AP Biology Unit 6: Gene Expression and Regulation MCQs – 30 Practice Questions with Answers (College Board CED)
Free AP Biology practice for Unit 6: Gene Expression and Regulation, covering DNA structure and replication, transcription, translation, gene regulation, mutations, and biotechnology, with a short explanation after every question.
DNA and RNA Structure and Replication
Q1. A DNA molecule is described as antiparallel. What does this mean?
- (A) The molecule has no 5-prime or 3-prime ends
- (B) One strand is made of RNA
- (C) The two strands run in opposite directions
- (D) The two strands have identical base sequences
Show Answer
Answer: C – In a DNA double helix one strand runs 5-prime to 3-prime while its partner runs 3-prime to 5-prime.
Q2. In a sample of double-stranded DNA, 22 percent of bases are adenine. What percentage are guanine?
- (A) 28 percent
- (B) 44 percent
- (C) 78 percent
- (D) 22 percent
Show Answer
Answer: A – Adenine pairs with thymine, so thymine is also 22 percent; the remaining 56 percent is split equally between guanine and cytosine, giving 28 percent guanine.
Q3. In the classic Meselson-Stahl experiment, bacteria grown in heavy nitrogen were transferred to light nitrogen and allowed one round of replication. What band pattern supported semiconservative replication?
- (A) One light band only
- (B) One intermediate band only
- (C) One heavy and one light band
- (D) One heavy band only
Show Answer
Answer: B – Semiconservative replication predicts hybrid molecules after one round, each with one old heavy strand and one new light strand, giving a single intermediate band.
Q4. During replication, one new strand is built continuously while the other is built in short fragments. What are those fragments called?
- (A) Okazaki fragments
- (B) Introns, the non-coding sequences spliced out of pre-mRNA
- (C) Codons
- (D) Primers
Show Answer
Answer: A – The lagging strand is synthesised discontinuously as Okazaki fragments, later joined by ligase.
Q5. Which enzyme unwinds the DNA double helix ahead of the replication fork?
- (A) DNA ligase
- (B) DNA polymerase
- (C) Helicase
- (D) RNA polymerase
Show Answer
Answer: C – Helicase breaks the hydrogen bonds between base pairs and separates the two strands so they can be copied.
Q6. A replication error replaces one base but the new codon still codes for the same amino acid. What is this mutation called?
- (A) Frameshift
- (B) Missense
- (C) Nonsense
- (D) Silent
Show Answer
Answer: D – Because the genetic code is degenerate, some base changes leave the amino acid unchanged, so the protein is unaffected.
Q7. A deletion of a single nucleotide occurs in the middle of a coding sequence. What is the most likely consequence?
- (A) No change to the protein
- (B) A frameshift that alters every codon downstream
- (C) A single amino acid change only, with every later codon read normally
- (D) Loss of the promoter
Show Answer
Answer: B – Deleting one base shifts the reading frame, so all codons after the deletion are read differently.
Q8. A point mutation changes a codon for an amino acid into a stop codon. What is this called, and what is its usual effect?
- (A) Silent; no effect on the protein
- (B) Frameshift; the reading frame shifts
- (C) Missense; one amino acid is swapped
- (D) Nonsense; translation stops early
Show Answer
Answer: D – A nonsense mutation creates a premature stop codon, so translation ends early and a shortened protein is usually produced.
Transcription and Translation
Q9. During transcription, RNA polymerase binds to which DNA region to begin copying a gene?
- (A) An intron
- (B) The terminator
- (C) The promoter
- (D) The ribosome binding site
Show Answer
Answer: C – The promoter is the DNA sequence where RNA polymerase attaches to start transcription.
Q10. In eukaryotic cells, which processing step removes non-coding sequences from a primary transcript?
- (A) Addition of a 5-prime cap
- (B) Addition of a poly-A tail
- (C) Splicing out introns
- (D) Binding of a ribosome
Show Answer
Answer: C – Splicing removes introns from pre-mRNA and joins the remaining exons to form mature mRNA.
Q11. A messenger RNA sequence reads 5-prime AUG CCC UAA 3-prime. Using the standard codon table, where AUG codes for methionine and UAA is a stop codon, how many amino acids will the resulting peptide contain?
- (A) One
- (B) Two
- (C) Three
- (D) Zero
Show Answer
Answer: B – Translation starts at AUG (methionine), adds proline from CCC, then stops at UAA, giving a two-amino-acid peptide.
Q12. Which molecule carries a specific amino acid to the ribosome and matches it to a codon by its anticodon?
- (A) Ribosomal RNA only
- (B) DNA polymerase
- (C) Messenger RNA
- (D) Transfer RNA
Show Answer
Answer: D – Each transfer RNA carries one amino acid and has an anticodon complementary to an mRNA codon.
Q13. A researcher treats cells with a drug that blocks ribosomes from moving along mRNA. Which process is directly stopped?
- (A) Transcription
- (B) Translation
- (C) Splicing
- (D) DNA replication
Show Answer
Answer: B – Ribosome movement along mRNA is the core of translation, so blocking it prevents protein synthesis while DNA and RNA synthesis continue.
Q14. The genetic code is described as nearly universal. What does this imply for biotechnology?
- (A) Only bacteria can translate mRNA, because eukaryotic ribosomes cannot read codons
- (B) A gene from one species can be expressed in another
- (C) All organisms have identical genomes
- (D) Mutations never occur in coding DNA
Show Answer
Answer: B – Because almost all organisms use the same codon-to-amino-acid rules, a gene transferred between species can usually be translated correctly.
Q15. An mRNA codon is changed from UUU to UUC. Both code for phenylalanine. A student claims the protein must change. What is the best evaluation?
- (A) Incorrect, because both codons specify the same amino acid
- (B) Correct, because UUC is a stop codon
- (C) Incorrect, because mRNA is never translated
- (D) Correct, because every base change alters the protein, even when the new codon is synonymous
Show Answer
Answer: A – Degeneracy of the code means several codons can specify the same amino acid, so this synonymous change leaves the protein unchanged.
Q16. Which statement best compares transcription in prokaryotes and eukaryotes?
- (A) Bacterial mRNA is made in the nucleus, then exported to the cytoplasm for translation
- (B) Only prokaryotes use RNA polymerase
- (C) Eukaryotic transcripts are spliced; bacterial transcripts are not
- (D) Only eukaryotes produce mRNA
Show Answer
Answer: C – Eukaryotic pre-mRNA undergoes splicing and other processing, whereas bacterial mRNA is typically translated directly without intron removal.
Regulation of Gene Expression and Cell Specialisation
Q17. In the lac operon of E. coli, what happens when lactose is present?
- (A) RNA polymerase is destroyed
- (B) The repressor binds the operator and blocks transcription
- (C) Lactose inactivates the repressor, allowing transcription
- (D) The operon is permanently deleted
Show Answer
Answer: C – Lactose acts as an inducer: it binds the repressor so it cannot block the operator, and the operon genes are transcribed.
Q18. In the trp operon, high levels of tryptophan lead to what outcome?
- (A) The operon genes are duplicated
- (B) Translation stops in all cells
- (C) The repressor is inactivated and genes switch on, as in the inducible lac operon
- (D) Tryptophan binds the repressor, which then blocks transcription
Show Answer
Answer: D – The trp operon is repressible: tryptophan acts as a corepressor that activates the repressor, switching off further tryptophan synthesis.
Q19. A liver cell and a skin cell from the same person contain the same genome but make different proteins. What best explains this?
- (A) The skin cell has lost liver genes during development, leaving a smaller genome
- (B) Different genes are expressed in each cell type
- (C) Liver cells have extra chromosomes
- (D) Skin cells do not transcribe DNA
Show Answer
Answer: B – Cell specialisation arises from differential gene expression: regulatory mechanisms switch genes on or off in different cell types despite an identical genome.
Q20. Addition of acetyl groups to histones generally has which effect on gene expression?
- (A) It tightens DNA packing and silences genes, as promoter methylation often does
- (B) It loosens chromatin and increases transcription
- (C) It deletes the gene
- (D) It has no effect on transcription
Show Answer
Answer: B – Histone acetylation loosens the association between DNA and histones, making the gene more accessible to transcription machinery.
Q21. Heavy methylation of a gene’s promoter region is most likely to result in what?
- (A) A change in the gene’s base sequence
- (B) Faster translation of its mRNA
- (C) Increased transcription of that gene, because methyl groups open chromatin for RNA polymerase
- (D) Reduced transcription or silencing of that gene
Show Answer
Answer: D – DNA methylation at promoters is associated with reduced gene expression, often as a stable, heritable silencing mark.
Q22. A small RNA molecule binds to an mRNA and prevents its translation. Which regulatory mechanism is this?
- (A) DNA replication
- (B) Splicing
- (C) Operon induction by lactose binding to a repressor protein
- (D) RNA interference by a regulatory RNA
Show Answer
Answer: D – Regulatory RNAs, including microRNAs, can bind target mRNAs and block translation or trigger their breakdown.
Q23. An enhancer sequence lies far from a gene but increases its transcription when a specific transcription factor binds. What does this show?
- (A) Transcription factors work only in prokaryotes
- (B) Transcription depends only on the promoter
- (C) Distant regulatory DNA can control gene expression
- (D) Enhancers code for proteins
Show Answer
Answer: C – Eukaryotic regulation uses enhancers and transcription factors that act over distance, often by looping DNA to contact the promoter.
Q24. A researcher measures mRNA levels for gene X: high in root cells, near zero in leaf cells, although the gene sequence is identical. What is the strongest conclusion?
- (A) Gene X is differentially regulated between cell types
- (B) Root cells have more DNA overall
- (C) mRNA levels do not reflect transcription
- (D) The leaf cells have mutated gene X
Show Answer
Answer: A – Identical gene sequences with different mRNA levels point to regulation of transcription, not to a change in the DNA itself.
Mutations and Biotechnology
Q25. A gel electrophoresis run separates DNA fragments from four samples. The smallest fragments will be found where?
- (A) In the loading buffer
- (B) Closest to the wells
- (C) Furthest from the wells, toward the positive electrode
- (D) Outside the gel
Show Answer
Answer: C – DNA is negatively charged and moves toward the positive electrode; smaller fragments move through the gel matrix faster and travel furthest.
Q26. In a forensic comparison, a suspect’s DNA band pattern matches the sample from a scene at every band, while two other suspects differ at several bands. What is the best interpretation?
- (A) The matching suspect cannot be excluded as the source
- (B) All three suspects are equally likely sources
- (C) Gel electrophoresis cannot compare DNA samples
- (D) The match proves guilt on its own
Show Answer
Answer: A – A full band match means the suspect cannot be excluded, but a match alone does not prove guilt, since it must be weighed with other evidence.
Q27. Why is polymerase chain reaction (PCR) used before analysing a tiny DNA sample?
- (A) It copies a target DNA sequence many times
- (B) It translates DNA into protein
- (C) It removes introns from DNA
- (D) It cuts DNA into fragments
Show Answer
Answer: A – PCR amplifies a specific DNA region through repeated cycles of heating and cooling with primers and DNA polymerase, producing millions of copies.
Q28. A plasmid used to clone a gene also carries a gene for antibiotic resistance. What is the purpose of that marker gene?
- (A) Only bacteria that took up the plasmid grow on antibiotic plates
- (B) It cuts the plasmid open
- (C) It stops bacterial growth entirely
- (D) It makes the plasmid larger
Show Answer
Answer: A – The resistance marker lets researchers select transformed bacteria: only cells carrying the plasmid survive on medium containing the antibiotic.
Q29. A scientist inserts a human gene into bacteria using a plasmid vector. Which step must happen first?
- (A) The bacteria are heated to destroy all DNA
- (B) The plasmid and gene are cut with the same restriction enzyme
- (C) The protein product is purified
- (D) The bacteria divide by meiosis
Show Answer
Answer: B – Cutting vector and insert with the same restriction enzyme produces compatible ends that can be joined by ligase to form recombinant DNA.
Q30. A mutation in a regulatory region increases transcription of a growth gene tenfold, but the protein sequence is unchanged. Which level of gene expression is affected?
- (A) DNA replication rate
- (B) Ribosome structure
- (C) Translation only
- (D) Transcriptional regulation
Show Answer
Answer: D – Because the change alters how much mRNA is made without changing the coding sequence, it acts at the level of transcriptional regulation.
Work through any questions you missed again, then revise those areas before moving on to natural selection.