Biochemistry Exam 1 Review Study Guide
Biochemistry Exam 1 Review Study Guide 3050
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This 16 page Study Guide was uploaded by Luke Holden on Friday February 5, 2016. The Study Guide belongs to 3050 at Clemson University taught by Dr. Srikripa Chandrasekaran in Winter 2016. Since its upload, it has received 111 views. For similar materials see Essential Elements of Biochemistry in Biochemistry at Clemson University.
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Date Created: 02/05/16
Study Guide for BCHM EXAM 1 1. Understand different types of bonds a. Covelant Bonds: These bonds share electrons with other atoms i. Polar Covelant: Very Different EN= Unequal Sharing ii. Nonpolar Covelant: Same EN= Equal Sharing b. Ionic bonds: These are attracted ions that remain close to each other in order to have their octet rule satisfied. These typically occur between nonmetals and metals c. Hydrogen Bonding: This occurs with the atoms FON. They are a special type of covalent bonding and they give water its many properties. d. Metallic bonding occurs between two metal atoms. 2. Understand the properties of water and how hydrogen bonding influences them a. Water can form bonds at 1 time. Two on the oxygen and two on the hydrogen. So water can bind with other water molecules because of hydrogen bonding. b. Makes H20 the universal solvent. This makes water form solvation spheres( spheres that form around individual solute atoms: c. Cohesion: This is the ability for water to stick together and form a relatively high surface tension in the liquid state. This is due to the hydrogen bonding being able to occur among other water molecules d. Adhesion: This is the ability of water to stick to other things such as the xylem and phloem of a plant stem. This capability allows water to travel up the root system and nurture the plant. e. Surface Tension: This is the ability of water to remain taught. f. Ice on water: Ice floats on water because as the phase of water changes from liquid to solid, lattices begin to from amongst the water molecules. These lattices in essence open up the space in between each molecule which make the solid denser than the liquid. g. High Melting point: This is because HBonds are the strongest Intramolecular force besides ionic bonding. i. Increase the solutes in water= Increase in boiling point. ii. Increase the solutes in water= Decrease the melting point=NaCl on roads 3. What is osmotic pressure? Be able to calculate osmotic pressure a. Osmotic pressure is the amount of pressure required to stop net flow of water across a membrane. Equation: i. Pi= osmotic pressure ii. =I=1(for this class) iii. M=Molarity iv. R=Gas Constant= 0.08206 atm*L/(mol*K) v. T=Temperature in Kelvins celcius to kelvin = c+273 4. Explain isotonic, hypotonic and hypertonic solutions a. Water will always go towards the most solute!! b. Isotonic: equal amount of solute entering and leaving the cell c. Hypotonic: Influx of water into the cell d. Hypertonic: release of water out of the cell 5. Be able to calculate pH or H+ ion concentration of a strong acid a. Since strong acids dissociate fully, then the pH is calculated using the –log(H) concentration. b. To calculate the H concentration given the pH, then do the inverse log! 10^(pH) c. Equations to know: i. 14= pH+pOH ii. pH= pKa +log (A/HA) iii. p=log 6. Understand the significance of buffers and know the HendersonHassalbalch equation a. Buffers are so important because they keep certain systems of the body at a relative pH. The HerndersonHassalbauch equation is defined as: A− ¿ HA b. ¿ pH=pKa+log¿ c. So your dream buffer, is when the pH =pKa, therefore A=HA 7. Define or describe the “Central Dogma” in relation to the storage & retrieval of genetic information, the creation of proteins and the expression of traits. a. It can be described as Replication Transcription Translation Expression of genetic material through proteins. 8. Illustrate and/or describe the general structural, physical and/or chemical characteristics of amino acids. a. The amino acid has 3 things: i. Carboxylic Acid ii. Amine Group iii. And an R group b. The R group defines the amino acid and will determine whether or not it will fall into the one of the four categories: Nonpolar, Polar, Acidic, and Basic. c. Amino Acids have special characters such as the ability to amphoteric which is tha lity to act as an acid or base, carry both a positive and negative charge on the same molecule. NonPolar: Methionine (Met) Cystine (Cys) C M Glycine (Gly) G Acidic: Basic: Proline (Pro) P Alanine (Ala) A Aspartic Acid Lysine (Lys) K Polar: (Asp) D Valine (Val) V Arginine (Arg) R Leucine (Leu) L Serine (Ser) S Glutamic Acid Histidine (His) H (Glu) E Tyrosine (Tyr) Y Isoleucine (Ile) I Phenylalanine Threonine (Thr) T (Phe) F Aspargine (Asn) Tryptophan (Trp) A W Glutamine (Gln) Q 9. Describe and/or illustrate the titration of an amino acid, including the pI’s or pK’s of ionizable groups. a. The pI of I oniable groups is the average of the two pKa’s and depending on whether or not it is acidic or basic determines what pKas to use. b. c. For glutamic acid, you will use the pKa values of pK1 and pK2. Average these two together. This is because at a pH of 1, all of the functional groups are protonated. When it reaches a pKa of 2.2, this when 50% of the protons on the carboxylic acid have let go of there protns and thus will hang the charge of th amino acid in the balance. When it reaches a pH of 4.3, then 50% of the protons on the R group have let go and thus pulls the chrge down another step. Finally when it reaches a pKa of 9.7, all of the protons have dissociated and the amino acid is now basic. You can create a table like this: pH Charge 1 +1 2.2 +1/0 4.3 0/1 9.7 1/2 d. For the basic amino acid lysine, you will average the two pKa values at the top end. pKa 2 and pKa 3. You start with a charge of +2 at a pH of 1 and then create a table just like acidic amino acid. e. pH f. Charge g. 1 h. +2 i. 2.2 j. +2/+1 k. 9.0 l. +1/0 m. 10.5 n. 0/1 o. 10. Determine the charge of the amino acid at a certain pH ( based on their pK values) a. To determine the charge you just simply make a charge and pH chart like the previous question. Once you find the pH of the amino acid you just go to the spot on the chart that has your pH value (i.e pH 7 on the basic amino acid = +1 charge 11. Detrmine at what pH an aminoacid would have a particular charge ( Based on pK values) a. Do the same thing as before except work backwards. 12. Identify and/or describe the structures & properties of the 20 standard amino acids. a. Refer to quizlet 13. Distinguish between essential vs. nonessential amino acids vs. nonstandard amino acids, including examples of the physiological roles. a. Essential amino acids are the amino acids that needed to be required through food. i. IleLeuLysMetPhyTryThrVal b. Nonessential amino acids are the amino acids that can be made within the cell c. Nonstandard amino acids These are mostly modified standard amino acids that have special biological roles. Some but few are involved with the protein. i. GABAinhibitory NT of the brain, involved in muscle relaxation, sleep, diminished emotional reaction and sedation. ii. SerotoninNT of the brain; modulates mood, appetite, sexual activity, aggression, body temperature, sleep, smooth muscle constriction. iii. Melatoninsecreted by the pineal gland during darkness; linked to circadian rhythms and sleepwake cycles. iv. Thyroxinsecreted by the thyroid; increases rates of chemical reactions and metabolism in almost all cells of the body. v. Indole3acetic acid– (IAA): major plant hormone, stimulates cell growth & elongation, rooting; inhibits axillary bud development. 14. Understand primary structure of proteins a. The primary structure of the protein is the simple order of the amino acids in a protein. 15. Define/describe different secondary structures and supersecondary structures. a. AHelix This is the structure of a protein such that it forms a helix due to intramolecular forces i. This structure is rigid and that is due to resonance and hydrogen bonding ii. Every 4 th amino acid there is going to be a hydrogen bond between the Nitrogen and hydrogen 4 amino acids away. This is then iii. This loss of electrons is stabilized by resonance iv. Proteins such as Gly and Pro due not do a helix 1. Gly R group is a simple hydrogen and theus the structure will be to felixable 2. Pro Too rigid. The big pentagon at the bond would not allow the bond to rotate to form a turn in the helix. In addition there is no N H possible. 3. MALEK b. BPleated Sheet i. This when a single polypeptide chain loops around and forms side by side each polypeptide chain. This causes hydrogen bonds to form between the amine and carbonyl groups. 1. Parallel Sheets This is when the R groups are on the same side and the single segment. This is when Cterminus and CTerminus Line up side by side 2. 3. AntiparallelNterminus and Cterminus line up. Also the hydrogen bonds are straight across. This makes antiparallel more stable than the parallel. c. Motif: This is a mixture of a helix and bpleated sheets. i. BaB motifThis is two beta pleat sheets outline and connected by and alpha helix. They are stabilized by hydrophobic interactions by the nonpolar side chains of both the sheets and helix. ii. Bmender This looks like a solar panel which has multiple beta pleated sheets all connected by loops. iii. Aa unit this is to alpha helices connected by a turn iv. B barrel this is a corkscrew looking barrel of beta pleated sheets v. Greek Key This beta pleated sheets, however, the 1 goes to the 3 which rd loops back to the second which goes to the fourth. vi. d. Domain This is a big motif i. Leucine zipper when it is open it has one function and when it is closed it has another. 16. Understand how to classify them in a Ramachandran plot. What are turns and loops in the context of secondary structures? a. Loops and turns are the connector portions of a protein. These little guys are about 4 amino acids long and are quite huge in the structure of a protein. i. B turn is a very short loop that is created by the hydrogen bond formed between the 1 amino acid in the turn, glycine and its bro proline which is 4 amino acids away. Remember how proline can’t do A helices? Same thing bro. Proline causes a big old fat turn to happen which is perfect for a loop. And glycine is so passive it is cool with just about anything. b. Ramachandran Plots This is a logarithmic graph that allows you to determine whether it is an a helix or a b pleated sheet. c. 17. Define “tertiary protein structure” and describe how hydrophobic & electrostatic interactions and hydrogen & covalent bonds can all be involved in stabilizing tertiary protein structure. a. 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. b. It has several distinctive features: i. The amino acids that were far away are now close to each other. ii. This tight packing causes water to be expelled and the polar and nonpolar amino acids have ability to interact forming electrostatic bonds. iii. 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. iv. 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. c. This relies on a crap ton of interactions in order to be folded correctly. i. 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.) ii. 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. iii. Hydrogen bonds ALL OVER THE PLACE. These form literally everywhere there IS NOT WATER. iv. 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. v. 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. vi. 18. Describe the nature of the “Anfinsen experiments”, and how or why they were important in protein folding. a. These experiments proved that when folding a protein you had to take a specific pathway primary secondary tertiary and quaternary. b. His steps: i. He denatured the protein using BME and urea which BME destroyed the disulfide bonds and the urea disrupts the hydrophobic interactions and thus causes the protein to be denatured. ii. The first time through, he removed the BME first and then the urea and only 1% activity returned iii. The second time through he removed urea first and then BME and almost ALL function returned to the protein iv. The point: This experiment proved that the protein had to be folded a certain way and in a certain order. When the BME was removed first, the protein was able to take several conformations different than those of its original function. Removing Urea allowed the protein to be folded correctly. 19. What are heatshock proteins and/or molecular chaperones, and what is their role in protein folding and turnover. a. Molecular chaperons These guys assist in protein folding i. They keep the unfolded protein from interacting during transition stages to prevent misfolding. 1. Hsp 70s These are proteins that stabilize the early protein. They have to binding sites, one on for ATP and the other for the protein. ii. They actually aid with the process of folding (this is the shake and bake) 1. Hsp 60s These guys are the big can for folding proteins. They have two big rings composed of 7 different subunits and will form on top of each other. This creates a kind of machine that is boss at folding proteins. First, the folded protein enters the top of the Hsp 60 complex into the GROEL rings. Then the GROES cap comes and tops of the can. 7ATP are hydrolyzed in the process which takes about 20 seconds. Then the protein is released along with 7 ADP molecules. iii. 20. Define “protein denaturation” and describe the main reagents or treatments that denature proteins. a. Protein Denaturation: This is when the application of some external factor cause the protein to un fold or misfold thus losing its function. b. Things that make it happen i. Strong Acid or Base ii. Organic Solvents iii. Detergents iv. Reducing Agents v. Salt Concentration vi. Heavy metal Ions vii. Temperature Changes viii. Mechanical Stress 21. Define “quaternary protein structure” and describe how or why it can be advantageous over tertiary structure in terms of protein function. 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?: i. It may be easier to synthesize 4 short polypeptide units rather than 1 long one ii. It is easier to repair one small subunit instead of a big one iii. This increases biological function!! 22. How is protein degradation facilitated? a. Protein degradation in eukaryotes is facilitated by the ubiquitin (U) and protesasome pathway i. U is activated and attached to the protein on lysine by the enzyme E1 which uses ATPAMP ii. Then E2, the conjugation enzyme, and E3, the target enzyme comes and specifically binds U to the target protein. iii. Polyubiquination occurs and a crap ton of U gets attached to the U iv. The U’s flag down the proteasome and then it destroys protein. v. 23. Know structures of collagen and its function a. Part of connective tissue (synethesized) b. Special protein to the bone and skin c. Lots of Gly and Pro d. Each collegen fiber for the part it is put in, has a different structure and thus a different function e. General Structure: i. You have three Left hand a helicies all wrapped together in a Right hand formation. This is one segment. You take other segments and wrap it the same way around the first. This creates a rope like structure, f. General Function: i. This is specific for each connective tissue matrix but the general function is structure. ii. Long repeated structures of GlyXY 24. Describe the structural and functional similarities & differences between hemoglobin and myoglobin, including the mechanisms of action of any physiological factors that can affect O2 binding. Hemoglobin Myoglobin Lets go of O2 quickly Stores O2 for muscles Found in blood Found in muscle tissue 2A and 2B 8A Less Affinity for O2 High Affinity for O2 Cooperative Oxygen binding the binding of 1 Gives Muscle its Red color O2 molecule will enhance the binding of another. Has 4 heme groups which contain 4 iorn Has only 1 Heme group atoms which provide the positive charge to oxygen’s negative charge Factors that effect: 1. Amount of O2 2. Amount of CO 3. Binding of first O2 When oxygenated, it is in its relaxed state. When deoxygenated, it is in its taut state 4. Extra Notes: a. Enzymes: biological catalyst that amplify the rate of the reaction with out being used in the reaction i. It does not change free energy but rather activation energy ii. A catalyst eliminates randomness. b. Vocab: i. Substrate reactant ii. Active site where substrate binds iii. Transtion state intermediate c. Two types of models for enzyme activity: Lock and Key Induced Fit: Hand in glove Insert the substrate and the enzyme works The substrate causes the enzyme to change structure to accommodate the substrate and thus will create the product. d. Classification of Enzymes i. 6 Classes: 1. Oxioreductase Addition and removal of Protons and electrons 2. TransferaseTransfer one group to another 3. Hydrolase cleaves bond in the presence of water 4. Lyaseremoval of a functional group without water 5. Isomerase converts it into the isomer 6. Ligase Mends something together e. Cofactors loosely bound non protein components that aid in catalytic reactions 25. Describe why it is important for enzymes to be regulated. a. Enzyme regulation is huge because if you under regulate it to where too much product is made then the reaction will over work and thus to much product could be detrimental b. If over regulated where there was too little product made, then that would be determential. 26. List and describe the various ways that enzymes can be regulated. 27. Distinguish between competitive, noncompetitive, and uncompetitive inhibition, and mixed inhibition, including their mechanisms, effects on enzyme kinetics, and changes in MichaeilisMenton & and LineweaverBurke plots. Factors Competitive NonCompetitive Uncompetitive (Mixed) Information Binds to active site Binds to both the Binds to the enzyme and prevents Enzyme and the substrate complex substrate from Enzyme substrate (only) binding complex Km Increases No Change Decreases (Increase or Decrease) Reversible Yes No No Vmax No Change Decrease Decrease 28. 29. Describe the various ways that enzymes can be regulated by covalent modification. a. This is the addition of a functional group to a protein with a covalent bond b. Examples: i. Phosphorylation / Dephosphorylation ii. OxidationReduction iii. Proenzyme: An inactive enzyme precursor (polypeptide) that must be proteolytically cleaved or hydrolyzed in order to become active. 30. Describe (and exemplify) what is meant by “rational drug design” and its applications. 31. Describe the general characteristics of allosteric enzymes. a. An allosteric enzyme is an enzyme that has two active binding sites one for the substrate and the other for the activator or the inhibitor. The enzyme can either be activated or inhibited. b. Fig. 6.22 32. Explain how and why cofactors, pH and temperature affect or regulate enzyme activities. a. Changes in pH can effect the enzyme by: i. Changing the ionization of the protein molecules ii. Changing the ionization of the substrate iii. Precipitate the protein b. Changes in temperature: i. Can effect the rate of reaction ii. Extreme heat or cold can cause the protein to denature c. Availability of the cofactors will also regulate the rate of the reaction 33. How is sicklecell hemoglobin different from normal (adult) hemoglobin, and how do these differences contribute to hemoglobin function. a. Normal hemoglobin is spherical. b. For the sickle cell, it has a Valine in the position for glycine at position 6 for the subunits. Since Valine is hydrophobic, it pushes the cell in creating a pocket and thus forming the famous sickle cell. c. Because of this, the sickle cell bumps through the capillaries and cause clots and decrease the oxygen transport. This gives them immunity from malaria however, which is cause by infecting healthy red blood cells 34. Describe & exemplify the various types of enzyme cofactors. 35. Describe some of the main objectives of enzyme kinetics studies. 36. What is the meaning and significance of Vmax & Km. 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 37. Distinguish between the MichaelisMenton and LineweaverBurke equations and plots, and their applications. a. The equations to remember: i. The M&M equation: IntialVelocity=Vmax[s] 1. Km+[s] 1 Km 1 1 2. v= Vmax x S Vmax 38. Distinguish between the following terms: enzyme unit, specific activity, turnover number, catalytic efficiency. a. Enzyme Unitthe amount of enzyme that produces 1 micromole of product per minute. (Katal) b. Specific Activitymicromoles of product formed per minute, per mg protein (i.e. micromol / min*mg). c. Turnover numberamount of molecules that can be produced per unit of time for 1 enzyme molecule i. Kcat=Vmax/[Enz] d. Catalytic efficiencyhow good the enzyme is in the presence of low [s]
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