Bacterial and Phage Genetics: Conjugation, Transduction and Transformation

In 1946 two strains of Escherichia coli K-12, each unable to make several nutrients, were mixed and spread on a plate that supplied none of them. A few colonies grew. Neither strain alone gave any.

An auxotroph is a mutant that needs a nutrient added because it cannot make it; the wild type is a prototroph. In the pair of strains that Davis later tested, one needed biotin and methionine, the other threonine, leucine and thiamine. A colony on bare minimal medium had working copies of all five genes, some from each parent. Joshua Lederberg and Edward Tatum, at Yale, read the result as recombination: one bacterium can inherit genes from another.

Why strains that each lack several abilities? A single mutant can revert, mutating back to the working allele, and a revertant looks like a recombinant. Needing several reversions in one cell makes that very unlikely, and each parent plated alone gave no colonies.

How does DNA get from one bacterium into another, and how can the way it moves become a map of the chromosome?

The idea: a piece of chromosome gets in by one of three routes

Bacteria make no zygote. An E. coli cell is haploid, with one circular chromosome, and what enters it is a piece of a donor’s chromosome. For a while the recipient carries two copies of that segment, its own and the donor’s. Such a partial diploid is called a merozygote. The donor piece is inherited only if recombination swaps it into the recipient’s chromosome, and on a circle that takes an even number of crossovers, at least two, one on each side.

Think of a train taking on a borrowed carriage: it stays only if the train is uncoupled and rejoined on both sides of it.

Three routes deliver the piece.

• Transformation: the recipient takes up free DNA from its surroundings. Some species, such as Streptococcus pneumoniae and Bacillus subtilis, do this naturally when they are competent, a physiological state that allows uptake. Griffith’s pneumococci and Avery, MacLeod and McCarty’s purified DNA belong to that lesson.
• Conjugation: a donor cell makes contact with a recipient and passes DNA to it directly.
• Transduction: a bacteriophage, a virus that infects bacteria, carries a piece of one host’s DNA inside its protein coat into the next cell it infects.

Two tests tell the routes apart. In 1950 Bernard Davis placed two strains in the arms of a U-shaped tube, separated by a filter that stopped cells but let medium pass, and moved the medium back and forth between them. Lederberg and Tatum’s strains then gave no recombinants, which argued against a filterable agent and pointed to contact between cells. The second test is deoxyribonuclease (DNase), an enzyme that destroys free DNA.

Route	Needs cell contact	Crosses a filter that stops cells	Stopped by DNase in the medium
Transformation	No	Yes	Yes
Conjugation	Yes	No	No
Transduction	No	Yes	No; the DNA is inside a phage coat

Conjugation: a chromosome that enters by the minute

Conjugation in E. coli depends on F, the fertility factor. F is a plasmid, a small DNA circle that replicates separately from the chromosome and carries genes for a pilus and for DNA transfer. A cell carrying F is F+; one without it is F−. In an F+ by F− cross, one strand of F is passed from a fixed point, the origin of transfer, and each cell makes the partner strand. Recipients become F+, but chromosomal genes move only rarely, because F transfers itself.

Now and then F integrates into the chromosome. Such a cell is an Hfr, for high frequency of recombination, because transfer starting at the origin within F now pulls the chromosome after it in a fixed order. Hfr strains were described in the early 1950s by Luca Cavalli-Sforza and by William Hayes, who are usually credited with independently isolating such strains. Mating pairs usually break apart before transfer ends, so a recipient gets only part of the chromosome, and since the rest of F would enter last, it rarely becomes Hfr or F+.

Élie Wollman and François Jacob, at the Institut Pasteur, turned that ordered entry into a map. They mixed Hfr donors with F− recipients, broke the pairs apart in a blender at intervals, and plated on medium where only recombinants grew and streptomycin killed the donors. Each donor gene appeared only in samples taken after a characteristic time. If transfer runs at a steady rate, the time of entry is a position. Jacob and Wollman estimated that the whole chromosome would take about 100 minutes to enter at a constant rate, which is why the E. coli map is still drawn from 0 to 100 minutes. Hfr strains with F at different sites, in either orientation, gave overlapping orders that closed into a circle.

A colony counts more than entry: the donor piece must also be integrated, and recipients carrying a recB mutation yield fewer recombinants in Hfr crosses.

Worked problem (a): interrupted mating. Hfr strain 1 transfers markers a, b, c and d into a streptomycin-resistant F− recipient. They first appear at a 8 min, b 14 min, c 20 min and d 31 min. Hfr strain 2 gives d 4 min, c 15 min, b 21 min and a 27 min. Assume transfer runs at the same steady speed in both strains, and that first appearance marks entry.

1. Strain 1 order, earliest first: a, b, c, d.
2. Strain 1 gaps: b minus a is 14 − 8 = 6; c minus b is 20 − 14 = 6; d minus c is 31 − 20 = 11 minutes.
3. Strain 2 order: d, c, b, a. Gaps: c minus d is 15 − 4 = 11; b minus c is 21 − 15 = 6; a minus b is 27 − 21 = 6 minutes.
4. Compare. Both give the order a-b-c-d with a-b 6, b-c 6 and c-d 11 minutes. Strain 2 reads the same map backwards: its origin lies on the far side of d, about 4 minutes before d, and it transfers in the opposite orientation.

map: a —6— b —6— c —11— d (distances in minutes)

The absolute times differ because each strain starts at its own origin. The gaps agree, and the gaps are the map.

Interrupt the mating at 0 min Hfr donor strain Hfr strain 1 Hfr strain 2

Markers entered so far

none

Gaps between them (min)

none yet

Each donor gene is recovered only from samples taken after its time of entry, so the order of entry gives the gene order and the gaps give the map in minutes; the gaps, not the absolute times, are the map. This is a sketch of worked problem (a), assuming steady transfer, with first appearance marking entry; a colony also needs the piece to be integrated, and the whole E. coli map runs 0 to 100 minutes, of which only the first 35 are shown.

Transduction: a phage that packs the wrong DNA

Norton Zinder, a graduate student with Lederberg at Wisconsin, looked for mating in Salmonella typhimurium. Genes passed even across a filter that stopped cells, and the agent resisted DNase. In 1952 Zinder and Lederberg traced it to a temperate phage, later named P22. A temperate phage can either kill its host or insert its genome into the host’s DNA as a prophage, a state called lysogeny.

Generalized transduction. As P22 fills new heads with DNA, it occasionally packs a fragment of host chromosome instead of its own genome. That particle injects the fragment into the next cell, where it can be recombined in. Any part of the host chromosome can be carried. P22 does this in Salmonella and P1 in E. coli.

Specialized transduction. Phage lambda in E. coli K-12 inserts at one attachment site, between the gal and bio genes. When the prophage is cut out inaccurately, it can take a neighbouring host gene with it. Lambda therefore transduces only genes next to that site, which M. L. Morse, Esther Lederberg and Joshua Lederberg reported for the galactose genes in 1956.

Cotransduction as a fine map. A P1 head holds about 100 kb of DNA, roughly 2 minutes of the 4.6-million-base-pair E. coli chromosome. Two genes ride in one particle only if they lie within that span, and the closer they are, the more often they do. Interrupted mating spans the whole chromosome; cotransduction resolves genes inside a 2-minute window. The cotransduction frequency is the fraction of transductants, selected for one donor marker, that also carry another.

Worked problem (b): cotransduction. P1 grown on a wild-type donor transduces an x y z triple mutant. Assume the donor carries the wild-type allele at all three genes and only x+ is selected. Of 500 x+ transductants, 350 are also y+ and 100 are also z+.

1. Cotransduction of y with x: 350/500 = 0.70, or 70%.
2. Cotransduction of z with x: 100/500 = 0.20, or 20%.
3. Both genes lie within one P1 fragment of x, and y lies closer to x than z does.
4. These two-factor numbers do not give the order: x-y-z and y-x-z both fit. Ordering needs three-factor classes, as in the design problem.

Why bacteria and phage made genetics fast

A haploid cell has no second allele to mask the first, so a mutant genotype shows at once. A culture holds billions of cells, and E. coli can divide about every 20 minutes in rich medium. Above all, selection does the searching: minimal medium, or an antibiotic, lets only the rare recombinant grow, so one cell in many millions is found without looking at the rest. Many resistance genes ride on plasmids and encode, for example, an enzyme that destroys the drug; plasmids became tools for moving genes in that lesson.

Phage are faster still. On a lawn of bacteria each infecting phage gives one clear spot, a plaque, so phage are counted as plaques, and plaque appearance is a phenotype. When two mutant phages infect one cell their genomes can recombine, and the progeny can be mapped as in a fly cross. Seymour Benzer used a selective host to pick out rare T4 recombinants among very large numbers of progeny, and mapped mutations within the rII region of that lesson.

Design problem

Order x, y and z with a three-factor transduction. Design the experiment, then interpret these counts from the cross in worked problem (b), with x+ selected and y and z scored.

Class	Count
x+ y+ z+	95
x+ y+ z	255
x+ y z+	5
x+ y z	145

One way to do it

Design. Grow P1 on the wild-type donor, infect the triple mutant, select x+ only, then test each colony for y+ and z+. Plate the recipient alone and the phage alone to rule out reversion and contamination.

Interpretation. The donor fragment carries x+ y+ z+, and an even number of crossovers must place it into the circular recipient chromosome. For each candidate order, count the crossovers needed for the rarest class, x+ y z+ (5).

• Order x-y-z: keeping donor x+ and z+ but recipient y needs crossovers outside x, between x and y, between y and z, and outside z. That is four.
• Order y-x-z: the same class needs one crossover between y and x and one beyond z. That is two, and should not be the rarest.
• Order x-z-y: the same class needs one crossover beyond x and one between z and y. Also two.

Only x-y-z makes the rarest class the one that needs four crossovers, so the order is x-y-z.

Check. y+ is 95 + 255 = 350 of 500 (70%) and z+ is 95 + 5 = 100 of 500 (20%), as in worked problem (b). The middle gene, y, cotransduces more often than the outer gene z, as the order predicts.