BCHM 3050 FINAL EXAM STUDY GUIDE
BCHM 3050 FINAL EXAM STUDY GUIDE 3050
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This 25 page Study Guide was uploaded by Luke Holden on Monday April 25, 2016. The Study Guide belongs to 3050 at Clemson University taught by Dr. Srikripa Chandrasekaran in Winter 2016. Since its upload, it has received 88 views. For similar materials see Essential Elements of Biochemistry in Biochemistry at Clemson University.
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Date Created: 04/25/16
Final Exam Study Guide Material from Exam1 (~8 questions) Given a dipeptide, be able to identify the aminoacids, and using a chart with pK values, be able to calculate the pI KNOW ALL OF THE AMINO ACIDS!!!!! o USE THIS QUIZLET TO HELP: https://quizlet.com/115307317/amino acidsforbchm3050flashcards/ Steps to identify the pI: o Label the R groups and COOH and NH3 groups that have pKa’s Remember the structure of an amino acid: Some R groups may not have a pKa value and that is OK. DO NOT FACTOR IN THE pKa of the NH2 GROUP AND THE COO GROUP THAT IS INVOLVED IN PEPTIDE BONDING o Make a pH to charge chart/table Look at the charges on the peptide and add up the charges to get the starting charge. (The amine groups will generally be the ones racking up on charge.) Then generate chart pH Charge 1 +2 2.4 (first pKa) +2/+1 ( Remember pKa is the pH at which that corresponding substituent has released ½ of its protons) 4.7 (Second +1/0 pKa) 9.8 (Third 0/1 pKa) o Once the table has been made, you can then determine the answer to multiple questions. What is the charge of the peptide at a certain pH Go to where the corresponding pH value would be in the chart and that will give you the charge What would the pH be at a certain charge Work backwards! Calculate the pI Find where the charge is zero and find the answer choice that has that charge in between the two pKa’s Given a dipeptide, be able to look at the charge distribution on the peptide and determine the pH at which a certain charge can be maintained Explained above. Just find the charge with the corresponding pH! Given a dipeptide, be able to tell what the charge of that peptide will be at a certain pH Explained above Solve pH problems; Given [H+] or [OH] concentration, tell me what the pH would be. pH=−log H ] For example if the [H]= 2.7 x 10 pH=−log H ] pH=−log[0.00000027] pH=6.57 Remember : 14= pH+ pOH Given the pH tell me what the [H+] or [OH] would be For Example: pH=5 pH=−log H ] 5=−log H ] 10 =10 log[H] 0.00005=[H] Using MichelisMenton (MM) and Lineweaver burke ( LWB) equation and plots, calculate different variables ( Km, Vo, Vmax, substrate concentration). a. Vmax= max velocity of the reaction b. Km= ½ the concentration of ½ the maximum velocity c. Both are constants that help us identify enzymes d. Km enzyme affinity for a substrate. i. Decrease the KmIncrease Substrate affinity Using LWB plot, be able to calculate values of Km and Vmax. Km= X intercept = -1/Km Vmax= Y intercept= 1/Vmax Just use algebra to find the values of Km and Vmax Know how the 3 different inhibitors work (competitive, uncompetitive, and non competitive). How do they change/affect Km and Vmax? Given MM or LWB plots, be able to identify the type of inhibitors. Know the difference between primary, secondary, tertiary and quarternary protein structures. Chemically how are they different from each other? ( Focus on the kind of bonds that are formed in each structure). Structure Type of Bonds Special Notes Primary Peptide bonds Just the order of the amino acids Secondary Peptide Bonds α and β stuff in here. Hydogen Bonds (For the Turns) Tertiary Hydrophobic interactions Multiple motifs but together. Ionic (electrostatic) interactions But still one polypeptide Hydrogen bonds Covalent Bonds Hydration Quaternary Two or more tertiary This is TWO or more structures are connected. polypeptides (generally in tertiary structure) Very high ordered function Extra Notes on Protein Structure: Tertiary Notes: Unique 3d structure of one long polypeptide. This is typically the end point for most proteins. This is when it is originally very unorganized and it undergoes protein folding. This makes it a highly organized molecule. It has several distinctive features: The amino acids that were far away are now close to each other. This tight packing causes water to be expelled and the polar and nonpolar amino acids have ability to interact forming electrostatic bonds. This contains domains! So specifically it is called a domain. however, when in tertiary structure (meaning it has taken on a 3d shape) it is called a fold. Most proteins are considered mosaic. It is like the janitor (protein) carries around many keys. Each key fits a lock in the school (specific system) but some keys go to his car or to his house which is not apart of the specific system. He is still able to interact with other things but most specifically his system he is assigned to. This relies on a crap ton of interactions in order to be folded correctly. Hydrophobic interactions This when the water insoluble molecules are driven into the center of the molecule expelling the water and thus creating solvation spheres in the center of the molecule. This increases the entropy of water. (However, some water stays in the center and forms 4 different bonds stabilizing the backbone. This frees up the molecule to move and be flexible.) Ionic (electrostatic) interactions These are interactions where water ahs been expelled thus reveling the + and – charges. When they bond it is called a SALT BRIDGE. These are good for adjacent subunits (adjacent polypeptide chains) in a complex proteins. Hydrogen bonds ALL OVER THE PLACE. These form literally everywhere there IS NOT WATER. Covalent Bonds These are created whenever you have the modifications after translation. But most prominent are the disulfide bonds that form as a result of covalent bonds. These are the big foundation proteins because they allow the protein to remain folded (to an extent) adverse conditions. Hydration water causes the protein to remain folding. This is kind of like a lot of helpers holding up the protein which allows it to be sturdy. This is also important because whenever a molecule comes to the active site of enzyme and the protein binds to the enzyme. Water is moved thus increasing the entropy of water and ultimately the bind of the enzyme to the substrate. Quaternary Notes: a. This is when two or more polypeptides (subunits) join together to form a larger more difficult task. b. Why is this better than tertiary structure?: ii. It may be easier to synthesize 4 short polypeptide units rather than 1 long one iii. It is easier to repair one small subunit instead of a big one iv. This increases biological function!! DO NOT MEMORIZE ALL THE DETAILS JUST THE OVER ARCHIG THEMES! Materials from Exam2 ( ~6 questions) Chromatin structure Be able to identify all purines and pyrimidines, Know difference between nucleoside and nucleotide, Know the functions of a phosphoester bond and a phosphodiester bond; Know Chargaff’s rule. 1. Nucleotides Include a nitrogenous base, a pentose sugar, and a phosphate. 2. Nucleosides Nitrogenous bases attached to sugars (Basically a nucleotide without a phosphate) 3. Nucleic Acids Linear polymers of nucleotides that function in the storage and expression of genetic information and pass it along to the next generation. This includes both Ribonucleic Acid (RNA) and Deoxyribonucleic Acid (DNA) 4. Purine Fused 6 membered + 5 membered heterocyclic rings of Carbon and Nitrogen, including Adenine and Guanine (also intermediates: Xanthine and Hypoxanthine) 5. Pyrimidine 6 membered heterocyclic rings or Carbon and Nitrogen, including Thymine (DNA), Cytosine, and Uracil (RNA) 6. Quizlet Link for purine and pyrimidine structures: https://quizlet.com/121133124/purine andpyrimidinesflashcards/?ne . Nucleosides Nucleotide The Bond Chart Type of Bond: Between what: Glycositic linkage Nitrogenous Base and Sugar Hydrogen Bond Between the Bases of two strands Phosphoester bond Between the phosphate and the sugar Phosohpdiester Between the two phosphates of two nucleotides Phosphoanhydride Between the 2 phosphates on a triphosphate VERY EASILY BROKEN!!! Chargaff’s Rule: [A]=[T] and [G]=[C] [A]= 30% what is [G] ? Answer= 20% DNA Replication: Know names and functions of all the enzymes that participate in DNA replication Enzyme Function What is it was defected? Helicase Breaks hydrogen bonds, Replication could not start breaks the helix unwinds because the DNA could not DNA at origin, and stays until open the end of replication Singlestrand binding protein Make sure the DNA strands Hydrogen bonds would don’t snap back together reform after helicase already opened them DNA polymerase 3 Puts down the nucleotides and Replication could not take carries out the replication place because nothing would be able to replicate the DNA. RNA Primase Puts down RNA primer so DNA polymerase 3 could not that DNA can bind put nucleotides together and nucleotides together carry out replication because it could not start DNA polymerase 1 Removes RNA stretches Then the lagging strand could (primers) and replaces them not be fully synthesized and with DNA then patched together by Ligase because the primer would still be there. Ligase Joins together Okazaki The lagging strand would not Fragments become one complete strand and would still be in pieces Topoisomerases Nick supercoiling DNA and DNA would supercoil and relax stress by allowing break. uncoiling Know Griffith’s experiment o o THE POINT: TRANSFORMATION!!! o Since bacteria was recovered in the mouse that died in the last column, that meant that some type of DNA uptake had to occur. Know Avery, McLeod, McCarthy’s conclusions o THE POINT: VERIFICATION OF GRIFFTHS EXEPRIEMENT BY GOING BACKWARDS o They thought, since there are so many theories going around about what his transforming agent is, let’s put them all to the test at once o If we put the bacteria in with different digestive enzymes, ones that would degrade only specific stuff, then the mouse that lives is the one that is the transforming agent. o DNAase= mouse lives Know Hershey and Chase experiments o o They said what is the transforming agent for the bacteriophage? o Their determinant was the centrifuge o So when the bacteria was centrifuged, the DNA fell to the bottom while the actual shells of the bacteriophage stayed in the solution. o They labeled DNA P32 and protein S35 of bacteriophages, two radioactive isotopes used for identification. o When they centrifuged they found that S35 was found to be in the solution (with the protein) and that P32 was in the supernatant (P32) o THE POINT: DNA IS THE TRANFORMING AGENT OF BACTERIA Know the difference between prokaryotic and Eukaryotic replication Eukaryotes Prokaryotes It occurs inside the cytoplasm. It occurs inside the nucleus. There is single origin of replication. Origin of replications are numerous. DNA polymerase III carries out both Initiation is carried out by DNA initiation and elongation. polymerase α while elongation by DNA polymerase δ and ε. DNA repair and gap filling are done by DNA polymerase I. The same are performed by DNA polymerase β. RNA primer is removed by DNA polymerase I. RNA primer is removed by DNA polymerase β. Okazaki fragments are large, 1000 2000 nucleotides long. Okazaki fragments are short, 100200 nucleotides long. Replication is very rapid, some 2000 bp per second. Replication is slow, some 100 nucleotides per second. DNA gyrase is needed. DNA gyrase is not needed. How did Meselson and Stahl prove that replication is semiconservative. If given a description of the experiment, tell me the percentage distribution of N14 v/s N15 DNA. Material from Exam3 (~10 questions) Transcription: Know the enzyme/s that participate in DNA transcription in eu and prokaryotes; Transcription factors (eukaryotes), RNA polymerase Prokaryotes only have RNA Polymerase II This does it all Eukaryotes have 3 different polymerases Compare and Contrast Prokaryotic and Eukaryotic Characteristic Prokaryotes Eukaryotic Polymerase RNA Poly: Can do it all! RNA poly I: large RNA rRNA RNA poly II: mRNA mRNA RNA poly III: tRNA small tRNA tRNA Initiation Factors σ factor further binds to the TF IIs bind to the promoter promoter Know Differences in transcription in eu and prokaryotes; Prokaryotes promoter at 35 & 10 bp, Rho factor termination, lac and trp operon Eukaryotes TATA (25 bp) CAAT (50) GC (80) box, Enhancers/silencers that modify gene expression (thousands of bp downstream binds to an activator, loops entire DNA and activator starts interacting with RNA polymerase and starts transcription),3 different types of RNA polymerases, mRNA processing (5’ capping, 3’ poly A tail, splicing to remove introns ) What are the three types of RNA processing that occur in eukaryotes? (capping, tailing, splicingknow all details in the powerpoints about these 3 processes) 3 Types of Processing o 5’ Cap addition of 7 methylguanosine Need to know chemically what is a cap or a tail 5’5’ addition (not and error there) with a triphosphate Fig. 18.24 (McKee & McKee) 3 nucleotides before are methylated at the 2’ OH group This cap serves as the recognition site for attachment and prevents degradation by exonucleases o 3’ polyaddenation ADDs 100250 AA Helps direct mRNA’s out of the cell Protection from degradation RNA Poly A Polymerase Polymerase Increase the length of the tail increases the longevity of the mRNA o Splicing of mRNA snRNP’s and splicesomes (exonuclease) premRNAs have nucleotide sequences that serve as the splicing signals snRNPS’s recognize the sites Multiple snRNP’s join together to form a splicesome (RNA ligase and Exonuclease) Know this Cuts out introns and stiches together Exons Occurs in the nucleus o Functions of introns Exon shuffling Exons correspond to different functional regions of a polypeptide Provide cross over recombination sites Help with recombination to promote function Help signal when mRNA is ready o Active Genes Transcribed many times After the 1 wave of ribosomes go through, the 2 and third wave comes in which increase the amount of the gene that can be expressed Need to know about lamp brush chromosomes Constantly transcribed Only in eukaryotic Transcription regulation Know the difference between hetero and eu chromatin. o Heterochromatin “tight” can’t be translated o EU chromatin “loose” chromatin that is translated What chemical modification causes hetero v/s euchromatin? o Add acetyl group to lysine or arginine on histones to loosen DNA and form EU chromatin o Remove acetyl group to tighten DNA and from heterochromatin What are some examples of chemical modifications that happen to DNA and histones during epigenesis? o Addition of methyl group (on DNA) can change the way DNA is read and modify the function o Addition/removal of acetyl group (on histone) can tighten or loosen DNA Know all the examples we discussed about transcriptional and posttranscriptional regulation? o Transcriptional Regulation Enhancers activators and Mediators Activator binds to enhancer and stimulates transcription Mediator proteins mediate interaction between enhancers and DNA Transcription initiation complex: TF, Activators, RNA poly II and mediator LCR Locus Control Region Example of enhancer region in DNA Depressors and Silencers Repressors bind to silencers Inhibits DNA o Posttranscriptional Regulation Removal of introns 5’ cap 3’ Poly A tail How is lac operon regulated in the presence and absence of glucose and lactose. Know what happens to repressors and regulators in all these conditions +Glu +Glu Glu Glu +Lac Lac +Lac Lac Production + +++ (50 times /+ as much) Repressor Bound to Bound to Bound to Bound to allolactose operator lactose operator cAMP and CAP None None Increase cAMP Increase and CAP cAMP and CAP Translation Know all the steps in translation. Focus on differences in translation between eu and prokaryotes. o Initiation: This is the first step when the mRNA strand arrives at the ribosome Formation of the initiation complex The shinedalgarno sequence: this is a purine rich sequence upstream from the AUG codon. Recognition of this sequence is crucial and is recognized by the 16s rRNA strand. o Consensus sequence: AGGAGG Sequence of events: o IF1 and IF 3 bind to the 30 s subunit IF1 blocks the A site until the first tRNA is in place IF3 blocks the 50s subunit from binding the 30s subunit. o 30s subunit binds to the shine dalgarno sequence o 30s subunit slides in place over the AUG start codon (5’3’) o Insure that the first AUG start codon is located o IF2 with GTP binds to the 30s Asite o IF1 is then displaced and the fMettRNA (ONLY PROK) (Insert Picture) o GTP is hydrolyzed which makes IF2 and 3 fall off o The 50s subunit then comes and binds the complex thus creating the characteristic ribosome o Elongation: The EfTu protein carries the charged tRNAs and then hyrolzes GTP to attach it to the codon in the A site. GTP is then rehposphrylated and the EFtu is then reattached with a charged tRNA. peptidyl transferase (in the large ribosomal subunit)then attahces the growing polypeptide chain to the charged tRNA in the A site. EFGGTP moves the ribosome in the 3’ direction which shifts the tRNA’s over a site. The energy released from the GTP causes the ribosome to change conformation. o Termination: The stop codon calls for the initiation factor which comes and bids to the A site This converts peptidyl transferase into hydrolyase which cuts the polypeptide away from the tRNA and Ribosome falls away o Multiple ribosomes bind to the same sequence of mRNA and forming polysomes which amplify the protein. Eukaryotes don’t uses the shine delgarno sequence and their subunits are 40s & 60s What are the roles of the P,A and E sites during translation? o P: binds the first tRNA carrying a methyl (fmet for prokaryotes) on the AUG start codon o A: site where all charged tRNA enter (other than the first) and line up with codon o E: uncharged tRNA exit the ribosome to go get another amino acid Mutations; Given a protein sequence or an mRNA or a sense/antisense DNA sequence be able to identify the type of mutation shown here, and how the mutation affects function. o Missense: This is where an incorrect Amino acid is inserted into the protein sequence. This can be detrimental as it can influence misfolding of the protein and thus altering its function o Nonsense: This is a mutation where a stop codon is sequenced and causes early termination of translation. Usually results in dysfunctional protein that is useless o Silent: This typically occurs during the wobble hypothesis where the third nucleotide of the codon is switched for another nucleotide. However, to our luck, they code for the same protein and thus the protein goes on about its life. You would not know these occur unless you looked at the genetic code. Materials from Exam 4 (~ 8 questions) Glycolysis: Know all enzymes in glycolysis, which steps release ATP/NADH, invest ATP, keep track of C atoms, classify enzymes into groups, and know inhibitors and activators of enzymes Steps of glycolysis (10 Step) Energy investment Phase (15) o Step 1 ATP + Glucose(Hexokinase) Glucose6PPhosphate Traps glucose inside the cell The addition of a phosphate causes a negative charge to be on the molecule which restricts it from leaving the cell Irreversiable Regulation possible= The product inhibits Hexokinase Inhibition= glucose 6Phosphate Glucose 6 Phosphate is a huge compound Glycogen synthesis Glycolysis Pentose Phosphate o Step 2 Glucose 6 Phosphate (Phosphoglucose) Fructose 6 Phosphate Reversible o Step 3 (Most Heavily Regulated) fructose6phosphate (phosphofructokinase 1) fructose 1,6 bisphosphate Irreversible Activators: o AMP o ADP o fructose 2,6 bisphosphate Inhibitor o ATP o Citrate o glucagon ATP Generally inhibits glycolysis while AMP + ADP to activate it ATP is allosterically inhibited o Step 4 (Super important) fructose 1, 6 bisphosphate (Aldolase) dihydroxyacetone phosphate+ glyceraldehyde 3 phosphate o Step 5 dihydroxyacetone phosphate (triose phosphate) Glyceraldehyde 3Phosphate Energy Harvest Phase o Step 6 Glyceraldehyde 3 phosphate+ NAD + P(Glyceraldehyde 3 Phosphate dehydrogenase) 1,3 bisphsophoglycerate Generation of a reduction agent NADH o 2 molecules of NADH generated Oxidative phosphorylation step!!!! NADNADH (low potential energyHigh potential energy) o Step 7 (Important Enzyme) 1,3 bisphosphoglycerate (Phosphoglycerate Kinase) 3 phosphoglycerate First ATP generated in glycolysis via substrate level phosphorylation All kinases transferase reversible Standard Free Energy of Hydrolysis 1,3 Bisphosphoglycerate 43.4 kJ/mol ATP 30.5 kJ/mol When these are broken down the extra 15 kJ are lost as heat o Step 8 3phosphoglycerate(phosphoglycerate) 2 Phosphoglycerate o Step 9 Phosphoglycerate (Enolase)Phosphopheolpyruvate Inhibitor: Flouride (toothpaste) o Step 10 phosphoenolopyruvate +ADP (pyruvate) pyruvate +ATP Activators Fructose 1,6 bisphosphate Irreversiable Inhibitor o ATP o acetyl CoA o Alanine Name of Enzyme Class of Substrate Products Irreversible? Inhibitor Activator? ATP Enzyme ? invested ? Hexokinase Transferase Glucose & ATP Glucose 6phosphate & Yes Glucose No Yes, 1 ADP 6 phoaphate Phosphoglucose Isomerase Glucose 6phosphate Fructose 6phosphate No No No No Isomerase Phosphofrutokinas Transferase Fructose 6phosphate Fructose 1,6bisphosphate Yes ATP & AMP, ADP, Yes , 1 e1 & ATP & ADP Citrate fructose 2,6 biphosphate Aldolase Liase Fructose 1,6 Dihydroxyacetone No No No No biphosphate phosphate & glyceraldehyde 3 phosphate Triose phosphate Isomerase Dihydroxyacetone Glyceraldehyde 3 No No No No Isomerase phosphate phosphate Glyceraldehyde 3 Oxyreductas Glyceraldehyde 3 1,3bisphosphoglycerate No No No No phosphate e phosphate & NADH dehydrogenase Phosphoglycerate Transferase 1,3 3phosphoglycerate & No No No No kinase bisphosphoglycerate ATP Phosphoglycerate Isomerase 3phosphoglycerate 2phosphoglycerate No No No No mutase Endolase Liase 2phosphoglycerate Phosphoenolopyruvate & No No No No H2O Pyruvate kinase Transferase phosphoenolopyruvate Pyruvate & ATP Yes ATP, Fructose No Acetyl 1,6 CoA, biphosphate alanine Krebs: Know all enzymes, which steps synthesize ATP, keep track of C atoms, classify enzymes into groups, know inhibitors and activators of enzymes, which step releases NADH/FADH2/CO2 Step 1 o Pyruvate(Pyruvate Dehydrogenase) 2 Acetyl Coa o This is the only reaction that occurs in both the mitochondrial matrix and the inner membrane Step 2 o Oaxalacetate(from the previous cycle) + Acetyl Coa (Citrate Synthase) Citrate Step 3 o Citrate(Aconitase) Isocitrate Step 4 o Isocitrate +NAD(Isocitrate Dehydrogenase)αketogluterate +NADH+CO2 Step 5 o Oaxalsuccinate+ NAD ( αKetogluterate dehydrogenase) Succinyl Coa +NADH +CO2 Step 6 o Succinyl Coa +GDP(Succinyl Coa synthase) Succinate o GTP phosphorylates ADP to ATP o WARNING, EVEN TOUGH THE ENZYME SAYS SYNTHASE, IT REMOVES COA!!!!! DON’T BE TRICKED Step 7 o Succinate +FAD(Succinate Dehydrogenase) fumerate +FADH2 Step 8 o Fumerate +H2O(Fumerase) Malate Step 9 o Malate +NAD(Malate Dehydrogenase) Oaxalacetate +NADH ETC: Know order of complexes, be able to arrange compounds in the order of reduction potential values, know inhibitors of these complexes, know ionophores and uncouplers and how they affect ETC How the ETC is set up: Glycolysis NADH: 2 NADH Krebs NADH and FADH 8 NADH 2 FADH Some more about those complexes: o Complex 1 called NADH dehydrogenase o Complex 2 Called Succinate Dehydrogenase o Complex 3 cytochrome b/c complex o Complex 4cytochrome c oxidase Electron buses: o ubiqinone This bus carries the electrons from complex 1 and 2 and carries them to complex 3. o Cytochrome c carries the electrons from complex 3 to complex 4. 4 Important Biomolecules that make up the complexes: o Flavoproteins contain flavin (FAD)(FMN) o FeS proteinnonheme proteins contain 24 fe atoms bound to protein via cysteine residues o Cytochromes heme proteins contain single Fe atoms bound to the heme o Ubiquinone (Coenzyme Q) nonprotein Lipophillic molecule The ETC and Standard Reduction potential Explained: o When you start at complex 1 you are essentially starting at the top of the electron tower and then going down the tower. o As you go down the tower, you will increase in standard reduction potential. o Therefore, the highest standard reduction potential is oxygen Some poisons inhibit ETC Amytal and Rotenone Inhibits the movement of electrons from NADH to complex 1 Antimycin Inhibits the transfer of electrons in complex 3 Carbon monoxide, Cyanide, and Azide inhibit the final transfer of electrons to oxygen Know the differences between aerobic respiration and fermentation Lactic Acid Fermentation A common theme throughout the regulation process is the regeneration of NADH and FADH2. The cell knows that Lactic Acid fermentation is slow and inefficient. Thus, it does everything it can to keep the ETC going since it substantially more effective. Pyruvate (lactate dehydrogenase)Lactate Glyceraldehyde 3 Phosphate(GlyceraldehydeP Dehydrogenase) Glycerate 1,3 bisphosphate Alcoholic Fermentation Note re Alcoholism: Ethanol stimulates synthesis of NADH in liver by ADH The extra NADH will inhibit glycolysis and fatty acid oxidation Acetylaldehyde Acetic Acid Fatty acid synthesis Result is accumulation of fat in the liver o loss of function cirrhosis
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