Biology by Bradford
Unit 6: Gene Expression and Regulation. Practice questions with scoring guidelines.
116 original AP-style questions on Unit 6, written to the CED learning objectives: 88 multiple-choice, 28 free-response. Below is a 21-mark practice set built from them, ready to assign as a unit check or homework, or to sit yourself and score against the guidelines. Print it, project it, or build a fresh one.
Topics in this unit
- 6.1 DNA and RNA Structure 4 questions
- 6.2 Replication 11 questions
- 6.3 Transcription and RNA Processing 11 questions
- 6.4 Translation 8 questions
- 6.5 Regulation of Gene Expression 53 questions
- 6.6 Gene Expression and Cell Specialization 7 questions
- 6.7 Mutations 15 questions
- 6.8 Biotechnology 21 questions
In the bank for Unit 6
- 88 multiple-choice
- 28 free-response
Every question is original and tagged to a learning objective and science practice.
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The practice set
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A single gene is present in every cell of a multicellular organism, yet it is transcribed only in liver cells and not in neurons. Investigation shows that liver cells contain a particular set of transcription factors that bind regulatory sequences of the gene, while neurons lack those factors. Which statement best accounts for the tissue-specific expression of this gene?
- Liver cells and neurons carry different alleles of the gene
- The gene is transcribed in all cells but translated only in liver cells
- The gene has been deleted from the DNA of the neurons
- The gene is transcribed only where its transcription factors are present
Which of the following is an example of regulating gene expression at the level of chromatin rather than transcription initiation?
- Binding of a repressor protein to the operator region of a bacterial operon to block RNA polymerase
- Degradation of an mRNA by complementary microRNAs before it can be translated on the ribosome
- Methylation of DNA and modification of histones that make a gene inaccessible to RNA polymerase
- Phosphorylation of a protein after translation to activate it in response to a signal
During development, a transcription factor present only in muscle precursor cells switches on a set of muscle-specific genes. A mutation prevents this factor from binding DNA. What is the most likely consequence?
- Every cell in the organism becomes a muscle cell, because the factor no longer restricts the genes
- The precursor cells fail to express muscle-specific genes and do not differentiate normally
- Muscle-specific genes are deleted from the precursor cells' genome, because they are no longer needed
- The genetic code is altered in muscle precursor cells only, so those cells translate the genes incorrectly
During DNA replication, an unwound template strand is read in the 3' to 5' direction while a new complementary strand is assembled. Which enzyme is directly responsible for adding free nucleotides one at a time in the 5' to 3' direction to build the new strand?
- RNA polymerase
- Helicase
- DNA polymerase
- Primase
To insert a human insulin gene into a bacterial plasmid, a researcher cuts both the human DNA and the plasmid with the same enzyme so that their ends match, then seals the fragments together into a single circular molecule. Which pair of tools cuts DNA at specific sequences and then joins the fragments together?
- DNA polymerase to cut and helicase to join
- Primase to cut and RNA polymerase to join
- Restriction enzymes to cut and ligase to join
- Gel electrophoresis to cut and PCR to join
In a population of bacteria, a random mutation alters a gene so that the protein normally targeted by an antibiotic no longer binds the drug. When the antibiotic is applied, cells carrying this mutation survive and reproduce while others die. Which statement best explains how this genotype change leads to the resistant phenotype?
- The antibiotic caused the mutation when needed
- The mutation changes the chromosome number
- The altered protein can no longer be inhibited
- The mutation stops the gene being transcribed
In a eukaryotic nucleus, about two meters of DNA must fit inside a cell only micrometers wide, yet it must stay organized so it can be distributed accurately during cell division. The DNA is wound around clusters of proteins to form repeating bead-like units. Which structures allow long DNA molecules to be packaged and passed intact to daughter cells?
- Nucleosomes formed by DNA wound around histone proteins
- Phospholipid bilayers that surround individual genes
- Ribosomes assembled from ribosomal RNA and protein
- Free-floating plasmids in the cytoplasm
The following table gives the amino acid or signal specified by several mRNA codons. Use it to answer the questions in this set.
A cell-free system is given the mature mRNA 5'-AUG UUU GAA AAG UGC UAA-3' along with ribosomes, tRNAs, and amino acids. Using the codon assignments in the table, translation begins at the start codon and proceeds until a stop codon is reached. How many amino acids will the completed polypeptide contain?
| mRNA codon | Amino acid / signal |
|---|---|
| AUG | Met (start) |
| UUU | Phe |
| GAA | Glu |
| GAG | Glu |
| AAG | Lys |
| UGC | Cys |
| UAG | Stop |
| UAA | Stop |
- 4
- 5
- 6
- 7
The coding (nontemplate) strand of a short gene reads 5'-ATG AAA TTT GGG TAA-3'. This gene is transcribed and the resulting mRNA is translated in a cell-free system. Using the codon assignments in the table, how many amino acids are in the completed polypeptide?
| mRNA codon | Amino acid / signal |
|---|---|
| AUG | Met (start) |
| AAA | Lys |
| UUU | Phe |
| GGG | Gly |
| UAA | Stop |
- 3
- 4
- 5
- 6
Researchers studied how a transcription factor called Activator X controls a target gene. They linked different versions of the gene's promoter to a luciferase reporter, which glows in proportion to how much the gene is transcribed. Cells received either a wild-type promoter or a mutant promoter in which the Activator X binding site was deleted, and each was tested with and without Activator X. Relative expression is reported as a percentage of the maximum signal measured in the experiment.
| Promoter construct | Activator X added? | Relative reporter expression (%) |
|---|---|---|
| Wild-type promoter | No | 12 |
| Wild-type promoter | Yes | 95 |
| Mutant promoter (binding site deleted) | No | 10 |
| Mutant promoter (binding site deleted) | Yes | 14 |
A multicellular animal develops from a single fertilized egg into many specialized cell types, such as muscle cells and nerve cells. Nearly all of the animal's somatic cells contain the same complete genome, yet different cell types make different sets of proteins.
A bacterial gene encodes an enzyme required for a metabolic pathway. Researchers study several different mutations that occur in the coding sequence of this gene and examine their effects on the enzyme and on the organism's phenotype.
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Each point is worth 1 point and is credited independently. Accept any one of the listed alternatives per point; ( ) marks optional wording and / separates interchangeable wording.
- D — Cell-type-specific combinations of transcription factors determine which genes are switched on, allowing one shared genome to give rise to many different cell types.
- C — Epigenetic packaging of DNA controls accessibility; the others act at transcription, post-transcription and post-translation.
- The factor is the switch, not the gene. Remove its ability to bind and the target genes stay silent, so the cell never acquires the protein set that makes it a muscle cell.
- C — DNA polymerase catalyzes the addition of complementary nucleotides to the growing strand in the 5' to 3' direction, whereas RNA polymerase transcribes RNA from DNA, helicase unwinds the helix, and primase lays RNA primers.
- C — Restriction enzymes cleave DNA at specific recognition sequences to leave complementary ends, and DNA ligase seals the fragments into a single recombinant plasmid.
- C — The DNA change alters the target protein's structure so the antibiotic cannot bind it, and this changed protein produces the survival (resistant) phenotype; the mutation is random, not caused by the drug.
- A — DNA wraps around histone proteins to form nucleosomes, the packaging units that compact chromatin into chromosomes for faithful transmission of hereditary information.
- B — AUG-UUU-GAA-AAG-UGC code for Met-Phe-Glu-Lys-Cys and UAA is a stop codon, so five amino acids are joined before termination.
- B — The mRNA matches the coding strand with U replacing T (AUG AAA UUU GGG UAA), coding Met-Lys-Phe-Gly before the UAA stop, so four amino acids are joined.
- Adding Activator X sharply increases expression, from about 12% to about 95%
- Activator X raises wild-type expression roughly eightfold
Total for part (A): 1 point
- With the binding site deleted, Activator X barely changes expression (10% to 14%) instead of the large wild-type increase
- Deleting the site eliminates the large activation seen in the wild-type promoter
Total for part (B): 1 point
- Expression would be as high as or higher than the wild-type-plus-Activator-X value
- Higher than 95%, at or near the maximum signal
Total for part (C): 1 point
- More binding sites let more Activator X bind, recruiting more RNA polymerase and further raising transcription
- Because Activator X increases transcription only when bound, adding a second site allows more activation and higher or equal expression
Total for part (D): 1 point
Total for question 10: 4 points
- Different cell types express (transcribe) different subsets of their shared genes, so they make different proteins — this is differential gene expression.
- Each cell type turns on a particular combination of genes while keeping others off, producing distinct proteins from one genome.
Total for part (A): 1 point
- Transcription factors bind regulatory DNA and switch specific genes on or off; the particular set of transcription factors present in a cell determines which genes are transcribed, defining that cell type.
- Cell-type-specific transcription factors activate the genes needed for that cell's identity and leave others inactive.
Total for part (B): 1 point
- The liver cell would begin transcribing muscle-specific genes and take on some muscle-cell characteristics/proteins.
- Muscle-specific genes would be activated, shifting the cell toward a muscle-like pattern of expression.
Total for part (C): 1 point
- Because the muscle genes are present in every cell and are switched on by muscle transcription factors, supplying those factors to a liver cell provides the signals needed to activate the same muscle genes.
- Gene expression is set by which transcription factors are present; adding muscle-cell factors to a liver cell gives it the regulators that turn on muscle genes, so it expresses them.
Total for part (D): 1 point
Total for question 11: 4 points
- Transcription copies the gene's DNA into mRNA (by RNA polymerase), and translation at the ribosome reads the mRNA codons while tRNAs bring the matching amino acids to build the enzyme's polypeptide
- DNA is transcribed into mRNA, then the mRNA is translated into a specific sequence of amino acids that folds into the enzyme
Total for part (A): 1 point
- A nonsense mutation creates a premature stop codon that truncates the protein so most of it is never made, while a missense mutation changes only one amino acid and often leaves much of the protein intact
- An early stop codon prevents most of the polypeptide from being synthesized, whereas a single amino-acid substitution may still allow a partly or fully functional enzyme
Total for part (B): 1 point
- No change to the amino acid sequence; the enzyme is unchanged
- The protein is unaffected because the altered codon still specifies the same amino acid
Total for part (C): 1 point
- The genetic code is redundant, so a synonymous codon still codes for the same amino acid; with the protein sequence unchanged the enzyme's structure and function, and therefore the phenotype, are unchanged
- Because the amino acid sequence is unchanged, the enzyme still folds and works normally, so the phenotype is not affected
Total for part (D): 1 point
Total for question 12: 4 points
Other AP Biology units
- Unit 1 Chemistry of Life 96
- Unit 2 Cell Structure and Function 105
- Unit 3 Cellular Energetics 112
- Unit 4 Cell Communication and Cell Cycle 104
- Unit 5 Heredity 130
- Unit 7 Natural Selection 149
- Unit 8 Ecology 131
All eight units → · IB Biology subtopics → · Open the exam maker →