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BIS103(001)
Abel
Biological Sciences BIS103-001 Winter Quarter 2007(CRN 35498)
Lectures: Tuesday/Thursday 3:10 – 4:30 p.m. (Social Sciences 1100)
Lectures will be recorded as MP3 files and uploaded to the class website (podcast). Media Equipment Service (2-3553).
Instructor: Dr. Steffen Abel, Associate Professor Phone: 752-5549 Email: [email protected] Office Hours: Fridays (9-11 am),
Location: 210 Asmundson Hall (Office); for larger groups of students, 242 Asmundson Hall (Conference Room).
Other appointments are available upon request. Teaching Rebecca Shipman Email: [email protected] Assistants: Office Hours: Tuesdays (2-3 pm) and Wednesdays (9:30-10:30 am)
Location: Asmundson Hall (Conference Room Annex, 2. Floor). Stephen Abreu Email: [email protected] Office Hours: Tuesdays (12-1 pm) and Thursdays (12-1 pm) Location: 139 Hunt Hall
Text: One of these three textbooks is REQUIRED:
• Biochemistry by Garrett & Grisham • Principles of Biochemistry by Lehninger, Nelson & Cox • Fundamentals of Biochemistry by Voet, Voet & Pratt
Booklet: A collection of visual aids specifically prepared for this class is avail-able for purchase at Campus Books, which you need to bring to every lecture. However, this booklet does not substitute for one of the required textbooks!
Class Websites: MyUCDavis (BIS103-001) Go to The “Real” BIS103 Website (http://www.plantsciences.ucdavis.edu/bis103) Exams: Your final grade will be calculated based on your performance in three
examinations: Midterm 1 (33.3%), Midterm 2 (33.3%), and the final test (33.3 %). Parts of the final exam will be cumulative. Regrade requests for midterms must be submitted in writing within one week after return of the tests and will only be considered if you have used permanent ink during the exam. Exams will be given at the assigned times only! No early finals will be given! Exceptions may be granted to students with documented verifi-cation of personal loss or sickness and if the instructor was contacted before the exam.
MIDTERM 1 Tentative: Thursday, January 25 (in class) MIDTERM 2 Tentative: Thursday, February 22 (in class) FINAL EXAM Tuesday, March 20, 8:00 - 10:00 a.m.
Class Context Class Context
General and General and Organic ChemistryOrganic Chemistry
BIS 102BIS 102 ““StructuralStructural”” BiochemistryBiochemistry (carbohydrates,proteins, lipids, nucleic acids, enzymology)
BIS 103BIS 103 ““FunctionalFunctional”” BiochemistryBiochemistry (intermediaryor primary metabolism)
Thermodynamics, redox reactions, nucleophiles, electrophiles, major organic compound classes,chemistry of carbonyls
Nutrition, Microbiology, Human Physiology, Neurobiology, Exercise Biology, Plant Biology, Pharmacology, Cell Biology,
etc.
• How is chemical energy stored?
• Why is there a need for metabolism?
• Directionality of metabolism
Gibb’s Free Energy (ΔG)
Reduction Potential (ΔE)
Lecture 1 TopicsLecture 1 Topics
One Reaction
Fuel or Food + O2 CO2 + H2O + ΔU
John Candy’s Metabolism
Many Reactions
Regeneration ReproductionCarbon Chemistry
Food
Action
Fuel Energy
Energy
Fossil FuelsFossil Fuels
Exhaust (COExhaust (CO22, H, H22O)O)
ΔU = q (Heat) + work (PΔV)
ΔU
HydrogenHydrogen
Chemical(Potential)
Energy
Chemical(Potential)
Energy Food Food
COCO22, H, H22O O
ΔΔU = q (Heat) U = q (Heat) + work (P+ work (PΔΔV)V) = = ΔΔH (Enthalpy)H (Enthalpy)
ΔΔH H ““trappedtrapped”” as ATPas ATP
In intact cells, P and V are assumed to be constant In intact cells, P and V are assumed to be constant
IntermediatesIntermediates
What is the purpose of metabolism?What is the purpose of metabolism?
• Provides building blocksbuilding blocks for regeneration/growth
• Energy conversion compatiblecompatible with C-based life
• Provides energyenergy (motion, transport, syntheses, heat)
All life forms areAll life forms arelikely based onlikely based oncarbon chemistrycarbon chemistry
Energy required to break a bond ((++ΔΔHH)
Energy released during bond formation ((––ΔΔHH)
X Y X Y+ΔH
–ΔH
How is energy stored? In chemical bonds.How is energy stored? In chemical bonds.
Bond Energy ~ Bond StrengthBond Energy ~ Bond Strength
The strength of a chemical bond depends on:
• relative electronegativitieselectronegativities (affinity for electrons)• distance of electrons from nuclei• number of electrons shared• nuclear charge
Table 1: Table 1: ElectronegativitiesElectronegativities of Biologically Important Elementsof Biologically Important Elements
Hydrogen (H) 2.22.2 (P) Phosphorus
Carbon (C) 2.62.6 (S) Sulfur
Nitrogen (N) 3.0
Oxygen (O) 3.4
p. 2
HH HH
HH OO
HH HH
HH OO
+ 436 kJ/mol
+ 463463 kJ/mol
– 436 kJ/mol
– 463463 kJ/mol
Bond
kJ/mol Bond kJ/mol Bond kJ/mol Bonds kJ/mol
C-H 413 C-S 259 C-N 293 C=N 615 C≡N 891 C-C 348 C=C 614 C≡C 839 C-O 358 C=O 799 C≡O 1,072 O=C=O 1,598 H-H 436 H-O 463 H-O-H 926 N-H 391 N-N 163 N=N 418 N≡N 941 N-O 201 N=O 607 S-H 339 S-S 266 S=S 418 S-O S=O 523 O-O 146 O=O 495 P-O 599
p. 2
Table 2: Bonds of Life (Enthalpies of Chemical Bonds)Table 2: Bonds of Life (Enthalpies of Chemical Bonds)
ΔΔHH of a reaction is the sum of bond energies sum of bond energies
consumedconsumed during breaking of all reactant bonds
andand of bond energies
releasedreleased during formation of all product bonds.
If ΔΔH > 0H > 0: EndoEndothermic thermic reactionIf ΔΔH < 0H < 0: ExoExothermicthermic reaction
A + 2B A + 2B 3C + D + 2E 3C + D + 2E
Example: Ethanol combustion/oxidationExample: Ethanol combustion/oxidation
C2H5OH + 3O2 2CO2 + 3H2O
O=C=O
O=C=O
O-H-OO-H-OO-H-O
O=OO=O
O=OH-C-C-O-H
H H
H H
ΔH = –1,255 kJ/mol (exothermic)
Cracking all bonds (+Cracking all bonds (+ΔΔH)H)(kJ/mol)(kJ/mol)
3,2343,234 1,4851,485
4,7194,719
Forming all Forming all newnew bonds (bonds (––ΔΔH)H)(kJ/mol)(kJ/mol)
–– 3,1963,196 –– 2,7782,778
–– 5,9745,974
Other ExamplesOther Examples
ΔH = – 2,639 kJ/mol
C6H12O6 + 6O2 6CO2 + 6H2O
GlucoseGlucose
2H2O
ΔH = – 485 kJ/mol
2H2 + O2
H HH H O O HHHH OO
HHHH OO
““Fuel ValuesFuel Values”” of some foods and fuels (kJ/g)of some foods and fuels (kJ/g)
CarbohydratesCarbohydrates 1717
ProteinsProteins 1717
LipidsLipids 3838
WoodWood 1818
CoalCoal 3232
Crude OilCrude Oil 4545
Hydrogen Hydrogen 142142
Good molecules for storing energy:Good molecules for storing energy:
carbon polymers, hydrocarbons, hydrogencarbon polymers, hydrocarbons, hydrogen
-C—C—C—C-
HHHH HH HH
HH HHHHHH
“Harvest” hydrogen and transportas a “biologically safe” form
NADNAD——HH ++ HH++
O=OO=O
HH22OO CCOO22
Most Most ––ΔΔHH(ATP)(ATP)
CarbohydratesCarbohydratesProteinsProteinsLipidsLipids
Metabolic OxidationMetabolic Oxidation
““High EnergyHigh Energy””SubstratesSubstrates
“Low Energy”Products
CatabolismCatabolism
CellularCellularMacromoleculesMacromolecules
AnabolismAnabolism
IntermediatesIntermediatesPrecursor MoleculesPrecursor Molecules
ΔΔHH(ATP)(ATP)
ΔΔHH(ATP)(ATP)
MotionTransport
Heat
FOODFOOD
Waste
• How is chemical energy stored?
• Why is there a need for metabolism?
• Directionality of metabolismDirectionality of metabolism
Gibb’s Free Energy (ΔG)
Reduction Potential (ΔE)
Lecture 1 TopicsLecture 1 Topics
Gibbs Free Energy (Gibbs Free Energy (ΔΔG)G)• ΔΔGG = = ΔΔH H ––TTΔΔSS
If ΔG = 0: System is at equilibrium
If ΔG < 0: Exergonic (“downhill” process)
If ΔG > 0: Endergonic (“uphill” process)
• Process can be driven by ––ΔΔHH, ++ΔΔSS, or both
7 Molecules7 Molecules 12 Molecules12 Molecules
ΔΔH: H: ––2,639 kJ mol2,639 kJ mol--11
ΔΔS: positiveS: positive(more disorder, fragmentation)
Example:Example:
C6H12O6 + 6O2 6CO2 + 6H2O
ΔΔG: G: –– 2,840 kJ mol2,840 kJ mol--11
See p. 5
Glucose + 6O2
6CO2 + 6H2O
OneStep
Glucose + 6O2
6CO2 + 6H2O
John Candy’s Metabolism
•• Alternative way to calculate Alternative way to calculate ΔΔGG
ΔΔG:G: Measure for the displacement of a reaction from its equilibrium (EQ)
See p. 5
Reactants [R] Reactants [R] Products [P]Products [P]Reversible ReactionsReversible Reactions
If ΔG = 0: Reaction is (already) at EQ
If ΔG < 0: Not (yet) at EQ: Forward reaction is favored
If ΔG > 0: Not (yet) at EQ: Reverse reaction is favored
ΔΔGGoo’’ determined at ““Standard ConditionsStandard Conditions””
• 25 oC (298 K)• 1 M (or 1 atm ) of R and P; or equimolar concentrations
• pH 7 (10-7 M H+)• 55.5 M H2O• 1 mM Mg2+ (if part of the reaction)
ΔΔG = G = RTlnRTlnQQ –– RTlnRTlnKKeqeq
–– RTlnRTlnKKeqeq = = ΔΔGGoo’’
See p. 5
ΔΔG = G = ΔΔGGoo’’ + + RTlnRTlnQQ
KKeqeq = [P]eqeq/[R]eq
QQ = [P]initialinitial/[R]initialinitial
oo
’’
Question: Under standard conditionsstandard conditions, how far areequimolarequimolar concentrations from equilibrium (ΔΔGGoo’’) ?
At thermodynamic equilibriumAt thermodynamic equilibrium
See p. 5
RRR PPP??
If Keq > 1 ΔGo’ < 0
If Keq < 1 ΔGo’ > 0
R
RRRR
If Keq = 1 ΔGo’ = 0 RRR PPP
PPPPP
PP
1,0001001010.10.010.001
KKeqeq
- 17.1- 11.4- 5.7
05.711.417.1
ΔΔGGoo’’ (kJmol(kJmol--11))
C2H5OH + 3O2 2CO2 + 3H2O ΔΔGGoo’’ = = ~~1,300 kJ mol1,300 kJ mol--11 KKeqeq = 10= 10220220
C6H12O6 + 6O2 6CO2 + 6H2O ΔΔGGoo’’ = = ––2,840 kJ mol2,840 kJ mol--1 1 KKeqeq = 10= 10500500
ΔΔGGoo’’ = = –– RT RT lnlnKKeqeq
ΔΔGGoo’’: useful as a general guidegeneral guide to predict directiondirectionof a process (reaction)
However, real conditions in cells often differfrom standard conditions!
•• Temperature: 37 Temperature: 37 ooCC (310 K)(310 K)
•• Cellular or initial concentrations of R and PCellular or initial concentrations of R and P(they are unlikely to be at equilibrium)
ΔΔGG:: ΔΔG = G = ΔΔGGoo’’ ++ RRTTlnlnQQ
See p. 5
QQ = [P]initialinitial/[R]initialinitial
ΔΔGGoo’’ = –RTlnKeqKeq
If QQ = KKeqeq Reaction is already at EQ ΔΔG = 0G = 0(initial P/R ratio equalsequals P/R ratio at equilibrium)
ΔΔG = RT G = RT lnlnQQ –– RT RT lnlnKKeqeq = RT = RT lnln QQ//KKeqeq
If QQ < KKeqeq Reaction is NOT yet at EQ ΔΔG < 0G < 0(relatively more Rmore R or less Pless P,
PP formation favored)
If QQ > KKeqeq Reaction is NOT yet at EQ ΔΔG > 0G > 0(relatively less Rless R or more Pmore P,R R formation favored)
See p. 5
Reduction Potential (Reduction Potential (ΔΔE)E)…… yet another way to express yet another way to express ΔΔGG……
• Metabolic reactions are often redoxredox reactions, involving transfer of electronstransfer of electrons from a donor to an acceptor
1. Direct transfer (ee--)2. Transfer of hydrogen (HH, or ee-- + HH++)3. Transfer of the hydride anion (HH——H, or HH-- ++ HH++)4. Direct combination with oxygen (X X + O=O 2O X)
• Electronegativities can predict direction of e- transfer
• Redox reactions can be written as two “half-reactions”(by conventionconvention: each is written in the direction of the reductiondirection of the reduction!)
AAoxidizedoxidized ++ BBreducedreduced AAreducedreduced ++ BBoxidizedoxidized
1. 1. AAoxidizedoxidized + e+ e-- AAreducedreduced
2. 2. BBoxidizedoxidized + e+ e-- BBreducedreduced
See p. 6
Question: Which “half-reaction” has the higher affinity for electrons at standard conditionsstandard conditions ?
By convention:
If e- flow from reference to test (“test” is stronger e- acceptor): EEoo > 0 (+V)> 0 (+V)
If e- flow from test to reference (“test” is weaker e- acceptor): EEoo < 0 (< 0 (-- V)V)
See p. 6
Reference H+ + e- H2 “half-reaction” (1M each)Electrode: (Eo= 0.00 V)
Test A+ + e- A “half-reaction” (1M each)Electrode: (Eo’= ??? V)
Table 3: Standard Reduction PotentialsTable 3: Standard Reduction Potentials
See p. 3
Excited (Chlorophyll a)Excited (Chlorophyll a)22** ~ ~ -- 1.001.00
Acetate + 2H+ + 2e- Acetaldehyde + H2O - 0.58
NAD+ + 2H+ + 2e- NADH + H+ - 0.32
Pyruvate + 2H+ + 2e- Lactate - 0.18
2H2H++ + 2e+ 2e-- HH22 (at standard conditions, 1M each, pH 0) 0.00(at standard conditions, 1M each, pH 0) 0.00
NO3- + 2H+ +2e- NO2- + H2O + 0.42
O2 + 4H+ + 4e- 2H2O + 0.82
(Chlorophyll a)(Chlorophyll a)22.+.+ + e+ e-- (Chlorophyll a)(Chlorophyll a)22 + 1.10+ 1.10
HalfHalf--reaction (written as reduction by convention)reaction (written as reduction by convention) EEoo (V)(V)
ee-- flowflow
Reduction Potential E for each Reduction Potential E for each ““HalfHalf--ReactionReaction”” (Nernst Equation)(Nernst Equation)
ΔE = EOxidant (A) – EReductant (B)
ΔΔE of E of RedoxRedox ReactionReaction ((AAoxox + B+ Bredred AAredred + B+ Boxox))
RTnFE = Eo’ + ln [e- Acceptor, Aox]
[e- Donor, Ared]
RTnFE = Eo’ + ln [e- Acceptor, Box]
[e- Donor, Bred]
1. Aoxidized + e- Areduced
2. Boxidized + e- Breduced
Relationship between ΔE and ΔG
ΔΔGG = – n F ΔΔE E ΔΔGGoo’’ = – n F ΔΔEEoo’’
ΔΔE E = = –– ΔΔGG//nFnF ΔΔEEoo’’ = = –– ΔΔGGoo’’/nF/nFSee p. 6