Microbial Genetics: Mutation, Gene Transfer, Repair, Regulation and Genomes

This chapter answers Section XL3.9, "Microbial Genetics", of XL3 Microbiology, one of the five optional sections of the Life Sciences (XL) paper: "Types of mutation; UV and chemical mutagens; Selection of mutants; Ames test for mutagenesis; Bacterial genetic system: transformation, conjugation, transduction, recombination, plasmids, transposons; DNA repair; Regulation of gene expression: repression and induction; Operon model; Bacterial genome with special reference to E. coli; Phage λ and its life cycle; Mutation in virus genomes, virus recombination and reassortment; Basic concept of microbial genomics; Cell cycle and cell division in lower eukaryotes." It teaches how mutations arise and are classed, how mutants are selected and how the Ames test reads mutagenicity, the three modes of horizontal gene transfer with homologous recombination, plasmids, transposons and DNA repair, operon control, the E. coli genome, phage λ, variation in RNA viruses, microbial genomics and the budding-yeast cell cycle. Its numericals are interrupted-mating map minutes, cotransduction frequency, recombination frequency, Ames revertant counts and mutation rate.

1. Mutation: types and how they arise

A mutation is a heritable change in the DNA sequence. A point mutation changes a single base pair. A base substitution is a transition (purine for purine or pyrimidine for pyrimidine: A↔G, C↔T) or a transversion (purine for pyrimidine). An insertion or deletion (indel) of bases not a multiple of three shifts the reading frame of every codon downstream.

  • Silent (synonymous): the codon changes but still codes the same amino acid (the genetic code is degenerate, chiefly at the third base), so the protein is unchanged.
  • Missense: the new codon specifies a different amino acid; the effect ranges from none to severe depending on the residue.
  • Nonsense: the codon becomes a stop codon, truncating the protein; usually strongly inactivating.
  • Frameshift: an indel that is not a multiple of three; downstream codons are all wrong and a stop usually appears early, so the protein is almost always dead.

Spontaneous mutations arise without an added agent: replication errors (a tautomeric shift lets a base mispair), deamination of cytosine to uracil (and of 5-methylcytosine to thymine, a mutational hotspot), depurination, and oxidative damage such as 8-oxoguanine. Induced mutations follow a mutagen. Rates are low, of the order of 10⁻⁶ to 10⁻⁹ per gene per generation, which is why large populations are grown to recover rare mutants.

Common mutagens and what they do
MutagenAction on DNAResult
UV lightForms cyclobutane pyrimidine (thymine) dimersBlocks replication; error-prone bypass gives mutations
Base analogue (5-bromouracil)Incorporated for thymine, then mispairs with guanineTransitions (AT↔GC)
Alkylating agent (EMS, nitrosoguanidine)Adds alkyl groups; O6-methylguanine pairs with thymineTransitions
Deaminating agent (nitrous acid)Deaminates C to U and A to hypoxanthineTransitions
Intercalating agent (acridine, ethidium)Wedges between stacked basesInsertions or deletions, so frameshifts

2. Selecting mutants and the Ames test

A selection lets only the wanted cells grow (antibiotic-resistant mutants on a drug plate), so one rare cell is recovered among billions. A screen inspects every colony. Replica plating (the Lederbergs) presses a velvet pad onto a master plate and transfers the colony pattern to several plates; a colony that grows on complete but not on minimal medium is an auxotroph. Because the mutant was identified only from plates it never touched, replica plating also proved that mutations arise before selection, not in response to it.

Auxotrophs cannot be selected directly because they fail to grow, so they are enriched by penicillin enrichment: in minimal medium lacking the needed nutrient, only prototrophs grow, and penicillin kills growing cells while sparing the non-growing auxotrophs, which are then recovered on complete medium.

The Ames test scores whether a compound is a mutagen. Histidine-requiring (his⁻) Salmonella typhimurium tester strains are plated on medium with only a trace of histidine; a mutagen raises the number of his⁺ revertant colonies far above the spontaneous background. A rat-liver S9 extract is added so that promutagens needing mammalian metabolism are activated. Tester strains carry extra lesions that sharpen the assay (a leaky cell wall, deleted excision repair, and a plasmid boosting error-prone repair), and different strains detect base-substitution versus frameshift mutagens. Mutagenicity correlates strongly with carcinogenicity, which is why the test is a cheap pre-screen.

3. Transformation and conjugation

Bacteria pass genes sideways between cells by three horizontal gene transfer routes. In transformation a cell takes up naked DNA released into the surroundings, provided it is competent. Griffith's 1928 pneumococcus experiment showed a heat-killed virulent strain could transform a live avirulent one, and Avery, MacLeod and McCarty showed in 1944 that the transforming principle is DNA.

Conjugation needs cell-to-cell contact through a pilus and transfers DNA from a donor to a recipient (Lederberg and Tatum, 1946). The donor carries the F (fertility) plasmid: an F⁺ cell transfers a copy of F to an F⁻ cell, making it F⁺, but transfers no chromosomal genes.

When F integrates into the chromosome the cell becomes an Hfr (high frequency of recombination). It transfers chromosomal genes in a fixed order starting from the origin of transfer; the strand usually breaks before the whole chromosome passes, so genes near the origin enter first and distal genes rarely. Breaking the mating pairs at timed intervals (interrupted mating, Wollman and Jacob) orders genes and maps them in minutes on the 100-minute E. coli map. Imprecise excision of F gives an F′ plasmid carrying a few chromosomal genes (sexduction), creating a partial diploid.

4. Transduction and homologous recombination

In transduction a bacteriophage carries host DNA from one cell to another. The two kinds differ in what can be moved:

  • Generalized transduction: during a lytic infection the phage occasionally packages a random fragment of host DNA instead of its own, so any gene can be transferred (phage P1 in E. coli, P22 in Salmonella; discovered by Zinder and Lederberg).
  • Specialized transduction: a temperate phage such as λ sits at a fixed attachment site, and imprecise excision of the prophage picks up only the adjacent genes (gal or bio for λ), so only those are transferred.

Two genes close enough to ride in one phage head are moved together: their cotransduction frequency falls as the distance between them rises, so it is itself a fine-scale map. After entry by any route, the incoming DNA is joined to the chromosome by homologous recombination: RecA catalyses the search for a matching sequence and the strand exchange, RecBCD prepares the DNA ends, and the crossover (Holliday junction) is resolved by RuvABC. RecA also triggers the SOS response (next section).

5. Plasmids, transposons and DNA repair

A plasmid is an extrachromosomal, self-replicating DNA molecule (a replicon), usually a small circle. Common classes are F (fertility), R (antibiotic-resistance genes, the main carrier of clinical drug resistance), Col (colicins) and virulence plasmids. Plasmids sharing one replication-control system are incompatible and cannot be stably kept in the same cell; conjugative plasmids carry the tra genes for their own transfer.

A transposon is a DNA segment that moves to new sites using its own transposase, flanked by inverted repeats and leaving a short target-site duplication. The simplest is an insertion sequence (IS), carrying only transposase; a composite transposon (Tn10) is a resistance gene flanked by two IS elements; a complex transposon (Tn3) encodes transposase, resolvase and resistance. Movement is conservative (cut-and-paste) or replicative (a copy is left behind). Transposons cause insertional mutations and spread resistance genes.

  • Photoreactivation: photolyase uses blue light to split pyrimidine dimers directly, with no excision.
  • Base-excision repair (BER): a DNA glycosylase removes a single damaged base (such as uracil), then an AP endonuclease, polymerase and ligase rebuild the site.
  • Nucleotide-excision repair (NER): the UvrABC excinuclease removes a short oligonucleotide carrying a bulky lesion such as a pyrimidine dimer; polymerase and ligase fill the gap.
  • Mismatch repair (MMR): MutS, MutL and MutH correct replication mismatches and tell the new strand from the template because the old strand is already Dam-methylated at GATC sites while the new one is not.
  • SOS response: heavy damage makes RecA activate cleavage of the LexA repressor, derepressing repair genes and switching on error-prone translesion synthesis, which lets replication pass a lesion at the cost of mutations.

6. Gene regulation: operons and the E. coli genome

Bacteria regulate transcription through operons, clusters of genes read from one promoter as a single polycistronic mRNA, controlled at an operator (Jacob and Monod). Control is negative when a repressor blocks transcription, positive when an activator promotes it. Induction turns a set on (an inducer inactivates a repressor; typical of catabolic operons), while repression turns a set off (a corepressor activates a repressor; typical of biosynthetic operons).

The lac operon (lacZYA, for using lactose) is inducible: with no lactose the LacI repressor sits on the operator; allolactose made from lactose binds LacI and releases it, so the genes switch on. A second, positive control imposes catabolite repression: the activator CAP (CRP) works only with cAMP, and cAMP is high only when glucose is absent. So the operon is expressed strongly only when lactose is present and glucose is absent.

The trp operon (trpEDCBA, for making tryptophan) is repressible: TrpR is inactive alone, but tryptophan is the corepressor that activates it, so plentiful tryptophan shuts the operon off. A second layer, attenuation, stops transcription early when tryptophan is ample, because a ribosome translating the leader's trp codons lets a terminator hairpin form. The E. coli genome itself is a single circular double-stranded DNA chromosome of about 4.6 Mbp with roughly 4,300 genes, supercoiled into a nucleoid with no nuclear membrane, replicated bidirectionally from one origin, oriC, often alongside plasmids.

7. Phage λ, RNA viruses, genomics and the eukaryotic cell cycle

Phage λ is a temperate phage that chooses between two fates. In lysogeny it integrates into the chromosome (between gal and bio) as a prophage, and the CI repressor keeps the lytic genes off and makes the cell immune to further λ. In the lytic path Cro wins and the phage multiplies and lyses the cell. The CI-versus-Cro switch is tipped toward lysogeny by the CII protein when the host is poorly nourished or multiply infected; DNA damage activates RecA, which triggers CI self-cleavage and induces the prophage into the lytic cycle.

RNA viruses vary quickly in three ways. Mutation is frequent because RNA-dependent RNA polymerase has no proofreading, so a virus exists as a cloud of variants (a quasispecies) and drifts antigenically. Recombination occurs when the polymerase switches templates during replication. Reassortment happens only in viruses with a segmented genome: when two strains co-infect a cell, progeny can inherit whole segments from each parent, which is how influenza (eight segments) makes the sudden antigenic shifts behind pandemics.

Microbial genomics reads whole genomes: Haemophilus influenzae was the first free-living organism fully sequenced (1995). Genomes are annotated for open reading frames, and comparative genomics defines a species' core and accessory (pan-) genome, while obligate symbionts show reduced genomes. In lower eukaryotes such as budding yeast (Saccharomyces cerevisiae) the cell cycle runs G1 → S → G2 → M; commitment at Start in late G1 and progression are driven by a cyclin-dependent kinase (Cdc28) whose cyclins rise and fall. The daughter grows as an asymmetric bud, whereas fission yeast divides by a medial septum into two equal cells.

Key takeaways

  • Point mutations are transitions (A↔G, C↔T) or transversions, read as silent, missense or nonsense; a non-triple indel is a frameshift. Spontaneous sources include deamination and replication error; UV makes pyrimidine dimers, base analogues and alkylating and deaminating agents give transitions, and intercalators give frameshifts.
  • Replica plating detects auxotrophs and showed mutations arise before selection; penicillin enrichment recovers auxotrophs by killing growing prototrophs. The Ames test scores his⁻ to his⁺ reversion in Salmonella with an S9 activation extract, taking mutagenicity as a proxy for carcinogenicity.
  • Three transfer routes: transformation (naked DNA into a competent cell), conjugation (F plasmid; an Hfr transfers the chromosome in time order, mapped in minutes), and transduction (phage; generalized moves any gene, specialized only genes by the att site). RecA then drives homologous recombination.
  • Plasmids (F, R, Col) are replicons governed by incompatibility; transposons (IS, composite Tn10, complex Tn3) move by transposase, conservatively or replicatively. Repair is photoreactivation, base- and nucleotide-excision repair, mismatch repair (new strand told from old by Dam-GATC methylation), and the error-prone SOS response.
  • Operons give polycistronic mRNA under negative or positive control: the inducible lac operon needs allolactose (off the repressor) and, for strong output, low glucose so cAMP-CAP can act; the repressible trp operon is shut by tryptophan as corepressor plus attenuation. E. coli has one circular ~4.6-Mbp chromosome of about 4,300 genes.
  • Phage λ chooses lysogeny (CI repressor, immunity) or lysis (Cro), with RecA-driven CI cleavage inducing the lytic cycle. RNA viruses vary by mutation (no RdRp proofreading), recombination (template switching) and reassortment (only segmented genomes, e.g. influenza). Budding yeast cycles G1-S-G2-M with Start and the Cdc28 CDK.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A single base substitution in a gene changes a codon that specified glutamine into a UAA codon. What class of mutation is this, and what is its usual effect on the protein?

    1. A nonsense mutation, which introduces a stop codon and truncates the protein
    2. A missense mutation, which replaces one amino acid with a different amino acid
    3. A silent mutation, which leaves the encoded amino acid sequence unchanged
    4. A frameshift mutation, which shifts the reading frame of all later codons
    Show answer

    Answer: A — A nonsense mutation, which introduces a stop codon and truncates the protein

    UAA is one of the three stop codons, so a sense codon turning into it ends translation early and truncates the protein: that is a nonsense mutation. Missense would give a different amino acid, not a stop; silent would keep the same amino acid; a frameshift needs an insertion or deletion, not a single substitution.

  2. Spontaneous deamination of cytosine in DNA produces uracil. If this uracil is not removed before the next round of replication, what change appears in the daughter DNA?

    1. A GC pair is converted to a TA pair as a transversion, because uracil pairs with thymine
    2. No change occurs, because uracil is read exactly as the original cytosine was read
    3. A GC pair is converted to an AT pair, a transition, because uracil pairs with adenine
    4. The reading frame shifts by one base, because uracil cannot be copied by the polymerase
    Show answer

    Answer: C — A GC pair is converted to an AT pair, a transition, because uracil pairs with adenine

    Uracil has the base-pairing behaviour of thymine, so it pairs with adenine; the former C (which paired with G) now directs incorporation of A, turning a GC pair into an AT pair. Each base is replaced by one of the same chemical type (the pyrimidine C becomes pyrimidine T and the purine G becomes purine A), so the change is a transition, not a transversion, and a single base change shifts no reading frame.

  3. Two mutagens are compared. One is incorporated into DNA in place of thymine and then occasionally mispairs with guanine; the other is a flat molecule that slips between stacked base pairs during replication. What kinds of mutation does each mainly cause?

    1. The base analogue causes frameshifts, and the intercalating agent causes base-substitution transitions
    2. The base analogue causes base-substitution transitions, and the intercalating agent causes frameshifts
    3. Both cause frameshifts, because both act only during DNA replication
    4. Both cause transitions, because both change the pairing of a single base
    Show answer

    Answer: B — The base analogue causes base-substitution transitions, and the intercalating agent causes frameshifts

    A base analogue such as 5-bromouracil substitutes for a normal base and later mispairs, so it changes one base pair to another: a transition. An intercalating agent wedged between bases adds or removes a base during replication, shifting the reading frame. Saying both do the same thing ignores that one works by mispairing and the other by distorting spacing.

  4. In a replica-plating experiment, colonies from a master plate are transferred by velvet to plates containing a bacteriophage, and resistant colonies grow at the same positions on independent replica plates. What does this parallel pattern establish?

    1. The phage directly instructs each exposed cell to become resistant at the very spot where it lands, so resistance is a response induced by that contact
    2. Resistance is a non-heritable physiological adaptation to the phage
    3. The velvet transfer itself creates the resistance mutations during pressing
    4. Resistance mutations were present before the phage was ever encountered, so selection reveals pre-existing mutants rather than inducing them
    Show answer

    Answer: D — Resistance mutations were present before the phage was ever encountered, so selection reveals pre-existing mutants rather than inducing them

    The same colonies are resistant on several replicas made from one master plate that never met the phage, so the mutation must have existed in that colony beforehand; the phage only selects it. If exposure induced resistance, resistant colonies would appear at random, unrelated positions on each plate rather than matching the master.

  5. How does an Hfr strain of E. coli differ from an ordinary F⁺ strain in what it transfers during conjugation?

    1. In an Hfr the F factor is integrated into the chromosome, so it transfers chromosomal genes in a fixed order, whereas an F⁺ transfers only the F plasmid
    2. An Hfr transfers only a copy of the free F plasmid, whereas an ordinary F⁺ cell transfers its entire chromosome to the recipient every single time the two cells mate
    3. Neither transfers any DNA; conjugation only exchanges cytoplasmic proteins between the cells
    4. An Hfr always transfers its entire chromosome so the recipient becomes a complete diploid
    Show answer

    Answer: A — In an Hfr the F factor is integrated into the chromosome, so it transfers chromosomal genes in a fixed order, whereas an F⁺ transfers only the F plasmid

    An F⁺ cell carries F as a free plasmid and passes a copy of F, making the recipient F⁺ but transferring no chromosomal genes. In an Hfr, F is integrated, so transfer begins at the integration site and carries chromosomal genes in order; the mating almost always breaks before the full chromosome passes, so the recipient is only a partial diploid, not a complete one.

  6. A phage can transfer essentially any chromosomal marker from donor to recipient at low frequency, with no preference for particular genes. Which mechanism is operating, and why is there no gene preference?

    1. Specialized transduction, because the prophage excises imprecisely from its fixed site and carries along only the host genes that lie immediately beside its attachment site
    2. Conjugation, because the F plasmid carries any chromosomal gene across the pilus
    3. Generalized transduction, because during lysis the phage sometimes packages random host DNA fragments instead of its own genome
    4. Transformation, because naked DNA fragments of every gene are taken up equally
    Show answer

    Answer: C — Generalized transduction, because during lysis the phage sometimes packages random host DNA fragments instead of its own genome

    Packaging random host fragments during the lytic cycle is generalized transduction, and because the mistake is indiscriminate any gene can be carried. Specialized transduction moves only genes next to a fixed att site, so it would show a strong preference. The transfer is phage-mediated, which rules out conjugation and transformation.

  7. Methyl-directed mismatch repair in E. coli must fix the newly made strand, not the template, after a replication error. How does the system know which strand is new?

    1. The newly synthesized strand is permanently methylated at its GATC sites immediately, so the repair machinery always removes whichever strand happens to carry the methyl marks
    2. The template strand is already methylated at GATC sequences by Dam, while the new strand is transiently unmethylated, so repair is directed to the unmethylated strand
    3. The template strand carries the mismatch, so the system simply replaces whichever strand holds the wrong base
    4. Both strands are identical, so the choice of strand is made at random each time
    Show answer

    Answer: B — The template strand is already methylated at GATC sequences by Dam, while the new strand is transiently unmethylated, so repair is directed to the unmethylated strand

    Dam methylase marks adenine in GATC on the parental strand soon after replication, but the daughter strand stays unmethylated for a short window; MutH nicks that unmethylated (new) strand, so the error made during copying is the one removed. The mismatch sits in the new strand by definition, so just replacing the strand with the wrong base is circular, and the new strand is not permanently methylated.

  8. Under which combination of sugars is the lac operon of E. coli transcribed at its highest rate, given that it is controlled by the LacI repressor and by cAMP-CAP activation?

    1. Lactose present and glucose present, because having both sugars available at once gives the cell the most energy with which to transcribe the operon
    2. Lactose absent and glucose absent, because starvation fully derepresses the operon
    3. Lactose absent and glucose present, because glucose is the preferred inducer of the lac genes
    4. Lactose present and glucose absent, so the repressor is off the operator and high cAMP lets CAP activate the promoter
    Show answer

    Answer: D — Lactose present and glucose absent, so the repressor is off the operator and high cAMP lets CAP activate the promoter

    Lactose (via allolactose) must be present to pull LacI off the operator, and glucose must be absent so cAMP is high and cAMP-CAP can activate the weak promoter; both conditions together give maximal transcription. With glucose present cAMP is low and CAP cannot help, and with no lactose the repressor stays bound, so those combinations give little or no expression.

  9. The trp operon of E. coli, which encodes the enzymes for making tryptophan, is switched off when tryptophan is plentiful. In the primary repressor-mediated control, how does tryptophan act?

    1. It acts as a corepressor, binding the inactive TrpR protein and converting it into an active repressor that then blocks transcription
    2. It acts as an inducer, binding and inactivating the repressor so that transcription of this biosynthetic operon is switched fully on
    3. It binds the promoter directly and physically prevents RNA polymerase from ever attaching to the operon's DNA
    4. It activates the cAMP-CAP complex, which represses the operon by bending the DNA sharply at the operator
    Show answer

    Answer: A — It acts as a corepressor, binding the inactive TrpR protein and converting it into an active repressor that then blocks transcription

    The trp operon is repressible: TrpR alone cannot bind the operator, but tryptophan acts as a corepressor that activates TrpR, so a surplus of the end product shuts off its own synthesis. Calling tryptophan an inducer describes the opposite, lac-style logic; the repressor binds the operator, not the promoter, and cAMP-CAP is the positive control of catabolic operons, not of trp.

  10. Which statement correctly describes cell division and its control in the budding yeast Saccharomyces cerevisiae?

    1. It divides by simple binary fission just like a bacterium, copying its chromosome without a true S phase and segregating it without a mitotic spindle
    2. It divides into two equal daughter cells by a central septum, which is the defining feature of budding yeast
    3. It divides by forming an asymmetric bud, and commitment to division occurs at Start in late G1 under a cyclin-dependent kinase
    4. It skips the cell cycle entirely and copies its DNA only during sporulation
    Show answer

    Answer: C — It divides by forming an asymmetric bud, and commitment to division occurs at Start in late G1 under a cyclin-dependent kinase

    Budding yeast is a eukaryote with a full G1-S-G2-M cycle; it buds asymmetrically and commits to a new cycle at Start in late G1, driven by the Cdc28 cyclin-dependent kinase. Binary fission and skipping S phase describe prokaryotes, not yeast, and division by a central septum into equal cells describes fission yeast, not the budding species named.

  11. Regarding the Ames test for mutagenicity, which of the following statements are correct?

    1. It measures reversion of histidine-requiring Salmonella to histidine-independence, so more revertants indicate a stronger mutagen
    2. A rat-liver S9 extract is added so that compounds needing mammalian metabolic activation can be detected
    3. A control plate without the test compound is used to measure the spontaneous revertant background
    4. A positive Ames result proves directly that the compound causes cancer in humans with certainty
    Show answer

    Answer: A — It measures reversion of histidine-requiring Salmonella to histidine-independence, so more revertants indicate a stronger mutagen; B — A rat-liver S9 extract is added so that compounds needing mammalian metabolic activation can be detected; C — A control plate without the test compound is used to measure the spontaneous revertant background

    The assay counts his⁺ revertants of a his⁻ tester, uses S9 to mimic liver activation of promutagens, and needs a no-compound control to set the spontaneous background. It is a mutagenicity screen that correlates with carcinogenicity, not a proof of human cancer causation, so claiming certain carcinogenicity in humans overreaches.

  12. Which of the following statements about bacterial plasmids and transposons are correct?

    1. R plasmids carry antibiotic-resistance genes and are a major vehicle for the spread of clinical drug resistance
    2. A transposon moves using a transposase and is bounded by inverted repeats, leaving a short target-site duplication where it inserts
    3. Two plasmids sharing the same replication-control system are incompatible and are not stably maintained together
    4. An insertion sequence carries antibiotic-resistance genes in addition to its transposase, which is what distinguishes it from a composite transposon
    Show answer

    Answer: A — R plasmids carry antibiotic-resistance genes and are a major vehicle for the spread of clinical drug resistance; B — A transposon moves using a transposase and is bounded by inverted repeats, leaving a short target-site duplication where it inserts; C — Two plasmids sharing the same replication-control system are incompatible and are not stably maintained together

    R plasmids spread drug resistance, transposons move by transposase within inverted repeats and duplicate the target site, and plasmids of one incompatibility group displace each other. An insertion sequence is the simplest element and carries only the transposase and its repeats; it is the composite transposon that adds a resistance gene between two IS elements, so that statement is reversed.

  13. Regarding sources of variation in RNA viruses, which of the following statements are correct?

    1. High mutation rates arise because RNA-dependent RNA polymerase lacks proofreading activity
    2. Reassortment, the swapping of whole genome segments, can occur only in viruses whose genome is segmented
    3. Recombination can occur when the polymerase switches from one RNA template to another during replication
    4. Reassortment requires only a single infecting virus strain, since each virion rearranges its own segments
    Show answer

    Answer: A — High mutation rates arise because RNA-dependent RNA polymerase lacks proofreading activity; B — Reassortment, the swapping of whole genome segments, can occur only in viruses whose genome is segmented; C — Recombination can occur when the polymerase switches from one RNA template to another during replication

    RNA polymerases without proofreading give high mutation rates, segment swapping (reassortment) is possible only for segmented genomes, and template switching produces recombination. Reassortment needs two different strains to co-infect one cell so that progeny can mix their segments; a single strain cannot reassort, so that statement is wrong.

  14. In an interrupted-mating experiment with an Hfr strain, the markers enter the recipient at these times after mating begins: azi at 10 minutes, lac at 18 minutes and gal at 27 minutes. What is the map distance, in minutes, between azi and gal?

    Numerical answer — type the value.

    Show answer

    Answer: 17

    In interrupted mating the time a marker first enters marks its map position, and the distance between two markers is the difference of their entry times. Here gal enters at 27 minutes and azi at 10 minutes, so the distance is 27 − 10 = 17 minutes. Using the lac time or adding the times instead of subtracting are the usual slips.

  15. Two markers lie 0.5 minutes apart on the E. coli map, and a generalized transducing particle carries a chromosome fragment of length 2.0 minutes. Using Wu's formula, cotransduction frequency = (1 − d/L)³, where d is the distance between the markers and L is the fragment length, compute the cotransduction frequency to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 0.42

    Substitute d = 0.5 and L = 2.0 into (1 − d/L)³: d/L = 0.5/2.0 = 0.25, so (1 − 0.25)³ = 0.75³ = 0.421875, which rounds to 0.42. The closer the markers (smaller d/L), the higher the cotransduction frequency, so forgetting to cube, or cubing before subtracting, gives the wrong value.

  16. A two-factor cross between two bacteriophage strains gives these progeny plaque counts: parental type one 420, parental type two 380, recombinant type one 120 and recombinant type two 80. Recombination frequency (%) = (total recombinants ÷ total progeny) × 100. What is the recombination frequency as a percentage?

    Numerical answer — type the value.

    Show answer

    Answer: 20

    Add all progeny: 420 + 380 + 120 + 80 = 1000. The recombinants are both recombinant classes: 120 + 80 = 200. Recombination frequency = 200 ÷ 1000 × 100 = 20%. Counting only one recombinant class (giving 12% or 8%) is the usual error; both recombinant genotypes must be summed.

  17. In an Ames test, a plate spread with a test compound plus S9 mix yields 362 his⁺ revertant colonies, while the solvent-only control plate yields 42 spontaneous revertant colonies. How many revertant colonies are attributable to the test compound?

    Numerical answer — type the value.

    Show answer

    Answer: 320

    The spontaneous background on the control plate must be subtracted from the total on the test plate: 362 − 42 = 320 revertants are induced by the compound. Reporting the raw 362 without subtracting the background overstates the mutagenic effect, which is the trap this control is designed to catch.

  18. A bacterial culture grows from a negligible inoculum to 2×10⁹ cells. Take the number of cell divisions to equal the final cell number. If the mutation rate at a particular locus is 5×10⁻⁹ per cell division, what is the expected number of newly arisen mutant cells? Give an integer.

    Numerical answer — type the value.

    Show answer

    Answer: 10

    The expected number of new mutants equals the mutation rate times the number of cell divisions. Taking divisions as the final cell number, that is 5×10⁻⁹ × 2×10⁹ = 10 mutant cells. Multiplying the rate by the generation count (about 31 doublings) instead of the number of divisions is the usual slip.