Exam Lectures for Molecular Biology
Exam Lectures for Molecular Biology BIOL 5344
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DNA and RNA Structure Wednesday, August 26, 2015 BIOL 5304 lecture 2 1 DNA is the genetic material of all living organisms. It is stored, copied, repaired, and passed on to progeny. RNA has many roles in the expression of genes. mRNA: the transcription product of DNA tRNA: the adapter molecules in translating mRNA into protein rRNA: ribosomes, the translation machinery snRNA: small nuclear RNA for splicing mRNA, etc. miRNA: microRNAs that silence genes lncRNA: long, noncoding RNAs-recently discovered Proteins play a variety of important roles. 2 The double-stranded structure of DNA Fig. 4-1 3 DNA and RNA are polymeric nucleic acids, composed of nucleotides linked by phosphodiester bonds Fig. 4-2 4 Relationship of bases, nucleosides and nucleotides Nucleotide Nucleoside H 5 A (Adenine) G (Guanine) are found in both DNA and RNA C (Cytosine) T (Thymine) is found in DNA U (Uracil) is found in RNA They differ by only a single methyl group on thymine. This difference will be seen as important when we later learn about DNA repair. The ring atoms of the bases are numbered in a standard way. The sugar atoms are numbered with primes (’) to distinguish those positions from the bases. 6 The 1’ position is bonded to a base, in a C-N glycosidic bond. The 2’ position is OH in ribose and is H in deoxyribose. Fig. 4-2 The 4’ O is used to form the ring with the 1’ C. The remaining -OH positions are 3’ and 5’. These are commonly linked to phosphates in both DNA and RNA. 7 Nucleic acids are built in a regular pattern: The 3’ OH of one nucleoside is linked to the 5’ OH of the next nucleoside via 5’ phosphodiester bonds. This means that any 3’ linear nucleic acid has 5’ a 5’ end and a 3’ end. Simply a consequence 3’ of the asymmetry of the deoxyribose. 5’ Fig. 4-3 8 The famous double helix of DNA is formed by 2 strands with an anti-parallel sense. The strands are wound around each other in a right-handed fashion, and interact via Fig. 6-1 hydrogen bonds between bases from opposite strands, and through base- stacking, primarily within one strand. The hydrogen bonds can only occur in the double helix between particular pairs of bases, otherwise the helix would be distorted. A-T making 2 Hydrogen bonds G-C making 3 Hydrogen bonds The double helix shown here is the B-form. The major and minor grooves are distinct. 9 Fig. 4-3 Fig. 4-4 Antiparallel, On the basis of size and H- hydrogen-bonded strands bonding complementarity, of DNA one purine must be paired with a particular pyrimidine. 10 Fig. 4-5 Tautomeric forms of the bases exist, but one of the two forms predominates. Watson-Crick base-pairing is only possible within the constraints of the double helix if the amino and keto tautomeric forms are present. These are the common forms. 11 The Watson-Crick base pairs have nearly the same size and shape. If the non-Watson-Crick A-C pair is similarly aligned, it lacks hydrogen bonding. Fig. 4-7 If hydrogen bonding were optimized, the sugars would be out of alignment. Fig. 4-6 12 Fig. 4-8 Base-ﬂipping refers to the rotation of a single base out of its normal position in the double helix. The remaining helix appears to be relatively undistorted. This ﬂexibility allows access of enzymes to the bases, e.g., during repair of damaged DNA. 13 3 well-deﬁned forms of DNA have been identiﬁed: A, B & Z B-form is the standard form that predominates inside cells. A-form represents a dehydrated B-form, which was produced in the lab. It is biologically signiﬁcant in that RNA-DNA hybrids take up a similar conformation. Z-form is a special conformation that requires a particular sequence of nucleotides. Unlike the others, it is a left-handed double helix. 14 Note: Handedness of helices Tilt of bases Width of helices Central channels Width of grooves (See next slide) Structural Tutorial The anatomy of A-, B-, and Z-DNA. Wing RM, Fratini AV, Kopka ML.BN, Science. 1982 Apr 30;216(4545):475-85. 15 A-form B-form Z-form B A Z Fig. 4-13 In A- and B-form DNA, all of the connections between base and sugar are anti. Z-form DNA requires a sequence of alternating purine and pyrimidine, in which the purines adopt the syn conformation, and the pyrimidines are anti. Alexander Rich 1924-2015 Molecular structure of a left- handed double helical DNA Wang AH, Quigley GJ, Kolpak FJ, Crawford JL, van Boom JH, van der Marel G, Rich A. Nature. 1979 Dec 13;282(5740):680-6. 18 Double-stranded DNA has a major groove and a minor groove. Fig. 4-10 Fig. 4-1 19 Double-stranded DNA has a major groove and a minor groove. How are the Fig. 4-10 grooves deﬁned? From the size of the angle from the sugar (C1’) to the helix axis to the other sugar (C1’). That’s why the major groove can be wide or narrow, deep or shallow. It depends upon the type of double helix. 20 The major groove of B-form DNA can be used for recognition. Proteins can Fig. 4-10 recognize speciﬁc nucleotide sequences through speciﬁc interactions with the edges of the bases. The bases project hydrogen- bond donors (D), hydrogen-bond acceptors (A), methyl groups (M), or non-hydrogen bonding hydrogens (H) into the major groove. 21 DNA strands can come apart in a process called denaturation. This will occur as the temperature is increased, and so it is also called melting. The re-association of two complementary strands (renaturation or annealing) can also occur. Short regions that are not complementary may cause a bulge, as shown here. 22 The melting of DNA can be detected by measuring the change in absorbance at 260 nm. The A 260increases as the DNA melts. The sharpness of the transition indicates that this is a cooperative process. The melting temperature (T ) m is deﬁned as the temperature of the midpoint of the change. Fig. 6-15 Melting of DNA must occur when DNA is replicated and when it is transcribed into RNA. Fig. 4-15 23 It is well-known that the melting temperature of natural DNA increases with the G+C %. Red (low ionic strength) and green (high ionic strength) inﬂuence the T difmerently, as shown below. (Higher T atmhigher ionic strength.) Fig. 4-16 This is not actually due to the difference in the number of hydrogen bonds between A-T (2) and G-C (3) base pairs. Rather, it is due to the generally stronger base-stacking interactions of duplexes containing GC pairs. 24 RNA Structure The 2’OH and the uracil base, not found in DNA, are shown in red. The 2’OH profoundly inﬂuences the structure of RNA. Although often described as a single-stranded nucleic acid, RNA is generally found in some sort of compact structure, Fig. 6-29 involving short Fig. 5-1 helices of 2 or 3 25 interacting strands. The higher order structure of RNA usually includes base- pairing, not only between A-U and G-C, but also between non Watson-Crick partners. In particular, the G-U pair is commonly found. Fig. 5-2 Fig. 5-6 26 Double helical characteristics of RNA Complementary regions form duplexes Noncomplementary regions form bulges, internal loops and hairpin loops. Fig. 5-3 27 Below is a schematic drawing of a tetra-loop, a variation on the stem-loop. Base-stacking interactions are important for its stability. (try to follow the chain) 28 This structure is called a pseudoknot. The central strand is base-paired with different regions. Fig. 5-5 29 More complex base-pairings are also common. Fig. 5-8 Fig. 6-34 On the right, the A-U interaction is a conventional double-helical base pair. The hydrogen bonding of the U on the left is an example of a tertiary interaction. 30 The so-called Hammerhead RNA is a self-cleaving molecule, and its discovery was an early indication of the catalytic powers of RNA. The crystal structure of an all-RNA hammerhead ribozyme: a A structural tutorial is on Scott WG, Finch JT, Klug A.catalytic cleavage. the website. 31Cell. 1995 Jun 30;81(7):991-1002. The cleavage site is indicated by an arrow (in each view). One region binds a Mg ion (red), which is in proximity to a nucleotide from a distant region, G12. It activates the 2’- OH oxygen to form a 2’-3’ cyclic phosphate, producing a broken chain with a 5’-OH. Fig. 5-13 Fig. 5-13 32 Yeast phenylalanyl-tRNA The backbone of the chain is colored ﬁrst cyan, then magenta. It starts in the upper left, then makes an all- cyan stem and loop, then continues to the bottom. Further refinement of tRNAPhe.cture of yeast Hingerty B, Brown RS, Jack A. 25;124(3):523-34.Sep 33 A segment of 5S rRNA Colors: A blue U gray G salmon C orange Mg green Can you detect any non-WC base pairs? structure of a 5S rRNA domain.n in the crystal Correll CC, Freeborn B, Moore PB, Steitz TA. Cell. 1997 Nov 28;91(5):705-12. 34 RNase P is a ribozyme that cleaves RNA molecules to form tRNA Sidney Altman (1939- ) showed that the RNA component of RNAse P was sufficient for the catalytic activity, thus demonstrating that it was a ribozyme. He was awarded the Nobel Prize in Chemistry in 1989 35 RNase P contains a protein component (cyan), and an RNA component (purple). It is shown here binding its substrate: a tRNA precursor (yellow) 36 Fig. 5-12 ▯ Supercoiling of DNA, Topoisomerases, Proteins, and Introduction to Genomes Friday, August 28, 2015 BIOL 5304 lecture 3 1 Chapter 4 : Double-stranded DNA molecules have topological states that can be distinguished visually by electron microscopy. Shown below are covalently-closed, circular DNA molecules. In the upper panel the DNA is relaxed. In the lower panel, the DNA is supercoiled. It appears to be highly twisted and more compact. Fig. 4-20 2 There are two interconvertible types of supercoiling: a. the telephone cord and b. the garden hose (interwound) (toroidal) Fig. 4-18 These situations arise commonly with (a) circular chromosomes (e.g., bacteria) and (b) linear ch3omosomes (nucleosomes). A mathematical analysis of topology,as it applies to DNA,was done in a series a papers in the 1970’s F.Brock Fuller 1927-2009 my Calculus I professor 4 Let’s examine the topological properties of covalently-closed, circular duplex DNA. If one travelled along one strand for one complete circuit of a plasmid, and returned to the starting point, one would go around the other strand a certain number of times (Lk). (where Lk is a whole number, and Lk ≥ 0). This is called the Linking number Lk Lk=1 Mathematically, the Linking number can be written as the sum of two other numbers, called the Twist (Tw) and the Writhe (Wr). Lk = Tw + Wr For dsDNA, twist is the number of times one strand goes around the other, in a local sense. For B-form DNA this would be the number of basepairs/10 (approx.). Writhe is the number of times one duplex goes around another. This is called supercoiling. This will be illustrated in class. 5 The Linking number of totally relaxed DNA is called Lk o If the DNA is totally B-form, this will be approximately equal to the number of base pairs divided by the number of bp per turn (i.e. ~10). (Remember, Lk must be a whole number.) So, in a relaxed DNA molecule, Wr ≅ 0 and Tw ≅ Lk =Lk o How does a covalently closed circular DNA become relaxed? Supercoils can only be removed by breaking a covalent bond in the nucleic acid chain. This can be done by enzymes that break phosphodiester bonds, such as a DNase, or a topoisomerase. In the lab, these bonds can be broken by exposure to high pH. 6 (a) is relaxed (b) is negatively (c) is partially DNA supercoiled DNA unwound DNA b㱻c Inter- conversion of Tw and Wr while Lk remains the same Right-handed DNA (e.g., B-form) has a positive Tw, by deﬁnition Fig. 4-17 7 Breaking one strand of duplex DNA is called nicking. Free rotation around a pivot point in the other strand allows the DNA to relax, and so the supercoils will be lost. So Lk→Tw. A single break per circular molecule will completely relax it. Fig. 6-19 8 Supercoiling can be quantiﬁed by the difference between the actual Linking number and the Linking number of the completely relaxed state: o ∆Lk = Lk - Lk If this number is less than zero, the molecule is negatively-supercoiled. In the example at the right, Lko would be 36, if the DNA has 10 bp/turn. So, ∆Lk = 32- 36 = -4 It is negatively supercoiled. 9 Superhelical density is deﬁned as o σ = ∆Lk / Lk In the example here σ = -4/36 = -.11 Circular DNA molecules from bacteria are typically negatively supercoiled, with σ ≈ -.06 What is the utility of negative supercoiling? Duplex molecules with negative supercoiling have a greater tendency to partially unwind. 10 This transition does not require the action of an enzyme, (although proteins exist that facilitate it), because Tw and Wr can interconvert without breaking or making covalent bonds. If 40 bp Fig. 4-17bc melt,Tw becomes 320/10=32 Bacteria that live at very high temperatures (80-100˚C) tend to have positively supercoiled DNA, presumably to counter thermal denaturation. They contain an enzyme, reverse gyrase, that introduces positive supercoils. Crystal structures of Thermotoga maritima reverse gyrase: inferences for the mechanism of positive DNA supercoiling. Nucleic Acids Res. 2013 Jan;41(2):1058-70. Ganguly A, Klostermeier D. 11 Now let’s look at enzymes that change the linking number of DNA molecules. There are 2 classes of Topoisomerases. Type II Topoisomerases cut both strands of DNA, then a segment is passed through the break, and the broken ends are resealed. The linking number of the DNA changes by 2 (typically decreases). Type II enzymes require the hydrolysis of ATP . This is thought to be necessary for the movement of the proteins to carry out the movement of the DNA strands. II topoisomerases: prevention of DNA in type double-strand breaks. Bates AD, Berger JM, Maxwell A. 10.1093/nar/gkr258. Epub 2011 Apr 27. Review.oi: Fig. 4-21 12 Possible mechanism of a Type II Topoisomerase, DNA Gyrase DNA gyrase generates negative supercoiling in steps of ∆Lk = -2 Crystal structure of the breakage-reunion domain of DNA gyrase. Shikotra N, Maxwell A, Liddington RC.V, Nature. 1997 Aug 28;388(6645):903-6. DNA gyrase is a heterotetramer of 2 A and 2 B subunits. 2 molecules of ATP are consumed per cycle. ATP hydrolysis drives conformational changes necessary for transport steps. 13 E. coli has 2 different Type II topoisomerases The second enzyme, topo IV (ParC), unlike DNA gyrase, can only convert positively supercoiled DNA to relaxed DNA. It appears to have a sophisticated mechanism for recognizing positive supercoils. The structural basis for substrate specificity in DNA topoisomerase IV. J Mol Biol. 2005 Aug 19;351(3):545-61. Berger JM. 14 Type I topoisomerase cuts one strand, and reseals it, causing the linking number to increase by 1. No ATP hydrolysis is required by these enzymes. Fig. 4-22 15 The mechanism of Type I topoisomerases involves a tyrosine that forms a covalent intermediate with one strand of the DNA. There is conservation of phosphoester bonds. In b, the protein-DNA intermediate is shown. Another subclass of Type I topo- isomerases forms covalent complexes at the 3’ end of DNA Fig. 4-24 16 A closer look at the E. coli Topoisomerase I (Type I) The central part of the polypeptide chain (magenta) could function as the gate, by opening at the bottom. Possible sites for the binding of single-strand DNA are shown by red nucleotides. Three-dimensional structure of the 67K N-terminal fragment of E. coli DNA topoisomerase I. Lima CD, Wang JC, Mondragón A. 138-46. 1994 Jan 13;367(6459): 17 A schematic view of how a Type I topoisomerase might pass the other strand through the break in the DNA chain. 18 Various roles of topoisomerases. Type I can do this job (decatenation) if one strand is nicked. Fig. 4-23 19 Topoisomers can be separated by gel electrophoresis. The rates of migration are also affected by the presence of ethidium bromide, which is commonly used to visualize DNA in gels. 20 Fig. 4-26 The ethidium ion intercalates between the base pairs in duplex DNA, causing an unwinding. Its brilliant ﬂuorescence allows easy detection of DNA by ultraviolet light. Fig. 4-28 Fig. 4-27 Now there are 36 bp per turn, so Tw has been reduced, and then Wr must increase. 21 Chapter 6. The Structure of Proteins A brief overview Proteins are composed of amino acids, which differ at the “R” group, known as the “side chain”. Fig. 6-1 22 Fig. 6-2 Classiﬁcation of Amino Acids 23 Peptide bond formation Amino acids are joined by the formation of peptide bonds. In general proteins are linear, unbranched chains of amino acids. One exception is that some cysteines form covalent, disulﬁde linkages with other 24 cysteines. The folded structure of a protein is deﬁned by its conformation. Conformation refers to the three angles of rotation associated with each amino acid in the polypeptide chain. In the backbone (or main chain) each amino acid contributes N-C-C. The two internal bonds have free rotation which is restricted by steric effects involving the side chains and the CO. The third bond is the peptide bond which is relatively ﬁxed in a trans, planar arrangement. 25 Fig.6-66 Disulﬁde bonds Fig. 6-5 Levels of protein structure Fig. 6-6 26 The two most common secondary structures are alpha- helices and beta- sheets. Most folded proteins will be composed of a series of such elements connected by short loops, or by longer irregular segments. In beta-sheets the strands can be parallel or anti-parallel. Both helices and sheets have regular patterns of hydrogen bonding between amino acids through the backbone. 27 Fig. 6-8 Yeast GCN4 transcription factor Two alpha-helices can align, either antiparallel, or parallel as shown for this case, to form a “coiled coil”. Notice that binding of the protein to the DNA (in the major groove) causes the helix to extend. 28 Fig. 6-8 Proteins are composed of domains A domain is a structural module This protein has 2 domains. Each is composed primarily of beta-strands. 29 Fig. 6-10a Three principal classes of protein folds Center: Left: Right: All alpha-helical All beta-strand Mixed alpha- fold fold (2 domains) helix, beta sheet 30 Fig. 6-11 Genome Organization The genome is the entire genetic material of an organism. The DNA sequences of the genomes from many organisms have been determined in recent years. The ﬁrst was the bacterium Haemophilus influenzae, published in July 1995 in the journal Science 4" ADVAOFEMENT SCIENCE CIENCE VOL.26449-604.00 6 About 1.67 x 10 bp About 1600 genes First we will look at how genes are organized in chromosomes. 31 The genetic material of cells is organized into chromosomes: They contain both nucleic acids and proteins. In eukaryotes, each is about 50% of the total mass. Most of the proteins are histones, but non-histone proteins also exist. The packaging of DNA into chromosomes serves several important functions: 1. Compact size 2. Protection from damage 3. Transmission to daughter cells 4. Organization of the genes Chromosome Comparisons Archaea and bacteria generally have one or only a few chromosomes and are haploid. Eukaryotes have multiple chromosomes, and most cells Fig. 8-1 are diploid. The analysis of genomes is new research area. It is possible to analyze genomes because there are now more than several thousand that have been sequenced. Below is a page from the web site of the National Center for Biotechnology Information (www.ncbi.nlm.nih.gov) ▯ Genomes and Nucleosome Structure Monday, August 31, 2015 BIOL 5304 Lecture 4 Relationship of genes, genomes and organisms: The size of an organism’s genome is roughly correlated with the complexity of the organism. The number of genes in a genome is also roughly correlated with the size of an organism. These relationships are not strictly followed, as the Tables that follow demonstrate. 3 The number of genes in a genome does not scale exactly with the size of the genome The complexity of an organism is more closely related to the number of genes, rather than the size of the genome. The density of genes varies over two orders of magnitude from bacteria to mammals. 5 What contributes to gene density? Intergenic regions, introns, and repeated sequences. Look at the RNA polymerase gene (in red) above, from four different genomes. Fig. 8-2 7 Genes are longer in more complex organisms because of introns, and also because of more regulatory sequences such as those for promoters. The initial transcript of an average human gene is 27 kb, while the average protein-coding region is only 1.3 kb, or about 5% of the transcript. Genes density is also lower in more complex organisms because of much larger intergenic regions. About 60% of the human genome is intergenic sequence. These regions in humans are about 25% unique and 75% repeated sequences. Breakdown of the human genome Fig. 8-4 Microsatellites are simple repeats such as CACACACACA that arise from replication errors. Genome wide repeats are 100-1000 bp repeats that are parts of transposons. The viral enzyme, reverse transcriptase can make DNA copies of host RNAs that can integrate into the host genome as a pseudogene, one that is not expressed because it lacks control elements. Fig. 8-5 In eukaryotic cells, chromosomal DNA is packaged into nucleosomes. 8 histones make up the core (2 copies of 4 types). The DNA, 147 bp, is wrapped around the histone core, 1.65 times. Fig. 8-18 A section of linker DNA ( 20-60 bp) connects the nucleosomes. The length of the linker varies according to species. 12 Insight into the structure of the nucleosome was ﬁrst revealed by nuclease digestion experiments. Extensive digestion by a non-speciﬁc nuclease generated fragments of 147 bp. A light digestion revealed a ladder of fragments at intervals of about 200 bp. These results suggest a core of DNA about 150 bp that cannot be digested, separated by a linker of about 50 bp that can be digested. EM studies also showed a bead-like structure. There are 5 abundant types of histones in most eukaryotic cells. Four are found in the core, and one is associated with the linker DNA. They are all basic proteins, with a high percentage of positively charged amino acids. Stoi.h 2 2 2 2 1 largest 14 The core histones have a conserved structure (the histone fold). They differ in their N-terminal and C-terminal extensions. They pair by crossing the long helices. Fig. 8-19 Fig. 8-20 Assembly pathway for the histone core of the nucleosome: H3-H4 tetramers, DNA wraps around H2A-H2B dimers join Notice the N-terminal tails of the histones are exposed. The N-terminal tails can be removed by proteases. This does not disrupt the structure of the nucleosomes. The rest of the histones are not accessible to proteases. Inside the cell, the N-terminal tails are covalently modiﬁed by several different enzymes, as part of a signaling scheme. This will be examined more carefully later. Fig. 8-21 The nucleosome is roughly two-fold symmetrical (dyad axis). Histones 3 and 4 have more contacts with the DNA. The DNA is held in a negatively supercoiled state in the nucleosome. Remember that this allows easy unwinding of duplex DNA for replication, transcription etc. In a closed circular DNA, each nucleosome would contribute about -1.2 to the Linking number. Fig. 8-22 18 DNA that interacts with H3-H4 is colored turquoise (a) H3-H4 binds the middle and the ends of the DNA. DNA that interacts with H2A-H2B is colored orange (b). Fig. 8-23 In 2007, the structure was solved at very high resolution, indicating that the DNA was somewhat distorted from the normal conformation. In particular there are some sharper corners, rather than all smooth curves. Part 1: DNAgolden coil H2Ared H2B violet H3 blue H4 cyan Part 2: DNAspaceﬁlling Histones wireframe Positive charges Lysines green Arginines yellow DNA stretching and extreme kinking in the nucleosome core. 20 Ong MS, Richmond TJ, Davey CA. The DNA is bent due to the many (~140) hydrogen bonds between the H3-H4 histones and the oxygen atoms of the phosphates in the minor groove. (Not many with the bases, why not?) The formation of this intermediate structure makes it easy for the H2A-H2B dimers to add, one to each side. Fig. 8-24 Close up of H3-H4 interactions with DNA. Most interactions are with the minor grooves (circled). The N-terminal tails of the histones help to guide the DNA as it wraps around the histones. They can be seen at speciﬁc locations. Fig. 8-26 Nucleosomes are only the ﬁrst step from duplex DNA to chromosomal structure. What else lies between? Previous work had shown that chromosomes are not uniformly structured. Heterochromatin is the more condensed, and densely stained variety. It is associated with low levels of gene expression. Euchromatin is more open, and extended. It is associated with higher levels of gene expression. Both contain nucleosomes. How do we get from nucleosomes to higher order chromatin structure? The linker histone H1, is also a small, basic protein. It binds to 2 distinct regions of DNA. In mammals there are 11 subtypes of histone H1. of the globular domain of histone H1: full assignment, tertiary structure, and comparison with the globular domain of histone H5. Cerf C, Lippens G, Ramakrishnan V, Muyldermans S, Segers A, Wyns L, Wodak SJ, Hallenga K. Fig. 8-27 25 Biochemistry. 1994 Sep 20;33(37):11079-86. The addition of linker histone H1 can be seen to impose higher order on nucleosome structure. 30-nm ﬁber Two models have been proposed for the 30-nm ﬁber. The solenoid model requires only a short linker DNA. Fig. 8-30a The zigzag model requires a longer linker DNA, one that crosses the center of the ﬁber. New results from 2005 suggested an even different arrangement. Fig. 8-30b Four connected nucleosomes were crystallized. The histones are colored: H2A, H2B, H3, H4 The nucleosomes alternate from left to right and back, in a zig-zag X-ray structure of a tetranucleosome and its implications for the chromatin fibre. Nature. 2005 Jul 7;436(7047):138-41.ond TJ. 29 This model was built from the previous picture to see what the ﬁber might look like. Direct model (left) Idealized model (right) These models would ﬁt the 30 nm ﬁbers. There is a higher order left-handed double helix: Consecutive nucleosomes alternate between one strand and the other. X-ray structure of a tetranucleosome and its implications for the chromatin fibre. Schalch T, Duda S, Sargent DF, Richmond TJ. From cryo-electron microscopy a 30-nm structure was seen that is similar to the previous tetra-nucleosome: a super-double helix The difference is that the microscopy is done on samples with linker histones, and so it might pack more normally. On the other hand, some researchers have proposed that there is not a single 30-nm structure of chromatin, but rather, the 10-nm ﬁbers are ﬂexible and can adopt many forms. Chromatin as dynamic 10-nm fibers. Maeshima K, Imai R, Tamura S, Nozaki T. Chromosoma. 2014 Jun;123(3):225-37. doi: 10.1007/s00412-014-0460-2 Cryo-EM study of the chromatin fiber reveals a double helix twisted by tetranucleosomal units. Song F, Chen P, Sun D, Wang M, Dong L, Lian31D, Xu RM, Zhu P, Li G. Top movie shows the linker DNA, which is stretched out straight between the nucleosomes, across the middle of the double helix. Bottom movie shows the interactions between nucleosomes in the stack, and the location of the H1 histones. Cryo-EM study of the chromatin fiber reveals a double helix twisted by Song F, Chen P, Sun D,. Wang M, Dong L, Liang D, Xu RM, Zhu P, Li G. Science. 2014 Apr 25;344(6182):376-80. doi: 10.1126/science.1251413. 32 Nucleosomes whose histones lack the N-terminal tails cannot form 30-nm ﬁbers. These parts of the histones might be necessary for interactions between nucleosomes. In this way, modiﬁcations of the N-terminal tails would inﬂuence nucleosome packing. Linker histones are also necessary Fig. 8-31 The nuclear scaffold is important in organizing higher order structures of nucleosomes, but the details are not clear. One model is shown below. Fig. 8-32 There are variants of the core histones that affect the structure and interactions of nucleosomes. CENP-A is a variant of H3 that attaches to the kinetochore during mitosis. Nucleosomes are dynamic structures For DNA replication the nucleosomes must generally come apart for the chromosome to be copied. For transcription, some regions of the chromosome must become accessible to the RNA polymerase and the general transcription factors. This necessitates a dynamic structure. Some sites are more accessible to DNA-binding proteins than are others. In spontaneous unwinding, site 1 will be exposed more often than site 2. Fig. 8-34 Large protein complexes carry out remodeling of nucleosomes in different ways. (a) sliding (c) H2A-H2B dimers can be (b) the entire unit of histones is exchanged transferred to a new segment of DNA. Fig. 8-35 new ﬁg A model for chromatin-remodeling using the RSC complex DNA is in black, the histones in gray. RSC binds and holds the histones, while pulling the DNA away. The helicase, Sth1, then moves along the DNA, causing “sliding”. Then RSC releases the DNA and the histones, allowing the nucleosome to reform, but at a new position. Mechanism of chromatin remodeling. Lorch Y, Maier-Davis B, Kornberg RD. Proc Natl Acad Sci U S A. 2010 Feb 23;107(8):3458-62. doi: 10.1073/pnas.1000398107. 39 A more detailed look at a mechanism for DNA-sliding induced by a nucleosomal-remodeling complex. Histones are held tight while the DNA is pulled. Fig. 8-36 Fig. 8-36 continued The DNA loop continues to move around the histone core as the DNA translocator continues to bind, move and release DNA There are many types of nucleosomal remodeling complexes These proteins contain domains that bind to histones on the basis of their covalent modiﬁcations 42 ▯ Modiﬁcation and Assembly of Nucleosomes and Synthesis of DNA Wednesday, September 2, 2015 BIOL 5304 lecture 5 Two models for nucleosome positioning. Positioning can hide or expose regions of DNA necessary for transcription to begin. This scheme generates a This scheme generates a nucleosome-free region of DNA. positioned-nucleosome. Fig. 8-37 A second mechanism for positioning involves the preference for A-T rich sequences to orient their minor groove towards the histones, and G-C the opposite way. Positioning can serve to make some sequences more accessible, and others less accessible. Fig. 8-38 Phosphorylation Modiﬁcations to the N-terminal Methylation tails of histones affect the Acetylation accessibility of nucleosomes. Modiﬁcations tend to decrease the positive charge of the histones. Acetylation of lysines is associated with increased transcriptional activity. Remember that the N-terminal tails are involved in interactions between nucleosomes, and so these covalent modiﬁcations are likely to change those interactions. Fig. 8-39 Acetylation of lysines neutralizes the positive charge. HAT: histone acetyltransferase HDAc: histone deacetylase Phosphoryation of serine or threonine generates a negative charge. Kinase Phosphatase Methylation of lysine or arginine does not decrease the charge. HMT: histone methyltransferase HDM: histone demethylase Fig. 8-40 If histones are acetylated they bind proteins with bromodomains. If they are methylated, they bind proteins with chromodomains. Such proteins are often components of large complexes involved in modifying histones, or in transcriptional activity. Fig. 8-41 Other domains that bind to histones according to their modiﬁcations: TUDOR domains: methylation PHD ﬁngers: methylation SANT domains: unmodiﬁed These domains will bind to particular sites of modiﬁcation. Nucleosome modifying enzymes can then bind to these complexes. Modiﬁed histones can recruit complexes that carry out the same modiﬁcation, leading to propagation of that modiﬁcation throughout nearby nucleosomes. Another example is the TFIID for the RNA polymerase II, which has a bromodomain. This enhances transcription of nucleosomes with acetylated histones. 8 Histone modiﬁcations are carried out by speciﬁc enzymes: Histone acetyl transferases (HATs) catalyze acetylation of lysine residues on histones. Histone deacetylases (HDAcs) catalyze removal of acetyl groups. Histone methyl transferases catalyze methylation of lysine residues on histones. Histone demethylases, discovered more recently, remove methyl groups from lysine residues. No arginine demethylase has been conﬁrmed yet. Methylation appears to be more speciﬁc, targeting single residues, while acetylation often occurs at numerous residues at once. Fig. 8-42 Chromatin remodeling and histone modiﬁcation work in conjunction to alter DNA accessibility. Shown here: DNA-binding protein 1 recruits a HAT, which acetylates nearby histones. This opens up the chromatin to a 10-nm ﬁber. Now a remodeling complex enters, and a new DNA-binding protein gains access to a segment of DNA. That could be to initiate transcription. What happens to nucleosomes during DNA replication? Are the histones disassembled? How are they efﬁciently reassembled? How are the covalent modiﬁcations maintained, if new histones are made? What happens to histones during replication? They are taken apart prior to DNA synthesis. H3-H4 tetramers tend to stay together and end up randomly in one duplex or the other. They contain most of the modiﬁcation sites, and so the memory is partially carried forward in both DNA duplexes. H2A-H2B dimers also Fig. 8-43 tend to stay together. How can memory of the modiﬁcation state be retained? Bromo- and chromo-domain proteins help to ensure that the original modiﬁcation state is restored. For example, here a bromodomain protein binds to the existing acetylated histones, then recruits HAT to acetylate nearby lysines. Fig. 8-44 Histone chaperones are proteins that prevent non-productive interactions of histones with DNA. CAF-1 only functions during replication, because it requires binding to the sliding clamp PCNA. The clamp is used to mark the newly synthesized DNA that needs to be assembled into nucleosomes. Histone assembly requires chaperones to bind the histones and deposit them on the DNA. CAF-1 also binds to PCNA (clamp). Fig. 8-45 Ch. 9 How is DNA replicated? After viewing the double-stranded DNA model of Watson and Crick, it is easy to see how DNA replication might work. Each strand is a template for synthesis of new strands. "It has not escaped our notice that the speciﬁc pairing we have postulated immediately suggests a possible copying mechanism for the genetic material." However, it was actually many years before details were known about this process. DNA replication is far more complicated than most scientists could have imagined 50 years ago. 16 The synthesis of DNA occurs in all life forms, and while some of the details vary, there are essential features that occur in most if not all examples. First, look at the chemical reaction: These are the 2 key substrates Fig. 9-1 17 Fig. 8-1 (a) the four nucleoside triphosphates (b) a primer-template junction: the primer with a 3’OH end is the actual substrate. A covalent bond forms there. The template serves to provide speciﬁcity for the reaction. It deﬁnes the sequence of the newly synthesized strand of DNA. The primer need only be long enough to ensure that the 3’OH is stably in the correct position to initiate the reaction. What is the reaction? 18 2 Steps: XTP + (XMP) → (XMP) + P-P n n+1 Fig. 9-2 P-P → 2P i 19 The ﬁrst step is somewhat favorable, and the second step is more favorable, in terms of free energy change (∆G). The sum of the two free energies corresponds to an 5 equilibrium constant of about 10 . That means that the reaction is essentially irreversible. Actually, danger to DNA is not from the reversibility of DNA synthesis, but rather from general hydrolysis. Simple hydrolysis of phosphoester bonds is a favorable reaction, and water is present in high concentrations. Fortunately, water is not kinetically competent to carry out this reaction. And DNA
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