), the repressor-corepressor complex binds the operator to shut off transcription when the end product is abundant. If the operator is mutated so the repressor cannot bind, transcription cannot be repressed — the operon is constitutively expressed even when corepressor levels are high. This is distinct from an inducible operon; both use repressors, but the regulatory logic is inverted.

Q187. During DNA replication, proofreading by DNA polymerase III reduces the error rate from approximately 1 in 10^5 to 1 in 10^7. If mismatch repair further reduces errors to 1 in 10^9, what is the approximate fold-improvement provided by mismatch repair alone after polymerase proofreading has already occurred?
A 100-fold
B 10,000-fold
C 1,000-fold
D 10-fold

After proofreading, the error rate is approximately 1 in 10^7 (one error per 10 million bases). Mismatch repair further reduces this to 1 in 10^9. The fold-improvement is 10^9 / 10^7 = 100-fold. Understanding layered fidelity mechanisms — intrinsic polymerase accuracy, 3'-to-5' proofreading, and post-replication mismatch repair — is important for understanding how the genome is maintained with such high accuracy.

Q188. A transcription factor contains a zinc finger domain and a transcriptional activation domain. A researcher creates a mutant version that retains the zinc finger domain but has a deletion in the activation domain. What would be the most likely effect on target gene expression?
A The mutant could bind DNA at the correct sites but would fail to activate transcription, potentially acting as a dominant negative inhibitor
B The mutant would activate transcription at higher levels because the deletion reduces steric hindrance
C The mutant would be unable to enter the nucleus due to misfolding of the entire protein
D The mutant would bind DNA more strongly and constitutively activate all target genes

Transcription factors have separable functional domains: a DNA-binding domain (here, the zinc finger) and an activation domain that interacts with co-activators or the basal transcription machinery. If the activation domain is deleted, the protein can still bind its target DNA sites through the intact zinc finger domain but cannot recruit the transcriptional machinery. By occupying binding sites without activating transcription, it blocks wild-type transcription factors — a dominant negative effect. This principle underlies many dominant-negative mutations studied in AP Biology.

Q189. Researchers find that in a particular cell type, a gene is transcribed at high levels but its protein product is nearly undetectable. Which combination of post-transcriptional mechanisms could best explain this discrepancy?
A Rapid mRNA degradation by miRNA and inefficient ribosome recruitment due to secondary structure in the 5' UTR
B A promoter mutation reducing RNA polymerase binding and increased histone acetylation
C Constitutive splicing removing all exons and increased tRNA availability
D Enhanced nuclear export of mRNA and increased ribosome biogenesis

The question asks about a disconnect between high transcription and low protein levels, which must be explained by post-transcriptional mechanisms. miRNA-mediated degradation or translational silencing can rapidly destroy mRNA, and structured 5' UTRs can block ribosome scanning and translation initiation — together these would prevent protein accumulation despite high transcription. Promoter mutations and histone changes affect transcription, not the post-transcriptional gap. Constitutive splicing removing all exons is not a real mechanism. Enhanced export would increase, not decrease, protein levels.

Q190. In an experiment testing epigenetic inheritance, cells with heavily methylated CpG islands in a gene's promoter are treated with a DNA methyltransferase inhibitor and allowed to undergo ten rounds of replication. What would be the expected distribution of methylation in the resulting cell population?
A Methylation would be progressively diluted with each replication, producing cells with decreasing methylation density over generations
B All daughter cells would immediately lose all methylation after the first replication
C Methylation would be unchanged because methyltransferase inhibitors only affect new CpG sites
D Methylation density would increase because inhibiting the enzyme causes a compensatory feedback response

During DNA replication, the parental methylation pattern serves as a template for maintenance methyltransferase (DNMT1), which re-methylates the newly synthesized strand at hemi-methylated CpG sites. When DNMT1 is inhibited, new strands are not re-methylated. After each replication, roughly half the daughter molecules inherit the old methylated strand, but new strands are unmethylated. Over many generations, the proportion of methylated copies is progressively halved (diluted), leading to a population with progressively lower average methylation — this is called passive demethylation.

Q191. Which of the following correctly describes the directionality of DNA synthesis during replication?
A DNA polymerase synthesizes new strands in the 3' to 5' direction
B DNA polymerase synthesizes new strands in the 5' to 3' direction
C DNA polymerase can synthesize in either direction depending on the template strand
D DNA polymerase synthesizes the leading strand 3' to 5' and the lagging strand 5' to 3'

DNA polymerase can only add nucleotides to the 3'-OH end of a growing strand, meaning synthesis always proceeds 5' to 3'. This constraint is why the lagging strand must be synthesized discontinuously as Okazaki fragments — both strands are still synthesized 5' to 3', but in opposite physical directions relative to the replication fork.

Q192. Which RNA molecule carries amino acids to the ribosome during translation?
A mRNA
B rRNA
C tRNA
D snRNA

Transfer RNA (tRNA) is the adaptor molecule that carries specific amino acids to the ribosome. Each tRNA has an anticodon that base-pairs with a complementary mRNA codon, and an amino acid attachment site at its 3' end. rRNA is a structural and catalytic component of the ribosome itself, while mRNA carries the genetic message being translated.

Q193. Which of the following is the function of the poly-A tail added to eukaryotic mRNA?
A It serves as the start signal for translation
B It helps protect the mRNA from degradation and aids in nuclear export
C It codes for a string of lysine residues at the end of the protein
D It acts as a splice site for intron removal

The poly-A tail — a string of adenine nucleotides added post-transcriptionally — protects mRNA from enzymatic degradation, facilitates export from the nucleus, and assists in translation initiation. It does not code for amino acids because it is part of the 3' UTR, downstream of the stop codon. Splice sites are specific sequences within pre-mRNA recognized by the spliceosome, unrelated to the poly-A tail.

Q194. In the lac operon system, which condition leads to the highest level of transcription of the structural genes?
A Lactose absent, glucose absent
B Lactose present, glucose absent
C Lactose absent, glucose present
D Lactose present, glucose present

Maximum transcription requires two conditions: the repressor must be inactive (lactose present — allolactose inactivates the repressor, freeing the operator) and the CAP-cAMP activator complex must be bound to the promoter (glucose absent — low glucose raises cAMP levels, activating CAP). When glucose is present, cAMP is low and CAP cannot activate transcription even if the repressor is off, resulting in only moderate transcription.

Q195. A researcher treats cells with a drug that prevents the addition of 5' methyl guanosine caps to newly synthesized mRNA. Which of the following outcomes is most likely?
A Pre-mRNA splicing will be blocked entirely
B Ribosomes will be unable to recognize and initiate translation of the mRNA
C The poly-A tail will also fail to be added, destabilizing mRNA
D Transcription will terminate prematurely at the 5' end of the gene

The 5' methylguanosine cap is recognized by initiation factors (eIF4E) that recruit the ribosome to the mRNA, making it essential for cap-dependent translation initiation. Without the cap, ribosomes cannot efficiently bind and begin translation. The cap also helps protect mRNA from 5' exonucleases, but its primary role in this context is ribosome recognition. Splicing and polyadenylation are separate processing events not directly dependent on the 5' cap.

Q196. During translation elongation, which site of the ribosome holds the growing polypeptide chain attached to a tRNA?
A A site (aminoacyl site)
B P site (peptidyl site)
C E site (exit site)
D Decoding center

The P site (peptidyl site) holds the tRNA bearing the growing polypeptide chain. The A site (aminoacyl site) accepts the incoming aminoacyl-tRNA carrying the next amino acid. After peptide bond formation transfers the polypeptide to the new tRNA in the A site, translocation moves that tRNA to the P site and the deacylated tRNA from the P site to the E site, where it exits the ribosome.

Q197. Chromatin remodeling complexes contribute to gene regulation primarily by:
A Directly editing the nucleotide sequence of promoter regions
B Altering the accessibility of DNA to transcription factors by repositioning nucleosomes
C Degrading mRNA transcripts produced from silenced genes
D Methylating the poly-A tail to prevent ribosome binding

Chromatin remodeling complexes use ATP hydrolysis to slide, eject, or restructure nucleosomes — the histone-DNA spools that compact chromatin. By repositioning nucleosomes, these complexes expose or conceal promoter sequences and transcription factor binding sites, thereby increasing or decreasing transcriptional activity. They do not directly edit DNA sequence, degrade mRNA, or modify the poly-A tail.

Q198. A mutation in the gene encoding the sigma factor of a bacterial RNA polymerase would most directly affect which step of transcription?
A Elongation rate along the template strand
B Termination at Rho-independent terminator sequences
C Promoter recognition and transcription initiation
D Addition of the 5' cap to newly synthesized RNA

The sigma factor is a dissociable subunit of bacterial RNA polymerase that confers promoter specificity — it recognizes the -10 and -35 consensus sequences upstream of the transcription start site, enabling initiation. A mutation disrupting sigma factor function would impair the holoenzyme's ability to locate and bind promoters, blocking transcription initiation. Elongation and termination involve the core enzyme, and 5' capping is a eukaryotic modification absent in bacteria.

Q199. A nonsense mutation near the beginning of a gene's coding sequence is generally more harmful than one near the end because:
A Nonsense mutations at the 5' end alter the promoter, preventing transcription entirely
B A premature stop codon early in the sequence produces a severely truncated, likely nonfunctional protein
C The ribosome cannot initiate translation when a stop codon is within the first 50 codons
D Nonsense mutations early in the gene trigger RNA interference pathways that destroy all mRNA copies

A nonsense mutation introduces a premature stop codon, truncating the polypeptide. When this occurs early in the coding sequence, the resulting protein is missing most of its functional domains and is almost certainly nonfunctional. A late-occurring premature stop codon produces a near-full-length protein that may retain partial function. The mutation affects the coding sequence, not the promoter, and while nonsense-mediated decay (NMD) can degrade the mRNA, the primary reason for greater harm is the severely truncated product.

Q200. MicroRNAs (miRNAs) regulate gene expression post-transcriptionally. Which of the following best describes their mechanism of action?
A miRNAs bind to the promoter of a target gene and recruit histone deacetylases to silence transcription
B miRNAs base-pair with complementary sequences in target mRNAs, leading to mRNA degradation or translational repression
C miRNAs act as competitive inhibitors of ribosomal RNA, slowing the overall rate of translation in the cell
D miRNAs insert into the coding sequence of target mRNAs, causing frameshift mutations that produce nonfunctional proteins

miRNAs are short (~22 nucleotide) non-coding RNAs that are incorporated into the RISC (RNA-Induced Silencing Complex). RISC uses the miRNA as a guide to find complementary or partially complementary sequences, typically in the 3' UTR of target mRNAs. Depending on the degree of complementarity, the mRNA is either cleaved and degraded or its translation is repressed. This is a sequence-specific, post-transcriptional mechanism distinct from transcriptional silencing or ribosome inhibition.

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Quick summary

This unit covers DNA replication, transcription, translation and gene regulation — essential concepts for AP Biology. Use our interactive study games to test your understanding, or review questions in traditional format below.

Key concepts
  • Dna replication
  • Transcription
  • Translation
  • Gene regulation
What you need to know

Key Concepts Breakdown

1 DNA Replication

DNA replication is semiconservative, meaning each new double helix contains one original strand and one newly synthesized strand. Students must know the roles of key enzymes (helicase, primase, DNA polymerase III, DNA polymerase I, ligase) and the directionality constraint (synthesis only 5'→3'). The leading strand is synthesized continuously; the lagging strand is synthesized in Okazaki fragments.

Key Points

  • Helicase unwinds the double helix at the replication fork; primase lays down an RNA primer to provide a free 3'-OH for DNA polymerase
  • DNA polymerase III adds nucleotides 5'→3', reading the template 3'→5'; it cannot initiate a new strand de novo
  • DNA polymerase I replaces RNA primers with DNA; ligase seals the nicks between Okazaki fragments
  • A mutation in the template strand is passed to one daughter cell; errors not caught by proofreading become heritable mutations
Example

A cell is replicating DNA and a researcher adds a drug that inhibits primase. Which process is most directly disrupted and why?

Explanation

Without primase, no RNA primers can be synthesized. Because DNA polymerase III requires a free 3'-OH group to begin adding nucleotides, it cannot start synthesis on either the leading or lagging strand. Replication halts entirely, not just on the lagging strand, because even the leading strand requires one primer at the origin.

2 Transcription

Transcription is the synthesis of a pre-mRNA from a DNA template by RNA polymerase, occurring in the nucleus in eukaryotes. Students must know the three stages (initiation, elongation, termination), the role of the promoter and transcription factors, and eukaryotic pre-mRNA processing (5' cap, poly-A tail, splicing out introns). The template strand is read 3'→5', producing an mRNA 5'→3'.

Key Points

  • RNA polymerase binds to the promoter (e.g., TATA box) with the help of transcription factors; no primer is needed
  • The non-template (coding) strand has the same sequence as the mRNA, except T is replaced by U
  • Pre-mRNA processing: 5' methylguanosine cap protects from degradation; poly-A tail added to 3' end; introns removed by spliceosomes
  • Alternative splicing of exons allows one gene to code for multiple proteins — a key concept for gene regulation questions
Example

A mutation changes the TATA box sequence in the promoter of a gene. Predict the most likely effect on transcription.

Explanation

The TATA box is a core promoter element where the transcription initiation complex assembles. If it is mutated, transcription factors cannot bind properly, and RNA polymerase is not recruited to the correct start site. The result is a significant reduction or complete elimination of transcription of that gene.

3 Translation

Translation is the synthesis of a polypeptide from an mRNA sequence by ribosomes, occurring in the cytoplasm. Students must understand the codon–anticodon interaction, the three ribosomal sites (A, P, E), and the sequential steps of initiation, elongation, and termination. The genetic code is degenerate (multiple codons per amino acid) and nearly universal.

Key Points

  • Ribosomes read mRNA codons 5'→3'; tRNA anticodons are antiparallel and complementary to the codon
  • Initiation: small ribosomal subunit binds mRNA at the start codon (AUG, Met); large subunit joins forming P and A sites
  • Elongation: aminoacyl-tRNA enters A site, peptide bond forms (peptidyl transferase activity of rRNA), ribosome translocates 5'→3'
  • Stop codons (UAA, UAG, UGA) recruit release factors, not tRNAs, causing the polypeptide to be released
Example

A point mutation changes codon 47 in an mRNA from GAA (Glu) to GAG. What is the most likely effect on the protein?

Explanation

Both GAA and GAG code for glutamic acid — this is a synonymous (silent) mutation due to the degeneracy of the genetic code. The amino acid sequence of the protein is unchanged, so protein function is most likely unaffected. This example tests whether students can use a codon table and understand degeneracy rather than assuming all mutations alter protein function.

4 Gene Regulation

Gene expression is regulated at multiple levels: transcriptional, post-transcriptional, translational, and post-translational. For the AP exam, students must know prokaryotic operon models (lac and trp operons) and eukaryotic regulation via transcription factors, enhancers, and epigenetic mechanisms (DNA methylation, histone modification). The core principle is that cells regulate which genes are expressed to respond to environmental signals.

Key Points

  • Lac operon (inducible): in the absence of lactose, a repressor blocks transcription; lactose (as allolactose) binds the repressor, causing it to detach and allowing transcription
  • Trp operon (repressible): when tryptophan is abundant, it acts as a corepressor binding the repressor protein, which then blocks transcription
  • Eukaryotic enhancers are DNA sequences that, when bound by activator proteins, increase transcription even from a distance via DNA looping
  • Epigenetic regulation: methylation of cytosine residues silences genes; acetylation of histones loosens chromatin and increases transcription — neither changes the DNA sequence
Example

E. coli is grown in a medium with both glucose and lactose present. Predict whether the lac operon genes will be highly expressed, and justify your answer.

Explanation

Even though lactose is present (which inactivates the repressor), the lac operon will NOT be highly expressed because glucose is also present. When glucose is available, cAMP levels are low, so the CAP activator protein cannot bind the promoter to stimulate transcription. This demonstrates that the lac operon requires both the absence of glucose (for CAP activation) and the presence of lactose (to relieve repression) to be maximally transcribed — a classic two-factor regulation question on the AP exam.

FAQ

Questions, answered.

What is Gene Expression and Regulation?

Gene Expression and Regulation is Unit 6 of AP Biology, covering DNA replication, transcription, translation and gene regulation.

How to study for AP Biology Unit 6?

Start with the Quick Summary above, review the Key Concepts, then test yourself with our interactive study games. Aim for 80%+ accuracy before moving on.

How many questions are in this unit?

This unit has 200 review questions, each with a written explanation, playable across 5 different game modes or readable in plain-text mode.