# BIOL 201 Study Guide - Winter 2018, Comprehensive Midterm Notes - Adenosine Triphosphate, Protein, Amplitude Modulation

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12 Oct 2018
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BIOL 201
MIDTERM EXAM
STUDY GUIDE
Fall 2018
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Living things convert energy from one form to another, but never created or destroyed!
Ex. Plants take energy from photon, convert to chemical energy through PS
Animals do NOT consume energy! (First law of thermodynamics!)
Ex. You can store chemical energy in the form of more biomass
Caveat: organisms store energy during growth
Conversions of chemical bonds into heat and work
~20% of genome devoted to metabolism
Number 2: organisms store energy temporarily during activity
Metabolism: the science of energy conversions
We’re converting a simple molecule into a complex organism
Seems to go against 2nd law of thermodynamics that disorder and entropy are always increasing in the universe!
Not all energy is equal! It can differ in its quality
Doesn’t work both ways
Potential E of mass on pulley is very high quality energy, has high amount of order
Energy in water tank low quality, due to random motion of molecules
Joule showed that increase in water temperature (thermal E) was directly related to height of mass spinning paddles from pulley
Are we getting something from nothing?
Consume high quality energy (food, chemical bonds) and convert it into low quality energy (heat and work)
Quality can never be spontaneously recovered!
Living organisms convert high quality energy into low quality energy!
Gibbs: defined total energy as ENTHALPY (H)
G= free energy or available energy, or H - “unavailable energy”
G =H - TS (total energy minus that which is no longer available
Entropy: disorder, number of states in which system exists (Microsystems)
The quality of energy depends on how much is “free”
Cells capture the free energy released by atp hydrolysis to create order
G = GFINAL - GINITIAL
G = H - T S **for an isolated system, H = 0
Entropy increases; S > 0
The change in Free Energy determines whether a process occurs spontaneously
G < 0 for spontaneous processes!
If not standardized, must convert to conditions of reaction under study
A
B
C
A
→ C
= A
→ B
+ B
→ C
Free Energy values are “standardized” ( G°’) where T=298K, P= 1 atm, pH = 7.0, [ ]’s = 1M
Animals consume ENERGY QUALITY
So how do you ever make it over a transition state?
Catalysts lower energy of transition state
Always have some reactants and products in equilibrium state
Delta G of transition state is positive?
Amount of free energy a molecule has fluctuates
Ex. Trk dimer can spontaneously undimerize, takes a few tries to form dimer
Interaction of surrounding molecules that collide with protein to alter
free energy state
If fluctuations exceed threshold, will drive formation of transition state
(can go in either direction!)
Probability of spontaneously acquiring this energy allows
calculation of rate constant
Signaling pathways are DYNAMIC and in EQUILIBRIUM
The energy of a molecule fluctuates!
Enzymes catalyze reactions by lowering the free energy
of transition states
Releasing Free Energy is not Always “Easy
Capture energy in glucose to drive ATP formation!
Glucose (chemical E) CO2+ H2O ( G < 0)
Can then use this energy via ATP hydrolysis
ADP + PiATP ( G > 0)
Organisms burn ATP to drive unfavorable reactions
Cells capture free energy in ATP
Creates an overall negative delta G
Coupling: using energetically favorable reactions to drive energetically unfavourable
reactions
Lecture 1: Introduction to Energy
January 12, 2018
5:44 AM
Section 1 Page 1
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Creates an overall negative delta G
Coupling: using energetically favorable reactions to drive energetically unfavourable
reactions
ATP can induce strain in protein structure
Every atom in structure moved away from energy minimum caused by protein folding,
raising overall energy of protein
If proteins is interacting with other reactant molecules, this strain can distort reactants
and catalyze reaction!
Ex glycogen phosphorylase: atp binding is distant from glycogen binding site, strain
rips apart glycogen and releases glucose
ATP binding, hydrolysis, phosphate release or ADP release can all trigger this step
ATP binding/hydrolysis can distort the 3D structure of protein
Reduction ; gaining an electron
Oxidation: losing an electron
Fuels are oxidized by metabolic enzymes
Electrons are carriers of energy: hot potato of metabolism
Redox reactions provide a basis for energy transduction
Harvester molecule of energy for all of metabolism
NAD+ may be useful in a wide range of therapies
Tuberculosis: Active form binds NADH and inhibits cell wall synthesis enzyme
Wlds mice have delayed degeneration, overproduction of NAD+
NAD+ is the principle electron acceptor in metabolic redox reaction
FAD is used when the available free energy could not reduce NAD
Enzyme that capture electrons have evolved to bind and orient molecules in
redox to perfectly catalyze reaction
Large machines to catalyze reactions of small molecules, precise positioning
of molecules required for proper reaction to occur
GAPDH brings NAD+into position to be reduced
Ultimately passed to oxygen
Electrons are passed to O2to generate the proton motive force in mitochondria
GFINAL = GINITIAL, GPRODUCTS = GREACTANTS, G = 0
But... Equilibrium is a dynamic state!
“Equilibrium” is the state at which Free Energy has been minimized
What about the other definition of equilibrium?
A + B
C + D
Rate forward= kf[A][B]
Rate reverse= kr[C][D]
At equilibrium: kf[A][B] = kr[C][D]
[C][D]/[A][B]= kf/kr= equilibrium constant, Keq.
All chemical reactions have a forward and reverse rate
Keq> 1 favors the forward reaction.
Energy can be released to drive a chemical reaction
G = GFINAL - GINITIAL
G = H - T S
Reactants
Products
H < 0 and S > 0
a.
H < 0 and | H| > |T S|
b.
H > 0 and |T S| > | H|
c.
When is G < 0?
G°’ can be calculated from initial conditions and Keq
G°’ = -RT ln Keq
G = G°’ + RT ln [products]/[reactants] (initial concentrations!)
G°’ = -1362 log Keq (in kcal/mol)
G = G°’ + 1362 log [products]/[reactants] (in cal/mol)
Equilibrium
The reaction will proceed in the left to right direction, producing a net increase in the
concentration of B
a.
the reaction will proceed in the right to left direction, producing a net increase in the
concentration of A
b.
the reaction is at equilibrium
c.
If the equilibrium constant for the reaction AB is 0.5 and the initial concentration of A is 25
mM and of B is 0 mM, then:
1.
f the equilibrium constant for the reaction AB is 0.5 and the initial concentration of A is 25
Here are some practice problems..
Section 1 Page 2
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