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.
Recall what stimuli can open and close gated channels and predict how these stimuli could affect the channel’s structure.
stimuli affecting gated channels: 1) changes in mbn potential, 2) changes in pH, 3) binding of a specific ligand 4) phosphorylation, 5) temperature, 6) mechanical pressure
affects on channel structure: mechanosensitive channels
Describe how an aquaporin can select for water transport without allowing H+ to pass through based on the channel structure along with the specific amino acids in the channel.
aquaporin selection for water transport: aquaportin channel contains two Asn residues that disrupt the hydrogen-bonding chain of passing H2O molecules: hydrophobic residues lining the channel force H2O into an orientation that hydrogen bonds with the asparagrines
the size of the channel is optimized for a single chain of H2O and a positvely charged Arg residue disfavors entry of H3O+
Define uniport, antiport, symport, primary active transport, secondary active transport, passive transport, facilitated diffusion, exocytosis, and endocytosis.
uniport: move a single substance across mbn, eg: glucose transporter allows glucose to move down its concentration gradient ; Glucose binding changes the conformation, so that the transporter opens on the opposite side of the mbn.
symport: moves two different substances across the mbn in the same direction, eg: Na+-glucose transporter uses the energy of Na + moving down its concentration gradient (into the cell) to drive glucose import into the cell agonist the [ glucose] gradient.
antiport: move two different substances across the mbn in opposite directions, eg: Na+/K+-ATPase uses the energy of ATP to move three Na+ out of the cell and two K+ into the cell against their concentration gradients,
primary active transport: the use of ATP to drive transport
secondary active transport: movement of an ion down its electrochemical gradient, to move another substance across the mbn
passive transport:
facilitated diffusion: using a protien to bind and move a molecule or ion across the mbn
exocytosis: AP at axon terminal –> VG Ca2+ to open –> Ca2+ concentration influx results in synaptic vesicles w/ the plasma mbns so that the NT Ach is released into the synaptic cleft (exocytosis) –> Ach binding to receptors on the surface of the muscle cell leads to muscle contraction. Signal is short lived b/c Ach remaining in the synaptic cleft is rapidly degenerated.
endocytosis:
Describe how the glucose transporter works.
glucose transporter fxns: NC 104
Describe the importance of the Na+,K+-ATPase for establishing Na+ and K+ gradients along with the steps in its mechanism.
Importance of Na+/K+ - ATPase for gradient establishment: NC 104
Describe how transport across membranes can occur through vesicle fusion.
transport across mbns via vesicle fusion: NC 104
Describe the structure of SNAREs along with how they facilitate vesicle fusion.
SNAREs fxn: the favorable formation of a four-helix coiled-coil pulls the vesicles into the target mbn
1) Zipping: as the vessicle approaches its target mbn, the SNAREs being zipping together from their N-termini , which draws the two mbns toward eachother to form trans-SNAREs complexes 2) Hemifusion: as docking proceeds, the increased curvature and lateral tension induce the approaching bilayer leaflets to fuse, exposing the bilayer interior 3) the two bilayer leaflets that where orgionally farthest apart are brought together to form a new bilayer 4) fusion pore formation: continuing SNARE-induced lateral tension causes mbn breakdown, resulting in the formation of a fusion pore; 5) the fusion pore expands as the now fused mbn relaxes –> cis-SNARE complexes.
Recall the general structures of RTK and GPCR receptors and describe how they are activated.
RTK:
GPCR: 7 transmbn helices with a ligand binding region on the extracellular face and a G protein binding region on the intracellular face.
Recall the steps for GPCR activation of a G-protein through the inactivation of the G-protein and reassociation of the G-protein with the receptor.
GPCR activation: Ligand binds –> conformation change in GCPR –> Ga releases GDP and binds GTP (switches on) –> Ga-GTP dissiciates from the Gby subunits (both Ga and Gby may activate other proteins). –> Ga has intrinsic GTPase activity and hydrolyses GTP to GDP+Pi (switches off) –> Ga-GDP reassociates with the Gby subunits. GPCRs are associated with G-proteins, heterotrimic intracellular proteins w/ a,b, and y subunits. The Ga subunits has a guanine nucelotide binding site
Describe the basic role of epinephrine on the body and the specific steps of the epinephrine signaling pathway from binding the β-adrenergic receptor through the activation of protein kinase A (PKA).
epi role in body & signal pathway: Epinephrine is an agoinsit for the B2 adrenergic receptor (a GPCR) effects: increase HR, vasodialtion of blood vessels in essential organs. mobilize fuels -> liver (glycogen) + fat tissue.
As long as GaGTP is present, AC will keep making more cAMP (amplification signaling). ATP–> cAMP + PPi, GaGTP activates adenylyl cyclase (AC), AC produces cAMP.
PKA (Protein Kinase A): has two regularoty subunits and each subunit binds 2 cAMPs, Once cAMP has bound , the catalytic subunits dissciate and can transfer a phosphate from ATP to a Ser or Thr side chain on target proteins involved in fuel metabolism.
Describe how specific steps in a signaling pathway are turned off.
signal pathway regulation: are not linear, they can have branching or cross talk as multiple pathways intersect, multiple steps allow multiple levels of regulation to fine tune the signaling response.
Turning off signaling pathways: Phosphodieasterase hydrolyse cAMP (PKA regulatory subunits reassociate with the catalytic subunits).
Process: Ga converts GTP to GDP + Pi and reassoicates with Gby; Arrestin binds the GCPR and prevents further activation of the G-protein.
Describe what a second messenger is along with its role in signaling pathways.
second messengers: cAMP its a signaling molecule that is procuded by AC ( Adenenylyl cyclase)
fxn: amplification of signaling)
Describe how a signaling pathway can amplify a signal.
signal amplification: making a signal stonger which recruits more protein kinase A and produces a greater response within a cell.
Recall the two ways that RTKs can activate a pathway following ligand binding.
RTKs pathway activation: Ligand binding to EC domain of an RTK dimer induces a conformation change in the cytosolic domains. Cytosolic domain of one monomer moves close enough to the cytosolic domain of the other monomer to phosphorylate Tyr resides (swithcing receptor on). –> RTK activation induces a kinase cascade where one kinase activates the next kinase in a chaon –> activation at each step amplifies the signal.
Some activated RTKs can recruit binding protein to transmit a signal, while other RTKs directly phosphorylate kinases.
Describe the role of Ras and the signaling events from growth factor binding through initiation of a kinase cascade.
Ras pathway: monomeric G-Protein (homologous to Ga that is activated by adaptor protiens once tha adapter recognizes phosphoTyr on an RTK with a growth factor ligand bound. Ras exchanges GDP for GTP ( activating RAS), Ras initiates a kinase cascade that regulates the expression of genes that are important for cell growth and division.
Recognize lipid hormones given a structure and describe how they activate intracellular receptors.
lipid homone stucture: many lipid hormones move inside the cell to activate intracellular receptors. hydrophobic character allows them to move through the mbn.
Phospholipids –> Phospholipase A2 (blocked by cortisol-like steroids)–> Arachidonate –> converted to compound
ligand activation of IC receptors is either autocrine ( same cell) or paracrine (neighboring cell) .
Recall the steps and enzymes involved in converting phospholipids to prostaglandins that cause pain and inflammation and decribe how cortisol and COX inhibitors can interfere with the pathway.
phospholipids to prostagladins steps: Arachidonate –> Cyclooxygenase (COX) (blocked by aspirin, ibuprofen (targets of postglandins which inhibit their effect) –> postglandins –> inflammatory and pain response.
COX enzymes then convert the arachidonate to prostaglandins that promote a pain and inflammatory response.
Describe how and why COX-2 specific inhibitors were designed to avoid some of the side effects associated with common NSAIDs like aspirin.
COX-2 specific inhibitors: Aspirin and ibuprofen inhibit both COX1 and COX2 –> reduced pain and inflammation through COX2 inhibition, but increase the risk of stomach ulcers through COX-1 inhibition. (if taken constantly)
Drugs such as Celebrex are designed as inhibitors of COX-2, but cannot bind COX-1 due to a steric clash. (for more chronic pain)
Recall the various naming conventions for sugars including how to indicate the number of carbons, the presence of an aldehyde or ketone, chirality, the type of ring, the anomer, and indicate whether or not the sugar is classified as a reducing sugar.
naming of sugars: use common name for sugars (eg glucose or can indicate the number of carbons and whether the sugar is an aldehyde or ketone (eg aldohexose) ex: Glyceraldehyde –> aldotriose; Dihyroxyactone –> Ketotriose; ribose–> aldopentose; Glucose –> aldohexose; Fuctose –> ketohexose).
number of carbons: when numbering the longest carbon chain, the anomeric carbon is assigned the lowest possible number. Anomeric carbon has two bonds with an electronegative element such as oxygen or nitrogen.
aldehyde or ketone;
chirality:
type of rings: 5 or 6 member rings; 5 membered ring = ‘furanose’ and 6 membered ring =’pyranose’
anomer: has two carbon bonds with an electronegative element ie: oxygen or nitrogen.
reducing sugars: Fehling’s test and Benedict’s test both Use Ca2+ solutions (deep blue) that are reduced to Cu+ (forming CuO2, a deep red ppt.) in the presence of aldehydes and alpha-hydroxyketones. Sugars that react with oxidizing agents are called reducing sugars, whereas those that do not are nonreducing sugars.
A sugar that has an anomeric carbon with a free hydroxyl in the ring form will be an aldehyde or an a-hydroxyketone in the open chain form and is a reducing sugar.
Draw the structure of D-glucose
D-glucose characteristics; in the standard Fischer orientation, a D-sugar has the -OH group to the right of the stereocenter furthest from the anomeric carbon
L-sugars have the -OH group to the left of anomeric carbon.
enantiomer of D-glucose is L-glucose. the C5 epimer of D-glucose is L-idose.
Identify enantiomers, epimers, and the significance of the anomeric carbon.
enantiomers: molecules that are non-superimposible mirro images of one another, most biological sugars are D-enantiomers.
epimers: diastereomers that differ only by the stereochemistry of a single carbon center (stereocenter). ex: D-glucose and D-galactose are C-4 epimers of one another.
signficance of anomeric carbon: standard Fischer projection of sugar has the anomeric carbon at the top.
Describe the chirality of carbons in a sugar and represent that chirality with both Fischer and Haworth projections.
chiratlity of carbons: Haworth projections are used to represent the stereochemistry of sugars in their ring form. In standard Haworth orientation, the ring oxygen is at the back and the anomeric carbon is to the right.
If the last stereocenter’s -OH group closes the ring, then a D-sugar will have the next carbon above the ring and an L sugar will have the next carbon below the ring. If the ring-forming hydroxyl is to the right in the standard Fischer projection, then the next carbon will be above the ring.
In solution, sugars are mostly in the ring form (although monosaccharides quickly convert btwn the ring and staight chain conformations), At a given amount, a soln of D-glucose ia ~ 36 % a-anomer and ~64 % b-anomer. ( has OH in eq postion in chair = more stable); A D-sugar with the anomeric hydroxyl below the ring is called the a-anomer (above the ring is b-anomer); an L-sugar with the anomeric hydroxyl below the ring is called the b-anomer (above the ring is the a-anomer).
Describe glycosidic bonds and systematically name disaccharides or trisaccharides.
glycosidic bonds: formed when the anomeric carbon is joined to another group (making the sugar non-reducing)
systemic naming of disaccharides or triglycerides: describing the glycosidic bond by noting the anomer along with which carbons are linked eg: sucrose is a Alpha-D-glucopyranosyl-91-2)-b-D-fructofuranose.
Describe the roles and structural features of homopolysaccharides such as starch (amylose and amylopectin), glycogen, cellulose, and chitin and discuss how the structure of each relates to its functioion
structural features of homopolysaccharides:
starch: glucose stored in plants, can be either an unbranched polymer called amylose (only a(1–>4) linkages or a branched form = amyliopectin that has a(1–>4) linkages along with a(1–> 6) branch points every ~ 30 resides.
glycogen: glucose stored in animals, similar structure to amylopectin, but branch points are every 8 to 12 residues.
cellulose: primary component of plant cell walls, straight chains with extensive H-bonding strengthening the fibrils. b(1–>4 ) linkage
chitin: primary component in exoskeletons; straight chains have extensive H-bonding b(1–>4) linkage of N-acetylglucosamine.
all are HOMOPOLYSACCHARIDES b/c they are made of the same monosaccharide residues.
Describe the roles and structural features of heteropolysaccharides such as glycosaminoglycans and peptidoglycan and discuss how the structure of each relates to its function.
struct fnx of: conncetive tissue such as cartilage, skin, and tendons include proteoglycans: structural proteins linked to glycosaminoglycans (GAGs). These gycosaminoglycans are heteropolysaccharides consisting of repeating uronic acid and hexosamine resides.
heteropolysaccharides: glycosaminoglycans, peptidogluycan: the polar groups on glycosaminoglycans attract H2O to help lubricate tissue, while negative charges repel each other upon compression to act as shock absorbers.
Peptidoglycan consists of N-acetylglucosamine (NAG) and N-acetylmuramic acid (NAM) repeating units attached to cross-linked tetrapeptides.
Describe the role of glycoproteins and recall how N- and O-linked oligosaccharides are attached to proteins.
roles of glycoproteins: Wide variety of oligosaccharides = modifications possible, b/c they can serve as unique cell ID’s
N and O -linked oligosaccharides on proteins: N-linked oligosaccharide is attached to an Asn side chain. O-linked oligosaccharide is attached to a Ser or Thr side chain. ( ecf on mbn is glycolated (from Ch 8). Wide variety of oligosaccharides = modifications possible, b/c they can serve as unique cell ID’s
Describe how the ABO blood groups are produced in the cell along with the biological basis for blood compatibility within these types.
ABO blood groups produced: Differences in a few active site residues of a glycosyltansferase result in a different sugar being added to the O antigen in A- and B- type individuals, O–> type indivduals dont have a fxnal version of this glycosyltransferase so no sugar is added. ABO blood type is determined by part of a O-linked oligosaccharide on the cell surface.
A, B and O blood types: A = anti-B antibodies in plasma, A antigens on RBC, Compatible A,O bloodtypes. B= anti-A antibodies in plasma, B-antigen in RBCs, Compatable B,O bloodtypes. AB = no antibodies, A and B antigens, AB+ is universal recipient, compatable with A,B,O,AB blood types. O= Anti-A and Anti-B Antibodies in Plasma, no antigens in RBC, compatable with O (O is the universal donor)
Define anabolism and catabolism and describe how energy is conserved or utilized in each process.
anabolism: building complex molecules at the expense of energy
catabolism: breaking down larger molecules
Recall that ingested biopolymers are hydrolyzed by intestinal enzymes to facilitate absorption.
bioploymers hydrolyzation by intestinal enzymes: Enzymes hydrolyze biopolymers into smaller products that may be absorbed by the intestine. energy 70-80 % come from nutrients main energy comes from sugar. W/o constant supply of energy –> break down glycogen and once glycogen is gradually depleated –> fat is then broken down. Starvation leads to break down of proteins –> N as energy.
Recall the order in which available fuels are utilized for energy in the body.
order of fuel availabilty:
Recall which class of proteins are degraded by the lysosome and recall which are degraded by the proteasome along with describing the steps for degradation by the proteasome.
proteins degraded by lysosomes: lysosomes are organells containing digestive enzymes that primarily breakdown extracellular or mbm proteins. increase pH of 7 prevents enzymes activation if they fall outside cell.
proteins degreaded by proteasomes: multisubunit protease that targets intracellular proteins. core regonizes which proteins are meant to be degraded and which aren’t based on specific structure
binding steps for degradation by proteasome: Ubiquitin ligase transfers a small protein, = ubiqutin to a Lys residue of the target protein. –> once four or more ubiquitins have been added, the end cap subunits of the proteasome recognize the target protein for degradation. Proteasome –> A chain of ubiquitin molecules targets a protein to the proteasome –> the polypeptide is unfolded as it enters the proteasome –> protease active sites inside the proteasome cleave the polypeptide into small peptides that diffuse away.
Proteins may be degraded b/c they are abnormal or b/c their concentration must be regulated
Describe the main physiological role and the consequence of a deficiency for the water-soluble vitamins: B1 (thiamine), B3 (niacin), and C (ascorbic acid).
Water soluable vitamins:
Vitamin C: an antioxidant that acts as a cofactor for the enzyme the hydroxylates proline residues in collagen. deficency = scurvy
Vitamin B1 = thiamine is a cofactor for the pyruvate dehydrogenase complex and a-ketoglutarate dehydrogenase; Vitamin B1 deficiency leads to beriberi –> characterized by weakness and leg swelling.
Vitamin B3 deficiency leads to pellagra, characterized by diarrhea, dermatitis, dementia, and death. (ordered from most to least funny –> 5 D’s),. Cells have high rates of metabolism.
Recall the properties (water soluble? # of electrons carried, mobile?) of the following redox cofactors: NAD+/NADH, NADP+/NADPH, Q/QH2, FMN/FMNH2, and FAD/FADH2.
Redox cofactor properties:
NAD+/NADH: Niacin is a precoursor for NAD(P)+/NAD(P)H. Vitamin B3 deficiency leads to pellagra, characterized by diarrhea, dermatitis, dementia, and death. (ordered from most to least funny –> 5 D’s),. Cells have high rates of metabolism.
NADP+/NADPH: Nicotinamide adenine dinucleotide (NAD+/NADH) is a water soluable 2e- carrier, the phosphorylated form (NADP+ / NADPH) is generally used as a redox cofactor in biosynthetic pathways.
Q/QH2: Ubiquinone or Coenzyme Q (Q or CoQ) is a lipid soluable 1(QH*) or 2e- (QH2) carrier.
FMN/FMNH2: Ubiquinone is a mobile electron carrier that often receives electron from an FADH2 or FMNH2 porstentic group, these flavins are dervied from Vitamin B2 (riboflavin). Flavins are prosthetic groups that can carry 1 or 2 e-
FAD/FADH2: Ubiquinone is a mobile electron carrier that often receives electron from an FADH2 or FMNH2 porstentic group, these flavins are dervied from Vitamin B2 (riboflavin). Flavins are prosthetic groups that can carry 1 or 2 e-
Recall the symbols for, and differentiate, chemical standard state and biochemical standard state along with solving calculations involving free energy change, the equilibrium constant, and/or the mass action ratio.
Free energy change: delta G not is free energy change under chemical standard state conditions : 298 K, 1 atm and all [1M].
free energy change under biochemical standard state conditions (same asv chemical standard state except [H+] = 1x10-7 and [H2O] =55.5 M
equillibrium constant: Keq = [C]ceq [D]deq / [A]aeq [B]b eq
mass action ratio: Q = [C]c [D]d / [A]a [B]b , used when the concentrations at equilibrium.
Recall what classes of molecules are used as energy currency by the cell.
classes of molecules used as energy currency by cell: energy may be stored in reduced cofactors, phosphoanhydride bonds (eg ATP) or thioester bonds (eg Acetyl-CoA).
Energy currency is used in chemical reactions to make the overall process favorable (negative delta G).
Describe what is meant by reversible and irreversible steps in a metabolic pathway along with how these steps affect the regulation of a pathway.
reversible steps in metabolic pathway: a metabolic pathway with a postive delta G
irreversible steps in metabolic pathway: a metabolic pathway with a negative delta G
effects on pathway regulation: metabolic pathways are often regulated at steps with a negative delta G (irreversible step), steps with a delta G near zero are reversible and the direction of the net reaction depends on the concentrations of reactants and prodcuts. some opposing pathways may share reversible enzymes (but require different enzymes at any irreversible steps). An irreversible step generally occurs early in a metabolic pathway (often the 1st step) in order to commit a metabolite to the pathway. Keq less than 1 => reactants are favored as Delta G not = would be (+) , little changes in the [ ] amt of rxns and products will change the favorably of the rxn.
For glycolysis, describe its purpose, where it occurs in the cell, and how it is regulated.
glycolysis: occurs in the cytosol of cells; two stages of glycolysis: 1) rxns 1-5 -> Glucose –> G-3-P 9 requires 2 ATPs) 2) reactions 6-10 G-3-P –> pyruvate (yeilds 4 ATP). Net yeild = 2 ATP + 2 NADH ; under anaerobic conditions (from muscles) : 2 NADH –> heat ; under aerobic conditions (from cytsol to mitochondrial matrix) : 2 NADH –> ~ 5 ATP
Describe why it is advantageous to have hexokinase phosphorylate glucose.
advantages of hexokinase phosphorylate glucose : Phosphorylating six carbon sugars prevents the sugars from leaving the cell and reduces the intracellular [glucose] so that the [glucose] gradient favors import. Hexokinase rxn is irreversible.
Describe how alanine screening, transition state analogs, radiolabels, and iodoacetate may be used to support a proposed enzyme mechanism.
alanine screening of active site Lys or Asp eliminates enzymatic activity. the gene for a protein is modified to test the importance of a specific residue. When the codon for Lys is changed to Ala, and activity of the expressed protein is measured, there will be not activity since Lys is essential to the form the Schiff base.
transition state analongs: enediolate transition state analogs, 2-phosphoglycolate and phosphoglycohydroxamate, bind more tightly to TIM than either GAP or DHAP.
radiolabels:
iodoacetate: the presence of an active site Cys is suggested by inactivation of the enzyme by iodoacetate, carboymethlcysteine is then isolated when the protein is separated into individual amino acids.
Recall the different fates of pyruvate and the circumstances for each fate.
Fates of pyruvate: pyruvate is reduced to lactate so the that NADH can be reoxidized to NAD+. NAD+ is then used in the GAPDH rxn to cont glycolysis. ( in animals)
( in yeast) -> Pyruvate is decarboxylated to acetalaldehyde which is then reduced to ethanol so that NADH can be reoxidized to NAD+.
Pyruvate dehydrogenase: pyruvate may be converted to acetyl-CoA which is then oxidized further in the citric acid cyle or used in lipid synthesis.
Pyruvate carboylase: Pyruvate may be converted to oxaloacetate which can be used for amino acid or glucose biosynthesis.
Describe how humans metabolize ethanol.
humans metabolizing methanol: ethanol = alcohol dehydrogenase ( NAD+ -> NADH,H+ is produced via reducing reaction)
acetaldehyde = avetaldehyde dehydrogenase , ( OH - NAD+ forms NADH, H+ ) Acetate; energy currency tells liver cells to produce collagen and fatty acids –> liver disease (cirrhosis & triglycerides).
For gluconeogenesis, describe its purpose, where it occurs in the cell, how it is regulated, which glycolysis reactions must be bypassed (and why), and recall the net reaction of the pathway.
gluconeogensis: energetically expensive. Costs : 4ATP + 2 GTP + 2 NADH, occurs primarily in the liver.
Steps 1, 3 , and 10 in gluconeo have a substantial drop in free energy => irreversible -> separate pathway in gluconeogenesis are bypassed which results in hydration of phosphates ( break phosphoester (via hydrolysis)
glycolysis regulation: high concentrations of glycolytic products such as ATP and PEP inhibition further glycolysis, while molecules that indicate a low energy charge in the cell such as AMP and ADP activate glycolysis.
F-2-6- bisphosphate is produced by PFK-2 ( as opposed to PFK-1, which is the glycolysis enzyme that makes F-1,6,-bisphosphate, aka FBP, step 3 glycolysis). F26BP is an allosteric activator of PFK-1 amd inhibitor of the gluconeogensis enzyme FBPase-1 . Ensures that the opposing pathways do not run simultaneously.
For glycogen synthesis and glycogenolysis, recall each step between glucose-6-phosphate and glycogen including the enzymes involved. Also, describe how branches are formed and removed. You don’t need to draw the exact structures of intermediates.
glycogen synthesis: phosphoglucomutase interconverts G6P and G1P –> G1P is activated by UTP to form UDP-glucose and PPi. the subsequent hydrolysis of PPi drives the rxn forward. (irreversible). 2 phosphates come off at and get hydrolyzed by inorganic pyrophosphate, 2 broken + 1 formed = favorable. –> 3) glycogen synthesis links glucose units via a(1-> 4) and UDP functions as a leaving group. Glycogen branching enzyme moved 7 residues from the main chain to form a new branch.
glycogenolysis: Glycogen phosphorylase use phosphorolysis to break a(1->4) bonds and produce G1P monomers. When a branch has only 4 glucose residues, a debranching enzyme moves 3 residues to the main chain and then cleaves the a(1->6) branch point. Phophorylated sugars cannot cross the plasma mbn so G6P is converted to glucose for export.
For the pentose phosphate pathway, recall the purpose, and the starting and end points if the cell needs NADPH and ribose, or if the cell only needs NADPH, or if the cell only needs ribose.
pentose phosphate pathway; Depending on the cellular requirements for ribose or NADPH, different paths may be used. Path 1: the oxidative path, irreversible. produces NADPH and ribulose-5-phosphate (Ru5P) which may be reversibly converted to ribose-5-phosphate (R5P) Path 2: the carbon rearrangement path. Two F6P + one GAP are rearranged through a series of reversible reactiosn to Ru5P. Path 3: of the cell needs NADPH but not ribose, then G6P may be converted to Ru5P through the oxidative path and Ru5P is converted to GAP and F6P by reversing the carbon rearrangement path.
Recall that cancer cells use glycolysis much more than other cells (Warburg effect). Consequently, glycolytic enzymes may be targeted in chemotherapy or diagnostics.
Cancer and the warburg effect : cancer cells have rates of glycolysis 10 x normal cells. inhibiting glycolysis harms cancer cells more than regular cells.
glycolytic enzyme target of chemotherapy: FdG is metabolized by HK, but cannot be metabolized by PGI.
For the pyruvate dehydrogenase complex, recall the net reaction, purpose, and the role of each cofactor and active site in the multienzyme complex.
pyruvate dehydrogenase complex: a multienzyme that links glycolsysis to the citric acid cycle.
Describe the advantages of multienzyme complexes and define substrate channeling along with describing how substrate channeling may be accomplished.
advantages of multienzyme complexes : 1) the product of one activate site quickly moves to the next active site. PDC is an example of substrate channeling where the intermediates are never released into solution. 2) easier to regulate all enymes together, 3) Minizes side rxns.
substrate channeling: PDC is an example of substrate channeling where the intermediates are never released into solution.
For the citric acid cycle, recall the net reaction, purpose, how and where it is regulated, and the location of the pathway in the cell.
citric acid cycle: occurs in the mitochondria, the mitochondrial matrix contains the pyruvate dehydrogenase complex and most of the enzymes of the citric acid cyle.
Acetyl-CoA enters the citric acid cycle from the pyruvate dehydrogenase complex, amino acid catabolism, or fatty acid oxidation, steps 1,3, and 4 are irreversible, step 6 has an FAD prosthetic group. FADH2 transferase 2e- to Q producing QH2 at this step.
acetyl-CoA + 3NAD+ +Q +GDP + Pi + 2H2O –> 2 CO2 + CoA + 3 NADH + QH2 + GTP.
Step 1: Citrate synthase + step 8 : malate dehydrogenase. ( favorable combined with favorable) the citrate synthase rxn has a lrg enough (negative) free energy change to pull the malate dehydrogenase rxn forward.
Regulation:
1) Substrate availabilty of acetyl-CoA and oxaloacetate (OAA), 2) Allosteric activation by Ca2+ (steps 3 and 4) and ADP (step 3), 3) Allosteric inhibition of isocirtrate dehydrogenase (step 3) by ATP. 4) Product inhibtion (NADH, citrate, and succinyl-CoA) , 5) Feedback inhibition (NADH and succinyl-CoA).
Key idea of citric acid cycle: Intermediates of the citric acid cycle serves a precoursors for a variety of anabolic patheays through cataplortic rxns (anaplerotic rxns replenish citric acid cycle intermediates).
Recall that the aconitase step of the citric acid cycle is an example of a stereospecific enzyme reaction and infer the products for similar reactions.
acontinase step of citric acid cylcle: Citrate is prochiral -> the hydroxyl group is always moved to the carbon originating from the oxaloacetate and not the carbon from the acetyl-CoA.
Recall how the malate dehydrogenase step proceeds in the forward direction despite an unfavorable standard free energy change for the reaction.
malate dehydrogenase step forward despite unfavorable standard free energy change for rxn: Step 1: Citrate synthase + step 8 : malate dehydrogenase. ( favorable combined with favorable) the citrate synthase rxn has a lrg enough (negative) free energy change to pull the malate dehydrogenase rxn forward.
Recall how radiolabels may be used to track carbons through the citric acid cycle (or metabolic pathways in general) and the fate of the carbons from acetyl-CoA and citrate in each round.
Radiolabels used to track carbons through citric acid cycle:
Carbons used to make CO2 originate in oxaloacetate, not acetyl-CoA, track carbons using radioactive isotopes to see where acetyl-CoA is used and where OAA is used via radiocarbon traces.
Define anaplerotic and cataplerotic reactions and describe how they affect flux through the citric acid cycle.
anaplerotic : consequence of catabolic process.; rxns that increase the number of intermediates,
cataplerotic: taking out/ reducing intermediates out of pathway, rxns that reduce # of intermediates.
Interpret and solve calculations involving free energy and reduction potentials.
intrepretations of calculations: e = reduction potential ( the affinity of a substance for electrons)
eo’ = standard reduction potential
electrons flow spontaneously from the substance with the lower reduction potential to the substance with the higher reduction potential.
Recall the basic architecture of mitochondria and where the complexes of oxidative phosphorylation are located.
struct of mitochondria: the inner mbn of mitochondria contains all the complexes for oxidative phosphorylation, the outer mbn is very porous, while the inner mbn is impermeable to most small molecules and ions intermembrane space = cytosol.
complexes of oxdiative phosphorylation: reduced cofactors donate electrons to an electron transport chain that generates a H+ gradient. this gradient is used to generate ATP.
Describe the purpose and thermodynamic driving force for the ATP/ADP transporter and the H+/Pi transporter.
thermodynaic driving force of ATP/ADP transporter:
H+/Pi transporter: ATP is produced in the matrix, so it must be exported to the cytosol while ADP is imported into the matrix in order to make more ATP.
ATP and ADP are moved using an antiporter driven by the inner mbn potential (negative on the matrix side of the mbn).
Pi is imported along with a H+ using a symporter driven by the H+ gradient (lower [H+] in matrix).
Describe the purpose, steps, and thermodynamic driving force for the malate/aspartate shuttle.
malate/aspartate shuttle: net effect is one fewer NADH in the cytosol and one additional NADH in the matrix. NADH is too lrg to transported efficently in the matrix.
For complex I (NADH dehydrogenase): recall what mobile electron carrier donates electrons to the complex, what mobile electron carrier carries electrons away from the complex, how many protons are translocated, and the major electron carriers within the complex. Electron carriers should be described in their correct oxidized or reduced form depending on the step.
NADH dehydrogenase: Complex I is a proton pump driven by NADH oxidation, in complex I (aka NADH dehydrogenase), 4H+ are translocated into the intermbn space for every 2 e- that pass through the complex.
NADH + H+ +Q –> NAD+ + QH2
Complex I , a H- ion is trasnferred from NADH to FMN. FMN then passes the electrons ( one at a time) to a series of 8-10 iron-sulfur clusters until they reach Q. NADH has to give 2e- at a time.
NADH is a water soluable, mobile 2e- carrier. Flavins are prosthetic group that can carry 1 and 2e-. Fe-S clusters are obligate 1e- carriers using cystines and inorganic sulfure to corrdinate Fe2+/Fe3+
Ubiquinone is a lipid soluable, mobile 1 or 2 e- carrier.
For complex II, recall that it is part of the citric acid cycle (succinate dehydrogenase) and reduces ubiquinone (via FADH2).
Complex II in citric acid cycle: a) from succinate dehydrongenase (complex II) in the citric acid cyle (succinate +FAD -> Fumarate + FADH2) , b) fatty acid oxidation (acyl-CoA dehydrogenase rxn), c) from glycerol-3-phosphate (mitochondrial dehydrogenase rxn.
For complexes III (cytochrome bc1) and IV (cytochrome c oxidase), recall what mobile electron carrier donates electrons to the complex, what mobile electron carrier carries electrons away from the complex, and how many protons are translocated in each complex. Electron carriers should be described in their correct oxidized or reduced form depending on the step.
Complex III: Ubiquinol (QH2) donates 2 electrons to complex III (aka cytochrome bc1), ultimately reducing cytochrome c (cty c) , a mobile (single) electron carrier, 4H+ are translocated into the intermbn space for every 2e- that pass through complex III.
Comples #4: Two cytochrome c proteins each donate 1 electron to complex IV (aka cytochrome c oxidase), ultimately reducing the terminal electron acceptor O2. 2H+ are translocated into the intermembrane space for every 2e- that pass through complex IV.
Recall that cytochromes are heme-containing proteins where the letter designation indicates the side groups on the heme and can only carry one electron at a time.
cytochromes heme-containing proteins: Cytochromes are proteins with a heme group that are obligate single electron carriers. a,b, and c denote the type of side groups on the heme.
oxidative phosphoralation is regulated only by the availability of NADH and QH2.
Describe the chemiosmotic model.
chemiosmotic model: A combination of the H+ imbalance 9due to e- transport), and the negative charges on the matrix side of the mbn, results in an electrochemical gradient that favors H+ moving into the matrix.
For complex V (ATP synthase), recall the general structure including the F1 portion, F0 portion, α, β, γ, a and c-subunits. Recall that rotation of the c-ring builds up strain in the γ subunit until it rotates 120o, forcing the three β subunits into one of three conformations: loose, tight, and open. Describe the mechanism of the complex and what happens during each of the β conformation shifts.
Complex V: (f1,F0ATP): The F1 soluable portion includes 3a and 3b subunits. –> Each of the b subunits has ATP synthase activity.
F0 mbn-spanning portion includes the a subunit, where H+ enter and exit, and a c-ring. The c-ring consists of 8-15 subunits that each bind a H+ to help rotate the ring.
Protons move down their gradieent by passing through ATP synthase
The y subunit extends from F0 to F1 and rotates 120 deg once enough strain has built up from the c-ring rotation. Experimentally, we observe 1 ATP produced for every 4 H+ translocated in the electron transport chain (no matter how many subunits are in the c-ring)
1) L: ADP+Pi are bound, 2) ATP is formed 3) O:ATP is released, 120 degree rotation of the y subunit forces the b-subunits into one of threee conformations. One full rotation of the the y subunits produces 3 ATP (1 ATP/b subunit)
Define P:O ratios and describe how they are used to calculate ATP yields for metabolic pathways assuming it takes about 4 H+ translocated to make 1 ATP.
P:O ratios: the ratio of phosphorylations (ADP+Pi –> ATP) to oxygen atoms reduced (1/2 O2 + 2H+ + 2e- –> H2O) is the P:O ratio.
use for ATP yeilds: NADH oxidation translocases 10 H+ (4H+ in comples I + 4H+ in complex III + 2H+ in complex III + 2H+ in complex IV , 10 H+ x( 1 ATP/4H+) = 2.5 ATP
QH2 oxidation translocases 6 H+ (4H+ in complex III + 2H+ in complex IV) 6H+ x (1ATP/ 4H+) =1.5 ATP
ie: the P:O ratio of NADH is 2.5 and of QH2 is 1.5
Recall that uncoupling oxidative phosphorylation (through uncoupling proteins or dinitrophenol ie. DNP) provides a path to bring H+ down their concentration gradient without passing through ATP synthase, so the free energy for this process generates heat.
uncoupling oxidative phosphorylation: Venturididin and oligomycin block the H+ channel in ATP synthase. blocking the H+ channel causes the electron transport chain to slow down as the H+ gradient gets too sleep (it doesnt stop electron transport completely since some H+ can leak across the mbn.
Dintrophenol (DNP) dissipates the electrochemical gradient and generates heat. uncouplers such as 2,4-dinitrophenol (DNP) carry H+ across the mitchondrial inner mbn (down the H+ gradient) w/o passing through ATP synthase.
Recall that cyanide blocks electron transport through complex IV, while venturicidin and oligomycin block H+ movement through ATP synthase and predict the effect of inhibiting each complex of oxidative phosphorylation.
Cyanide effects: O2 consumption indicates that electrons are moving through the electron transoprt chain; Succinate produces QH2 ( at complex II) and QH2 is a source of eletrons in the electron transport chain
ATP synthesis requires ADP + Pi and indicates that H+ are moving into the matrix through ATP synthase.
Cyanide (CN-) blocks electron transport through complex IV, shutting down the entire electron transport chain and preventing the formation of a H+ gradient.
Recall that succinate may be added to mitochondria to generate QH2 and that O2 consumption indicates electron transport is working, while ATP synthesis implies protons are moving through ATP synthase.
succinate added to mitochondria: Succinate produces QH2 ( at complex II) and QH2 is a source of eletrons in the electron transport chain
Recall the basic architecture of chloroplasts and where the complexes of photosynthesis are located.
strcture/fxn of chloroplasts: Chloroplasts have inner mbn that surrounds the stroma (where the soluable enzymes are located)
Within the stroma, flattened vesicles = thylakoids that contain the light harvesting complexes, the ETC and ATP synthase, teh thylakoid lumen has a low pH (high [H+] concentration and the H+ gradient drives ATP synthase.
Recall that chlorophylls and carotenoids are examples of pigments that absorb light at various wavelengths depending on both pigment structure and the protein environment of the pigment.
chlorophylls: pigment that can absorb light at various wavelengths
carotenoids: pigment that can absorb light at various wavelengths
Photosynthetic pigments absorb light. The pigments are bound to proteins and the differing protein enviroments affect the wavelength of light absorbed.
Absorbing a photon promotes an electron to a higher energy level (excited state) .
Describe the different ways that an excited molecule can lose energy and how these are relevant in photosynthesis.
excited molecule energy loss: heat, light (fluroescence), excitation transfer, photooxidation
relevance to photosynthesis: photosynthetic organisms have light-harvesting complexes consisting of pigments bound to proteins, The reaction center is a protein with a special pair of chlorophylls that has a lower excited state than the surrounding antenna chorlophylls.
Describe how antenna pigments funnel light energy (via exciton transfer) to a reaction center consisting of a special pair of chlorophylls. The excited reaction center has a low reduction potential allowing it to reduce the next electron carrier in the transport chain. An oxidized reaction center has a very high reduction potential allowing it to be spontaneously reduced by other electron carriers (water in the case of P680+ or plastocyanin in the case of P700+).
antenna pigments funnel light energy to reaction center: antenna chlorophyll surround a rxn center and pass the energy of an absorbed photon from chlorophyll to chlorophyll until the energy is trapped by the reaction center.
The excited reaction center has a low reduction potential allowing it to reduce the next electron carrier in the transport chain. An oxidized reaction center has a very high reduction potential allowing it to be spontaneously reduced by other electron carriers (water in the case of P680+ or plastocyanin in the case of P700+).
For the light reactions (cytochrome b6f, photosystems I and II), recall what electron carrier donates electrons to the complex, what electron carrier carries electrons away from the complex, and how many protons are translocated. For photosystem II, recall that this is where the oxygen-evolving complex is located. For photosystem I, describe the purpose of (and the fate of ferredoxin in) cyclic and noncyclic electron flow. Electron carriers should be described in their correct oxidized or reduced form depending on the step.
light reactions: the series of events from the oxidation of H2O through the production of NADPH and ATP. An electron transport chain generates a H+ gradient that drives ATP synthase.
Photosystem II: P680 + hv –> P680* Excited P680* reduces phephytin (Pheo). P680* + Pheo –> P680+ + Pheo- Pheo- then passes an e- to a bound pastoquinone, PQa- ; PQa transfers an electron to the mobile plastoquinone: PQb; PQa - + PQb –> PQa + PQb-. Photosystem II has a P680 rxn center that is excited by photon energy, lowering the P680 reduction potential (e) so that it can transfer an electron to pheophytin.
P680* has a high enough reduction potential that it is spontaneously reduced back to P680 by H2O in the oxygen-evolving complex (OEC) of PSII , PQBH2 carries 2e- away from PSII and toward cytochrome b6f. Overall, 4 Photons produce 4 photooxidations and 2PQBH2. 4H+ are effectivley translocated into the thylakoid lumen.
photosystem I: The movement of 4e- through cytochrome b6f translocates 8H+into the thylakoid lumen and reduces 4 plastocyanins ( a mobile e- carrier); plastocyanin carries an electron to another photosystem (PSI) . Electrons move through PSI and reduce the mobile e- carrier ferredoxin.
Free energy can be conserved by reducing NADP+ –> NADPH (noncyclic e- flow) or in the H+ gradient by transferring the e- back to cytochrome b6f (cyclic e- flow)
Recall the Z-scheme plots reduction potentials for the main electron carriers in photosynthesis and indicate the order of electron transfers.
Z-scheme plots reduction potentials: H2O –> P680–> P680* –> Plastoquinone–> Cytochrome b6f –> plastocyanin –> P700 –> P700* –> Ferredoxin –> NADP+
Recall that CF1CF0-ATP synthase has a similar structure and function to mitochondrial ATP synthase, but is driven by a much larger pH gradient since thylakoid membranes are permeable to ions that prevent establishment of a potential across the membrane.
C1F CF0-ATP synthase driving factors: CF1CF0-ATP synthase has a similar structure and function to mitochondrial ATP synthase, but is driven by a much larger pH gradient since thylakoid membranes are permeable to ions that prevent establishment of a potential across the membrane.
ATP synthase driving factors: The thylakoid mbns is permeable to some ions so chloroplasts need a much larger pH gradient to compensate dor the lack of a charge difference across the mbn.
Rxns occur: Mitochondrion, Chloroplast, bacterium
Describe the purpose of RuBisCO, its net reaction, and how it is regulated by pH.
RuBisCO: ribulose Bisphosphate carboylase/oxygenase; regulated by pH, when the light rxns are occuring the stroma pH increases to activate rubisco. otherwise, rubisco is inactive to allow the cell to conserve ATP and NADPH
Recall the purpose of the Calvin cycle, the net reaction, that it requires ATP and NADPH, and that glyceraldehyde-3-phosphate is siphoned off for various anabolic pathways.
Calvin Cycle: RuBisCO fixes CO2 to form 3PG, which is then converted to GAP at the expense of ATP and NADPH.
GAP is converted to Ru5P, Ru5P is then phosphorylated by ATP to get the RuBisCO substrate RuBP, the dark (light indep rxn occur in stroma)
For every 6 GAP produced, one is removed from the cycle to serve as a precoursor for various biomolecules. The other five GAPs have their carbons rearranged to form three 5-carbon ribulose sugars.
Describe the digestion, absorption, and transport of lipids.
digestion + absorption: bile salts emulsify dietary fats in small intestine, intestinal lipases degrade triglycerides, fatty acids and other breakdown products are taken up by the intestinal mucosa and converted into triglycerols, triacylglyerols are incorporated, with cholesterol and apoliproteins, into chylomicrons, chylomicrons move through the lymphatic system and bloodstream to tissues, lipoprotein lipase, activated by apoC-II in the capillary, converts triacylglycerols to fatty acids and glycerol, fatty acids enter cells, fatty acids oxidized as fuel or reesterified for storage.
Transport: Chylomicrons transport dietary triglycerols from the intestine to adipose tissue and transport cholesterol to the liver, Low density lipoproteins (LDL or bad cholesterol) transport cholesterol to various tissues. high LDL –> high risk for athereosclerosis.
High density lipoproteins HDL / good cholesterol, trasnport excess cholesterol back to the liver (HDL associated with decrease risk of athereosclerosis.
Very low density lipoproteins (VLDL) transport triacylglycerols from the liver to other tissues, intermediate density lipoproteins (IDL) from as VLDLs lose triacylglycerols.
Recall the general composition and structure of lipoproteins and describe how the composition changes as you increase density. Recall the roles of the different types of lipoproteins and their relative densities and amounts of protein, triacylglycerol and cholesterol.
lipoproteins: struct and fxn –> Lipoproteins have a highly hydrophobic core of triacylglycerols and cholesteryl esters surrounded by proteins and amphipathic lipids such as cholesterol and phospholipids
Roles of lipoproteins: fatty acids are actived for catabolism: linked to CoA via a high energy thioester bond at the expense of 2ATP equivalents
Recall that fatty acids are removed from triacylglycerols by lipases and are activated to form acyl-CoA at the expense of two ATP equivalents. Acyl-CoA is then imported to the matrix by the carnitine transporter.
fatty acid removal: Long chain fatty acid (> 20 carbons) and branched fatty acids are oxidized in the perioxisome instead of the mitochondrial matrix.
Acyl-CoA is too large to move into the matrix. Instead, the acyl group is esterified to carnitine for import to the mitochondrial matrix from the cytosol.
Recall the net reaction for a round of β-oxidation. Recall that the last round of β-oxidation produces two acetyl-CoA molecules. Indicate how ATP yield changes when you have odd-chain or unsaturated fatty acids.
b-oxidation: Each round of beta ocidation consists of 4 rxns, there are 2 main oxidation steps
The last round splits a 4-carbon acyl-CoA into two acetyl-CoAs. Each round of b-oxidation removes two carbons (as acetyl-CoA) from the activated end of the acyl-CoA.
Each double bond in a fatty acid will reduce the energy yeild by 1 QH2 or 1 NADH depending on the position of the double bond. Odd-chainn fatty acids have approximately the same energy yeild (per carbon) as even-chain fatty acids.
Recall the purpose and location of the carnitine transporter along with what molecules are transported by the carnitine transporter.
carnitine transporter :
Recall that branched fatty acids and fatty acids that are 20 carbons or longer are oxidized in the peroxisome.
branched fatty acid & fatty acids that are 20 carbons or longer are oxidized in the peroxisome.
Recall that the citrate transporter removes one acetyl-CoA from the matrix and produces one acetyl-CoA in the cytosol.
citrate transporter : ACC is activates by citrate which produces acetyl-CoA
Recall that acetyl-CoA carboxylase converts acetyl-CoA and HCO3- to malonyl-CoA at the expense of 1 ATP. Know the net reaction for fatty acid synthesis. Recall that elongases can extend the length of fatty acids and that desaturases introduce double bonds to fatty acids.
fatty acid synthesis net rxn: Fatty acid sythesis (lipogensis) occurs in the cytosol, Acetyl-CoA is carboxylated in the cytosol by acetyl-CoA carboxylase (ACC) to from malonyl-CoA; glucagon/epinephrine signaling inhibits this step by phosphorylating ACC, insulin signaling acctivates this step by dephosphorylating ACC.
FA Synth is a multienzyme with 6 different active sites that catalyzes 7 rxns, the usual product of fatty acid synthase is palmitate (16:0), elongases can extend palmitate to make longer fatty acids anddesaturases can introduce double bonds in fatty acids, Acetyl-CoA is extended, two carbons at a time ( from malonyl-CoA), by the fatty acid synthase multifxnal enzyme.
Recall that fatty acid synthase has multiple active sites and a flexible linker to move the product of one reaction to the next active site via substrate channeling. Describe why this is more efficient than synthesizing separate enzymes.
multiple active sites and flexible linker advantages: uses sibstrate channeling: acyl carrier protein (ACP) swings intermediates btwn active sites.
Acetyl-Cys –> FA synthase ( Malonyl-CoA + 2 NADPH to CoA + 2NADP+ + CO2) –> 4 Carbon acyl-Cys (Malonyl-CoA + 2 NADPH to CoA + 2NADP+ + CO2) –> 6 carbon acyl-Cys –> 5 times (Malonyl-CoA + 2 NADPH to CoA + 2NADP+ + CO2) –> 16 carbon acyl- ACP –> FA synthase (thioesterase) –> Palmitate (16:0) + ACP.
Describe how and where fatty acid synthesis and oxidation are regulated.
fatty acid synthesis regulation: Acetyl-CoA carboxylase (ACC) is the primary regulation point for FA metabolism
oxidation regulation: ACC is activated by citrate which produces acetyl-CoA
ACC is inhibited by fatty acids ( the product of the pathway)
the product of ACC, malonyl-CoA, inhibits acylation of carnitine ( so that fatty acid biosynthesis and fatty acid oxidation dont occur simultaneously).
Calculate the ATP yield for the complete oxidation of saturated fatty acids.
ATP yeild for complete oxidation:
Calculate the ATP cost for the complete synthesis of a fatty acid from cytosolic acetyl-CoA.
ATP cost:
Recall the purpose of ketogenesis along with the substrates and products of ketogenesis.
Ketogenesis: The CNS uses glucose as its primary fuel (fatty acids do not cross the blood-brain barrier efficiently)
Ketogenesis occurs when blood [glucose] os low and glycogen is depleated (fasting)
the liver responds by generating glucose through gluconeogenesis and using acetyl-CoA to make the ketone bodies acetoacetate and 3-hydroxybutyrate. Ketone bodies are converted back to acetyl-CoA in the CNS.
Recall that cholesterol is synthesized from acetyl-CoA and that the pathway can be inhibited at its rate-limiting step (HMG-CoA reductase) by competitive inhibitors (eg. Lipitor) called statins.
cholesterol synthesis: Assembled from acetyl-CoA units
pathway inhibtion: the pathway can be inhibited at its rate-limiting step (HMG-CoA reductase) by competitive inhibitors (eg. Lipitor) called statins.
rate limiting step: HMG-CoA reductase
competitive inhibtor: Lipitor) called statins.
Describe why statin inhibition of cholesterol synthesis will ultimately lower circulating LDL levels.
effects of statin on cholesterol and LDL levels: Statins mimic HMG-CoA and inhibit HMG-CoA reductase,
HMG-CoA reductase is the rate limiting step of cholesterol biosynthesis, cells that cannot make enough cholesterol compensate by increasing production of LDL receptors, more LDL receptors results in more endocytosis of LDLs, fewer circulating LDLs reduces the risk of atherosclerosis.
Describe the nitrogen cycle along with the role of diazotrophs in the cycle.
Nitrogen cycle: N2 –> nitrogen fixation (nitrogenase) –> NH4+ –> nitrification NO2- (reversible via nitrate reductase) –> NO3- –> nitrification (reversible via nitrate reductase) –> N2 via denitrification.
Diazotrophs are the only organisms taht produce the enzyme nitrogenase (fixates N2)
N2 + 8H+ + 8e- + 16 ATP + 16 H2O –> 2 NH3 + H2+ 16 ADP + 16 Pi.
Recall that the nitrogenase reaction is very energetically expensive and thus the enzyme is not produced by organisms that can obtain fixed nitrogen in their diet.
nitrogenase rxn: aKG + NH4+ NADPH + ATP –> Glu + NADP+ + ADP + Pi
fixed nitrogen in diet of organisms: a combination of glutamine synthetase and glutamate synthase rxns results in the formation of an animo acid using ammonium as the nitrogen source.
nitrogenase reaction is very energetically expensive and thus the enzyme is not produced by organisms that can obtain fixed nitrogen in their diet.
Describe how the glutamine synthetase and glutamate synthase reactions combine to allow the entry of nitrogen into biomolecules in many organisms.
glutamine synthetase and glutamate rxn combo: a combination of glutamine synthetase and glutamate synthase rxns results in the formation of an animo acid using ammonium as the nitrogen source.
Recall that amino acids serve as nitrogen carriers in our body and ultimately provide the nitrogen for other biomolecules.
amino acids that serve as nitrogen carriers: Nitrogen may be incorporated into other biomolecules by incorporating the amino acid, or through transaminase rxns,
transaminase enzymes swap the carbonyl group of alpha-keto acids with the amino group of amino acids.
Identify transaminase reactions and deduce reactants or products given the other species.
transaminase rxn: transaminase enzymes swap the carbonyl group of alpha-keto acids with the amino group of amino acids. Ex: glutamate + Pyruvate rxn
Define essential and nonessential amino acids.
essential AAs: necessary in diet
nonessential AAs: not necessary in diet, only animals have biosynthetic pathways for some nonessential amino acids,
Recall that the nitrogen in nitrogenous bases comes from amino acids.
nitrogen in Nitrogenous bases from AAs: many signaling molecules dervied from AAs ex: Epi and Norepi Catechols, Serotonin, melatonin, Tyrosine, Dopamine,
vasodilator (NO) derived from arginine.
AAs are nitrogen precoursors of nitrogenous bases.
Recall that inosine monophosphate (IMP) is a precursor for purine nucleotides and that UMP is a precursor for pyrimidine nucleotides.
IMP = purine nucs
UMP = precursor for pyrimidine nucleotides
Recall how nucleotides are catabolized to nitrogenous bases and describe the fate of both purine and pyrimidine nitrogenous bases.
fates of purine and pyrimidine nb’s: Sodium urate has poor solubility and can crystallize in joints (gout) or form kidney stones.
Purines are converted to uric acid (urate) and extreted in urine.
Pyrimidine catabolism produces intermediates of fatty acid metabolism, in humans, nitrogen from pyrimidine catabolism ends up in the urea cycle.
Explain the relationship between gout and purine degradation.
gout and purine degradation: Sodium urate has poor solubility and can crystallize in joints (gout) or form kidney stones.
Purines are converted to uric acid (urate) and extreted in urine.
Identify the amino acid precursor for signaling molecules given the structure of the signaling molecule.
aa precoursor for signaling molecules:
structure of signaling molecule:
Define glucogenic and ketogenic amino acids.
glucogenic : catabolized to precoursors for gluconeogenesis such as pyruvate and oxaloacetate
ketogenic: catabolized to acetyl-CoA for use in ketogenesis or fatty acid synth.
Describe the different ways ammonotelic, ureotelic, and uricotelic organisms excrete nitrogen along with the advantages of each method.
ammonotelic: excrete nitrogen as ammonia
ureotelic: excrete nitrogen as urea
uricotelic organsisms: uric acid requires very little water to produce a paste but requires more energy to make. Using less water is an advantage to birds as it means less weight, using less water is an adv to animals in the desert as there is less water loss, uric acid/ urate is not very toxic so it can build up in eggs.
Recall that the urea cycle occurs primarily in the liver and that amino acids carry nitrogen to the cycle and donate amino groups to alpha-ketoacids to form Asp and Glu that then donate the amino groups to make urea.
urea cycle: amino acids transfer their amino groups to alpha-ketoglutarate and oxaloacetate to form Glu and ASP, Asp and Glu donate their amino groups to ultimately form urea, incorporating NH4+ into urea requires some energy investment, but excretion only requires ~ 0.05 L or H2O/gram of nitrogen extreted
Recall that urea is polar and requires substantial water to eliminate in urine.
urea: polar and requires substantial water to eliminate in urine.
Explain why certain metabolic pathways might be expressed in particular cell types or localized in different parts of the cell.
metabolic pathway expression: compartmentalized in different cell types, or in different regions of a given cell
Liver –> Fed state –> glycolysis, Fasted state –> gluconeogenesis
Kideny –> gluconeogenesis
Adipose tissue –> fat storage (fed state), fat break down (lipolysis) in fasted state.
Muscles –> glycolysis, store energy.
Recall the steps and purpose of the Cori cycle.
Cori Cycle: under anaerobic conditions, muscles produce lactate which is exported to the liver for gluconeogenesis. Liver then exports the glucose back to the mucles.
Define hormones and describe the role of insulin, glucagon, and epinephrine in fuel metabolism. Describe the physiological conditions that stimulate release of these hormones and recall what type of cells produce each of these hormones.
hormones: substances produced by one tissue that effect the fxn of other tissues throughout the body
insulin: released by b-cells of pancreatic islets in response to high blood [glucose]. it signals fuel abundance promoting fuel storage while limiting the release of stored fuels.
glucagon: released by a-cells of pancreatic islets in response tp low blood [glucose]. it signals stored fuel release
epinephrine: released by adrenal medula in response to fight or flight situations.
epi and glucagon have similar effects on liver cells (promote gluconeogeneis and glycogen breakdown), muscle cells do not have glucagon receports or the enzymes for gluconeogenesis, in muscle cells, epi promotes glycolysis and glycogen breakdown.
conditions which stimulate release of hormones:
types of cells that produce each hormones:
Describe the effect of these hormones on glucose import (via GLUT4), glycogen synthase, glycogen phosphorylase, triacylglycerol synthesis and breakdown, glycolysis, gluconeogenesis, fatty acid synthesis, and β-oxidation and predict their effect on catabolic or anabolic pathways.
effect of hormones on glucose (GLUT4): insulin receports initates a signaling cascade with various intracellular effects including fusion of vesicles, containg GLUT 4 transporter, w/ plasma mbn
glycogen synthase: insulin signaling activates phosphatases that dephosphorylate both glycogen synthase (activating it)
glycogen phosphorylase: insulin signaling activates phosphatases that dephosphorylate glycogen phosphorylatse (inhibiting it)
triacylglycerol synthesis and breakdown:
glycolysis:
gluconeogenesis:
fatty acid synthesis:
b-oxidation:
effects on catabolic pathways:
effects on anabolic pathways:
Describe the causes and treatment of type I and II diabetes along with the acute and long-term effects.
Causes and treatments of Type I and II diabetes:
A disorder of fuel metabolism characterized by high blood [glucose], affects ` 10 % of Americans,
Type I: juvenile onset –> no insulin: autoimmune disease that kills pancreatic b-cells, treated with insulin injections
Type II: adults onset –> insulin resistance: underlying cause unknown, but closely linked to obesity and inactivity, treated with lifestyle changes (improve diet/exersize), drugs that lower blood [glucose], or bariatric surgery.
Hyperglycemia leads to long term affects such as cataracts, kidney failure, and cadio changes, acute affects include hypoglycemia (if too much insulin is injected) and ketoacidosis.
Describe the steps of DNA replication and the role of the following components: helicase, single-strand binding protein (SSB), primase, DNA polymerase (I, III, δ, and ε), telomerase, ligase, and RNase H.
Steps of DNA replication: DNA is replicated in a semiconservative manner (one exsisting strand is present in new model. Proteins associate with a specific DNA sequence = origin of replication to open up an intital replication bubble.
Helicase: Uses ATP to unwind the DNA helix
single stranded binding protein: bind the single-stranded regions to prevent reannealing or nucleasse digestion of the DNA. Provides the reg struct so that sequences dont fold over
DNA polymerase (I, III, …, … ): DNA Polymerase adds deoxynucleases to the free 3’OH of an esisting strand. DNA Pol III is the main polymerase used for replication. DNA pol delta does lagging strand synthesis and DNA pol epsilon does the leading strand. ( Most DNA polymerases have a proofreading fxn: the polymerase slows down due to the bulge causes by the mispairing and a 3’–> 5’ exonuclease removes the misincorporated nucleotide. Sinse DNA polymerase extends DNA in the 5’–>3’ direction, one strand is synthesized continuously ( leading strand) and the other strand is syntheszied in segements (Okazaki fragments (lagging strand).
primase: introduces a short RNA primer that is complementary to the template stand. A 5’ –> 3’ exonuclease such as RNase H or DNA pol I removes the RNA primer from each Okazaki fragment.
telomerase: the end shortening problem: linear DNA gets shorter adter a round of replication as the 5’ RNA primer cannot be replaced by DNA. The soln: telomerase adds 5’ TTAGGG repeasts to the 3’ ends of human DNA. A reverse transcriptase that carries an RNA template that codes for a new repeat while base pairing with the previous repleat. Aster a found of replication, the shortened chromosome is missing repetitive DNA rather than essential genetic information. Cells that replicate indefinietly have telomerase activity, most cells stop expressing telomerases to protect against cancer.
RNase H : A 5’ –> 3’ exonuclease such as RNase H or DNA pol I removes the RNA primer from each Okazaki fragment.
Define leading strand, lagging strand, and Okazaki fragments and explain why they are necessary.
leading strand: 5’–> 3’ direction strand (leading strand) DNA ( uninterrupted)
lagging strand: strand that is synthezied in fragments.
Okazaki fragments: smaller individual strands that synthesize the lagging strand.
Describe the structure and role of telomeres in cells.
telomeres struct and fxn: Linear DNA is susceptible to end-joining by the cell’s DNA repair machinery or degredation by exonucleases. Ends of chromosomes adopt special structures called telomeres where a repleating sequencing of DNA folds into a loop and binds protective proteins.
Draw the arrow pushing mechanism of the DNA polymerase reaction.
mech for DNA polymerase rxn:
Describe the basic structure and role of a nucleosome.
nucleosome struct and fxn: an enzyme the degrades nucleic acids,
Describe the proofreading activity of a polymerase and explain how it is accomplished along with the consequence for the fidelity of replication.
proofreading activity of a polymerase: the polymerase slows down due to the bulge causes by the mispairing and a 3’–> 5’ exonuclease removes the misincorporated nucleotide. Sinse DNA polymerase extends DNA in the 5’–>3’ direction, one strand is synthesized continuously ( leading strand) and the other strand is syntheszied in segements (Okazaki fragments (lagging strand).
accomplishment with fidelity of replication:
Describe the structure and purpose of heterochromatin and euchromatin.
heterochromatin struct and fxn: is tightly packed (eg 30 nm fibre) and less transcriptionally active
euchromatin struct and fxn: is lightly packed DNA (eg. beads-on-a-string) that is heavily transcribed
Describe the characteristics and origins of cancer.
Characteristics: uncrontrolled cell division, results from accumulated mutations or modifications of DNA,
origins: multiple dieases within multiple origins, Mutations may be induced by environmental factors (eg, sun smoke), genetic predisposition (eg. BRCA1), viral infection 9eg HPV, hep B) or spontaneous mutations.
Recall how carcinogens such as UV light, benzopyrene, methylating agents, and reactive oxygen species induce DNA mutations.
UV light: promotes pyrimidine dimers (most frequently btwn thymines)
benzopyrene : is oxidized in the liver and then forms an adduct with N2 of guanine, distorts helix, benzopyrene causes G–> T transversions.
methylating agents: disrupt hydrogen bonding , transition G to A (purine to purine transition) ;
reactive oxygen species:
Define transversion and transition point mutations.
transversion mutations:
point mutations: a transversion is a mutation leading to a switch btwn purine and pyrimidine.
a transition is a purine or pyrimidine to pyrmidine mutation.
ex: guanine –> 8-oxoguanine caused by reactive oxygen species such peroxide and results in a G–> T transversion. a
Describe how abasic sites and deaminations are generated in DNA.
abasic sites: thousands of bases are spontanrously hyrolyzed from ribose in our body each day , breaking glycosidic bond and taking off base pair bair, purines have more abasic site.
deaminations: cytosine undergoes a demination rxn with water, H2O leaves behind an oxygen.
Explain why DNA incorporates thymine nucleotides instead of uracil.
thymine vs uracil: DNA uses thymine instead of uracil so taht, if demination occurs, uracil is recognized as an error and can be removed.
Describe how the following repair mechanisms work and how they recognize errors in DNA: base excision repair, nucleotide excision repair, mismatch repair, nonhomologous end-joining, homologous recombination, and alkyltransferases.
base excision repair: cells that have specific glycosylases that recognize common erris, such as T-G pairs or uracil, and remove the incorrect bases. Ex: damaged base –> DNA glycosylase –> abasic site –> endonuclease –> Free hydroxyl group –> DNA polymerase + DNA ligase –> second phosphodiester bond with bp restored by enzyme leading sequence
nucleotide excison repair: errors that significantly distort the helix are repaired by removing a segment of DNA surrounding the distortion. Ex: damaged nucleotide –> helicase–> endonuclease –> segment separation in need of repair –> DNA polymerase + DNA ligase –> repaired DNA strand. (Change of purine to pyrimidine.
mismatch repair: scans newly synthesized stand and exices any segments with misincorporated nucleotides., Older strand is methylated ,
nonhomologous end-joining: broken DAN is recognized by Ku proteins, Ku proteins recruit a nuclease, that tims some nucleotides, and a DNA polymerase that adds some nucleotides, ligase connects the backbone , inherently mutagenix , radiation or free radicals can induce double-stand breaks in DNA.
homologous recombination: 2nd chromosome is used as a template to repair the broken chromosone. Eg: An exonuclease toms the 5’ ends at the break, –> One of the 3’ ends invades a homologous segment of double stranded DNA –> DNA polymerase extends that 3’ –> Resolution of the crossed -over structure yeilds two DNA segments with no breaks.
alkyltransferases: transfer the methyl group to a cysteine (permanently inactivating the enzyme)
Describe how restriction enzymes work and predict the products of a given DNA sequence exposed to specific restriction enzymes (including any overhangs/sticky ends).
restriction enzymes : Type II restriction enzymes are isolated from bacteria and cut double stranded DNA within a defined recognition sequence that is usually palindromic.
Complementray sticky ends allow reannealing and ligation of DNA fragments. recognition sites are directional (5’ to 3’).
product prediction of given DNA sequence:
Describe the role of restriction enzymes in prokaryotes and explain how organisms protect their genome from digestion.
role of restriction enzymes in prokaryotes: bacteria methylate restrtiction enzyme sites within their own genomic DNA to prevent cleavage, but the restriction enzymes will degreade viral DNA,
organisms protecting genome from digestion:
Describe the features of cloning plasmids and explain how they may be engineered to express recombinant proteins efficiently.
cloning patterns of plasmids: Plasmids enginerred as cloning vectors , have origin of replication, have an antibiotic resistance gene to select for bacteria carrying the plamid, expression vectors also include transcriptional and translational control sequences, have a polylinker region (multiple cloning site) , a set of unique restriction enzyme sites that allow directional insertion of DNA fragments.
plasmid engineering for recombinant protein expression: cut the gene and the vector using restriction enzymes that generate the same sticky ends, use gel electrophoresis to sparate unwanted fragments, use complementary sticky ends to anneal the gene inside vector,
Cloning involves the recombination of DNA. A polypeptide od interest can be expressed by inserting a gene into a vector that may be replicated by the host organism.
Describe the steps and reagents for Illumina sequencing, PCR, and the CRISPR-Cas9 gene editing system along with the applications and limitations of each technique.
steps of reagents for Illumina sequencing:
PCR: is a technique that amplifies any DNA sequence provided some sequence information is known on either side of the target DNA to be amplified, uses Cloning , making mutants, identifying pathogens. Consisits of three steps that are repeated until the desired degree of amplification is achieved, Need template DNA, dNTPs, primers that flank the sequence to be amplified, Mg2+ buffer, and a thermostable polymerase (eg. Pfu or taq).
A thermocycler quickly changes the temp of the rxn mixture , 1) ` 95 deg cel –> DNA is dentured , 2) ~60 deg cel –> Primers anneal , 3) ~ 72 deg C –> DNA olymerase enxtends the primers, DNA is effectively doubled each cycle. Primers do not need to be perfectly complementary and can introduce restriction enzyme sites or change codons (site-directed mutagenesis) as the primer is incorporated into the amplified product.
CRISPR-Cas9 gene editing system: Bacteria transcribe the viral segments and teh Cas9 protein uses the resulting guide RNAs to locate and destroy any complementary DNA sequences that might be present in an invading virus.
Targeted genes may be inactivated by Cas9 as nonhomoglous end-joining (NHEJ) is inherently mutagenic,resulting in a nonfuctional protein product. Genes may be edited, by homology-directed repair, (HDR) after Cas9 cleavage, if replacement DNA is introduced, the cell will repair the cut DNA based on the replacement DNA sequence (that is modified to produce a desired protein sequence).
plasmids that express Cas9, and contain DNA sequences that will produce guide RNAs against specific genes, may be used in research applications.
Describe how sequencing may be applied to genome length DNA.
sequecing application to genone length DNA: Developed to sequence thousands of DNA fragments togethe.
1) DNA polymerase incorporates a complementary nucleotide into teh new DNA chain and unreacted nucleotides are washed away 2) the incorporated nucleotide is detected by its fluroescnece, then the fluroeescent group is removed 3) a new soln of nucleotides is introduced.
Genomic DNA fragmented into manageable pieces and known sequences are attached to the ends of each fragment,
Fragments are linked to solid support and amplifed in place so that a given spot on the solid support includes thousands of indentical copies of the fragment.
Each cycle, dATP, dCTP, dGTP, and dTTP are added with each deoxynucleotide connected to a different fluorescent group . The complementary deoxynucleotide may be added to the 3’ end of a primer but the fluorescent group prevents further extension.
At each spot, the fluorescent color indicates which base was added to the growing strand opposite the unknown fragment and a computer records the identity of the nucleotide.
Fluroescent group are chemically cleaved and washed away leaving a free 3’ OH group before beginning the next cycle.
Each spot corresponds to a different unknown fragment that may be sequenced by washing the same set of reagents over all spots in each cycle.
Over lapping sequences from different fragments may be assesmbled to sequence entire genomes.
Explain the role of RNAP in E. coli and RNAPI, RNAP II, and RNAP III in eukaryotes.
RNAP in E.Coli in euks: have single RNA polymerase (RNAP)
RNAPI: Produceds ribosomal RNA (rRNA), rRNA includes structural and catalytic components of the ribosome (~80% of cellular RNA)
RNAP II: produces messenger RNA (mRNA), mRNA codies for polypeptides ( ~5% of cellular RNA)
RNAP III: produces rRNA and transfer RNA (tRNA), tRNAs deliver amino acids during translation ( ~15% of cellular RNA)
RNA polymerase synthesizes new RNA stands from a DNA template.
Define the following transcription terms: upstream, downstream, template/non-template strand, coding/non-coding strand, sense/antisense strand, +/- strand, exon, and intron.
upstream: from 5’ to 3’ , direction from which polymerase/ribosime has come.
downstream: 3’ to 5’ direction of transcription
template: strand of DNA that is continuosly trasncribed
non-template stand: strand of DNA that is non-continuously trasncribed
non-coding/antisense/negative/ template: The DNA strand that has a sequence complementary (except for the replacement of U with T) to the transcribed RNA; it is the template strand. Also called the antisense strand.
coding/sense/positive/ coding: The DNA strand that has a sequence complementary (except for the replacement of U with T) to the transcribed RNA; it is the template strand. Also called the antisense strand.
exon: A portion of a gene that appears in both the primary and mature mRNA transcripts.
intron: A portion of a gene that is transcribed but excised by splicing prior to translation.
Recall how DNA sequences are numbered relative to transcription (ie. initiation site is +1, 1st upstream base is -1, etc.)
DNA sequences numbering: a gene segment of DNA that codes for a protein (Via mRNA) or another fxnal RNA, transcription produces RNA in the 5’ to 3’ direction w/ no primer required.
Describe how histones may be covalently modified and recall that this affects gene expression at the modified region.
histones covalent modification: may increase or repress transcription
affects on gene expression at modified region: acetylation of histone Lys residues typcially loosens the DNA-histone interaction and promotes transcription, other histone modifications, ie phorphorylation, methylation, and ubiquitination have varying effects on chromatin structure and transcription that depend on the specific residue being modified.
Recall the core promoters, and their approximate location, that help locate RNAPII to the transcription start site in eukaryotes.
core promoters: Euks have a variety of upstream and downstream elements that may vary from gene to gene. general transcription factors bind some combination of these elements and recruit RNAPII to the start site.
Describe the role of TFIIA, TFIIB, TFIID, TBP, TFIIE, TFIIF, and TFIIH in eukaryotic transcription initiation.
Eukaryotic transcription initiation requires 6 highly conserved general transcription factors
General transcription factors (GFTs) help RNAPII locate the transcription start site and open up the transcription bubble.
TFIIA: assits TBP
TFIIB: helps recruit TFIIF-RNAPII to the transcription start site.
TFIID: includes the TATA-bindning protein (TBP) and TBP-associated factors (TAFs) that help recurit and locate Inr and DPE
TBP: works with TFIIA
TFIIE: recruits TFIIH which acts as a helicase to open up the transcription bubble
TFIIF: increases RNAPII’s affinity for the start site, enlongate the mRNA product, and bind the nontemplate strand to stabilize the transcription bubble.
TFIIH: and mediator help phosphorylate the C-terminal domain of RNAPII
Describe the role of mediator complex, activators, repressors, silencers, and enhancers in the regulation of transcription.
mediator complex: Activators are connected to the transcription complex in eukaryotes using a protein complex
activators: proteins that increase the ability of RNAP to transcribe a gene
repressors: proteins that decrease activity of RNAP
silencers & enhacers: DNA sequences that help regulate transcription
Recall how the structural features of RNAPII are involved in elongating and proofreading transcripts.
RNAPII struct involved in elongation and proofreading transcripts: Phosphorylation of the C-terminal domain (CTD) allows RNAP II to dissociate from mediator and some of the general transcription factors so that it can make the full-length transcript.
Describe the two ways transcription may be terminated in E. coli.
transcription termination: transcription may be intrinsically terminated when RNAP transcribes a region that pulls apart the RNA-DNA hybrid helix when the RNA base pairs with itself to form a hairpin structure.
Intrinsic transcription , spontaneously essembled and tugs on DNA –> break off, force from favorable base pairing U&A makes it easier for RNA to be released.
Transciption may be terminated by a Rho protein factor that specifically binds a C-rich region on the transcript and bumps RNAP off the template.
Define operons are and explain why they are advantageous to prokaryotes.
operons: consisit of several genes under the control of one promoter that are transcribed as one mRNA. ( tryptophan = needs to express all 6 genes ).
Eukaryotes: related genes may be grouped together under the control of the same regulatory elements but are transcribed separately. ( ingest tryptophan).
advantages to prokaryotes:
Describe the purpose of 5’ caps, 3’ poly (A) tails, and splicing in eukaryotic transcriptional processing.
5’ caps fxn: eukaryotic mRNAs are capped at the 5’ end by linking a guanosine residue to the emerging mRNA via a 5’-5’ triphosphate linkage.
3’ poly (A) tails fxn: Eukaryotic mRNAs are polyadenylated at the 3’ end. poly(A) binding proteins protect the mRNA from degredation and help control the mRNA lifetime.
splicing in eukaryotic transcriptional processing: introns are removed and exons are spliced together to form the mature mRNA product.Eukaryotic mRNA is transcribed with exons (expressed regions) that code for the polypeptide sequence, separated by introns (intragenic regions which are later removed).
Describe how the cell determines the locations of exons and splices them together.
cell determination of loc of exon splicing:
Splicing occurs at conserved sequences at the exon-intron junction.
Splicing occurs through two transesterification rxns catalyzed by an RNA-protein complex = spliceosome or catalyzed by the RNA itself.
Most human structural genes undergo altnerative splicing where different processing of the transcript produces variant mature mRNAs.
Recall that RNA can act as a ribozyme since it has catalytic functional groups and can adopt specific 3D structures. Note that some ribozymes are pure RNA, while others have protein components (that are not involved in catalysis). Some ribozymes involve self-cleavage reactions and thus cannot do multiple turnovers.
rRNA: RNA can act as a ribozyme since it has catalytic functional groups and can adopt specific 3D structures. Note that some ribozymes are pure RNA, while others have protein components (that are not involved in catalysis). Some ribozymes involve self-cleavage reactions and thus cannot do multiple turnovers.
RNA generally has multiple functional groups and adopts unique 3-dimesional conformations just like a protein enzyme.
Define codons and translate a DNA or RNA sequence to an amino acid sequence given a codon table.
codons: nucleotides grouped in sets of 3 , codon sequence in a particular reading frame corresponds with polypeptide sequence.
recgonized by three-base anticodons in tRNA, mRNA is read by the ribosome in the 5–> 3 prime direction and protein synth proceeds from n-term to c-term, AUG is both the start codon and codes for methionine , UAG, UAA and UGA are stop (nonsense/termination) codons.
DNA or RNA translation:
Define initiator codon and stop/nonsense codon.
initiator codon: AUG is both the start codon and codes for methionine
stop/nonsense codon:UAG, UAA and UGA are stop (nonsense/termination) codons.
Describe how changes in gene sequence may lead to frameshift mutations or point mutations (including nonsense, missense, silent, conservative, and non-conservative point mutations)
frameshift mutations: insertions or deletions of a number of nucleotide (that isnt evenly divisible by 3.
point mutations: involve a single base substitution,
nonsense: the new codon codes for stop
missense: the new codon codes for a different amino acid,
silent: new codon codes for the same amino acid
conservative missense mutations:(the new codon codes for a similar amino acid (eg. Leu –> IIe) or nonconservative ( the new codon codes for a dissimilar amino acid (eg.Glu –> Val)
non-conservative point mutations: there are more codons for the most common amino acids, codons for the same or similar amino acids are grouped togther.
Describe how the codon table is organized in a way to minimize the effect of point mutations.
codon table organization:
Describe the general role and structural features of tRNAs.
tRNA struct and fxn: serve as adapters btwn mRNA and amino acids, teh tRNA anitcodon base pairs with the codon to read the mRNA while the CCA 3’ acceptor stem is linked to the appropriate amino acid.
Describe the wobble hypothesis and explain how it affects the number of tRNAs required to read all the codons as well as why changes in the 3rd codon position frequently code for the same amino acid.
wobble hypothesis: the 5’ anticodon postion experiences some flexibility (wobble) in its hydrogen bonding with the 3’ codon position, some organisms have less than 61 tRNAs as some tRNA anticodons can read more than one codon, changes in the 3’ codon base usually code for the same amino acid.
Define charged tRNA and describe the steps and enzyme required to charge the tRNA.
charged tRNA: aminoacyl-tRNA synthetases (aaRS) are enzymes that add the correct amino acid to a given tRNA. Each aaRS recognizes a specific amino acid along with all associated tRNAs for that amino acid.
steps and enzymes required to charge tRNA:
Recall that the ribosome is ribozyme and consists of large and small subunits composed of both RNA and protein.
ribosome =ribozyme –> struct: protein-RNA complex that catalyzes the formation of peptide bonds
Describe what happens in the A-site, P-site, and E-site of the ribosome.
A-site: amino-acyl site binds the incoming aminoacyl-tRNA
P-site: binds the tRNA with the growing polypeptide chain (charged)
E-site: exit site transiently binds teh exisiting deacylated tRNA (uncharged)
Describe how the ribosome locates the correct translation start site and all the steps and protein factors required for initiation, elongation, and termination of translation in prokaryotes.
ribosme locates correct translation start site steps: initation factor 2 (IF-2) brings the charged initatior tRNA to the start codon in the P-site ( the A site is blocked by IF-1).
if the correct codon-anticodon match is made, the ribosiome will shift its conformation to interact with the first two base-pairs and promote hydrolysis of the GTP bound to EF-Tu, EF-Tu GDP departs.
Draw the arrow-pushing mechanism for the ribosome’s peptidyl transferase reaction.
mech for ribosome’s peptidyl transferase rxn: the amino acid in the A site attacks the ester bond in the P site –> EF-G (another G protein) bumps the peptidyl-tRNA from the A site to the P site (and the deacylated tRNA from the P site to the E site). –> Ribsome recycling factor (RRF), EF-G and IF-3 remove the deacylated tRNA and separate the ribosomal subunits in preparation for another round of peptide synthesis.
Describe the steps for the synthesis of secreted and membrane proteins.
steps for synthesis of secreted and mbn proteins:
the signal recognition particle (SRP) recognizes an N-terminal amino acid sequence (pos charged residue followed by several hydrophobic residues) and stops translation.
SRP targets proteins to the ER for eventual secreation or for directional insertion to organelle mbns or the plasma mbn,
Vesicles move from the ER to the Golgi complex and then from the Golgi to the target mbn to deliver proteins,
Proteins may be modified inside the ER and the golgi appatatus (glycosylation and disulfide formation).
