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AP Biology Unit 6 Practice Questions: Gene Expression and Regulation

10 original exam-style questions on Gene Expression and Regulation. Answer each one to see the explanation — no account needed.

Question 1 of 10 · Eukaryotic Transcription Regulation

A researcher finds that a gene is transcribed at high rates in all cell types regardless of signaling conditions. Which of the following regulatory features most likely explains constitutive (always-on) gene expression?
  1. A. The gene's promoter contains TATA boxes and GC boxes that are always accessible to RNA polymerase II and basal transcription factors.
  2. B. The gene's introns contain repressor binding sites that prevent gene silencing.
  3. C. The gene's coding region is hypermethylated, keeping chromatin in an open configuration.
  4. D. The gene is located on the X chromosome and is protected from inactivation.
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Correct answer: A

Constitutively expressed housekeeping genes typically have accessible, CpG-island-associated promoters with core promoter elements (TATA box, GC boxes) that bind basal transcription factors and RNA Pol II without requiring regulatory signals. DNA methylation in coding regions (option C) is generally associated with silencing, not activation.

Question 2 of 10 · Epigenetic Regulation

Researchers used chromatin immunoprecipitation (ChIP) to study histone modification patterns near the promoters of three genes in cancer cells compared to normal cells. H3K27me3 is a repressive mark (gene silencing); H3K4me3 is an activating mark (gene expression).

Histone modification enrichment at gene promoters (fold-enrichment relative to control)
GeneH3K27me3 (Normal)H3K27me3 (Cancer)H3K4me3 (Normal)H3K4me3 (Cancer)
CDKN2A (tumor suppressor)1.08.36.21.1
MYC (proto-oncogene)0.90.81.59.4
BRCA1 (DNA repair)1.16.95.80.9
Which of the following best explains how increased H3K27me3 at the CDKN2A promoter could silence the gene without changing the DNA sequence?
  1. A. H3K27me3 recruits enzymes that degrade the CDKN2A mRNA after transcription.
  2. B. H3K27me3 methylates cytosine residues in the CDKN2A coding sequence, introducing stop codons.
  3. C. H3K27me3 causes chromatin compaction, making the CDKN2A promoter inaccessible to transcription factors and RNA polymerase.
  4. D. H3K27me3 blocks translation of CDKN2A mRNA at the ribosome.
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Correct answer: C

Trimethylation of H3K27 (a lysine on histone H3) is associated with Polycomb repressive complex activity, leading to chromatin compaction (heterochromatin) that physically blocks access of transcription machinery to the promoter. Histone methylation is not the same as DNA methylation and does not alter the DNA sequence or directly affect mRNA translation.

Question 3 of 10 · Lac Operon and Prokaryotic Gene Regulation

In the lac operon of E. coli, the repressor protein is normally bound to the operator, blocking transcription. When lactose is present, it is converted to allolactose, which acts as an inducer. Which of the following best describes how allolactose enables transcription of the lac genes?
  1. A. Allolactose binds to the repressor protein, causing a conformational change that reduces its affinity for the operator, allowing transcription to proceed
  2. B. Allolactose binds to RNA polymerase, enhancing its affinity for the lac promoter and initiating transcription
  3. C. Allolactose directly degrades the repressor protein through proteolysis, permanently inactivating gene repression
  4. D. Allolactose binds to the operator, physically displacing the repressor and exposing the promoter to RNA polymerase
Show answer and explanation

Correct answer: A

Allolactose is an allosteric inducer — it binds the lac repressor and induces a conformational change that greatly reduces the repressor's DNA-binding affinity, causing it to dissociate from the operator and allowing RNA polymerase to transcribe the lac genes. Choice D incorrectly describes allolactose binding to the operator rather than the repressor protein.

Question 4 of 10 · Prokaryotic vs Eukaryotic Gene Regulation

Which of the following best describes a key difference between transcription in prokaryotes and eukaryotes?
  1. A. Prokaryotic transcription uses RNA polymerase, whereas eukaryotic transcription uses DNA polymerase.
  2. B. In eukaryotes, transcription and translation are spatially separated (nucleus and cytoplasm), while in prokaryotes they can occur simultaneously.
  3. C. Eukaryotic genes lack promoters, while prokaryotic genes have complex promoter sequences.
  4. D. Eukaryotic mRNA is translated directly without any post-transcriptional processing.
Show answer and explanation

Correct answer: B

Eukaryotic transcription occurs in the nucleus and the mature mRNA must be exported to the cytoplasm for translation, making these two processes physically and temporally separated. Prokaryotes lack a nuclear membrane, so ribosomes can translate mRNA while it is still being transcribed (coupled transcription-translation). Both domains use RNA polymerase for transcription.

Question 5 of 10 · Meiosis and Independent Assortment

A genetics student performed a series of crosses to investigate the inheritance of two traits in pea plants: seed color (yellow, Y dominant over green, y) and seed texture (round, R dominant over wrinkled, r). The student crossed two true-breeding strains to obtain an F1 dihybrid, then performed a testcross with a homozygous recessive plant. The progeny counts and chi-square analysis are shown below.

Testcross Progeny: F1 (YyRr) × yyrr
PhenotypeObserved (O)Expected (E)(O−E)2/E(O - E)^2 / E
Yellow, Round2182001.62
Yellow, Wrinkled1862000.98
Green, Round1922000.32
Green, Wrinkled2042000.08
**Total**800800χ2=3.00\chi^2 = 3.00

Critical value for chi-square at p=0.05p = 0.05 with 3 degrees of freedom: χcrit2=7.815\chi^2_{crit} = 7.815

The gene for seed color resides on chromosome 1 and the gene for seed texture resides on chromosome 7 of the pea genome. Which of the following cellular events most directly produces the four gamete types (YR, Yr, yR, yr) in approximately equal proportions?
  1. A. DNA replication during the S phase preceding meiosis, which produces identical sister chromatids for each allele on both chromosomes
  2. B. Random orientation of homologous chromosome pairs at the metaphase I plate, which independently distributes non-homologous chromosomes into gametes
  3. C. Crossing over between chromosomes 1 and 7 during prophase I, which physically transfers alleles between the two non-homologous chromosomes
  4. D. Cytokinesis after meiosis II, which partitions cytoplasm so that each gamete receives exactly one allele at each locus
Show answer and explanation

Correct answer: B

Random alignment of bivalents at metaphase I means the orientation of chromosome 1 is independent of chromosome 7; this random assortment produces the four gamete classes in equal frequency when the two genes are on non-homologous chromosomes. Choice C is incorrect because crossing over occurs between homologous chromosomes (not between non-homologous chromosomes 1 and 7), so it does not generate the four gamete classes observed here.

Question 6 of 10 · Prokaryotic vs Eukaryotic Gene Regulation

A research team studied the expression of the lac operon in E. coli under four experimental conditions. They measured beta-galactosidase activity as a proxy for lac operon expression.

Beta-galactosidase activity under different conditions (relative units)
ConditionLactose presentGlucose presentBeta-galactosidase Activity
1NoNo10 (baseline)
2YesNo980 (high)
3NoYes8 (very low)
4YesYes75 (moderate)
The lac operon is a prokaryotic regulatory mechanism. Which of the following correctly distinguishes a key difference between prokaryotic and eukaryotic gene regulation?
  1. A. Prokaryotes use operons to coordinate related genes in a single transcriptional unit, while eukaryotic genes are typically regulated individually from separate promoters.
  2. B. Eukaryotic genes cannot be turned off because they lack repressor proteins.
  3. C. Prokaryotic transcription occurs in the nucleus, allowing mRNA to be processed before translation.
  4. D. Only eukaryotes use transcription factors to regulate gene expression.
Show answer and explanation

Correct answer: A

Operons allow prokaryotes to regulate functionally related genes (e.g., lac genes) as a single polycistronic transcript. Eukaryotic genes are generally regulated individually with complex enhancers and transcription factors acting on each gene separately. Prokaryotic transcription occurs in the cytoplasm, not the nucleus, and prokaryotes do use regulatory proteins similar to transcription factors.

Question 7 of 10 · Frameshift Mutations and Reading Frame Disruption

A frameshift mutation is caused by the insertion of a single adenine nucleotide at position 10 in the following mRNA sequence. Which of the following best describes the molecular consequence of this insertion on the resulting protein?
  1. A. Only the amino acid encoded by the codon containing position 10 will be changed; downstream amino acids remain unaffected
  2. B. The insertion will be corrected by the ribosome's proofreading function before translation is complete
  3. C. All amino acids from the insertion site to the end of the protein will be encoded by a shifted reading frame, producing an altered sequence that likely includes a premature stop codon
  4. D. The protein will be longer than normal because the additional nucleotide adds one amino acid to the sequence
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Correct answer: C

A single-nucleotide insertion shifts the reading frame for all downstream codons, producing a novel amino acid sequence from the insertion point onward — this altered reading frame almost invariably encounters an out-of-frame stop codon, resulting in a truncated, nonfunctional protein. Choice A describes the effect of a missense mutation (single base substitution), not a frameshift.

Question 8 of 10 · Types of Mutations and Their Effects on Proteins

A single base substitution in a gene changes the codon UCA (serine) to UGA (stop codon). Which of the following best predicts the effect of this mutation on the protein product?
  1. A. The mutation will be silent, producing a protein identical in sequence to the wild-type because of codon degeneracy
  2. B. The mutation will substitute one amino acid for another, potentially changing the protein's tertiary structure
  3. C. The mutation will cause a frameshift that alters every amino acid downstream of the mutation site
  4. D. The mutation will produce a truncated, likely nonfunctional protein that terminates prematurely at the mutation site
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Correct answer: D

Replacing a sense codon with a premature stop codon (UGA) is a nonsense mutation — the ribosome terminates translation at that point, producing a truncated polypeptide that typically lacks critical functional domains. Choice C describes a frameshift mutation caused by an insertion or deletion, not a single base substitution.

Question 9 of 10 · Types of Mutations

A frameshift mutation is caused by the insertion of a single nucleotide in the middle of a coding sequence. Which of the following best predicts the effect of this mutation on the protein?
  1. A. The reading frame shifts for all codons downstream of the insertion, likely producing a completely different and truncated protein.
  2. B. Only the amino acid at the insertion site is changed; all downstream amino acids remain the same.
  3. C. The insertion duplicates one codon, adding a single extra amino acid to the protein chain.
  4. D. The mutation is silent because the genetic code is redundant and can accommodate extra nucleotides.
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Correct answer: A

A single-nucleotide insertion shifts the reading frame for all codons 3' of the insertion, changing every downstream amino acid and usually encountering a premature stop codon — producing a severely altered, typically nonfunctional protein. Missense mutations change only one amino acid; frameshifts affect the entire downstream sequence.

Question 10 of 10 · Types of Mutations

Students analyzed the effect of a single nucleotide substitution on the amino acid sequence of a protein. The original DNA template strand sequence and mutant sequences are shown. The genetic code table is provided for reference.

Original and mutant DNA template strand sequences (5'→3' shown for coding strand; template strand is complementary)
SequenceCodon region (coding strand 5'→3')Amino acid encoded
Original...GAA...Glutamate (Glu)
Mutant 1...GGA...Glycine (Gly) — missense
Mutant 2...GAG...Glutamate (Glu) — synonymous
Mutant 3...TAA...STOP — nonsense
Which mutation would most likely have the greatest negative effect on protein function?
  1. A. Mutant 1, because a missense mutation changes the amino acid, potentially altering protein folding.
  2. B. Mutant 2, because synonymous mutations always disrupt protein structure.
  3. C. Mutant 3, because a nonsense mutation introduces a premature stop codon, truncating the protein.
  4. D. All three mutations have equal effects because only one nucleotide is changed in each.
Show answer and explanation

Correct answer: C

A nonsense mutation introduces a premature stop codon, producing a truncated, almost certainly nonfunctional protein — typically the most severe effect. Missense mutations may or may not significantly affect function depending on location, while synonymous mutations (Mutant 2) produce the same amino acid and usually have no effect.

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