AP Biology · Unit 6 · 12–16% of Exam ⚡ SPRINT MODE
Gene Expression
& Regulation
Tied for the highest exam weight. The FRQ favorite: trace a DNA mutation all the way to a phenotype change. Master the central dogma directions, lac operon 4 states, and every mutation type — these alone cover half this unit's exam points.
Semiconservative ReplicationTemplate vs Coding Strand
mRNA ProcessingCodon / Anticodon
Lac OperonEpigenetics
Silent / Missense / Nonsense / FrameshiftPCR / Gel
⚡ Quick Glance — All Topics at a Glance
| Topic | Priority | Exam Format | Key Trap / Must-Know |
|---|---|---|---|
| 6.1–2 DNA Structure & Replication | ★★★ | MCQFRQ | Semiconservative; DNA pol only adds 5'→3'; lagging strand = Okazaki fragments + ligase |
| 6.3 Transcription & RNA Processing | ★★★ | MCQFRQ | RNA pol reads TEMPLATE strand 3'→5'; mRNA same as CODING strand (T→U). Introns OUT, Exons IN |
| 6.4 Translation | ★★★ | MCQFRQ | AUG = start (Met); UAA/UAG/UGA = stop (no amino acid); anticodon is ANTIPARALLEL to codon |
| 6.5 Gene Regulation | ★★★ | MCQFRQ | Lac operon: 4 states. Transcription/translation COUPLED in prokaryotes only |
| 6.6 Cell Specialization | ★★ | MCQ | All cells have same genome; differential gene expression → cell identity. HOX genes = body plan |
| 6.7 Mutations | ★★★ | MCQFRQ | Point mutation = substitution (never frameshift). Frameshift = insertion or deletion ONLY |
| 6.8 Biotechnology | ★★★ | MCQData | Gel: small = travels FURTHER (toward +). PCR: 95°C denature → 55°C anneal → 72°C extend |
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Topics 6.1–6.2
DNA Structure & Replication
The Central Dogma — Know Every Arrow
DNA
→
Replication
DNA
→
Transcription
RNA
→
Translation
Protein
Some viruses
(e.g. retroviruses):
RNA→DNA via
reverse transcriptase
(e.g. retroviruses):
RNA→DNA via
reverse transcriptase
Core Concept
🔁 Semiconservative Replication
- Each new DNA molecule: one parental strand + one new strand
- Proven by Meselson-Stahl experiment (1958): ¹⁵N/¹⁴N density labeling
- After 1 round in ¹⁴N: ALL hybrid (intermediate density) → not conservative
- After 2 rounds: half hybrid + half light → confirms semiconservative
- Occurs in S phase of cell cycle (before mitosis)
Enzyme Roles — Memorize
⚙️ Replication Enzymes
- Helicase: unwinds double helix, breaks H-bonds at replication fork
- Topoisomerase: relieves supercoiling tension ahead of helicase
- Primase (RNA pol): synthesizes short RNA primer (provides free 3'–OH)
- DNA Polymerase: adds nucleotides 5'→3'; proofreads; CANNOT start de novo
- Ligase: seals nicks between Okazaki fragments on lagging strand
Strand Synthesis
➡ Leading vs. Lagging Strand
- DNA pol only synthesizes 5'→3' — this is non-negotiable
- Leading strand: template 3'→5'; pol moves toward fork; continuous synthesis; 1 primer
- Lagging strand: template 5'→3'; pol moves away from fork; discontinuous → Okazaki fragments; multiple primers; ligase joins fragments
- Okazaki fragments: short DNA segments (~100–200 nt) each initiated by separate RNA primer
🎯 Exam Sniper
- MCQ (Meselson-Stahl): "DNA was labeled with ¹⁵N, then replicated once in ¹⁴N. The density of the new DNA molecules would be..." → intermediate (hybrid) — one strand ¹⁵N, one strand ¹⁴N = semiconservative evidence
- MCQ: "Why is an RNA primer necessary for DNA replication?" → DNA polymerase cannot initiate a new strand — it can only add nucleotides to an existing 3'–OH group. Primase (an RNA polymerase) can start de novo
- MCQ: "A drug inhibits ligase. Which strand is most severely affected?" → Lagging strand — it consists of many Okazaki fragments that cannot be joined into a continuous strand without ligase
Topic 6.3
Transcription & RNA Processing
The Most-Tested Direction Rule in AP Bio
Template strand
3'— T A C G G A T C A —5' ← RNA pol reads this direction
Coding strand
5'— A T G C C T A G T —3' ← same sequence as mRNA (T→U)
mRNA produced
5'— A U G C C U A G U —3' ← identical to coding strand, U instead of T
The Rules
📋 Transcription Directions
- RNA polymerase reads template strand 3'→5'
- RNA is synthesized 5'→3'
- mRNA sequence = same as coding strand (except T→U)
- RNA pol binds promoter (TATA box) → recruits transcription factors
- RNA pol does NOT need a primer (starts de novo — unlike DNA pol)
- No proofreading (more error-prone than DNA replication)
Eukaryotic Only — High Exam Yield
✂️ mRNA Processing (Pre-mRNA → mRNA)
- 5' cap (modified guanosine): added first; protects mRNA from degradation; aids ribosome attachment
- Poly-A tail (poly-adenylate, ~200 A's): added to 3' end; protects mRNA; aids export from nucleus
- RNA splicing: introns (INtervening sequences) removed; exons (EXpressed) joined together
- Introns OUT, Exons IN — mnemonic: "EXons are EXpressed; INtrons go IN the trash"
- Alternative splicing: different exon combinations → different proteins from one gene → proteome diversity exceeds genome
Prokaryote vs. Eukaryote
🔬 Key Differences
- Prokaryotes: no nucleus → transcription and translation are coupled (simultaneous); no RNA processing; polycistronic mRNA (multiple genes per mRNA)
- Eukaryotes: transcription in nucleus, translation in cytoplasm → NOT coupled; mRNA processed before export; monocistronic mRNA
- RNA polymerases: prokaryotes have 1; eukaryotes have 3 (RNA pol I, II, III) — Pol II makes mRNA
🎯 Exam Sniper
- MCQ (top hit): "The template strand of a gene is 3'-TACGGA-5'. What is the sequence of the mRNA transcribed from it?" → Read template 3'→5', build mRNA 5'→3' with RNA bases: 5'-AUGCCU-3'
- MCQ: "A drug prevents the addition of the 5' cap to eukaryotic mRNA. What is the most likely effect?" → mRNA degraded faster → less protein produced; also poor ribosome attachment → reduced translation
- MCQ: "Two proteins with different functions are encoded by the same gene. This is possible because of..." → Alternative splicing — different combinations of exons produce different mRNA and thus different proteins
- FRQ: "Explain why transcription and translation occur simultaneously in E. coli but not in human liver cells." → E. coli is prokaryotic, has no nuclear membrane — ribosomes attach to mRNA as it's being transcribed. Human liver cells are eukaryotic — transcription in nucleus, mRNA must be processed and exported; translation occurs separately in cytoplasm
💣 Trap Alert
- ❌ RNA polymerase reads the template strand (not the coding/sense strand). The mRNA is complementary to the template, which means it has the same sequence as the coding strand (except U instead of T)
- ❌ Introns are spliced OUT; Exons are kept. Many students get these backwards
- ❌ Coupled transcription/translation = prokaryotes only (no nuclear envelope). Eukaryotes spatially separate the two processes
Topic 6.4
Translation
Translation Machinery
🏭 Three RNAs, One Goal
- mRNA: carries genetic message from nucleus; read 5'→3' in triplets (codons)
- tRNA: adaptor molecule; anticodon (3 nt) pairs with complementary mRNA codon; carries specific amino acid at 3' end (CCA)
- rRNA: structural and catalytic component of ribosomes; rRNA catalyzes peptide bond formation (ribozyme)
- Ribosomes: 3 sites: A (aminoacyl — incoming tRNA), P (peptidyl — growing chain), E (exit — leaves)
Codons to Memorize
🔑 Key Codons
- AUG = start codon; codes for Methionine (Met); translation initiates here — note: the initial Met may be removed during post-translational processing in the finished protein
- UAA, UAG, UGA = stop codons; do NOT code for any amino acid; release factors (not tRNAs) recognize them → polypeptide released
- Genetic code is redundant (multiple codons for one AA) but unambiguous (one codon = one AA)
- Genetic code is universal across all life → evidence of common ancestry
- Anticodon reads codon antiparallel: mRNA 5'-AUG-3' pairs with tRNA anticodon 3'-UAC-5'
Three Stages
⚙️ Initiation → Elongation → Termination
- Initiation: ribosome assembles at AUG; initiator tRNA (Met) enters P site
- Elongation: aminoacyl-tRNA enters A site; peptide bond forms (dehydration); ribosome translocates 5'→3' (moves one codon); tRNA exits at E site; cycle repeats
- Termination: stop codon in A site; release factor binds; polypeptide released; ribosome disassembles
- One mRNA can be translated by multiple ribosomes simultaneously (polysome/polyribosome)
🎯 Exam Sniper
- Codon table MCQ (every year): Given a codon table and a mRNA sequence, identify the amino acid sequence. Always read mRNA 5'→3', in triplets starting from AUG
- MCQ: "A tRNA has the anticodon 3'-UAC-5'. Which amino acid does it carry?" → Read anticodon antiparallel → mRNA codon is 5'-AUG-3' = Met. Anticodons are always written 3'→5' (antiparallel to mRNA)
- MCQ: "What happens when a ribosome reaches UAG on the mRNA?" → A release factor (not a tRNA) binds → polypeptide is released → ribosome disassembles. No amino acid is added for stop codons
- FRQ: "The genetic code is described as universal. What does this suggest about the evolution of living organisms?" → All organisms use the same genetic code → they share a common ancestor in which the current code became established. This is powerful evidence for universal common descent
Topic 6.5
Regulation of Gene Expression
Prokaryotic Regulation — Lac Operon (All 4 States)
No Lactose / Glucose Present
🔴 Operon OFF
Repressor protein binds operator → blocks RNA pol → no transcription. Glucose present → low cAMP → CAP inactive → no positive activation. Result: no lactose-metabolizing enzymes made
Lactose Present / Glucose Present
🟡 Operon LOW
Allolactose binds repressor → repressor leaves operator → RNA pol can proceed. BUT glucose present → low cAMP → CAP inactive → low transcription. Some enzymes made but at low level
Lactose Present / No Glucose
🟢 Operon MAX ON
Allolactose removes repressor. No glucose → high cAMP → cAMP binds CAP → CAP-cAMP binds activator site → strongly activates transcription. Maximum enzyme production
No Lactose / No Glucose
🔴 Operon OFF
Repressor binds operator (no allolactose to remove it) → no transcription possible regardless of CAP status. Even though cAMP is high (CAP active), repressor blocks RNA pol. Operon stays off — no substrate to metabolize
Lac Operon Components
🧬 Know the Parts
- Promoter: RNA pol binding site (+ CAP activator site)
- Operator: repressor binding site (between promoter and genes)
- Structural genes: lacZ, lacY, lacA — encode lactose-metabolizing enzymes (β-galactosidase, permease)
- Repressor (lacI): binds operator when no lactose → blocks transcription
- Allolactose: the actual inducer; a metabolite of lactose; binds repressor → repressor leaves operator. (It is allolactose — not lactose itself — that binds the repressor; lactose availability leads to allolactose production)
- CAP + cAMP: positive regulator; activates transcription when glucose absent (high cAMP)
Eukaryotic Gene Regulation
🎛 Multiple Levels of Control
- Chromatin remodeling: histone acetylation → looser chromatin → ↑ transcription; histone deacetylation → tighter → ↓ transcription
- DNA methylation: methyl groups on cytosine → gene silencing; inherited through cell division (epigenetics)
- Transcription factors: proteins that bind promoters/enhancers → recruit or block RNA pol
- Enhancers/silencers: DNA sequences far from gene → activator/repressor proteins bind → loop to promoter → modulate transcription
- miRNA/siRNA: small RNAs that bind complementary mRNA → block translation or cause degradation (post-transcriptional regulation)
🎯 Exam Sniper
- MCQ (top hit): "E. coli is grown in medium with lactose but no glucose. What is the state of the lac operon?" → Maximally ON: allolactose removes repressor AND no glucose → high cAMP → CAP activated → highest transcription level
- MCQ: "A mutation permanently inactivates the lac repressor. What happens to lac gene expression?" → Genes are constitutively expressed (always on) regardless of lactose presence — the repressor can never bind the operator
- FRQ: "Explain how DNA methylation regulates gene expression in eukaryotes." → Methylation of cytosine residues in gene promoter regions attracts proteins that compact chromatin → blocks transcription factor and RNA pol access → gene silenced. Methylation patterns are heritable (epigenetic) and can be reversed
- MCQ: "miRNA prevents expression of a target gene without changing the DNA sequence. At what level does this regulation occur?" → Post-transcriptional — miRNA degrades mRNA or blocks ribosome → protein not made, but DNA/mRNA sequence unchanged
Topic 6.6
Gene Expression & Cell Specialization
Core Concept
🔬 Differential Gene Expression
- All cells in an organism have the same genome (same DNA)
- Different cell types express different subsets of genes → different proteins → different functions
- Cell identity determined by transcription factor combinations active in that cell
- Differentiation: stem cell → specialized cell via progressive gene activation/silencing
- Totipotent: can become any cell type (early embryo cells, zygote)
- Pluripotent: can become most cell types (embryonic stem cells)
- Multipotent: can become limited cell types (adult stem cells)
Body Patterning
🧬 HOX Genes & miRNA
- HOX genes: master regulatory genes; encode transcription factors that activate/repress other genes; determine body segment identity (which genes active where along body axis)
- HOX genes highly conserved across animals → evidence of common ancestry
- Mutation in HOX gene → homeotic transformation (e.g., leg growing where antenna should be in Drosophila)
- miRNA/siRNA: post-transcriptional silencers; complement and degrade target mRNA or block translation; essential for normal development
🎯 Exam Sniper
- MCQ: "A liver cell and a muscle cell contain the same DNA sequence, yet produce different proteins. What explains this?" → Differential gene expression — different transcription factors active in each cell type activate different genes from the same genome
- MCQ: "HOX genes are found in organisms as different as flies and humans, with very similar sequences. What does this suggest?" → Common evolutionary ancestry — these highly conserved genes likely existed in a common ancestor and were maintained because mutations would be lethal (too important to change)
Topic 6.7
Mutations
Mutation Types — Master This Table
| Type | What Changes | Effect on Protein | AP Example |
|---|---|---|---|
| Silent | Nucleotide substitution → different codon → same amino acid (code redundancy) | ✅ No change — protein unchanged | GAA → GAG (both Glu); least harmful |
| Missense | Nucleotide substitution → different codon → different amino acid | ⚠️ May change protein shape/function (if critical region) | Sickle cell: GAG→GTG (Glu→Val) in hemoglobin — changes RBC shape |
| Nonsense | Nucleotide substitution → codon becomes premature stop codon | ❌ Truncated (shorter) protein — usually non-functional | CAA → UAA → stop; protein cut short |
| Frameshift (insertion) | 1–2 extra nucleotides inserted → shifts reading frame for all downstream codons | ❌ Completely different amino acid sequence after insertion point → usually non-functional | ATGCGA → ATGCCGA (all codons after insertion changed) |
| Frameshift (deletion) | 1–2 nucleotides deleted → shifts reading frame for all downstream codons. Original: AUG-CGA-UUC-GAA→ Del 1 nt: AUG-GAU-UCG-AA → all codons after deletion changed | ❌ Completely different amino acid sequence after deletion point → usually non-functional | Cystic fibrosis ΔF508: 3-nt deletion (in-frame, no frameshift but removes one AA) |
⚠ AP EXAM NOTE: Insertion or deletion of 3 nucleotides (or multiples of 3) does NOT cause a frameshift — it adds or removes amino acids but keeps the rest of the reading frame intact. Only insertions/deletions of 1, 2, 4, or 5 nucleotides (non-multiples of 3) cause frameshifts.
Mutation Severity — General Guidance
📊 Which Mutations Are Usually More Disruptive?
- Frameshifts and nonsense mutations are often most disruptive — they affect all downstream codons or truncate the protein
- Missense mutations in the active site / binding site tend to be more damaging than missense mutations in non-critical regions
- Silent mutations typically have no effect on protein
- Severity always depends on context: location in the gene, whether the affected region is functionally critical, and whether the amino acid change alters charge/polarity
- A missense near the C-terminus may be harmless; a nonsense near the N-terminus may eliminate nearly the entire protein
Prokaryotic Gene Transfer
🦠 Horizontal Gene Transfer
- Transformation: bacteria take up free DNA from environment (e.g., Griffith's experiment)
- Transduction: bacteriophage accidentally carries bacterial DNA from one cell to another
- Conjugation: direct cell-to-cell DNA transfer via pilus (F plasmid); spreads antibiotic resistance rapidly
- Note: Transposition (jumping genes) is beyond AP core scope — focus on the three above
🎯 Exam Sniper
- FRQ (top pattern): "A single base substitution in the coding region of a gene changes codon 6 from GAG to GUG. Predict the effect on the protein and explain." → Missense mutation — changes amino acid 6 from Glu (glutamic acid) to Val (valine). If this position is critical (e.g., active site of enzyme, oxygen-binding site of hemoglobin), protein function may be lost. This is the sickle cell mutation
- MCQ: "Which mutation type would most severely affect ALL amino acids downstream of the mutation?" → Frameshift (insertion or deletion of non-multiple of 3) — shifts reading frame for every codon after the mutation point
- MCQ: "A mutation changes a codon from AUG to GUG at position 1. What is the effect?" → Eliminates start codon → no initiation of translation → no protein produced at all. First codon is critical!
A mutation converts the 4th codon in a gene from CGA (Arg) to UGA. Which of the following best predicts the effect on the resulting protein?
- (A) The protein will have a different amino acid at position 4, changing its shape slightly
- (B) The mutation shifts the reading frame and produces a completely different protein
- (C) The protein will be truncated — translation terminates at codon 4 and a very short, likely non-functional polypeptide is made
- (D) No protein is made because the start codon is affected
Answer: (C) — CGA → UGA is a single nucleotide substitution (point mutation), specifically a nonsense mutation because UGA is a stop codon. Translation will terminate at codon 4 (the first 3 amino acids are Met + 2 others, then stop). The resulting polypeptide (only 3 amino acids) is far too short to fold into a functional protein. This is NOT a frameshift (no insertion/deletion), NOT a missense (not a different amino acid), and the start codon (AUG, codon 1) is unaffected (D).
Topic 6.8
Biotechnology
PCR — Polymerase Chain Reaction
~95°C
Denaturation
Heat breaks H-bonds between complementary strands → DNA double helix separates into 2 single strands
~55°C
Annealing
Temperature lowered → short DNA primers bind (anneal) to complementary sequences flanking the target region
~72°C
Extension
Taq polymerase (heat-stable; from Thermus aquaticus) extends from primers, synthesizing new DNA strands 5'→3'
Each cycle doubles the DNA → 2ⁿ copies after n cycles → 1 billion copies from 1 template in ~30 cycles
Gel Electrophoresis — Read This Right
Gel Orientation
→ + (positive)
Key Rules:
🔼 Wells at the TOP — DNA loaded here
➡ DNA migrates toward positive electrode (negatively charged phosphate backbone)
📏 Smaller fragments = travel FURTHER (less impeded by agarose matrix)
📏 Larger fragments = stay near top (more impeded)
📌 Ladder (standard): known-size fragments in one lane — used to estimate unknown fragment sizes
Applications: DNA fingerprinting, RFLP analysis, PCR product verification, protein separation (SDS-PAGE)
🔼 Wells at the TOP — DNA loaded here
➡ DNA migrates toward positive electrode (negatively charged phosphate backbone)
📏 Smaller fragments = travel FURTHER (less impeded by agarose matrix)
📏 Larger fragments = stay near top (more impeded)
📌 Ladder (standard): known-size fragments in one lane — used to estimate unknown fragment sizes
Applications: DNA fingerprinting, RFLP analysis, PCR product verification, protein separation (SDS-PAGE)
Genetic Engineering
✂️ Restriction Enzymes & Cloning
- Restriction enzymes: cut DNA at specific palindromic sequences → create "sticky ends" (complementary overhangs)
- Recombinant DNA: insert foreign gene into vector (plasmid) using same restriction enzyme → sticky ends anneal → ligase seals
- Bacterial transformation: recombinant plasmid introduced into bacterial host → bacteria produce foreign protein (e.g., human insulin)
- Selectable markers (antibiotic resistance) identify transformed bacteria
Modern Tools
🔬 Other Key Biotechnologies
- DNA sequencing: determines exact nucleotide order of DNA (Sanger method uses dideoxy nucleotides)
- CRISPR-Cas9: guide RNA directs Cas9 nuclease to specific sequence → cuts DNA → allows editing; precise, efficient
- Bioinformatics: comparing gene/protein sequences across species → identifies conserved regions, evolutionary relationships
- Gene therapy: delivering functional gene copies to correct genetic disorders
🎯 Exam Sniper
- Data MCQ (top hit): "In a gel electrophoresis result, bands A and B are shown. Band A is closer to the wells; Band B is farther from the wells. Which DNA fragment is larger?" → Band A (closer to wells = larger, slower migration)
- MCQ: "PCR requires Taq polymerase rather than ordinary DNA polymerase. Why?" → Taq polymerase is heat-stable — it survives the 95°C denaturation step. Regular DNA polymerase would denature and lose function at this temperature
- MCQ: "A researcher wants to produce human insulin in bacteria. What would be the FIRST step in creating a recombinant bacterium?" → Isolate the human insulin gene → cut with restriction enzymes → insert into plasmid vector using same restriction enzyme + ligase → transform into bacteria
⚡ FRQ Template
The Molecular Chain of Effects — AP's Favorite FRQ
Mutation → Molecular → Cellular → Organismal Effect Chain
DNA mutation
(substitution/
insertion/
deletion)
(substitution/
insertion/
deletion)
→
transcription
Changed
mRNA codon
sequence
mRNA codon
sequence
→
translation
Changed
amino acid
sequence
amino acid
sequence
→
folding
Changed
protein
shape
protein
shape
→
function
Changed/
lost protein
function
lost protein
function
→
phenotype
Changed
phenotype /
disease
phenotype /
disease
FRQ Scoring Guide — These Steps = Points
- Step 1 — Identify mutation type: substitution (silent/missense/nonsense) vs. insertion/deletion (frameshift or in-frame). Explain why
- Step 2 — mRNA change: State the changed codon(s) in the mRNA (complement of template strand)
- Step 3 — Amino acid change: Use codon table to identify new amino acid (or stop)
- Step 4 — Protein structure: Explain why the new amino acid disrupts structure (if it's in active site, changes charge/polarity/shape, breaks key bond)
- Step 5 — Function change: Enzyme can't bind substrate / protein can't fold properly / signal peptide lost
- Step 6 — Phenotype: How does this manifest in the organism (disease, loss of trait, death)
- AP FRQs award points for each identifiable step — even if you get one step wrong, subsequent correct reasoning can still earn credit
Practice
Sprint Practice — Mixed Questions
A segment of double-stranded DNA has the following sequence:
Strand 1: 5'-ATGCCTGAATCC-3'
Strand 2: 3'-TACGGACTTAGG-5'
If Strand 2 is the template strand, what is the sequence of the mRNA transcribed, and what is the first amino acid encoded?
- (A) mRNA: 3'-UACGGACUUAGG-5'; first amino acid is Met
- (B) mRNA: 5'-AUGCCUGAAUCC-3'; first amino acid is Met (AUG)
- (C) mRNA: 5'-UACGGACUUAGG-3'; first amino acid is Tyr
- (D) mRNA: 3'-AUGCCUGAAUCC-5'; no amino acid is encoded
Answer: (B) — RNA polymerase reads the template strand (Strand 2) in the 3'→5' direction: 3'-TACGGACTTAGG-5'. Building the complementary RNA 5'→3', replacing T with U: 5'-AUGCCUGAAUCC-3'. The first codon is AUG = start codon, encoding Methionine (Met). Note: the mRNA sequence is identical to Strand 1 (the coding/sense strand), but with U instead of T. This is always the case — mRNA = coding strand sequence (U for T).
E. coli cells are placed in a medium containing glucose only (no lactose). A researcher then switches the medium to contain lactose only (no glucose). Which of the following best describes what happens to lac operon transcription during the transition?
- (A) Transcription stays off because glucose is more efficient than lactose as an energy source
- (B) Transcription increases dramatically — the allolactose inducer removes the repressor AND absence of glucose raises cAMP, activating CAP
- (C) Transcription turns on at a low level because the repressor is removed but CAP is not activated
- (D) Transcription stays off because the operator is permanently blocked when glucose is present
Answer: (B) — In the new medium (lactose only, no glucose): (1) Lactose is metabolized to allolactose → allolactose binds repressor → repressor cannot bind operator → RNA polymerase can access structural genes. (2) Absence of glucose → adenylyl cyclase active → cAMP levels rise → cAMP binds CAP protein → CAP-cAMP binds activator site near promoter → stimulates RNA polymerase binding. Both negative control (repressor removed) AND positive control (CAP activated) are simultaneously favorable → maximum transcription of lac genes → E. coli rapidly produces β-galactosidase and permease to metabolize lactose.
A researcher digests a 6,000 bp circular plasmid with restriction enzyme EcoRI and runs the products on a gel. The gel shows two bands: one at approximately 4,000 bp and one at approximately 2,000 bp. Which band appears closer to the wells (top of the gel), and what does this result tell you about the plasmid?
- (A) The 2,000 bp band is near the wells; the plasmid has two EcoRI cut sites
- (B) The 4,000 bp band is near the wells; the plasmid has exactly two EcoRI cut sites producing fragments of 4,000 bp and 2,000 bp
- (C) Both bands are at the same distance; EcoRI cuts all DNA equally
- (D) The 4,000 bp band is near the wells; the plasmid has one EcoRI cut site that linearizes it into a 4,000 bp piece
Answer: (B) — Larger fragments travel more slowly and stay closer to the wells (top). The 4,000 bp fragment = closer to wells; 2,000 bp fragment = farther from wells (migrated faster). Two bands from one circular plasmid means EcoRI cuts the plasmid at exactly 2 sites (2 cuts on a circular molecule produces 2 linear fragments). The two fragments sum to 4,000 + 2,000 = 6,000 bp = total plasmid size (consistent). If only 1 cut site, we'd see one band (linearized plasmid, 6,000 bp).
⚠ Trap Alert
Unit 6 High-Frequency Exam Traps
- 🧬RNA polymerase reads the TEMPLATE strand — mRNA = coding strand (with U)RNA polymerase reads the template (antisense) strand 3'→5' and synthesizes mRNA 5'→3'. The mRNA has the same sequence as the NON-template (coding/sense) strand, except T is replaced by U. Students who confuse which strand is being read produce the wrong mRNA sequence on every problem.
- ✂️Introns OUT, Exons IN — not the other way aroundIntrons are INtervening sequences that are spliced OUT of the pre-mRNA and degraded. Exons are EXpressed sequences that are retained and joined in the mature mRNA. Memory: "EXons are EXpressed; INtrons go IN the trash." Alternative splicing uses different combinations of exons to produce protein diversity.
- 🔄Coupled transcription/translation = prokaryotes ONLYIn bacteria (no nucleus), ribosomes begin translating mRNA while it is still being transcribed. In eukaryotes, transcription happens in the nucleus; mRNA is processed (5' cap, poly-A tail, splicing) then exported; translation happens in the cytoplasm. These are physically separated in eukaryotes.
- ⛔Stop codons (UAA, UAG, UGA) do NOT encode amino acidsStop codons signal termination of translation. Release factor proteins (not aminoacyl-tRNAs) recognize them. No amino acid is added when a ribosome reaches a stop codon — the polypeptide is simply released. Writing an amino acid for a stop codon is a classic exam error.
- ⚡Point mutations are substitutions — they NEVER cause frameshiftsFrameshifts ONLY occur with insertions or deletions of nucleotides (indels). A point mutation replaces one nucleotide with another → one codon changes → at most one amino acid change → no shift in reading frame. Only adding or removing nucleotides shifts the triplet reading frame.
- 📊Gel electrophoresis: smaller = farther from wells (not larger)DNA migrates toward the positive electrode (DNA is negatively charged). Smaller fragments move faster and travel farther from the wells. Larger fragments are more impeded by the gel matrix and stay closer to the wells. "The smaller the faster, the bigger the braker."
- 🌿Not all mutations are harmful — mutations are required for evolutionSilent mutations have no effect on protein. Missense mutations may be neutral, beneficial, or harmful depending on the amino acid change and context. Sickle cell heterozygotes have a survival advantage in malaria regions. Mutations are the ultimate source of genetic variation that evolution acts upon — without them, evolution is impossible.
✓ Last-Min Checklist
Pre-Exam 10-Minute Checklist
Click each item to confirm before exam day.
DNA Replication (6.1–6.2)
- Semiconservative: each daughter DNA = 1 parental + 1 new strand (Meselson-Stahl proof)
- DNA pol only synthesizes 5'→3'; needs RNA primer (primase) to start
- Leading strand: continuous; lagging strand: Okazaki fragments → sealed by ligase
Transcription & Translation (6.3–6.4)
- RNA pol reads template strand 3'→5'; builds mRNA 5'→3'; mRNA = coding strand sequence (T→U)
- Eukaryotic mRNA processing: 5' cap + poly-A tail + splicing (introns OUT, exons IN)
- Coupled transcription/translation = prokaryotes only
- AUG = start (Met); UAA, UAG, UGA = stop (no amino acid, release factor)
- tRNA anticodon pairs antiparallel with mRNA codon; ribosome sites: A → P → E
- Genetic code is universal → evidence of common ancestry
Gene Regulation (6.5–6.6)
- Lac operon: lactose (allolactose) removes repressor; no glucose → high cAMP → CAP activated → max transcription
- Eukaryotes: chromatin remodeling, DNA methylation, transcription factors, enhancers/silencers, miRNA
- All cells have same DNA; differential gene expression → cell specialization
Mutations & Biotechnology (6.7–6.8)
- Silent = same AA; Missense = different AA; Nonsense = premature stop; Frameshift = insertion/deletion shifts reading frame
- Point mutations (substitutions) NEVER cause frameshifts — only indels do
- PCR: 95°C denature → 55°C anneal primers → 72°C Taq extends. Doubles DNA each cycle
- Gel electrophoresis: small = travels FURTHER; large = stays near wells (top)
- Horizontal gene transfer in prokaryotes: transformation, transduction, conjugation
⚡ Final Sprint Strategy for Unit 6
- Top FRQ format: "A mutation changes nucleotide X in gene Y → trace to phenotype." Use the 6-step chain: DNA change → mRNA codon → amino acid → protein structure → function → phenotype. Each step = points
- Strand direction problems: Always establish which is the template strand first. mRNA = coding strand sequence (T→U). Anticodon = antiparallel to codon (3'→5' on tRNA)
- Lac operon shortcut: Two questions to ask: (1) Is lactose present? Yes → repressor removed → operator open. (2) Is glucose absent? Yes → cAMP high → CAP on → max transcription. Both yes = MAX ON; both no = OFF
- Connections: Unit 1 (nucleotide structure); Unit 4 (signal transduction → gene expression); Unit 5 (mutations → heredity); Unit 7 (natural selection acts on phenotype, which comes from gene expression)