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Biology 1002B Lectures 19-26

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Biology 1002B
Tom Haffie

Lecture 19: Molecular Homology Determining if two structures are homologous is of central importance to evolutionary biology - how does one prove homology? • Lecture Outcomes: Identify: 1. Strategies for determining if features are homologous • Comparative genomics 1.Sequence genome (Cheap! ~$1000) 2.Genome annotation: ascribing biological function (meaning) to sequenced genome.(Bioinformatics) • Mostly computer automated. • Gene prediction, finds regulatory elements, biological function ascribed through similarity searches. 3.Protein-coding gene prediction: • Computer algorithms detect: promoter elements, ‘intron/exon boundaries’ and other conserved DNA motifs. • Predicts number of protein coding sequences, (see question 2) 4.Align sequences: with protein similarity we can infer structural and functional similarity and evolutionary homology. • BLAST: Local. (doesn’t force alignment), faster, etc (see question 4) (Considers exon shuffling/Alternative splicing) • CLUSTAL: Global. Forces alignment . start at beginning of protein coding frame, trying to align all bases (might introduce a gap though, considering exons (but doesn’t consider exon shuffling/Alternative splicing) 5.Determine homology: High similarity = homologous, Same amino acid sequence homologous • E-value = probability… lower e-value = higher probability of homology. 2. Sequences detected by annotation programs to detect open reading frames (orf) • Computer programs detect ‘intron-exon boundaries, splice out Introns from sequence, thus deducing the protein sequence. Once deduced there may be multiple open reading frames … The program looks for longest open reading frame (ie. which sequence gives the longestprotein) (this is how we get 15,000 predicted proteins of Chlamy.) 3. Characteristics that are, and are not, common between homologous genes • SIMILAR DNA sequences, SAME protein sequences (amino acid) why? Because amino acids are redundant. Some single bases that are different do not affect the amino acid sequence. 4. Usefulness of blast analysis of sequences in genebank at ncbi • Doesn’t try to force alignment, finds smaller areas of high similarity. Most common, and fastest! 5. Reasons why amino acid sequence comparisons are more informative than nucleotide sequence comparisons 1. More information in an amino acid sequence of same length (see question 7…each are 1 length ie. G=1) 2. The genetic code of protein is redundant (more than 1 codon can code for same amino acid!) 3. DNA databases are too big! (contain a lot of junk DNA, protein database are more specific and defined) 6. Mathematical relationship among total information, # of symbols, # letters in the alphabet • I = Total info in message,G = Length of message = # of symbols, N = # letters in the alphabet 𝑮 ∙ 𝐥𝐧(𝒏) 𝑰 = 𝐥𝐧(𝟐) 7. Relative number of bits of information in a single nucleotide vs. Single amino acid I = Total info in message,G = Length of message, N = Length of alphabet Single nucleotide Single amino acid I = Total info in message = ? I = Total info in message = ? G = Length of message = 1 nucleotide G = Length of message = 1 amino acid N = Length of alphabet = 4 possible nucleotides (A/T/G/C) N = Length of alphabet = 20 possible amino acids 𝐺 ∙ ln(𝑛) 1ln(4) 𝐺 ∙ ln(𝑛) 1ln(20) 𝐼 = = = 2 𝐼 = = = 4.32 ln(2) ln(2) ln(2) ln(2) 8. Relationship between e-value and likelihood of homology • When comparing sequences the higher the similarity the higher the probability of homology. E -vulue is liked to this probability factor in that the lower the e-value, the higher the chance of homology. • If e-value = 0, then very high chance for homology… It would be illogical to say the organisms aren’t homologous. 9. Mentalfloss: • What’s the information content in a sequence that consists of a single amino acid. i. See question 7 • Think about underlying reasons to explain that two proteins have weak global alignment (clustal) but strong local alignment (blast). i. Exon shuffling Lecture 20: Molecular Convergence Proteins do not have to be homologous to have similarities in structure/function. • Lecture Outcomes: Identify: 1. Synonymous vs. Nonsynonymous mutations • Synonymous: different nucleotide sequence, same amino acid sequence (neutral/ do not affect phenotype). 1.Silent substitution • Nonsynonymous: different nucleotide sequence and different amino acid sequence. 1.Replacement substitution 2. Characteristics of the neutral theory of molecular evolution • The fact that many mutations have no affect to protein/phenotype supports the idea of ‘Neutral Theory of Evolution’. • # 𝑜𝑓 𝑁𝑜𝑛𝑠𝑦𝑛𝑜𝑛𝑦𝑚𝑜𝑢𝑠 𝑚𝑢𝑡𝑎𝑡𝑖𝑜𝑛𝑠 (𝑏𝑒𝑡𝑤𝑒𝑒𝑛 2 𝑠𝑝𝑒𝑐𝑖𝑒𝑠) 𝛼 𝑇𝑖𝑚𝑒 𝑠𝑖𝑛𝑐𝑒 𝑠𝑝𝑒𝑐𝑖𝑒𝑠 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑑 1.Can use this characteristic to develop a “Molecular Clock” (Substitutions vs. Yrs. Since Divergence) 2.Mutations occur at a constant rate due to the fact that they are mostly neutral. • 3. Relationship between frequency of amino acid substitutions in given proteins vs. Time since common ancestor • Constant rates (which keys into Neutral Theory on the fact that mutations can me neutral.) • Different Rates for different Proteins (keys into constraint). 4. Relative rates of accumulation of synonymous vs. Non-synonymous mutations • Nonsynonymous Mutations (silent) higher rate than Synonymous Mutations (replacement). (less constraint on silent mutations) 5. Variables that affect the rate of evolution of a particular protein • Some are more sensitive than others (less variability possible that won’t be deleterious) • (Constant rates of divergence had been hard to explain based of selection theory since with that theory mutations could be either bad or good, ther were no ‘silent mutation’s within the context of the theory. • Ex. Histone H4 sequence (used for chromosome structuring) has much stronger constraint than 𝛼-globin sequence (hemoglobin protein) since Histone H4 sequence doesn’t have many locations for possible synonymous mutations. 1.Higher constraint = low rate 2.Low constraint = higher rate 6. Deduce time of divergence given number of amino acid changes in particular protein • Use graph. Since rate is constant, then # of a.a changes between two organisms is directlyrelated to time of divergence. 7. Characteristics of the "molecular" clock • Strong tool to use along with geological record, for defining divergence more definitively than phenotype. 1.Based on Neutral Hypothesis • See Questions 5 and 6. 8. Regions of two unrelated proteins that would be expected to be similar if they were the products of convergent evolution • Localized regions of similarity. 9. Function of lysozyme • Breaks down peptidoglycan wall of bacteria. 10. Characteristics of ruminant organisms that enable them to extract energy from cellulose • Digestive Lysozyme enzyme that had been evolutionarily structured to survive within cow stomach (low pH lows, or no bonding location for pepsin/protease to break it down). • Can digest grass (cellulose) (shows convergence). 11. Role of lysozyme in digestive physiology of ruminants, langur monkeys and hoisan birds • Much more divergence in Langur vs. Baboon, than Cow solely on convergence • No common ancestor with lysozyme. 12. Characteristics that distinguish "digestive" lysozyme from "conventional" lysozyme • See question 10. Lecture 21: Experimental Evolution Evolution experiments with E. coli shows how novel traits can evolve • Assigned Independent Study Readings or Activities: o Watch the short video and look over attached paper (Nature 2012). • Independent Study Outcomes: 1. From the paper: Looking at Figure 1, what is meant by Potentiation, Actualization and Refinement a. Bacterial populations went through three successive evolutionary steps: • ‘Potentiating’ mutations (of unclear nature) were required for cells to acquire… • ‘Actualizing’ mutations consisted of a specific rearrangement of a few genes and that allowed some growth — although poor — on citrate in the presence of oxygen. • ‘Refining’ mutations involved duplications of the rearranged DNA sequence, were needed for robust growth under such conditions. 2. Look back at your respiration notes...where does glucose and citrate come in to cellular respiration? a. Glucose: beginning of cellular respiration, in cytoplasm (GLycolosis) b. Citrate: Krebs cycle, in mitochondrial matrix… Citrate, an intermediate in the Krebscycle. After the pyruvate dehydrogenase complex forms acetyl-CoA, from pyruvate and five cofactors (thiamine pyrophosphate, lipoamide, FAD, NAD+, and CoA), citrate synthase catalyzes the condensation of oxaloacetate with acetyl CoA to form citrate. • Lecture Outcomes: 1. Characteristics of model systems that can be used for experimental evolution • Using controlled experiments comparing two current species to see probability of homology. • Use species with very short generation time. 1.Ecoli are a very good use since we can freeze (cryopreserve) each generation, and if needed unfreeze them again to continue to grow them. 2. Origins of genetic novelty (variation) 1) Spontaneous mutation  Random  Mostly either bad (deleterious) or neutral (silent mutation, since amino acid sequences are redundant.  Rarely good... But would be selectively advantaged and is the source of genetic novelty . 2) Gene duplication/Amplification  3 fates (see Q.3) 3) Gene rearrangement  Taking a promoter from one gene and replacing another gene’s promoter with it.  Similar to sub functionalization, except the gene with its new promoter got the promoter from another gene instead of getting the new promoter from a mutation.  New gene expression. 3. Possible fates of duplicated genes • Neo functionalization. 1.Copy produced  mutation in the copied gene − New genetic novelty − Only 1 gene essential, so the copied gene experiences less selective pressure than the original required gene. This copied gen becomes mutated faster due to the weak selective pressure aroused. • Copied gene is lost. 1.Happens 99% of the time 2.No effect in copied gene. • Sub functionalization. 1.Copy produced  New Gene expression (gene is kept, only an additional expression arises) − New genetic novelty − Changed regulatory element (gene expression) 4. Relative impact of selection on duplicated genes • Less selective pressure on duplicated genes, unless altered in some way to change the phenotype or gene expression. 5. Design of lenski's long term evolutionary experiment (lee) with e. Coli • One cell  12 copies  12 identical populations  everyday transfer 0.1 ml cells (5million cells) into fresch media (containing nutrients that would normally last for 8 hours). 1.Every 500 generations (75 days) freeze (cryopreserve population) 2.Generation time of 3.63 hours 3.Asexual (keeping things simple = no recombination) (binary fission) • Mutation: Spontaneous mutation, gene duplication, and gene rearrangement possible. 4.Population size is huge: good since higher chance new alleles may come about. 6. Value of cryopreservation to lee • Can see how evolved if subjected to some form of selective pressure. • Can compare generations to see evolutionary change caused by some form of selective pressure. 7. Where citrate enters metabolism • Krebs cycle, after acytl-CoA (which comes from pyruvate in cytosol) enters the mitochondrial matrix it converts to citrate, and is used for Krebs cycle. 8. Role of glucose limitation in lee experiment • Selective pressure on Ecoli to find some way tofind an additional carbon source (citrate) to survive longer ie. Ecological opportunity. 9. How to determine if cit+ phenotype arose from one single mutation or was dependent on previous mutations (contingency)? +  We replay the evolution in the Ara-3 line (which had the Cit mutation).  Can trace the Cit + based on previous generations that are frozen + • Found that between 32,000 and 33,000 Cit had been slightly present (Actualization) (1 gene rearrangement). + • Then from 33,000 and on the Cit had become used in very high amounts (Refinement) (multiple gene copies with the new promoter sequence that allows Cit +transporter to be active in O 2 environments)  After replaying evolution in the Ara-3 line (using ancestors) it was found that there had been no capacity + th to generate Cit genes before the 20,000 generation. This shows that it had not been a single mutation, and that it had relied on another prior mutation. + • If it were a single mutation dependant phenotype then one would find the Cit come about in any one of the generations. 10. Genetic changes giving rise to potentiation, actualization and refinement of cit+ phenotype + • Potentiation: potential for Cit phenotype to arise ie. generation ~20,000. • Actualization: single gene duplation and sub-functionization mutation giving rise to a new promoter, that is expressed even when O 2 is present. + • Refinement: multiple copies of the same gene that gave rise to slight Cit phenotype, so that it is now expressed in a large amount. 11. Result of "replaying" evolution of cit+ phenotype + • Replaying experiments showed similar mutations giving rise to Cit phenotype, but the sequences varied ever so slightly (ie. different boundaries of amplification of the Cit + line). 12. Why cit+ lines do not drive cit- lines to extinction • The Cit lines did not drive out Cit lines because it was found that the Cit lines out competed the Cit lines + in using glucose as an energy source. (ie. take away citrate, and the Cit - lines where more abundant than the Cit +lines. + 13. Does the Cit phenotypical Ecoli deserve to be called a new species? • No, not really since the Cit lines weren’t driven to extinction. Lecture 22: The Elysia/Vaucheria system What is an animal doing with a photosynthetic gene in its nucleus? • Assigned Independent Study Readings or Activities: o As preparation for the final exam the instructor in the video makes a number of small errors with regards to genes, proteins, photosynthesis. Can you pick them up? o • Independent Study Outcomes: 1. Identify a correct PCR reaction. (see PCR.pdf) a. See notes • Lecture Outcomes: 1. Location of psbo gene in photoautotrophic organisms • Nucleus (although it is a chloroplast gene) 2. Location and role of psbo gene product in photosynthetic electron transport • The gene codes for a protein in the photosystem II of the chloroplast ETC, it is responsible for electron donation to H 2 in the making of reduced ½ O .2 3. Purpose of molecular size markers in electrophoresis • To identify the approximate size of a molecule run on a gel(ie. approx. bp length) 4. Interpretation of agarose gel data • Represents amplified DNA (the DNA had been amplified by PCR) 5. Mechanism of polymerase chain reaction • Amplification of specific DNA 1.Taq DNA polymerase (heat stable 72 oC) 2.2 Gene specific primers (1-30 bp long) • The longer the more specific 3.Monomers of DNA to be used by polymerase (ie. dATP, dTTP, dCTP, dGTP) 6. Role of thermal cycling in polymerase chain reaction • To continually separate the binding DNA double strands, in order to be copied again. 7. Role of primers in polymerase chain reaction • To attach to the appropriate gene sequence 8. Role of Taq polymerase in polymerase chain reaction • It is a DNA polymerase that is stable in high temperatures (doesn’t denature when we want temperatures to be ~72oC). 9. How to make primers to amplify an unknown gene sequence • Goto genBank • The gene is bound to be known in other sequenced organisms. • Make degenerate primers (allowing every possible combination) for both of the DNA sequences (ie. the complementary strand too). 10. Why it is necessary to make degenerate primers for PCR • To consider that there are different DNA sequences for a particular amino acid sequence (ie. psbO) 11. How to use RTPCR to amplify particular mrna • Use a poly-T-tail (complmentary to all mRNA poly-A tails) as a primer for DNA reverse transcriptase • All mRNA in cytoplasm is sequenced into DNA using poly-T tail and DNA reverse transcriptase. • Using primers for the cDNA of the mRNA it will bind to it and basically PCR is carried out, amplifying the cDNA of the mRNA, thus showing if a particular mRNA is expressed in the cell. 12. Usefulness of Northern blots in addition to standard RT -PCR • To show the expression of the mRNA, in terms of cDNA on the blottage. 13. Mechanism and result of secondary endosymbiosis • Primary endosymbiosis: Primitive eukaryote engulfs cyanobacteria and in turn the cyanobacteria develops into a chloroplast for the cell (with 2 membranes). • Secondary endosymbiosis: after Primary endosymbiosis, another eukaryote engulfs the cell, the mitochondria of the engulfed cell breaks down, and the engulfed cell becomes a chloroplast with 4 membranes. 14. Role of tripartite target sequences in localizing proteins to sub-organellar compartments (ie. Thylakoid lumen) • The role of tripartite target sequences is to enter the 4 membranes of the chloroplast of a cell undergone secondary endosymbiosis. The first portion of the sequence is for entering the first 2 outer membranes , the second portion of the sequence is for entering the next 2 membranes of the chloroplast and the last portion is for entering the thylakoid membrane to enter the thylakoid lumen. Lecture 23: Oxygen and Aging The paradox of aerobic life is that while oxygen is essential for life it is also potentially toxic • Assigned Independent Study Readings or Activities: o Chapter 6.7d • Independent Study Outcomes: 1. What are the major forms of reaction oxygen and how are they formed (see Figure 6.25). a. Superoxide, Hydrogen peroxide, Hydroxyl radical i. Conversion of O2 to water by four electron reduction. These reaction oxygen are intermediates of the process of O2  H2O 2. What reactions are catalyzed by the two enzymes Superoxide Dismutase and Catalase. a. Superoxide Dismutase = SOD … see figure 6.25! • Lecture Outcomes: 1. Role of cytochrome oxidase in production of ROS • Not a major source of ROS! It is the last complex to receive the e from the original NADH molecules (from Complex I: NADH Dehydrogenase). The Cytochrome Oxidase (Complex IV) stores the electrons until a total of 4 e-are stored, which is then all at once transferred to the O 2olecule (terminal electron acceptor) to - create 2 H 2O molecules (Donates 4 e to O2 at a time so that no partially reduced forms of Oxygen is made). 2. Role of Q (quinone) pool in production of ROS • Because the Q pool is basically a bridge from Complex I Complex II, it is responsible for transferring the e-taken from the NADH in NADH Dehydrogenase (Complex I) to the Cytochrome complex (Complex II)(ie. Q becomes reduced, then oxidized in the transfer). However, since there are a lot of O 2 molecules in the matrix of the mitochondria, these O 2molecules and Complex III are competing for the electron. Which if O 2gets it, superoxide (a ROS) is produced. 3. How defects in mitochondrial function might lead to disease 1) Mutation in an ETC Complex protein/gene, which would promote electron leakage, hence ROS production is increased. (Ex. Parkinson’s Disease: slight Complex 1 mutation) 2) Mutation in one of the Antioxidant enzymes (SoD1 or Catalase) leading to the inability to get rid ofall the inevitably produced ROS. (Ex. ALS (Amyotrophic lateral sclerosis): slight SoD1 enzyme mutation) 4. Human diseases associated with mitochondrial dysfunction • ALS (Amyotrophic lateral sclerosis): slight SoD1 enzyme mutation • Parkinson’s Disease: slight Complex I mutation 5. How defects in cytochrome complex lead to increased oxygen toxicity in C. Elegans (worm) • Cytochrome complex (Complex III) (slight protein mutation ie. too much would be lethal) becomes not as efficient as the wildtype worms. CC mutant percent survival is much more susceptible to an increase in O 2 (since higher electron leakage/ lower efficiency/ a lot of O 2 in matrix = more ROS made) which is linked to higher ROS. The mutants also have a much lower life expectancy. 6. Relationship between mitochondrial ROS and ageing • See question 5. • 7. Evidence in support of ROS ageing hypothesis • Parkinson’s disease (Complex I), ALS disease (SoD1), Centenarians (mitDNA). • Also the worm with mutant
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