The lac Operon: How Bacteria Regulate Genes

Give E. coli a broth with both glucose and lactose and it does not eat them together. It uses up the glucose, pauses, and only then starts on the lactose.

Jacques Monod, studying bacterial growth around 1940, called this two-step growth diauxie. During the pause the cells make the proteins that handle lactose, which they did not make while glucose lasted. So a bacterium chooses which genes to use from what is in its medium. This lesson is about how, in the lac operon of E. coli and, for contrast, the trp operon.

The switch: a repressor that lets go

Three genes for using lactose sit side by side on the E. coli chromosome: lacZ encodes β-galactosidase, which splits lactose into glucose and galactose; lacY encodes the permease that carries lactose in; lacA encodes a transacetylase. RNA polymerase transcribes all three from one promoter into one mRNA, read into three separate proteins. An mRNA that carries several genes like this is called polycistronic, and a set of genes transcribed together under one control is an operon. (Promoters and RNA polymerase are in the lesson on transcription.)

Next to the operon is a fourth gene, lacI, with its own promoter. It encodes the lac repressor, a protein that binds a short stretch of DNA called the operator, which lies at the start of transcription. Bound there, the repressor keeps RNA polymerase from transcribing the genes. This is negative control: the regulator’s job is to stop.

Picture a hand holding a light switch off; lactose arrives and the hand lets go. The real mechanism: when lactose gets in, β-galactosidase converts a little of it to allolactose, which joins the same two sugars by a different bond. Allolactose is the natural inducer. It binds the repressor and changes the protein’s shape, which lowers its affinity for the operator, so the repressor comes off and transcription starts. A molecule that changes a protein’s shape by binding at a separate site is an allosteric effector (the lesson on protein binding and allostery).

Two traps. Lactose itself is not the inducer; allolactose is. And in the laboratory geneticists usually use IPTG, a synthetic compound that binds the repressor but is not broken down by β-galactosidase, so its level stays steady. An inducer of this kind is called gratuitous.

There is a loop here: lactose needs the permease and β-galactosidase, yet both come from the operon it switches on. Repression leaks, so an uninduced cell can hold a few copies of each protein.

One repressor on one operator is a simplification. The repressor is a tetramer, and the lac region has three operators, O1, O2 and O3. One tetramer can hold two operators at once, looping the DNA between them. Geneticists measure the strength of repression as a ratio:

repression factor = enzyme activity with inducer ÷ enzyme activity without inducer

In chromosomal reporter strains, O1 alone gave a repression factor of about 20, and removing both auxiliary operators reduced repression more than 50-fold (Oehler and colleagues, 1990). For the genetics below, one operator is enough.

The partial diploid: telling a site from a product

The genetics came before the biochemistry. Some mutants made β-galactosidase with or without inducer; they are called constitutive. Some mapped in lacI (written I⁻) and some in the operator (Oᶜ, operator-constitutive). In a haploid cell the two look the same. How do you tell a broken repressor gene from a broken binding site?

You put two copies of the region in one cell. The F factor, the plasmid that lets E. coli mate (see bacterial and phage genetics), can carry the lac region; such a plasmid is written F′ lac. A cell with its own chromosomal lac region plus an F′ lac is a partial diploid, or merodiploid: diploid for those genes only. In 1959 Arthur Pardee, François Jacob and Monod brought lac genes into recipient cells by mating, and Jacob, Monod and colleagues went on to use F′ lac partial diploids to compare mutations on the same DNA molecule (in cis) or on different ones (in trans). This is the cis-trans logic of complementation and pathways.

The rule: a gene product diffuses, so it acts in trans, on both copies; a site such as the operator never leaves its DNA, so it acts only in cis.

Worked problem: four partial diploids

Predict β-galactosidase activity without and with IPTG for cells grown without glucose, so cAMP and CAP are available. Each genotype is written chromosome / F′, and every lac gene not listed is wild type. Assume Z⁻ makes no active enzyme, I⁻ makes no working repressor, Iˢ (a superrepressor) makes a repressor that binds the operator but cannot bind inducer, and Oᶜ cannot be bound by any repressor. For each, ask two questions: where is the working Z⁺, and what controls the operator next to it?

(a) I⁻ Z⁺ / F′ I⁺ Z⁻. The only Z⁺ is on the chromosome, beside a normal operator. The chromosome makes no repressor, but the F′ I⁺ does, and repressor diffuses to the chromosomal operator. No IPTG: off. IPTG: on. Inducible. I⁻ is recessive.

(b) Oᶜ Z⁻ / F′ O⁺ Z⁺. Both copies are I⁺. The Oᶜ copy is transcribed all the time, but its Z⁻ makes nothing active. The only Z⁺ sits beside a normal operator, which the repressor holds until IPTG arrives. Inducible.

(c) Oᶜ Z⁺ / F′ O⁺ Z⁻. Now the Z⁺ is in cis with Oᶜ, which no repressor can bind. Active enzyme is made with or without IPTG. Constitutive. Compare (b): the same two mutations, swapped between the copies, give a different answer. That is the mark of a cis-acting site. Oᶜ is called cis-dominant: dominant, but only over genes on its own DNA.

(d) Iˢ Z⁺ / F′ I⁺ Z⁺. Two working Z⁺ genes, both beside normal operators. IPTG pulls the normal repressor off, but the Iˢ repressor cannot bind IPTG, and because it also diffuses, it stays on both operators. Uninducible. Iˢ is dominant to I⁺.

An operator acts only on its own DNA molecule; a repressor acts on every operator in the cell.

Chromosome

Repressor gene I⁺ normal repressor I⁻ no working repressor Iˢ superrepressor I⁻ᵈ dominant-negative Operator O⁺ normal Oᶜ constitutive lacZ Z⁺ active enzyme Z⁻ no active enzyme

F′ lac plasmid

Plasmid present (partial diploid) absent (haploid) Repressor gene I⁺ normal repressor I⁻ no working repressor Iˢ superrepressor I⁻ᵈ dominant-negative Operator O⁺ normal Oᶜ constitutive lacZ Z⁺ active enzyme Z⁻ no active enzyme

inducer present (IPTG, which needs no β-galactosidase) no glucose, so CAP and cAMP are available. Lactose itself would induce only through β-galactosidase.

Or load

No IPTG

With IPTG

Behaviour

An operator acts only on its own DNA molecule; a repressor acts on every operator in the cell. Every lac gene not listed is wild type, and “little or none” allows for the leak the lesson describes. This is a sketch of the logic.

One more class completes the set. Some I⁻ mutations are dominant. Written I⁻ᵈ, they make subunits that still join tetramers but cannot bind DNA, so tetramers that mix them with normal subunits from the other copy fail too. Such a mutation is called dominant-negative.

What was the repressor made of? In 1961 Jacob and Monod did not know; Monod later wrote that it had seemed logical to assume a polyribonucleotide. In 1966 Walter Gilbert and Benno Müller-Hill isolated it, following its binding of radioactive IPTG, and it was a protein; in 1967 they showed it binds DNA carrying the lac operator.

CAP and cAMP: the second input

The repressor asks whether lactose is here. Diauxie shows the cell also asks whether something better is. The lac promoter is weak on its own. Full transcription needs an activator, the catabolite activator protein (CAP, also called CRP, the cAMP receptor protein). CAP binds cyclic AMP, and CAP with cAMP bound attaches to a site just upstream of the lac promoter. There it contacts RNA polymerase, at the C-terminal part of the polymerase’s α subunit, and helps the polymerase bind and start. This is positive control.

In the classic account, cAMP is higher when glucose is scarce, so CAP is active, and lower when glucose is plentiful; how much that change matters in living cells is debated. Glucose does not bind CAP. A second glucose effect also matters, and in E. coli it is a major one: while glucose is being taken up, a protein of the glucose transport system binds the lac permease and blocks lactose entry. This is inducer exclusion. Less lactose in means less allolactose, so more repressor stays on.

Put the two inputs together for a wild-type cell:

Lactose	Glucose	lac transcription	Why
Absent	Present	Very low	Repressor bound; little active CAP
Absent	Absent	Very low	CAP cannot overcome a bound repressor
Present	Present	Reduced	Less active CAP, and inducer exclusion
Present	Absent	High	Repressor off and CAP–cAMP on

Keep this apart from your own cells, where cAMP acts mainly through protein kinase A (cAMP and PKA), not through a CAP.

The trp operon: a repressor that needs a partner, and a brake on the ribosome

The trp operon of E. coli encodes the enzymes that make tryptophan, and its logic runs the other way: the genes should be off when tryptophan is plentiful. The trp repressor alone binds the operator poorly. Tryptophan binds the repressor and lets it bind the operator, so the end product shuts down its own pathway. The lac operon is inducible; the trp operon is repressible.

A second control, worked out mostly in Charles Yanofsky’s laboratory, acts after transcription has started. Transcription can stop in a leader region before the first structural gene, at a point called the attenuator. The leader encodes a peptide of 14 amino acids with two tryptophan codons in a row. Its RNA can fold into different hairpins. One, called 3:4, followed by a run of U’s, is a terminator. Another, 2:3, prevents 3:4 from forming and lets transcription continue.

Which hairpin forms depends on a ribosome. In bacteria there is no nucleus, so a ribosome starts translating the leader while RNA polymerase is still transcribing just ahead of it (the genetic code and translation).

• Tryptophan plentiful. Charged tryptophan tRNA is easy to find, and the ribosome moves past the two tryptophan codons onto region 2. With region 2 covered, regions 3 and 4 pair, the terminator forms, and polymerase stops.
• Tryptophan scarce. The ribosome stalls at the tryptophan codons, short of region 2. Region 2 pairs with region 3, the terminator cannot form, and polymerase continues into the structural genes.

In a test tube with no ribosomes, only about 5 percent of transcripts read through the E. coli trp attenuator (Lee and Yanofsky, 1977). In this model the ribosome is the sensor, turning the supply of charged tRNA into a choice of RNA fold, which needs translation coupled to transcription. In the nucleus of a eukaryotic cell the two are separated, so attenuation of this kind does not work there (eukaryotic gene regulation takes up what does).

Putting the rules to work: classifying a new mutant

Design problem

You isolate a new constitutive lac mutant, m. Design partial diploids that tell whether it is a recessive I⁻, a dominant-negative I⁻ᵈ, or an operator Oᶜ.

One way to do it

Start with m Z⁺ / F′ I⁺ O⁺ Z⁻: the only working Z⁺ is beside m, and the F′ supplies a normal repressor gene.

• If expression becomes inducible, m is a recessive I⁻: the normal repressor from the F′ diffuses to the chromosomal operator.
• If it stays constitutive, m is either I⁻ᵈ or Oᶜ. An Oᶜ ignores the repressor in cis, and an I⁻ᵈ poisons the tetramers made with F′ subunits.

To separate those two, move the working Z⁺ to the other copy: m Z⁻ / F′ I⁺ O⁺ Z⁺.

• If m is Oᶜ, expression is inducible: the Oᶜ copy’s Z⁻ makes nothing active, and the F′ Z⁺ sits beside a normal operator.
• If m is I⁻ᵈ, expression stays constitutive: the defective subunits act in trans and spoil repressor at the F′ operator too.

Move the reporter gene and see whether the defect moves with it: a defect that stays on its own DNA is a site, and one that reaches the other copy is a product.