Chromosome Organization and Molecular Structure
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Chromosome Organization and Molecular Structure

Course Number: BIO 325 49180, Fall 2010

College/University: University of Texas

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Chromosome Organization and Molecular Structure Introduction Chromosomes are the structures that contain the genetic material. They are complexes of DNA and proteins. The genome comprises all the genetic material that an organism possesses. In bacteria, it is typically a single circular chromosome. In eukaryotes, it refers to one complete set of nuclear chromosomes. The main function of the genetic material is to...

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Organization Chromosome and Molecular Structure Introduction Chromosomes are the structures that contain the genetic material. They are complexes of DNA and proteins. The genome comprises all the genetic material that an organism possesses. In bacteria, it is typically a single circular chromosome. In eukaryotes, it refers to one complete set of nuclear chromosomes. The main function of the genetic material is to store information required to produce an organism; the DNA molecule does that through its base sequence. DNA sequences are necessary for: Synthesis of RNA and cellular proteins. Replication of chromosomes. Proper segregation of chromosomes. Compaction of chromosomes so they can fit within living cells. Viral Genomes Viruses are small infectious particles containing nucleic acid surrounded by a capsid of proteins (see fig. 10.1). For replication, viruses rely on their host cells; i.e., the cells they infect. Most viruses exhibit a limited host range; they typically infect only specific types of cells of one host species. Bacteriophages may also contain a sheath, base plate and tail fibers (see fig. 9.4). Viral genomes are relatively small and are composed of DNA or RNA (see table 10.1). A viral genome, also termed the viral chromosome, is the genetic material of the virus; the genome can be DNA or RNA, single-stranded or double-stranded, circular or linear. Viral genomes vary in size from a few thousand to more than a hundred thousand nucleotides. 1 Viral genomes are packaged into the capsid in an assembly process. During an infection process, mature viral particles need to be assembled. Viruses with a simple structure may self-assemble; genetic material and capsid proteins spontaneously bind to each other (e.g., Tobacco mosaic virus see fig. 10.2). Complex viruses, such as T2 bacteriophages, undergo a process called directed assembly. Complex virus assembly requires proteins that are not part of the mature virus itself; noncapsid proteins usually have two main functions: Carry out the assembly process (e.g., scaffolding proteins that are not part of the mature virus). Act as proteases that cleave viral capsid proteins; this yields smaller capsid proteins that assemble correctly. Bacterial Chromosomes The bacterial chromosome is found in a region called the nucleoid (see fig. 10.3); the nucleoid is not membrane-bounded, so the DNA is in direct contact with the cytoplasm. Bacterial chromosomes contain a few thousand gene sequences that are interspersed with other functionally important sequences (see fig. 10.4). Bacterial chromosomal DNA is usually a circular molecule. A typical chromosome is a few million nucleotides in length; e.g., Escherichia coli has about 4.6 million base pairs and Haemophilus influenzae has about 1.8 million base pairs. Most bacterial species contain a single type of chromosome, but it may be present in multiple copies. Several thousand different genes are interspersed throughout the chromosome; structural gene sequences (encoding proteins) account for the majority of bacterial DNA; nontranscribed DNA sequences between adjacent genes are termed intergenic regions. One origin of replication is required to initiate DNA replication. Repetitive sequences (which may play a role in a variety of genetic processes including DNA folding, DNA replication, gene regulation, and genetic recombination) may be interspersed throughout the chromosome. 2 The formation of chromosomal loops helps make the bacterial chromosome more compact (see fig. 10.5). To fit within the bacterial cell, the chromosomal DNA must be compacted about a 1000fold; this involves the formation of loop domains. The number of loops varies according to the size of the bacterial chromosome and the species; E. coli has 50-100 with 40,000 to 80,000 bp of DNA in each. DNA supercoiling further compacts the bacterial chromosome (see figs. 10.6 and 10.7). Because the two strands within DNA already coil around each other, the formation of additional coils due to twisting forces is referred to as DNA supercoiling. Different conformations that result from supercoiling are referred to as topoisomers. Chromosome function is influenced by DNA supercoiling. We will not go into the details of supercoiling, but it is worth mentioning some basic facts. Chromosomal DNA in bacteria is negatively supercoiled and negative supercoiling has two major effects: It helps in the compaction of the chromosome (see fig. 10.6), and It creates tension that may be released by DNA strand separation (see fig. 10.8). The control of supercoiling in bacteria is accomplished by two main enzymes: DNA gyrase (also termed DNA topoisomerase II) introduces negative supercoils (see fig. 10.9); it can also relax positive supercoils when they occur. DNA topoisomerase I relaxes negative supercoils. The competing action of these two enzymes governs the overall supercoiling of bacterial DNA. The ability of gyrase to introduce negative supercoils into DNA is crucial for bacteria to survive; therefore, blocking the function of this enzyme is a way to cure or alleviate bacterial diseases. Two main classes of drugs inhibit gyrase and other bacterial topoisomerases: quinolones and coumarins (these do not inhibit eukaryotic topoisomerases). An example of a quinolone is Ciprofloxacin (Cipro), which is used in the treatment of anthrax, among other diseases. 3 Eukaryotic Chromosomes Eukaryotic species contain one or more sets of chromosomes; each set is composed of several different linear chromosomes. The total amount of DNA in eukaryotic species is typically greater than that in bacterial cells. Chromosomes in eukaryotes are located in the nucleus. To fit in the nucleus, the chromosomes must be highly compacted; this is accomplished by the binding of many proteins. The DNA-protein complex is termed chromatin. The sizes of eukaryotic genomes vary substantially (see fig. 10.10a). In many cases, this variation is not related to complexity of the species; for example, there is a two-fold difference in the size of the genome in two closely related salamander species (see figs. 10.10b and 10.10c). The difference in the size of the genome is not because of extra genes; rather, it is due to the accumulation of repetitive DNA sequences (these do not encode proteins). Eukaryotic chromosomes have many functionally important sequences including genes, origins of replication, centromeres, and telomeres (see fig. 10.11). A eukaryotic chromosome contains a long, linear DNA molecule. Three types of DNA sequences are required for chromosomal replication and segregation: origins of replication, centromeres, and telomeres. Genes are located between the centromeric and telomeric regions along the entire chromosome; a single chromosome usually has a few hundred to several thousand genes. In lower eukaryotes (such as yeast) genes are relatively small; they contain primarily the sequences encoding the polypeptides (i.e., very few introns are present). In higher eukaryotes (such as mammals) genes are long; they tend to have many introns of lengths from less than 100 to more than 10,000 bp. The genomes of eukaryotes contain sequences that are unique, moderately repetitive, or highly repetitive. Sequence complexity refers to the number of times a particular base sequence appears in the genome. 4 Unique or non-repetitive sequences. Found once or a few times in the genome. Includes structural genes as well as intergenic areas. In humans, make up roughly 40% of the genome. Moderately repetitive sequences. Found a few hundred to a few thousand times. Includes genes for rRNA and histones, origins of replication, transposable elements. Highly repetitive sequences. Found tens of thousands to millions of times; each copy is relatively short (a few nucleotides to several hundred in length). Some sequences are interspersed throughout the genome; e.g., Alu family in humans. Other sequences are clustered together in tandem arrays; e.g., AATAT and AATATAT sequences in Drosophila; these are commonly found in the centromeric regions. Sequence complexity can be evaluated in a renaturation experiment. Renaturation studies have proven useful in understanding genome complexity. A renaturation study involves the following steps (see fig. 10.12a): Double-stranded DNA is broken into small pieces. The pieces are melted into single strands by heat treatment. The temperature is lowered, allowing the renaturation of complementary DNA strands. The rate of renaturation of complementary DNA strands provides a way to distinguish among unique, moderately repetitive, and highly repetitive sequences; the renaturation rate of a particular DNA sequence depends on the concentration of its complementary partner. 5 Highly repetitive DNA will be the fastest to renature because there are many copies of complementary sequences. Unique sequences will be the slowest to renature; it takes added time for these sequences to find each other. The kinetics of renaturation can be described in a (relatively) simple mathematical formula: C C0 1 = 1 + k2C0t Where: C = the concentration of single-stranded DNA at a later time, t. C0 = the concentration of single-stranded DNA at time zero. k2 = the second-order rate constant for renaturation. In this equation, C/C0 is the fraction of DNA that is still in single-stranded form after a given length of time. A renaturation experiment can provide quantitative information about the complexity of DNA sequences. In the experiment shown in Figure 10.12b, human DNA was sheared into small pieces (~ 600 bp each), subjected heat, to then allowed to renature at a lower temperature. The rates of renaturation can be represented in a plot of C/C0 versus C0t; this is known as the C0t curve, called a cot curve. Result: renaturation of highly repetitive DNA is fast, renaturation of moderately repetitive DNA is intermediate, and renaturation of unique DNA is slow (see fig. 10.12b). Eukaryotic chromatin must be compacted to fit within the cell. If stretched end to end, a single set of human chromosomes will be more than 1 meter long; the cells nucleus is only 2 to 4 m in diameter; therefore, the DNA must be tightly compacted to fit. The compaction of linear DNA in eukaryotic chromosomes involves interactions between DNA and various proteins. Proteins bound to DNA are subject to change during the life of the cell; these changes affect the degree of chromatin compaction. 6 Linear DNA wraps around histone proteins to form nucleosomes, the repeating structural unit of chromatin (see fig. 10.13). The repeating structural unit within eukaryotic chromatin is the nucleosome. It is composed of double-stranded DNA wrapped around an octamer of histone proteins; the octamer is composed two copies each of four different histones (see fig. 10.13a). Overall structure of connected nucleosomes resembles beads on a string; this structure shortens the DNA length about seven-fold. Histone proteins are basic; they contain many positively-charged amino acids such as lysine and arginine that bind with the phosphates along the DNA backbone. Histone proteins have a globular domain and a flexible, charged amino terminus or tail. There are five types of histones (see fig. 10.13): H2A, H2B, H3 and H4 are the core histones two of each make up the octamer. H1 is the linker histone it binds to linker DNA and to nucleosomes, but not as tightly as the core histones (see fig. 10.13c). Nonhistone proteins play a role in the organization and compaction of the chromosome. Experiment 10A: The repeating nucleosome structure is revealed by digestion of the linker region. The model of nucleosome structure was proposed in 1974 by Roger Kornberg. Kornberg based his proposal on various observations about chromatin: biochemical experiments, X-ray diffraction studies, electron microscopy images. Markus Noll decided to test Kornbergs model via the following procedure: Digest DNA with the enzyme DNase I. Accurately measure the molecular weight of the resulting DNA fragments. The rationale is that the linker DNA is more accessible than the core DNA to the DNase I; thus, the cuts made by DNase I should occur in the linker DNA. 7 The hypothesis. The experiment by Markus Noll tests the beads-on-a-string model of chromatin structure. If the model is correct, DNase I should cut in the linker region thereby producing DNA pieces that are about 200 bp long. Testing the hypothesis (see fig. 10.14). Starting material nuclei from rat liver cells. Incubate the nuclei with low, medium, and high concentrations of DNase I. Extract the DNA. Load the DNA into a well of an agarose gel and run the gel to separate the DNA fragments according to size. Visualize the DNA fragments by staining the DNA with ethidium bromide and fluorescing with UV light. The data and interpreting the data (see fig. 10.14). At a high DNase concentration, the entire sample of chromosomal DNA was digested into fragments of approximately 200 bp in length the length of DNA containing one nucleosome. At a low concentration, some of the linker regions were not cut resulting in fragments containing two or more nucleosomes. The results strongly support the nucleosome model for chromatin structure. Nucleosomes become closely associated to form a 30 nm fiber (see figs. 10.15 and 10.16). Nucleosomes associate with each other to form a more compact structure termed the 30 nm fiber; histone H1 plays a role in this compaction. At moderate salt concentrations, H1 is removed; the result is the classic beads-on-astring morphology. At low salt concentrations, H1 remains bound; beads associate together into a more compact morphology. The 30 nm fiber shortens the total length of DNA another seven-fold over the basic beads-on-a-string structure. 8 Structure of the 30 nm fiber has proven difficult to determine, because the DNA conformation may be substantially altered when extracted from living cells. Two models have been proposed: solenoid model and three-dimensional zigzag model (see figs. 10.16b and 10.16c). Chromosomes are further compacted by anchoring the 30 nm fiber into radial loop domains along the nuclear matrix (see fig. 10.17). Wrapping DNA around histone octamers to form the nucleosome beads-on-a-string structure followed by formation of the 30 nm fiber shorten the DNA about 50-fold. A third level of compaction, the formation of radial loop domains, involves interaction between the 30 nm fiber and the nuclear matrix, which is composed of two parts the nuclear lamina, a collection of fibrous proteins that line the inner nuclear membrane, and an internal nuclear matrix of proteins. Matrix-attachment regions (MARs), also called scaffold-attachment regions (SARs), which contain high percentages of A and T bases, bind to the nuclear matrix and create radial loops (see fig. 10.17d). The attachment of radial loops to the nuclear matrix is important in two ways: 1) it plays a role in gene regulation, and 2) it serves to organize the chromosomes within the nucleus (each chromosome in the nucleus is located in a discrete and nonoverlapping chromosome territory see fig. 10.18). The compaction level of interphase chromosomes is not completely uniform (see fig. 10.19). Euchromatin forms less condensed regions of chromosomes that are transcriptionally active; these are regions where 30 nm fiber forms radial loop domains. Heterochromatin forms tightly compacted regions of chromosomes that are generally transcriptionally inactive (in general); in these regions the radial loop domains are compacted even further. There are two types of heterochromatin: Constitutive heterochromatin regions that are always heterochromatic and permanently inactive with regard to transcription. Facultative heterochromatin regions that can interconvert between euchromatin and heterochromatin (e.g., the Barr body, which is formed in female mammals when one of the two X chromosomes is converted to a heterochromatic state during embryonic development of somatic cells). 9 The histone code controls chromatin compaction. The compaction level of even euchromatin is too high for transcription factors and RNA polymerase to easily access and transcribe genes. Chromatin remodeling changes chromatin structure; it regulates the ability of transcription factors to access genes. Histone core protein tails are modified; more than 50 different enzymes modify tails by means such as acetylation, methylation and phosphorylation, all of which are covalent changes (see fig. 10.20). Histone modifications 1) directly influence interactions between nucleosomes, and 2) provide binding sites that are recognized by proteins. The histone code hypothesis proposes that the pattern of modification is a code specifying alterations in chromatin structure. Condensin and cohesin promote the formation of metaphase chromosomes. The levels of compaction leading to a metaphase chromosome are shown in fig. 10.21. DNA double helix nucleosomes 30 nm fibers radial loop domains further compaction of radial loops formation of a scaffold from the nuclear matrix and further compaction of all radial loops the metaphase chromosome As cells enter M phase, the level of compaction changes dramatically. By the end of prophase, sister chromatids are entirely heterochromatic and two parallel chromatids have an overall diameter of 1,400 nm. These highly condensed metaphase chromosomes undergo little gene transcription. In metaphase chromosomes the radial loops are highly compacted and stay anchored to a scaffold, which is formed from the nuclear matrix; histones are needed for the compaction of radial loops (see fig. 10.22). Condensin and cohesin are multiprotein complexes that help form and organize metaphase chromosomes. Condensin and cohesin are two completely distinct complexes, but both contain a category of proteins called SMC proteins (acronym = Structural Maintenance of Chromosomes see fig. 10.23). SMC proteins use energy from ATP and catalyze changes in chromosome structure. Together with topoisomerases, SMC proteins have been shown to promote major changes in DNA structure. 10 Condensin plays a critical role in chromosome condensation (see fig. 10.24). Prior to M phase, condensin is found outside the nucleus. As M phase begins, condensin coats the individual chromatids as euchromatin is converted to heterochromatin. Cohesin plays a critical role in sister chromatid alignment (see fig. 10.25). After S phase and until the middle of prophase, sister chromatids remain attached to each other along their lengths. By middle to late prophase, cohesin located along the arms of the chromosome is released allowing the arms to separate. At anaphase, cohesin molecules that remained at the centromere are degraded by a protease allowing the sister chromatids to separate. Please see the Conceptual Summary and Experimental Summary for Chapter 10 on pages 266 and 267. This lecture outline was prepared from Genetics: Analysis and Principles, by Brooker, 2009 (3rd edition). It contains phrases and entire sentences taken verbatim from that source, and is in no way meant to represent original work by Mark Bierner. 11

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V 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 81V.1.1 123PLK (Harmonic Series) Show thatn X1 k=1klog n:Deduce that the series estimateP11 k=1 kdoes not converge. Hint. Use thek+11 kZk1 dx: xTherefore, for nn X1 k=1Soluti
UCLA - MATH - 132
VI 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 81VI.1.1 1 23P L K 111 LLL Find all possible Laurent expansions centered at 0 of the following functions. 1 1 (a) z21 z (b) z+1 (c) (z2 1)(z2 4) z Solution (a) In the region 0 < jz j < 1,
UCLA - MATH - 132
VII 1 2 3 4 5 6 7 81 X X X X X X2 X X X X X X X3 X X X X X X X4 X X X X X X5 X X X X X X6 X X X X X X78910 11 12 13 14 15 16 17 18 19 X X X XXXX X XXX X1VII.1.1 Evaluate the following residues. (d) Res sin2z ; 0 (a) Res z21 ; 2i +4 z (b) Res
UCLA - MATH - 132
VIII 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 81VIII.1.1 123PL K LLL Show that z 4 + 2z 2 z + 1 has exactly one root in each quadrant.SolutionVIII.1.1Set p (z ) = z 4 +2z 2 z +1, and compute 4 arg p (z ) along the three segments