📖 ContentsChapter 9
Genetics · Chapter 9

Gene Regulation and Epigenetics

Every cell in your body carries the same genome, yet a neuron and a liver cell behave like different organisms. The difference lies in which genes they use, when, and how much. This chapter explains the switches, dials and memory systems that decide gene expression, from bacterial operons to the chromatin marks and small RNAs that shape development and disease.

By the end of this chapter you can

  • Distinguish cis-acting elements from trans-acting factors and classify a regulatory system as positive or negative, and inducible or repressible.
  • Name the levels at which gene expression is controlled and estimate how mRNA synthesis and decay set steady state levels.
  • Explain how the lac operon responds to lactose and glucose, including the roles of the lac repressor, allolactose, CAP and cAMP.
  • Predict enzyme expression in partial diploids carrying lacI, operator and structural gene mutations.
  • Describe repression and attenuation in the trp operon and predict the effect of leader mutations.
  • Explain how eukaryotic activators, enhancers, Mediator and insulators control transcription, and interpret a reporter gene deletion experiment.
  • Describe how histone modifications, chromatin remodeling and DNA methylation establish heritable states, including genomic imprinting and X inactivation.
  • Read a bisulfite sequencing result and calculate percent methylation.
  • Trace the biogenesis of siRNAs and microRNAs, identify a miRNA seed match, and compare their modes of silencing.

Section 1Principles of gene regulation

Why regulate genes at all?

Making RNA and protein is expensive. A bacterium that made all of its roughly 4,300 proteins at full speed all the time would waste energy on enzymes it does not need. Multicellular organisms face a second problem: they must build many cell types from one genome and keep each type stable through many divisions. Gene regulation solves both problems.

Genes expressed at a fairly steady level in nearly every cell, such as those for ribosomal proteins and glycolytic enzymes, are called housekeeping genes, and their expression is constitutive. Other genes are regulated: switched on or off, or turned up or down, in response to signals.

The levels of control

Gene expression is a pipeline, and a cell can intervene at any stage. In eukaryotes the main control points are:

  1. Chromatin accessibility: tightly packed nucleosomes and heterochromatin hide DNA.
  2. Transcription initiation: the most common and most economical control point.
  3. RNA processing: alternative splicing and polyadenylation change which protein is made.
  4. RNA export and localization.
  5. mRNA stability: half lives range from minutes to days, set by 3′ UTR sequences, RNA binding proteins and microRNAs.
  6. Translation: initiation factors, regulatory proteins and small RNAs control ribosome loading.
  7. Protein modification, localization and degradation, for example phosphorylation or ubiquitin tagging for the proteasome.

In bacteria, where transcription and translation are coupled, most regulation acts on transcription initiation, transcription termination and translation initiation. Bacterial mRNAs often last only 1 to 3 minutes, so protein supply follows transcription closely.

Nucleus Cytoplasm Chromatin (gene) Pre-mRNA Mature mRNA mRNA in cytoplasm Protein Active protein 1 2 3 4 5 6 7 Chromatin access Transcription Processing, splicing Export mRNA decay Translation Modification, degradation
Figure 9.1 Control points in eukaryotic gene expression. Transcription initiation (2) is the most common control point, but every step from chromatin access (1) to protein turnover (7) can be regulated.

Cis-acting elements and trans-acting factors

A cis-acting element is a DNA (or RNA) sequence that controls only the gene on the same molecule: promoters, operators, enhancers, silencers and 3′ UTR sequences are binding sites, not products. A trans-acting factor is a diffusible product, usually a protein or RNA, that can act on targets anywhere in the cell, including on another chromosome. Repressors, activators, sigma factors and microRNAs are trans-acting.

The distinction is testable in a partial diploid (merodiploid) carrying two copies of a regulatory region, one on the chromosome and one on a plasmid. A defective trans-acting factor is usually rescued by a wild type copy elsewhere, because the wild type product diffuses to both targets. A defective cis-acting site is not rescued; it affects only the genes physically linked to it.

Positive and negative control; induction and repression

Regulatory systems are classified along two independent axes.

The logic fits physiology. Catabolic enzymes are worth making only when their food is present, so catabolic operons are usually inducible; biosynthetic enzymes are worth making only when their product is scarce, so biosynthetic operons are usually repressible. The signal usually works allosterically, binding the regulatory protein away from its DNA binding domain and changing its DNA affinity.

Type of controlRegulatory proteinEffect of small moleculeClassic example
Negative inducibleRepressor binds operator without signalInducer binds repressor and releases it from DNAlac operon (lacI repressor, allolactose)
Negative repressibleRepressor cannot bind DNA aloneCorepressor binds repressor and enables DNA bindingtrp operon (TrpR, tryptophan)
Positive inducibleActivator inactive aloneInducer activates the activatorara operon (AraC with arabinose); CAP with cAMP at lac
Positive repressibleActivator active aloneSignal molecule inactivates the activatorLess common; some amino acid pathways
Study tip. Separate the two questions every time. “Does the protein turn transcription off or on when it is bound?” gives negative or positive. “Does the small molecule turn the pathway on or off?” gives inducible or repressible. The lac repressor and the trp repressor are both negative regulators; one system is inducible and the other repressible because their small molecules have opposite effects on DNA binding.

Operons, regulons and global control

Bacteria often cluster genes for one pathway into an operon: a set of genes transcribed from a single promoter into one polycistronic mRNA and controlled by shared regulatory sites. The lac operon carries three genes for lactose use, and the trp operon carries five genes for tryptophan synthesis. Operons guarantee that enzymes needed together are made together.

A single regulator can control many operons scattered around the chromosome, a set called a regulon. Alternative sigma factors act globally: in E. coli, σ70 directs RNA polymerase to most housekeeping promoters, while heat shock raises σ32, which switches on chaperones and proteases. Some mRNAs regulate themselves: a riboswitch is a structured 5′ UTR segment that binds a metabolite directly and changes shape to end transcription early or hide the ribosome binding site, so regulation does not always require a protein.

How fast can expression change? Synthesis, decay and steady state

The amount of an mRNA or protein reflects a balance between synthesis and destruction. If a molecule is made at a constant rate s and decays with first order kinetics, the decay constant is k = ln 2 ÷ t1/2. At steady state synthesis equals decay, so N = s ÷ k. Doubling the half life therefore raises the level as much as doubling transcription does. The time to reach a new steady state depends only on the half life: halfway in one half life, 75 percent in two, and about 97 percent in five. Short lived mRNAs let cells respond quickly.

Study tip. Worked problem: A bacterial mRNA is transcribed at 12 molecules per minute and has a half life of 2.0 minutes. (a) What is its steady state number per cell? (b) If a regulatory RNA shortens the half life to 0.5 minutes without changing transcription, what is the new steady state level? (c) How long does it take to get about 97 percent of the way to the new level?
Step 1: Find the decay constant. k = ln 2 ÷ t1/2 = 0.693 ÷ 2.0 = 0.347 per minute.
Step 2: Steady state N = s ÷ k = 12 ÷ 0.347 = 34.6, about 35 molecules.
Step 3: With t1/2 = 0.5 minutes, k = 0.693 ÷ 0.5 = 1.386 per minute, so N = 12 ÷ 1.386 = 8.7, about 9 molecules. A fourfold shorter half life gives a fourfold lower level (34.6 ÷ 8.7 = 4.0).
Step 4: Approach to the new level takes about five of the new half lives: 5 × 0.5 = 2.5 minutes.
Answer: about 35 molecules at first, about 9 after destabilization, with the new level essentially reached in 2.5 minutes.
Key point

Gene regulation saves resources and creates cell identity. Cis-acting sites are sequences that affect only linked genes, while trans-acting factors are diffusible products. Systems are classified as negative or positive (repressor versus activator) and inducible or repressible (what the small molecule does). Expression levels reflect both synthesis and decay, and the half life sets how quickly a cell can change course.

Section 2The lac and trp operons

The lac operon: genes and sites

The lactose operon of E. coli, worked out by François Jacob and Jacques Monod around 1960, was the first gene regulatory system understood in molecular detail. Lactose is a disaccharide of galactose and glucose. The operon carries three genes in the order lacZ, lacY, lacA:

Upstream of the structural genes lie the cis-acting sites. The promoter (P) binds RNA polymerase. The main operator (O1) overlaps the transcription start, so a bound repressor blocks polymerase; auxiliary operators O2 (in lacZ) and O3 (upstream) let one repressor tetramer bind two operators and loop the DNA, tightening repression. Just upstream of the promoter is the site for the catabolite activator protein (CAP). The nearby lacI gene, with its own constitutive promoter, encodes the tetrameric lac repressor.

Negative control by the repressor

Without lactose, the repressor binds the operator and blocks transcription. Repression is not absolute: a few molecules of permease and β-galactosidase escape, which the cell needs to import the first lactose and make the first allolactose. Allolactose binds the repressor and lowers its affinity for the operator, so it falls off and expression rises roughly a thousandfold. When lactose runs out, the repressor rebinds and the short lived mRNA is degraded within minutes.

Researchers often induce the operon with IPTG, a gratuitous inducer that binds the repressor like allolactose but is not broken down, so its concentration stays constant. X-gal, a colorless galactoside that β-galactosidase cleaves to a blue product, shows whether lacZ is active.

Positive control by CAP and cAMP

With both glucose and lactose present, E. coli uses glucose first and expresses lac only weakly, called catabolite repression. Two mechanisms produce it.

The result is a logic gate. High transcription requires two conditions at once: lactose present (repressor off) and glucose absent (CAP-cAMP on, permease active).

GlucoseLactoseRepressorcAMP and CAPlac transcription
PresentAbsentBound to operatorLow cAMP, CAP not boundOff (basal trace)
AbsentAbsentBound to operatorHigh cAMP, CAP boundOff (basal trace)
PresentPresentReleasedLow cAMP, CAP not bound; inducer exclusionLow
AbsentPresentReleasedHigh cAMP, CAP boundHigh (fully induced)
A. No lactose: repressed lacI CAP site P O lacZ lacY lacA Rep Pol Repressor tetramer on operator blocks RNA polymerase lacI made constitutively B. Lactose present, glucose absent: induced lacI P O lacZ lacY lacA CAP Pol CAP + cAMP One polycistronic mRNA (lacZ, lacY, lacA) Rep allolactose releases repressor
Figure 9.2 The lac operon. (A) Without lactose the repressor occupies the operator. (B) Allolactose pulls the repressor off; when glucose is low, CAP with cAMP binds upstream and recruits RNA polymerase, and the three genes are transcribed as one mRNA.

Regulatory mutations and the partial diploid test

Jacob and Monod deduced the logic of the operon from mutants, which remain the basis of many problems.

MutationDefectPhenotype in haploidBehavior in partial diploid
lacZ− or lacY−Nonfunctional enzymeThat enzyme absentRecessive; a wild type copy elsewhere supplies the enzyme
lacI−Repressor cannot bind operatorConstitutive (on without inducer)Recessive; wild type repressor diffuses and acts in trans
lacIS (superrepressor)Repressor cannot bind inducerUninducible (always off)Dominant in trans; it binds every operator and never leaves
lacI−d (dominant negative)Defective subunits poison mixed tetramersConstitutiveDominant over lacI+
lacOc (operator constitutive)Operator cannot bind repressorConstitutiveCis-dominant: only genes on the same DNA are constitutive
lacP−Promoter cannot bind polymeraseNo expressionCis-acting: only linked genes are silent

To solve a partial diploid problem, treat the cell as a pool of diffusible products and a set of fixed DNA molecules. Repressor made from either copy of lacI goes everywhere. Each operator and promoter controls only the structural genes on its own DNA. Work one structural gene at a time: find every functional copy of that gene, then ask whether the operator and promoter next to it will allow expression with and without inducer.

Study tip. Worked problem: An E. coli strain has the chromosome genotype I+ Oc Z− Y+ and carries an F′ plasmid with I− O+ Z+ Y−. Glucose is absent. Predict β-galactosidase and permease activity with and without lactose.
Step 1: Repressor. The chromosome makes functional repressor (I+); the plasmid makes none (I−). Repressor is diffusible, so functional repressor is available to every operator in the cell.
Step 2: β-galactosidase. The only working lacZ is on the plasmid, next to an O+ operator. Repressor binds that operator without inducer and leaves with inducer. So β-galactosidase is inducible: absent without lactose, present with lactose.
Step 3: Permease. The only working lacY is on the chromosome, next to Oc, which repressor cannot bind. So permease is made whether or not lactose is present: constitutive.
Answer: β-galactosidase inducible; permease constitutive. Notice that the Oc mutation affects only the gene in cis with it, while the I− defect is fully covered by the I+ copy in trans. Had the chromosome carried IS instead of I+, β-galactosidase would be uninducible (absent in both conditions), but permease would still be constitutive because Oc ignores even the superrepressor.

The trp operon: a repressible system

The tryptophan operon contains five genes, trpE, trpD, trpC, trpB, trpA, encoding the enzymes that make tryptophan. The trp repressor, encoded by the separate trpR gene, binds DNA poorly on its own. Tryptophan acts as a corepressor: it binds the repressor dimer and repositions its helix-turn-helix motifs to fit the operator, blocking transcription. When tryptophan is scarce, the repressor drops off. Repression alone changes expression about 70 fold.

Attenuation: fine tuning by coupled transcription and translation

Charles Yanofsky found that even trpR mutants lowered operon expression about eightfold when tryptophan was plentiful. The explanation is attenuation, premature termination controlled by translation of a 162 nucleotide leader (trpL) between the operator and trpE. Its RNA has four regions that can pair in alternative ways:

Because region 2 cannot pair with both 1 and 3 at once, and region 3 cannot pair with both 2 and 4, the structures are mutually exclusive. Region 1 also encodes a 14 amino acid leader peptide that contains two tryptophan codons in a row. Because bacteria translate mRNA while it is still being made, the position of the ribosome on the leader decides which hairpin forms.

Repression (about 70 fold) and attenuation (about 8 to 10 fold) multiply to give a total range of roughly 600 fold. Attenuation senses charged tRNA, while repression senses free tryptophan. Leaders rich in the relevant codons also control the histidine, leucine and phenylalanine operons. Attenuation cannot work in eukaryotes, where transcription and translation are separated by the nuclear envelope.

High tryptophan Low tryptophan 1 2 34 3:4 terminator UUUU Ribosome Ribosome reaches leader stop, covers region 2: transcription ends 1 23 2:3 antiterminator 4 trpE... Ribosome stalled atTrp codons Region 4 stays single stranded: polymerase continues into trpE
Figure 9.3 Attenuation of the trp operon. The position of the ribosome on the leader decides whether the 3:4 terminator or the 2:3 antiterminator forms. Charged tRNATrp lets the ribosome pass the Trp codons and cover region 2, which ends transcription.
Study tip. Worked problem: Predict trp operon expression (high or low) in each case. (a) Both leader Trp codons are changed to Ala codons, cells grown with low tryptophan but plenty of alanine. (b) Region 3 is mutated so it cannot pair with region 4, high tryptophan, trpR−. (c) The leader AUG is destroyed, low tryptophan, trpR−.
Step 1 (a): The ribosome no longer needs tRNATrp in the leader. Alanyl tRNA is plentiful, so the ribosome does not stall, covers region 2, and 3:4 forms. Attenuation now responds to alanine, not tryptophan: expression stays low even in tryptophan starvation (repression is relieved, so expression is attenuated but not repressed).
Step 2 (b): Without a 3:4 terminator, polymerase cannot be stopped at the attenuator, and with no repressor nothing blocks initiation. Expression is high and constitutive.
Step 3 (c): With no translation, 1:2 forms, freeing 3 to pair with 4. Termination occurs even though tryptophan is low. Expression is low.
Answer: (a) low, (b) high, (c) low. The trick in every case is to ask where the ribosome sits and which pairings are left available.
Connection. Lac operon parts power modern biotechnology. Insulin, growth hormone and many vaccine antigens are produced in bacteria from genes placed behind lac derived promoters and switched on with IPTG once the culture is dense. Blue and white screening with X-gal, which detects whether a cloned insert has disrupted lacZ, is still taught in nearly every molecular biology lab.
Key point

The lac operon is negatively controlled by a repressor that allolactose removes and positively controlled by CAP-cAMP, so it is strongly expressed only with lactose and without glucose. Partial diploids distinguish trans-acting lacI products from cis-acting operators. The trp operon is repressed by TrpR plus tryptophan and further tuned by attenuation, in which ribosome stalling at Trp codons in the leader chooses between antiterminator and terminator hairpins.

Section 3Eukaryotic transcription factors and enhancers

The parts of a eukaryotic gene's control region

A typical protein coding gene in a eukaryote is transcribed by RNA polymerase II, which cannot recognize a promoter by itself. Its control region has three layers.

The general transcription factors assemble at the core promoter to form the preinitiation complex. TFIID, containing the TATA binding protein (TBP), binds first; TFIIA and TFIIB stabilize it, TFIIF brings RNA polymerase II, and TFIIE and TFIIH join last. TFIIH opens the DNA and phosphorylates the polymerase C-terminal domain (CTD), releasing polymerase into elongation. These factors give only low, basal transcription; high levels need activators.

Activators have separable domains

A eukaryotic activator usually has at least two functional parts that can be separated and swapped like modules:

Modularity was shown by domain swaps with the yeast activator GAL4, which binds the UASG near the galactose genes. The GAL4 DNA binding domain fused to another protein's activation domain still activates UASG genes; the DNA binding domain alone binds but activates nothing. This modularity underlies the yeast two hybrid screen and the GAL4/UAS system used in fruit flies.

Without galactose, GAL4 is already bound to the UAS, but the protein Gal80 masks its activation domain. Galactose binds the sensor Gal3, which pulls Gal80 away. The signal acts on the activator, not on DNA binding.

DNA binding motifs

MotifStructureExamples
Helix-turn-helix (and homeodomain)Two α helices joined by a short turn; one, the recognition helix, lies in the major grooveBacterial lac and trp repressors, CAP; homeodomain proteins such as Hox factors
Zinc finger (C2H2)About 30 amino acids folded around a zinc ion held by two cysteines and two histidines; often several fingers in tandem, each reading about 3 bpSp1, TFIIIA, CTCF
Nuclear receptor zinc finger (C4)Two zinc modules each held by four cysteines; bind as dimers to paired half sitesGlucocorticoid, estrogen, thyroid hormone and vitamin D receptors
Basic leucine zipper (bZIP)Leucine every seventh residue forms a coiled coil for dimerization; basic region grips DNAFos and Jun (AP-1), CREB
Basic helix-loop-helix (bHLH)Two helices joined by a loop form the dimer interface; basic region binds DNAMyoD, Myc and Max

Enhancers and how they work at a distance

An enhancer is a cluster of binding sites for activators, usually 100 to 1,000 bp long, that raises transcription of a target gene. Enhancers have three defining properties that distinguish them from promoter elements:

  1. They work at a distance, from a few hundred bases to more than a megabase away.
  2. They work in either orientation; flipping an enhancer usually leaves its activity intact.
  3. They work upstream, downstream or within introns of the gene they control.

Enhancers act by DNA looping. Activators at the enhancer contact the promoter through Mediator, a coactivator complex of about 25 to 30 subunits that bridges activators and RNA polymerase II. Activators also recruit histone acetyltransferases such as CBP and p300 and ATP dependent remodeling complexes such as SWI/SNF that slide or evict nucleosomes. Many genes have several enhancers, each active in a different tissue, so a mutation in one enhancer can alter one tissue alone. Point mutations in the ZRS enhancer, about a megabase from SHH, cause extra digits without other effects of SHH loss.

Silencers are the mirror image: binding sites for repressors. Eukaryotic repressors work in several ways: they compete with activators for overlapping sites, they mask activation domains (as Gal80 does), they block contact with Mediator, or they recruit corepressors that bring in histone deacetylases and methyltransferases.

Enhancer Activators Mediator TATA Pol II GTFs Gene Transcription Insulator (CTCF) Loop can span from a few kilobases to more than a megabase
Figure 9.4 Enhancer action by DNA looping. Activators bound to a distant enhancer contact Mediator, which recruits and stimulates RNA polymerase II and the general transcription factors (GTFs) at the core promoter. Insulators bound by CTCF mark the boundaries of the domain an enhancer can reach.

Insulators and topological domains

What stops an enhancer from activating the wrong gene? Insulators (boundary elements) block enhancer-promoter communication when placed between them and stop the spread of heterochromatin. In vertebrates most are bound by the zinc finger protein CTCF, which with the ring shaped cohesin complex organizes the genome into topologically associating domains (TADs) of roughly 100 kb to 1 Mb. Deleting a TAD boundary can let an enhancer reach a gene it normally never touches, a cause of some human limb malformations.

Combinatorial control and signals

Humans have roughly 1,600 sequence specific transcription factors, far fewer than the number of expression patterns cells display. The answer is combinatorial control: each enhancer is active only when the right combination of factors is present. A small set of master regulators can therefore specify many cell types; four factors (Oct4, Sox2, Klf4 and c-Myc) can even reprogram skin fibroblasts into induced pluripotent stem cells.

Signals reach transcription factors in several ways. Lipid soluble steroid hormones enter the cell: cortisol releases the glucocorticoid receptor from the chaperone Hsp90, and the receptor moves to the nucleus, dimerizes and binds hormone response elements. Other signals act through kinases; protein kinase A phosphorylates CREB, letting it recruit CBP/p300. The signal changes the activity, location or partners of a factor, not the DNA.

Reporter genes and the dissection of enhancers

Enhancers are mapped with reporter genes. A candidate regulatory region is cloned upstream of a gene whose product is easy to measure, such as luciferase (light) or green fluorescent protein (fluorescence), and the construct is introduced into cells. Deleting or mutating pieces of the region and measuring the reporter reveals which pieces matter. Moving a fragment to new positions or inverting it tests whether it behaves as an enhancer.

Study tip. Worked problem: A 2.0 kb region upstream of a liver gene is split into four 500 bp fragments, A (farthest upstream) to D (next to the promoter), and tested with a luciferase reporter in liver cells. Relative activities: full region 100; minimal promoter alone 5; deletion of A 97; of B 6; of C 94; of D 52; B alone on the minimal promoter 88; B inverted 90; B placed 3 kb downstream 84. In kidney cells the full region gives 5. Interpret.
Step 1: Fold activation of the full region: 100 ÷ 5 = 20 fold.
Step 2: Deleting B drops activity to baseline (6), so B is necessary. Deleting A or C changes little. Deleting D halves activity (52 ÷ 100), a modest proximal effect.
Step 3: B alone gives 88 ÷ 5 = 17.6 fold, so B is sufficient for most activity.
Step 4: B works inverted and downstream, the defining properties of an enhancer.
Step 5: No activity in kidney (5) means the enhancer is tissue specific, presumably because its activators are made only in liver.
Answer: B is a liver specific enhancer, necessary and sufficient for most of the 20 fold activation; D holds a weaker proximal element.
Connection. Most disease associated variants found by genome wide association studies lie outside protein coding sequence, and many fall in enhancers. A common variant in an intronic enhancer of BCL11A lowers that repressor's expression in red cell precursors and raises fetal hemoglobin. That finding inspired a gene editing therapy for sickle cell disease and β thalassemia that disrupts the same erythroid enhancer, reactivating fetal hemoglobin without affecting BCL11A in other tissues.
Key point

RNA polymerase II needs general transcription factors to initiate and activators to reach high levels. Activators are modular, with separate DNA binding and activation domains built on motifs like helix-turn-helix, zinc fingers, leucine zippers and helix-loop-helix. Enhancers work at a distance and in either orientation by looping to the promoter through Mediator and by recruiting chromatin modifiers, insulators confine their reach, and combinations of factors give each cell type its own expression pattern.

Section 4Chromatin, DNA methylation and epigenetics

Nucleosomes as gatekeepers

In the nucleus, about 147 bp of DNA wraps around each nucleosome, an octamer with two copies each of histones H2A, H2B, H3 and H4, and histone H1 binds the linker. Nucleosomes block many factors from reaching DNA. Euchromatin is relatively open and contains most active genes; heterochromatin is condensed and largely silent. Constitutive heterochromatin at centromeres stays condensed in all cells, while facultative heterochromatin, such as the inactive X, is condensed only in some.

Cells change chromatin in two ways. ATP dependent chromatin remodelers (SWI/SNF and related families) use ATP to slide nucleosomes along DNA, eject them or exchange histone variants. Histone modifying enzymes add or remove chemical groups on the flexible N-terminal tails of histones, which protrude from the nucleosome.

Histone modifications

The best studied modifications are acetylation and methylation of lysines. Histone acetyltransferases (HATs) add acetyl groups to lysines, which neutralizes their positive charge, loosens the tails' grip on negatively charged DNA and creates docking sites for proteins with bromodomains. Histone deacetylases (HDACs) remove them. Acetylation is therefore a mark of active chromatin. Methylation does not change charge; instead, the effect of a methyl mark depends on which residue carries it and how many methyl groups (one, two or three) are present, because different reader proteins recognize each one. Proteins with chromodomains bind methylated lysines. The idea that combinations of marks are read by specific proteins is called the histone code hypothesis.

MarkWhere it is foundUsual effectWriter and reader examples
H3K27ac, H3K9acActive promoters and enhancersActivationWritten by CBP/p300; read by bromodomain proteins
H3K4me3Promoters of active genesActivationWritten by trithorax (MLL/COMPASS) complexes
H3K4me1Enhancers, active or poisedMarks enhancersMLL3 and MLL4
H3K36me3Bodies of transcribed genesAssociated with elongationSETD2, which travels with Pol II
H3K27me3Silent developmental genesRepression (facultative)Written by Polycomb PRC2 (EZH2); read by PRC1
H3K9me3Pericentromeric heterochromatin, repeatsRepression (constitutive)Written by SUV39H1; read by HP1

Repressive marks can spread: HP1 binds H3K9me3 and recruits more SUV39H1, which methylates the next nucleosome. This explains position effect variegation in fruit flies. When an inversion moves the white gene next to centromeric heterochromatin, silencing spreads into it in some cells but not others, and each cell's state is inherited, giving a mottled eye. Polycomb and trithorax complexes form a similar memory system that keeps Hox genes off or on in the segments where they were first set.

DNA methylation

In mammals the major DNA modification is 5-methylcytosine, made by adding a methyl group to carbon 5 of cytosine, almost always where the C is followed by a G (a CpG dinucleotide; the p denotes the phosphate between them). Because CpG is palindromic, a methylated site has methyl groups on both strands. About 70 to 80 percent of CpGs in the human genome are methylated. The exceptions are CpG islands, stretches of about 0.5 to 2 kb rich in G and C and in CpGs, found at roughly 60 to 70 percent of gene promoters and usually unmethylated. Methylation of a promoter CpG island silences the gene, both by blocking binding of some activators and by recruiting methyl-CpG binding proteins such as MeCP2, which bring in HDACs and H3K9 methyltransferases.

Three enzymes establish and maintain the pattern:

Methylated cytosines are mutation hot spots. Deamination of cytosine gives uracil, which repair recognizes as foreign, but deamination of 5-methylcytosine gives thymine, a normal base, leaving a G-T mismatch that is often repaired the wrong way. This is why CpGs are depleted to about one quarter of their expected frequency and why C-to-T changes at CpGs are common in human disease.

Active (open) Silent (closed) Ac Unmethylated CpGs Pol II Activator Acetylated tails (H3K27ac), H3K4me3 at the promoter MeCP2 HDAC HP1 Methylated CpGs (filled), H3K9me3 and H3K27me3, compact nucleosomes Maintenance methylation after replication 5' ..CmG..CmG.. 3'3' ..GmC..GmC.. 5' parental ..CmG..CmG..new ..G C..G C.. new ..GmC..GmC.. replicate Hemimethylated DNMT1 restores (with UHRF1)
Figure 9.5 Chromatin states and their inheritance. Active promoters have spaced, acetylated nucleosomes and unmethylated CpG islands; silent ones have compact nucleosomes, methylated CpGs read by MeCP2, and repressive histone marks. After replication, DNMT1 recognizes hemimethylated CpGs (m marks a methylated C) and methylates the new strand, so the pattern is copied.

What makes a mark epigenetic?

In its strict sense, epigenetics means changes in gene function that are heritable through cell division without a change in DNA sequence. DNA methylation qualifies because of DNMT1 copying, and some histone marks qualify through reader-writer loops (HP1 with SUV39H1, PRC2 binding its own H3K27me3). Most acetylation turns over in minutes and reflects current activity rather than memory.

Genomic imprinting

For most genes, the maternal and paternal copies are expressed equally. For about 100 to 200 mammalian genes, one copy is silenced depending on which parent it came from. This genomic imprinting is set by methylation at imprinting control regions (ICRs) during gamete formation, maintained through all somatic divisions of the offspring, then erased and reset in the offspring's own germ cells according to that individual's sex.

At the Igf2-H19 locus, Igf2 (a fetal growth factor) is expressed from the paternal allele and H19 (a noncoding RNA) from the maternal allele; both share enhancers downstream of H19. On the maternal chromosome the imprinting control region between them is unmethylated and binds CTCF, an insulator that blocks the enhancers from Igf2, so they activate H19. On the paternal chromosome the region is methylated, CTCF cannot bind, the enhancers reach Igf2, and H19 is silenced.

Study tip. Worked problem: In mice, Igf2 is expressed only from the paternal allele. A null allele, Igf2m, makes mice small when it is the expressed copy. (a) A heterozygous male (Igf2+/Igf2m) is crossed to a wild type female. What fraction of offspring are small? (b) A heterozygous female is crossed to a wild type male. What fraction are small? (c) A small offspring from cross (a), if female, is mated to a wild type male. What fraction of her offspring are small?
Step 1 (a): Each offspring receives one paternal allele, Igf2+ or Igf2m with probability 1/2 each. The paternal copy is the expressed one, so offspring receiving Igf2m from the father are small: 1/2.
Step 2 (b): The paternal allele is always Igf2+ from the wild type father, so every offspring expresses a working copy. The maternal Igf2m is silenced anyway. Fraction small: 0.
Step 3 (c): The small female carries Igf2m from her father. In her germ line the imprint is erased and reset to the maternal pattern, so any Igf2m she transmits is silent in her offspring. The father is wild type. Fraction small: 0.
Answer: 1/2, 0, 0. Imprinted traits follow the sex of the transmitting parent, not the sex of the offspring, and the imprint is reset every generation.

Imprinting explains two disorders of chromosome 15q11 to q13. Prader-Willi syndrome (neonatal hypotonia, later overeating and obesity, intellectual disability) results from loss of paternally expressed genes, by a paternal deletion or maternal uniparental disomy (both chromosome 15s from the mother). Angelman syndrome (severe intellectual disability, absent speech, seizures, happy demeanor) results from loss of the maternally expressed UBE3A gene in neurons, by a maternal deletion, paternal uniparental disomy or a UBE3A mutation. The same deletion gives different diseases depending on the parent of origin.

X inactivation

Mammals achieve dosage compensation by silencing all but one X in each somatic cell, as Mary Lyon proposed in 1961. Early in the embryo each cell chooses one X at random. That X expresses the long noncoding RNA Xist, which coats it in cis and recruits Polycomb complexes; the chromosome gains H3K27me3 and promoter methylation and condenses into a Barr body. The choice is inherited by the cell's descendants, making every female a mosaic. The number of Barr bodies is the number of X chromosomes minus one.

The orange and black patches of a calico cat show this mosaic: the two alleles of an X linked coat gene are expressed in different patches. Carriers of X linked disorders may show patchy signs, or symptoms if inactivation is skewed toward the normal X. About 15 percent of human X linked genes escape inactivation, which helps explain why Turner (45,X) and Klinefelter (47,XXY) syndromes have phenotypes.

Measuring methylation: bisulfite sequencing

Ordinary sequencing cannot distinguish C from 5-methylcytosine. In bisulfite sequencing, sodium bisulfite deaminates unmethylated cytosine to uracil, read as T after PCR, while 5-methylcytosine stays C. Non-CpG cytosines are rarely methylated in somatic cells and should all convert, a built in check on the treatment.

Study tip. Worked problem: The reference sequence of a promoter fragment is 5′-ACGTTCGACCGA-3′. After bisulfite treatment and PCR, one clone reads 5′-ACGTTTGATCGA-3′. (a) Which CpGs were methylated in this molecule? (b) Did the conversion work? (c) Ten clones are sequenced; CpG 1 is methylated in 9, CpG 2 in 2 and CpG 3 in 7. What is the average methylation of this region?
Step 1: Number the bases. Reference: A1 C2 G3 T4 T5 C6 G7 A8 C9 C10 G11 A12. The CpGs are C2-G3 (CpG 1), C6-G7 (CpG 2) and C10-G11 (CpG 3). C9 is followed by C, so it is not a CpG.
Step 2: Compare the clone: position 2 is still C, so CpG 1 was methylated. Position 6 reads T, so CpG 2 was unmethylated. Position 10 is still C, so CpG 3 was methylated.
Step 3: Position 9, a non-CpG C, reads T, so the bisulfite conversion worked.
Step 4: Across clones, methylated CpG observations = 9 + 2 + 7 = 18 of 3 × 10 = 30 possible. 18 ÷ 30 = 0.60.
Answer: (a) CpGs 1 and 3; (b) yes, the non-CpG C converted; (c) 60 percent average methylation, though CpG 2 (20 percent) behaves very differently from CpGs 1 (90 percent) and 3 (70 percent).

Epigenetics in development, environment and disease

Most methylation is erased and rebuilt twice in mammalian life, in the early embryo (sparing imprints) and in primordial germ cells (including imprints), which is why most environmentally induced marks are not passed on. A well studied exception is the agouti viable yellow (Avy) mouse: a retrotransposon upstream of the agouti gene drives yellow coat and obesity when unmethylated, and feeding pregnant mothers methyl donors (folate, choline, vitamin B12) shifts offspring toward the methylated, brown, lean state.

Cancer cells show genome wide hypomethylation together with hypermethylation of CpG island promoters of tumor suppressor and repair genes; methylation silencing of MLH1 causes many sporadic colorectal cancers. Because these changes are reversible, DNMT inhibitors (azacitidine, decitabine) and HDAC inhibitors (vorinostat) are used as drugs. Mutations in the machinery itself cause disease: Rett syndrome results from mutations in MECP2, which reads methylated DNA.

Connection. A child with developmental delay and a 15q11 to q13 deletion on a chromosome analysis could have either Prader-Willi or Angelman syndrome. A methylation test of the region tells them apart and also detects uniparental disomy and imprinting defects that a deletion test would miss: a pattern showing only maternal methylation points to Prader-Willi, and only paternal methylation to Angelman.
Key point

Chromatin controls access to DNA. Acetylation and H3K4me3 mark active genes; H3K27me3 (Polycomb) and H3K9me3 (HP1) mark silent ones. CpG methylation silences promoters and is copied after replication by DNMT1 at hemimethylated sites, giving true epigenetic memory. Imprinting uses parent specific methylation at control regions, X inactivation uses the Xist RNA, and bisulfite sequencing reads methylation by converting only unmethylated C to T.

Section 5Regulation by RNA: RNAi and microRNA

The discovery of RNA interference

In 1998 Andrew Fire and Craig Mello found that injecting C. elegans with double stranded RNA (dsRNA) matching a gene silenced it far more strongly than sense or antisense RNA, at only a few molecules per cell. They called this RNA interference (RNAi). It is an ancient defense against viruses and transposons, and its machinery is shared with endogenous regulators, the microRNAs. The first microRNA, lin-4, had been found in worms by Victor Ambros and Gary Ruvkun in 1993.

Shared machinery: Dicer, Argonaute and RISC

Small RNAs of 20 to 30 nucleotides guide proteins to complementary targets. Two enzymes are central:

siRNAs and microRNAs compared

Small interfering RNAs (siRNAs) come from long dsRNA, such as a viral replication intermediate, a transposon transcript read in both directions, or an experimenter's synthetic duplex. They usually pair perfectly with their target, and AGO slicing cuts the target in the middle of the paired region, after which the fragments are degraded.

MicroRNAs (miRNAs) come from the cell's own genes. Their biogenesis has several steps:

  1. RNA polymerase II transcribes a long primary miRNA (pri-miRNA), capped and polyadenylated, containing one or more hairpins. Many miRNAs sit in introns of protein coding genes.
  2. In the nucleus, the Microprocessor, made of the RNase III enzyme Drosha and its partner DGCR8, cuts out a hairpin of about 60 to 70 nucleotides, the precursor miRNA (pre-miRNA).
  3. Exportin-5, powered by Ran-GTP, carries the pre-miRNA to the cytoplasm.
  4. Dicer removes the loop, leaving a duplex of about 22 bp.
  5. One strand is loaded into Argonaute to form the active miRISC.

In animals, a miRNA usually pairs only partly with its target, in the 3′ UTR. The key region is the seed, nucleotides 2 to 8 from the miRNA 5′ end. Without full pairing, AGO does not slice; instead miRISC recruits GW182, which brings in deadenylases, represses translation and leads to decay. The effect on one target is modest, but each miRNA has hundreds of targets, and more than 60 percent of human protein coding genes carry conserved sites. Plant miRNAs usually pair nearly perfectly and direct slicing.

FeaturesiRNAmiRNA (animal)piRNA
SourceLong dsRNA (virus, transposon, experimental)Endogenous hairpin genesSingle stranded precursors from piRNA clusters
Processing enzymesDicerDrosha-DGCR8, then DicerDicer independent
Length21 to 23 ntAbout 22 nt24 to 31 nt
Partner proteinArgonaute (AGO2)Argonaute (AGO1 to 4)PIWI proteins
Pairing with targetPerfectPartial; seed nucleotides 2 to 8Extensive
Main outcomeTarget cleavageTranslational repression and mRNA decayTransposon cleavage and silencing chromatin
Main roleAntiviral defense, research tool, drugsTuning of endogenous genes, developmentProtecting the germ line genome
Nucleus Cytoplasm Pol II transcription pri-miRNA (cap, poly(A)) Drosha + DGCR8 pre-miRNA hairpin Exportin-5 Dicer 22 bp duplex AGO (RISC) guide strand kept Seed match in 3' UTR repression, deadenylation (animal miRNA) Perfect pairing slicing by AGO2 (siRNA, plant miRNA)
Figure 9.6 The microRNA pathway. Drosha-DGCR8 crops the pri-miRNA in the nucleus, Exportin-5 exports the pre-miRNA, and Dicer produces a duplex whose guide strand loads into Argonaute. The degree of pairing with the target decides between repression with decay and direct cleavage.

Finding miRNA targets

Target prediction begins by finding 3′ UTR sequences complementary to seed nucleotides 2 to 8, written antiparallel as the reverse complement of the seed. Such a site is a 7mer; one that also has an A across from miRNA nucleotide 1 is an 8mer, the most effective. Predictions must be tested with reporters.

Study tip. Worked problem: The mature let-7a miRNA is 5′-UGAGGUAGUAGGUUGUAUAGUU-3′. (a) What is its seed? (b) Write the 7mer and 8mer target sites as they appear in an mRNA, 5′ to 3′. (c) Which 3′ UTR contains a site: UTR-1 5′-...GAGGUAG...-3′, or UTR-2 5′-...AACUACCUCAU...-3′?
Step 1: Number from the 5′ end: U1 G2 A3 G4 G5 U6 A7 G8. The seed (nucleotides 2 to 8) is 5′-GAGGUAG-3′.
Step 2: The complement of each seed base, in order 2 to 8, is C U C C A U C. Because the strands are antiparallel, the mRNA reads this in reverse: 5′-CUACCUC-3′. Check by pairing: mRNA C (5′ end) pairs with G8, U with A7, A with U6, C with G5, C with G4, U with A3, and C (3′ end) with G2. All pairs are Watson-Crick.
Step 3: The 8mer adds an A on the 3′ side, across from nucleotide 1: 5′-CUACCUCA-3′.
Step 4: UTR-1 contains the seed sequence itself, which is the same sense as the miRNA, not its complement; it cannot pair. UTR-2 contains CUACCUCA, a full 8mer site.
Answer: (a) GAGGUAG; (b) 7mer CUACCUC, 8mer CUACCUCA; (c) UTR-2. The most common error is copying the seed directly instead of writing its reverse complement.

Biological roles and other regulatory RNAs

MicroRNAs often control developmental timing. lin-4 and let-7 switch worm larvae between stages, and human let-7 represses oncogenes such as RAS. Some miRNAs act as oncogenes (the miR-17-92 cluster) and others as tumor suppressors (miR-15a and miR-16-1, deleted in many chronic lymphocytic leukemias).

PIWI-interacting RNAs (piRNAs) silence transposons in the germ line by cleaving their transcripts and guiding DNA and H3K9 methylation to their copies. Long noncoding RNAs (over 200 nucleotides), such as Xist, act as scaffolds that bring chromatin modifiers to particular loci.

RNAi as a tool and a therapy

RNAi made reverse genetics easy: feeding worms bacteria that express dsRNA silences the matching gene. In mammalian cells long dsRNA triggers an interferon response, so researchers use synthetic 21 nucleotide siRNAs or short hairpin RNAs (shRNAs). RNAi gives a knockdown rather than a knockout, and partial seed matches cause off target effects.

siRNA drugs, chemically stabilized and delivered to the liver in lipid nanoparticles or attached to GalNAc (which binds a hepatocyte receptor), are now approved: patisiran silences transthyretin in hereditary transthyretin amyloidosis, and inclisiran silences PCSK9 to lower LDL cholesterol with dosing about twice a year.

Connection. RNAi is also used in agriculture. Crops can be engineered to produce dsRNA that matches an essential gene of an insect pest; when the insect eats the plant, the dsRNA silences its gene. Corn producing dsRNA against a western corn rootworm gene is one example, and sprayable dsRNA products are being developed. Because RNAi depends on sequence matching, these approaches can target a pest species while sparing unrelated insects.
Key point

Small RNAs guide Argonaute proteins to complementary targets. siRNAs from long dsRNA pair perfectly and direct cleavage; miRNAs, processed by Drosha-DGCR8 and Dicer from endogenous hairpins, pair through their seed (nucleotides 2 to 8) in 3′ UTRs and cause translational repression and decay; piRNAs with PIWI proteins guard the germ line against transposons. Target sites are the reverse complement of the seed, and RNAi now serves as both a research tool and a class of drugs.

The chapter in one breath

  • Cis-acting sites affect only linked genes; trans-acting factors diffuse and act anywhere.
  • Negative control uses repressors and positive control uses activators; inducible systems are turned on by a signal and repressible systems turned off.
  • The lac operon is induced by allolactose removing the repressor and activated by CAP-cAMP when glucose is low.
  • In partial diploids, lacI− is recessive, lacIS is dominant and Oc is cis-dominant.
  • The trp operon is repressed by TrpR with tryptophan and fine tuned by attenuation, in which ribosome stalling chooses between 2:3 and 3:4 hairpins.
  • Eukaryotic activators have modular DNA binding and activation domains and work through Mediator and chromatin modifiers.
  • Enhancers act at a distance, in either orientation, by DNA looping; insulators bound by CTCF limit their reach.
  • Histone acetylation and H3K4me3 mark active chromatin; H3K27me3 and H3K9me3 mark silent chromatin.
  • CpG methylation is set by DNMT3A/B, copied by DNMT1, and removed by TET enzymes; bisulfite converts only unmethylated C.
  • Imprinting and X inactivation use methylation and noncoding RNAs to silence one copy of a gene or chromosome.
  • siRNAs direct cleavage, miRNAs repress through seed pairing, and piRNAs silence transposons.

Words worth knowing

TermWhat it means
Cis-acting elementA DNA or RNA sequence that regulates only the gene on the same molecule.
Trans-acting factorA diffusible product, usually a protein or RNA, that can regulate genes anywhere in the cell.
OperonA group of bacterial genes transcribed from one promoter into a single mRNA under shared control.
InducerA small molecule that turns on transcription, often by releasing a repressor from DNA.
CAPCatabolite activator protein; with cAMP it binds near the lac promoter and helps RNA polymerase bind.
Partial diploidA bacterium carrying a second copy of some genes, usually on an F′ plasmid, used to test dominance and cis or trans action.
AttenuationControl by premature termination of transcription, governed by translation of a leader sequence.
EnhancerA cluster of activator binding sites that raises transcription from a distance in either orientation.
MediatorA large coactivator complex that bridges activators and RNA polymerase II.
InsulatorA boundary element, often bound by CTCF, that blocks enhancer action or heterochromatin spread across it.
PolycombProtein complexes that maintain silencing of developmental genes, depositing H3K27me3.
CpG islandA CG rich stretch, often at a promoter, that is usually unmethylated.
EpigeneticDescribing a change in gene function that is heritable through cell division without a change in DNA sequence.
Genomic imprintingExpression of only one allele of a gene depending on which parent transmitted it.
XistA long noncoding RNA that coats and silences the inactive X chromosome.
Bisulfite sequencingA method that converts unmethylated cytosine to uracil so methylated sites can be read.
DicerAn RNase III enzyme that cuts double stranded RNA into 21 to 23 bp small RNA duplexes.
ArgonauteThe protein that holds a small guide RNA and carries out silencing within RISC.
MicroRNAAn endogenous 22 nucleotide RNA that represses target mRNAs, mainly through seed pairing in the 3′ UTR.
Seed regionNucleotides 2 to 8 of a miRNA, the main determinant of target recognition.

Check yourself

Predict β-galactosidase expression, with and without lactose and with no glucose, in the partial diploid IS O+ Z+ / F′ I+ Oc Z−.
Uninducible (off in both conditions). The only working lacZ sits next to O+, and the superrepressor binds that operator and cannot be removed by inducer. The Oc operator is constitutive, but its linked lacZ is defective.
Why is lac transcription low when both glucose and lactose are present?
Catabolite repression. Glucose keeps cAMP low, so CAP does not activate the weak lac promoter, and unphosphorylated EIIAGlc inhibits lactose permease (inducer exclusion), limiting allolactose formation.
A mutation deletes region 2 of the trp leader. With high tryptophan in a trpR− strain, is the operon expressed?
Little expression, and it no longer responds to tryptophan levels. Without region 2 no antiterminator can form, so 3:4 terminator forms at any tryptophan level; transcription is attenuated even during starvation.
An mRNA has a half life of 30 minutes and a steady state of 60 molecules. If its transcription rate doubles, what is the new steady state, and roughly how long does it take to get there?
120 molecules, in about 2.5 hours. Steady state is proportional to synthesis rate, and the approach takes about five half lives (5 × 30 minutes).
How would you show that a DNA fragment is an enhancer rather than part of the promoter?
Test position and orientation independence. Place it on a reporter at different distances, upstream and downstream, and in reverse orientation; an enhancer still stimulates transcription in each case.
What would happen to the yeast galactose genes in a mutant lacking Gal80?
Expressed without galactose. GAL4 is already bound to the UAS; Gal80 normally masks its activation domain, so without Gal80 the genes are constitutive (glucose repression aside).
Why are DNA methylation patterns copied through cell division while most histone acetylation is not?
DNMT1 copies hemimethylated sites. Symmetric CpG methylation leaves a template on the parental strand after replication, which UHRF1 and DNMT1 use to methylate the new strand. Acetylation turns over rapidly and has no such templating mechanism.
A child has a 15q11 to q13 deletion on the chromosome inherited from the mother. Which syndrome is expected, and which gene is responsible?
Angelman syndrome, from loss of UBE3A. UBE3A is expressed only from the maternal allele in neurons; the paternal copy is silenced, so a maternal deletion leaves no working copy.
How many Barr bodies are expected in cells of a 47,XXY individual and of a 48,XXXY individual?
One and two. All X chromosomes but one are inactivated, so the number of Barr bodies equals the number of X chromosomes minus one.
A miRNA's seed (nucleotides 2 to 8) is 5′-AGCACUU-3′. Write its 7mer target site in the mRNA, 5′ to 3′.
5′-AAGUGCU-3′. Take the complement of each seed base (UCGUGAA) and reverse it to read 5′ to 3′: AAGUGCU. Check: the 5′ A of the site pairs with the 3′ U of the seed.