biochemistry review session i bryan mitton [email protected]
TRANSCRIPT
BIOCHEMISTRY REVIEWBIOCHEMISTRY REVIEWSession ISession I
Bryan MittonBryan Mitton
[email protected]@ucmail.uc.edu
Biochemistry is almost over!Biochemistry is almost over!
Today’s ReviewToday’s Review
• 1) Amino Acids and Proteins
• 2) DNA and RNA
• 3) Glycolysis, Krebs Cycle, and ETC.
• Plus a 5 minute break between each section.
Amino AcidsAmino Acids
• You need to know the basic structure of each AA, but not the pKa’s.
• A few AA facts:– Hydrophobicity is a function of the positional entropy
of water. (Virtually always on test.) – The only imino amino acid = _______?
• Be able to calculate the isoelectric of any amino acid. – Try Histidine: pKa1 = 1.82
pKa2 = 6.0 pKa3 = 9.17 Guaranteed Q.
Isoelectric point Isoelectric point
• The “isoelectric point” of an AA or protein is the pH at which there is NO NET CHARGE.
A B C D
pKa1 pKa2 pKa3
6.01.82 9.17
Isoelectric point Isoelectric point
Uncharged form.
So average the pKa values around it. (9.17 + 6)/2 = 7.6
6.0 9.171.82
DefinitionsDefinitions
• Primary Structure– Linear order of Amino Acids in a chain.
• Secondary Structure– Comprised of beta pleated sheets, beta turns, alpha
helices. • Tertiary Structure
– How the secondary structures arrange themselves with respect to each other.
• Quaternary Structure– Subunit-subunit interactions.
• What are the major physical forces that hold each “structure” together?
ForcesForces• Primary Structure: Covalent• Secondary Structure: Hydrogen Bonding• Tertiary Structure: Hydrophobic Forces, Hydrogen
Bonding, Salt Bridges, Van der Waals Forces, and Disulfide Bonds. – The strongest covalent bonds are disulfide bonds.– The strongest non-covalent bonds are salt bridges.– The force that contributes the most to tertiary structure is
HYDROPHOBIC forces. • Hydrophobic residues put in core of protein and dictate stability.
• Quaternary Structure: Same as tertiary. • Q: What AA is very likely to be found at beta turns?
– A: Proline, as its imino structure allows for a tight turn.
Practice Q’s. Practice Q’s.
• In the following peptide bond sketch, which atoms are coplanar?
• In an alpha helix, how many AA residues are there per turn? How “long” is one turn (the “pitch”)?
• What are prion diseases a result of?
• Which atoms in a peptide bond are coplanar?
The C and N are both sp2 hybridized and so adopt a trigonal planar arrangement.
•In an alpha helix, how many AA residues are there per turn?
Answer: 3.6 Amino acids, for a length of 5.4 Angstroms. The carbonyl of the 1st residue hydrogen bonds with the amino group of the fourth.
• What causes prion diseases?• Prion diseases result from accumulation of
protein misfolding. • The misfolded molecule is dubbed “PrPSc”.• The misfolding of a “PrPc” molecule initiates a
cascade of further misfolding…• PrPSc induces other properly folded to misfold.
This polymerizes, causing cell damage + disease.
ProteinsProteins
• 3 Proteins you need to know about:– Hemoglobin (myoglobin too)– Collagen– Elastin
Hemoglobin (Hb)Hemoglobin (Hb)
• Myoglobin = 1 hemoglobin chain (almost).• Myoglobin and each hemoglobin chain contains
a “heme” group. – Heme sits in an apolar pocket in the middle of each
chain of hemoglobin/myoglobin.– Heme is metabolized to bilirubin, the buildup of which
causes of jaundice. • Heme = 1 iron atom plus a porphyrin ring.
– Porphyrin ring = 4 pyrrole groups + 4 methyl, 2 vinyl, 2 propionates stuck onto it.
– Porphyrin ring coordinates with 4 Fe+2 orbitals via nitrogen atoms.
Heme GroupHeme Group
Blue = NitrogenBlack = CarbonRed = Iron
Oxygen bindingOxygen binding
• The 5th coordination position of Fe+2 is with a histidine.
• HIS 93 = F8. This is the Proximal Histidine.
• Oxygen will be at the 6th spot.
• The Distal His is near where the oxygen binds. This is E7, or His 64.
• The distal is present to DECREASE Fe+2 AFFINITY FOR CARBON MONOXIDE.
• Q: What happens if Fe+2 turns into Fe+3?– A: It binds to water, becoming
“methemoglobin”.
NomenclatureNomenclature• Adult hemoglobin is normally an 22 tetramer.
This is called “HbA.”
• Also, 2% of total blood Hb is 22. This is “HbA2”.
• Fetally, here is the progression:22 22 22 plus 22
• A question about this was on my board exam:
• Which one is “fetal hemoglobin”?
A: alpha 2 gamma 2.
Fetal vs. Adult HbFetal vs. Adult Hb• Important difference between gamma and beta
chains:– BPG binds in a pocket that forms in the middle of all
four chains when Hb is in the “taut” form. • Recall: TAUT = low affinity for oxygen, so usually NO oxygen
bound. RELAXED = high affinity for oxygen, so usually oxygen is bound to Hb. More on this later.
– When bound, BPG lowers the affinity of Hb for oxygen because it stabilizes the taut form of Hb.
– Gamma chain – a Serine is replaced with a Histidine, so BPG doesn’t bind to fetal Hb very well.
– Thus, fetal Hb has HIGHER affinity for oxygen, because BPG doesn’t bind to it and Hb remains in a relaxed conformation.
BPG ImportanceBPG Importance
• Q: When happens to plasma BPG concentration at high altitude, and why?
• A: Its concentration increases in the blood, so that hemoglobin spends more time in the taut conformation and lets go of oxygen more easily.
• Remember: It lowers oxygen affinity by stabilizing the taut conformation of Hb.
CooperativityCooperativity
• Cooperativity – Once the first oxygen is bound to Hb, it is easier for the other 3 to bind.
P50 = 27 torr
CooperativityCooperativity
• Compare the curves for myoglobin and hemoglobin. • In the absence of BPG, Hb oxygen affinity curve looks
like that for myoglobin.
• Hill coefficient = 2.8 for Hb. Any Hill coefficient >1 implies functional cooperativity among the molecule’s subunits.
• What is the Hill coefficient for myoglobin?
Hb OxygenationHb Oxygenation
• When taut form gets oxygenated, Fe is pulled forward in heme group. – This pulls on His 92 (the proximal Histidine, or F8)
and this ends up breaking a hydrogen bond between Val 98 and Tyr 145.
• Breaking Val 98 – Tyr 145 bond has two effects:– 1) An H-bond between His 146 – Asp 94 is broken.– 2) An H-bond between His 146 and a Lysine on the
alpha chain is broken.
• As for the His 146 – Asp 94 bond….
Bohr EffectBohr Effect
• Breaking the His 146 – Lysine bond is how the beta chain tells the alpha chain it has picked up an oxygen.
• The His 146 – Asp 94 bond is responsible for the Bohr effect.
• So, uh, what was the Bohr Effect again?– Hemoglobin’s oxygen affinity is dependent
upon the local pH and carbon dioxide level.
• It works like this:– If this bond is intact, Hb adopts Taut form.– If this bond is broken, Hb adopts the Relaxed
form.
• Thus, when oxygenated Hb enters a low pH zone, the His 146 gets protonated. It then forms the bond with Asp 94.
• When this bond is made, Hb will switch to the Taut form, and let go of oxygen.
• So protonation of His 146 is the key.
• Q: List the 3 major variables/molecules that affect the affinity of Hb for oxygen.
• #1 factor: pH
• #2 factor: pCO2
• #3 factor: BPG
Which graph is the one withLOWER oxygen affinity?
If we RAISE the following, WHICH WAY will the curve shift?
BPG pHpCO2
HbHb
• Molecule can carry carbon dioxide, oxygen, and protons.
• 10% of all carbon dioxide in blood bound to 1st amino acid of Hb, Valine.
• Other 90% is carried as bicarbonate. The proton formed is part of Bohr effect.
• What enzyme catalyzes this reaction?
Sickle Cell AnemiaSickle Cell Anemia• The main problem in sickle cell anemia is a point
mutation….– Glutamate 6 is substituted with what residue?– [This was also on my board exam!]
• When Hb is deoxygenated, this residue is exposed and causes polymerization of Hb. – The chains form and deforms the cell, giving it the
sickle shape. – Having this mutation offers resistance to Plasmodium
falciparum, the bug that causes malaria.
• Called HbS. Two bad chains = SS, one bad one good = AS (heterozygote).
Remember – Glutamic Acid 6 is mutated to Valine, and causes polymerization of Hb.
ThalassemiaThalassemia
Thalassemia - Not enough chains produced by cell, so chains accumulate.
Thalassemia – Not enough chains produced, and chains accumulate.
• This disease is usually caused by a problem with splicing… the mRNA isn’t spliced correctly, so it gets destroyed.
Bone marrow expands in skull to make more RBC, so you get this “crew-cut appearance”.
Again… thalassemia means you don’t make chains.
• That’s all for Hb!
• On to collagen.
CollagenCollagen
• The Amino Acid Sequence = Gly-X-Y. • Gly = 33%, Pro = 20%, 10% = Ala. 5% = lysine. • Hydroxyproline and hydroxylysine are also part
of primary structure.– Prolylhydroxylase and lysylhydroxylase both require
Ascorbic Acid, Vitamin C to work. – These enzymes modify the individual polypeptides
before they wrap up into the triple helix.
• Without Vitamin C, what disease do you get?
Scurvy, characterized by spontaneous bleeding from joints and hair follicles.
Collagen – The VocabCollagen – The Vocab• The confusing nomenclature of collagen:
– One collagen polypeptide has a helical structure. This is the “minor helix”.
– Three polypeptides (three minor helices) wrap up to form the “triple helix”.
– A triple helix gets cleaved after it is exported from the cell at both the the N and C termini. After trimming, it is called “tropocollagen”.
– Tropocollagens line up to form fibrils. – Fibrils line up to form the overall structure.
• Enzyme lysyl oxidase forms lysine cross-links in fibrils. Also requires vitamin C.
• Disulfide bonds form at both N and C termini:– At C termini, the disulfides form to help line up
the 3 minor helices.– At N termini, they form to stop intracellular
fibrinogenesis.
ElastinElastin
• Weird amino acids in it: desmosine, etc.
• Has “coiled-coil regions.”
• No hydroxyproline or lysine.
• 1/3 = ala + val.
• Elastin is made of tropoelastin monomers.
• Take a 5 minute break!
II – DNA and RNAII – DNA and RNA
DNA and RNADNA and RNA
• Some facts:– Purines = Adenine + Guanine– Pyrimidines = Thymine + Cytidine + Uracil. – Uracil found only in RNA.
• In double stranded DNA, G-C base pairing is stronger than T-A base pairing. Why?
• Chagraff’s rule problems: (T/F)
• In dsDNA:• If T>35%, then G>15%.• Answer: F. • If T>35%, then A+T>70%, and G<15%.• If A=15% of the bases in one strand, G must =35% of
the bases in the whole ds molecule. • Answer: F. Who knows how many A’s there are in the
other strand. • If G+C = 40%, then T+G = 50%• Answer: T
– In ssRNA, if T=24%, then A=24% – Answer: F. It is single stranded, so there is no
relationship. – Lieberman = Good Teacher
• Answer: F
– He’s a GREAT teacher!!!
DNA and RNA synthesisDNA and RNA synthesis
• RNA and DNA are made in the 5’ to 3’ direction. – Be careful: The template strand is read 3’ to 5’.
Strands always run antiparallel.
• Drugs that stop HIV replication have some modification of the 3’ OH group, such that phosphodiester bonds cannot be made after viral incorporation of these nucleotides. So, these drugs terminate viral replication.
• DNA Replication is semi-conservative. – The “parent” strands are separated from each other,
and after replication each parent strand is base-paired with newly-synthesized DNA.
ReplicationReplication
• Replication begins at ori sites and proceeds bidirectionally.
• Helicase first unravels the DNA. Topoisomerase relieves tension developed during the unraveling at the other end.
• Leading and lagging strands form at each replication fork…
• The following will demonstrate the names of the enzymes involved in prokaryotic DNA synthesis and the order in which they act.
Bacterial ReplicationBacterial Replication
• Ligase connects the 5’ end of primer/DNA to the 3’ end of completed DNA.
• Telomerase takes care of ends. • Proofreading:
– DNA pol I, II, and III all have 3’ to 5’ exonuclease activity.
– DNA pol I is the only one with 5’ to 3’ exonuclease activity.
– The parental DNA strand is identified by methylation, so that the wrong strand doesn’t get changed.
• There are 3 DNA error correction systems:– Mismatch repair– Base Excision Repair– Nucleotide Excision Repair.
Mismatch Repair SystemMismatch Repair System
• In E. Coli, 3 Mut proteins recognize the mismatch.– 1 of them makes a “nick” in the DNA 5’ to the
mistake on the unmethylated strand. – This is called “endonuclease activity”. – Exonucleases come in and remove a large
piece of the DNA, and then Pol III fills in the space with the correct base pairs.
Base Excision RepairBase Excision Repair
• 2 Spontaneous Events. (Cell G0 phase)– 1) Sometimes A or G just “falls off”; the bond between
the nucleotide and the riboses is spontaneously borken.
– 2) Sometimes the amine group falls off of cytosine, leaving a uracil behind.
• The uracil gets cut out when the cell detects this.
• Either way, a “blank” spot is left behind; it is called the abasic site.
• AP Endonuclease comes in and nicks the DNA 5’ to the abasic site. DNA pol I then comes in, excises the bases, and fills it in with good bases.
Nucleotide Excision RepairNucleotide Excision Repair
• The system that fixes “thymine dimers”, which usually come of UV light exposure. – Bacteria use “UvrABC endonuclease” to detect and
nick the DNA near the thymine dimer. – Pol I then excises the DNA section and
simultaneously fills in gap.
• Xeroderma Pigmentosa – humans with an inability to sufficiently repair DNA damage, especially thymine dimers, among other things. Prone to skin cancer, etc.
• There are at least 7 proteins associated with our DNA NER system… Xp-1, 2, 3…
• On to transcription and translation…
TranscriptionTranscription
• Begins at the promoter, often called the “TATA box”, which is usually 10 bp away from first exon. – First base transcribed is the “+1” base, the one
right beside it is “-1”.
• The sequence that the RNA polymerase reads is the Template strand. – The template strand is the same thing as the:
• NON-CODING strand. • ANTI-SENSE strand.
1) The mRNA will have the same sequence as the DNA coding strand.
2) The mRNA will have the complementary sequence to the template strand.
TranscriptionTranscription
• Q: There are 3 types of RNA polymerases: Polymerase I, II, and III. What type of RNA do they each transcribe?
• A: I = rRNA II = mRNA III = tRNA– 90% of all RNA is “t” or “r”.
TranscriptionTranscription
• In addition to the TATA box, two other DNA sequences affect how often pol jumps on the promoter.
• 1) Upstream regulatory elements… <200 BP away. (URE)
• 2) Enhancers/silencers… anywhere in the entire genome.
TranscriptionTranscription
• Pol proceeds along DNA, making RNA 5’ to 3’. • In bacteria, RNA pol II is a holoenzyme.
– It loses the subunit when it binds DNA. – Without , pol cannot find promoter. – The strand elongates, processively, like DNA pol III.
• Termination of transcription: either rho-dependent or independent. – If rho-independent, the mRNA being made forms a
hairpin… UUUUUUU-AAAAAAAA. – If rho-dependent, the protein “rho” comes in and binds
to the RNA to physically remove it from the DNA.
TranscriptionTranscription
• Bacteria have operons – several genes in a row get transcribed to make a polycistronic message.
• Eukaryotes process out mRNAs individually. • mRNA processing involves:
– 5’-5’ 7-methyl-Guanine cap on 5’ end. – Intron splicing.– Poly A tail– Something is modified at the “Beginning, middle and
end.”
5’-5’ 7-methyl-Guanine cap5’-5’ 7-methyl-Guanine cap
• This cap is necessary for the mRNA to get out of the eukaryotic nucleus.
• Only mRNA gets capped.
• Has a weird 5’ 5’ linkage… 3 phosphates between the nucleosides.
Poly A tailPoly A tail
• The Poly A tail is made by Poly-A tail polymerase.– The tail is not coded for in the genome.– It is attached to the mRNA when transcription
is over. – Poly A tail is attached 10 to 30 bp after a
“AAUAAAAA” sequence.
Splicing – the removal of intronsSplicing – the removal of introns
• The 5’ end of the intron to be removed is called the splice donor. 3’ end is the splice acceptor.
• 1) 5’ end of the intron is cleaved. • 2) This is stuck onto an A residue about 20 bp in
front of acceptor site. This makes a strange 5’-2’ bond. This is called the lariat.
• 3) 3’ end of intron cleaved, and exons are joined.
• All of this is done by snRNPs (small nuclear ribonucleoproteins).
The 2’5’ bond forms here.
Lariat
Processing. Processing.
• mRNAs are capped and tailed BEFORE they are spliced.
• Introns can exist anywhere in the immature mRNA… before or after start or stop codon.
• Note that in prokaryote, transcription and translation are simultaneous; in eukaryote, processing occurs first in nucleus, then moves to cytosol.
• Find the intron:
TranscriptionTranscription
• The genetic code is degenerate – more than one codon codes for the same AA.
• 64 possible codons (4^3), but only 20 AA’s.
• Think: Many codons for one AA, but only one AA for a codon.
Translation InitiationTranslation Initiation
• Translation begins when a translation initiation factor (TIF) binds the first tRNA and a GTP. – This whole complex binds the mRNA first.– Next, the smaller ribosome subunit joins. – Last, the larger ribosome subunit binds and the GTP
is hydrolyzed.
• The first codon sits in the ribosomal P site, and the second sits in the A site.
• Overall: Transcription initiation burns 1 GTP.
Translation InitiationTranslation Initiation
• First AA is usually Methionine, as its sequence is AUG, the “start” codon. This has a formyl group in bacteria, but not in eukaryotes.
• Hence, it’s called N-formyl Methionine.
mRNA ElongationmRNA Elongation
• Attaching an amino acid to a tRNA costs 1 GTP.• Moving the ribosome down one codon also costs
1 GTP (elongation factors use it). • tRNA enters the ribosomal A site with its amino
acid attached. • At the P site, the previous AA gets covalently
bonded to the first by the enzyme peptidyl transferase.
• What is the polarity of the growth of the peptide chain?
• The polypeptide chain grows from N terminal to the C terminal.
• Lame rhyming mnemonic for directionality:
• “5 to 3, N to C”
WobbleWobble
• Wobble = the tRNA anticodon (the 3 nucleotides of the tRNA that bind to the mRNA codon) can bind a few different sequences.
• The wobble base is the 3’ end of codon, and the 5’ end of the tRNA.
• Which one is the wobble base position?
• tRNA can use Inosine as a base too.
• Don’t memorize what wobbles with what… just understand that last slide.
Stop!Stop!
• A stop codon exists, but there is no tRNA or amino acid for it.
• Ribosome simply disassembles. • Termination costs 1 GTP, used by “termination
factors”.
• To make a polypeptide “n” amino acids long, it costs:
2n+2 GTP.
Lac OperonLac Operon
Under no lactose conditions, the I gene will be transcribed and translated, and the “I” protein binds the Operator site. With this protein bound, polymerase cannot move beyond the operator. No X, Y, or Z will be expressed.
Lac OperonLac Operon
• When lactose is present, it binds to the I protein. • The I protein now cannot bind the operator, and
polymerase can transcribe the whole operon. • The rate of transcription can be determined by cAMP
levels. When cAMP binds to CRP (cAMP-Receptor Protein), or CAP (same thing), this complex strongly increases the affinity of pol II for the promotor.
Lac OperonLac Operon
• cAMP levels increase as glucose decreases. Transcription rate of lacZ decreases in this order: Lactose alone, Lac + Glucose present, Glucose alone present.
Lac OperonLac Operon
• Predict the results of these operon manipulations in the presence of lactose:– Overactive adenylyl cyclase.– High cAMP, so lots of lacZ made. – Mutated P site. – Polymerase cannot bind, so no lacZ. – Mutant lacZ.– No lacZ made.
• Take a 5 minute break!
III – Glycolysis, Krebs cycle, and III – Glycolysis, Krebs cycle, and the ETCthe ETC
GLYCOLYSISGLYCOLYSIS
• For all of metabolism:– Focus on regulation. – Focus on rate-limiting steps. – I’ll give you the facts most likely to be
tested.
11stst Step… Hexokinase Step… Hexokinase
• Hexokinase: Glucose Glucose-6P– Works by induced-fit.– Burns one ATP at a time.– Irreversible enzyme. – Negatively regulated by G6P.– Phosphorylation traps Glucose in cell, so it cannot
diffuse back out. – Helped by GLUCOKINASE, which catalyzes the same
reaction.• Glucokinase has lower affinity for glucose, so:
– 1) Has a high Km– 2) Only works when Hexokinase is overburdened– 3) “Pushes” glycolysis forward.
33rdrd Step. PFK-1 Step. PFK-1
• PFK-1: Fructose-6P Fructose-1,6BP
• Major regulated step of glycolysis.– Regulation of PFK-1 was on my board exam.
• Irreversible. Burns 1 ATP.• It is activated by • AMP and F26BP• And inhibited by • ATP and citrate. • THINK: ATP, citrate, vs. AMP and F26BP
Side reaction – F26BPSide reaction – F26BP
• PFK-2: F6P Fructose-2,6BP. – Fructose-2,6-bisphosphatase reverses this
reaction. – When PFK-2 is phosphorylated, PFK-2 is
OFF, and F26BPase is ON. – Opposite for dephosphorylation.– Muscle isozyme not phosphorylatable, so this
regulation only happens in the liver.
Side reaction – F26BPSide reaction – F26BP
• If PFK-2 is phosphorylated, it is off, and
F-2,6BPase will be on.
• Levels of F-2,6BP will drop.– F-2,6BP normally activates PFK-1.
• So, in the liver, phosphorylation slows down glycolysis.
99thth Reaction-Pyruvate Kinase Reaction-Pyruvate Kinase
• Pyruvate Kinase – Liver form. – PEP and F16BP activate– ATP and Alanine inhibit. – Also phosphorylated to be…– OFF
• Since phosphorylation slows down PFK-1, phosphorylation of PK ought to also slow glycolysis.
• Muscle form… not phosphorylatable. Just like PFK-2!!
Phosphorylatable Enzymes…Phosphorylatable Enzymes…
• Enzymes all get phosphoryated after glucagon or epinephrine activates adenylyl cyclase, which then turns on PKA.
• PKA phosphorylates the relevant enzymes.
• When do things get phosphorylated?– When you are hungry.– When you are exercising. – When you are afraid.
• Q: Name the enzymes that catalyze the 3 irreversible steps of glycolysis.
• Q: Name the 3 major regulation points of glycolysis.
• Hint: They are the same.
– Hexokinase– PFK-1– Pyruvate Kinase
• Phosphorylation is important… it slows down glycolysis in the liver.
• What is the net yield of ATP from 1 molecule of glucose?
• Glucose + 2NAD+ 2 pyruvate + 2NADH + 2ATP
• Carbon labeling…• Carbons 1 and 6 of glucose will become the
“top” carbons of pyruvate.• 2 and 5 are the middle carbons.• 3 and 4 will become the ones on bottom.
Fructose and GalactoseFructose and Galactose
• Fructokinase: F F1P– Next, F1P is cut into Dihydroxyacetone phosphate
and glyceraldehyde. – Triose kinase phosphorylates glyceraldehyde to G3P.
These slide right into glycolysis.
• Fructose bypasses major regulated step of glycolysis!!!!! – Phosphate wasting due to substrate level
phosphorylations.– Lactic acidosis also occurs because glycolysis is
running extremely fast, which consumes NAD+. • Lactate is made from pyruvate to regenerate NAD+.
Fructose and GalactoseFructose and Galactose
• Galactose is made into Gal-1P by Galactokinase.
• Then an enzymes flips an OH group around to make it into Glucose 1P via UDP mechanism.
• Therefore, galactose does not bypass the major regulated step of glycolysis.
The Krebs CycleThe Krebs Cycle
• Once Again, focus on regulation.
• Still responsible for names, products made by each reaction.
Inside the Mitochondrial MatrixInside the Mitochondrial Matrix
• First, pyruvate pumped into mitochondria via H+ symport or citrate antiport.
• That’s why/how citrate feeds back to stop glycolysis!
• Pyruvate goes to AcCoA irreversibly by Pyruvate DH, creating NADH.
• Cofactors of Pyruvate DH:– NAD– FAD– TPP (Vitamin B1, Thiamine Tri-Phosphate)– 2 Lipoic Acids
Pyruvate DHPyruvate DH
• Regulated so tightly because we cannot ever use the AcCoA carbons to directly make glucose again.
• Enzyme ON: CoA, NAD+, AMP• Enzyme OFF: AcCoA, NADH, ATP• How do all these factors work?
– There is a kinase and a phosphatase on this huge complex… when the above regulators are high/low, they simply influence the kinase and the phosphatase activity.
• Ultimately: Phosphorylated = OFF• Dephosphorylated = ON
Pyruvate DHPyruvate DH
• Has to have:– NAD– FAD– TPP (Vitamin B1, Thiamine Tri-Phosphate)– 2 Lipoic Acids
• Is turned on by:– CoA, NAD+, AMP– De-Phosphorylation
• Is turned off by:– AcCoA, NADH, ATP– Phosphorylation
Krebs cycleKrebs cycle
• No particular “regulation”…
• There is usually little oxaloacetate present in the mitochondria, so this limits how fast the Krebs cycle proceeds forward.
• Isocitrate DH, alpha-ketoglutarate DH, and malate DH all make NADH.
• Succinate DH makes FADH2.
Electron Transport ChainElectron Transport Chain
• High energy electrons move from complex to complex, driving Hydrogen into the intermembrane space.
• The hydrogen gradient drives ATP production. Called: Proton-Motive Force
III
IVIII
NADNADH FADH
FADSuccinate
Fumarate
CoQe-Cyt C
H+ H+H+
ADP
ATP
H+
H2O
O2
Intermembrane Space
Matrix
Selected Inhibitors of ETCSelected Inhibitors of ETC
• DNP is an uncoupling agent.– It bonds to hydrogens in the intermembrane space,
diffuses to the matrix, and lets them go. – It ruins the H gradient, so electron transfer occurs in
the absence of ATP production.
• Oligomycin blocks ATP synthase. • Cyanide blocks complex IV. • Rotenone blocks Complex I. • Antimycin A blocks Complex III.
• Remember: 2.5 ATP from NADH
• 1.5 ATP from FADH2
• This is because they enter the ETC at slightly different locations.
Summing it upSumming it up
• Hexokinase = G6P• PFK – 1 = AMP and F26BP (ON)• ATP and citrate. (OFF)• PFK-2 = Phosphorylation (OFF)• Pyruvate Kinase = PEP and F16BP (ON)• ATP and Alanine (OFF)• Pyruvate DH = CoA, NAD+, AMP (ON) • AcCoA, NADH, ATP (OFF)• Via Phosphorylation
GOOD LUCK!GOOD LUCK!(See you in Pharmacology next year!)(See you in Pharmacology next year!)
…Stay tuned for Chris Brubaker at 1 PM in this room.