Genome Structure, Chromatin and The Nucleosome

I. Introduction

A. Introduction: The Human Genome Project

1. Since the first gene was sequenced in 1972, Molecular Biologists have wished to take a systematic approach to understanding the human genome

2. From the idea of the systematic approach, the Human Genome Project was inaugurated in 1990

a. Billed as Biologys first big project

b. Projected time frame to finish the project was 15 years

3. A Molecular Biologists defines genome as the total set of DNA molecules present in either an organism, a cell or an organelle

4. When the Human Genome and Human Genome project is being discussed, the total set of DNA molecules in the cell is being looked at

a. Chromosomes (24 types)

b. Mitochondrial DNA

5. The goal of the Human Genome project was to acquire fundamental information about our genetic makeup

a. How many genes are present in the human genome

b. How many genes are actually similar to each other in DNA sequence

c. How many base pairs are present in the Human Genome

d. How many base pairs are present in a chromosome

e. How many base pairs are not part of genes (intervening sequences)

f. Understand how genes interact with each other

6. Although the basic goals of the project are listed above, the major justification for a project lasting 13 years and costing $2.7 Billion were the possible medical benefits

a. Allow for more accurate diagnosis of inherited disorders

b. Provide a framework for the development of new therapies

c. Personalized medicine: Wide-scale application of mutation screening which better allows a move from treating advanced symptoms of disease to preventative disease

7. With personalized medicine comes significant ethical, legal and societal implications-~20% of the $2.7 Billion spent went into this category

B. Introduction: Human Genome Project Findings

1. The Human genome consists of roughly 3 Billion base pairs spread across 46 chromosomes (23 pairs)

2. Human chromosomes can range in size from 50 million to 300 million base pairs

a. The smallest chromosome is chromosome 21

b. The largest chromosome is chromosome 1

3. The Human Genome has roughly 20,000 genes (protein coding only)

a. Chromosome 1 has the most genes at about 3000

b. The Y chromosome has the least number of genes having only 70-200 genes

c. Chromosome 5 has lowest gene density

d. Chromosome 19 has the highest gene density

4. The 20,000 genes is only a rough estimate and was determined through using computer based methods

a. Genes that encode proteins

b. Evidence of evolutionarily conserved gene sequences

5. The human genome does not have many more genes than other less complex organisms

a. C. elegans has 20,000 genes and a genome size of 100 million base pairs

b. Drosophila has 14,000 genes and a genome size of 165 million base pairs

6. The reason why humans can get away with so few genes is that on average one human gene encodes not just one protein but actually three

7. Introduction: Human Genome Project Findings: Supplemental Figure

II. Fitting the Human Genome Into The Cell

A. Fitting the Human Genome Into The Cell: Introduction

1. Fitting the 3.0 billion base pair genome into a human cell nucleus is a difficult proposition for the cell

2. If you were to stretch out the 3.0 billion bases, one would find that the DNA molecule would be about 2m long

a. A human cell nucleus is going to be approximately 10 m in diameter

b. Represents a packing ratio of 1000 10,000 fold

3. Another way of looking at this is if we were to take a string the length of the Willis Tower, and fit it into a tennis ball, how easy would it be to do that?

4. Somehow, the cell must have a way to actually package all of the DNA into the nucleus

5. The way in which eukaryotic DNA is packaged in the cell nucleus is one of the wonders of macromolecular structure

G. Michael Blackburn, Nucleic Acids in Chemistry and Biology (1990) p. 65

6. In order to fit the large human genome into a relatively small nucleus human cells have an intricate mechanisms for packaging the DNA

a. Dividing the genome into linear chromosomes

b. Winding the DNA around proteins called histones

B. Fitting the Human Genome Into The Cell: Dividing the Genome Into Chromosomes

1. By definition, a chromosome is a structure of a long DNA molecule and its associated proteins, which carries all or part of the hereditary information for an organism

2. All organisms and viruses carry their genomes in the form of chromosomes

3. The number of chromosomes an organism has varies, and does not correlate with the complexity of the organism (see table 7-1)

4. In general, most prokaryotes have one circular chromosome

5. Most eukaryotes have multiple linear chromosomes of varying size and copy number

a. Most eukaryotes have 2 copies of each chromosome (homologs)

b. Some eukaryotes may have only one copy of each chromosome

c. In rare cases, some eukaryotes may have more than two copies of each chromosome (Tetrahymena)

C. Fitting the Human Genome Into The Cell: The Purpose of Dividing The Genome Into Chromosomes

1. Dividing the genome into chromosomes is actually very important for all organism

a. Allows for easier compaction to fit the DNA into the nucleus (or nucleoid)

b. Protects the DNA from damage

c. Provides an easy way to transmit the DNA to daughter cells when a cell divides

2. For sexually reproducing organisms, having chromosomes allows for easy transmission of half the genome to each gamete in a process called meiosis

3. By diving the genome into chromosomes gives the cell the ability to control DNA structure, and by proxy control gene expression

a. Allows some genes to be expressed

b. Allows some genes to not be expressed

D. Fitting the Human Genome Into The Cell: The Human Chromosomal Content

1. Humans have a total of 46 chromosomes, with two copies of each chromosome

2. Humans have 23 pairs of chromosomes

3. Each member of a chromosome pair (homolog) will have the exact same genes

4. However, each member will not necessarily have the same sequence (variant or allele) for each gene

a. In cases where both chromosomes have the same sequence of a specific gene, the individual is a homozygote for that gene

b. In cases where both chromosomes have different sequence for a specific gene, the individual is a heterozygote for that gene

D. Fitting the Human Genome Into The Cell: The Human Chromosomal Content and Sickle Cell Anemia (Supplemental Figure)

E. Fitting the Human Genome Into The Cell: The Human Chromosomal Content and Sickle Cell Anemia (Supplemental Figure)

F. Fitting The Genome Into The Cell: Chromosome Structure

1. Eukaryotic chromosomes have several important features

a. Allow for greater DNA stability

b. Allow for efficient gene expression

c. Allow for control of gene expression

2. Each end of a linear chromosome is capped with a telomere for maintenance of chromosome stability

3. Throughout each chromosome reside origins of replication, which direct the start of DNA synthesis

4. Each eukaryotic chromosome will have a more or less centrally located centromere, which is important for chromosome segregation in mitosis/meiosis

a. An elaborate protein complex known as the kinetochore binds the centromere

b. Microtubules will in turn interact with the kinetochore to pull the chromosomes to opposite poles of the cell during mitosis

5. The centromere divides the chromosome into arms

a. p-arm is the short arm (petit)

b. q-arm is the long arm

6. Fitting The Genome Into The Cell: Chromosome Structure (Supplemental Figure)

7. It is not the norm to have chromosomes with multiple centromeres

8. However, this does occasionally happen in humans

9. Having a chromosome with two centromeres creates a problem during mitosis as each centromere may be getting pulled in opposite directions leading to chromosomal breakage

10. Having no centromeres is also not the norm in human populations, and results in completely random segregation during mitosis

III. Chromatin Formation

A. Winding The DNA Into Chromatin: Introduction

1. Although dividing the genomic DNA into chromosomes helps with packaging, it clearly is not enough

2. In addition to having chromosomes, the chromosomal DNA is wound around proteins to form chromatin

3. Chromatin is the genomic DNA and its associated proteins

4. By winding the chromosomal DNA around proteins, it is very easily fit into the nucleus of the cell

5. To form chromatin, the genomic DNA must be wound around a specific structure called a nucleosome

6. Each chromosome will have many nucleosomes associated with it

7. Each nucleosome contains the following

a. Chromosomal DNA

b. Histone proteins around which the DNA is wound

8. The nucleosome allows for compaction of the DNA about 6-fold

9. This is only the first stage in compaction, as the genome needs to be compacted 100-10,000 fold to efficiently fit into the nucleus

B. Winding The DNA Into Chromatin: The Histone Proteins

1. Each nucleosome consists of a core of eight histone proteins and the chromosomal DNA wound around them

2. The histone proteins were discovered back in 1928 by Albrecht Kossel, who isolated from goose erythrocytes

a. Discovered that they were basic (have a positive charge)

b. Since then histones have been discovered in many other organisms

3. Eukaryotic nucleosomes generally have a total of 5 abundant histones divided into two sub-categories

a. Core histones, which show strong conservation

b. Linker histones

4. The Core histones, form the core of the nucleosome, and will form an octamer (complex of 8 protiens) and are found in the cell in roughly equal amounts

a. Histone H2A

b. Histone H2B

c. Histone H3

d. Histone H4

5. The core histones to be basic,

a. Rich in arginine

b. Rich in lysine

6. The protein core of a nucleosome is a disk-shaped structure that assembles in an order fashion only in the presence of DNA (discovered through in vitro experiments)

7. Without DNA, the core histones will create intermediate assemblies, which are mediated by the conserved histone fold domain (discovered through in vitro experiments

B. Winding The DNA Into Chromatin: The Core Histone Proteins Come Together To Form an Octomer

1. In order to form the core, the Histone H3 and Histone H4 proteins form a central tetramer

a. H3 and H4 proteins form heterodimers

b. Two H3/H4 heterodimers interact to form the tetramer

2. Once the tetramer H3/H4 tetramer forms, it binds to DNA to start the winding process (~147 bp of DNA)

3. Next, Histones H2A and H2B form heterodimers which then bind onto each end of the H3/H4 tetramer

4. The DNA that is wound around the core histones is considered the core DNA and is wound ~1.65 times around the octomer like thread around a spool

5. The fifth histone present in a nucleosome is the linker histone

6. There are two possible types of linker histones

a. Histone H1

b. Histone H5 (only found in avian erythrocytes)

7. Compared to the core histones, the linker histones are larger, with a molecular weight of greater than 20 kD

8. Histone H1 is half as abundant in the cell as the core histones

9. Histone H1 is not part of the core nucleosome particle, instead it binds to the linker DNA

10. If one counts the DNA wound around the core nucleosomes and the linker DNA which is associated with Histone H1 there are about 180 bp (in humans)

11. The role of the linker Histone H1 is to induce tigher wrapping of the DNA around the nucleosome

12. Histone H1 binding relative to the nucleosome is different than the core histones

a. Only one Histone H1 protein binds

b. Histone H1 binding sites are located asymetrically with respect to the nucleosome

13. The two Histone H1 binding sites are as follows:

a. The linker DNA at one end of the nucleosome (only linker DNA on one side of the nucleosome is protected from mnase digestion)

b. At the mid-point of the 147 bp associated with the core histone complex

14. In the end, the whole nucleosome includes the following

a. Core histones and the DNA that is wrapped around them

b. Linker histone, and the DNA that is covered by it

C. Winding The DNA Into Chromatin: Determining How Many Base Pairs of DNA Are Associated With A Nucleosome

1. In order to figure out the amount of DNA wound around a nucleosome, the enzyme Micrococcal nuclease was used

a. Micrococcal nuclease will digest free double stranded DNA

b. Micrococcal nuclease will not digest double stranded DNA associated with proteins

2. To perform the Mnase experiment chromatin was isolated and subjected to treatment with a low concentration of micrococcal nuclease

3. Once treatment was finished, protein components were removed, leaving only the DNA that was associated with the nucleosomes

4. The DNA was then run on an agarose gel

5. On the agarose gel, they saw DNA bands at multiples of 180 bp

6. Winding The DNA Into Chromatin: Determining How Many Base Pairs of DNA Are Associated With A Nucleosome

7. As we saw before, the 180 bp refers to the amount of DNA wound around the core nucleosome and the linker DNA associated with Histone H1

8. In order to figure out how much DNA was wound around the core nucleosome, the chromatin was exposed to more extensive micrococcal nuclease treatment

a. The linker Histone (Histone H1) is not tightly bound to the DNA

b. Extensive treatment with Mnase results in the release of Histone H1

c. The release of Histone H1 allows for the Mnase to in turn degrade the rest of the DNA Mnase was bound to

9. After extensive treatment only a 147 base pair band was seen-indicative of the amount of DNA wound around the core nucleosome

D. Winding The DNA Into Chromatin: The Winding of the DNA Around The Core Histones

1. Although the nucleosome is not perfectly symmetryical, it does have a twofold axis of symmetry called the dyad axis

a. Imagine the nucleosome as a clock, with the midpoint of the 147 bp of DNA at the 12 oclock position

b. This places the ends of the DNA at the 11 oclock and 1 oclock positions

2. Throughout the nucleosome, there are 14 distinct sites of contact between the core histones and the DNA

3. An interaction occur each time the minor groove of the DNA faces the histone octamer

4. Winding The DNA Into Chromatin: The Winding of the DNA Around the Core Histones

5. The core histone proteins bind the DNA through H-bonds in a sequence non-specific manner (minor groove binding)

a. Majority of H-bonds are between the Histones and the oxygens in the phospho-diester backbone

b. Few H-bonds are formed between the histones and the nitrogenous bases

6. In order to get the DNA wound around a nucleosome, it must be bent

7. In general DNA is a rigid molecule due to the fact the fact the negatively charged phosphates in the backbone (resists bending)

8. The total number of Hydrogen bonds between the core histones and the DNA is twice as many as a typical sequence specific DNA binding protein

a. Core histones interact with the DNA through 40 Hydrogen bonds

b. Typical Sequence specific DNA binding proteins interact with the DNA through 20 hydrogen bonds

9. Two aspects of the core histone-DNA interaction result in allowing the rigid DNA to bend

a. Increased hydrogen bonding

b. Basic nature of the histones masks the negative charge of the phosphate groups

IV. Packaging The Genome

A. Packaging The Genome: Introduction

1. Winding the DNA around nucleosomes only compacts the DNA 6 fold-well short of the needed 1000-10,000 fold compaction that is necessary

2. To further look at DNA compaction, chromatin was carefully isolated from nuclei and viewed by electron microscopy using a low salt extraction

3. In 1975, Chambons lab published electron micrographs of the eukaryotic genome revealed the existence of uniformly sized particles with a repeating pattern

4. The genome looked like Beads-on-a-string by electron microscopy

a. The beads represent DNA wrapped around the histone core octamer

b. The string represents the free DNA helix between nucleosomes

c. Note: these methods of extraction are non-physiological and therefore, it is unknown whether this conformation exists in vivo-however it is likely this conformation is not present in vivo

B. Packaging The Genome: The 30 nm Fiber

1. The binding of Histone H1 not only increases how tight the DNA is wound around the nucleosome, but also stabilizes higher order chromatin structures

2. If the chromatin is studied In the test tube, as salt concentrations are increased, the addition of Histone H1 results in the nucleosomal DNA forming a 30 nm fiber

3. Chromatin in situ experiments using electron microscopy also further revealed that the arrays of nucleosomes formed a more compact fiber

a. Revealed the 30 nm fiber (30 nm in diameter)

b. Difficult to maintain the 30 nm structure upon purification and so the structure remains not well understood, as it is present in vivo

4. Two possible models have been proposed for the structure of the 30 nm fiber-based on high salt extractions

a. Selenoid model

b. Zig-Zag model

5. The solenoid model involves six consecutive nucleosomes arranged per turn of the higher order structure

6. This model was initially supported by both electron microscopy and X-ray diffraction studies

a. The 30 nm fiber has a helical pitch of ~11 nm which is the diameter of the nucleosome disc

b. Suggests that the 30 nm fiber is composed of nucleosome discs stacked on edge in the form of a helix

c. Flat surfaces on either side of the core histone octamer face each other

d. Linker DNA is buried on the inside of the helix, but never passes through the central axis-means having shorter linker DNA favors this conformation

7. The zig-zag ribbon structure twists and supercoils, and through X-ray crystallography has been seen in transcriptionally active cells

a. The zig-zag model is based on the zig-zag pattern of nucleosomes formed upon Histone H1 addition in vitro

b. Requires the linker DNA to pass through the central axis in straight form

c. Longer linker DNA favors this conformation

8. Recent data has suggested that perhaps the Zig-zag model is the physiological model as compared to the once thought to be the physiologically relevant solenoid model

a. Solenoid model does not form at physiological salt concentrations (150 nM)-probably not relevant in situ

b. Controversy is still unresolved because different eukaryotic species have different characteristic lengths of linker DNA those that have longer linker DNA may have their chromatin present in the zig-zag conformation and those with shorter linker DNA may have their chromatin present in the solenoid conformation

C. Packaging The Genome: Interactions Between Core Nucleosomes In Formation Of The 30 nm Fiber

1. Although Histone H1 addition is required in vitro to stabilize the 30 nm fiber other interactions are necessary for its formation

2. The core nucleosomes must also be able to interact with each other

3. The core nucleosomes interact with each other through the presence of the amino-terminal tails of the core histones (note: the amino-terminal tails have been shown to have no role in DNA binding)

4. If core histones lacking the amino-terminal tails are used in the in vitro chromatin assay, then the 30 nm fiber fails to form

5. The 3-dimensional crystal structure of the nucleosome shows that each of the amino-terminal tails of Histones H2A, H3 and H4 interacting with adjacent nucleosome cores in the crystal lattice

6. The core Histone H4 has a positively charged tail, which can interact with a negatively charged portion of the histone fold domain Histone H2A

7. The negatively charged portion of histone H2A is conserved, but plays no role in DNA binding

8. It is thought that this interaction is particularly important for 30 nm fiber formation

D. Packaging The Genome: Formation of Loop Domains To Increase Compaction

1. Building the 30 nm fiber increases the compaction of the DNA by approximately 40 fold

2. This is still significantly less than the 1000-10,000 fold compaction ratio we need

3. Additional folding of the 30 nm fibers are required to compact the DNA even further

4. The exact nature of the folded loop structure remains unclear

5. However, one popular model proposes that the 30 nm fiber forms loops of 40-90 kb of DNA that are held together at their bases by the nuclear scaffold

a. Nuclear scaffold is made of proteins

b. Nuclear scaffold gives the nucleus structure

c. The true nature of the scaffold is still unknown

6. Two types of proteins have been identified to be part of the nuclear scaffold

a. Topoisomerase II (Topo II)

b. SMC proteins

7. The 30 nm fibers are tethered to the nuclear scaffold at their base

8. The loops extend out and away from the base

E. Packaging The Genome: Heterochromatin and Euchromatin

1. Not all of the DNA in the cell is equally packaged

a. Some DNA is more tightly packaged

b. Some DNA is more loosely packaged

2. Early studies of the chromosomes were performed using DNA dyes-these divided chromosomal regions into two categories, depending on how tightly the DNA was packaged

a. Euchromatin

b. Heterochromatin

3. Heterochromatin is more densely packaged and thus more readily stained

4. Euchromatin has a more open structure and thus stained more poorly with dyes

5. (Supplemental Figure) Packaging The Genome: Heterochromatin and Euchromatin

6. The difference between heterochromatin and euchromatin structure is how the nucleosomes are packaged into higher order structures

a. 30 nm fiber

b. Loop domains

7. In heterochromatin the nucleosomes readily assemble into highly organized higher-order chromatin structures

8. In euchromatin, the nucleosomes are found to be in much less organized assemblies

9. Heterochromatin and euchromatin have significant roles in the cell

10. The state of the chromatin in any one region of a chromosome/genome greatly affects gene expression

11. The euchromatin is much less densely packed

a. Genes in euchromatic regions are accessible by the machinery that drives gene expression

b. Genes in euchromatic regions can be expressed

12. The heterochromatin is more tightly packed

a. Genes in heterochromatic regions are not accessible by the machinery that drives gene expression

b. Genes in heterochromatic regions are not expressed

13. In the human body, heterochromatic and euchromatic regions are different for each different cell type

V. Regulation of Chromatin Structure

A. Regulation of Chromatin Structure: Introduction

1. Chromatin structure is not static within the nucleus of the cell

2. Chromatin structure changes dynamically to allow for short-term changes in gene expression

3. The dynamic nature of the DNA binding to the histone core structure is biologically relevant as sequence specific DNA binding proteins strongly perfer to bind histone free DNA

4. As a result of intermittent, spontaneous unwrapping, a protein can gain access to its DNA binding site with a probability of 1 in 50-1 in 100,000

B. Regulation of Chromatin Structure: Where Sequence Specific DNA Binding Proteins Bind The DNA

1. This change in winding either allows or blocks binding of a variety of sequence specific DNA binding proteins that promote gene expression

a. Tighter winding blocks binding

b. Looser winding allows binding

2. There are preferred sites on the nucleosome free DNA where sequence specific DNA binding proteins bind

3. If the more central the binding site is in when the DNA is wound, the less accessible it will be when the DNA is unwound

4. If the binding site is closer to the ends when the DNA is wound, the more accessible the binding site will be when the DNA is unwound

C. Regulation of Chromatin Structure: Histone Modification Affects DNA Winding

1. Modification of the histones can have a great affect on how tightly the DNA is wound around the histone core

2. Since the DNA is normally tightly wrapped around the histone core, most modifications promote a loosening of the DNA from the histone core

3. There are three major modifications of the tails of the core histones that change the ability of the DNA to wind

a. Acetylation

b. Methylation

c. Phosporylation

4. Methylation can either work to wind the DNA tighter or looser depending on the location of the methylation site

5. Phosphorylation generally acts to wind the DNA tighter to allow for condensation of the chromosomes during mitosis

D. Regulation of Chromatin Structure: Histone Acetylation

1. The role of acetylation in vivo is to promote gene expression by allowing for unwinding the DNA wound around the nucleosome

2. Acetyl groups (C2H3O2) promotes unwinding of the DNA around the nucleosome in two ways

a. Acetyl groups are bulky and can displace DNA

b. Acetyl groups have a negative charge, which reduces the positive charge on the histones allowing the DNA to become unwound

3. Acetyl groups are placed on the core histones by an enzyme known as acetyl-transferase

4. Generally lysines within the amino-terminal tails of the core histone are targeted for acetylation by acetyl-transferase

5. As an example, acetylation of lysines 8 and 16 of Histone H4 result in an unwinding of DNA

6. This unwinding of DNA results in increased gene expression due to binding of proteins that promote active gene expression

7. In reverse, deacetylation results in tighter winding of the DNA and an inhibition of gene expression

E. Regulation of Chromatin Structure: Histone Acetylation and Colon Cancer

1. About 10 years ago, several papers were published that suggested both histone and hyperacetylation and hypoacetylation are important in the neoplastic process depending on the target gene involved

2. Effects of histone acetylation on the expression of p21 is critically important in regards to the formation of colon cancer

3. Colon cancer is a cancer that forms in the lower part of your digestive system, the large intestine

4. Most colon cancers start as benign clumps of cells (tumors) called polyps

5. In time some polyps become malignant and form cancers

6. In general most colon cancer patients are over 50 years of age

7. There is a higher incidence of colon cancer among African Americans as compared to the general population

8. The initial symptoms of colon cancer generally only affect the lower digestive tract

a. A change in bowel habits lasting more than a several weeks

b. Rectal bleeding or blood in the stool

c. A feeling the bowel does not properly empty

d. Weakness and fatigue

e. Unexplained weight loss

9. On a molecular scale, high levels of DNA damage have been seen in colon carcinomas

10. Increased intake dietary fiber has long been linked to a decreased chance of contracting colon cancer

11. It is thought that dietary fiber can induce the expression of the p21 gene

12. Expression of p21 results in promoting cell cycle arrest which can allow the DNA repair machinery to repair the DNA

13. In the intestine, dietary fiber is broken down by resident bacteria to SFCAs (short chain fatty acids) along with carbon dioxide, water, methane and hydrogen gases

a. Acetate

b. Propionate

c. Butyrate

14. Butyrate has been shown to inhibit HDACs in the p21 gene region

15. This leaves core histones in this region hyperacetylated

a. The DNA remains in an unwound or loose conformation

b. The p21 gene is highly expressed

c. The process of colon carcinogenesis is in part inhibited