Mobile Elements, Gene Regulation & Viral Genetics

Mobile Elements, Gene Regulation & Viral Genetics

Part 1 of 2 — Mobile Elements, Defense & Regulation

Where this fits. The Bacterial Genetics & Gene Transfer lesson covered the bacterial genome, mutation and selection, and the three routes of horizontal transfer between cells. This lesson moves DNA within one cell (transposons), shows how bacteria defend against foreign DNA (restriction-modification and CRISPR) and how they regulate genes (operons), then turns to the genetics of viruses.

Transposable Elements ("Jumping Genes")

Transposons are DNA segments that move from one site to another within a cell: chromosome to plasmid, plasmid to plasmid, or to a new chromosomal site. Stacked with plasmids, which move genes BETWEEN cells, they spread resistance.

• Insertion sequence (IS): the simplest element, a transposase gene flanked by short inverted repeats that the transposase recognizes as the element's ends.
• Composite transposon: cargo genes, often antibiotic resistance, flanked by two IS elements. The flanking IS elements supply transposase and ends, so the whole block hops as one unit.
• Cut-and-paste (conservative) transposition excises the element and reinserts it elsewhere, leaving a break at the donor site. Replicative transposition copies it, leaving one copy behind and adding one at the target.

• Target-site duplication: the transposase makes staggered cuts in the target, and host repair fills the gaps, so short direct repeats of target DNA flank every inserted element.
• Replicative transposition passes through a cointegrate (donor and target fused) that a resolvase separates; Tn10 (tetracycline resistance) is the classic composite transposon.
• Integrons capture circular resistance-gene cassettes with an integrase and stack them behind one shared promoter, often riding on transposons and plasmids.
• Barbara McClintock discovered transposable elements in maize, decades before they were found in bacteria.

Genetic consequences of an insertion:

1. Insertional inactivation: landing in a coding sequence knocks the gene out. Transposons are a natural mutagen, and labs use them for insertional mutagenesis, since the resistance marker makes every insertion selectable.
2. Polar effects: an insertion in an upstream gene of an operon can silence the downstream genes of the same polycistronic transcript, though those genes are intact.
3. Mobilization: a resistance gene can move from the chromosome onto a conjugative plasmid, then leave the cell by conjugation.
4. Rearrangement: recombination between two copies of an IS element can delete, invert or fuse DNA, and IS homology is how the F plasmid integrates into the chromosome to form an Hfr donor.

Restriction-Modification: Innate Defense

A restriction endonuclease cuts DNA at a short recognition sequence, often a palindrome (EcoRI cuts GAATTC). Its partner methyltransferase methylates the same sequence in the host's own DNA, and the enzyme cuts only unmethylated sites. Incoming phage DNA, unmethylated, is cut and degraded; the chromosome is spared.

A rare phage genome that is methylated before it is cut escapes, and its progeny, now modified, infect that strain efficiently until they meet a strain with a different system. The system recognizes any DNA carrying the site and keeps no memory of past infections: it is innate. Its enzymes became the cutting tools of molecular cloning.

CRISPR-Cas: Adaptive Immunity in a Single Cell

A CRISPR array is a row of identical repeats separated by unique spacers, each a short fragment copied from a past invader (phage or plasmid).

1. Acquisition: after surviving an infection, the cell inserts a new spacer from the invader's DNA into the array.
2. Expression: the array is transcribed and cut into short CRISPR RNAs, one spacer each.
3. Interference: each RNA guides a Cas nuclease to DNA with the matching sequence, which is cleaved.

The memory is acquired, sequence-specific and heritable: daughter cells inherit the array, so the immunity is passed on. That makes CRISPR adaptive, unlike restriction-modification. Reprogramming the guide RNA turned it into a genome-editing tool (Biotechnology lesson).

• Targets must sit next to a short PAM motif that the Cas protein recognizes. The array's own spacers lack a PAM, which keeps the cell from cutting its own CRISPR locus.

Operons: Regulation You Can Map with Mutants

An operon is a set of genes transcribed from one promoter into one polycistronic mRNA, controlled by an operator next to the promoter.

lac (inducible, catabolic): lacZ (β-galactosidase), lacY (permease) and lacA, with the lacI repressor gene expressed from its own promoter.

• Negative control: the repressor binds the operator; allolactose, made from lactose, binds the repressor and releases it.
• Positive control: low glucose → high cAMP → the CAP-cAMP complex binds near the promoter and recruits RNA polymerase.

Mutant logic. A merodiploid (partial diploid) carries an F′ plasmid, an F fertility plasmid that picked up a stretch of chromosome, so it has two copies of those genes. Merodiploids test which elements act in cis and which in trans:

A cell that cannot make cAMP (no adenylyl cyclase) expresses lac poorly even with lactose and no glucose; adding cAMP rescues it, while a CAP-deficient mutant is not rescued.

• The lacI$^s$ "super-repressor" cannot bind inducer: the operon is uninducible, and lacI$^s$ is dominant over lacI$^+$.

trp (repressible, anabolic): the trp repressor is inactive alone. Tryptophan is the corepressor: when it is abundant it binds the repressor, which then binds the operator and shuts the operon off. The product turns off its own synthesis.

• Attenuation adds a second brake: ribosomes translating a leader peptide rich in Trp codons decide, by stalling or not, whether the leader RNA folds into a terminator hairpin.

Catabolic operons are inducible (the substrate turns them ON); anabolic operons are repressible (the product turns them OFF).

Worked Examples — Mobile DNA and Regulation

Problem 1. A plasmid that cannot replicate in E. coli carries a transposon with a kanamycin-resistance gene. Cells are plated on kanamycin, and survivors are replica-plated onto minimal medium. One mutant has lost three enzymes of one pathway.

• (a) What do the kanamycin plates select?
• (b) What do the minimal plates find?
• (c) How does the resistance gene help once a mutant is found?
• (d) Why could one mutant lose three enzymes at once?

(a) Insertions. The plasmid cannot replicate, so a cell stays kanamycin-resistant only if the transposon hopped into its chromosome. The kanamycin plates select for insertions.

(b) Auxotrophs. Replica plating stamps every colony, in the same positions, onto minimal medium (salts plus a carbon source). Colonies that grow on the master but fail on minimal are auxotrophs: they need a nutrient supplied because the insertion knocked out a biosynthetic gene.

(c) A built-in tag. The resistance gene now tags the mutated gene, so it can be mapped or cloned by selecting for kanamycin resistance.

(d) A polar effect. If the insertion hit the first gene of an operon, it can also silence the downstream genes, so one insertion can cost a whole pathway. Transposon = mutagen + built-in selectable tag.

Problem 2. Predict the β-galactosidase level with no lactose, then with lactose (no glucose), for each strain below; every gene not named as mutant is wild type.

• (a) lacI$^-$ / F′ lacI$^+$: what are the two levels?
• (b) lacO$^c$ lacZ$^-$ / F′ lacO$^+$ lacZ$^+$: what are the two levels?
• (c) A strain that cannot make cAMP: what are the two levels, and what rescues it?
• (d) What single rule predicts every merodiploid?

(a) Repressor in trans. The F′ makes repressor that diffuses to both operators (trans), so expression is normal and inducible: off, then on.

(b) Operator in cis. The constitutive operator controls only its own, dead lacZ. The working lacZ$^+$ sits behind a normal operator, so expression is inducible: off, then on.

(c) No cAMP. Lactose releases the repressor, but without cAMP there is no CAP activation: off, then low. Adding cAMP to the medium restores high induction.

(d) Cis or trans. Proteins (repressor, CAP) act in trans; DNA sites (operator, promoter) act in cis. Locate each mutation: is it in a gene for a diffusible product, or in a site on the DNA? That answer alone predicts the merodiploid.

Problem 3. A phage stock grown on strain X forms many plaques on X but very few on Y. Phage picked from one of those rare plaques on Y now forms many plaques on Y but few on X.

• (a) What difference between X and Y explains the first result?
• (b) How do the rare phage escape on Y?
• (c) Why do the escapees now grow on Y but poorly on X?
• (d) Did the phage's sequence change?

(a) Different R-M systems. X and Y carry different restriction-modification systems. Phage grown on X carry X's methylation pattern, so X's enzyme spares them while Y's enzyme cuts most of them.

(b) Methylated in time. Rare genomes are methylated by Y before they are cut, and they escape.

(c) A new mark. Progeny from that escape carry Y's methylation, not X's, so they now grow well on Y and are restricted by X.

(d) Modification, not mutation. No: its protection is a methylation mark set by the last host it grew in. A change that flips back after one passage through a different host is modification; a mutation would persist on every host.

Transposons, Defense & Operons 🎯

Key Takeaways — Part 1

• Mobile DNA. Transposons move DNA within one cell; composite ones carry cargo, often resistance, between two IS elements.
• Innate vs. adaptive defense. Restriction enzymes cut unmethylated copies of their recognition sequence; CRISPR guides Cas to one remembered invader.
• Insertion consequences. Knockout, a polar effect on downstream genes, and mobilization of resistance onto plasmids.
• Cis and trans. Repressor proteins act in trans, so lacI$^-$ is recessive; operators act in cis, so lacO$^c$ is cis-dominant.

Common traps

• Methylation marks for attack. Methylation protects the host's own sites; the restriction enzyme cuts only unmethylated DNA.
• Glucose raises cAMP. Glucose lowers cAMP, so CAP-cAMP activates lac only when glucose is absent.
• O$^c$ acting in trans. An operator controls only the genes on its own DNA, so O$^c$ beside a dead lacZ gives no constitutive enzyme.
• Tryptophan as an inducer. Tryptophan binds and activates the trp repressor as a corepressor; it does not release it.

Low-yield extras

Target-site duplications flank inserted elements; replicative transposition via a cointegrate and resolvase; Tn10; integrons; McClintock's maize discovery; the PAM motif; the lacI$^s$ super-repressor; trp attenuation.

Mobile Elements, Gene Regulation & Viral Genetics

Part 2 of 2 — Viral Genetics & MCAT Integration

Phage life cycles and the Baltimore classes are taught in the Viruses lesson; this part covers the genetics built on them. Its closing integration table also builds on the Bacterial Genetics & Gene Transfer lesson, sorting the between-cell transfer routes taught there alongside this lesson's within-cell moves.

Prophages and Induction

A temperate phage can integrate its genome into the host chromosome as a prophage, which is copied passively with the chromosome. A phage repressor keeps the lytic genes off and makes the lysogen immune to superinfection by the same phage. Induction: DNA damage, such as UV light, leads to destruction of the repressor; the prophage excises, enters the lytic cycle, and the cell lyses. If excision is imprecise at that moment, host genes beside the insertion site ride along into the next cell the phage infects: specialized transduction (phage-carried transfer of the host genes beside the prophage).

• The damage sensor is the host SOS response: activated RecA promotes self-cleavage of the phage repressor.

Hershey-Chase: DNA Is the Genetic Material of Phage

T2 phage were grown with $^{35}$S, which labels protein (methionine and cysteine; DNA contains no sulfur), or with $^{32}$P, which labels DNA (its phosphate backbone; protein contains almost no phosphorus). After infection, a blender sheared the attached phage coats off the cells, and centrifugation pelleted the cells while the empty coats stayed in the supernatant.

$^{32}$P went into the cells and turned up in progeny phage; $^{35}$S stayed outside. The material that enters and is inherited is DNA. It is the phage counterpart of the Avery, MacLeod and McCarty experiment, in which only DNase destroyed the pneumococcal "transforming principle".

Mapping and Counting Genes in Phage

• Phage crosses: infect cells with two mutants at once (high multiplicity). The genomes recombine inside the cell, and recombinant frequency rises with the distance between the mutations, which gives fine-scale maps of mutation sites inside one gene.
• Complementation test: put two recessive mutants in one cell (co-infection, or a merodiploid in bacteria). If the phenotype is wild type, they complement: each genome supplies the product the other lacks, so the mutations are in different genes. If the phenotype is mutant, they fail to complement and are almost certainly in the same gene.
• Recessive only, with a cis control: a dominant mutation gives the mutant phenotype even beside a working copy, so the test works only for recessive mutations. Its control is the cis arrangement (both mutations on one genome, a wild-type copy on the other), which must look wild type.
• Complementation groups: every mutation that fails to complement the others joins one group, and each group is one gene. Counting groups counts the genes the screen has hit.

• Benzer's rII system: rII mutants of T4 plaque on E. coli B but not on E. coli K(λ). Crossing two rII mutants on B and plating on K selects only wild-type recombinants. Recombinants come in two reciprocal classes and only wild type is scored, so total recombination is twice the wild-type frequency: 0.001 wild-type plaques means 0.2 map units. The complementation unit was named the cistron.
• A rare exception: intragenic complementation, in which two mutant subunits of one multimeric protein can together form a partly working protein.

Why Viruses Mutate So Fast

RNA viruses copy their genomes with an RNA-dependent RNA polymerase, and retroviruses with reverse transcriptase. Most of these enzymes lack 3′→5′ proofreading (coronaviruses are a rare exception), so their error rates run thousands of times or more above cellular DNA replication, which is proofread and then mismatch-repaired. DNA viruses use proofreading DNA polymerases and are far more stable.

• An infected patient carries a swarm of related variants (a quasispecies), so drug-resistant variants exist before treatment. This is the fluctuation-test logic: resistance mutations arise at random during growth, before the selecting agent arrives, rather than in response to it.
• Combination therapy, as for HIV, is the answer: a variant resistant to all drugs at once is far rarer, because independent resistance mutations multiply (1 in $10^4$ for one drug × 1 in $10^5$ for another ≈ 1 in $10^9$ for both).
• In surface proteins, the same error rate drives antigenic drift.

• Order-of-magnitude rates: RNA polymerases and reverse transcriptase make about one error per $10^{4}$ to $10^{5}$ nucleotides copied; proofread and repaired cellular DNA replication, about one per $10^{9}$ to $10^{10}$.

Recombination and Reassortment

• Recombination: two strains in one cell let the polymerase switch templates and produce a hybrid genome; retroviruses package two RNA copies, so reverse transcriptase can switch between them.
• Reassortment: influenza's genome is 8 separate RNA segments; a cell infected by two strains can package segments from both parents into one virion.
• Antigenic drift = point mutations accumulating in hemagglutinin (HA) and neuraminidase (NA), so last year's antibodies bind less well: seasonal epidemics, annual vaccine updates.
• Antigenic shift = reassortment brings in a whole new HA or NA segment, often from an avian strain. The population has no immunity: pandemic risk. Only segmented viruses can shift.

Integration: Which Mechanism Moved the Gene?

Worked Examples — Experiment Reading

Problem 1. In one patient, a carbapenem-resistance gene first sits on a Klebsiella chromosome between two copies of an insertion sequence. Later the same block is on a large plasmid with pilus genes and two other resistance genes, in Klebsiella and then in E. coli.

• (a) Which mechanism moved the block from the chromosome to the plasmid?
• (b) Which mechanism carried it from Klebsiella into E. coli?
• (c) What lab results would confirm the between-cell step?
• (d) Why could treating with one of the other two drugs still enrich carbapenem resistance?

(a) Transposition. A resistance gene between two copies of an insertion sequence is a composite transposon. It changed location inside one cell, chromosome → plasmid, which is transposition, a within-cell move.

(b) Conjugation. The plasmid is an R plasmid: resistance genes plus transfer (pilus) genes, so it can cross species through a mating bridge. Independent mutation cannot explain an identical block in two species on one plasmid.

(c) The two controls. Transfer would be unaffected by DNase, which rules out naked DNA, and blocked by a filter that keeps the cells apart, because conjugation needs contact.

(d) Co-selection. Any one of the three drugs selects for the whole plasmid. Separate within-cell moves (transposition) from between-cell moves (conjugation, transduction, transformation): most resistance passages chain one of each.

Problem 2. Five recessive mutants with the same phenotype are tested in all pairs. Wild type (complementation) appears for every pair EXCEPT 1 + 3 and 2 + 4.

• (a) Which mutants must share a gene?
• (b) Where does mutant 5 belong?
• (c) How many genes are there?
• (d) Can two mutants in one group ever give wild-type progeny, and what would that measure?

(a) Group the failures. Failure to complement puts mutations in the same gene: 1 and 3 form one group, 2 and 4 another.

(b) On its own. Mutant 5 complements every other mutant, so it is a third group by itself.

(c) Three genes. Three complementation groups = three genes: {1, 3}, {2, 4} and {5}. Draw the grid, circle the failures, and group by them; mutants that complement everything stand alone.

(d) Rarely, by recombination. Two mutants within one group can still give rare wild-type progeny by recombination; that tests the distance between sites, not gene identity.

Problem 3. Isolate A (year 1) and isolate B (year 2) differ only by 7 amino acid changes scattered through HA. Isolate C (year 3) has an HA segment that matches a bird strain far better than any human strain; its other seven segments match isolate B.

• (a) What kind of change separates A from B, and what season should follow?
• (b) What kind of change separates B from C?
• (c) What public-health risk does isolate C carry?
• (d) Which kind of genome does each change require?

(a) Antigenic drift. Scattered point changes in one surface protein = antigenic drift. Existing immunity still partly works, so expect a typical season and an updated vaccine.

(b) Antigenic shift. A whole foreign HA segment on a human-strain background = reassortment = antigenic shift.

(c) Pandemic potential. The population has little immunity to the new HA, so C carries pandemic potential.

(d) Segments for shift. Any RNA virus can drift, but only a segmented genome allows the B → C step; an unsegmented RNA virus could only drift and recombine. Whole segments swapped = shift; small changes accumulating = drift.

Problem 4 (low-yield). Two rII mutants are crossed on E. coli B, and the progeny are plated on E. coli K(λ), where only wild type grows. Wild-type plaques are 0.001 of the total progeny.

• (a) Why does each crossover give only one visible recombinant?
• (b) What is the map distance between the two mutations?

(a) One class seen. Each crossover makes one wild-type and one double-mutant recombinant, and only wild type grows on K(λ).

(b) Double it. Total recombinants are 2 × 0.001 = 0.002, or 0.2 map units.

Viral Genetics & Integration 🎯

Key Takeaways — Part 2

• Prophage and induction. A phage repressor keeps the prophage silent; DNA damage leads to the repressor's destruction, and lysis.
• Hershey-Chase. $^{32}$P (DNA) entered the cells and reached progeny; $^{35}$S (protein) stayed outside → DNA is the genetic material.
• Complementation vs. recombination. Rescue in nearly every co-infected cell means different genes; rare heritable wild type is recombination.
• Drift vs. shift. Unproofread RNA copying drives drift in HA and NA; a swapped whole segment is shift, with pandemic risk.

Common traps

• Shared phenotype, shared gene. Mutants with one phenotype can lie in different genes; only failure to complement groups them.
• Any virus can shift. Only segmented genomes such as influenza's reassort; unsegmented viruses can only drift and recombine.
• Resistance waits for the drug. Error-prone copying makes resistant variants before treatment, which is why combination therapy works.
• Complementation tests dominant mutants. It works only for recessive mutations, with the cis arrangement as the control.

Low-yield extras

SOS/RecA-driven repressor cleavage; Benzer's rII system, the cistron, and doubling the wild-type recombinant frequency to get map distance; intragenic complementation; order-of-magnitude error rates.

Next up: the Antimicrobials & Resistance lesson — drug targets, resistance pathways, and treatment principles.