Chapter 8 : Genome structure, chromatin and the nucleosome. (Watson)

Effects of histone tail modifications

(a) The effect on the association with nucleosome-bound DNA. Unmodified and methylated histone tails are thought to associate more tightly with nucleosomal DNA than acetyiated histone tails, (b) Modification of histone tails creates binding sites for chromatin-modifying enzymes.

How is DNA is rapidly repackaged into nucleosomes in an ordered series of events.

Although the replication of DNA requires the nucleosome disassembly, the DNA is rapidly repackaged into nucleosomes in an ordered series of events. As discussed above, the first step in the assembly of a nucleosome is the binding of an H3-H4 tetramer to the DNA. Once the tetramer is bound, two H2A-H2B dimers associate to form the final nucleosome. Hi joins this complex last, presumably during the formation of higher-order chromatin assemblies.

What is CAF-I ?

CAF-1 (chromatin assembly factor-1) is a complex, including Chaf1a (p150), Chaf1b (p60) and p50 subunits that assembles histone tetramers onto replicating DNA in vitro, This complex is histone chaperone involved in creating cellular memory of somatic cell identity - cellular differentiation. CAF-1 mediates the first step in nucleosome formation by tetramerization and depositing newly synthesized histone H3/H4 onto DNA rapidly behind replication forks[4] Several studies have shown that the interaction between CAF-1 and PCNA, which stabilizes CAF-1 at replication forks, is important for CAF-1's role in nucleosome assembly

Question 6. How does the sister chromatid cohesion ensure that each daughter cell receives one copy of each chromosome?

Cohesion holds sister chromatids together during S phase and early stages of mitosis. During late mitosis (anaphase), cohesion is eliminated so that the microtubules attached to kinet- ochores that assemble at the centromere separate sister chroma- tid pairs into the daughter cells.

Heterochromatin

Heterochromatin was characterized by dense staining with a variety of dyes and a more condensed appearance, whereas euchromatin had the opposite characteristics, staining poorly with dyes and having a relatively open structure. As our molecular understanding of genes and their expression advanced, it became clear that heterochro- matic regions of chromosomes had very limited gene expression. heterochromatin is associated with particular chromosomal regions, including the telomere and the centromere, and is important for the function of both of these key chromosomal elements. As we shall learn when gene expression is discussed, keeping a gene turned off can be just as important as turning a gene on.

Question 10. Name the types of bonds that occur between histone proteins and DNA and the region of DNA where these bonds form. Are these interactions sequence-specific? Explain why or why not.

Hydrogen bonds form primarily between the his- tone proteins and the phosphodiester backbone near the minor groove and additionally between the bases of the minor groove. These interactions are not sequence-specific. DNA throughout the genome wraps around the histone proteins. Proteins that interact with the minor groove of DNA are much less likely to interact in a sequence-specific manner. In contrast, interactions with the major groove of the DNA generally make sequence-spe- cific interactions (Chapter 4).

solenoid model

In the solenoid model, the nucleosomal DNA forms a superhelix containing approximately six nucleosomes per turn (Fig. 8-30a). This structure is supported by both electron microscopy and X-ray diffraction studies, which indicate that the 30-nm fiber has a helical pitch of ~11 nm. This distance is also the approx- imate diameter of the nucleosome disc, suggesting that the 30-nm fiber is composed of nucleosome discs stacked on edge in the form of a helix

Fig 8-43 , explain process of inheritance of histones

Inheritance of histones after DNA replication. As the chromosome is replicated, histones that were associated with the parental chromosome are differently distributed. The histone H3-H4 tetramers are randomly transferred to one of the two daughter strands but do not enter into the soluble pool of H3-H4 tetramers. Newly synthesized H3-H4 tetramers form the basis of the nucleosomes on the strand that does not inherit the parental tetramer. In contrast, H2A and H2B dimers are re- leased into the soluble pool and compete for H3-H4 association with newly synthe- sized H2A and H2B. As a consequence of this type of distribution, on average, every second H3-H4 tetramer on newly synthe- sized DNA will be derived from the parental chromosome. These tetramers will include all of the modifications added to the paren- tal nucleosomes. The H2A-H2B dimers are more likely to be derived from newly synthe- sized protein.

Do all bacteria need to maintain their DNA in a negatively supercoiled state? What is reverse gyrase?

Not all bacteria need to maintain their DNA in a negatively supercoiled state. Bacteria that prefer to grow at very high temperatures (>80°C) must expend energy to prevent their DNA from unwinding due to thermal dena- turation. These organisms have a different topoisomerase called reverse gyrase. Consistent with its name, reverse gyrase increases the linking num- ber of relaxed DNA in the presence of ATP. By keeping the genome positively supercoiled, reverse gyrase counteracts the effect of thermal dena- turation that would ordinarily result in many regions of the genome being unwound.

FIGURE 8-35 Nudeosome movement catalyzed by nudeosome-remodeling activities. How does DNA move?

Nucleosome movement by sliding along a DNA molecule exposes sites for DNA-binding proteins. (Middle) Nudeosome-remodeling complexes can also eject a nucleosome from the DNA creating larger nucleosome-free regions of DNA. (Bottom) A subset of nudeosome-remodeling complexes cata- lyzes the exchange of H2A/H2B dimers with either unmodified or variant H2A/H2B dimers (e.g., H2A-X). Transfer and sliding !

What is PCNA?

Proliferating cell nuclear antigen (PCNA) is a DNA clamp that acts as a processivity factor for DNA polymerase δ in eukaryotic cells and is essential for replication. this factor forms a ring around the DNA duplex and is responsible for holding DNA polymerase on the DNA during DNA synthesis. After the polymerase is finished, PCNA is released from the DNA polymerase but still encircles the DNA. In this condition, PCNA is available to interact with other proteins. CAF-I associates with the released PCNA and assembles H3-H4 preferentially on the PCNA-bound DNA. Thus, by associating with a component of the DNA replication machi- nery, CAF-I is directed to assemble nucleosomes at sites of recent DNA replication.

bromodomains, chromodomains, TUDOR domains, and PHD (for plant homeodomain) fingers do what?

Protein domains called bromodomains, chromodomains, TUDOR domains, and PHD (for plant homeodomain) fingers specifically recognize modified forms of histone tails. Bromodomain-containing pro- teins interact with acetyiated histone tails, and chromodomain-TUDOR domains and PHD-finger-containing proteins interact with methylated his- tone tails. Yet another protein domain, called a SANT domain, has the oppo- site property. SANT-domain-containing proteins interact preferentially with unmodified histone tails. Consistent with these protein domains being important for interpreting histone modifications, in many instances pro- teins containing these domains specifically recognize the modified form of only one of the many possible sites of histone modification. For example, the protein HPl contains a chromodomain that will bind to methylated lysine 9 of histone H3 but not to any other site of histone methylation.

Question 13. Review Box 8-2, Figure 1. For each of the DNAs described below, predict where the DNA would migrate on an agarose gel. Use the image of a gel below as a guide.

Question 13. A. The DNA would migrate like the top band shown in the gel for relaxed DNA. B. TheDNAwouldmigratelikethetopbandshowninthegelfor relaxed DNA. C. The DNA would migrate like the middle band shown in the gel for supercoiled DNA. D. The DNA would migrate like the bottom band shown in the gel for supercoiled DNA (more supercoiled than part C).

Question 7. For a diploid human cell, state how many copies of each chromosome are present in each cell (or soon to be daughter cell). Start of mitosis End of mitosis Start of meiosis End of meiosis I End of meiosis II

Start of mitosis, four copies; end of mitosis, two cop- ies; start of meiosis, four copies; end of meiosis I, two copies; end of meiosis II, one copy.

Histone chaperones

Studies of nucleosome assembly under physiological salt concentrations identified factors required to direct the assembly of histones onto the DNA. These factors are negatively charged proteins that form complexes with either H3-H4 or H2A-H2B dimers (see Table 8-8) and escort them to sites of nucleosome assembly. Because they act to keep histones from interacting with the DNA nonproductively, these factors have been referred to as his- tone chaperones

How do the histone chaperones direct nucleosome assembly to sites of new DNA synthesis?

Studies of the histone H3-H4 chaperone CAF-I reveal a likely answer. Nucleosome assembly directed by CAF-I requires that the target DNA be replicating. Thus, replicating DNA is marked in some way for nucleosome assembly.Interestingly, this mark is gradually lost after rep- lication is completed. Studies of CAF-I-dependent assembly have deter- mined that the mark is a ring-shaped sliding clamp protein called PCNA.

Heterochromatin vs Euchromatic

The difference between heterochromatin structure and euchromatin structure is how the nucleosomes in these different chromosomal regions are (or are not) assembled into larger assemblies. It has become clear that hetero- chromatic regions are composed of nucleosomal DNA assembled into

Gyrase

This is accomplished by a specialized topoisomerase called gyrase that has the ability to introduce negative superhelicity into relaxed DNA by reducing the linking number. For example, in E. coli cells, gyrase action results in the genome having an average superhelical density of approximately -0.07. The addition of negative supercoils into otherwise relaxed DNA is an energy-requiring reaction. Consistent with this, gyrase requires ATP to introduce negative supercoils. In the absence of ATP, gyrase can only relax DNA (e.g., reduce the linking number of positively super- coiled DNA).

site of the translocation

This loop is proposed to propagate on the surface of the histone octamer until it reaches the other end of the nucleosomal DNA. Although this loop movement could potentially proceed in either direction, it is thought that other interactions between the nudeosome-remodeling complex and the nucleosomal DNA prevent propagation toward the proxi- mal DNA linker (which would result in no change in nucleosome position- ing). Importantly, this approach does not demand that all interactions between the histone octamer and nucleosomal DNA be broken simultane- ously

Solenoid v/s Zigzag

Two models for the 30-nm chromatin fiber. In each panel, the left-hand view shows the side of the fiber, and the right-hand view shows a view down the central axis of the fiber, (a) The solenoid model. Note that the linker DNA does not pass through the central axis of the superhelix and that the sides and entry and exit points of the nucleosomes are relatively inaccessible, (b) The "zigzag" model. In this model, the linker DNA frequently passes through the central axis of the fiber, and the sides and even the entry and exit points are more accessible.

Question 11. Explain why stored negative superhelicity from packaging DNA into nucleosomes is advantageous for cellular functions.

When nucleosomes are disassembled, the released DNA becomes negatively supercoiled. In this state, DNA unwinding is favored. The cellular processes of DNA replica- tion, transcription, and recombination require DNA unwinding. So when nucleosomes are removed during one of these proc- esses, the involved proteins can both gain access to the DNA and more easily unwind the DNA.

Euchromatic

euchromatic regions showed higher levels of gene expression, suggesting that these different structures were connected to global levels of gene expression.

How to map nucleosome location accurately. (Process)

it is important to isolate the cellular chromatin and treat it with the appropriate amount of micrococcal nuclease with minimal disruption of the overall chromatin structure. This is typically achieved by gently lysing cells while leaving the nuclei intact. The nuclei are then briefly treated (typically for 1 min) with several different concentrations of micrococcal nuclease, a protein small enough to diffuse rapidly into the nucleus. The goal of the titration is for micrococcal nuclease to cleave the region of interest only once in each cell. Once the DNA has been digested, the nuclei can be lysed, and all of the protein can be removed from the DNA. The sites of cleavage (and, more importantly, the sites not cleaved) leave a record of the protein bound to DNA. To identify the sites of cleavage in a particular region, it is necessary to create a defined end point for all of the cleaved frag- ments and exploit the specificity of DNA hybridization. To create a defined end point, the purified DNA from each sample is cut with a restriction enzyme known to cleave adjacent to the site of interest. After separation by size using agarose gel electropho- resis, the DNA is denatured and transferred to a nitrocellulose membrane such that its position in the gel is retained. A labeled DNA probe of specific sequence is then hybridized to the nitrocellulose-bound DNA (this is called a Southern blot )

Lysine and arginine do .. to ATT

methylation

Serine and threonine do to ATT( Amino terminal tail)

phosphorylation

how does nucleosomal DNA sliding catalyzed by nucleosome- remodeling complexes work?

(a) The model pro- poses that a DNA translocating domain of the ATP-hydrolyzing subunit of the nucle- osome-remodeling complex binds the nucleosomal DNA two helical turns from the central dyad (e.g., at position 52 out of the total of 1 47 bp associated with the nucleosome). Other subunits of the nucleosome remodeling complex bind tightly to the histones. The illustration shows each of the contacts between the DNA and the histones from the dyad to the closest unbound DNA (one contact per helical turn, seven of the 1 4 total), (b) Using the ATP-dependent DNA translocating activity, the nucleosome-re- modeling complex first pulls the DNA from the nearest linker domain into the nucleosome. This breaks the five histone-DNA contacts between the ATP-hydrolyzing subunit and the linker DNA (broken contacts are shown in black, intact contacts in white) and creates a loop of DNA on the opposite side of the translocase domain, (c) The broken contacts re-form with thetranslocated DNA( positions 1 -5), leaving the loop of DNAnextto the ATP-hydrolyzing subunit (disrupting the contacts at position 6). (d) To remove the loop of DNA, the model proposes thatthe loop moves like a "wave" across the surface of the histones, breaking one or two contacts at a time (first contact 6 and then 7, etc.) until all of the contacts have re-formed with the appropriate amount of DNA between them, at which point the excess DNA is no longer present within the histone-associated DNA and the nucleosome has shifted its position on the DNA. (e) After the loop of DNA has propagated to the distal linker, and the nucleosome has shifted its position on the DNA.

look at pg.256 for Gel fig, and question

A. The histone deacetylase binds nucleosome bound-DNA (lanes 1, 2, 3, and 4 compared to lane 5). Assuming the his- tone deacetylase is a monomer, two deacetylases are capable of binding the nucleosome-bound DNA at the same time (two higher migrating bands in lanes 2, 3, and 4). The histone deacetylase seems to recognize nucleosomes that are methy- lated at lysine 36 of histone H3 better than unmethylated nucleosomes (lanes 1 and 2 vs. 3 and 4). B. Based on this data, the histone deacetylase likely includes a chromodomain to interact with methylated histone H3.

Lysine does...to the amino terminal tail?

Acetylation

How does chromatin assembly factors facilitate the assembly of nucleosomes?

After the replication fork has passed, chromatin assembly factors chaperone free H3-H4 tetramers (e.g., CAF-I) and H2A-H2B dimers (NAP-1) to the site of newly replicated DNA. Once at the newly replicated DNA, these factors transfer their associated histone to the DNA. CAF-I is recruited to the newly replicated DNA by interactions with DNA sliding clamps. These ring-shaped, auxiliary replication factors encircle the DNA and are released from the replication machinery as the replication fork moves.

Question 8. For humans, what cells undergo mitosis? What cells undergo meiosis?

All cells that grow and divide (somatic cells and germ cells) use mitosis. Only cells (germ cells) that produce egg and sperm cells go through meiosis.

Zigzag model

An alternative model for the 30-nm fiber is the "zigzag" model (Fig. 8-30b). This model is based on the zigzag pattern of nucleosomes formed upon Hi addition. In this case, the 30-nm fiber is a compacted form of these zigzag nucle- osome arrays. A recent X-ray structure of a single DNA molecule participating in four nucleosomes and biophysical studies of the spring-like nature of iso- lated 30-nm fibers support the zigzag model. Unlike the solenoid model, the zigzag conformation requires the linker DNA to pass through the central axis of the fiber in a relatively straight form (see Fig. 8-30b). Thus, longer linker DNA favors this conformation. Because the average linker DNA varies between different species (see Table 8-4), the form of the 30-nm fiber may not always be the same, and both forms of the 30-nm fiber may be found in cells depending on the local linker DNA length.

How does inheritance of parental H3 H4tetramers facilitates the inheritance of chromatin states?

As a chromosome is replicated, the distribution of the parental H3-H4 tetramers results in the daughter chromo- somes receiving the same modifications as the parent. The ability of these modifications to recruit enzymes that perform the same modifications facilitates the propagation of the modification to the two daughter chro- mosomes. For simplicity, acetylation is shown on the core regions of the histones. In reality, this modification is generally on the amino- terminal tails. The ability of these modified histones to recruit enzymes that add similar modifications to adjacent nucleosomes (see the discussion of modified histone-binding domains above) provides a simple mechanism to main- tain states of modification after DNA replication has occurred (Fig. 8-44). Such mechanisms are likely to play a critical role in the inheritance of chromatin states from one generation to another. fig 8-44

Why do nucleosomes alter the topological state of the DNA they include?

As described in Chapter 4, there are two forms of writhe that can contribute to the formation of supercoiled DNA: toroi- dal and interwound (also referred to as plectonemic). The wrap- ping of DNA around the histone octamer is a form of toroidal writhe. The handedness of the writhe controls whether it intro- duces positive or negative supercoils (i.e., increases or decreases the linking number of the associated DNA). For toroidal writhe, left-handed wrapping induces negative superhelicity (for inter- wound writhe, the opposite is true; right-handed pitch is associ- ated with negative superhelicity). Thus, the left-handed toroidal wrapping of DNA around the nucleosome reduces the linking number of the associated DNA. For this reason, nucleosomes preferentially form with DNA that has negative superhelical density. In contrast, assembling nucleosomes on DNA that has positive superhelical density is very difficult.

Some Nucleosomes Are Found in Specific Positions: Nucleosome Positioning

Because of their sequence-nonspecific and dynamic interactions with DNA, most nucleosomes are not fixed in their locations. But there are occasions when restricting nucleosome location, or positioning nucleosomes as it is called, is beneficial. Typically, positioning a nucleosome allows the DNA- binding site for a regulatory protein to remain in the accessible linker DNA region. In many instances, such nucleosome-free regions are larger to allow the binding sites for multiple regulatory proteins to remain accessi- ble. For example, the regions upstream of active transcription start sites are frequently associated with large nucleosome-free regions. Nucleosome positioning can be directed by DNA-binding proteins or particular DNA sequences.

Nucleosome Arrays Can Form More Complex Structures: What is the 30-nm Fiber

Binding of Hi stabilizes higher-order chromatin structures. In the test tube, as salt concentrations are increased, the addition of histone Hi results in the nucleosomal DNA forming a 30-nm fiber. This structure, which can also be observed in vivo, represents the next level of DNA compaction. Importantly, the incorporation of DNA into this fiber makes the DNA less accessible to many DNA-dependent enzymes (such as RNA polymerases).

Question 12. Which protein domain(s) recognize the acetylation of histone amino-terminal tails? Which protein domain(s) recog- nize methylated histone amino-terminal tails?

Bromodomain recognizes acetylation. Chromodo- mains, TUDOR-domains, and PHD fingers recognize methyla- tion. (SANT domains recognize unmodified histone tails.)

How do the fragment sizes reveal the location of positioned nu- cleosomes?

DNA associated with positioned nucleosomes will be resistant to micrococcal nuclease digestion, leaving an ~1 60- 200-bp region of DNA that is not cleaved. This will appear as a gap in the ladder of DNA bands detected on the Southern blot. The location of these gaps reveals the position of the nucleosomes adjacent to the restriction site/labeled DNA probe.

Question 1 . List at least three properties that differ between the chromosome makeup in E. coli compared to human cells

E. coli cells have one chromosome, whereas hu- mans (Homo sapiens) have 22 plus the sex chromosomes. E. coli cells only keep the DNA in the haploid state, whereas human cells have a diploid set of chromosomes. The E. coli chromosome is circular, whereas the human chromosomes are linear. The size of the human genome is almost 700 larger than the genome for E. coli cells.

Question 5. Explain why each chromosome in a eukaryotic cell contains multiple origins of replication but includes one and only one centromere.

Eukaryotic chromosomes include multiple origins of replication so that DNA polymerases can complete the duplica- tion of these larger chromosomes in a timely manner. A single centromere per chromosome is the "handle" by which sister chromosomes are moved and separated into the two daughter cells. Without a centromere, one daughter cell may not receive a chromosome or may receive both sister chomosomes. A chro- mosome with more than one centromere per chromosome risks attaching to both mitotic spindles and being pulled to both poles leading to chromosome breakage.

Nucleosome-Remodeling Complexes Facilitate Nucleosome Movement, ( what are nucleosome - remodeling complexes?)

In addition to the intrinsic dynamics shown by the nucleosome, the stability of the histone octamer-DNA interaction is influenced by large protein com- plexes called nucleosome-remodeling complexes. These multiprotein com- plexes facilitate changes in nucleosome location or interaction with the DNA using the energy of ATP hydrolysis. There are three basic types of nucleosome changes mediated by these enzymes (Fig. 8-35). All nucleo- some-remodeling complexes can catalyze the "sliding" of DNA along the surface of the histone octamer. A subset of nucleosome-remodeling com- plexes can catalyze a second, more extreme change in which a histone octamer is ejected into solution or "transferred" from one DNA helix to another. Finally, some of these enzymes can facilitate the exchange of the H2A/H2B dimer within a nucleosome with variants of the dimer (e.g., H2A.X/H2B exchanged for H2A/H2B at double-strand breaks)

Inheritance of histones after DNA replecation

In experiments that differentially labeled old and new histones, it was found that the old histones are present on both of the daughter chromosomes (Fig. 8-43). Mixing is not entirely random, however. H3-H4 tetramers and H2A-H2B dimers are composed of either all new or all old histones. Thus, as the replication fork passes, nucleosomes are broken down into their component subassemblies. H3-H4 tetramers appear to remain bound to one of the two daughter duplexes at random and are never released from DNA into the free pool of histones. In contrast, the H2A-H2B dimers are released and enter the local pool, available for new nucleosome assembly.

What kind of DNA do nucleosomes prefer?

Nucleosomes prefer to bind bent DNA. Specific DNA sequences can position nucleosomes. Because the DNA is bent severely during association with the nucleosome, DNA sequences that position nucleosomes are intrinsically bent. A:T base pairs have an intrinsic tendency to bend toward the minor groove and G:C base pairs have the opposite tendency. Sequences that alternate between A:T- and G:C-rich sequences with a periodicity of ~5 bp will act as preferred nucleosome-binding sites.

How does histone modification alter nucleosome function?

One obvious change is that acetylation and phosphorylation each acts to reduce the over- all positive charge of the histone tails; acetylation of lysine neutralizes its positive charge (Fig. 8-41). This loss of positive charge reduces the affinity of the tails for the negatively charged backbone of the DNA. More impor- tantly, modification of the histone tails affects the ability of nucleosome arrays to form more repressive higher-order chromatin structure.

If nucleosomes store negative superhelicity in eukaryotic cells, what serves the equivalent function in prokaryotic cells?

The answer for many prokaryotic organisms is that the entire genome is maintained in a nega- tively supercoiled state.

Question 2. Explain where the chromosomal DNA is located in prokaryotic versus eukaryotic cells.

The chromosomal DNA is located within the nucle- oid in prokaryotic cells. The chromosomal DNA is located in the nucleus for eukaryotic cells. The nucleus, unlike the nucleoid, is membrane-bound and typically occupies a small fraction of the cell volume.

Nucleosomes Are Assembled Immediately after DNA Replication

The duplication of a chromosome requires replication of the DNA and the reassembly of the associated proteins on each daughter DNA molecule. The latter process is tightly linked to DNA replication to ensure that the newly replicated DNA is rapidly packaged into nucleosomes.

Question 4. Intergenic sequences make up >60% of the human genome. Where do these intergenic sequences come from and what are some of their functions?

The intergenic sequences may have arisen from transposition events. They may encode miRNAs, may serve as regulatory sequences for transcription, or may simply be non- functional sequences such as pseudogenes.

Structure of histone tail modifications ( lysine, arginine, ect) Know acronyms

The molecular structure of the small molecule histone modifications and the class of enzyme responsible are illustrated (histone acetyl transferase [HAT]; histone deacetylase [HDAc]; histone methyltransferase [HMT]; histone demeth- ylase [HDM]). Only the affected amino acid is shown. Currently there is no known histone demethylase for arginine methylation, suggesting that these marks are only lost when the histone is removed from the DNA.

Question 9. Describe the components of a nucleosome.

The nucleosome consists 147 base pairs of DNA that is wrapped around an octamer of histone proteins. The histone is composed of protein two H2A, two H2B that are together creating a dimer. There are also two H3 and two positively charged histone H4 proteins that form tertadimer. In a nucleosome, about 147 base pairs of DNA are wrapped around an octamer of histone proteins. Each nucleo- some includes two H2A, two H2B, two H3, and two positively charged histone H4 proteins. (H1 is not part of the nucleosome.)

How do modifications of the histone amino-terminal tails alters the function of chromatin?

The sites of known his- tone modifications are illustrated on each histone. Although the types of histone modifications continue to grow, for simplicity, only sites of acetylation, methylation, phosphorylation, and ubiquitinylation are shown. The majority of these modifications occur on the tail regions, but there are occasional modifications within the histone fold (e.g., methylation of lysine 79 of histone H3).

Question 3. Does genome size correlate directly with organism complexity? Explain your reasoning

There is a rough relationship between genome size and organism complexity. Generally, the more complex eukar- yotes have larger genomes than prokaryotes do. The relationship does not always hold true because of different proportions of repetitive sequences. Several examples exist in which a more complex organism has a smaller genome than a simpler organ- ism (e.g., Homo sapiens vs. Fritillaria assyriaca). Accordingly, genome size is a poor predictor of organism complexity. In con- trast, gene number shows a better correlation with organism complexity.

How to duplicate a chromosome?

To duplicate a chromosome, at least half of the nucleosomes on the daughter chromosomes must be newly synthesized. Are all of the old his- tones lost and only new histones assembled into nucleosomes? If not, how are the old histones distributed between the two daughter chromosomes? The fate of the old histones is a particularly important issue given the effects that histone modification can have on the accessibility of the resulting chro- matin. If the old histones were lost completely, then chromosome duplica- tion would erase any "memory" of the previously modified nucleosomes. In contrast, if the old histones were retained on a single chromosome, that chromosome would have a distinct set of modifications relative to the other copy of the chromosome.

Nuclear Scaffolding

Together, the packaging of DNA into nucleosomes and the 30-nm fiber results in the compaction of the linear length of DNA by ~40-fold. This is still insufficient to fit 1-2 m of DNA into a nucleus ~10 -5 m across. Addi- tional folding of the 30-nm fiber is required to compact the DNA further. Although the exact nature of this folded structure remains unclear, one pop- ular model proposes that the 30-nm fiber forms loops of 40-90 kb that are held together at their bases by a proteinaceous structure referred to as the nuclear scaffold (Fig. 8-32). A variety of methods have been developed to identify proteins that are part of this structure, although the true nature of the nuclear scaffold remains mysterious.