AP Biology Gene Expression and Regulation — Worked Answer Explanations

Unit 6 · 12 questions explained

Below is a complete answer key for our AP Biology Gene Expression and Regulation practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Gene Expression and Regulation practice test and come back here to review, or head back to the Gene Expression and Regulation unit overview.

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Question 1 · Easy
During DNA replication, which enzyme synthesizes new DNA strands by adding nucleotides complementary to the template strand?
- A
  Helicase
  Why not A: Helicase unwinds the double helix by breaking hydrogen bonds; it does not synthesize DNA.
- B
  DNA polymeraseCorrect
- C
  Primase
  Why not C: Primase synthesizes short RNA primers, not the bulk of new DNA.
- D
  Ligase
  Why not D: Ligase joins Okazaki fragments by sealing nicks in the sugar-phosphate backbone; it does not synthesize new DNA strands.
Explanation
DNA polymerase is the enzyme that catalyzes the addition of deoxyribonucleotides to the 3′ end of a growing strand, using an existing strand as a template. It requires a free 3′-OH group (provided by a primer) and reads the template 3′→5′ while synthesizing the new strand 5′→3′. Multiple DNA polymerases exist; in prokaryotes, DNA Pol III is the primary replicative enzyme.

Key takeaway
DNA polymerase synthesizes new DNA strands by adding nucleotides in the 5′→3′ direction, using a template strand.
Question 2 · Easy
In the central dogma of molecular biology, which of the following correctly describes the typical flow of genetic information?
- A
  Protein → RNA → DNA
  Why not A: This reverses the flow; information moves from DNA outward to protein, not from protein back to DNA.
- B
  DNA → RNA → ProteinCorrect
- C
  RNA → DNA → Protein
  Why not C: RNA → DNA occurs in retroviruses (reverse transcription), which is an exception to the standard central dogma.
- D
  DNA → Protein → RNA
  Why not D: Proteins are not used as templates to make RNA; transcription always uses DNA as its template.
Explanation
The central dogma, proposed by Crick, describes the standard flow of genetic information: DNA is transcribed into mRNA (transcription), and mRNA is translated into protein (translation). DNA also replicates itself. Exceptions exist (e.g., reverse transcription in retroviruses), but the canonical path is DNA → RNA → Protein.

Key takeaway
Genetic information normally flows from DNA to RNA to protein, as described by the central dogma.
Question 3 · Easy
A ribosome is reading an mRNA strand and reaches the codon 5′-UAA-3′. What happens at this point in translation?
- A
  A tRNA carrying the amino acid tyrosine enters the A site
  Why not A: UAA is a stop codon; no tRNA recognizes stop codons (release factors do instead).
- B
  A release factor binds the A site, and the completed polypeptide is releasedCorrect
- C
  The ribosome shifts back one nucleotide and re-reads the codon
  Why not C: Ribosomes do not shift back; they read codons in the 5′→3′ direction and stop at stop codons.
- D
  mRNA is degraded immediately to recycle the nucleotides
  Why not D: mRNA degradation is a separate regulatory process; it is not triggered instantly at stop codon recognition.
Explanation
UAA, UAG, and UGA are the three stop codons (nonsense codons). No tRNA anticodon matches stop codons. Instead, protein release factors (RF1, RF2 in prokaryotes; eRF1 in eukaryotes) recognize stop codons in the A site, triggering hydrolysis of the polypeptide-tRNA bond at the P site and release of the completed polypeptide. The ribosome then dissociates from the mRNA.

Key takeaway
Stop codons recruit release factors (not tRNAs) that trigger polypeptide release and ribosome disassembly.
Question 4 · Easy
The lac operon in E. coli is repressed under which of the following conditions?
- A
  Lactose present, glucose absent
  Why not A: When lactose is present and glucose is absent, the repressor is inactive and CAP is active — the lac operon is maximally expressed.
- B
  Lactose absent, glucose presentCorrect
- C
  Lactose present, glucose present
  Why not C: When both are present, the repressor is inactivated by allolactose, but high glucose keeps cAMP low so CAP is not active — expression is low but not fully repressed.
- D
  Lactose absent, glucose absent
  Why not D: Without lactose, the repressor is active (the operon is repressed); but when glucose is also absent, cAMP is high and CAP is active — yet the repressor overrides, so full expression requires both lactose present and glucose absent.
Explanation
The lac operon encodes enzymes for lactose metabolism. It is controlled by two regulatory mechanisms: (1) the lac repressor, which blocks transcription when lactose (specifically allolactose) is absent; (2) catabolite repression (CAP-cAMP), which maximally activates the operon when glucose is absent (high cAMP). When lactose is absent, allolactose cannot inactivate the repressor, so the repressor binds the operator and blocks transcription — the operon is repressed regardless of glucose levels. Repression is most stringent when lactose is absent and glucose is present.

Key takeaway
The lac operon is repressed when the inducer (allolactose) is absent; without lactose, the repressor blocks transcription.
Question 5 · Medium
During transcription in eukaryotes, which of the following modifications occurs to pre-mRNA before it exits the nucleus?
- A
  Introns are translated into regulatory peptides before being removed
  Why not A: Introns are removed from RNA, not translated; their sequence is excised by spliceosomes.
- B
  Addition of a 5′ cap, a 3′ poly-A tail, and splicing of intronsCorrect
- C
  Reverse transcription to produce a complementary DNA (cDNA) copy
  Why not C: Reverse transcription is not a normal nuclear modification; it is used by retroviruses and as a laboratory technique.
- D
  Translation of the first exon to begin protein synthesis in the nucleus
  Why not D: In eukaryotes, translation occurs in the cytoplasm after mRNA export; it does not begin in the nucleus.
Explanation
Eukaryotic pre-mRNA undergoes three major processing steps in the nucleus before export: (1) Addition of a 5′ 7-methylguanosine (m7G) cap, which protects mRNA and helps ribosome binding during translation; (2) Addition of a poly-A tail (~150–250 adenine nucleotides) at the 3′ end, which protects mRNA from degradation and aids export; (3) RNA splicing by spliceosomes, which removes introns (non-coding sequences) and joins exons (coding sequences) to form mature mRNA.

Key takeaway
Eukaryotic pre-mRNA is processed via 5′ capping, 3′ poly-A tail addition, and intron splicing before export to the cytoplasm.
Question 6 · Medium
Polymerase chain reaction (PCR) requires a template, primers, dNTPs, and a heat-stable DNA polymerase (Taq). Which of the following is the correct sequence of steps in one PCR cycle?
- A
  Annealing → Denaturation → Extension
  Why not A: Denaturation must come first to separate the double-stranded template; primers cannot anneal to double-stranded DNA.
- B
  Denaturation → Annealing → ExtensionCorrect
- C
  Extension → Denaturation → Annealing
  Why not C: Extension requires primers already annealed to single-stranded template; it cannot precede denaturation and annealing.
- D
  Denaturation → Extension → Annealing
  Why not D: Primers must anneal before extension; extension (new strand synthesis) cannot occur on double-stranded DNA.
Explanation
PCR amplifies DNA through repeated thermocycling. Each cycle consists of three temperature-controlled steps: (1) Denaturation (~94–98°C): heat separates the double-stranded DNA template into two single strands; (2) Annealing (~50–65°C): temperature is lowered to allow short primer sequences to bind complementarily to each template strand; (3) Extension (~72°C): Taq polymerase synthesizes new DNA strands from the 3′ end of each primer, copying the target region. After ~30 cycles, the target sequence is amplified exponentially (~2³⁰ fold).

Key takeaway
PCR cycles through denaturation, annealing, and extension to exponentially amplify a target DNA sequence.
Question 7 · Medium
In gel electrophoresis of DNA fragments, a researcher loads four samples and observes that one fragment migrated the farthest from the well. What does this indicate about that fragment?
- A
  It has the highest molecular weight and is the largest fragment
  Why not A: Larger fragments migrate more slowly and remain closer to the well; smaller fragments migrate farther.
- B
  It is the smallest fragment in the sampleCorrect
- C
  It has the most negative charge and is most attracted to the positive electrode
  Why not C: DNA is uniformly negatively charged because of the phosphate backbone; charge per unit mass is essentially constant for all DNA fragments.
- D
  It contains more guanine-cytosine base pairs than the other fragments
  Why not D: GC content affects melting temperature but does not significantly affect migration rate in standard gel electrophoresis.
Explanation
In agarose gel electrophoresis, DNA fragments migrate through the gel matrix under an electric field toward the positive electrode (DNA is negatively charged). Smaller fragments migrate faster and farther because they experience less friction passing through the gel pores. Larger fragments are impeded by the matrix and stay closer to the wells. Fragment size is determined by comparing migration distance to a DNA ladder (standard fragments of known size).

Key takeaway
In gel electrophoresis, smaller DNA fragments migrate farther from the well because they move more easily through the gel matrix.
Question 8 · Medium
A point mutation changes a codon from UUU (phenylalanine) to UUA (leucine) in a translated protein. Which of the following best describes this mutation?
- A
  Silent mutation, because the protein is unchanged
  Why not A: A silent mutation preserves the same amino acid (same protein); UUU→UUA changes phenylalanine to leucine, so the protein IS changed.
- B
  Missense mutation, because one amino acid is substituted for anotherCorrect
- C
  Nonsense mutation, because a stop codon is introduced
  Why not C: A nonsense mutation introduces a stop codon (UAA, UAG, UGA); UUA codes for leucine, not a stop codon.
- D
  Frameshift mutation, because a nucleotide is inserted or deleted
  Why not D: A frameshift results from insertion or deletion of nucleotides that shift the reading frame; a single base substitution (point mutation) does not cause a frameshift.
Explanation
Mutations are classified by their effect on the protein. A missense mutation changes a single nucleotide such that the codon specifies a different amino acid (UUU → UUA changes phenylalanine to leucine). A silent mutation changes the codon but results in the same amino acid (due to degeneracy of the genetic code). A nonsense mutation converts an amino acid codon to a stop codon. A frameshift inserts or deletes nucleotides, shifting the reading frame of all downstream codons.

Key takeaway
A missense mutation substitutes one amino acid for another; silent mutations preserve the amino acid due to codon degeneracy.
Question 9 · Medium
Scientists use bacterial transformation to introduce a gene of interest into E. coli using a plasmid vector. The plasmid contains an ampicillin resistance gene (amp^R) and a lacZ gene (which turns colonies blue on X-gal plates) into which the gene of interest has been inserted. If transformation is successful and the insert disrupts lacZ, which of the following correctly describes the transformed bacteria?
- A
  Ampicillin sensitive, white colonies on X-gal plates
  Why not A: If the bacteria have the plasmid, they carry amp^R and are ampicillin resistant; sensitivity would indicate no plasmid.
- B
  Ampicillin resistant, white colonies on X-gal platesCorrect
- C
  Ampicillin resistant, blue colonies on X-gal plates
  Why not C: Blue colonies indicate intact lacZ; inserting the gene of interest into lacZ disrupts it, giving white colonies (blue-white screening).
- D
  Ampicillin sensitive, blue colonies on X-gal plates
  Why not D: Ampicillin sensitivity means no plasmid uptake; such bacteria would not be transformed and would be killed on ampicillin plates.
Explanation
In blue-white screening with insertional inactivation: (1) The plasmid carries amp^R (selection marker) and lacZ (reporter). (2) The gene of interest is cloned into the multiple cloning site within lacZ, disrupting it. (3) Bacteria are plated on ampicillin + X-gal plates. (4) Bacteria without the plasmid die (ampicillin kills them). (5) Bacteria with intact plasmid (no insert) express functional lacZ → cleave X-gal → blue colonies. (6) Bacteria with recombinant plasmid (insert in lacZ) cannot produce functional β-galactosidase → cannot cleave X-gal → white colonies. Transformants with the insert are therefore ampicillin resistant and white.

Key takeaway
Blue-white screening identifies successful insertional cloning: recombinant bacteria are ampicillin resistant (have plasmid) but white (lacZ disrupted by insert).
Question 10 · Hard
Transcription factors bind to specific DNA sequences in the promoter or enhancer regions of eukaryotic genes. A mutation abolishes the binding of an activator transcription factor to an enhancer 10,000 bp upstream of the gene's transcription start site. Which of the following best predicts the effect on gene expression?
- A
  No effect, because enhancers must be adjacent to the promoter to function
  Why not A: Enhancers can act over tens of thousands of base pairs through DNA looping, so distance from the promoter does not prevent their function.
- B
  Gene expression decreases or is abolished because the activator can no longer stimulate RNA polymerase II recruitmentCorrect
- C
  Gene expression increases because the enhancer normally inhibits the gene
  Why not C: An activator transcription factor increases transcription; its loss decreases expression. Silencers/repressors would inhibit expression.
- D
  The mutation affects only transcription in embryonic cells because enhancers are only active during development
  Why not D: Enhancers operate in many cell types and developmental stages; their tissue specificity depends on which transcription factors are expressed, not on a universal developmental restriction.
Explanation
Enhancers are cis-regulatory elements that can dramatically increase transcription even when located thousands of base pairs from the promoter. They function by binding activator transcription factors, which then interact with the transcription initiation complex at the promoter through DNA looping (chromatin looping). Loss of enhancer-activator binding reduces or eliminates the upregulation of transcription, decreasing RNA polymerase II recruitment and reducing gene expression. This is distinct from a promoter mutation, which affects the basal transcription machinery.

Key takeaway
Enhancers activate transcription over long distances via DNA looping; loss of activator binding to an enhancer reduces transcription of the target gene.
Question 11 · Hard
CRISPR-Cas9 is used to introduce a double-strand break in a gene. In most mammalian cells, the break is repaired by non-homologous end joining (NHEJ). Which of the following best describes a likely molecular outcome of NHEJ repair?
- A
  The break is repaired precisely using the sister chromatid as a template, restoring the original sequence
  Why not A: Precise template-based repair describes homology-directed repair (HDR), not NHEJ; HDR requires a homologous template.
- B
  Small insertions or deletions (indels) are introduced at the cut site, often causing frameshift mutationsCorrect
- C
  The entire chromosome containing the cut is degraded to prevent genome instability
  Why not C: NHEJ ligates the broken ends; it does not degrade the chromosome.
- D
  The Cas9 protein becomes integrated into the genome at the cut site
  Why not D: Cas9 is a protein, not a nucleic acid; proteins are not integrated into the genome.
Explanation
NHEJ (non-homologous end joining) is the predominant DNA double-strand break repair pathway in mammalian cells during G1 and G2 phases. It joins the broken ends without requiring a homologous template. However, NHEJ is error-prone: DNA ends may be processed by exonucleases or polymerases before ligation, resulting in insertions or deletions (indels) at the cut site. If the indel shifts the reading frame of the coding sequence, it produces a frameshift mutation that typically introduces premature stop codons, truncating or disrupting the protein. This is how CRISPR-Cas9 is commonly used for gene knockout.

Key takeaway
NHEJ repair of CRISPR-induced double-strand breaks often introduces indels at the cut site, frequently causing frameshift mutations that knock out the gene.
Question 12 · Hard
Researchers studying gene regulation find that a eukaryotic gene is highly expressed in liver cells but completely silenced in neurons despite both cell types having identical genomic DNA. Which combination of mechanisms could best explain this tissue-specific expression pattern?
- A
  Different DNA sequences in liver vs. neuron genomes encoding different versions of the gene
  Why not A: Both cell types are derived from the same zygote and have identical genomic DNA; differential expression is not due to DNA sequence differences.
- B
  Liver-specific activator transcription factors and neuron-specific DNA methylation silencing the gene's promoterCorrect
- C
  Alternative splicing that produces a functional protein only in liver cells
  Why not C: Alternative splicing can produce different protein isoforms but does not explain complete transcriptional silencing in neurons.
- D
  Ribosomes in neurons are unable to translate liver-specific mRNAs
  Why not D: Ribosome structure is largely conserved and does not confer tissue-specific translational discrimination based on mRNA source.
Explanation
Cell differentiation is driven by gene regulation, not genomic sequence changes. Tissue-specific expression results from: (1) Cell-type-specific transcription factors — liver cells express activators (e.g., HNF transcription factors) that bind enhancers of liver-specific genes, recruiting coactivators and RNA Pol II; (2) Epigenetic modifications — DNA methylation (particularly CpG methylation at promoters) and histone modifications (e.g., H3K27me3 from Polycomb repressors) create stable transcriptional silencing in neurons. These epigenetic marks are heritable through cell division within the lineage, maintaining the silenced state in all neurons. Together, cell-type-specific transcription factors and epigenetic regulation explain why identical genomes produce different gene expression profiles.

Key takeaway
Tissue-specific expression from identical genomes is achieved through cell-type-specific transcription factors and epigenetic silencing mechanisms like DNA methylation.