Lesson 3: Gene Expression, Regulation and BiotechnologyAP Deep Dive

Gene regulation questions separate students who memorised the lac operon diagram from students who can run the operon like a circuit — feeding it mutant genotypes and predicting outputs. Biotechnology questions do the same for gels and maps. This lesson trains both, plus the eukaryotic layers the main course only sketched.

The lac operon as a logic circuit

Three components matter: the repressor (gene I, a diffusible protein — it acts in *trans*, anywhere in the cell), the operator (O, a DNA site — it acts only in *cis*, on its own DNA molecule), and the structural genes (Z codes for β-galactosidase). Allolactose (from lactose) inactivates the repressor. Now run the mutants:

The discriminating exam move is the partial diploid. Pair I⁺ Oᶜ Z⁺ with I⁻ O⁺ Z⁺ on a second DNA molecule: the Z gene next to Oᶜ is ON (its operator ignores the repressor supplied in trans), while the Z gene next to O⁺ is OFF without lactose because the I⁺ repressor reaches it through the cell. That single experiment proves the operator is cis-acting and the repressor trans-acting — and questions reproduce it constantly. CAP–cAMP adds the second input: even with lactose present, glucose high means cAMP low, CAP inactive, and transcription only weak — the operon is an AND gate (lactose present AND glucose absent).

Biotechnology: numbers and maps

Worked example 1 — transformation efficiency. Cells are mixed with 0.25 µg of plasmid DNA, recovered in a total volume of 10 mL, and 0.1 mL is plated, yielding 120 colonies. Total transformants = 120 × (10 / 0.1) = 12,000. Efficiency = 12,000 / 0.25 µg = 48,000 transformants per µg. The two standard errors: forgetting the plating fraction (using 120 directly), and dividing by the volume plated instead of the DNA mass.

Worked example 2 — building a restriction map. A circular plasmid is 4.8 kb. Enzyme A cuts once; enzyme B cuts twice. Digests: A alone → one 4.8 kb band; B alone → 3.0 and 1.8 kb; A + B → 2.2, 1.8 and 0.8 kb. Reasoning: A's single cut linearises the circle (one band, full length). B's two sites divide the circle into 3.0 and 1.8 fragments. In the double digest the 1.8 fragment survives intact — so A does not cut inside it — while the 3.0 fragment is split into 2.2 + 0.8, placing A's site 0.8 kb from one B site (and 2.2 from the other). Every map question is this arithmetic: fragments must sum to the total, and a fragment that disappears in the double digest has been cut.

Worked example 3 — PCR bookkeeping. Starting from 200 template molecules, ideal PCR gives 200 × 2ⁿ copies after n cycles: after 25 cycles, 200 × 2²⁵ ≈ 6.7 × 10⁹. Two refinements earn top marks: exact-length products (bounded by both primers) only appear from cycle 2 onward, because first-cycle products run off the template's end; and real reactions plateau as primers/dNTPs deplete and polymerase activity falls — exponential forever is a model, not a bench result.

Eukaryotic regulation: the extra layers

Prokaryotic-style switches are only the ground floor. Eukaryotes add: chromatin state (acetylation of histones loosens DNA and permits transcription; methylation of DNA typically silences — and can be inherited through mitosis, which is epigenetics); enhancers and transcription factors acting over long distances by DNA looping; alternative splicing, letting one gene produce many proteins (the champion, the fly Dscam gene, can generate tens of thousands of isoforms); small RNAs — miRNAs and siRNAs guide the RISC complex to complementary mRNAs, blocking translation or triggering destruction, the basis of RNA-interference experiments; and mRNA lifetime control via the poly-A tail and untranslated regions. The unifying exam sentence: cells with identical genomes differ because they express different subsets of genes, regulated at every step from chromatin to protein degradation.

Common mistakes at the top band

• "The operator codes for the repressor." The operator is a binding site, not a gene; I codes for the repressor.
• "Lactose binds the operator." Allolactose binds the *repressor*, changing its shape so it releases the operator.
• Gel direction and distance. DNA is negatively charged and runs to the positive electrode; smaller fragments travel farther. Every year, strong students invert one of these under pressure.
• "A mutation in the promoter changes the amino acid sequence." Promoters are not transcribed into the protein; such mutations change *how much* protein is made, not its sequence.
• Treating PCR's 2ⁿ as exact from cycle 1, or claiming PCR creates DNA "from nothing" — it needs template, primers, nucleotides, polymerase and thermal cycling.

AP Exam Experience

1. A bacterium carries a super-repressor mutation (Iˢ) with normal operator and structural genes (Iˢ O⁺ Z⁺). β-galactosidase will be produced

• (A) only when lactose is present
• (B) only when lactose is absent
• (C) at all times
• (D) at no time, because the repressor cannot be removed from the operator even by allolactose

Answer: (D). The Iˢ repressor binds the operator normally but its allolactose-binding site is defective, so lactose can never trigger release. The operon is permanently off — the mirror image of the constitutive I⁻ and Oᶜ mutants.

2. In gel electrophoresis of a DNA digest, the band that travels farthest from the wells contains fragments that are

• (A) the longest, because they carry the most charge
• (B) the shortest, because they move most easily through the gel matrix
• (C) the most negatively charged per molecule
• (D) circular rather than linear

Answer: (B). Charge per unit length is essentially constant for DNA, so separation is by sieving: short fragments thread the gel pores fastest and travel farthest toward the positive electrode.

3. A single DNA molecule is subjected to 30 cycles of ideal PCR. The approximate number of copies of the target region produced is

• (A) 30
• (B) 900
• (C) 2³⁰ ≈ 10⁹
• (D) 30² ≈ 10³

Answer: (C). Each cycle doubles the target (approximately, once exact-length products dominate from cycle 2). 2³⁰ ≈ 1.07 × 10⁹.

Free-response question. A circular plasmid of 6.0 kb carries a gene for ampicillin resistance. Enzyme X cuts the plasmid at one site, enzyme Y at two sites. Digests give: X alone — one band of 6.0 kb; Y alone — bands of 4.0 and 2.0 kb; X + Y — bands of 3.4, 2.0 and 0.6 kb.

(a) Explain why the X-only digest shows a single band even though the plasmid was cut. (b) Determine the position of the X site relative to the Y sites, justifying your answer with the fragment sizes. (c) Bacteria are transformed with the plasmid. The transformation mix (total volume 10 mL) used 0.50 µg of plasmid DNA; 0.1 mL plated on ampicillin agar gives 240 colonies. Calculate the transformation efficiency in transformants per µg. (d) A second plate, spread with the same volume of cells that received no plasmid, also shows 3 colonies on ampicillin agar. Give one explanation for these colonies and state how they affect the efficiency calculation in (c), if at all.

Scoring guideline.

• (a) 1 point. A single cut in a circular molecule produces one linear molecule of the full 6.0 kb, hence one band. Answers must connect circularity to the single full-length fragment; "the enzyme cut once" alone does not earn the point.
• (b) 2 points. 1 point for identifying that the 2.0 kb Y-fragment is unchanged in the double digest, so X does not cut within it. 1 point for the placement: the 4.0 kb Y-fragment is split into 3.4 + 0.6, so the X site lies inside the 4.0 kb fragment, 0.6 kb from one Y site (3.4 kb from the other). Fragment sums must check: 3.4 + 2.0 + 0.6 = 6.0 kb.
• (c) 2 points. 1 point for total transformants = 240 × (10 / 0.1) = 24,000. 1 point for efficiency = 24,000 / 0.50 µg = 48,000 transformants/µg. Omitting the plating-fraction correction (answering 480/µg) earns neither point — the correction is the skill being tested.
• (d) 2 points. 1 point for a valid explanation: spontaneous ampicillin-resistant mutants in the culture, or contamination, or satellite colonies feeding on locally degraded antibiotic (accept any one, plausibly stated). 1 point for the effect: strictly, the background (≈3 colonies per plate ≈ 300 in the total volume ≈ equivalent to ~600 transformants) should be subtracted, slightly lowering the true efficiency — or, acceptably argued, it is negligible (about 1%) relative to 24,000 and within experimental error, provided the student quantifies the comparison rather than ignoring the colonies entirely.

Reading a Sanger sequencing gel

Dideoxy (Sanger) sequencing turns the same logic as a digest gel into a reading task. Four reactions each contain one ddNTP, which terminates synthesis wherever that base is incorporated; the products are run in four lanes (G, A, T, C) and separate by size, shortest at the bottom. You read the sequence of the synthesised strand from the bottom up — smallest fragment first — by noting which lane holds each successive band. Suppose the bands from the bottom read: lane A, lane C, lane G, lane A, lane T, lane C. The synthesised strand is 5′-ACGATC-3′ (in the order synthesised), and the template strand is its complement, 3′-TGCTAG-5′. Two habits prevent the standard errors: read bottom-to-top (the shortest product was terminated earliest, nearest the primer), and remember the gel shows the *new* strand — if the question asks for the template or the gene sequence, you must complement it (and respect antiparallel direction). Modern machines replace lanes with four fluorescent colours in one capillary, but exam gels still use the four-lane logic, and the reading rules are identical.