Chapter 7 Lecture Notes: Genome Structure, Chromatin and the Nucleosome
I. The Human Genome Project
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 Biology’s 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)
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
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 14,00 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
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 μm 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 Sears 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: The Chromosome
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
b. Some eukaryotes may have only one copy of each chromosome
c. In rare cases, there eukaryotes may have more than two copies of each chromosome
C. Fitting the Human Genome Into The Cell: The Purpose of Chromosomes
1. Dividing the genome into chromosomes is actually very important for all organism
a. The chromosome is a compact form of the DNA that will readily fit inside the cell/nucleus
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. The way the chromosomes are organized also gives the cell the ability to 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 will have the exact same genes
4. However, each member will not necessarily have the same sequence (variant or allele) for each gene
5. In cases where both chromosomes have the same sequence of a specific gene, the individual is a homozygote for that gene
In cases where both chromosomes have different sequence for a specific gene, the individual is a heterozygote for that gene
E. Fitting the Human Genome Into The Cell: The Human Chromosomal Content and Sickle Cell Anemia (Supplemental Figures)
F. Fitting The Genome Into The Cell: Chromosome Structure
1. Eukaryotic chromosomes have several important features
a. For stability
b. To allow for Biological Processes to occur
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. It is not the norm to have chromosomes with multiple centromeres
7. However, this does occasionally happen in humans
8. 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
9. Having no centromeres is also not the norm in human populations, and results in completely random segregation during mitosis
III. Winding the DNA Into Chromatin
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
a. Isolated from goose erythrocytes
b. Discovered that they were basic (have a positive charge)
3. Eukaryotic nucleosomes generally have 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)
a. Histone H2A
b. Histone H2B
c. Histone H3
d. Histone H4
5. To allow for the core histones to be basic, they are rich in arginine and lysine
6. The protein core of a nucleosome is a disk-shaped structure t hat assembles in an order fashion onl in the presence of DNA
7. Without DNA, the core histones will create intermediate assemblies, which are mediated by the conserved histone fold domain
C. Winding The DNA Into Chromatin: The Histone Proteins Come Together To Form A Nucleosome
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
D. Winding The DNA Into Chromatin: The Histone Proteins Come Together To Form A Nucleosome
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
E. 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 experiment,isolated chromatin was first 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 mutliples of 180 bp
6. As we saw before, the 180 bp refers to the amount of DNA wound around the core nucleosome and the linker DNA
7. 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
8. After extensive treatment only a 147 base pair band was seen-indicative of the amount of DNA wound around the core nucleosome
F. Eukaryotic Genome Structure: Histone H1 Linker Function Function