regulation of gene expression in eukaryotes

385

Key Questions

• What are the molecularmechanisms of gene regulationin eukaryotes?

• How do eukaryotes generatemany different patterns of geneexpression with a limited numberof regulatory proteins?

• What role does chromatin play ineukaryotic gene regulation?

• What are epigenetic marks andhow do they influence geneexpression?

The MSL complex enhances gene expression on the X chromosome. The MSL complex(indicated by orange coloring) binds only to the X chromosome in male Drosophila. This imageis an indirect immunofluorescence staining of a chromosomal spread from a salivary gland ofa male larva exposed to MSL1 antiserum. [From J. Lucchesi, W. Kelly, and B. Panning, “ChromatinRemodeling in Dosage Compensation,” Annu. Rev. Genet. 39, 2005, 615–651.]

Regulation of Gene Expression in Eukaryotes

11

The cloning of Dolly, a sheep, was reported worldwide in 1996. Dolly developedfrom adult somatic nuclei that had been implanted into enucleated eggs (eggs

with the nuclei removed). More recently, cows, pigs, mice, and other mammalshave been cloned as well with the use of similar technology (Figure 11-1). The suc-cessful cloning of Dolly was a great surprise to the scientific community becausethe cloning of mammals from somatic cells was thought to be impossible. A reasonfor the initial skepticism was that the formation of male and female gametes(sperm and egg cells) was known to include sex-specific modifications to therespective genomes that resulted in sex-specific patterns of gene expression. Assuch, Dolly is symbolic of how far we have progressed in understanding aspects ofeukaryotic gene regulation such as the global control of gene expression exempli-fied by gamete development. However, for every successful clone, including Dolly,there are many more, perhaps hundreds of embryos that fail to develop into viableprogeny. The extremely high failure rate underscores how much remains to bedeciphered about eukaryotic gene regulation.

Outline

11.1 Transcriptional regulation ineukaryotes: an overview

11.2 Lessons from yeast: the GALsystem

11.3 Dynamic chromatin andeukaryotic gene regulation

11.4 Enhancers: cooperativeinteractions, combinatorialcontrol, and chromatinremodeling

11.5 Genomic imprinting

11.6 Chromatin domains and theirinheritance

In this chapter, we will examine gene regulation ineukaryotes. In many ways, our look at gene regulation will bea study of contrasts. In bacteria, you learned how the activi-ties of genetic switches were often governed by single activa-tor or repressor proteins and how the control of sets of geneswas achieved by their organization into operons or by theactivity of specific � factors (see Chapter 10). Initial expecta-tions were that eukaryotic gene expression would be regu-lated by similar means. In eukaryotes, however, most genesare not found in operons. Furthermore, we will see that theproteins and DNA sequences participating in eukaryotic generegulation are more numerous. Often, many DNA-bindingproteins act on a single switch, with many separate switchesper gene, and the regulatory sequences of these switches areoften located far from promoters. A key additional differencebetween bacteria and eukaryotes is that the access to eukary-otic gene promoters is restricted by chromatin. Gene regula-tion in eukaryotes requires the activity of large protein com-

plexes that promote or restrict access to gene promoters by RNA polymerase. Thischapter will provide an essential foundation for understanding the spatiotemporalregulation of gene expression that choreographs the process of developmentdescribed in Chapter 12.

The biological properties of each eukaryotic cell type are largely determined by theproteins expressed within it. This constellation of expressed proteins determinesmuch of the cell’s architecture, its enzymatic activities, its interactions with itsenvironment, and many other physiological properties. However, at any given timein a cell’s life history, only a fraction of the RNAs and proteins encoded in itsgenome are expressed. At different times, the profile of expressed gene productscan differ dramatically, both in regard to which proteins are expressed and at whatlevels. How are these specific profiles generated?

As one might expect, if the final product is a protein, regulation could beachieved by controlling the transcription of DNA into RNA or the translation ofRNA into protein. In fact, gene regulation takes place at many levels, including at the mRNA level (through alterations in splicing or the stability of the mRNA)and after translation (by modifications of proteins). However, most regulation isthought to take place at the level of gene transcription; so, in this chapter, the pri-mary focus is on the regulation of transcription. The basic mechanism at work isthat molecular signals from outside or inside the cell lead to the binding of regula-tory proteins to specific DNA sites outside of protein-encoding regions, and thebinding of these proteins modulates the rate of transcription. These proteins maydirectly or indirectly assist RNA polymerase in binding to its transcription initia-tion site—the promoter—or they may repress transcription by preventing the bind-ing of RNA polymerase.

Although bacteria and eukaryotes have much of the logic of gene regulation incommon, there are some fundamental differences in the underlying mechanismsand machinery. Both use sequence-specific DNA-binding proteins to modulate thelevel of transcription. However, eukaryotic genomes are bigger and their range ofproperties is larger than those of bacteria. Inevitably their regulation is more com-plex, requiring more types of regulatory proteins and more types of interactionswith the adjacent regulatory regions in DNA. The most important difference is

11.1 Transcriptional Regulation in Eukaryotes: An Overview

386 Chapter 11 • Regulation of Gene Expression in Eukaryotes

The first cloned mammal

FIGURE 11-1 The first cloned mammalwas a sheep named Dolly. [PHOTOTAKE/Alamy.]

that eukaryotic DNA is packaged into nucleosomes, forming chromatin, whereasbacterial DNA lacks nucleosomes. In eukaryotes, chromatin structure is dynamicand is an essential ingredient in gene regulation.

In general, the ground state of a bacterial gene is “on.” Thus, RNA polymerasecan usually bind to a promoter when no other regulatory proteins are around tobind to the DNA. In bacteria, transcription initiation is prevented or reduced if thebinding of RNA polymerase is blocked, usually through the binding of a repressorregulatory protein. Activator regulatory proteins increase the binding of RNApolymerase to promoters where a little help is needed. In contrast, the groundstate in eukaryotes is “off.” Therefore, the transcriptional machinery (includingRNA polymerase II and associated general transcription factors) cannot bind tothe promoter in the absence of other regulatory proteins (Figure 11-2). In manycases, the binding of the transcriptional apparatus is not possible, owing to theposition of nucleosomes near the promoter. Thus, chromatin structure usually hasto be changed to activate eukaryotic transcription. The structure of chromatinaround activated or repressed genes within cells can be quite stable and inheritedby daughter cells. The inheritance of chromatin structure is a form of inheritancethat does not directly entail DNA sequence.

The unique features of eukaryotic transcriptional regulation are the focus ofthe rest of this chapter. Some differences from transcriptional regulation in bacte-ria were already noted in Chapter 8:

1. In bacteria, all genes are transcribed into RNA by the same RNApolymerase, whereas three RNA polymerases function in eukaryotes. RNA


Promoter

Ground state: on Ground state: off

Repressed state: off Active state: on

Operator Codingregion

TATA

RNA pol II

RNA pol

EnhancerTranscriptionfactors

Activatorprotein

Repressorprotein

BACTERIAL EUKARYOTIC

TATA

Overview of transcriptional regulation FIGURE 11-2 In bacteria, RNApolymerase can usually begintranscription unless a repressorprotein blocks it. In eukaryotes,however, the packaging of DNAwith nucleosomes preventstranscription unless otherregulatory proteins are present.These regulatory proteinsexpose promoter sequences byaltering nucleosome density orposition. They may also recruitRNA polymerase II more directlythrough binding.

polymerase II, which transcribes mRNAs, was the focus of Chapter 8 and will be the only polymerase discussed in this chapter.

2. RNA transcripts are extensively processed during transcription ineukaryotes; the 5� and 3� ends are modified and introns are spliced out.

3. RNA polymerase II is much larger and more complex than its bacterialcounterpart. One reason for the added complexity is that RNA polymerase IImust synthesize RNA and coordinate the special processing events unique to eukaryotes.

Multicellular eukaryotes may have as many as 25,000 genes, severalfold morethan the average bacterium. Moreover, patterns of eukaryotic gene expression canbe extraordinarily complex. That is, there is great variation among genes in when agene is on (transcribed) or off (not transcribed) and in how much transcript needsto be made. For example, one gene may be transcribed only during early develop-ment and another only in the presence of a viral infection. Finally, the majority ofthe genes in a eukaryotic cell are off at any one time. On the basis of these consid-erations alone, eukaryotic gene regulation must be able to

1. ensure that the expression of most genes in the genome is off at any onetime while activating a subset of genes; and

2. generate thousands of patterns of gene expression.

As you will see later in the chapter, mechanisms have evolved to ensure thatmost of the genes in a eukaryotic cell are not transcribed. Before considering howgenes are kept transcriptionally inactive, we will focus on the second point: Howare eukaryotic genes able to exhibit an enormous number and diversity of expres-sion patterns? The machinery required for generating so many patterns of genetranscription in vivo has many components, including both regulatory proteinsand cis-acting regulatory sequences. The first set of proteins comprises the largeRNA polymerase II complex and the general transcription factors that you learnedabout in Chapter 8. To initiate transcription, these proteins interact with DNAsequences called promoter-proximal elements near the promoter of a gene. Thesecond group of protein components consists of specific transcription factors thatbind to cis-acting regulatory sequences in the DNA called enhancers or upstreamactivating sequences (UAS’s). These regulatory sequences may be located aconsiderable distance from gene promoters. Generally speaking, promoters andpromoter-proximal elements are bound by transcription factors that affect theexpression of many genes. Enhancers are the targets of more specific transcriptionfactors that control the regulation of smaller subsets of genes. Often, an enhancerwill act in only one or a few cell types in a multicellular eukaryote.

For RNA polymerase II to transcribe DNA into RNA ata maximum rate, multiple cis-acting regulatory elementsmust play a part. The promoters, promoter-proximal ele-ments, and enhancers are all targets for binding by differenttrans-acting DNA binding proteins. Figure 11-3 is a schematicrepresentation of the promoter and promoter-proximal se-quence elements. The binding of RNA polymerase II to thepromoter does not produce efficient transcription by itself.Transcription requires the binding of general transcriptionfactors to additional promoter-proximal elements that arecommonly found within 100 bp of the transcription initiation

site of many (but not all) genes. One of these elements is the CCAAT box, and oftenanother is a GC-rich segment farther upstream. The general transcription factorsthat bind to the promoter-proximal elements are expressed in most cells, and sothey are available to initiate transcription at any time. Mutations in these sites can


GGGCGG TATACCAAT� – 200 bp �–100 bp

–30 bp

PromoterPromoter-proximalelements

mRNAGC-rich

box

Promoter-proximal elements precede the promoter of a eukaryotic gene

FIGURE 11-3 The region upstream ofthe transcription start site in highereukaryotes contains promoter-proximalelements and the promoter.

have a dramatic effect on transcription, demonstrating how important they are. Anexample of the consequences on transcription rates of mutating these sequenceelements is shown in Figure 11-4.

To modulate transcription, regulatory proteins possess one or more of the fol-lowing functional domains:

1. A domain that recognizes a DNA regulatory sequence (the protein’s DNA-binding site)

2. A domain that interacts with one or more proteins of the transcriptionalapparatus (RNA polymerase or a protein associated with RNA polymerase)

3. A domain that interacts with proteins bound to nearby regulatorysequences on DNA such that they can act cooperatively to regulatetranscription

4. A domain that influences chromatin condensation either directly orindirectly

5. A domain that acts as a sensor of physiological conditions within the cell

Much of the strategy of eukaryotic transcriptional control hinges on how spe-cific transcription factors control the access of general transcription factors andRNA polymerase II. Eukaryotic gene regulatory mechanisms have been discoveredthrough both biochemical and genetic approaches. The latter has been advancedin particular by studies of the single-celled yeast Saccharomyces cerevisiae (see theModel Organism box). Several decades of research have been a source of manyinsights into general principles of how eukaryotic transcriptional regulatory pro-teins work and how different cell types are generated. We’ll examine two yeastgene regulatory systems in detail: the first concerns the galactose-utilization path-way; the second is the control of mating type.


3.5

3.0

1.0

0

Rel

ativ

e tr

ansc

riptio

n le

vel

GCCACACCC ATATAAGGCCAATC

Promoter-proximal elements are necessary for efficient transcription

FIGURE 11-4 Point mutations in the promoter and promoter-proximal elements hindertranscription of the β-globin gene. Point mutations throughout the promoter region wereanalyzed for their effects on transcription rates. The height of each line represents thetranscription level relative to a wild-type promoter or promoter-proximal element (1.0). Only the base substitutions that lie within the three elements shown change the level oftranscription. Positions with black dots were not tested. [From T. Maniatis, S. Goodbourn, and J. A. Fischer, “Regulation of Inducible and Tissue-Specific Gene Expression,” Science 236, 1987, 1237.]

To make use of extracellular galactose, yeast imports the sugar and converts it intoa form of glucose that can be metabolized. Several genes—GAL1, GAL2, GAL7, andGAL10—in the yeast genome encode enzymes that catalyze steps in the biochemi-cal pathway that converts galactose into glucose (Figure 11-5). Three additionalgenes—GAL3, GAL4, and GAL80—encode proteins that regulate the expression ofthe enzyme genes. Just as in the lac system, the abundance of the sugar determines

11.2 Lessons from Yeast: The GAL System


Saccharomyces cerevisiae, or budding yeast, has emergedin recent years as the premier eukaryotic genetic system.Humans have grown yeast for centuries because it is anessential component of beer, bread, and wine. Yeast hasmany features that make it an ideal model organism. As aunicellular eukaryote, it can be grown on agar plates and,with a life cycle of just 90 minutes, large quantities can becultured in liquid media. It has a very compact genomewith only about 12 megabase pairs of DNA (comparedwith almost 3000 megabase pairs for humans) containingapproximately 6000 genes that are distributed on 16 chro-mosomes. It was the first eukaryote to have its genomesequenced.

The yeast life cycle makes it very versatile for labora-tory studies. Cells can be grown as either diploid or hap-loid. In both cases, the mother cell produces a bud contain-ing an identical daughter cell. Diploid cells either continueto grow by budding or are induced to undergo meiosis,which produces four haploid spores held together in anascus (also called a tetrad). Haploid spores of opposite mat-ing type (a or `) will fuse and form a diploid. Spores of thesame mating type will continue growth by budding.

Yeast has been called the E. coli of eukaryotes becauseof the ease of forward and reverse mutant analysis. Toisolate mutants by using a forward genetic approach, hap-loid cells are mutagenized (with X rays, for example) andscreened on plates for mutant phenotypes. This procedureis usually done by first plating cells on a rich medium onwhich all cells grow and by copying, or replica plating, thecolonies from this master plate onto replica plates contain-ing selective media or special growth conditions. (See alsoChapter 15.) For example, temperature-sensitive mutantswill grow on the master plate at the permissive tempera-ture but not on a replica plate at a restrictive temperature.Comparison of the colonies on the master and replicaplates will reveal the temperature-sensitive mutants. Usingreverse genetics, scientists can also replace any yeast gene(of known or unknown function) with a mutant version(synthesized in a test tube) to understand the nature of thegene product.

Model Organism Yeast

Electron micrograph of budding yeast cells. [SciMAT/PhotoResearchers.]

The life cycle of baker’s yeast. The nuclear alleles MATa and MATadetermine mating type.

a�Fusion

(n)(n)(2n)

aa

�

(2n)

(n)

Ascus

�(n)

�(n) a(n)

MitosisMitosis

Meiosis

Culturecolony

Mitosis

+

(2n)

(n)(n)

Culturecolony

a /a

a /aa /a

the level of gene expression in the biochemical pathway. In yeast cells growing inmedia lacking galactose, the GAL genes are largely silent. But, in the presence ofgalactose (and the absence of glucose), the GAL genes are induced. Just as for thelac operon, genetic and molecular analyses of mutants have been key to under-standing how the expression of the genes in the galactose pathway is controlled.

The key regulator of GAL gene expression is the Gal4 protein, a sequence-specific DNA-binding protein. Gal4 is perhaps the best-studied transcriptional reg-ulatory protein in eukaryotes. The detailed dissection of its regulation and activ-ity has been a source of several key insights into the control of transcription ineukaryotes.

Gal4 regulates multiple genes through upstream activatingsequencesIn the presence of galactose, the GAL1, GAL2, GAL7, and GAL10 genes are induced1000-fold or more. In GAL4 mutants, however, they remain silent. Each of thesefour genes has two or more Gal4-binding sites located 5� (upstream) of its pro-moter. Consider the GAL10 and GAL1 genes, which are adjacent to each other andtranscribed in opposite directions. Between the GAL1 transcription start site andthe GAL10 transcription start site is a single 118-bp region that contains four Gal4-binding sites (Figure 11-6). Each Gal4-binding site is 17 base pairs long and isbound by one Gal4 protein dimer. There are two Gal4-binding sites upstream ofthe GAL2 gene as well, and another two upstream of the GAL7 gene. These bind-ing sites are required for gene activation in vivo. If they are deleted, the genes aresilent, even in the presence of galactose. These regulatory sequences are enhancersthat are also referred to as upstream activating sequences. The presence of en-hancers located at a considerable linear distance from a eukaryotic gene’s pro-moter is typical.

Message The binding of sequence-specific DNA-binding proteins to regions outside the promoters of target genes is a common feature of eukaryotic transcriptionalregulation.

39111.2 Lesson from Yeast: The GAL System

Galactose (extracellular)

Gal2

Galactose (intracellular)

Gal1

Galactose-1-phosphate

Gal7

UDP-galactose

Gal10

UDP-glucose

Gal7

Glucose-1-phosphate

Glycosis

The Gal pathway

FIGURE 11-5 Galactose is convertedinto glucose-1-phosphate in a series ofsteps. These steps are catalyzed byenzymes (Gal1, and so forth) encoded bythe structural genes GAL1, GAL2, GAL7,and GAL10.

Chr II Chr XIIGAL7

UAS UASUAS

GAL10 GAL1 GAL2

Gal4

5′ 5′3′ 3′

Transcriptional activator proteins bind to UAS elements in yeast

FIGURE 11-6 The Gal4 protein activates target genes through upstream-activating-sequence(UAS) elements. The Gal4 protein has two functional domains: a DNA-binding domain (redsquare) and an activation domain (orange oval). The protein binds to specific sequencesupstream of the promoters of Gal-pathway genes. Some of the GAL genes are adjacent (GAL1,GAL10), whereas others are on different chromosomes. The GAL1 UAS element contains fourGal4-binding sites.

The Gal4 protein has separable DNA-binding and activation domainsAfter Gal4 is bound to the UAS element, how is gene expression induced? A dis-tinct domain of the Gal4 protein, the activation domain, is required for regula-tory activity. Thus, the Gal4 protein has at least two domains: one for DNA bindingand another for activating transcription. A similar modular organization has beenfound to be a common feature of other DNA-binding transcription factors as well.

The modular organization of the Gal4 protein was demon-strated in a series of simple, elegant experiments. The strategy wasto test the DNA binding and gene activation of mutant forms of theprotein in which parts had been either deleted or fused to otherproteins. In this manner, whether a part of the protein was neces-sary for a particular function could be determined. To carry outthese studies, experimenters needed a simple means of assayingthe expression of the enzymes encoded by the GAL genes. Theexpression of GAL genes and other targets of transcription factorsis typically monitored by using a reporter gene whose expressionis easily tracked. Often, the reporter gene is the lacZ gene of E. coli,which can act on substrates whose products are easily measured bytheir bright color or fluourescence. Another common reportergene is the gene that encodes the green fluorescent protein of jelly-fish, which, as its name suggests, is easily tracked by the light that itemits. The coding region of one of these reporter genes and a pro-moter are placed downstream of a UAS element from a GAL gene.Reporter expression is then a read-out of Gal4 activity in cells.

When a form of the Gal4 protein lacking the activation domainis expressed in yeast, the binding sites of the UAS element are occu-pied, but no transcription is stimulated (Figure 11-7). The same istrue when other regulatory proteins lacking activation domains,such as the bacterial repressor LexA, are expressed in cells bearingreporter genes with their respective binding sites. The more inter-esting result is obtained when a form of the Gal4 protein lackingthe DNA-binding domain is grafted to the LexA DNA-bindingdomain; the hybrid protein now activates transcription from LexA-binding sites (Figure 11-7). Further “domain swap” experimentshave revealed that the transcriptional activation function of theGal4 protein resides in two small domains about 50 to 100 aminoacids in length. These domains are separable from those used inthe dimerization of the protein, DNA binding, and interactionwith the Gal80 protein (see next). The activation domain helpsrecruit the transcriptional machinery to the promoter, as we willsee in Section 11.3. This highly modular arrangement of activity-regulating domains is found in many transcription factors.

Gal4 activity is physiologically regulatedHow does Gal4 become active in the presence of galactose? Key clues came fromanalyses of mutations in the GAL80 and GAL3 genes. In GAL80 mutants, the GALstructural genes are active even in the absence of galactose. This result suggeststhat the normal function of the Gal80 protein is to somehow inhibit GAL gene

Message Many eukaryotic transcriptional regulatory proteins are modular proteins,with separable domains for DNA binding, activation or repression, and interaction withother proteins.


ONGal4 site

Gal4

lacZ

(a) The complete Gal4 dimer

DNA-bindingdomain

Activationdomain

OFFGal4 site

lacZ

(b) Gal4 lacking the activation domain

OFFLexA site

lacZ

(c) LexA lacking the activation domain

DNA-bindingdomain

ONLexA site

lacZ

(d) Gal4–LexA hybrid

LexADNA-bindingdomain

Gal4 activationdomain

Transcriptional activator proteins are modular

FIGURE 11-7 Transcriptional activatorproteins have multiple, separable domains. (a) The Gal4 protein has two domains andforms a dimer. (b) The experimental removalof the activation domain shows that DNAbinding is not sufficient for gene activation. (c) Similarly, the bacterial LexA protein cannotactivate transcription on its own, but, whenfused to the Gal4 activation domain (d), it canactivate transcription through LexA-bindingsites. [After J. Watson et al., Molecular Biology ofthe Gene, Fifth Edition, copyright © 2004,Benjamin Cummings.]

expression. Conversely, in GAL3 mutants, the GAL structural genes arenot active in the presence of galactose, suggesting that Gal3 normallypromotes expression of the GAL genes.

Extensive biochemical analyses have revealed that the Gal80 pro-tein binds to the Gal4 protein with high affinity and directly inhibitsGal4 activity. Specifically, Gal80 binds to a region within one of theGal4 activation domains, blocking its ability to promote the transcrip-tion of target genes. The role of the Gal3 protein is to release Gal4 fromits inhibition by Gal80 in the presence of galactose. Gal3 is a sensor andinducer. When Gal3 binds galactose and ATP, it undergoes an allostericchange that promotes binding to Gal80, which in turn causes Gal80 torelease Gal4, which is then able to activate transcription of its targetgenes. Thus, Gal3, Gal80, and Gal4 are all part of a switch whose state isdetermined by the presence or absence of galactose (Figure 11-8). Inthis switch, DNA binding by the transcriptional regulator is not thephysiologically regulated step (as is the case in the lac operon and bac-teriophage �); rather, the activity of the activation domain is regulated.

Gal4 functions in most eukaryotesIn addition to its action in yeast cells, Gal4 has been shown to be able to activatetranscription in insect cells, human cells, and many other eukaryotic species. Thisversatility suggests that biochemical machinery and mechanisms of gene activationare common to a broad array of eukaryotes and that features revealed in yeast aregenerally present in other eukaryotes, and vice versa. Furthermore, because of theirversatility, Gal4 and its UAS elements have become favored tools in genetic analysisfor manipulating gene expression and function in a wide variety of model systems.

Now we look at how activators and other regulatory proteins interact with thetranscriptional machinery to control gene expression.

Activators recruit the transcriptional machineryIn bacteria, activators commonly stimulate transcription by interacting directlywith DNA and with RNA polymerase. In eukaryotes, activators generally work in-directly to recruit RNA polymerase II to gene promoters through two major mech-anisms. First, activators can interact with subunits of the protein complexes havingroles in transcription initiation. Second, activators can recruit proteins that modifychromatin structure, allowing RNA polymerase II and other proteins access to theDNA. Many activators, including Gal4, have both activities. We’ll examine therecruitment of parts of the transcriptional initiation complex first.

Recall from Chapter 8 that the eukaryotic transcriptional machinery containsmany proteins that are parts of various subcomplexes within the transcriptionalapparatus that is assembled on gene promoters. One subcomplex, transcription fac-tor IID (TFIID), binds to the TATA box of eukaryotic promoters through the TATA-binding protein (TBP; see Figure 8-12). Gal4 binds to TBP at a site in its activation

Message The ability of Gal4, as well as other eukaryotic regulators, to function in avariety of eukaryotes indicates that eukaryotes generally have the transcriptionalregulatory machinery and mechanisms in common.

Message The activity of eukaryotic transcriptional regulatory proteinsis often controlled by interactions with other proteins.

39311.2 Lesson from Yeast: The GAL System

Gal80

OFFUAS

InactiveGal4

+ Galactose+ Gal3

GAL1

ONUAS

ActiveGal4

GAL1

Transcriptional activator proteins may be activated by an inducer

FIGURE 11-8 Gal4 activity is regulatedby the Gal80 protein. (Top) In the absenceof galactose, the Gal4 protein is inactive,even though it can bind to sites upstreamof the GAL1 target gene. Gal4 activity issuppressed by the binding of the Gal80protein. (Bottom) In the presence ofgalactose and the Gal3 protein, Gal80undergoes a conformational change andreleases the Gal4 activation domain,permitting target gene transcription.

domain, and, through this binding, it recruits the TFIID com-plex and, in turn, RNA polymerase II to the promoter (Figure11-9). The affinity of this interaction correlates well withGal4’s potency as an activator. Gal4 also interacts with thelarge Mediator complex, which directly interacts with RNApolymerase II to recruit it to gene promoters. The Mediatorcomplex is an example of a coactivator, a term applied to aprotein or protein complex that facilitates gene activation bya transcription factor but that itself is neither part of the tran-scriptional machinery nor a DNA-binding protein.

The ability of activators to bind to upstream DNAsequences and to interact with proteins that bind directly orindirectly to promoters helps to explain how transcriptioncan be stimulated from more distant regulatory sequences(see Figure 11-9).

A second mechanism for influencing gene transcription in eukaryotes modifies thelocal chromatin structure around gene regulatory sequences. To fully understandhow this mechanism works, we need to first review chromatin structure and thenconsider how it can change and how these changes affect gene expression.

The recruitment of transcriptional machinery by activators may appear to besomewhat similar in eukaryotes and bacteria, with the major difference being inthe number of interacting proteins in the transcriptional machinery. Indeed, lessthan a decade ago, many biologists pictured eukaryotic regulation simply as a bio-chemically more complicated version of what had been discovered in bacteria.However, this view has changed dramatically as biologists have considered theeffect of the organization of genomic DNA in eukaryotes.

Compared with eukaryotic DNA, bacterial DNA is relatively “naked,” makingit readily accessible to RNA polymerase. In contrast, eukaryotic chromosomes arepackaged into chromatin, which is composed of DNA and proteins (mostly his-tones). As mentioned briefly in Chapter 2, the basic unit of chromatin is the nucle-osome, containing about 150 bp of DNA wrapped twice around a histone octamer(Figure 11-10). The histone octamer is composed of two subunits of each of thefour histones: histone 2A, 2B, 3, and 4. Nucleosomes can associate into higher-order structures that further condense the DNA. The packaging of eukaryoticDNA into chromatin means that much of the DNA is not readily accessible to reg-ulatory proteins and the transcriptional apparatus. Thus, whereas prokaryoticgenes are generally accessible and “on” unless repressed, eukaryotic genes are inac-cessible and “off” unless activated. Therefore, the modification of chromatin struc-ture is a distinctive feature of eukaryotic gene regulation.

One can imagine several ways to alter chromatin structure. For example, onemechanism might be to simply move the histone octamer along the DNA. In the1980s, biochemical techniques were developed that allowed researchers to deter-mine the position of nucleosomes in and around specific genes. In these studies,chromatin was isolated from tissues or cells in which a gene was on and comparedwith chromatin from tissue where the same gene was off. The result for most genesanalyzed was that nucleosome positions changed, especially in a gene’s regulatoryregions. Thus, which DNA regions are wrapped up in nucleosomes can change:

11.3 Dynamic Chromatin and Eukaryotic Gene Regulation

Message Eukaryotic transcriptional activators often workby recruiting parts of the transcriptional machinery to genepromoters.


UAS

TATA GAL genes

Mediator

RNA polymerase II

Gal4

TFIID

TBP

Transcriptional activator proteins recruit the transcriptional machinery

FIGURE 11-9 Gal4 recruits thetranscriptional machinery. The Gal4protein, and many other transcriptionalactivators, binds to multiple proteincomplexes, including the TFIID andMediator complexes, that recruit RNApolymerase II to gene promoters. Theinteractions facilitate gene activationthrough binding sites that are distant fromgene promoters. [After J. Watson et al.,Molecular Biology of the Gene, Fifth Edition,copyright © 2004, Benjamin Cummings.]

nucleosome positions can shift on the DNA from cell to cell and over the life cycleof an organism. Transcription might be repressed when the promoter and flankingsequences are wound up in a nucleosome and inaccessible to RNA polymerase II.Activation of transcription would thus require the blocking nucleosome to be reor-ganized by nudging the histones or removing them entirely. Conversely, when generepression is necessary, histone octamers may shift into a position that preventstranscription. The changing of nucleosome position is referred to as chromatinremodeling. Now, chromatin remodeling is known to be an integral part ofeukaryotic gene expression, and great advances are being made in determining theunderlying mechanism(s) and the regulatory proteins taking part. Here, again,genetic studies in yeast have been pivotal.

Chromatin-remodeling proteins and gene activationTwo genetic screens in yeast for mutants in seemingly unrelated processes led to thediscovery of the same gene whose product plays a key role in chromatin remodel-ing. In both cases, yeast cells were treated with agents that would cause mutations.In one screen, these mutagenized yeast cells were screened for cells that could notgrow well on sucrose (sugar nonfermenting mutants, snf). In another screen, muta-genized yeast cells were screened for mutants that were defective in switching theirmating type (switch mutants, swi; see Section 11.4). Many mutants for different lociwere recovered in each screen, but one mutant gene was found to cause both phe-notypes. Mutants at the so-called swi2/snf2 locus (“switch–sniff”) could neither uti-lize sucrose effectively nor switch mating type.

What was the connection between the ability to utilize sugar and the abilityto switch mating types? The Snf2–Swi2 protein was purified and discovered to bepart of a large, multisubunit complex called the SWI–SNF complex that can repo-sition nucleosomes in a test-tube assay if ATP is provided as an energy source(Figure 11-11). In some situations, the multisubunit SWI–SNF complex activatestranscription by moving nucleosomes that are covering the TATA sequences and,in this way, facilitates the binding of RNA polymerase II. The SWI–SNF complex isthus a coactivator.


(a)

11 nm

30 nm

2 nm

Nucleosomes:the basic unit of chromatin

Chromatin fiber of packednucleosomes

Short region ofDNA double helix

The structure of chromatin

FIGURE 11-10 (a) The nucleosome in decondensed and condensed chromatin. (b) Chromatin structure varies along the length of a chromosome. The least-condensedchromatin (euchromatin) is shown in yellow, regions of intermediate condensation are inorange and blue, and heterochromatin coated with special proteins (purple) is in red. [(b) From P. J. Horn and C. L. Peterson, “Chromatin Higher Order Folding: Wrapping Up Transcription,”Science 297, 2002, 1827, Fig. 3. Copyright 2002, AAAS.]

(b)

Nucleosomeremodeling

Chromatin remodeling exposes regulatory

sequences

FIGURE 11-11 The histone octamerslides in response to chromatin-remodeling activity (such as that of theSWI–SNF complex), in this caseexposing the DNA marked in red. (SeeFigure 11-15 for details on howSWI–SNF is recruited to a particularDNA region). [After J. Watson et al.,Molecular Biology of the Gene, Fifth Edition,copyright © 2004, Benjamin Cummings.]

Gal4 also binds to the SWI–SNF complex and recruits the chromatin-remodelingcomplex to activated promoters. Yeast strains containing a defective SWI–SNFcomplex show a reduced level of Gal4 activity. Why might an activator use multipleactivation mechanisms? There are at least two reasons understood at present. Thefirst is that the accessibility of target promoters may change at different stages ofthe cell cycle or in different cell types (in multicellular eukaryotes). For example,during mitosis, when chromatin is more condensed, genes are less accessible. Atthat stage, Gal4 must recruit the chromatin-remodeling complexes, whereas, atother times, such recruitment might not be required to activate gene expression.

A second reason is that many transcription factors act in combinations to con-trol gene expression synergistically. We will see shortly that this combinatorialsynergy is a result of the fact that chromatin-remodeling complexes and the tran-scriptional machinery are recruited more efficiently when multiple transcriptionfactors act together.

Histones and chromatin remodelingLet’s look at the nucleosome more closely to see if anypart of this structure could carry the information neces-sary to influence nucleosome position or nucleosomedensity or both.

A histone code As already stated, most nucleosomes arecomposed of an octamer made up of two copies each ofthe four core histones. Histones are known to be the mostconserved proteins in nature; that is, histones are almostidentical in all eukaryotic organisms from yeast to plantsto animals. This conservation contributed to the view thathistones could not take part in anything more compli-cated than the packaging of DNA to fit in the nucleus.However, recall that DNA with its four bases also was con-sidered too “dumb” a molecule to carry the blueprint forall organisms on Earth.

Figure 11-12 shows a model of nucleosome structurethat represents contributions from many studies. Of noteis that the histone proteins are organized into the coreoctamer with their amino-terminal ends protruding fromthe nucleosome. These protruding ends are called histonetails. Since the early 1960s, specific lysine residues in thehistone tails have been known to be able to be covalentlymodified by the attachment of acetyl and methyl groups.These reactions take place after the histone protein hasbeen translated and even after the histone has been incor-porated into a nucleosome.

There are now known to be at least 150 different his-tone modifications that require a wide variety of molecules in addition to theacetyl and methyl groups already mentioned (for example, phosphorylation andubiquitylation).

Histone acetylation, deacetylation, and gene expression The acetylation reac-tion is the best-characterized histone modification:

Message Chromatin can be dynamic; nucleosomes are not necessarily in fixedpositions on the chromosome. Chromatin remodeling changes nucleosome density orposition and is an integral part of eukaryotic gene regulation.


Acetylation

Methylation

H2B

H2A

H4

H4

H3

H2A

H3

Modified histone tails protrude from the nucleosome

FIGURE 11-12 Nucleosome structureshowing seven of the eight histones andmost but not all of their tails. The sites ofposttranslational modifications such asacetylation and methylation are shownfor one histone tail. In fact, all the tailscontain such sites.

Note that the reaction is reversible, which means that acetyl groups can beadded and removed from the same histone residue. With 44 histone lysine residuesavailable to accept acetyl groups, the presence or absence of these groups can carrya tremendous amount of information. For this reason, the covalent modification ofhistone tails is said to be a histone code. Scientists coined the expression histonecode because the covalent modification of histone tails is reminiscent of the geneticcode. For the histone code, information is stored in the patterns of histone modifi-cation rather than in the sequence of nucleotides. With more than 150 known his-tone modifications, there are a huge number of possible patterns and their effectson chromatin structure and transcriptional regulation are just beginning to be deci-phered. To add to this complexity, the code is likely not interpreted in precisely thesame way in all organisms. For now, let’s see how the acetylation of histone aminoacids influences chromatin structure and gene expression.

Evidence had been accumulating for years that the histones associated with thenucleosomes of active genes are rich in acetyl groups (said to be hyperacetylated),whereas inactive genes are underacetylated (hypoacetylated). The enzyme re-sponsible for adding acetyl groups, histone acetyltransferase (HAT), proved verydifficult to isolate. When it was finally isolated and its protein sequence deduced, itwas found to be an ortholog of a yeast transcriptional activator called GCN5 (mean-ing that it was encoded by the same gene in a different organism). Thus, the con-clusion was that GCN5 is a histone acetyltransferase. It binds to the DNA in the reg-ulatory regions of some genes and activates transcription by acetylating nearbyhistones. Various protein complexes that are recruited by transcriptional activatorsare now understood to possess a HAT activity.

How does histone acetylation facilitate changes in gene expression? Thereappear to be at least two mechanisms for doing so. First, the addition of acetylgroups to specific histone residues can alter the interaction in a nucleosomebetween the DNA and a histone octamer so that the octamer ismore likely to slide along the DNA to a new position. Second,histone acetylation, in conjunction with other histone modifica-tions, influences the binding of regulatory proteins to the DNA.The bound regulatory protein may take part in one of severalfunctions that either directly or indirectly increase the fre-quency of transcription initiation.

Like other histone modifications, acetylation is reversible,and histone deacetylases (HDAT’s) also have been identified.Such proteins play key roles in gene repression. For example, inthe presence of galactose and glucose, the activation of GALgenes is prevented by the Mig1 protein. Mig1 is a sequence-specific DNA-bindingrepressor that binds to a site between the UAS element and the promoter of theGAL1 gene (Figure 11-13). Mig1 recruits a protein complex called Tup1 that con-tains a histone deacetylase and that represses gene transcription. The Tup1 com-plex is an example of a corepressor, which faciliates gene repression but is notitself a DNA-binding repressor. The Tup1 complex is also recruited by other yeastrepressors, such as MATα2 (see page 400), and counterparts of this complex arefound in all eukaryotes.

Message In most cases examined, histone acetylation and deacetylation promoteand repress gene transcription, respectively. These activities are recruited to genes bysequence-specific activators and repressors.

NH3�

CH3

CoA C

O

S

�

Amino group at end of lysine side chain Acetyl CoA

Acetyl group

NH

CH3

C

O


Tup1Mig1

OFFMig1site

UAS

Gal4

GAL1

Histone deacetylation can turn off gene transcription

FIGURE 11-13 Recruitment of arepressing complex leads to repressionof transcription. In the presence ofglucose, GAL1 transcription is repressedby the Mig1 protein, which binds to asite between the UAS and the promoterof the GAL1 gene. Mig1 recruits the Tup1repressing complex, which recruits ahistone deacetylase, turning genetranscription off. [After J. Watson et al.,Molecular Biology of the Gene, Fifth Edition,copyright © 2004, Benjamin Cummings.]

The development of a complex organism requires that transcription levels be regu-lated over a wide range. Think of a regulation mechanism as more like a rheostatthan an on-or-off switch. In eukaryotes, transcription levels are made finelyadjustable by the clustering of binding sites into enhancers. Several different tran-scription factors or several molecules of the same transcription factor may bind to

adjacent sites. The binding of these factors to sites that arethe correct distance apart leads to an amplified, or super-additive, effect on activating transcription. When an effectis greater than additive, it is said to be synergistic.

The binding of multiple regulatory proteins to themultiple binding sites in an enhancer can catalyze the for-mation of an enhanceosome, a large protein complex thatacts synergistically to activate transcription. In Figure 11-14,you can see how architectural proteins bend the DNA topromote cooperative interactions between the other DNA-binding proteins. In this mode of enhanceosome action,transcription is activated to very high levels only when allthe proteins are present and touching one another in justthe right way.

To better understand what an enhanceosome is andhow it acts synergistically, let’s look at a specific example.

The a-interferon enhanceosomeThe human β-interferon gene, which encodes the antiviral protein interferon, isone of the best-characterized genes in eukaryotes. It is normally switched off butis activated to very high levels of transcription on viral infection. The key to theactivation of this gene is the assembly of transcription factors into an enhanceo-some about 100 bp upstream of the TATA box and transcription start site. Theregulatory proteins of the β-interferon enhanceosome all bind to the same face ofthe DNA double helix. Binding to the other side of the helix are several architec-tural proteins that bend the DNA and allow the different regulatory proteins totouch one another and form an activated complex. When all of the regulatoryproteins are bound and interacting correctly, they form a “landing pad,” a high-affinity binding site for the protein CBP, a coactivator protein that also recruitsthe transcriptional machinery. The large CBP protein also contains an intrinsichistone acetylase activity that modifies nucleosomes and facilitates high levels oftranscription.

Although the β-interferon promoter is shown without nucleosomes in Figure11-14, the enhanceosome is actually surrounded by two nucleosomes, called nuc 1and nuc 2 in Figure 11-15. One of them, nuc 2, is strategically positioned over theTATA box and transcription start site. However, the binding of GCN5, another coac-tivator, is now known to actually precede CBP binding. GCN5 acetylates the twonucleosomes. After acetylation, the activating transcription factors recruit the coac-tivator CBP, the RNA pol II holoenzyme, and the SWI–SNF chromatin-remodelingcomplex. SWI–SNF is then positioned to nudge the nucleosome 37 bp off the TATAbox, making the TATA box accessible to the TATA-binding protein and allowingtranscription to be initiated.

Cooperative interactions help to explain several perplexing observations aboutenhancers. For example, they explain why mutating any one transcription factor orbinding site dramatically reduces enhancer activity. They also explain why the dis-tance between binding sites within the enhancer is such a critical feature. Further-more, enhancers do not have to be close to the start site of transcription, as is the

11.4 Mechanism of Enhancer Action


DNA-bending proteins

RNA pol II

CBP

Enhanceosomes help recruit the transcriptional machinery

FIGURE 11-14 The β-interferonenhanceosome. In this case, thetranscription factors recruit acoactivator (CBP), which binds both tothe transcription factors and to RNApolymerase II, initiating transcription.[After A. J. Courey, “Cooperativity inTranscriptional Control,” Curr. Biol. 7, 2001,R250–R253, Fig. 1.]

example shown in Figure 11-15. One characteristic of en-hancers is that they can activate transcription when they arelocated at great distances from the promoter (>50 kb), eitherupstream or downstream from a gene or even in an intron.

The control of yeast mating type:Combinatorial interactionsThus far, we have focused in this chapter on the regulationof single genes or a few genes in one pathway. In multicellu-lar organisms, distinct cell types differ in the expression ofhundreds of genes. The expression or repression of sets ofgenes must therefore be coordinated in the making of par-ticular cell types. One of the best-understood examples ofcell-type regulation in eukaryotes is the regulation of mat-ing type in yeast. This regulatory system has been dissectedby an elegant combination of genetics, molecular biology,and biochemistry. Mating type serves as an excellent modelfor understanding the logic of gene regulation in multicellu-lar animals.

The yeast Saccharomyces cerevisiae can exist in any ofthree different cell types known as a, `, and a/` (see Chap-ter 2). The two cell types a and ` are haploid and containonly one copy of each chromosome. Although the two hap-loid cell types cannot be distinguished by their appearancein the microscope, they can be differentiated by a number ofspecific cellular characteristics, principally their mating type(see the Model Organism box on page 390). An ` cell matesonly with an a cell and secretes an oligopeptide pheromone,or sex hormone, called ` factor that arrests a cells in the cellcycle. A cell of the a type mates only with an ` cell andsecretes a pheromone, called a factor, that arrests ` cells.The diploid a/` cell does not mate, is larger than the ` anda cells, and does not respond to the mating hormones.

Genetic analysis of mutants defective in mating hasshown that cell type is controlled by a single genetic locus,the mating-type locus, MAT. There are two alleles of theMAT locus: haploid a cells have the MATa allele, haploid ` cells have the MATa

allele, and the a/` diploid has both alleles. Although mating type is under geneticcontrol, certain strains switch their mating type, sometimes as frequently as everycell division. We will examine the basis of switching later in this chapter, but, first,let’s see how each cell type expresses the right set of genes.

DNA-binding proteins combinatorially regulate the expressionof cell-type-specific genesHow does the MAT locus control cell type? Genetic analyses of mutants that can-not mate have identified a number of structural genes that are separate from the

Message Eukaryotic enhancers can act at great distances tomodulate the activity of the transcriptional apparatus.Enhancers contain binding sites for many transcription factors,which bind and interact cooperatively. These interactions resultin a variety of responses, including the recruitment ofadditional coactivators and the remodeling of chromatin.

39911.4 Enhancers: Cooperative Interactions, Combinatorial Control, and Chromatin Remodeling

+

SWI–SNF

CBP

RNA pol II

The TATA-binding protein (TBP)binds to the newly exposed TATAbox, allowing transcription to begin.

The coactivator CBP binds, recruiting RNA pol II.

The enhanceosome formsa binding site for GCN5, whichbinds and adds acetyl groups to nuc 1, 2.

SWI–SNF nudges aside nuc 2.

Enhanceosome

nuc 1 nuc 2

GCN5complex

GCN5

GCN5

CBP

RNA pol II

CBP

SWI–SNF

TBP

Enhanceosomes recruit chromatin remodelers

FIGURE 11-15 The β-interferonenhanceosome acts to movenucleosomes by recruiting the SWI–SNFcomplex.

MAT locus, but their protein products are required for mating. One group of struc-tural genes is expressed only in the ` cell type (`-specific genes), and another set isexpressed only in the a cell type (a-specific genes). The different alleles of theMAT locus encode different regulatory proteins that control which of these sets ofstructural genes is expressed in each cell type. In addition, a regulatory protein notencoded by the MAT locus, called MCM1, plays a key role in regulating cell type.

The simplest case is the a cell type (Figure 11-16a). The MATa locus encodes asingle regulatory protein, a1. However, this regulatory protein has no effect in hap-loid cells, only in diploid cells. In a haploid a cell, the regulatory protein MCM1turns on the expression of the structural genes needed by an a cell, by binding toregulatory sequences within a-specific gene promoters.


a1MAT locus

Expressedregulatoryproteins

a1

a-specificgenes ON

(a) a cell (b) α cell

a1

(c) a/α cell

α2 α1 α2 α1

α2

α2 α2

α2α1a1

MCM1 MCM1MCM1

MCM1OFF

MCM1

α2 α2OFF

MCM1

α-specificgenes OFF

α1ON

MCM1OFF

Haploid-specificgenes ON ON

α2OFF

a1

Combinations of regulatory proteins control cell types

FIGURE 11-16 Control of cell-type-specific gene expression in yeast. Thethree cell types of S. cerevisiae aredetermined by the regulatory proteinsa1, α1, and α2, which regulate differentsubsets of target genes. The MCM1protein acts in all three cell types andinteracts with α1 and α2.

In an ` cell, the `-specific structural genes must be transcribed, but, in addi-tion, the MCM1 protein must be prevented from activating the a-specific genes.The DNA sequence of the MATa allele encodes two proteins, α1 and α2, that areproduced by separate transcription units. These two proteins have different regula-tory roles in the ` cell, as can be demonstrated by analyzing their DNA-bindingproperties in vitro (Figure 11-16b). The α1 protein binds in concert with the MCM1protein to a discrete DNA sequence controlling several `-specific genes. Thus, α1 isan activator of `-specific gene expression. The α2 protein represses transcription ofthe a-specific genes. It binds as a dimer, with MCM1, to sites in DNA sequenceslocated 5� of a group of a-specific genes and acts as a repressor.

In a diploid yeast cell, all three regulatory proteins encoded by the MAT locusare expressed (Figure 11-16c). What is the result? The a1 protein encoded by MATahas a part to play at last. The a1 protein can bind to α2 and alter its binding speci-ficity such that the a1–α2 complex does not bind to a-specific genes. Rather, thea1–α2 complex binds to a different sequence found upstream of another set ofgenes, called haploid specific, that are expressed in haploid cells but not diploidcells. In diploid cells, then, α2 exists in two forms: (1) as an α2–MCM1 complex thatrepresses a-specific genes and (2) in a complex with a1 that represses haploid-spe-

cific genes. The different binding partners determine which specific DNA se-quences are bound and which genes are regulated by each α2-containing complex.The regulation of different sets of target genes by the association of the same tran-scription factor with different binding partners plays a major role in the generationof different patterns of gene expression in different cell types within multicellulareukaryotes.

Enhancer-blocking insulatorsA regulatory element, such as an enhancer, that can act over tens of thousands ofbase pairs could interfere with the regulation of nearby genes. To prevent suchpromiscuous activation, regulatory elements called enhancer-blocking insulatorshave evolved. When positioned between an enhancer and a promoter, enhancer-blocking insulators prevent the enhancer from activating transcription at that pro-moter. Such insulators have no effect on the activation of other promoters that arenot separated from their enhancers by the insulator (Figure 11-17). Several modelshave been proposed to explain how an insulator could block enhancer activity onlywhen placed between an enhancer and a promoter. Many of the models, like theone shown in Figure 11-18, propose that the DNA is organized into loops containing

Message In yeast and in multicellular eukaryotes, cell-type-specific patterns of geneexpression are governed by combinations of interacting transcription factors.

40111.4 Enhancers: Cooperative Interactions, Combinatorial Control, and Chromatin Remodeling

FIGURE 11-17 Enhancer-blockinginsulators prevent gene activation whenplaced between an enhancer and apromoter. [After M. Gaszner and G. Felsenfeld,“Insulators: Exploiting Transcriptional andEpigenetic Mechanisms,” Nat. Rev. Genet. 7,2006, 703–713.]

FIGURE 11-18 The proposal is thatenhancer-blocking insulators (EB) createnew loops that physically separate apromoter from its enhancer (E). [After M. Gaszner and G. Felsenfeld,“Insulators: Exploiting Transcriptional andEpigenetic Mechanisms,” Nat. Rev. Genet. 7,2006, 703–713.]

ON OFF

Promoter 2 Promoter 1Enhancer-blockinginsulator

Enhancer

××

Enhancer-blocking insulators prevent enhancer activation

Promoter 1 Promoter 2

OFF

ON

Enhancer

Enhancer-blockinginsulator

××

Model for how enhancer-blocking insulators might work

active genes. According to this model, insulators act by moving a promoter into anew loop, where it is shielded from the enhancer.

As you will see next, enhancer-blocking insulators are a fundamental compo-nent of a phenomenon called genomic imprinting.

The phenomenon of genomic imprinting was discovered almost 20 years ago inmammals. In genomic imprinting, certain autosomal genes have unusual inheri-tance patterns. For example, an igf2 allele is expressed in a mouse only if it isinherited from the mouse’s father—an example of maternal imprinting becausea copy of the gene derived from the mother is inactive. Conversely, a mouse H19allele is expressed only if it is inherited from the mother; H19 is an example ofpaternal imprinting because the paternal copy is inactive. The consequence ofparental imprinting is that imprinted genes are expressed as if there were onlyone copy of the gene present in the cell even though there are two. Hence,imprinting is an example of monoallelic inheritance. Importantly, no changesare observed in the DNA sequences of imprinted genes; that is, the identical genecan be active or inactive in the progeny, depending on whether it was inheritedfrom mom or dad.

If the DNA sequence of the gene does not correlate with activity, what does?The answer is that that the DNA in the regulatory regions of imprinted genes ismethylated in a sex-specific manner in the development of gametes. DNA methyla-tion usually results from the enzymatic addition of methyl groups to the carbon-5position of a specific cytosine residue.

Both DNA methylation marks and histone modificationmarks can be stably inheritable from one cell generation tothe next. We will see later in the chapter how such marks arethought to be duplicated in the course of DNA replication.For now, suffice it to say that such heritable alteration, inwhich the DNA sequence itself is unchanged, is called epi-genetic inheritance, and the alterations (including bothDNA methylation and histone modifications) are called epi-genetic marks.

Let’s turn again to the mouse ifg2 and H19 genes to seehow imprinting works at the molecular level. These twogenes are located in a cluster of imprinted genes on mousechromosome 7. There are an estimated 100 imprinted genesin the mouse, and most are found in clusters comprising

from 3 to 11 imprinted genes. (Humans have most of the same clustered imprintedgenes as those in the mouse.) In all cases examined, there is a specific DNA methy-lation pattern for each gene copy of an imprinted gene. For the ifg2–H19 cluster, aspecific region of DNA lying between the two genes (Figure 11-19) is methylated inmale germ cells and unmethylated in female germ cells. This region is called theimprinting control region (ICR). Only the unmethylated (female) ICR can bind aregulatory protein called CTCF. When bound, CTCF acts as an enhancer-blockinginsulator that prevents enhancer activation of Igf2 transcription. However, theenhancer in females can still activate H19 transcription. In males, CTCF cannot

NH2

CH3

1

4

2

3

6

5

O

CC

NC

N

C

Methyl group

Cytosine

Methyltransferase

NH2

1

4

2

3

6

5

O

CC

NC

N

C

11.5 Genomic Imprinting


FIGURE 11-19 Genomic imprinting inthe mouse. The imprinting controlregion (ICR) is unmethylated in femalegametes and can bind a CTCF dimer,forming an insulator that blocksenhancer activation of Igf2. Methylation(M) of the ICR in male germ cellsprevents CTCF binding, but it alsoprevents the binding of other proteinsto the H19 promoter.

M M M M

� Maternal allele

Igf2 H19

ICR

CT

CF

EnhancerOFF >50 Kb ON

� Paternal allele

Igf2 H19

ICR EnhancerON >50 Kb OFF

Genomic imprinting requires insulators

bind to the ICR and the enhancer can activate Igf2 transcription (recall thatenhancers can act at great distances). The enhancer cannot activate H19, however,because the methylated region extends into the H19 promoter. The methylatedpromoter cannot bind proteins needed for the transcription of H19.

Thus, we see how an enhancer-blocking insulator (in this case, CTCF bound topart of the ICR) prevents the enhancer from activating a distant gene (in this case,Igf2). Furthermore, we see that the CTCF-binding site is methylated only in chro-mosomes derived from the male parent. The methylation of the CTCF-binding siteprevents CTCF binding in males and permits the enhancer to activate Igf2.

Note that parental imprinting can greatly affect pedigree analysis. Because theinherited allele from one parent is inactive, a mutation in the allele inherited fromthe other parent will appear to be dominant, whereas, in fact, the allele is ex-pressed because only one of the two homologs is active for this gene. Figure 11-20shows how a mutation in an imprinted gene can have different outcomes on thephenotype of the organism if inherited from the male or from the female parent.

Many steps are required for imprinting (Figure 11-21). Soon after fertiliza-tion, mammals set aside cells that will become their germ cells. Imprints areremoved or erased before the germ cells form. Without their distinguishing markof DNA methylation, these genes are now said to be epigenetically equivalent. Asthese primordial germ cells become fully formed gametes, imprinted genes re-ceive the sex-specific mark that will determine whether the gene will be active orsilent after fertilization.

40311.5 Genomic Imprinting

FIGURE 11-20 A mutation (represented byan orange star) in gene A will have no effectif inherited from the male. Abbreviations: M, methylation; ICR, imprinting controlregion. [After S. T. da Rocha and A. C. Ferguson-Smith, “Genomic Imprinting,” Curr. Biol. 14, 2004,R646–R649.]

FIGURE 11-21 How Igf2 and H19 aredifferentially imprinted in males and females.

M

� A B

� A BICR

No mutations

M

� A B

� A BICR

Mutation in imprinted gene

M

� A B

� A BICR

OUTCOME UNAFFECTED

OUTCOME AFFECTED

Unusual inheritance of imprinted genes

Male

Two linked genes,one active, one silent

Homologouschromosomes

Igf2 H19 Igf2 H19

Oocyte

Primordialgerm cells

Primordialgerm cells

Gametes

Fertilization anddevelopment

Female

Sperm

Silent allele

Active allele

Imprintserased

Imprintsinitiated

Propagationof imprints

1

2

3

Steps required for imprinting

But what about Dolly and other cloned mammals?Genomic imprinting leads to what many thought would be a requirement for theparticipation of male and female germ cells in mammalian embryo development.That is, male and female gametes contain different subsets of imprinted genes sothat the embryo will have a full complement of active imprinted genes. Why thenare mammals such as Dolly and, more recently, cloned pigs, cats, dogs, and cowsthat were derived from somatic nuclei able to survive and even flourish? After all,as already noted, the mutation of even a single imprinted gene can be lethal or canlead to serious disease.

At this point, scientists do not understand why the cloning of many mam-malian species has been successful. However, despite these successes, cloning isextremely inefficient in all species tested. For most experiments, a successful cloneis an exceedingly rare event, requiring hundreds, even thousands, of attempts. Onecould argue that the failure of most cloned embryos to develop into viable organ-isms is a testament to the importance of the epigenetic mechanisms of gene regu-lation in eukaryotes. As such, it illustrates how knowledge of the complete DNAsequence of all genes in an organism is only a first step in understanding howeukaryotic genes are regulated.

Thus far, we have looked at how genes are activated in a chromatin environment.However, as stated at the beginning of this chapter, most of the genes in eukaryoticgenomes are off at any one time. Let’s now turn to those vast regions of thegenome that are transcriptionally inactive. One of the most useful models forunderstanding mechanisms that maintain the inactivity of genes concerns the con-trol of yeast mating type and mating-type switching.

Mating-type switching and gene silencingHaploid yeast cells are able to switch their mating type. Genetic analyses of certainmutants that either could not switch or could not mate (they were sterile) weresources of key insights into mating-type switching. Among the switch mutantswere several mutant loci including the HO gene and the HMRa and HMLa genes.Further study revealed that the HO gene encodes an endonuclease, an enzymethat cleaves DNA, required for the initiation of switching. It was also found thatthe HMRa and HMLa loci contain “cassettes” of unexpressed genetic informationfor the MATa and MATα mating types, respectively. The HMR and HML loci arethus referred to as “silent” cassettes.

The HO endonuclease initiates the mating-type switch by inserting a double-strand break at the MAT locus. The interconversion of mating type then takesplace by a type of recombination between the segment of DNA (a cassette) fromone of the two unexpressed loci and the MAT locus. The result is the replacementof the old cassette at the MAT locus with a new cassette. The resulting mating typeis either the MATa or the MATα type, depending on which cassette is at the MATlocus (Figure 11-22). The inserted cassette is actually copied from the HML orHMR locus. In this manner, the switch is reversible because the information for thea and a cassettes is always present at the HMR and HML loci and never lost.

Normally, the HMR and HML cassettes are “silent.” However, in SIR mutants(silent information regulators), silencing is compromised such that both a and ainformation is expressed. The resulting mutants are sterile. The Sir2, Sir3, and Sir4proteins form a complex that plays a key role in gene silencing. Sir2 is a histonedeacetylase that facilitates the condensation of chromatin and helps lock up HMRand HML in chromatin domains that are inaccessible to transcriptional activators.

11.6 Chromatin Domains and Their Inheritance


Gene silencing is a very different process from gene repression; silencing is a posi-tion effect that depends on the neighborhood in which genetic information islocated. You will learn more about position effects later, in the section on position-effect variegation in the fruit fly Drosophila melanogaster.

In summary, there are two distinct levels in the control of yeast mating type.First, the regulation of a DNA rearrangement controls the array of regulatoryproducts synthesized within the cell. Second, the DNA-binding activities of theseregulatory proteins (a1, α1, and α2) control the batteries of structural genesexpressed within each cell type. These two levels form a hierarchy: the genes of thefirst level control the activation of genes on the second level, which in turn controlthe activation of the structural genes. These structural genes encode the proteinshaving roles in the actual mating process and the biology of each cell type. As weshall see in regard to animals in Chapter 12, the genetic control of developmentalprocesses is often hierarchical: networks of regulatory genes set up the cell- andtissue-specific expression of proteins that mediate cell behavior and function.

Heterochromatin and euchromatin comparedLet’s return to the silent cassettes HML and HMR to understand why gene silencingis a very different process from gene repression and what is meant by a genomicneighborhood. To do so, it is important to note that chromatin is not uniform overall chromosomes; certain domains of chromosomes are bundled in highly con-densed chromatin called heterochromatin. Other domains are packaged in less-condensed chromatin called euchromatin (see Figure 11-10b). Chromatin conden-sation also changes in the course of the cell cycle. The chromatin of cells enteringmitosis becomes highly condensed as the chromosomes align in preparation for celldivision. After cell division, regions forming heterochromatin remain condensed


FIGURE 11-22 S. cerevisiae chromosome IIIencodes three mating-type loci, but only thegenes at the MAT locus are expressed. HMLencodes a silent cassette of the ` genes, andHMR encodes a silent cassette of the a genes.Copying of a silent cassette and insertionthrough recombination at the MAT locusswitches mating type.

(a)

HMLα MATa HMRa

Silent SilentActive

a mating type

(b) HMLα is copied into the MAT locus

(c) HMRa is copied into the MAT locus

HMLα MATα HMRa

Silent

a mating type

HMLα MATa HMRa

Silent

a mating type

Mating-type switching is controlled by recombination of DNA cassettes

especially around the centromeres and telomeres (called constitutive heterochro-matin), whereas the regions forming euchromatin become less condensed.

Geneticists first suspected a limited role for the influence of chromatin struc-ture on gene regulation early in the history of genetics. At that time, they noticedthat heterochromatic DNA contained few genes, whereas euchromatin was richin genes. But what is heterochromatin if not genes? Most of the eukaryoticgenome is composed of repetitive sequences that do not make protein or struc-tural RNA—sometimes called junk DNA (see Chapter 14). Thus, the denselypacked nucleosomes of heterochromatin were said to form a “closed” structurethat was inaccessible to regulatory proteins and inhospitable to gene activity. Incontrast, euchromatin, with its more widely spaced nucleosomes, was proposed toassume an “open” structure that permitted transcription. The existence of openand closed regions of chromatin was also suggested as a reason that recombinationfrequencies are 100- to 1000-fold higher in euchromatin compared with hete-rochromatin. Euchromatin, with its more open conformation, was hypothesized tobe more accessible to proteins needed for DNA recombination.

Position-effect variegation in Drosophila reveals genomicneighborhoodsThe geneticist Hermann Muller first discovered an interesting genetic phenome-non while studying Drosophila: there exist chromosomal neighborhoods that cansilence genes that are experimentally “relocated” to adjacent regions of the chro-mosome. In these experiments, flies were irradiated with X rays to induce muta-tions in their germ cells. The progeny of the irradiated flies were screened forunusual phenotypes. A mutation in the white gene, near the tip of the X chromo-some, will result in progeny with white eyes instead of the wild-type red color.Some of the progeny had very unusual eyes with patches of white and red color.Cytological examination revealed a chromosomal rearrangement in the mutantflies: present in the X chromosome was an inversion of a piece of the chromosomecarrying the white gene (Figure 11-23). Inversions and other chromosomal rear-rangements will be discussed in Chapter 16. In this rearrangement, the white gene,which is normally located in a euchromatic region of the X chromosome, nowfinds itself near the heterochromatic centromere. In some cells, the heterochro-matin can “spread” to the neighboring euchromatin and silences the white gene.Patches of white tissue in the eye are derived from the descendants of a single cellin which the white gene has been epigenetically silenced and remains silencedthrough future cell divisions. In contrast, the red patches arise from cells in whichheterochromatin has not spread to the white gene, and so this gene remains activein all its descendants. The existence of red and white patches of cells in the eye ofa single organism dramatically illustrates two features of epigenetic silencing. First,that the expression of a gene can be repressed by virtue of its position in the chro-mosome rather than by a mutation in its DNA sequence. Second, that epigeneticsilencing can be inherited from one cell generation to the next.

Findings from subsequent studies in Drosophila and yeast demonstrated thatmany active genes are silenced in this mosaic fashion when they are relocated toneighborhoods (near centromeres or telomeres) that are heterochromatic. Thus,the ability of heterochromatin to spread into euchromatin and silence genes is afeature common to many organisms. This phenomenon has been called position-effect variegation (PEV). It provides powerful evidence that chromatin structure

Message The chromatin of eukaryotes is not uniform. Highly condensedheterochromatic regions have fewer genes and lower recombination frequencies thando the less-condensed euchromatic regions.


is able to regulate the expression of genes—in this case, determining whether geneswith identical DNA sequence will be active or silenced.

Genetic analysis of PEV reveals proteins necessary for heterochromatin formationTo find out what proteins might be implicated in the establishment of heterochro-matin, geneticists isolated mutations at a second chromosomal locus that eithersuppressed or enhanced the variegated pattern (Figure 11-24). Suppressors of varie-gation [called Su(var)] are genes that, when mutated, reduce the spread of hete-rochromatin, meaning that the wild-type products of these genes are required forspreading. In fact, the Su(var) alleles have proved to be a treasure trove for scientistsinterested in the proteins that are required to establish and maintain the inactive,heterochromatic state. Among more than 50 Drosophila gene products identified bythese screens was heterochromatin protein-1 (HP-1), which had previously beenfound associated with the heterochromatic telomeres and centromeres. Thus, itmakes sense that a mutation in the gene encoding HP-1 will show up as a Su(var)allele because the protein is required in some way to produce or maintain hete-rochromatin. Another Su(var) gene was found to encode a methyltransferase that

Message Active genes that are relocated to genomic neighborhoods that areheterochromatic may be silenced if the heterochromatin spreads to the genes.


FIGURE 11-23 Chromosomalrearrangement produces position-effectvariegation. Chromosomal inversionplaces the wild-type white allele close toheterochromatin. The spread ofheterochromatin silences the allele. Eyefacets are white instead of the wild-typered wherever the allele has been silenced.[After J. C. Eissenberg and S. Elgin,Encyclopedia of Life Sciences. NaturePublishing Group, 2001, p. 3, Fig. 1.]

Heterochromatin spreads

Centromere

Chromosome white+

Telomere

white+ geneexpressed Wild-type eye

Eye is a mixtureof red and whitefacets.

white+ geneexpressed

white+ genesilent

Red facet

White facet

white+

white+

Inversion places white+ close to heterochromatin.

Gene silencing is caused by the spread of heterochromatin

adds methyl groups to a specific amino acid residue in the tail of histone H3(called histone H3 methyltransferase or HMTase). One of the reactions catalyzedby HMTase is shown here:

Proteins similar to HP-1 and HMTase have been isolated in diverse taxa, suggestingthe conservation of an important eukaryotic function.

We have seen that actively transcribed regions are associated with nucleo-somes whose histone tails are hyperacetylated and that transcriptional activatorssuch as GCN5 encode a histone acetytransferase activity. As heretofore discussed,acetyl marks can also be removed from histones by histone deacetylases. Similarly,chromatin made up of nucleosomes that are methylated at lysine 9 of H3 (calledH3meK9) and bound up with HP-1 protein contain epigenetic marks that are asso-ciated with heterochromatin. Scientists are now able to separate heterochromatinand euchromatin and analyze differences in histone modifications and bound pro-


FIGURE 11-24 Mutations were used toidentify genes that suppress, Su(var), orenhance, E(var), position-effect variegation.[After J. C. Eissenberg and S. Elgin, Encyclopediaof Life Sciences. Nature Publishing Group, 2001,p. 3, Fig. 1.]

Drosophila eye(translocated white+) E(var)

Su(var)

Second-sitemutations that affect

the spreading ofheterochromatin

Spreadingenhanced.More white+

are silenced.

Spreadingsuppressed.Fewer white+

are silenced.

Some genes enhance or suppress the spread of heterochromatin

HMTase

NH3+

+H3N

(CH2)4

C H

COO–

Lysine

N+

+H3N

(CH2)4

H3C

C

HH

H

COO–

Monomethyl lysine

HMTase

N+

+H3N

(CH2)4

H3C CH3

C

H

H

COO–

Dimethyl lysine

HMTase

N+

+H3N

(CH2)4

H3C CH3CH3

C H

COO–

Trimethyl lysine

teins. The procedure used, chromatin immunoprecipitation (ChIP), isdescribed in Chapter 20.

Figure 11-25 illustrates how, in the absence of any barriers, hete-rochromatin might spread into adjoining regions in some cells butnot in others and inactivate genes. It could be what is happening tothe white gene of Drosophila when it is translocated near the domainof heterochromatin associated with the chromosome ends. But canthe spread of heterochromatin be stopped? One can imagine that thespreading of heterochromatin into active gene regions could be disas-trous for an organism because active genes would be silenced as theyare converted into heterochromatin. To avert this potential disaster,the genome contains DNA elements called barrier insulators thatprevent the spreading of heterochromatin by creating a local envi-ronment that is not favorable to heterochromatin formation. Forexample, a barrier insulator could bind HATs and, in doing so, makesure that the adjacent histones are hyperacetlyated. A model for howa barrier insulator might act to “protect” a region of euchromatinfrom being converted into heterochromatin is shown in Figure 11-26.

Silencing an entire chromosome: X-chromosomeinactivationThe epigenetic phenomenon called X-chromosome inactivation hasintrigued scientists for decades. In Chapter 16, you will learn about the effects ofgene copy number on the phenotype of an organism. For now, it is sufficient toknow that the number of transcripts produced by a gene is usually proportional tothe number of copies of that gene in a cell. Mammals, for example, are diploid andhave two copies of each gene located on their autosomes. For the vast majority ofgenes, both alleles are expressed. However, this is not possible for the sex chromo-somes. As discussed in Chapter 2, the number of the X and Y sex chromosomes dif-fers between the sexes, with female mammals having two X chromosomes andmales having only one. The mammalian X chromosome is thought to containabout 1000 genes. Females have twice as many copies of these X-linked genes andwould otherwise express twice as much transcript from these genes as males do ifthere were not a mechanism to correct this imbalance. (Not having a Y chromo-some is not a problem for females, because the very few genes on this chromosomeare required only for the development of males.) This dosage imbalance is cor-rected by a process called dosage compensation, which makes the amount ofmost gene products from the two copies of the X chromosome in females equiva-lent to the single dose of the X chromosome in males. In mammals, this equiva-lency is accomplished by random inactivation of one of the two X chromosomes ineach cell at an early stage in development. This inactive state is then propagated to


FIGURE 11-25 The spread ofheterochromatin into adjacenteuchromatin is variable. In four geneticallyidentical diploid cells, heterochromatinspread enough to knock out a gene insome chromosomes but not others.Heterochromatin and euchromatin arerepresented by orange and green spheres,respectively. [After M. Gaszner and G.Felsenfeld, “Insulators: Exploiting Transcriptionaland Epigenetic Mechanisms,” Nat. Rev. Genet. 7,2006, 703–713.]

FIGURE 11-26 In this model, barrierinsulators recruit enzymatic activities suchas histone acetyltransferase (HAT) thatpromote euchromatin formation. The letter“M” stands for methylation and the letters“Ac” for acetylation. [After M. Gaszner and G. Felsenfeld, “Insulators: ExploitingTranscriptional and Epigenetic Mechanisms,”Nat. Rev. Genet. 7, 2006, 703–713.]

OFF

OFF

OFF

ON

OFF

OFF

OFF

ON

Heterochromatin may spread farther in some cells than in others

AcAcAc Ac

Heterochromatin Euchromatin

HP-1

M M M M

HMTase HAT

Barrierinsulator

Barrier insulators stop the spread of heterochromatin

all progeny cells. (In the germ line, the second X chromosome becomes reactivatedin oogenesis). The inactivated chromosome, called a Barr body, can be seen in thenucleus as a darkly staining, highly condensed, heterochromatic structure.

Two aspects of X-chromosome inactivation are relevant to a discussion ofchromatin and the regulation of gene expression. First, most of the genes on theinactivated X chromosome are silenced, and the chromosome has epigeneticmarks associated with heterochromatin including methylation of H3 at lysine 9and hypermethylation of its DNA. Second, genes on the inactivated chromosomeremain inactive in all descendants of these cells. Because the DNA sequence itselfis unchanged, this heritable alteration is an example of epigenetic inheritance.

Interestingly, although diverse taxa exhibit dosage compensation, the compen-sation mechanism can differ dramatically. For example, in fruit flies, the expres-sion of genes on the X chromosome is compensated not by inactivating one of thetwo X’s in females, but instead, by doubling the expression of the genes on the oneX in the male (Figure 11-27). This mechanism is characterized by the binding of aRNA–protein complex, called MSL, along the entire length of the X chromosomein males (see illustration on page 385). One of the components of the MSL com-plex is a histone acetyltransferase. Recall that acetylated histones are a main fea-ture of active chromatin. Thus, the function of the MSL complex appears to be toadd acetyl groups to histones. MSL stands for male-specific lethal, and the complexwas so named because genetic screens for mutations lethal to males identified itscomponents.

The inheritance of epigenetic marks and chromatin structureEpigenetic inheritance can be defined operationally as the inheritance of chro-matin states from one cell generation to the next. What this inheritance means isthat, in DNA replication, both the DNA sequence and the chromatin structure arefaithfully passed on to the next cell generation. However, unlike the sequence ofDNA, chromatin structure can change in the course of the cell cycle when, forexample, transcription factors modify the histone code, causing local changes innucleosome position or nucleosome density or both.

As mentioned in Chapter 7, the replisome not only copies the parental strandsbut also disassembles the nucleosomes in the parental strands and reassembles

Message For most diploid organisms, both alleles of a gene are expressedindependently. X inactivation and genomic imprinting are examples of monoallelicexpression. In these cases, epigenetic mechanisms silence one copy of an entirechromosome or of a single chromosomal locus, respectively.


FIGURE 11-27 Dosage compensation canbe achieved by doubling the expression ofthe male X chromosome (hypertranscription),by X inactivation, or by halving the expression of both female X chromosomes(hypotranscription).

Hypertranscription (Drosophila)

X inactivation (mammals)

Hypotranscription (C. elegans)

Female Male

1 1+ = 2

1 = 1

+ = 112

12

X

X

X

X X Y

No Y

YX

X X X

Different mechanisms of dosage compensation

FIGURE 11-28 In replication,old histones (purple) with theirhistone codes are distributedrandomly to the daughterstrands, where they direct thecoding of adjacent newlyassembled histones (orange) toform complete nucleosomes.

Nucleosome

Replication

Newly synthesizedhistones, no histone code

Histones with code

Inheritance of chromatin states

them in both the parental and the daughter strands. This process is accomplishedby the random distribution of the old histones from existing nucleosomes todaughter strands and the delivery of new histones to the replisome. In this way, theold histones with their modified tails and the new histones with unmodified tailsare assembled into nucleosomes that become associated with both daughterstrands. The code carried by the old histones most likely guides the modification ofthe new histones (Figure 11-28).

The inheritance of DNA methylation is better understood. Semiconservativereplication generates daughter helices that are methylated on one of their twostrands (the parental strand). The unmethylated strands are methylated by DNAmethyltransferases that have a high affinity for these so-called hemimethylatedsubstrates and are guided by the methylation pattern on the parental strand (Fig-ure 11-29). Thus, the information inherent in the histone code and the existingDNA methylation patterns serve to reconstitute the local chromatin structure thatexisted before DNA synthesis and mitosis.

Message Chromatin structure is inherited from cell generation to cell generationbecause mechanisms exist to replicate the DNA along with the associated epigeneticmarks.

411Summary

FIGURE 11-29 After replication, thehemimethylated dinucleotide CG (shown as CpG)residues are fully methylated. The parental strandsare black, and the daughter strand is red. Theletter “M” represents the methyl group on the C nucleotide. [After Y. H. Jiang, J. Bressler, and A. L.Beaudet, “Epigenetics and Human Disease,” Annu. Rev.Genomics Hum. Genet. 5, 2004, 479–510.]

MethylatedMCpGGpCM

DNAreplication

DNAmethyltransferase

MCpGGpC

CpGGpCM

MCpGGpCM

MCpGGpCM

A model for the inheritance of DNA methylation

SummaryMany aspects of eukaryotic gene regulation resemble the reg-ulation of bacterial operons. Both operate largely at the levelof transcription, and both rely on trans-acting proteins thatbind to cis-acting regulatory target sequences on the DNAmolecule. These regulatory proteins determine the level oftranscription from a gene by controlling the binding of RNApolymerase to the gene’s promoter.

There are three major distinguishing features of the con-trol of transcription in eukaryotes. First, eukaryotic genes pos-sess enhancers, which are cis-acting regulatory elements lo-cated at sometimes great linear distances from the promoter.Many genes possess multiple enhancers. Second, these en-hancers are often bound by more transcription factors thanare bacterial operons. Multicellular eukaryotes must generatethousands of patterns of gene expression with a limited num-ber of regulatory proteins (transcription factors). They do sothrough combinatorial interactions among transcription fac-tors. Enhanceosomes are complexes of regulatory proteinsthat interact in a cooperative and synergistic fashion to pro-

mote high levels of transcription through the recruitment ofRNA polymerase II to the transcription start site.

Third, eukaryotic genes are packaged in chromatin.Gene activation and repression require specific modifica-tions to chromatin. The vast majority of the tens of thou-sands of genes in a typical eukaryotic genome are turned offat any one time. Genes are maintained in a transcription-ally inactive state through the participation of nucleosomes,which serve to compact the chromatin and prevent the bind-ing of RNA polymerase II. The position of nucleosomes andthe extent of chromatin condensation are instructed by thehistone code, the pattern of posttranslational modificationsof the histone tails. The histone code is an epigenetic markthat, along with the methylation of cytosine bases, can bealtered by transcription factors. These factors bind to regula-tory regions and recruit protein complexes that enzymati-cally modify adjacent nucleosomes. These large multisubunitprotein complexes use the energy of ATP hydrolysis to movenucleosomes and remodel chromatin.


The existence of epigenetic phenomena such as geneticimprinting and X-chromosome inactivation demonstratesthat eukaryotic gene expression can be silenced withoutchanging the DNA sequence of the gene. Another epige-netic phenomenon, position-effect variegation, revealed theexistence of repressive heterochromatic domains that areassociated with highly condensed nucleosomes and containfew genes. Barrier insulators maintain the integrity of the

genome by preventing the conversion of euchromatin intoheterochromatin.

DNA replication faithfully copies both the DNA se-quence and the chromatin structure from parent to daugh-ter cells. Newly formed cells inherit both genetic informa-tion, inherent in the nucleotide sequence of DNA, andepigenetic information, which is in the histone code andthe pattern of DNA methylation.

Key Terms

activation domain (p. 392)

Barr body (p. 410)

barrier insulator (p. 409)

chromatin remodeling (p. 395)

coactivator (p. 394)

constitutive heterochromatin (p. 406)

corepressor (p. 397)

DNA methylation (p. 402)

dosage compensation (p. 409)

enhanceosome (p. 398)

enhancer (p. 388)

enhancer-blocking insulator (p. 401)

epigenetic inheritance (p. 402)

epigenetic mark (p. 402)

epigenetic silencing (p. 406)

euchromatin (p. 405)

gene silencing (p. 405)

genomic imprinting (p. 402)

hemimethylation (p. 411)

heterochromatin (p. 405)

heterochromatin protein-1 (HP-1) (p. 407)

histone code (p. 397)

histone deacetylase (HDAT) (p. 397)

histone tail (p. 396)

hyperacetylation (p. 397)

hypoacetylation (p. 397)

maternal imprinting (p. 402)

Mediator complex (p. 394)

monoallelic inheritance (p. 402)

paternal imprinting (p. 402)

pheromone (p. 399)

position-effect variegation (PEV) (p. 406)

promoter-proximal element (p. 388)

reporter gene (p. 392)

synergistic effect (p. 398)

upstream activating sequence (UAS)(p. 388)

BASIC PROBLEMS

1. What analogies can you draw between transcriptionaltrans-acting factors that activate gene expression ineukaryotes and the corresponding factors in bacteria?Give an example.

2. Contrast the states of genes in bacteria and eukaryoteswith respect to gene activation.

3. Predict and explain the effect on GAL1 transcription,in the presence of galactose alone, of the followingmutations:

a. Deletion of one Gal4-binding site in the GAL1 UASelement.

b. Deletion of all four Gal4-binding sites in the GAL1UAS element.

c. Deletion of the Mig1-binding site upstream of GAL1.

d. Deletion of the Gal4 activation domain.

e. Deletion of the GAL80 gene.

f. Deletion of the GAL1 promoter.

g. Deletion of the GAL3 gene.

4. How is the activation of the GAL1 gene prevented inthe presence of galactose and glucose?

5. What are the roles of histone deacetylation and histoneacetylation in gene regulation, respectively?

6. An ` strain of yeast that cannot switch mating type isisolated. What mutations might it carry that wouldexplain this phenotype?

7. What genes are regulated by the α1 and α2 proteins inan ` cell?

8. What are Sir proteins? How do mutations in SIR genesaffect the expression of mating-type cassettes?

9. What is meant by the term epigenetic inheritance? Whatare two examples of such inheritance?

Problems

413Problems

10. What is an enhanceosome? Why could a mutation inany one of the enhanceosome proteins severely reducethe transcription rate?

11. Why are mutations in imprinted genes usually dominant?

12. What features distinguish an epigenetically silencedgene from a gene that is not expressed, owing to analteration in its DNA sequence?

13. What mechanisms are thought to be responsible forthe inheritance of epigenetic information?

14. What is the fundamental difference in how bacterialand eukaryotic genes are regulated?

15. Why is it said that transcriptional regulation in eukary-otes is characterized by combinatorial interactions?

16. The following diagram represents the structure of a genein Drosophila melanogaster; blue segments are exons, andyellow segments are introns.

a. Which segments of the gene will be represented inthe initial RNA transcript?

b. Which segments of the gene will be removed byRNA splicing?

c. Which segments would most likely bind proteinsthat interact with RNA polymerase?

CHALLENGING PROBLEMS

17. The transcription of a gene called YFG (your favoritegene) is activated when three transcription factors (TFA,TFB, TFC) interact to recruit the coactivator CRX. TFA,TFB, TFC, and CRX and their respective binding sitesconstitute an enhanceosome located 10 kb from thetranscription start site. Draw a diagram showing howyou think the enhanceosome functions to recruit RNApolymerase to the promoter of YFG.

18. A single mutation in one of the transcription factors inProblem 17 results in a drastic reduction in YFG tran-scription. Diagram what this mutant interaction mightlook like.

19. Diagram the effect of a mutation in the binding site forone of the transcription factors in Problem 17.

20. How does an epigenetically silenced gene differ from amutant gene (a null allele of the same gene)?

21. What are epigenetic marks? Which are associated withheterochromatin? How are epigenetic marks thoughtto be interpreted into chromatin structure?

A

Enhancer Promoter

B C D E F G H I J K L

Enhancer

22. You receive four strains of yeast in the mail and theaccompanying instructions state that each strain con-tains a single copy of transgene A. You grow the fourstrains and determine that only three strains express theprotein product of transgene A. Further analysis revealsthat transgene A is located at a different position in theyeast genome in each of the four strains. Provide anhypothesis to explain this result.

23. In Neurospora, all mutants affecting the enzymes car-bamyl phosphate synthetase and aspartate transcar-bamylase map at the pyr-3 locus. If you induce pyr-3mutations by ICR-170 (a chemical mutagen), you findthat either both enzyme functions are lacking or onlythe transcarbamylase function is lacking; in no case isthe synthetase activity lacking when the transcarbamy-lase activity is present. (ICR-170 is assumed to induceframeshifts.) Interpret these results in regard to a pos-sible operon.

24. You wish to find the cis-acting regulatory DNA ele-ments responsible for the transcriptional responses oftwo genes, c-fos and globin. Transcription of the c-fosgene is activated in response to fibroblast growth factor(FGF), but it is inhibited by cortisol (Cort). On theother hand, transcription of the globin gene is notaffected by either FGF or cortisol, but it is stimulatedby the hormone erythropoietin (EP). To find the cis-acting regulatory DNA elements responsible for thesetranscriptional responses, you use the following clonesof the c-fos and globin genes, as well as two “hybrid”combinations (fusion genes), as shown in diagram 1.The letter A represents the intact c-fos gene, D repre-sents the intact globin gene, and B and C represent thec-fos–globin gene fusions. The c-fos and globin exons (E)and introns (I) are numbered. For example, E3(f) is thethird exon of the c-fos gene and I2(g) is the secondintron of the globin gene. (These labels are provided tohelp you make your answer clear.) The transcriptionstart sites (black arrows) and polyadenylation sites (redarrows) are indicated.

Ap p

p

E1(f)

E1(g)

I1(f)

I1(g) I2(g)

I2(f)

E2(f)

E2(g) E3(g)

E3(f)

p

p

p

p

p

B

C

D

Diagram 1.


You introduce all four of these clones simultaneouslyinto tissue-culture cells and then stimulate individualaliquots of these cells with one of the three factors. Gelanalysis of the RNA isolated from the cells gives the following results. The levels of transcripts produced

from the introduced genes in response to various treat-ments are shown; the intensity of these bands is pro-portional to the amount of transcript made from a par-ticular clone. (The failure of a band to appear indicatesthat the level of transcript is undetectable.)

a. Where is the DNA element that permits activationby FGF?

b. Where is the DNA element that permits repressionby Cort?

c. Where is the DNA element that permits inductionby EP? Explain your answer.

Clone

A

Notreatment FGF Cort EP

B

C

D

Diagram 2.

regulation of gene expression in eukaryotes

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