--- title: "AP Biology Unit 6 Practice Test" description: "A timed AP Biology Unit 6 (Gene Expression and Regulation) practice test in the AP format, drawn from 2,875 multiple-choice questions and 127 FRQs written for the unit. The standard version takes about 60 minutes." canonical: "https://fiveable.me/ap-bio/unit-test/6" type: "unit-test" subject: "AP Biology" unit: "Unit 6" --- # AP Biology Unit 6 Practice Test ## Overview A timed AP Biology Unit 6 (Gene Expression and Regulation) practice test in the AP format, drawn from 2,875 multiple-choice questions and 127 FRQs written for the unit. The standard version takes about 60 minutes. ## Test Format - 20 Multiple Choice Questions (30 minutes) - 1 FRQ – FRQ 1 – Interpreting and Evaluating Experimental Results (Long) (15 minutes) - 1 FRQ – FRQ 3 – Scientific Investigation (Short) (15 minutes) ## Unit FRQs by Type - [FRQ 1 – Interpreting and Evaluating Experimental Results (Long)](/ap-bio/frqs/experimental-results?unit=6): 35 questions - [FRQ 2 – Interpreting and Evaluating Experimental Results with Graphing (Long)](/ap-bio/frqs/graphing?unit=6): 9 questions - [FRQ 3 – Scientific Investigation (Short)](/ap-bio/frqs/scientific-investigation?unit=6): 16 questions - [FRQ 4 – Conceptual Analysis (Short)](/ap-bio/frqs/conceptual-analysis?unit=6): 19 questions - [FRQ 5 – Analyze Model or Visual Representation (Short)](/ap-bio/frqs/model-analysis?unit=6): 27 questions - [FRQ 6 – Analyze Data (Short)](/ap-bio/frqs/data-analysis?unit=6): 21 questions ## Practice Questions ### Question 1 During DNA replication, the lagging strand is synthesized in discontinuous fragments known as Okazaki fragments. Which structural constraint of DNA polymerase necessitates this discontinuous synthesis? - (A) It adds nucleotides to the 3' end, requiring synthesis away from the fork. - (B) It adds nucleotides to the 5' end, requiring synthesis away from the fork. - (C) It adds nucleotides to the 3' end, requiring synthesis toward the fork. - (D) It adds nucleotides to the 5' end, requiring synthesis toward the fork. **Answer: (A)** Explaining the mechanism of polymerase reveals the cause of discontinuity. Because polymerase can only extend the 3' end, synthesis on the lagging strand must proceed away from the opening replication fork, requiring repeated restarts. ### Question 2 At a replication fork, a student models the leading strand as continuous and the lagging strand as a series of fragments. Template directions are described in the direction the fork opens. Which statement best explains how the antiparallel templates cause this difference in synthesis? - (A) DNA polymerase adds nucleotides to the 3' hydroxyl, so the template running 5' to 3' requires discontinuous synthesis. - (B) DNA polymerase adds nucleotides to the 5' phosphate, so the template running 3' to 5' requires discontinuous synthesis. - (C) DNA polymerase adds nucleotides to the 3' hydroxyl, so the template running 3' to 5' requires discontinuous synthesis. - (D) DNA polymerase adds nucleotides to the 5' phosphate, so the template running 5' to 3' requires discontinuous synthesis. **Answer: (A)** Connecting the enzyme's directional limitation (5' to 3' synthesis) with the antiparallel template orientation explains why one strand must be synthesized in fragments (Okazaki fragments) moving away from the replication fork. ### Question 3 During DNA replication, the lagging strand is synthesized in discontinuous fragments. Which reasoning best justifies the claim that this pattern results from the antiparallel structure of DNA? - (A) DNA polymerase adds nucleotides only to the 3' end, forcing synthesis away from the replication fork on one strand. - (B) DNA polymerase adds nucleotides only to the 5' end, forcing synthesis away from the replication fork on one strand. - (C) DNA polymerase adds nucleotides only to the 3' end, forcing synthesis toward the replication fork on both strands. - (D) DNA polymerase adds nucleotides only to the 5' end, forcing synthesis toward the replication fork on both strands. **Answer: (A)** DNA polymerase adds nucleotides only to the 3' end, so both new strands grow 5' to 3'. The lagging-strand template runs 5' to 3' in the direction the fork opens. Its new strand must therefore be synthesized in short segments away from the fork. ### Question 4 A researcher observes replication forks in a bacterium and notes that while one new DNA strand grows continuously, the other grows in short segments. This asymmetry in synthesis occurs because: - (A) DNA polymerase synthesizes in the 5' to 3' direction on antiparallel templates. - (B) DNA polymerase synthesizes in the 3' to 5' direction on antiparallel templates. - (C) DNA polymerase synthesizes in the 5' to 3' direction on identical templates. - (D) DNA polymerase synthesizes in the 3' to 5' direction on identical templates. **Answer: (A)** Synthesizing the interaction between enzyme directionality and helix structure explains the lagging strand. Since DNA strands are antiparallel and polymerase only works 5' to 3', one strand must be made in fragments moving away from the fork. ### Question 5 A student models DNA replication and hypothesizes that the antiparallel nature of DNA requires discontinuous synthesis on one strand. If this hypothesis is correct, what pattern should appear when analyzing the newly synthesized strands? - (A) One strand forms continuously while the other forms in short fragments. - (B) Both strands form continuously in the 5' to 3' direction without fragments. - (C) One strand forms continuously while the other forms in fragments oriented 5' to 3' and 3' to 5' alternately. - (D) Both strands form in short fragments due to the antiparallel structure requiring multiple initiation points. **Answer: (A)** Predicting the replication pattern requires connecting the antiparallel structure to the 5' to 3' synthesis constraint. Since polymerases only work in one direction, the lagging strand must be built in discontinuous segments (Okazaki fragments). ### Question 6 Bacteria grown in heavy nitrogen (N-15) are transferred to light nitrogen (N-14), so all newly synthesized DNA contains N-14. After one round of replication, the intact double-stranded DNA forms a single intermediate-density band. In the semiconservative model, each daughter DNA molecule contains one parental strand and one newly synthesized strand. In the dispersive model, every daughter strand contains interspersed segments of parental and newly synthesized DNA, with the segments distributed evenly among molecules. Both models predict the observed first-round band. Which follow-up investigation would best distinguish the two models? - (A) Allow a second round of replication in N-14 and compare the number and positions of the DNA density bands. - (B) Repeat the density analysis on the first-round sample with equipment that resolves smaller density differences. - (C) Measure the total amount of DNA before replication and after one complete round of replication. - (D) Compare the proportions of adenine, thymine, guanine, and cytosine before and after the second round of replication. **Answer: (A)** After the second round, semiconservative replication produces two bands: half the DNA molecules have one heavy and one light strand, and half have two light strands. Dispersive replication produces one band closer to the light-DNA position because every molecule contains about one-quarter heavy DNA. These different predictions let the second-round results distinguish the models. ### Question 7 Eukaryotic DNA is tightly packaged into chromatin within the nucleus. Which statement best explains how histone proteins structurally enable this compaction? - (A) Positively charged histones bind to negatively charged DNA phosphate groups. - (B) Negatively charged histones bind to positively charged DNA phosphate groups. - (C) Hydrophobic histone cores bind to hydrophilic DNA nitrogenous base pairs. - (D) Uncharged histone tails bind to negatively charged DNA deoxyribose sugars. **Answer: (A)** The electrostatic attraction between the positive amino acids in histones and the negative phosphate backbone of DNA allows for tight wrapping and nucleosome formation. ### Question 8 A strain of bacteria acquires antibiotic resistance from a neighboring cell without undergoing cell division. Which statement best explains the structure responsible for this genetic transfer? - (A) A small, circular extra-chromosomal DNA molecule called a plasmid. - (B) A small, circular extra-chromosomal DNA molecule called a chromosome. - (C) A small, circular extra-chromosomal RNA molecule called a plasmid. - (D) A small, linear extra-chromosomal DNA molecule called a pilus. **Answer: (A)** Plasmids are small, circular DNA molecules separate from the main bacterial chromosome that can carry and transfer genes like those for antibiotic resistance. ### Question 9 A specific virus uses RNA rather than DNA as its genomic material. Which statement best explains a structural characteristic of this viral genome compared to a typical eukaryotic genome? - (A) It contains ribose sugar and uracil bases instead of deoxyribose and thymine. - (B) It contains deoxyribose sugar and uracil bases instead of ribose and thymine. - (C) It contains ribose sugar and thymine bases instead of deoxyribose and uracil. - (D) It contains deoxyribose sugar and thymine bases instead of ribose and uracil. **Answer: (A)** RNA structure is defined by the presence of ribose sugars and uracil, whereas DNA contains deoxyribose and thymine. ### Question 10 Genetic analysis reveals the presence of DNA inside the mitochondria of a plant cell. Which statement best explains the structural similarity of this DNA to a prokaryotic genome? - (A) It exists as a circular DNA molecule that lacks histone association. - (B) It exists as a circular DNA molecule that maintains histone association. - (C) It exists as a circular DNA molecule that is packaged with prokaryotic-type proteins. - (D) It exists as a circular DNA molecule that contains histone-like proteins similar to eukaryotic chromosomes. **Answer: (A)** Mitochondrial DNA resembles bacterial genomes by being circular and generally lacking the histone packaging found in eukaryotic nuclear DNA. ## Review the Unit - [6.1 DNA and RNA Structure ](/ap-bio/unit-6/dna-rna-structure/study-guide/29u5wVp2a9rSDLfLGBlD) - [6.2 DNA Replication](/ap-bio/unit-6/replication/study-guide/dWnyvQBkJXbdCZAXGCfQ) - [6.3 Transcription and RNA Processing](/ap-bio/unit-6/transcription-rna-processing/study-guide/yDYxXivMePm9Kw0GI5jC) - [6.4 Translation](/ap-bio/unit-6/translation/study-guide/U6N7DadIQajK0Z25lHSh) - [6.5 Regulation of Gene Expression](/ap-bio/unit-6/regulation-gene-expression/study-guide/P4c8bkdRD2cOlW0cbWpB) - [6.6 Gene Expression and Cell Specialization](/ap-bio/unit-6/gene-expression-specialization/study-guide/g4BjZXbPu9cSt2JcU9pU) - [6.7 Mutations](/ap-bio/unit-6/mutations/study-guide/WIuGA11Yy2RsVq8JpSnt) - [6.8 Biotechnology](/ap-bio/unit-6/biotechnology/study-guide/9xwtV4SAygOIewEHrjGK) ## Other Unit Tests - [Unit 1: Chemistry of Life](/ap-bio/unit-test/1) - [Unit 2: Cells](/ap-bio/unit-test/2) - [Unit 3: Cellular Energetics](/ap-bio/unit-test/3) - [Unit 4: Cell Communication and Cell Cycle](/ap-bio/unit-test/4) - [Unit 5: Heredity](/ap-bio/unit-test/5) - [Unit 7: Natural Selection](/ap-bio/unit-test/7) - [Unit 8: Ecology](/ap-bio/unit-test/8) ## About This Document Canonical Fiveable pages are available as Markdown at the same path plus `.md`. - [llms.txt](https://fiveable.me/llms.txt): index of Fiveable's sections and URL patterns - [llms-full.txt](https://fiveable.me/llms-full.txt): complete subject and unit listing - [MCP server](https://fiveable.me/mcp): call Fiveable as tools instead of fetching pages (`https://fiveable.me/api/mcp`) - [MCP server for AP teachers](https://fiveable.me/mcp/teachers): a teacher's classes, assignments and AP-rubric grading (`https://fiveable.me/api/mcp/teacher`)