Lectures 20&21 (reupload)
Lectures 20&21 (reupload) BCM 475 - M001
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Popular in Biochemistry
BCM 475 - M001
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Lecture 1019 Evolution and Bioinformatics p 173191 Structure and Function Bovine ribonuclease and human ribonuclease are structurally and functionally similar Angiogenin possesses a structure similar to that of ribonuclease indicating a possible evolutionary relationship human ribonuclease and human angiogenin are paralogs Sequence Comparison Methods Door to understanding the function and mechanism of newly obtained sequences Utilized for the determination of evolutionary pathways Homology Homologous molecules homologs come from a common ancestor Homologs 1 Paralogs 2 Orthologs Paralogs homologs that are present within one speciesquot but differ in their detailed biochemical functionsquot Orthologs homologs that are present within different species and have very similar or identical functionsquot Understanding the homology between molecules can reveal the evolutionary history of the molecules as well as information about their functionquot Detecting Homology Similarity in amino acid sequences 9 common ancestor 9 similar structure 9 similar functions 9 similar mechanisms Globins Class of proteins including myoglobin and hemoglobin Sequence alignment method utilized to detect common sequences in hemoglobin and myoglobin Comparing the Amino Acid Sequences of aHemoglobin and Myoglobin Sequence alignment 1 Slide amino acid sequences of hemoglobin and myoglobin past each other one amino acid at a time 2 Count the number of matched residues or sequence identitiesquot aHemoglobin and myoglobin 23 sequence identities 3 Analyze graph plotting number of matches vs alignment Sequence alignment techniques must take into consideration insertions or deletions that may have occurred By shifting one sequence of comparison by a certain number of residues and subsequently comparing to an adjacent sequence one can detect different amino acid identities matched residues By introducing a gap into one of the sequences the identities found in both alignments will be representedquot Alignment with gap insertion The greater number of gaps the increased number of sequence identities Because an excessive number of gaps can lead to an unreasonably high number of matched amino acid residues mathematical methods have been developed to compensate for such cases Eg a scoring system that counts each identity as 10 points and each gap as 25 points Determining the Statistical Significance of Sequence Alignments Not all sequence alignments correlate to homology It is the order of the residues within their sequences that implies a evolutionary relationship between themquot Steps for determining the statistical significance of sequence alignments STUDENNEquot Shu le or randomly rearrange one of the two sequences of comparison Repeat sequence alignment Calculate a new alignment score for shuf ed sequence Repeat process Create a histogram depicting the number of alignments for many shuf ed sequences plotted against the alignment score Unshu led sequences with a significantly greater alignment score than those of other sh u led sequences indicate the statistical significance of the sequence alignments Eg The alignment score for unshuf ed alphahemoglobin and myoglobin is substantially greater than any of these scores for shuf ed sequences strongly suggesting that the sequence similarity is significantquot and hence that alpha hemoglobin and myoglobin are homologous Detecting Distant Evolutionary Relationships Substitution matrices enable the detection of evolutionary relationships between proteins that have undergone amino acid substitutions Substitution types 1 Conservative substitution A substitution in which one amino acid is replaced with another that is similar in size and chemical propertiesquot No significant effect on protein structure or function Common amino acid change Nonconservative substitution A substitution in which an amino acid is replaced by one that is structurally dissimilarquot Singlenucleotide substitution Common amino acid change Steps for comparing sequences with adjustments for substitutions that may have occurred 1 Examine the substitutions that have actually taken place in evolutionarily related proteinsquot 2 Derive a substitution matrix from data obtained in step one Substitution matrix A substitution matrix describes a scoring system for the replacement of any amino acid with each of the other 19 amino acidsquot Large positive score in substitution matrix 9 common amino acid substitution Large negative score in substitution matrix 9 rare amino acid substitution Eg Blosum62 Look at Figure 69 Blosum62 Substitution matrix Depicts conservative and nonconservative substitutions with positive and negative scores respectively This scoring system detects homology between less obviously related sequences with greater sensitivity than would a comparison of identities onlyquot Alignment of Identities Only Versus the Blosum62 To detect homology between leghemoglobin and human myoglobin two different scoring systems were used 1 Identitybased scoring system and 2 Blosum62 According to the identitybased scoring system there is a 1 in 20 chance of statistical significance existing for the aligned sequences According to Blosum62 there is a 1 in 300 chance of statistical significance existing for the aligned sequences Blosum62 strengthens the conclusion that homology exists between leghemoglobin and human myoglobin Rules of Thumb for Sequence Analysis Sequence identities greater than 25 9 most likely homologous sequences Sequence identities less than 15 9 alignment alone is unlikely to indicate statistically significant similarityquot Sequence identities between 15 and 25 9 requires further analysis Note the lack of a statistically significant degree of sequence similarity does not rule out homologyquot Databases Searching databases for homologous sequences is often the first step performed in identifying a newly obtained protein sequence The Basic Local Alignment Search Tool BLAST is used to align newly obtained amino acid sequences with other known sequences Investigators in 1995 were able to identify likely functions for more than half the proteins within this bacterium Haemophilus in uenzae organism solely by sequence comparisonsquot via BLAST ThreeDimensional Structure and Homology Because threedimensional structure is much more closely associated with function than is sequence tertiary structure is more evolutionarily conserved than is primary structurequot Eg While the tertiary structures of hemoglobin myoglobin and leghemoglobin are conserved each possess a heme group that binds oxygen this conservation is not apparent in the primary structures of the globins Eg Actin and heat shock protein 70 Hsp70 were found to be noticeably similar in structure threedimensional despite only 156 sequence identityquot Threedimensional structure can be used to evaluate sequence alignments by the following steps 1 Identify highly conserved functionally significant residues within related proteins Eg identify the histidine residue important for the heme group s function that is conserved in hemoglobin myoglobin and leghemoglobin 2 Generate a sequence template with the highly conserved key residues Sequence template a map of conserved residues that are structurally and functionally important and are characteristic of particular families of proteinsquot By generating a sequence template one can recognize new family members that might be undetectable by other meansquot Sequence Alignment of Internal Repeats Repeated motifs can be detected by aligning sequences with themselvesquot Convergent Evolution The process by which very different evolutionary pathways lead to the same solutionquot Eg chymotrypsin vs subtilisin Both enzymes possess a catalytic triad in their active sites The enzymes however are not homologous Chymotrypsin is structurally composed mostly of beta sheets while subtilisin is composed of many alphahelices making it unlikely that these two proteins are homologous Comparison of RNA Sequences Comparison of homologous RNA sequences can serve as a starting point for identifying homology and RNA secondary structures Eg a comparison of RNA sequences in a part of ribosomal RNA taken from a variety of speciesquot Figure 620 For each of the different sequences compared the bases in positions 9 and 22 as well as several of the neighboring positions retain the ability to form Watson Crick base pairs even though the identities of the bases in these positions varyquot The comparison of the RNA sequences enabled the identification of the possible RNA secondary structure Evolutionary Trees Figure 62 1 Constructed from aligned sequences In evolutionary trees the length of the branch connecting each pair of proteins is proportional to the number of amino acid differences between the sequencesquot Evolutionary trees can be calibrated by comparing the deduced branch points with divergence times determined from the fossil recordquot Fossil records enable scientists to estimate the time of evolutionary divergence Eg an evolutionary tree for globin depicts how the duplication leading to the two chains of hemoglobinquot occurred 350 millions years ago while fossil records indicate how the jawless fish such as the lamprey which diverged from bony fish approximately 400 millions years ago contain hemoglobin built from a single type of subunitquot Modern Techniques Utilized to Understand Molecular Evolution Biochemistry techniques used to understand evolution 1 Polymerase chain reaction PCR and 2 Combinatorial chemistry PCR Utilized to amplify ancient DNA for sequencing Was used to amplify mitochondrial DNA from a Neanderthal fossil Investigators have completely sequenced the mitochondrial genomequot from the Neanderthal and comparison of the DNA sequences of Neanderthal with those from Homo sapiens via an evolutionary tree revealed that the Neanderthal is not on the line of direct descent leading to Homo sapiens but instead branched off earlier and then became extinctquot Evolution in the Laboratory Essential processes of evolution 1 The generation of a diverse populationquot 2 The selection of members based on some criterion of fitnessquot 3 Reproduction to enrich the population in these morefit membersquot Nucleic acids can undergo all three processes mentioned above and combinatorial chemistry can be used to synthesize nucleic acid molecules Before proteins existed did RNA molecules function as catalysts To answer scientists performed the following experiment 1 A randomized pool of RNA molecules was synthesized by combinatorial chemistry 2 The newly synthesized RNA molecules were passed through an ATP affinity column for the selection of ATPbinding molecules 3 The ATP affinity column was washed with excess ATP 4 ATPbinding RNA molecules were eluted off the column 5 The ATPbinding RNA molecules were allowed to replicate by reverse transcription into DNA amplification by PCR and transcription back into RNAquot 6 The new population was subjected to additional rounds of selection for ATP binding activityquot 7 Replication and selection repeated 8 Final RNA products that bound ATP with high affinity were isolated 9 Final RNA products were characterized NMR revealed how RNA forms a pocket in which ATP can fit and bind an evolved ATPbinding RNA molecule Lecture 1023 Genetic Information I p 109121 Gene Expression Transcription Translation DNA 9 RNA 9 Protein Nucleic Acids Linear polymers Monomer unit within polymer9 nucleotide One nucleotide consists of the following 1 A sugar 2 A phosphate 3 One of four bases Eg deoxyribonucleic acid DNA and ribonucleic acid RNA Nucleic Acid Structure Sugarphosphate backbone Structural support for nucleic acids Sugars linked by phosphodiester bridges 3 hydroxyl group of the sugar moiety of one nucleotide is esterified to a phosphate group which is in turn joined to the 5 hydroxyl group of the adjacent sugarquot Each phosphodiester bridge is negatively charged and therefore less prone to hydrolytic attack than other esters this resistance is crucial for maintaining the integrity of information stored in nucleic acidsquot A polynucleotide backbone contains 6 rotatable single bonds per unit Bases Carry genetic information Adenine A 9 purine Guanine G 9 purine Cytosine C 9 pyrimidine Thymine T 9 pyrimidine Uracil U 9 pyrimidine RNA vs DNA DNA Sugar 9 deoxyribose 2 carbon atom of sugar lacks oxygen 2 carbon atom has two hydrogen atoms The absence of a hydroxyl group at the 2 position of DNA enables this nucleic acid to be less susceptible to hydrolytic attack and hence more stable in carrying genetic information than RNA Bases A T and C G RNA Sugar 9 ribose 2 carbon atom of sugar contains a hydroxyl group Bases A U and C G Nucleoside Base sugar Nonphosphorylated DNA nucleoside units 1 Deoxyadenosine 2 Deoxyguanosine 3 Deoxycytidine 4 T hymidine RNA nucleoside units 1 Adenosine 2 Guanosine 3 Cytidine 4 Uridine N9 of a purine or N1 ofa pyrimidine is attached to C1 of the sugar by an N glycosidic linkagequot the Bglycosidic linkage with a base is above the plane of the sugar The glycosyl bond glycosidic bond in a nucleoside can rotate 180 degrees to go from an anti conformation to a syn conformation and vice versa Nucleotide Nucleoside joined to one or more phosphoryl groups by an ester linkagequot Nucleotide triphosphates Nucleosides bound to three phosphoryl groups Monomers of DNA and RNA 5 nucleotide Nucleoside 5 phosphate Compound with a phosphoryl group bound to C5 of sugar Eg adenosine 5 triphosphate ATP Structure of a DNA Chain DNA is conventionally written from the 5 end to the 3 end Polarity exists in a DNA chain Free 5 OH group Free 3 OH group DNA Length DNA molecules are very long polymers that are tightly compacted in cells Figure 48 lysed E coli cell Eg human DNA has a total contour length of greater than 1 meter the average contour length of the 46 chromosomal DNA molecules is greater than 2 cm WatsonCrick Model of DoubleHelical DNA The doublehelix facilitates DNA replication Features Two helical polynucleotide chains are coiled around a common axis with a right handed screw sensequot Antiparallel chains with opposing directionality The hydrophilic sugarphosphate backbone is located on the exterior of the double helix while the hydrophobic bases are located inside the helical structure The bases are nearly perpendicular to the helix axisquot in BDNA Adjacent bases are separated by 34 angstromsquot About 10 bases per turn of righthanded helix of BDNA Diameter of the double helix 20 angstroms WatsonCrick Base Pairs Base pairs bind via hydrogen bonding For DNA A T and C G For RNA A U and C G Chargaff s rule GC AT 1 Contributions to the Stability of the DoubleHelical Structure Base stacking The stacking of base pairs in a doublehelical structure leads to the stability of the helix via 1 The hydrophobic effect Hydrophilic sugarphosphate backbone located on outside of helix is exposed to water while hydrophobic bases are protected from an aqueous environment by lying inside the double helix 2 Van der Waals forces These forces are present between stacked base pairs and further stabilize a double helix Due to correlated motion of pi electrons Hydrogen bonds Responsible for holding base pairs together Although the hydrogen bonds holding base pairs together are fairly weak they stabilize the helix because of their large numbers in a DNA or RNA moleculequot Electrostatic interactions Mutual repulsion of negatively charged phosphate groups is relieved by having the phosphates on the exterior of the double helixquot Counterion condensation Occurs between adjacent pairs of phosphates on the same chain each cation binding site is occupied 80 in the presence of 150 mM Na to 98 in the presence of physiological concentration of Mg2quot DoubleStranded DNA has 10 Distinct Stacking Interactions Basestacking is a more significant contributor to the stability of the doublehelical structure 052 kcalmol than hydrogen bonding The stacking of C and G base pairs is more favorable than the stacking of A and T base pairs and the basestacking of CG is more significant than the three hydrogen bonds between C and G base pairs two hydrogen bonds connect A and T base pairs Structural Forms of DNA Aform helix Form for doublestranded RNA Form of dehydrated DNA Righthanded double helix Antiparallel strands Shape broadest helix type Shorter than BDNA Anti glycosidic bond Tilt of base pairs from perpendicular to helix axis 19 degrees Bform helix Most common form for DNA Righthanded double helix Antiparallel strands Shape Intermediate helix type Narrower and longer than ADNA Anti glycosidic bond Tilt of base pairs from perpendicular to helix axis 1 degree Biochemical Basis for the Structural Differences Between the A and B Form of DNA Different sugar puckers Aform 9 C3 carbon atom lies above the approximate plane defined by the four other sugar nonhydrogen atoms conformation called C3 endoquot Bform 9 each deoxyribose is in a CZ endo conformation in which C2 lies out of the planequot Another helix type in addition to A and B Shape narrowest helix type Lefthanded double helix Zigzagged phosphates in backbone Alternating anti and syn glycosidic bonds Different DNA Structures Circular DNA Supercoiled DNA Compact DNA Supercoiling may hinder or favor the capacity of the double helix to unwind and thereby affect the interactions between DNA and other moleculesquot Structures of SingleStranded Nucleic Acids Eg stemloop structure created when two complementary sequences within a single strand of DNA or RNA come together to form doublehelical structuresquot The Double Helix and the Transfer of Genetic Information Because a particular base on one strand of DNA is always paired with a specific base on an adjacent strand the sequence of bases of one strand of the double helix precisely determines the sequence of the other strandquot DNA is replicated via semiconservative replication The Melting and Annealing of the Double Helix A disruption of the hydrogen bonds holding DNA base pairs together will lead to the separation of the double helix into its two component strands Melting is the dissociation of the double helixquot Melting temperature Tm of DNA 9 temperature at which half the helical structure is lostquot Tm can be calculated from the change in free energy values of the component stacking interactions An allornone transition is a good model at any given temperature fully formed duplexes and separated single strands coexistquot Only a small amount of partiallymelted helices are ever presentquot Helicases not heat are responsible for melting DNA double helices inside cells Observing DNA melting DNA absorbs light the greatest at a wavelength of 260 nm the average wavelength for the four bases Hypochromism Increased base stacking SS DS 9 decreased absorbance of ultraviolet light Because doublehelical DNA absorbs light less effectively than singlestranded DNA one can observe the melting of DNA by monitoring the absorption of light particularly at 260 nm Annealing The process by which complementary DNA strands recoil into a double helix when the temperature is lowered below Tm Note Quotations indicate text obtained directly from textbook or lecture notes References Berg Jeremy John Tymoczko and Lubert Stryer Biochemistry 7th ed WH Freeman 2012 1 246 Print 10
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