📖 ContentsChapter 5
Genetics · Chapter 5

Bacterial and Viral Genetics

Bacteria and viruses have no meiosis, yet they exchange genes constantly, and that exchange is why antibiotic resistance spreads through a hospital in months and why a new influenza strain can appear almost overnight. The same tiny genomes gave geneticists their sharpest tools: billions of offspring in a single tube, selection for one rare recombinant among millions, and maps fine enough to resolve neighboring base pairs. This chapter shows how bacteria and phages move DNA, how each route of transfer becomes a mapping method, and how viral genomes, including retroviruses, are built and copied.

By the end of this chapter you can

  • Describe the organization of a bacterial nucleoid and the main classes of plasmids, and write bacterial genotypes and phenotypes in standard notation.
  • Explain how auxotrophs, resistance mutants and conditional mutants are isolated by replica plating, and how the fluctuation test showed that mutations arise before selection.
  • Distinguish F−, F+, Hfr and F′ cells and predict the outcome of each mating.
  • Build a circular gene map from interrupted mating data collected with several Hfr strains.
  • Compare transformation, generalized transduction and specialized transduction, and use cotransformation and cotransduction frequencies to order genes and estimate distances.
  • Contrast the lytic and lysogenic cycles and calculate recombination frequencies from phage crosses scored by plaque type.
  • Explain Benzer's rII analysis, separating the complementation test from the recombination test, and order mutations by deletion mapping.
  • Classify viruses by genome type with the Baltimore scheme and trace the retrovirus life cycle from reverse transcription to integration.

Section 1Bacterial genomes, plasmids and mutants

Why bacteria are such powerful genetic organisms

A culture of Escherichia coli in rich medium doubles about every 20 minutes, so a single cell becomes more than a billion cells overnight. Bacteria are haploid, which means every mutation shows its phenotype at once, with no dominant allele hiding a recessive one. And because you can spread a hundred million cells on one Petri plate and allow only a rare type to grow, you can find events that happen once in a million or once in a billion cells.

The bacterial chromosome

Most bacteria carry a single, circular, double stranded DNA chromosome. The E. coli K-12 chromosome is about 4.6 million base pairs (4.6 Mb) and encodes roughly 4,300 genes. It is compacted into an irregular region called the nucleoid. There is no nuclear membrane. Compaction comes from negative supercoiling, introduced by DNA gyrase (a type II topoisomerase), and from nucleoid associated proteins such as HU, H-NS and IHF that bend and bridge DNA.

Bacterial genomes are gene dense: about 88 percent of the E. coli chromosome codes for protein, introns are essentially absent, and functionally related genes are often clustered into operons that are transcribed as one messenger RNA. Replication begins at a single origin, oriC, and proceeds in both directions until the two forks meet in the terminus region opposite the origin.

Plasmids

Plasmids are small DNA molecules, usually circular and double stranded, that replicate independently of the chromosome. Each has its own origin of replication, which also sets its copy number: some plasmids are kept at one or two copies per cell under tight control, while others reach dozens or hundreds. Plasmids are generally not essential for growth under ordinary conditions, but the genes they carry can be decisive when conditions change. An episome is a plasmid that can either replicate on its own or integrate into the chromosome; the F factor is the classic example.

Plasmid classWhat it carriesWhy it matters
F (fertility) factorAbout 100 kb; tra genes for the sex pilus and DNA transfer, an origin of transfer (oriT)Makes a cell a donor in conjugation; can integrate to form an Hfr strain
R (resistance) plasmidsOne or more antibiotic resistance genes, often on transposons, plus transfer genesSpread multidrug resistance between cells and even between species
Col plasmidsGenes for colicins (proteins that kill related strains) and immunity to themCompetition among bacteria
Virulence plasmidsToxins, adhesins, secretion systemsTurn a harmless strain into a pathogen (for example, enterotoxigenic E. coli)
Connection. Many R plasmids are conjugative: they carry their own transfer machinery, so a single plasmid can move a block of five or six resistance genes from a harmless gut bacterium into a pathogen in one step. Every unnecessary course of antibiotics, in medicine or livestock, selects for cells that already carry such plasmids.

Notation for genotypes and phenotypes

Bacterial geneticists use a compact notation. Genotypes are written as three lowercase italic letters followed by an uppercase letter for the specific gene: leuA, lacZ, trpE. A superscript plus means the wild type allele (leu+, able to make leucine) and a superscript minus means a mutant allele (leu−, unable to make leucine). strr means streptomycin resistant. Phenotypes are written in roman type with the first letter capitalized: Leu+, Lac−, Strr.

Kinds of bacterial mutants

A prototroph is a wild type strain that grows on minimal medium: water, salts, a carbon source such as glucose and nothing else. From those raw materials it builds every amino acid, nucleotide and vitamin it needs. An auxotroph has lost the ability to make one or more of these and grows only if the missing nutrient is supplied. A his− auxotroph, for example, grows on minimal medium plus histidine but not on minimal medium alone. Complete medium contains everything, so both prototrophs and auxotrophs grow on it.

Finding mutants: selection, screening and replica plating

Resistance mutants are easy to find by direct selection: spread 108 cells on a plate containing the antibiotic, and only resistant cells form colonies. Auxotrophs are harder, because the mutant grows on fewer media than the wild type, not more. Joshua and Esther Lederberg solved this in 1952 with replica plating. Cells are spread on a complete medium master plate, giving a few hundred colonies. A sterile velvet pad pressed onto the master picks up cells from every colony and can then print that exact pattern onto a series of fresh plates. Any colony present on the master and on complete medium but absent from minimal medium is an auxotroph, and its position tells you where to pick it from the master. Printing onto minimal medium supplemented with single nutrients then identifies exactly which nutrient it needs.

Master plateCompleteMinimalMinimal + His 444 velvet pad no growth Colony 4 grows on complete medium and on minimal + histidine, but not on minimal alone. Conclusion: colony 4 is a his⁻ auxotroph; pick it from the master plate.
Figure 5.1 Replica plating. The velvet transfers every colony in its original position, so a colony that is missing from one medium but present on the master can be traced back and recovered. Supplemented plates identify which nutrient the auxotroph needs.
Study tip. Worked problem: Five colonies from a master plate are replicated onto five media. Colony A grows on complete, minimal + Leu, and minimal + Leu + Thr, but not on minimal or minimal + Thr. Colony B grows only on complete and minimal + Leu + Thr. Colony C grows everywhere. What is each genotype? Step 1, read colony C: it grows on minimal alone, so it is a prototroph, leu+ thr+. Step 2, read colony A: it needs leucine (grows whenever Leu is present) but never needs threonine, so it is leu− thr+. Step 3, read colony B: neither supplement alone rescues it, but both together do, so it is a double auxotroph, leu− thr−. Answer: C is prototrophic, A is a leucine auxotroph, B requires both leucine and threonine. Check: a colony that grows on complete but on none of the supplemented plates needs something you did not test.

Do mutations arise before or after selection? The fluctuation test

In the early 1940s it was an open question whether phage T1 caused resistance in the few cells that survived it, or whether resistant mutants arose spontaneously and the phage merely revealed them. Salvador Luria and Max Delbrück answered with the fluctuation test (1943). They grew many small, independent cultures from a few cells each, then plated each entire culture with phage.

The logic is simple. If the phage induced resistance at the moment of plating, every culture would have about the same small chance per cell, and the counts would vary little from culture to culture, like samples from one flask. If mutations arose at random during growth, a mutation early in the history of one culture would leave a huge resistant clone (a "jackpot"), while most cultures would have few or none. The independent cultures showed enormous fluctuation, with many zeros and a few jackpots, while the samples from the single flask were uniform. Mutations are spontaneous and random; selection only reveals them.

Study tip. Worked problem (the P0 method): You grow 20 independent cultures, each to a final population of 2 × 108 cells, and plate each on T1 phage. Eleven cultures give no resistant colonies at all. Estimate the mutation rate. Step 1, find the fraction with no mutants: P0 = 11/20 = 0.55. Step 2, mutations per culture follow a Poisson distribution, so P0 = e−m, where m is the mean number of mutational events per culture. Then m = −ln(0.55) = 0.598. Step 3, divide by the number of cell divisions, which is approximately the final number of cells: rate ≈ 0.598 / (2 × 108) = 3.0 × 10−9. Answer: about 3 × 10−9 mutations per cell per division. Using zeros avoids the jackpot problem: you count mutational events, not mutant cells.
Key point

Bacteria are haploid, fast and countable, with one circular chromosome in a nucleoid plus optional plasmids that carry fertility, resistance, toxin or degradative genes. Auxotrophs, resistance and conditional mutants are found by selection and by replica plating, and the fluctuation test proved that mutations arise at random before selection rather than in response to it.

Section 2Conjugation and mapping by interrupted mating

The discovery of gene exchange

In 1946 Joshua Lederberg and Edward Tatum mixed two E. coli auxotrophs. One strain was met− bio− thr+ leu+; the other was met+ bio+ thr− leu−. Neither grew on minimal medium alone, but the mixture produced prototrophic colonies at about 1 in 107 cells. Double reversion could not explain them. Bernard Davis then placed the two strains on opposite sides of a U-tube separated by a filter that let medium and large molecules through but not cells. No prototrophs formed, showing that exchange required cell to cell contact. This contact dependent transfer is conjugation.

The F factor and the three mating types

Conjugation depends on the F (fertility) factor, a plasmid of about 100 kb. A cell carrying F is F+; a cell lacking it is F−. The tra genes of F encode the sex pilus, which pulls an F− cell close, plus the transfer machinery. Transfer begins when a relaxase nicks one strand at the origin of transfer (oriT). That single strand, 5′ end first, threads into the recipient while the donor replaces it by rolling circle replication; the recipient synthesizes the complementary strand. The result of an F+ × F− mating is two F+ cells. F spreads rapidly, yet F+ cells rarely transfer chromosomal genes.

About once in 104 to 105 F+ cells, the F factor integrates into the chromosome by recombination between short homologous sequences (often insertion sequences) on F and on the chromosome. The result is an Hfr (high frequency of recombination) cell. When an Hfr cell conjugates, transfer still starts at oriT, but now the strand that follows is chromosomal DNA. Genes beside the integration site enter first, then genes farther away, in a fixed order and direction set by where and in which orientation F inserted. Transferring the entire 4.6 Mb chromosome takes about 100 minutes at 37 °C, and the mating pair almost always breaks apart before that. The part of F that lies behind oriT is transferred last, so the recipient of an Hfr mating nearly always remains F−.

Occasionally F excises from the Hfr chromosome imprecisely and carries a piece of neighboring chromosome with it. This plasmid is an F′ (F prime) factor. An F′ transfers efficiently like F, and the recipient becomes a partial diploid, or merozygote, carrying two copies of the captured genes.

DonorState of FWhat is transferredRecipient afterwardFrequency of chromosomal recombinants
F+Free plasmidThe F plasmid onlyBecomes F+Very low (about 10−7, from rare Hfr cells in the culture)
HfrIntegrated in the chromosomeChromosomal genes in fixed order from oriT; rarely all of FUsually stays F−High for early genes, falling for later genes
F′Free plasmid carrying some chromosomal genesF plus the captured genesBecomes F′, a partial diploidCaptured genes transferred at high frequency

Recombination in the recipient

The donor fragment that enters the recipient is linear and cannot replicate on its own. To be inherited, its genes must be recombined into the recipient's circular chromosome, and that requires an even number of crossovers, at least two, one on each side of the gene. A single crossover would open the circle into a linear molecule that cannot survive. The donor genes that do not recombine in are degraded. The recipient therefore needs a functional RecA protein.

Interrupted mating

In the mid 1950s Élie Wollman and François Jacob turned the ordered transfer of Hfr strains into a mapping method. They mixed an Hfr strain sensitive to streptomycin (strs) with an F− recipient that carried several auxotrophic markers and was streptomycin resistant (strr). At intervals they removed samples and broke the mating pairs apart in a kitchen blender, then plated the cells on media containing streptomycin to kill the donors (counterselection) and lacking a particular nutrient to select for recipients that had received a donor allele.

Each donor gene first appeared among recombinants at a characteristic time of entry, and the frequency of recombinants for that gene then rose to a plateau. Genes that entered early reached higher plateaus than genes that entered late, because the longer the transfer runs, the more likely the pair is to break apart spontaneously. This decline is called the gradient of transfer. The time of entry measures position: the E. coli chromosome is mapped in minutes, from 0 to 100, with the thr locus placed at 0. One minute corresponds to roughly 46 kb.

01020304050 Time of mating before blending (minutes) 0306090 % recombinants abcd 8122026 Later genes enter later and reach lower plateaus
Figure 5.2 An interrupted mating experiment. Each curve starts at that gene's time of entry (small numbers on the axis); the order of entry gives gene order, and the differences between entry times give distances in minutes. Plateaus fall for later genes because mating pairs break apart spontaneously, the gradient of transfer.

Building a circular map from several Hfr strains

Different Hfr strains have F inserted at different places and in different orientations, so each transfers the chromosome starting from a different point and in one of two directions. No single Hfr can map the whole chromosome, because pairs break apart long before 100 minutes. But the orders from several strains overlap, and when you align them the only arrangement consistent with all of them is a circle.

StrainFirst gene (min)Second (min)Third (min)Fourth (min)
Hfr 1a (5)b (20)c (27)d (43)
Hfr 2h (5)a (20)b (35)c (42)
Hfr 3e (5)d (17)c (33)b (40)
Hfr 4g (10)h (20)a (35)b (50)
Hfr 5f (9)e (20)d (32)c (48)
Study tip. Worked problem: Use the five Hfr strains in the table to build a complete circular map with distances. Step 1, find distances between neighbors by subtracting entry times within one strain. Hfr 1: a to b = 20 − 5 = 15, b to c = 27 − 20 = 7, c to d = 43 − 27 = 16. Hfr 3: e to d = 17 − 5 = 12. Hfr 5: f to e = 20 − 9 = 11. Hfr 4: g to h = 20 − 10 = 10, h to a = 35 − 20 = 15. Step 2, check for agreement: Hfr 2 gives h to a = 15, a to b = 15, b to c = 7; Hfr 3 gives d to c = 16 and c to b = 7; Hfr 5 gives e to d = 12 and d to c = 16. All agree, so the data are consistent. Step 3, string the genes together: a, b, c, d, e, f, then g and h, returning to a. Hfr 3 and Hfr 5 list the same genes in the reverse direction, which simply means F is inserted in the opposite orientation. Step 4, the only gap is f to g. The known intervals sum to 15 + 7 + 16 + 12 + 11 + 10 + 15 = 86 minutes, and the circle is 100 minutes, so f to g = 100 − 86 = 14 minutes. Answer: the circular order is a, b, c, d, e, f, g, h with intervals 15, 7, 16, 12, 11, 14, 10 and 15 minutes. Placing a at 10 minutes gives b 25, c 32, d 48, e 60, f 71, g 85, h 95.
Key point

Conjugation moves DNA through direct contact, driven by the F factor. F+ cells transfer only F; Hfr cells, with F integrated, transfer chromosomal genes linearly from oriT in a fixed order; F′ plasmids carry a few chromosomal genes and create partial diploids. Interrupted mating converts the order and time of entry into a map in minutes, and overlapping Hfr orders reveal a circular chromosome of 100 minutes.

Section 3Transformation and transduction

Transformation: taking up naked DNA

Transformation is the uptake of free DNA from the surroundings and its incorporation into the recipient's genome. It was the first form of bacterial gene transfer discovered. In 1928 Frederick Griffith found that heat killed virulent (smooth, encapsulated) Streptococcus pneumoniae could convert living harmless (rough) cells into virulent ones inside a mouse. In 1944 Oswald Avery, Colin MacLeod and Maclyn McCarty purified the "transforming principle" and showed that it was destroyed by DNase but not by protease or RNase. That experiment identified DNA as the genetic material.

Some species are naturally competent: they switch on a set of genes, usually in response to crowding or stress, that builds a DNA uptake machine. Streptococcus pneumoniae, Bacillus subtilis and Neisseria gonorrhoeae are classic examples. One strand of bound DNA enters and pairs with the homologous region of the chromosome with the help of RecA, forming a heteroduplex, and mismatch repair or the next round of replication resolves it into one recipient chromosome and one transformed chromosome. Because the entering strand is linear, at least two crossovers are again needed, one on each side of the marker.

E. coli is not naturally competent, so laboratories make it artificially competent, either by chilling cells in calcium chloride and giving a brief heat shock, or by electroporation, a short high voltage pulse that opens transient pores.

Mapping by cotransformation

A transforming fragment is only a small piece of the chromosome, typically a few tens of kilobases. Two genes can be transferred together on one fragment, an event called cotransformation, only if they lie close together. The frequency of cotransformation therefore measures linkage: the closer two genes, the more often they travel together. Two tests are used.

Study tip. Worked problem: In B. subtilis, DNA from an a+ b+ c+ donor is added to an a− b− c− recipient. Single transformation frequencies are 1 × 10−3 for each gene. Double transformants appear at: a+b+ 3 × 10−4, b+c+ 2 × 10−4, a+c+ 4 × 10−5. Are the genes linked, and what is their order? Step 1, compute the frequency expected if two genes were transferred on separate fragments: (1 × 10−3) × (1 × 10−3) = 1 × 10−6. Step 2, compare: every observed double frequency is 40 to 300 times higher than 10−6, so all three genes are close enough to be carried together on one fragment. Step 3, order them: a and c are cotransformed least often (4 × 10−5), so they are the outside genes. Answer: all three are linked, in the order a, b, c, with a closer to b than c is to b (because a with b is the most frequent pair).

Transduction: phages as couriers

Transduction is gene transfer from one bacterium to another carried inside a phage particle. Zinder and Lederberg discovered it in 1952 with phage P22 in Salmonella: transfer crossed a U-tube filter and resisted DNase. There are two types, and they differ in what the phage packages.

Generalized transduction

In generalized transduction, any part of the host chromosome can be transferred. Phages such as P1 (E. coli) and P22 (Salmonella) degrade the host chromosome into fragments during a lytic infection and package DNA into heads by a headful mechanism: they fill the head with as much DNA as fits. Occasionally, about once in every thousand or so particles, a head is filled with a piece of host DNA instead of phage DNA. Such a transducing particle contains no phage genes. It injects its bacterial DNA into the next cell it infects, and that cell survives because no phage genes enter. If the donor fragment recombines into the recipient chromosome by two crossovers, the result is a stable transductant (complete transduction). A fragment that persists without recombining or replicating gives abortive transduction.

A P1 head holds about 100 kb, roughly 2 percent of the E. coli chromosome, or about 2 minutes of the map. Two genes can be cotransduced only if they lie within that distance, and the closer they are, the more often they are.

Specialized transduction

In specialized transduction, only genes next to a prophage's integration site can be transferred. The classic example is phage lambda (λ), which integrates at a single att site between the gal and bio genes of E. coli. When a lysogen is induced, the prophage normally excises precisely. Rarely (about once in 105 to 106 excisions), the excision cut is in the wrong place, and the circle that loops out contains some host DNA on one side (gal or bio) and lacks a corresponding piece of phage DNA on the other. The resulting particle, written λdgal (defective, carrying gal) or λdbio, can inject the host gene into a new cell. A lysate that contains such particles transduces gal or bio at high frequency, but never transduces genes elsewhere on the chromosome.

Generalized (P1, P22) Specialized (lambda) 1. Host chromosome is cut into pieces any gene, e.g. leu+ 2. Headful packaging, rarely of host DNA normal phage transducing particle 3. Recipient gains leu+ by two crossovers Markers within about 2 minutes cotransduce 1. Prophage sits between gal and bio galbio prophage 2. Rare faulty excision loops out gal λdgal gal gained, some λ lost 3. Only neighbors of att are carried High frequency for gal or bio; never for genes elsewhere
Figure 5.3 Two kinds of transduction. A generalized transducing particle carries a random piece of host DNA and no phage genes, so any marker can be moved. A specialized transducing phage forms by faulty excision of a prophage and carries only the host genes beside its attachment site.

Mapping by cotransduction

Cotransduction frequency is used exactly like cotransformation. You select for one donor marker and score how many of the selected transductants also inherited a second, unselected marker. Close genes are cotransduced often; genes more than one transducing fragment apart are never cotransduced. For P1, Wu derived a useful approximation relating cotransduction frequency C to map distance d (in minutes) when the fragment length L is 2 minutes:

C = (1 − d / L)3

Three factor transduction crosses resolve gene order without any distance formula. Select for one marker and score the other two among the transductants. The class that requires four crossovers rather than two will be by far the rarest, and it identifies which gene is in the middle.

Study tip. Worked problem 1: P1 grown on an a+ b+ c+ donor transduces an a− b− c− recipient. You select a+ transductants and score 400 of them: a+b+c+ 70; a+b+c− 130; a+b−c− 196; a+b−c+ 4. Find the cotransduction frequencies and the gene order. Step 1, a with b: classes carrying b+ are 70 + 130 = 200, so 200/400 = 0.50. Step 2, a with c: classes carrying c+ are 70 + 4 = 74, so 74/400 = 0.185. Step 3, b is cotransduced with a more often than c is, so b is closer to a. Step 4, confirm with the rarest class, a+b−c+ (4). If the order is a, b, c, getting a+ and c+ from the donor while keeping the recipient's b− in between needs four crossovers, so this class should be rarest, and it is. Answer: the order is a, b, c. Worked problem 2: convert the a to b cotransduction frequency to minutes with Wu's formula (L = 2). Step 1, 0.50 = (1 − d/2)3. Step 2, take the cube root: 0.501/3 = 0.794, so 1 − d/2 = 0.794. Step 3, d/2 = 0.206, so d = 0.41 minute. Answer: a and b are about 0.4 minute (roughly 19 kb) apart.
FeatureConjugationTransformationGeneralized transductionSpecialized transduction
VehicleDirect contact through a pilusFree DNA in the mediumPhage head (P1, P22)Phage (lambda)
Blocked by DNase?NoYesNoNo
Needs cell contact?YesNoNoNo
Amount transferredUp to the whole chromosomeUsually tens of kbAbout 100 kb (2 min) for P1A few genes next to att
Which genesAny, in order from oriTAnyAnyOnly those beside the prophage site
Mapping measureTime of entry (minutes)Cotransformation frequencyCotransduction frequencyNot used for mapping
Connection. Transduction shaped some of the most dangerous bacteria in medicine. Cholera toxin and diphtheria toxin genes are carried by prophages, as are the Shiga toxins of enterohemorrhagic E. coli are carried by lambda-like prophages. Antibiotics that trigger the SOS response can induce those prophages and increase toxin release.
Key point

Transformation moves naked DNA (DNase sensitive) and is mapped by cotransformation. Generalized transduction packages a random host fragment of about 2 map minutes by headful packaging, and cotransduction frequency measures closeness; the rarest class in a three factor cross marks the middle gene. Specialized transduction arises from faulty prophage excision and moves only genes beside the attachment site.

Section 4Bacteriophage genetics and fine structure mapping

Phage life cycles

Bacteriophages (phages) are viruses that infect bacteria. A phage such as T4 has a protein head holding its double stranded DNA genome (about 169 kb) and a tail that attaches to receptors on the cell surface and injects the DNA. In the lytic cycle, the phage takes over the cell: early genes shut down host functions and degrade host DNA, middle and late genes replicate phage DNA and build heads and tails, the DNA is packaged, and a lysozyme-like enzyme bursts the cell. At 37 °C, a T4 infection lasts about 25 minutes and releases a burst of roughly 100 to 200 new phages.

Temperate phages such as lambda can choose. After infection, lambda either enters the lytic cycle or enters the lysogenic cycle, in which its 48.5 kb genome integrates into the host chromosome at att by site specific recombination, catalyzed by phage integrase. The integrated DNA is a prophage, and the host cell, now a lysogen, divides normally and copies the prophage along with its own DNA. The lambda cI repressor keeps the lytic genes silent and also makes the lysogen immune to superinfection by another lambda. When the host DNA is damaged (for example by ultraviolet light), activated RecA stimulates self cleavage of the cI repressor, the prophage excises (integrase plus excisionase), and the lytic cycle begins: this is induction.

Plaques: how phage genetics is scored

Phages are counted by the plaque assay. Phages are mixed with a large excess of bacteria in soft agar and poured over a plate. Bacteria grow into a cloudy lawn, but wherever a single phage infected a cell, successive rounds of infection and lysis clear a small circular zone, a plaque. Each plaque comes from one phage, so plaques measure the titer:

titer (plaque forming units per mL) = number of plaques / (volume plated in mL × dilution)

Phage mutants change plaque appearance or host range. Rapid lysis (r) mutants of T2 and T4 make large, sharp edged plaques, while wild type (r+) makes small, fuzzy plaques. Host range (h) mutants can infect bacterial strains that resist the wild type. Phage crosses use mixed infection: each cell receives both genotypes, which recombine as they replicate.

Study tip. Worked problem: Bacteria are infected with h− r+ and h+ r− phages. The progeny produce 1,000 plaques that can be classified: h+ r− 410, h− r+ 390, h+ r+ 105, h− r− 95. What is the recombination frequency between h and r? Step 1, identify the parentals. They are the input genotypes and the two largest classes: h− r+ and h+ r− (390 + 410 = 800). Step 2, identify the recombinants: h+ r+ and h− r− (105 + 95 = 200). Step 3, RF = 200 / 1,000 = 0.20. Answer: 20 percent, or 20 map units. Check: the two recombinant classes are reciprocal products and should be roughly equal (105 and 95 are).

Benzer and the fine structure of the rII region

In the 1950s, Seymour Benzer asked how finely recombination could divide a gene. He used the rII mutants of phage T4. The trick was a pair of hosts:

That asymmetry created an extraordinarily sensitive selection. After a cross between two rII mutants, plating on K12(λ) reveals only the rare wild type recombinants, while plating on B counts every progeny phage. Because one plate can hold progeny from a whole lysate, Benzer could detect one wild type recombinant among hundreds of millions of progeny, enough to measure recombination between mutations only one or a few base pairs apart.

Two different tests: complementation and recombination

Benzer used two tests that are easy to confuse, so keep them separate.

The complementation test asks whether two mutations are in the same functional unit. Cells of K12(λ) are coinfected with two rII mutants. If the two mutations are in different genes, each phage supplies the product the other lacks, the infection succeeds and the culture lyses: they complement. If both mutations are in the same gene, neither phage makes a working product and nothing happens. This is a trans test (the mutations are on different DNA molecules). Its control is the cis arrangement (both mutations on one molecule, the other molecule wild type), which always gives wild type function. Benzer called the unit defined by the cis-trans test a cistron, and found that the rII region contains two: rIIA and rIIB.

The recombination test asks whether two mutations are at different sites, and how far apart. Cells of E. coli B (on which all rII mutants grow) are coinfected, the progeny are collected, and the lysate is plated on both hosts. Wild type recombinants are one of two reciprocal products; the other (the double mutant) cannot be seen on K12(λ), so the frequency of wild type is doubled:

RF = 2 × (titer on K12(λ)) / (titer on B)

Mutations at the identical site never give wild type recombinants, and mutations in the same cistron can still recombine if they are at different sites.

QuestionComplementation (cis-trans) testRecombination test
What it asksAre the two mutations in the same gene?Are the two mutations at different sites, and how far apart?
Host for infectionK12(λ), restrictiveB, permissive; progeny then plated on both hosts
Positive resultLysis (growth) means different genesWild type plaques on K12(λ) mean different sites
Depends on recombination?No: protein products work together in one cellYes: needs a crossover between the two sites
Unit definedCistron (gene)Site (as small as one base pair)
Study tip. Worked problem: Two rII mutants are crossed in E. coli B. The lysate is diluted 10−6 and 0.1 mL is plated on B, giving 300 plaques. The same lysate is diluted 10−2 and 0.1 mL is plated on K12(λ), giving 60 plaques. Find the recombination frequency. Step 1, titer on B: 300 / (0.1 × 10−6) = 3 × 109 phages per mL (all progeny). Step 2, titer on K12(λ): 60 / (0.1 × 10−2) = 6 × 104 per mL (wild type recombinants only). Step 3, fraction wild type = 6 × 104 / 3 × 109 = 2 × 10−5. Step 4, double it for the invisible reciprocal double mutant: RF = 4 × 10−5. Answer: 0.004 percent, or 0.004 map units, so these mutations are very close together. The most common slip is forgetting the factor of 2.

Deletion mapping

Rather than crossing thousands of mutants pairwise, Benzer used deletion mutants, which are recognizable because they never revert. A point mutation crossed with a deletion can give wild type recombinants only if the point mutation lies outside the deleted segment; if the site is missing from the deletion phage, there is no wild type copy of that stretch to recombine in. A panel of overlapping deletions therefore divides the region into segments, and a single spot test against the panel assigns each new mutation to a segment. Fine mapping by recombination is then needed only among the few mutations within that segment.

rII gene divided into segments by overlapping deletions 123456 Segment D1D2D3D4 missingmissingmissingmissing No wild type recombinants = the point mutation lies inside that deletion
Figure 5.4 Deletion mapping. Red bars are the DNA missing from each deletion mutant. A point mutation that fails to give wild type recombinants with D1 and D2, but does with D3 and D4, must lie in segment 2, the only segment covered by D1 and D2 and by neither D3 nor D4.
Study tip. Worked problem: Using the deletions in Figure 5.4, five point mutants were crossed with each deletion (+ means wild type recombinants appeared, 0 means none). m1: D1 0, D2 0, D3 +, D4 +. m2: D1 +, D2 0, D3 0, D4 +. m3: D1 +, D2 +, D3 0, D4 0. m4: D1 +, D2 0, D3 +, D4 +. m5: D1 0, D2 +, D3 +, D4 +. Assign each to a segment. Step 1, write each deletion's segments: D1 = 1 to 2, D2 = 2 to 4, D3 = 4 to 6, D4 = 5 to 6. Step 2, for each mutant, find the segment inside every "0" deletion and outside every "+" deletion. m1: inside D1 and D2 means segment 2. m2: inside D2 and D3 means segment 4. m3: inside D3 and D4 means segment 5 or 6. m4: inside D2 only means segment 3. m5: inside D1 only means segment 1. Answer: m5 in 1, m1 in 2, m4 in 3, m2 in 4, m3 in 5 or 6. Separating segments 5 and 6 needs another deletion that ends between them.

What the rII work showed

Benzer mapped more than 2,000 rII mutations to about 300 sites. Recombination could separate mutations at neighboring sites, so the gene is not indivisible: the units of mutation and recombination are much smaller than the gene, ultimately single base pairs. He also found mutational hotspots. The map was linear, like DNA.

Key point

Phages grow lytically or, if temperate, integrate as prophages under repressor control. Phage genotypes are scored by plaque type, and recombination frequency in a mixed infection equals recombinant plaques over total. Benzer's rII system separated two questions: complementation (same gene or not, the cis-trans test) and recombination (same site or not), and deletion mapping showed that a gene is a linear array of hundreds of separable sites.

Section 5Viral genomes and retroviruses

What a virus is, genetically

A virus is a genome wrapped in a protein coat (the capsid), sometimes surrounded by a lipid envelope taken from the host membrane. Viruses have no ribosomes of their own, but their genomes vary enormously. Viral genomes can be DNA or RNA, single stranded or double stranded, linear or circular, in one piece or split into segments. They range from about 2 kb to more than a megabase.

The Baltimore classification

David Baltimore grouped viruses by the route their genome takes to messenger RNA (mRNA, defined as the plus strand), which predicts which enzymes each virus must carry.

ClassGenomeRoute to mRNAExamples
IDouble stranded DNATranscribed like host genes (usually in the nucleus)Herpesviruses, adenoviruses, poxviruses, papillomaviruses, phage T4
IISingle stranded DNAConverted to dsDNA, then transcribedParvoviruses, phage ΦX174
IIIDouble stranded RNAViral RNA dependent RNA polymerase copies the minus strandRotavirus, reoviruses
IVSingle stranded RNA, plus senseGenome itself serves as mRNAPoliovirus, rhinoviruses, coronaviruses, hepatitis C, Zika
VSingle stranded RNA, minus sensePackaged viral polymerase copies the genome into mRNAInfluenza, measles, mumps, rabies, Ebola, RSV
VISingle stranded RNA, plus sense, with a DNA intermediateReverse transcribed to DNA, integrated, then transcribedRetroviruses: HIV, HTLV-1, Rous sarcoma virus
VIIPartly double stranded DNA, with an RNA intermediateTranscribed to a pregenomic RNA that is reverse transcribedHepatitis B virus

Note that the naked genome of a plus strand RNA virus is infectious by itself, because host ribosomes can translate it directly, whereas minus strand and double stranded RNA viruses must carry their polymerase inside the virion. RNA dependent RNA polymerases and reverse transcriptases generally lack proofreading, so RNA viruses mutate far faster than DNA organisms, roughly 10−4 to 10−5 errors per nucleotide per replication compared with about 10−9 to 10−10 for cellular DNA replication. Coronaviruses (about 30 kb) are an exception: they encode a proofreading exonuclease.

Viral recombination and reassortment

Viruses exchange genes in two ways. When two related viruses infect the same cell, their polymerases can switch templates, producing recombinant genomes. Viruses with segmented genomes can also swap whole segments, a process called reassortment. Influenza A has eight minus strand RNA segments, including the ones encoding the surface proteins hemagglutinin (HA) and neuraminidase (NA). Point mutations in HA and NA cause antigenic drift (hence yearly vaccines); reassortment between, say, human and avian strains gives a new HA or NA, antigenic shift, the source of pandemics.

Study tip. Worked problem: A cell is infected with a human influenza strain and an avian strain, each with eight segments. If each progeny virion packages one copy of each segment and picks the human or avian version at random, how many genotypes are possible, and what fraction of progeny carry the avian HA segment with all seven other segments from the human strain? Step 1, each of 8 segments has 2 choices, so the number of genotypes is 28 = 256. Step 2, the specified genotype requires one particular choice at each segment: (1/2)8 = 1/256. Answer: 256 possible genotypes; about 1/256, or 0.39 percent, of progeny have exactly that combination.

The retrovirus genome

Retroviruses reverse the usual flow of genetic information: their RNA genome is copied into DNA, which integrates permanently into the host chromosome. Each virion carries two copies of a plus strand RNA genome of about 7 to 10 kb (HIV-1 is about 9.7 kb), so retroviruses are effectively diploid. Every retrovirus has three core genes:

HIV also carries regulatory and accessory genes such as tat, rev and nef.

The retroviral life cycle

  1. Entry. Env binds the receptor (CD4 plus the coreceptor CCR5 or CXCR4 for HIV), the envelope fuses with the cell membrane, and the core enters the cytoplasm.
  2. Reverse transcription. Reverse transcriptase, an RNA dependent and DNA dependent DNA polymerase with an RNase H domain that degrades the RNA in RNA DNA hybrids, makes a double stranded DNA copy of the genome. It is primed by a host transfer RNA (tRNALys3 for HIV). Two template "jumps" generate identical long terminal repeats (LTRs) at the two ends of the DNA. Switching between the two RNA copies also causes recombination.
  3. Integration. The viral DNA enters the nucleus, and integrase cuts the host DNA with a staggered cut and joins the viral ends to it. Repair of the gaps creates a short target site duplication (5 bp for HIV) flanking the provirus. The provirus is now a permanent part of that cell's genome and of all its descendants.
  4. Expression. Host RNA polymerase II transcribes the provirus from a promoter in the 5′ LTR; the 3′ LTR supplies the polyadenylation signal. Spliced transcripts make Env; full length RNA makes Gag and Gag-Pol and serves as new genome.
  5. Assembly and maturation. Two genome RNAs, Gag and Gag-Pol assemble at the membrane and bud off with an envelope. The viral protease then cleaves the polyproteins into mature proteins, converting the particle into an infectious virion.
host cell nucleus 1. virion binds receptor, fuses, enters 2. Reverse transcription RNA dsDNA with LTRs (gold) 3. Integraseinserts provirus host chromosome provirus 4. Pol II transcribes from 5′ LTR 5. RNA genomes, Gag, Gag-Pol, Env 6. Bud, then protease matures the virion
Figure 5.5 The retrovirus life cycle. The key genetic events are reverse transcription (RNA to double stranded DNA flanked by long terminal repeats) and integration, which makes the provirus a permanent part of the host genome that is transcribed by host RNA polymerase II.
Study tip. Worked problem: HIV-1 has a genome of 9,700 nucleotides, and its reverse transcriptase makes about 3 × 10−5 errors per nucleotide per replication cycle. (a) How many new mutations does an average genome acquire per cycle? (b) What fraction of new genomes carry no new mutation? (c) An untreated patient produces about 1010 virions per day. About how often does each possible single base substitution arise? Step 1, mean mutations per genome = 9,700 × 3 × 10−5 = 0.29. Step 2, mutations per genome follow a Poisson distribution, so P(0) = e−0.29 = 0.75. Step 3, there are 9,700 × 3 = 29,100 possible single base substitutions. Mutations made per day are 1010 × 0.29 = 2.9 × 109, so each possible substitution arises about 2.9 × 109 / 29,100 = 1 × 105 times per day. Answer: (a) 0.29; (b) about 75 percent have no new mutation, so about 25 percent carry at least one; (c) every single substitution, including any that confers drug resistance, arises roughly 100,000 times per day.
Connection. Each step of the retroviral cycle is a drug target. Nucleoside and non-nucleoside reverse transcriptase inhibitors block reverse transcription, integrase inhibitors block insertion of the provirus, and protease inhibitors block maturation. Because a single resistance mutation arises in a patient every day (see the worked problem), drugs are given in combinations of three or more, so a virus would need several specific mutations at once to escape. The same integrating machinery, stripped of viral genes, is used in lentiviral vectors that deliver therapeutic genes, including the receptor genes used in CAR T cell therapy.

Retroviruses, oncogenes and our own genome

Retroviruses gave cancer genetics its first oncogene. Rous sarcoma virus, which causes tumors in chickens, carries v-src, a gene that the virus captured long ago from a normal chicken gene, c-src, a proto-oncogene that encodes a tyrosine kinase. The viral copy is mutated and overexpressed, so it drives uncontrolled growth. Other retroviruses carry no oncogene but cause cancer by insertional activation, integrating beside a proto-oncogene such as MYC so that the strong LTR promoter or enhancer drives it.

When a retrovirus infects a germ line cell, the provirus can be inherited like any other gene. About 8 percent of the human genome consists of such endogenous retroviruses, almost all inactivated by mutation. Related LTR retrotransposons move within the genome by the same copy, reverse transcribe and integrate mechanism.

Key point

Viral genomes come in every chemical form, and the Baltimore classes sort them by their route to mRNA. RNA viruses mutate fast because their polymerases do not proofread, and segmented viruses such as influenza reassort. Retroviruses carry two plus strand RNA genomes with gag, pol and env; reverse transcriptase makes DNA flanked by LTRs, and integrase inserts it as a permanent provirus, a mechanism exploited by cancer causing viruses, by retrotransposons and by gene therapy vectors.

The chapter in one breath

  • Bacteria are haploid with one circular chromosome in a nucleoid; plasmids such as F, R and Col replicate independently and carry optional but often decisive genes.
  • Auxotrophs need a supplement that prototrophs make for themselves; replica plating finds them, and the fluctuation test showed that mutations arise randomly before selection.
  • In conjugation, F+ cells transfer only F; Hfr cells transfer chromosome in a fixed order from oriT; F′ cells create partial diploids.
  • Interrupted mating maps genes by time of entry in minutes; overlapping Hfr orders reveal a 100 minute circular map.
  • Transformation moves free DNA (DNase sensitive); cotransformation frequency measures linkage.
  • Generalized transduction packages any host fragment of about 2 minutes; cotransduction frequency and the rarest class of a three factor cross give distance and order.
  • Phages grow lytically or, if temperate, as prophages held silent by repressor; phage crosses are scored by plaque phenotypes.
  • Benzer's rII work separated the complementation (cis-trans) test from the recombination test and used deletion mapping to show that a gene contains hundreds of separable sites.
  • The Baltimore classes sort viruses by route to mRNA; RNA viruses mutate fast, and segmented viruses reassort.
  • Retroviruses reverse transcribe their RNA to DNA with LTRs and integrate as a provirus, the basis of HIV therapy, viral oncogenes and endogenous retroviruses.

Words worth knowing

TermWhat it means
NucleoidThe region of a bacterial cell that holds the compacted chromosome, with no surrounding membrane.
PlasmidA small, usually circular DNA molecule that replicates independently of the chromosome.
PrototrophA wild type strain that grows on minimal medium because it makes all its own building blocks.
AuxotrophA mutant that cannot make a required nutrient and grows only when it is supplied.
Replica platingTransferring colonies from a master plate to other media with a velvet pad, preserving their positions.
Fluctuation testLuria and Delbrück's comparison of independent cultures that showed mutations arise before selection.
ConjugationTransfer of DNA from donor to recipient bacterium through direct cell contact.
F factorThe fertility plasmid that encodes the sex pilus and transfer machinery.
Hfr strainA donor with F integrated into its chromosome, which transfers chromosomal genes in a fixed order.
F′ factorAn F plasmid that has picked up a segment of chromosomal DNA during excision.
TransformationUptake and incorporation of free DNA from the environment.
Generalized transductionPhage mediated transfer of any part of the host genome packaged in place of phage DNA.
Specialized transductionTransfer of only the host genes beside a prophage, after faulty excision.
Cotransduction frequencyThe fraction of transductants for one marker that also inherited a second marker.
Lytic cyclePhage infection that ends with the cell bursting and releasing new phages.
LysogenyA state in which a temperate phage genome is integrated as a prophage and kept silent.
ProphageA phage genome integrated into the host chromosome.
CistronThe unit of function defined by the cis-trans complementation test; a gene.
ComplementationRestoration of wild type function when two mutant genomes are in one cell, showing the mutations are in different genes.
Deletion mappingLocating point mutations by testing whether they can recombine with a panel of deletions.
Baltimore classificationA scheme that groups viruses into seven classes by how their genome produces mRNA.
ReassortmentExchange of whole genome segments between viruses infecting one cell.
Reverse transcriptaseA viral enzyme that makes DNA from an RNA template.
ProvirusRetroviral DNA integrated into a host chromosome.
Long terminal repeat (LTR)A repeated sequence at each end of retroviral DNA containing the promoter and polyadenylation signals.

Check yourself

A colony grows on complete medium and on minimal medium plus arginine, but not on minimal medium. What is its genotype for arginine synthesis, and how was it probably identified?
Answer. It is an arginine auxotroph, arg−. Replica plating from a master plate onto minimal and supplemented minimal media reveals colonies that grow only when arginine is supplied.
In a fluctuation test, 30 independent cultures are plated on a drug and 9 have no resistant colonies. Each culture reached 5 × 108 cells. Estimate the mutation rate.
Answer. P0 = 9/30 = 0.30, so m = −ln(0.30) = 1.20 mutations per culture. Rate ≈ 1.20 / (5 × 108) = 2.4 × 10−9 per cell per division.
Why does the recipient in an Hfr × F− mating almost always remain F−, while the recipient in an F+ × F− mating becomes F+?
Answer. In an Hfr, the F sequences needed to make the recipient a donor lie behind the whole chromosome and would enter last, after about 100 minutes; mating pairs break long before that. A free F plasmid is transferred completely in a few minutes.
Hfr A transfers p (6 min), q (14), r (30). Hfr B transfers s (4), r (12), q (28), p (36). What is the gene order and the distance from r to s?
Answer. The order is p, q, r, s. Hfr A gives p to q = 8 and q to r = 16; Hfr B gives the same intervals in reverse (28 − 12 = 16, 36 − 28 = 8), showing F is inserted in the opposite orientation, and r to s = 12 − 4 = 8 minutes.
Why must a donor fragment entering by conjugation, transformation or transduction undergo an even number of crossovers?
Answer. The incoming DNA is linear and the recipient chromosome is circular. A single crossover would open the circle into a linear molecule that cannot be maintained; two (or four) crossovers swap a segment while keeping the circle intact.
A gene transfer experiment is unaffected by DNase and still works when donor and recipient are separated by a filter that blocks cells. Which mechanism is operating?
Answer. Transduction. Transformation would be blocked by DNase, and conjugation requires cell contact; DNA inside phage heads is protected from DNase and can pass the filter.
With P1 (L = 2 minutes), two markers are cotransduced at a frequency of 0.216. How far apart are they?
Answer. 0.216 = (1 − d/2)3. The cube root of 0.216 is 0.6, so 1 − d/2 = 0.6, d/2 = 0.4 and d = 0.8 minute.
A lambda lysate made by inducing a lysogen transduces gal at low frequency and never transduces trp. Explain both observations.
Answer. Lambda integrates at att between gal and bio, so only those neighbors can be picked up by faulty excision, which is rare. trp lies far away, so it is never included. This is specialized transduction.
Two rII mutants fail to lyse K12(λ) when coinfected, but a cross between them in B yields wild type plaques on K12(λ). What do you conclude?
Answer. They do not complement, so they are in the same cistron (rIIA or rIIB). They do recombine to give wild type, so they are at different sites within that cistron.
Why do RNA viruses and retroviruses evolve drug resistance so much faster than bacteria, and why does combination therapy help?
Answer. Their polymerases lack proofreading, and huge populations mean every single mutation arises often. With three drugs, escape requires several specific mutations in one genome at once, the product of small probabilities, which is very unlikely.