FINAL EXAM Biochem Flashcards
Classify animals, plants, protists, fungi, archaea and bacteria as prokaryotes or eukaryotes and identify the distingushing features of prokaryotic and eukaryotic cells.
Prokaryotes: bacteria and archea; features: unicellular organisms without a nucleus, few proteins associated with genome due to simplicity of the cell, replicate + adapt over small scale periods of time.
Eukaryotes: Animals, plants, and fungi (Kingdoms of eukarya); features: have more complex cellular organization with membrane enclosed organells that have specialized fxn’s.
Identify the domaines of life that include unicellular or multicellular organisms.
bacteria and archaea = unicellular
eukarya = multicellular organisms.
Recognise/ label biologically relevant organic conpounds and functional groups and identify the polarity and geometry of molecule
amine: RNH2 or RNH3+; R2NH or R2NH2+; R3N or R3NH+ (amino group is functional group)
alcohol: ROH fxnal group is hydroxyl group
thiol : RSH fxnal group is sulfhydryl group
ether: ROR fxnal group is ether linkage.
aldehyde: R-C=O-R(H) fxnal group (carbonyl group, acyl group)
ketone: R-C=O-R fxnal group (carbonyl group, acyl group)
Carboxylic acid R-C=O-OH or R-C=O-O(-),fxnal group: carboxyl or carboxylate groups.
Ester: R-C=O-OR fxnal groups: ester linkage.
Amide: R-C=O-HN2, R-C=O-NHR, R-C=O-NR2, fxnal group : amido group
Imine: R=NH or R=NH2+, R=NR or R=NHR+, fxnal group: imino group.
Phosphoric acid ester: fxnal group phosphoester linkage
Diphosphoric acid ester: fxnal group diphosphoryl group, pyrophosphoryl group.
recognize and describe the classes of biologcal monomer and related polymers
Bio monomers: amino acids, monosaccharides, nucleotides, lipids.
Bio polymers: polypeptides, polysaccharides, nucleic acids, lipids have no true polymer they just aggregate via non covalent interactions
AA’s: typically has N atom, generically has carboxylate group, side chain, amino group. Ex: Asparagrine, Cysteine
Monosaccharaides: have 1:1 ratio of carbon:oxygen ex: C6 H12O6 or C6 H13NO5
Nucleotides: Nitrogenous base and sugar. ex: ATP, DNA, RNA
Lipids: high ratio of carbon to oxygen/nitrogen/ phosphorus ex: C16H31O2 or C27H46O. anthropathic contain polar and nonpolar region that define behavior in solution.
Define polypeptide
protein
and residue
Polypeptide: polymers of amino acids linked together by peptide bonds.
Protein: functional unit consisting of one or more polypeptides.
Residue: a monomer that has been incorporated into a polymer. Monomers are covalently linked together to form different macromolecules. (typically via covalent bonds)
Describe the major and minor roles of biopolymers
major roles: proteins: carry out metabolic reactions & support cellular structures ; Nucleic acids: encode information, polysaccharides: store energy and support cellular structures.
minor roles: proteins: store energy : if emergent situation due to starvation body reverts to elastrophoric methods.; nucleic acids: carry out metabolic rxns & support cellular structures. polysaccharides: encode information (fxn used when cells have unique identifiers on the cell surface.
Distinguish open systems, isolated systems and closed systems
open systems: one in which energy can be transferred between the system and its surroundings.
isolated systems: a thermodynamic system that cannot exchange either energy or matter outside the boundaries of the system
closed systems: one that cannot transfer energy to its surroundings
Define Gibbs Free energy, enthalpy, and entropy and solve equations involving these values
Gibbs: Change in entropy + enthalpy, unfavorable reactions may be coupled with favorable reactions so that the overall process is spontaneous. Pi is inorganic phosphate, Glucose to glucosr 6 phosphate rxn coupled with ATP break down rxn –> net favorable rxn
Enthalpy: a measure of the heat energy of a reaction
Entropy: thermodynamic measure of randomness throughout a system.
Define endergonic, exergoinc, endothermic and exothermic.
endergonic: energy is conserved and used within the reaction (energy is necessary for the reaction to occur, non spontaneous process, positive delta G)
exergonic: energy is produced by the reaction, delta G is negative and reaction is spontaneous
endothermic: heat energy is used by the system, non spontaneous
exothermic: heat energy is released by the system, spontaneous.
Solve for the delta G of a coupled reaction given the delta G of individual reactions
practice
Determine if a process is spontaneous ( favorable/exergonic) based on its delta G
If delta G < 0, then a process is ‘spontaneous’, ‘favorable’ or ‘exergonic’
If delta G > 0, the a process is ‘nonspontaneous’, ‘unfavorable’ or ‘endergonic’
Identify whether a molecule is being oxidized or reduced in a reaction.
oxidation: loss of electrons, qualitatively look at whether the molecule gains oxygen, loses hydrogen, or goes from a single bond to a double bond. more favorable and can drive the synthesis of monomers and macromolecules.
reduction: gain of electrons, qualitatively look at whether the molecule loses oxygen, gains hydrogen, or goes from a double bond to a single bond.
ex: CO 2 least reduced most oxidized, Methane most reduced/ least oxidized.
Define catabolism and describe how the free energy of catabolic reactions may be conserved.
catabolism: breaking down larger molecules, involves oxidation steps (favorable), free energy may be conserved in the formation of NTPs (eg ATP or GTP) or reduction of cofactors.
ex: NADP+ + H+ 2e - –> NADPH / Q + 2H+ + 2e- –> QH2
Define anabolism and describe how the celll makes anabolic reactions thermodynamically favorable.
anabiolism: building complex molecules at the expense of energy. ex: ATP , NADPH
Describe how the molecular structure of water affects its physical properties as well as its interactions with other molecules.
Water’s bent structure results in a polar molecule that is strongly attracted to neighboring water molecules.
Hydrogen bonds ( dipole dipole IMF) occur when an H atom in a molecule, bound to a small highly electronegative atom with lone pairs of electrons, is attracted to the lone pair in another molecule
dipole allows water to interact w/ charged hydrophilic groups due to favorable electrostatic interactions
Describe and order by strength covalent bonds, ionic interactions, hydrogen bonds and van der waals interactions (dipole-dipole, dipole induced dipole, and london dispersion forces).
Strength order: Covalent bonds > ionic interaction > hydrogen bonding > Van der Waals : dipole-dipole > dipole induced dipole > London dispersion (london dispersion forces are surface area dependent (more important with larger molecules and atoms – > the larger the surface area, the higher the boiling point.
Define amphipathic (amphiphilic) molecules and describe how they behave in aqueous solutions.
amphipathic molecules: aggregate (stick together) in water with the charged heads interacting with water and the hydrophobic nonpolar tails staying away from water
Define bronsted lowry acids, bases, conjugate acids, conjuate bases and neutral pH.
B-L acid : a proton donor
B-L base : a proton acceptor
CA : the acid that results from the B-L base
CB : the base that results from the B-L acid
neutral pH : has a value of 7 on 0-14 acid scale.
explain the hydrophobic effect
a process where nonpolar regions clusters together to maximize the entropy of the surrounding water molecules.
driving force in hydrophobic molecules: the more disordered lipid molecules the more favorable the reaction, there is an increase in entropy w/in the system.
intrepret and draw titration curves
at the flat curve, the acid and base are at equilirium
Describe a buffered solution and discuss its effective range
Buffered solutions need to be enough of both weak acid and congjugate base present to be effective. A buffer is a weak acid/base solution that resists changes in pH. useful when there is a +/- 1 pH/pKa value.
Describe the equilibrium of bicarbonate buffering system in the blood and explain how it maintians the blood pH close to 7.40
acidosis: a condition where blood pH is too low
Alkalosis: a condition where blood pH is too high.
Normal conditions; H+ + HCO3- = H2CO3 = H2O + CO2
Excess acid: H+ + HCO3- –> H2CO3 = H2O + CO2 ; H+ + HCO3- = H2CO3 –> H20 + CO2 ; H+ + HCO3 = H2CO3 = H2O + CO2
Insufficent acid: H+ + HCO3- = H2CO3 <– H2O +CO2 ; H+ + HCO3- <– H2CO3 = H2O + CO2 ; H+ + HCO3- = H2CO3 = H2O + CO2
Define acidosis and alkalosis and eplain how kidney and lung function can affect blood pH
Acidosis: a condition where blood pH is too low
Alkalosis: a condition where blood pH is too high.
Metabolic processes generate H+ , Kidney and lung fxn helps maintin a blood pH of 7.4.
Kidneys ususally retain HCO3- while eliminating H+ to prevent acidosis
Lungs breathe faster to raise blood pH and slow breathing to lower blood pH.
Carbonoic anhydrase speeds up the interconversion of HCO3- + H+ with H2O + CO2.
Label given structes as purine or pyrimidine nitrogenous bases
purine: Adenine, Guanine
pyrimidine: Cytosine, Thymine, Uracil
Classify thymine, uracil, adenine, and guanine as pruines or pyrimidines
thymine - pyrimidine
adenine- purine
uracil - pyrimidine
guanine- purine
Name and distinguish nitrogenous bases, nucleotides and nucleosides
NB’s : A, T, U,G,C
nucleoside break down: adenine + ribose sugar –> adenosine, adenine + deoxyribose sugar –> deoxyriboadenosine , guanine + ribose sugar –> guanosine, cytosine + ribose sugar –> cytidine , uracil + ribose sugar –> uridine, thymine + ribose sugar –> thymidine
Nucleoside is a nitrogenous base (ie: adenine) attached to a ribose sugar.
Nucleotide; a nuceloside with on to three phosphates attached.
A phosphodiester bond links the 3’ carbon of (deoxy) ribose to the 5’ carbon of the next nucleotide’s (deoxy)ribose.
Differentiate between RNA and DNA
DNA: deoxyribonucleic acid , double stranded
RNA: ribonucleic acid, single stranded
Recall that the base pairing rules for DNA and RNA along with the number of H-bonds in each base pair
DNA: 3 and 2 bps ; Chargffs rules: In doubled stranded DNA: the % A = % T and the %G = %C.
RNA: 2 bps only
Recall the handedness, strand direction, and dimensions of the Watson and crick double heliz and recognize the major and minor grooves
DNA : double helix that is 3 to 5 A ( angstromes) long., the two DNA strands are antiparallel and the resulting helix is right- handed.
Contains A, B and z forms; contains major and minor grooves.
DNA strands may be separated (denuatured) by heating. ( separation of strands by breaking weak forces holding them together.
Recall that a DNA duplex may be separated ( denatured) by heat in vitro and that the melting point corresponds to the point at which half of the DNA duplexes are denatured
DNA duplex separation heating ( separation of stands by breaking weak forces holding them together)
The melting temp of DNA is sequence and length dependent.
Recall that DNA denaturation may be monitored by UV absorbance due to the hyperchromic effect: UV absorbance increases with the amount of DNA denatured.
DNA denaturation occurs when UV light absrobance due to hyperchromatic effect. The more UV absorbance, the more DNA is denatured
Recall that the stability of the DNA double helix is mostly dependent on base-stacking interactions and the melting temperatures of DNA duplexes increases with the number of GC base pairs ( due to the greater stacking energy ) and length
DNA stability and stacking length; The melting temperature of DNA is sequence and length dependent GC base pairs contribute more to base stacking interactions than AT base pairs.
Descibe the central dogma of biology and predict the possible amino acids sequences of a polypeptide given an mRNA sequence ( no need to memorize codon specific amino acids
Central dogma: DNA is replicated –> transcription –> RNA –> translation –> protein.
Codons are three nucleotide sequences that specify an amino acid or stop translation. The codon sequence, in a particular reading frame, corresponds to the polypeptide sequence.
predict how mutations at the gene level can affect the sequence and function of a protein.
Normal hemoglobin b gene, Hemoglobin b protein, Normal hemoglobin b gene, hemoglobin protein.
Sickel cell anemia is a genetic disease arising from an A to T point mutation in the DNA that alters the amino acid sequence and properties of the translated protein.
Recall the 1-and 3 letter codes for the common 20 amino acids along with their names and structures.
20 amino acids: Hydrophobic amino acids: Alanine (Ala, A); Valine (Val,V); Phenylananine (Phe, F); Tryptophan (Trp, W); Leucine (Leu, L); Isoleucine (IIe, I); Methionine (Met, M).
Polar amino acids: Serine (Ser, S); Threonine (Thr, T); Tyrsoine (Tyr, Y); Cysteine (Cys, C); Asparagine (Asn, N); Glutamine (Gln, Q); Histidine (His, H); Glycine (Gly, G);
Charged amino acids: Asparatae (Asp ,D) ; Glutamate (Glu, E); Lysine (Lys, K); Arginine(Arg, R).
Describe the differences between “L-“ and “D-“ amino acids and recalll which are used in proteins
L :
D :
Use in proteins:
Estimate the hyrdophicity of an amino acid and predict how it contributes to protein feeding
Hydrophobicity of an amino acid: the hydrophobic effect is the driving force for protein folding and results in the burial of most hydrophobic resides in the interior of proteins.
Polar amino acids: polar residues often reside on the surface of proteins or hydrogen-bonded in the interior.
Charged resides tend to reside on the surface of a protein or as an ion-pair (salt-bridge) in the interior.
Define the “isoelectric point” (pl) of a molecule and identify the charge of a molecule above and below its pl.
isoelectric point: the pH at which the molecule has no net charge, if the solution pH is above the pI, then the molecule will be negatively charged, if the solution pH is below the pI, then the molecule will be postively charged.
Draw the structure of a polypeptide bond and illustrate how its formed by a condensation reaction of two amino acids.
polypeptide bonds consist of :
Define “N-term” and “C-term” of a polypeptide and recall that polypeptide sequences are usually written from N-term to C-term.
N-term: amino terminus
C-term: carboxyl terminus
Peptides are synthesized from the amino terminus (N-term) to the carboxy terminus (C-term) and sequences are usually written in that direction.
Within a polypeptide, only the N-term, C-term, and certain amino acid side chains are ionizable. ex: His, Glu, Asp, Cys ( C-term).
Tyr, Lys, Arg (N-term)
Recall that the peptide bond has partial double bond character and cannot rotate significantly. The other backbone bonds, N-Ca (Phi) and Ca-C( W) may rotate to sterically allowed positions
N-Ca (phi) & Ca-C(W) : bonds that may rotate to angles to avoid steric clashes
Ca-C(W) : peptide bond that has a partial double-bond character and does not rotate.
sigma bonds can only twist to a few angles avoiding steric clash.
Recall that cystines may form disulfide bond crosslinks if they are oxidized and that outside of the cytsol is generally an oxidizing environment, while inside the cytosol is a reducing environment.
disulfide bonds and cross links: disulfide bonds may form between nearby Cys side chains, either within a polypeptide or btwn two polypeptide, in an oxidizing environment. (outside the cytosol).
describe how different forms of chromatrography, including ion exchange (either anion exchange or cation exchane), gel filtration (aka size exclusion) , affinity chromatorgraphy (including His-tagged proteins), are used to separate proteins. You should be to describe the properties of the columns, the chemical basis for separation, and predict how the proteins would interact with and elute from each type of column
Types of Chromatography:
IEC : uses a stationary phase containing charged groups to bind proteins of opposite charge. Contains beads: DEAE and CM ( Cation exachange column) , buffer = nonstationary, moves through, separates based on affinity of charge and stationary = affinity for charge of beads , non sationary = low affinity –> move with buffer solution off column. A column with postively charged resin, ex DAEA (anion exchange column), will bind negatively charged proteins, while postively charged proteins elute first. tightly bound proteins can be eluteed by increasing salt concentration or by changing the pH of the buffer alter the charge of the bound proteins ,
Gel Purification: size exclusion chromatorgraphy, SEC uses a stationary phase with small pores and channels in the beads that separate proteins based on size. Columns have a fractioonation range for separation. eg: 10-150 kDa, proteins larger than this range are excluded, from the pores and channels, and elute together at the void volume ( the volume of buffer, surrounding the column beads, that must be displaced by buffer running over the column). Proteins within the fractionation rage are separated from one another as they diffuse into different numbers of pores and channels depending on the size of the protein. Proteins smaller than this range elute together at the total pore volume (they can diffuse into all pores and channels).
Affinity chromatography: takes advantage of the natural bininding properties of a protein (or of an engineered tag that extends the N or C term of the protein of interest. A protein with a His-tag ( 6 consecutive histadine residues) will specifically bind Ni2+-resin (this is not an ion exchange effect).
SDS-PAGE: Ususally proteins are boiled in SDS before loading into well. SDS is a detergent that disrupts protein folding (removing shapre effects when migrating though the gel). , SDS also imparts a uniform negative charge to the protein (masking any intristic protein charge), the acrylamide polymer forms a tangled network that slows down proteins moving through the gel. Smaller proteind migrate through the gel faster than large proteins.
Describe how SDS-PAGE is used to look at what proteins are present in a sample. Explain why SDS is used and how gel electrophoresis works.
SDS-PAGE Fxns : Ususally proteins are boiled in SDS before loading into well. SDS is a detergent that disrupts protein folding (removing shapre effects when migrating though the gel). , SDS also imparts a uniform negative charge to the protein (masking any intristic protein charge), the acrylamide polymer forms a tangled network that slows down proteins moving through the gel. Smaller proteind migrate through the gel faster than large proteins.
Describe the four levels of protein structure
Primary: Sequence of amino acids
Secondary: localized conformation of polypeptide backbone
Tertiary: 3 D struct of an entire polypeptide, including all its side chains
Quatrinary: the spatial arrangment of polypeptide chains in a protein with multiple subunits.
Desrcibe the structural features of alpha helicies and beta sheets
alpha helicies: 3.6 residies/turns, nth carboonyl oxygen H-bonds to n+4th NH.
Beta sheets: hydrogen bonding btwn the backbones of multiple beta-stands forms a stable beta-sheet. the side chains and carbonyls of consecutive resides alternate directions.
Define ‘domains’ in the context of proteins
Domains: regions of a polypeptide that fold independently (ie: have their own hydrophobic core and have their own functions
Describe the stabilizing foreces for secondary, tertiary and quaternary structures
Primary stabilizing structs: Hydrophobic resides on surface, interior charged resides tend to form ion pairs, major contributor = hydrophobic effect. Neighboring cysteines can form disulfide bonds within polypeptides or btwn separate polypeptides. Proteins that are misfolded may be targeted for degradation or refolded
Secondary stabilizing structs side chains and carbonyls of consecutive residues alternate directions
Tertiary stabilizing structs hydrophobic effect
Quat stabilizing structs hydrophilic resides
explain how protein misfolding may lead to certain diseases and describe the basic properties of amyloid diseases and prion diseases
misfoling results in diseases: Certain proteins can adopt a stable misfold conformation with higher beta sheets content. these misfolds proteins may aggregate into amyloid fibers with a beta sheet core. ex: alzhiemers disease and piron diseases such as scrapie and Creutzfeld-Jakob disease.
prion diseases are caused by infections proteins. the misfolded protein (PrPsc) induced natively folded protein (PrPc) to adopt the misfolded conformations that are prone to aggregation
Describe the roles and general structure of myoglobin and hemoglobin in the body and the role of heme prosthetic group in these proteins
myoglobin: protein that carries oxygen in muscles; has a similar secondsary and tertiary structure to Hb alpha and Hb beta, but has different sequence
hemoglobin: protein that carries oxygen in the blood, heterotetramer = 4 subunits that are not identical
fxn of heme: a prostethic oxygen binding group, distal histadine prevents superoxide interactions, 6th oxygen interacts w/ ion -> interacts with another His, proximal His directly interacts w/ iron.
Describe cooperativity in the context of protein binding and explain the physiological importance of hemoglobin being cooperative
coopertivity: binding at one site influences binding at the other sites, usually postive, binding at one site aids or helps binding at another site. fractional saturation of venus and arterial pressure.
importance of hemoglobin being cooperative: O2 binding at one subunit increases the affinity for O2 at the remaining subunits. The sigmoidal binding curve is characteristic of cooperative binding.
myoglobin -> high affinity for oxygen and would have a hard time letting go.
hemoglobic -> b/c of its high fractional context and lower affinity for oxygen, and can deliver higher amounts of oxygen,
Define what R-state and T-state refer to in the context of hemoglobin and protein binding in general
R-state: has a high affinity for O2 and is the conformation favored by oxygenated hemoglobin, Hb behavior is modeled well by considering Hb as existing in equilibrium btwn all four subunits in the T-state and all four subunits in teh R-state. W/o O2, the equilibrium heavily favors T-state whearas at high pO2 the equilibrium heavily favors R-state hemoglobin.
T-state: low affinity for O2 and is the conformation favored by deoxygenated hemoglobin, 35 residues are in contract at each of the a1b1 and a2b2 interfaces, while 15 residues are in contact at each of the a1b1 and a2b2 interfaces.
Explain how oxygen binding promotes the R-state conformation and why the T-state is favored when pO2 is low.
oxygen binding promotes the T state: in the absence of O2, T-state is favored due to the formation of salt bridges involving the C-term resides of the alpha and beta subunits., ionic interactions stabilize + and - state struct making it more favorable.
Recall why hemoglobin does not have any intermediate conformations between R and T states
hemoglobin = no intermediate states b/c intermediate conformations btwn R-state and T-state are disfavored due to steric clashes. Formation of the Fe-O2 bond is very favorable and pulls the Fe2+ into the plane of the porphyrin ring (shortening the Fe-N bonds).
Describe the relevant chemical equilibriums for the Bohr effect as well as the physiolocial importance of the Bohr effect
chemical equillibriums: Co2 + H2O = HCO3- + H+ ; HbH+ + O2 = HbO2+H+ ; N-term = 7.4; protonated N term –> +1 charge => allows participation ionic interactions -> stabilize T-state.
Bohr effect: Hb binding H+ favors unloading O2 to the tissue that needs O2 the most. Muscle metabolism causes a pH drop in surrounding tissues, increasing [ H+] protonates the a-subunit N-term and the b-subunit His 146 side chain, favor the salt-bridge formation that stabilizes T state.
Describe how BPG affects hemoglobin as well as the physiological importance of BPG at altitude and in fetal hemoglobin
BPG affects hemoglobin: BPG binds and stabilizes the T-state of Hb. BPG binds in the central cavity of T-state Hb and contributes to further salt-bridge formation; w/o BPG, Hb would bind O2 too tightly to unload a significant fraction to the tissues.
Physio importance of BPG at high alts: BPG increases to adapt to hight altitudes (the lower aterial pO2 at altitude is compensated by a lower fractional saturation at the venous pO2.
Fetal Hb has a2y2 subunits. Fetal hemoglobin has stonger affinity for O2 relative to adult Hb
binds BPG poorly, favors movement of O2 across the placental mbn. ( left shift in curve for affinity)
Recall the mutation that leads to sickle cell anemia and the consequences of the mutation on hemoglobin, red blood cells, and homozygotes and heterozygotes carrying the mutation.
consequences of Hb mutation , rbc: sickle cell anemia is a genetic disease arising from a Glu6Val mutation on Hb beta, Sickeled cell may block small blood vessles, limiting O2 delivery and causing pain, organ damage, and destruction of red blood cells. Sickling is triggered bu factors that promote T state Hb (eg. high altitude and dehydration).
A hydriophobic pocket is exposed on the b-subunit when Hb is in teh T-state. The pocket binds the exposed Val in neighboring Hb. Aggregation of Hb produces long, ridgid strands of Hb inside the red blood cells, these strands push on the cell mbn, deforming the cell into the sickle shape.
homozygotes with mutation:
heterozygotes with mutation:
Describe the general structures and roles of globular filamentous actin
Gen struct: Microfilaments, Intermediate filaments - intermediate diameter, microtubules - widest diameter, widely distributed. Actin monomers (‘globular actin’ or ‘G-actin’ polymerized to form ‘filamentous actin’ or ‘F-actin’ microfilaments
Fxns: microfilaments help determine cell shape, allow some cells to move, and are part of the contractile apparatus in mucles. Actin = structure protein that can divide and form microfilaments.
Mysoin: conventional myosin (myosin II) is a motor protein that helps contract actin filaments in muscles. dimer = two subunits, a helix, two alpha helixes wrapped around eachother.
Describe the steps of myosin movement, the physiological importance of the movement, and the filament associated with myosin.
Steps of myosin movement: a myosin head bound to an actin subunit of the thin filament, ATP binding alters the conformation of the myosin head so taht it releases actin. –> rapid hydolysis of ATP to ADP + Pi triggers a conformational change that rotates the myosin lever and increases the affinity of myosin for actin –> Myosin binds to an actin subunit farterh along the thin filament –> Binding to actin causes Pi and then ADP to be released. AS the rxn products exit, the myosin lever returns to its original postion, this causes thin filaments to move relative to the thick filaments (the power stroke). –> ATP replaces the lost ADP to repeat the rxn cycle.
physio importance:
Filaments associated:
Describe the structure and role of keratin and explain how its amino acid sequence relates to its function
struct of keratin: Two long a helices wrapped around one another to form a coiled coil. Each helix has a repeating seven amino acid sequence where the 1st and 4th residues of each repeat are hydrophobic and form the interface btwn helices
Keratin is a structural protein that assembles into intermediate filaments bundles and is present in hair, wool, nails, claws, quills, hooves, and the outer layer of skin.
fxn of keratin:
aa sequence related to fxn: Diaulfide bonds btwn dimers help strengthen keratin microfibrils
Describe the activation energy for a reaction and predict the effect of an enzyme on the thermodynamics and kinetics of the forward and reverse reactions.
Activation energy of rxn: Ea or Delta G++ = the energy to go from the ground state to teh transition state., barrier can delay rxn from occuring due to high activation energy.
effects of enzymes on thermodynamics: the sign of Delta G rxn (not Delta G ++) determines whether a process is spontaneous or not, the magnitude of delta G++ determines the rate of the rxn. Catalysts increased the rate of rxn by lowering delta G++ ( delta D rxn is unchanged).
kinetics of the forward and reverse rxns
Define substrate and active site and describe the forces that allow an enzyme to specifically bind a substrate.
substrate: Substrate binding occurs in a pocket or cleft on the surface of the enzyme. the substrate is held in place by electrostatic, hydrogen bonding, Van der Waals, and hydrophobic interactions. binding pocket is not fully formed until the protein is more flexible –> strains towards the stransition state.
active site: name usually descibes fxn, enzyme commision (EC) number assigns unique four level number eg (EC) to classify a rxn, we must consider both the forward and reverse direction, classification gepends only on the net rxn (not the mechanism).
Forces allowing enzyme substrate binding: enzymes speeds up both forward and reverse rxns.
Classify a given reaction as oxidoreductase, transferase, hydrolase, lyase, isomerase, or ligase.
oxireductase: either have a redox cofactor present or disulfide formation ex: NAD+/NADH, NADP+/NADPH, FAD/FADH2, FMN/FMHN2, Q/QH2
transferase: swap fxnal groups btwn subrates
hydrolase: use water to break a bond or condense to eliminate water
lyase: break bonds w/o the use of redox activity or water and produce an extra double bond (or ring) in the products
isomerase: rearrange fxnal goups in a substrate, but keep the chemical formula the same
ligase: use ATP energy to connect two other substrates.
Define: cofactor, coenzyme, cosubstrate, prosthetic group, apoenzyme, and holoenzyme.
cofactor: a small organic molecule or metal ion that is required for enzymatic activity
coenzyme: are organic cofactors
cosubstrate: are coenzymes that tansiently assoicate woth then enzyme
prosthetic group: are coenzymes that are tightly associated with the enzyme
Apoenzyme: an enzyme w/o its required cofacotrs (inactive)
holoenzyme: an enzyme w/ its required cofactors (active)
Describe the general mechanisms by which an enzyme can facilitate catalysis including acid-base catalysis, covalent catalysis, metal ions, proximity and orientation effects, transition state stabilization.
Acid-base: H+ is transferred btwn an enzyme and the substrate ; RNASE A is an example of an acid-base catalysis with histadine residues transferring H+ with the RNA substrate
covalent catalysis: a stable intermediate is formed when the enzyme covalenty binds to the substrate. Mech: acetoacetate decarboxylation –> nonenzymatic path (CO2 product) –> enoate formation –> H+ addition –> acetone formation. Acetoacetate with RNH2 in and OH- out creates a schiff base formation (imine) CO2 product –> rearragment intramolecular reaction –> H+ in –> imine –> OH in and RNH2 out –> acetone
metal ions: can stabilize negative charges that form in the transition state. eg. decarboxylation of dimethyloxaloacetate, can sheild charges that might repel the attacking group, can promote nucleophilic attacks through ionization of water., can participate in redox reactions.
proximity and orientation effects: Reactants must come (promxinity) with the proper spatial relationship (orientation) for the reaction to occur
transition state stabilization: The more tightlly an enzyme binds to the transition state, relative to the substrate, the greater the rate acceleration of the catalyzed rxn.
Explain how and why acid-base catalyzed reactions are very pH dependent.
pH dependency: RNase A is an ex of acid-base catalysis with histidine residues trasnferring H+ with the RNA substrate; A stable intemediate is formed when the enzyme covalently binds to the substrate
Describe the lock and key and induced fit models of enyzme-substrate binding.
lock and key fit models: specificty pocket only binds certain amino acids, positioning the peptide bond for cleavage by catalytic triad.
Describe the catalytic triad is and explain the role each residue plays in catalysis.
Catalytic triad: Asp, His, Ser
fxns of residues in catalysis: breaksdown proteins, Prancreatic proteases, subtilisin, catalytic triad
Describe how the specificity pocket of serine proteases determines what peptide bonds are broken and predict the specificity of proteases based on the structure of the specificity pocket.
pocket specificity of serine proteases and peptide bonds:
Define convergent and divergent evolution and infer how proteins are related based on their structure.
convergent evolution:
divergent evolution:
protein relations based on strucutre :
Recognize that although many serine proteases are evolutionarily related, others result from convergent evolution.
serine proteases evolutionary relations:
Recall the purpose and net reaction of serine proteases such as chymotrypsin.
goals of chymotrypsin rxn: Interaction of Ser and His generates a strongly nucleophilix alkoxide ion on serine; the ion attaccks the peptide carbonyl group, forming a tetrahedral acyl-enzyme. This accompanied by formation of a short lived negative charge on the carbonyl oxygen of the substate, which is stabilized by hyrdrogen bonding in the oxyanion hole.
Define zymogens and explain why they are necessary and how they are converted to active enzymes.
zygomens: are secreted into the small intestine and cleaved by other proteases so that they acquire a conformation where the specificity pocket and oxyanion hole are available for catalysis,
fxns : inhibitors in the blood stream and pancreas bind the proteases active site to inhibit any proteases that are active outside the small intestines; synthesized as inactive precursors
Describe how reaction velocity depends on substrate concentration.
velocity depends on substrate concentration: the hyperbolic shape of the rate dependence on [S] is due to the formation of an enzyme-substrate complex.
Recall the symbols for, and describe what is meant by, reaction velocity, rate of reaction, Michaelis constant (KM), turnover number (kcat), specificity constant (catalytic efficiency), maximal velocity (Vmax), substrate, Michaelis-Menten complex, and product, along with solving calculations involving these variables.
reaction velocity: (v) = d[P]/dt = -d[S]/dt
rate of reaction: (v) = d[P]/dt = -d[S]/dt
Michaelis constant (KM): k-1+k2/k1, the substrate concentration that gives Vo=Vmax/2
turnover number (kcat): Kcat/Km is the specificity constant ( a measure of catalytic efficiency)
specificity constant (catalytic efficiency):
maximal velocity (Vmax): Vmax=k2[E]t where k2=kcat =turnover and [E] is the total enzyme concentration; k2(or kcat) represents the number of catalytic cycles that each active site undergoes per unit time.
substrate:
Michaelis-Menten complex: hyperbolic relationship velocity as a fxn of substrate concentration
product:
Describe the assumptions for Michaelis-Menten kinetics and how they affect kinetic measurements.
assumptionso of M-M kinetics: only valid when 1) measurements are made before much product has formed (so we can ignore E+P– ES and ignore product inhibition. E+S k1=k-1 ES –> K2 E+P, 2) there is only one rxn substarate, 3) the rxn occurs in a single step, 4) binding is not cooperative.
affect on kinetic measurements:
Derive and interpret the Lineweaver-Burk (reciprocal) equation and plots for kinetic data.
L-B equation and plots:1/observed rate vs 1/[S], its easier to estimate the Vmax on a lineweaver-burk reciprocal plot
Describe how an irreversible inhibitor affects an enzyme.
irriversibe inhibitors: a reaagent that permanently associates w/ an enzyme inactivating it, is an irreversible inhibitor.
For reversible inhibition (competitive, uncompetitive, noncompetitive, mixed, allosteric) describe how each type affects kinetic data, and explain how the data is interpreted in terms of inhibitor interacting with enzyme.
competitive: competes with the substrate for binding at the enxyme’s activue sites; alpha is the degree of competitive inhibition, Ki is the inhibitor constant ( the dissociation constant for EI= E+I, a low Ki indicates a good inhibition (tight binding) Vmax stays the same, Km increasses by a factor of alpha, best competitive inhibitors minic the transition state for the rxn. inosine: KI = 310-4 M , 1,6-dihydroinosine KI= 1.510-13 M ( very strong competitive inhibitor fortune enzyme.
uncompetitive: only bind to the enzyme-substrate complex, Lowers Km and Vmax by the same factor (a’)
noncompetitive: is a special case of mixed inhibition where a=a’ >1, binds away from the enzyme active site and does not affect substrate binding.
mixed: affects both Vmax abd Km, a mixture of competitive inhibition.
allosteric : enzymes exhibit cooperativity , binding at one site affects the substrate affinity at other sites. inhibitors shift the sigmoidal curve to the right. regulators affect the activity of multisubunit enzymes by stabilizing the R-state (high activity stabilized by ATP, higher velocity, regulation purine and pyrimidine nucleotide production or T-state (low activity stabilized w/ CTP lower, stabilizes inactive conformation of substrate rxn.
Interpret or calculate the degree of inhibition and inhibitor constant for an inhibitor interacting with an enzyme.
degree of inhibition : uncompetitive : alpha prime is degree of uncomp inhibition a’ = km w/o inhibition / (apparent) km with inhibitor or a’= Vmax w/o inhibitor /(apparent) Vmax w/ inhibitor
pure competitive: a > 1 a’=1;
mixed : a > 1 and a’>1
Recall that competitive inhibitors frequently resemble the substrate or product of a reaction and that competitive inhibitors that mimic the transition state frequently bind with very high affinity to an active site (which can be exploited in rational drug design).
competitive inhibitors and substrate/product or a rxn: competitive inhibitors frequently resemble the substrate or product of a reaction and that competitive inhibitors that mimic the transition state frequently bind with very high affinity to an active site (which can be exploited in rational drug design).
Predict how allosteric effectors stabilize the R-state or T-state of an enzyme.
allosteric effector stabilization of R/ T states: regulators affect the activity of multisubunit enzymes by stabilizing the R-state (high activity stabilized by ATP, higher velocity, regulation purine and pyrimidine nucleotide production or T-state (low activity stabilized w/ CTP lower, stabilizes inactive conformation of substrate rxn.
Describe how enzyme activity may be regulated without the use of inhibitors.
enzyme activity regulation w/o inhibitors: 1) rate of synthesis and/or degradation, 2) Localization 3) release of co-activators such as Ca2+ 4) Covlalent modification (eg. phosphylation or zygomen cleavage.
Recall the basic HIV life cycle including the roles of HIV reverse transcriptase, HIV integrase, and HIV protease and explain how therapeutics inhibit the reverse transcriptase and protease enzymes (note reverse transcriptase inhibition is discussed in box 20.A on p. 530).
effects of HIV: HIV reverse transcriptase is very error prone (it is not as specific as most polymerases when incorporating a nucleotide) Viruses have high mutation rates –> increase resustance development to treatments. The error-prone nature of HIV reverse transcriptase means it may incorporate chain-terminator nucleotide drugs such as AZT and ddC. AZT and ddC prevent DNA polymerase from incorpating additional nucleotides, human polymerases are more discriminating and rarely incorporate these drugs. Nevirapine is a noncompetitive inhibitor that binds to a hydrophobic pocket away from the active site.
HIV-1 protease cleavage sites (cleavage occurs during maturation); HIV protease has some unique recognition sites that aren’t recogized by human proteases.
Saquuinavir and ritonavir are competitive inhibitors of the HIV protease that mimic the transition state of the rxn.
For fatty acids, recall the systematic and abbreviated naming systems (but common names are not required), along with how to draw or name a given fatty acid. Also describe what is meant by “delta end”, “omega end”, and “omega-3 fatty acid”.
systemic naming of fatty acids… : Unsaturated fatty acids have at least one double bond in the hydrocarbon tail. Water soluable end (delta end), fat soluable end (omega end)
The symbole name gives the ratio of the number of carbons to the number of double bonds, as chain length increases, so does the melting point (mp) of the fatty acid; tempurature must increase to overcome the great van der Waals interactions.
Symbol names: eg 6,9,12-octadecrienoate is 18:8n-6 or 18:3delta 6,9,12 (if cis or trans is not explicitly identified, assume cis for biological fatty acids n-6 or n-3 gives the postion of the last double bond ( where n is the number of carbons in the fatty acid), n-3 fatty acids are omega-3 fatty acids, n-6 fatty acids. Additional bonds are always spaced three carbons apart.
Describe how the packing and resulting intermolecular forces affect the melting points of fatty acids in terms of length and units of unsaturation.
packing and imf’s affects: IMFs –> SA increases due to chain lengths + van der waals interactions increase mp important in how they are placed + interactions w/ other bio molecules
unsaturated biological fatty acids usually have cis double bonds. The highly ordered packnig of fatty acid chains is disrupted by the presence of cis double bonds. low m.p of unsaturated fatty acids < higher m.p saturated fatty acids (within the biological range of fatty acid lengths), each additional double bond lowers Tm
Fatty acids are usually carried as glycerol esters (triacylglycerols) rather than free fatty acids. polar character ester linkages but overall molecule is nonpolar.
Recall that most biological unsaturated fatty acids have cis double bonds and describe the advantage of more saturated or unsaturated fatty acids in terms of maintaining membrane fluidity.
cis db’s:
advantages of more saturated v unsaturated fatty acids :
Describe the role of cholesterol in modulating membrane fluidity and as a precursor for steroid hormones along with being able to recognize the four-ring steroid nucleus of these molecules.
cholesterol fxn: Cholesterol and other steriods have a four-ring core sturecture, cholesterol is a major component of animal membranes that helps maintain fluidity and integrity. cholesterol also serves as a precoursor for steroid hormones, vitamin D, and bile salts. stored in hydrophobic form –> as cholesterol ester (hydroxyl group + lipid)
Recall the function and common structure of triacylglycerols (triglycerides).
tryglycerides:
Define vitamin and give a brief description of the main physiological role and the consequence of a deficiency for the lipid soluble vitamins.
vitamins: essential molecules that we cannot synthesize and must obtain in our diet.
Vitamin A: important for visual perception, deficiency results in night blindness. Lipid soluable.
Vitamin D: important for calcium homeostasis, deficency results in rickets, a disease characterized by deformed bones and stunted growth
Vitamin E: had antioxidant activity; hydrophobic, partions into mbns reacts w/ reactive oxygen species preventing modification of fatty acids tails –> mbn modification.
Vitamin K: a cofactor for an enzyme that modifies glutamate residues on blood clotting proteins, defiviency in excessive/prolonged bleeding , lipid soluable
Describe the common structural features and the role of glycerophospholipids and sphingolipids (you do not need to know the exact structure or names of specific membrane lipids, just what the structures have in common).
glycerolphospholipids: a glycerol backbone with fatty acids esterified to the first two carbons and a polar phosphate derivative head group attached to the third carbon
sphingolipids: consist a (dihydro)sphingosine backbone with a single fatty acid attached via an amide linkage and a sugar or polar phosphate derivative head group.
Describe why two-tailed lipids, such as glycerophospholipids and sphingolipids, form extended lipid bilayers, but most other lipids cannot.
two tailed lipids form extended lipid bilayers because: mbn lipids must have the correct geometry to stack correctly without forming voids
Describe and recognize the different types of proteins associated with membranes, including integral membrane proteins, peripheral membrane proteins, and the different types of lipid-linked protein (myristoylated, palmitoylated, prenylated, and GPI-linked).
integral proteins: have hydrophobic peptide regions embedded in the lipid bilayer and these proteins reqiure strong detergents to isolate
mbn proteins:
peripheral mbn proteins: associate with the polar head groups of membrane lipids or with the exposed regions of other mbn proteins, may be isolated with other mild solutions
lipid-linked proteins: insert a hydrophobic anchor into the mbn, can be 1) myristoyl group ( C 14:0) attached to an N-term Gly via an amide linkage; 2) a palmitoyl group ( C 16:0) reversibly attached to a Cys side chain via a thioester linkage. 3) an isoprenoid group linked to a C-term Cys via a thioether linkage 4) a glycosylphosphatidylinsitol (GPI) group linked to the C-term of the protein
Recall that integral membrane proteins often span the hydrophobic regions of the lipid bilayer with transmembrane helices or beta barrels.
integral mbn proteins span hydrophobic regions of bilayer beta barrels or transmbn helices:
Describe the fluid mosaic model.
fluid mosaic model: proposes that lipids and proteins may freely move laterally but cannot flip-flop through the layer by themselves, the cytoskeleton may be restrict some proteins to particular regions of the membranes. eg: Protein A is anchored to a cytoskeleton filament. Protein B is fenced in by cytoskeleton filaments.
Describe features that give lipid bilayer polarity/asymmetry and explain how lipids are moved from one leaflet of the bilayer to the other.
features of lipid bilayer: mbns have polarity. inside of a mbn is different, in both lipid and protein composition from the outside, both lipids and proteins facing the extracellular space are frequently glycosylated, integral mbn proteins are always oriented in the same direction ; translocases are proteins that move mbn lipids from one side of the mbn to the other.
lipid movment form leaflet to leaflet:
Recall the approximate concentrations of Na+ and K+ inside and outside the cell along with the typical membrane potential of a cell. SALTY BANANA !!!!!
Na+ concentrations: high outside, low concentrations inside
K+ concentrations: high inside, low concentrations outside
negatively charged (-70 mV),
Recall that the free energy for moving a substance across the membrane depends on the concentration gradient (ie. moving down a concentration gradient contributes favorably to a spontaneous process) and, if the substance is charged, depends on the membrane potential as well (ie. moving toward opposite charges contributes favorably to a spontaneous process).
free energy for a moving substance across mbn: the plasma mbn is mostly impermeable to Na+ but allows some K+ to leak across the mbn; K+ movement reaches equilibrium with a high [K+] inside the cell (favoring K+ efflux) and a negative mbn potential ( favoring K+ influx), this is the major contribution to the cell’s resting mbn potenial
Describe how and why the movement of Na+ and K+ during an action potential alters the membrane potential.
movment of Na+ and K+ across mbn during AP: stimulus orginates where the axon leaves the neuronal cell body. stimulus opens Na+ channels in the mbn –> VG Na+ channels open Na+ influx results in depolarization from -70 to +55 mV, closing of VG Na C’s results in opening of VG K+ channles results in efflux of K+ restoring resting potential –> occurs multiple times @ nodes of ranvier as AP propagates down axon.
Describe the basic structure of a porin and how its structure affects its selectivity
porin structure: proteins that form an open channel in the mbn of bacteria and certain organells to allow the passage of small solutes.
effects on selectivity: certains organells to allow the passage of small solutes
Describe the basic structure of a K+ channel and how this structure allows it to select for K+ transport over Na+ transport.
K+ channel strucutre and selectivity: K+ channels are much more selective for K+ than Na+, the last third of K+ channels is a selectivity filter that is too narrow for large ions and too wide for smaller ions such as Na+, when K+ moves into the filter, it looses favorable interactions w/ water, but gains favorable interactions with the residues lining the filter (selectivty filter) –> smaller ions (eg. Na+) do not interact favorably w/ the selectivity filter as the energy for desolvation cannot be compensated by interactions with the selectivity filter.
