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University of Houston
BCHS 3304: General Biochemistry I, Section 07553
Spring 2003 1:00-2:30 PM Mon./Wed. AH 101
Instructor:
Glen B. Legge, Ph.D., Cambridge UK
Phone: 713-743-8380
Fax: 713-743-2636
E-mail: glegge@uh.edu
Office hours:
Mon. and Wed. (2:30-4:00 PM) or by appointment
353 SR2 (Science and Research Building 2)
Research Interests:
Molecular mechanisms of cell adhesion, using primarily multi-
dimensional nuclear magnetic resonance (NMR) spectroscopy.
1
Lecture 1 Summary
• Life arose from simple organic molecules
• Compartmentalization gave rise to cells
• All cells are either prokaryotic or
eukaryotic
• Eukaryotic cells contain membrane-bound
oragnelles
• Three domains based on phylogeny
– Archea, bacteria, and eukarya
• Natural selection directs evolution
Thermodynamics and chemical
equilibria
• Lecture 2 1/15/2003
• Chapter 1 Voet, Voet and Pratt
Thermodynamics:
Allows a prediction as to the spontaneous nature of
a chemical reaction:
Will this reaction proceed in a forward direction as the
reaction is written:
A + B C
Will A react with B to form C or not?
Is this reaction going from higher to lower energy?
ENERGY not necessarily just heat!!!
Definitions:
System: a defined part of the universe
a chemical reaction
a bacteria
a reaction vessel
a metabolic pathway
Surroundings: the rest of the universe
Open system: allows exchange of energy and matter
Closed system: no exchange of matter or energy.
i.e. A perfect insulated box.
Units
Direction of heat flow by definition is most important.
q = heat absorbed by the system from surroundings
If q is positive reaction is endothermic
system absorbs heat from surroundings
If q is negative exothermic
system gives off heat.
A negative W is work done by the system on the
surroundings i.e expansion of a gas
First Law of Thermodynamics: Energy is Conserved
Energy is neither created or destroyed.
In a chemical reaction, all the energy must be accounted for.
Equivalence of work w and energy (heat) q
Work (w) is defined as w = F x D (organized motion)
Heat (q) is a reflection of random molecular motions (heat)
∆U = Ufinal - Uinitial = q - w
Exothermic system releases heat = -q
Endothermic system gains heat = +q
∆U = a state function dependent on the current properties only.
∆U is path independent while q and w are not state functions
because they can be converted from one form of energy to the
other. (excluding other forms of energy, e.g. electrical, light
and nuclear energy, from this discussion.)
I have $100 in my bank account, I
withdrew $50, now I have $50. I could
have added $100 and withdrew $150.
U (money in the account) is same but the
path is different.
Path: is a process by which energy changes in a system
Enthalpy (H)
At constant pressure w = P∆V + w1
w1
= work from all means other than pressure-volume work.
P∆V is also a state function.
By removing* this type of energy from U, we get enthalpy or
‘to warm in’
*remember the signs and direction
H = U + PV
∆H = ∆U + P∆V = qp -w + P∆V = qp - w1
Enthalpy (H)
When considering only pressure/volume work
∆H = qp - P∆V + P∆V = qp
∆H = qp when other work is 0
qp is heat transferred at constant pressure.
In biological systems the differences between
∆U and ∆H are negligible (e.g. volume changes)
The change in enthalpy in any hypothetical reaction
pathway can be determined from the enthalpy
change in any other reaction pathway between the
same products and reactants.
This is a calorie (joule) “bean counting”
Hot Cold
Which way does heat travel? This directionality, is
not mentioned in First Law
Other examples
Considering the random thermal motion of molecules, why at
any particular time doesn’t all the atoms in the air in this room
end up into the upper right corner of the room?
or
When a driver jumps into the water, energy from that person is
dissipated to the water molecules. Have you ever seen the
water molecules act in unison and retransfer this energy back
to the driver and push him out of the water?
In all cases Energy is conserved.
• Disorder increases
• Can we put a numerical
value on disorder? Yes,
most definitely!!
• N identical molecules in a
bulb, open the stop cock you
get 2N
equally probable ways
that N molecules can be
distributed in the bulb.
Gas on its own will expand to the available volume.
But gas molecules are indistinguishable.
Thus, only N+1 different states exist in the bulb or 0, 1, 2,
3, 4,…., (N-1) molecules in left bulb
WL = number of probable ways of placing L of the N
molecules in the left bulb is
The probability of occurrence of such a state
(Is its fraction of the total number of states)
1)!-(NL!
N!
WL =
N
LW
2
=
N = 4 particles, 2 orientations or 24
= 16 different
arrangements of distinguishable particles
1 {
4 {
6* {
4 {
1 {
} N+1
different
states
* The most probable state is the one with the most arrangements
The state that is the most probable is one with the highest
value of WL= N/2 in one bulb and N/2 in the other.
The probability that L is equal to N/2 increases when N is
large
For N=10 the probability that L lies within 20% of N/2 = .66
N=50 probability L lies within 20% of N/2 = .88
N=1023
, L = 1
For a mole of gas in the two bulbs 6.0221 x 1023
The probability of a ratio of 1 in 1010
(one in ten billion) difference or
10,000,000,000 vs. 10,000,000,001 on one side to the other is
10-434
= 0
The reason that direction occurs is not by the laws of
motion, but the aggregate probability of all other states is
So low!! or insignificant that only the most probable state
will occur.
Entropy
for N = 1023
This number is greater than the number of atoms in the
universe!!
NLn2
2
N 10W ≈
22
22
107
5x10
10W x
=
Possible arrangements
LnWkS b=
Each molecule has an inherent amount of energy which
drives it to the most probable state or maximum disorder.
kb or Boltzman’s constant equates the arrangement
probability to calories (joules) per mole.
Entropy is a state function and as such its value depends
only on parameters that describe a state of matter.
Entropy (S)
measure the degree of randomness
The process of diffusion of a gas from the left bulb initially W2 = 1 and
S = 0 to Right (N/2) + Left (N/2) at equilibrium gives a :
∆S that is (+) with a constant energy process such that ∆U = 0
while ∆S>0
This means if no energy flows into the bulbs from the outside
expansion will cool the gas! Conservation of Energy says that
the increase in Entropy is the same as the decrease in thermal
(kinetic) energy of the molecules!!
Entropy is the arrow of time
The entropy of the universe is always increasing.
What does this mean in a closed system?
What does this mean in an open system?
What if the universe starts to collapse, does time go
backwards?
It is difficult or (impossible) to count the number of
arrangements or the most probable state!
So how do we measure entropy?
It takes 80 kcal/mol of heat to change ice at zero °C to water at zero °C
80,000 = 293 ev’s or entropy units
273
A Reversible process means at equilibrium during the change. This is
impossible but makes the calculations easier but for irreversible process
∫≥∆
final
inital
T
dq
S
T
q
S≥∆
At constant pressure we have changes in qp (Enthalpy) and
changes in order - disorder (Entropy)
A spontaneous process gives up energy and becomes more
disordered.
G = H - TS
Describes the total usable energy of a system
A change from one state to another produces:
∆G = ∆H - T∆S = qp - T∆S
If ∆G is negative, the process is spontaneous
∆S ∆H
+ - All favorable
at all temperature
spontaneous
- - Enthalpy favored.
Spontaneous at
temperature below
T = ∆H
∆S
+ + Entropy driven,
enthalpy opposed.
Spontaneous at
Temperatures above
T = ∆H
∆S
- + Non-spontaneous
STP
Standard Temperature and Pressure and at 1M concentration.
We calculate ∆G’s under these conditions.
aA + bB cC + dD
We can calculate a G for each component
(1)
(2)
combining (1) and (2)
(3)
b
__
a
__
d
__
c
__
Gb-Ga-GdGcG +=∆
RTln[A]G-G
o
aa =
ba
dc
o
[B][A]
[D][C]
RTlnGG +∆=∆
Now if we are at equilibrium or ∆G = 0
Then
ba
dc
o
[B][A]
[D][C]
RTlnG0G +∆==∆
ba
dc
o
[B][A]
[D][C]
RTlnG −=∆
eq
o
RTlnKG −=∆
So what does ∆Go
really mean?
OR
Keq
Go
G
ba
dc
o
[B][A]
[D][C]
RTlnGG +=
If Keq = 1 then ∆G = 0
Go
equates to how far Keq varies from 1!!
Keq can vary from 106
to 10-6
or more!!!
∆Go
is a method to calculate two reactions whose Keq’s are different
However
The initial products and reactants maybe far from their equilibrium
concentrations
so
ba
dc
[B][A]
[D][C]
Must be used
RT
G
b
eq
a
eq
d
eq
c
eq
o
[B][A]
[D][C]
Keq
∆−
== e
Le Chatelier’s principle
Any deviation from equilibrium stimulates a process which restores
equilibrium. All closed systems must therefore reach equilibrium
What does this mean?
Keq varies with 1/T
If Keq varied with temperature things would be very unstable.
Exothermic reactions would heat up causing an increase in Keq
generating more heat etc….
Fortunately, this does not happen in nature.
R
S
T
1
R
H-
lnK
oo
eq
∆
+





=
R = gas constant for a 1M solution
Plot lnKeq vs. 1/T ( remember T is in absolute degrees Kelvin)
lnKeq R
H- o
∆
= Slope
R
So
∆
T
1
= Intercept
Van’t Hoff plot
Keq ∆G
o
(kJ·mole-1
106
-34.3
104
-22.8
102
-11.4
101
-5.7
100
0.0
10-1
5.7
10-2
11.4
10-4
22.8
10-6
34.3
The Variation of Keq with ∆Go
at 25 o
C
∑ ∑∆∆=∆ )(reactantsG-(products)GG o
f
o
f
o
The f = for formation.
By convention, the free energy of the elements is taken
as zero at 25 o
C
The free energies of any compound can be measured as
a sum of components from the free energies of
formation.
Standard State for Biochemistry
Unit Activity
25 o
C
pH = 7.0 (not 0, as used in chemistry)
[H2O] is taken as 1, however, if water is in the Keq equation
then [H2O] = 55.5
eqK′ o
G′∆ G′∆
The prime indicates Biochemical standard state
Most times
oo
GG ∆=′∆
However, species with either H2O or H+
requires consideration.
For A + B C + D + nH2O
[A][B]
O][C][D][H
RTln-G
n
2o
=∆
[A][B]
[C][D]
RTln-KRTln-G eq
o
=′=′∆
This is because water is at unity. Water is 55.5 M
and for 1 mol of H2O formed:
O]RTln[HGG 2
oo
n+∆=′∆
-1oo
molkJ9.96GG •+∆=′∆
For: A + B C + HD
H+
+ D-
Where a proton is in the equation
K
[HD]
]][D[H
K
-+
=
( )
o
ooo
]Hln[RT
K
][H-1
lnRT-GG +
+
+∆=′∆
M10][H -7
o =+
This is only valid for
Coupled Reactions
A + B C + D ∆G1 (1)
D + E F +G ∆G2 (2)
0G1 ≥∆If reaction 1 will not occur as written.
However, if 2G∆ is sufficiently exergonic so 0GG 21 ≤∆+∆
Then the combined reactions will be favorable through
the common intermediate D
A + B + E C + F + G ∆G3 0GGG 213 ≤∆+∆=∆
As long as the overall pathway is exergonic,
it will operate in a forward manner.
Thus, the free energy of ATP hydrolysis, a
highly exergonic reaction, is harnessed to
drive many otherwise endergonic biological
processes to completion!!
Life obeys the Laws of Thermodynamics
• Living organisms are open systems
• Living things maintain a steady state
• Enzymes catalyze biological reactions
Water and elements in life processes
• Lecture 3 1/22/2003

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Lecture 2

  • 1. University of Houston BCHS 3304: General Biochemistry I, Section 07553 Spring 2003 1:00-2:30 PM Mon./Wed. AH 101 Instructor: Glen B. Legge, Ph.D., Cambridge UK Phone: 713-743-8380 Fax: 713-743-2636 E-mail: glegge@uh.edu Office hours: Mon. and Wed. (2:30-4:00 PM) or by appointment 353 SR2 (Science and Research Building 2) Research Interests: Molecular mechanisms of cell adhesion, using primarily multi- dimensional nuclear magnetic resonance (NMR) spectroscopy. 1
  • 2. Lecture 1 Summary • Life arose from simple organic molecules • Compartmentalization gave rise to cells • All cells are either prokaryotic or eukaryotic • Eukaryotic cells contain membrane-bound oragnelles • Three domains based on phylogeny – Archea, bacteria, and eukarya • Natural selection directs evolution
  • 3. Thermodynamics and chemical equilibria • Lecture 2 1/15/2003 • Chapter 1 Voet, Voet and Pratt
  • 4. Thermodynamics: Allows a prediction as to the spontaneous nature of a chemical reaction: Will this reaction proceed in a forward direction as the reaction is written: A + B C Will A react with B to form C or not? Is this reaction going from higher to lower energy? ENERGY not necessarily just heat!!!
  • 5. Definitions: System: a defined part of the universe a chemical reaction a bacteria a reaction vessel a metabolic pathway Surroundings: the rest of the universe Open system: allows exchange of energy and matter Closed system: no exchange of matter or energy. i.e. A perfect insulated box.
  • 7. Direction of heat flow by definition is most important. q = heat absorbed by the system from surroundings If q is positive reaction is endothermic system absorbs heat from surroundings If q is negative exothermic system gives off heat. A negative W is work done by the system on the surroundings i.e expansion of a gas
  • 8. First Law of Thermodynamics: Energy is Conserved Energy is neither created or destroyed. In a chemical reaction, all the energy must be accounted for. Equivalence of work w and energy (heat) q Work (w) is defined as w = F x D (organized motion) Heat (q) is a reflection of random molecular motions (heat)
  • 9. ∆U = Ufinal - Uinitial = q - w Exothermic system releases heat = -q Endothermic system gains heat = +q ∆U = a state function dependent on the current properties only. ∆U is path independent while q and w are not state functions because they can be converted from one form of energy to the other. (excluding other forms of energy, e.g. electrical, light and nuclear energy, from this discussion.)
  • 10. I have $100 in my bank account, I withdrew $50, now I have $50. I could have added $100 and withdrew $150. U (money in the account) is same but the path is different. Path: is a process by which energy changes in a system
  • 11. Enthalpy (H) At constant pressure w = P∆V + w1 w1 = work from all means other than pressure-volume work. P∆V is also a state function. By removing* this type of energy from U, we get enthalpy or ‘to warm in’ *remember the signs and direction H = U + PV ∆H = ∆U + P∆V = qp -w + P∆V = qp - w1
  • 12. Enthalpy (H) When considering only pressure/volume work ∆H = qp - P∆V + P∆V = qp ∆H = qp when other work is 0 qp is heat transferred at constant pressure. In biological systems the differences between ∆U and ∆H are negligible (e.g. volume changes)
  • 13. The change in enthalpy in any hypothetical reaction pathway can be determined from the enthalpy change in any other reaction pathway between the same products and reactants. This is a calorie (joule) “bean counting” Hot Cold Which way does heat travel? This directionality, is not mentioned in First Law
  • 14. Other examples Considering the random thermal motion of molecules, why at any particular time doesn’t all the atoms in the air in this room end up into the upper right corner of the room? or When a driver jumps into the water, energy from that person is dissipated to the water molecules. Have you ever seen the water molecules act in unison and retransfer this energy back to the driver and push him out of the water? In all cases Energy is conserved.
  • 15. • Disorder increases • Can we put a numerical value on disorder? Yes, most definitely!! • N identical molecules in a bulb, open the stop cock you get 2N equally probable ways that N molecules can be distributed in the bulb. Gas on its own will expand to the available volume.
  • 16. But gas molecules are indistinguishable. Thus, only N+1 different states exist in the bulb or 0, 1, 2, 3, 4,…., (N-1) molecules in left bulb WL = number of probable ways of placing L of the N molecules in the left bulb is The probability of occurrence of such a state (Is its fraction of the total number of states) 1)!-(NL! N! WL = N LW 2 =
  • 17. N = 4 particles, 2 orientations or 24 = 16 different arrangements of distinguishable particles 1 { 4 { 6* { 4 { 1 { } N+1 different states * The most probable state is the one with the most arrangements
  • 18. The state that is the most probable is one with the highest value of WL= N/2 in one bulb and N/2 in the other. The probability that L is equal to N/2 increases when N is large For N=10 the probability that L lies within 20% of N/2 = .66 N=50 probability L lies within 20% of N/2 = .88 N=1023 , L = 1 For a mole of gas in the two bulbs 6.0221 x 1023 The probability of a ratio of 1 in 1010 (one in ten billion) difference or 10,000,000,000 vs. 10,000,000,001 on one side to the other is 10-434 = 0
  • 19. The reason that direction occurs is not by the laws of motion, but the aggregate probability of all other states is So low!! or insignificant that only the most probable state will occur. Entropy for N = 1023 This number is greater than the number of atoms in the universe!! NLn2 2 N 10W ≈ 22 22 107 5x10 10W x = Possible arrangements
  • 20. LnWkS b= Each molecule has an inherent amount of energy which drives it to the most probable state or maximum disorder. kb or Boltzman’s constant equates the arrangement probability to calories (joules) per mole. Entropy is a state function and as such its value depends only on parameters that describe a state of matter. Entropy (S) measure the degree of randomness
  • 21. The process of diffusion of a gas from the left bulb initially W2 = 1 and S = 0 to Right (N/2) + Left (N/2) at equilibrium gives a : ∆S that is (+) with a constant energy process such that ∆U = 0 while ∆S>0 This means if no energy flows into the bulbs from the outside expansion will cool the gas! Conservation of Energy says that the increase in Entropy is the same as the decrease in thermal (kinetic) energy of the molecules!!
  • 22. Entropy is the arrow of time The entropy of the universe is always increasing. What does this mean in a closed system? What does this mean in an open system? What if the universe starts to collapse, does time go backwards?
  • 23. It is difficult or (impossible) to count the number of arrangements or the most probable state! So how do we measure entropy? It takes 80 kcal/mol of heat to change ice at zero °C to water at zero °C 80,000 = 293 ev’s or entropy units 273 A Reversible process means at equilibrium during the change. This is impossible but makes the calculations easier but for irreversible process ∫≥∆ final inital T dq S T q S≥∆
  • 24. At constant pressure we have changes in qp (Enthalpy) and changes in order - disorder (Entropy) A spontaneous process gives up energy and becomes more disordered. G = H - TS Describes the total usable energy of a system A change from one state to another produces: ∆G = ∆H - T∆S = qp - T∆S If ∆G is negative, the process is spontaneous
  • 25. ∆S ∆H + - All favorable at all temperature spontaneous - - Enthalpy favored. Spontaneous at temperature below T = ∆H ∆S + + Entropy driven, enthalpy opposed. Spontaneous at Temperatures above T = ∆H ∆S - + Non-spontaneous
  • 26. STP Standard Temperature and Pressure and at 1M concentration. We calculate ∆G’s under these conditions. aA + bB cC + dD We can calculate a G for each component (1) (2) combining (1) and (2) (3) b __ a __ d __ c __ Gb-Ga-GdGcG +=∆ RTln[A]G-G o aa = ba dc o [B][A] [D][C] RTlnGG +∆=∆
  • 27. Now if we are at equilibrium or ∆G = 0 Then ba dc o [B][A] [D][C] RTlnG0G +∆==∆ ba dc o [B][A] [D][C] RTlnG −=∆ eq o RTlnKG −=∆ So what does ∆Go really mean? OR
  • 28. Keq Go G ba dc o [B][A] [D][C] RTlnGG += If Keq = 1 then ∆G = 0 Go equates to how far Keq varies from 1!!
  • 29. Keq can vary from 106 to 10-6 or more!!! ∆Go is a method to calculate two reactions whose Keq’s are different However The initial products and reactants maybe far from their equilibrium concentrations so ba dc [B][A] [D][C] Must be used
  • 30. RT G b eq a eq d eq c eq o [B][A] [D][C] Keq ∆− == e Le Chatelier’s principle Any deviation from equilibrium stimulates a process which restores equilibrium. All closed systems must therefore reach equilibrium What does this mean? Keq varies with 1/T If Keq varied with temperature things would be very unstable. Exothermic reactions would heat up causing an increase in Keq generating more heat etc….
  • 31. Fortunately, this does not happen in nature. R S T 1 R H- lnK oo eq ∆ +      = R = gas constant for a 1M solution Plot lnKeq vs. 1/T ( remember T is in absolute degrees Kelvin) lnKeq R H- o ∆ = Slope R So ∆ T 1 = Intercept Van’t Hoff plot
  • 33. ∑ ∑∆∆=∆ )(reactantsG-(products)GG o f o f o The f = for formation. By convention, the free energy of the elements is taken as zero at 25 o C The free energies of any compound can be measured as a sum of components from the free energies of formation.
  • 34. Standard State for Biochemistry Unit Activity 25 o C pH = 7.0 (not 0, as used in chemistry) [H2O] is taken as 1, however, if water is in the Keq equation then [H2O] = 55.5 eqK′ o G′∆ G′∆ The prime indicates Biochemical standard state
  • 35. Most times oo GG ∆=′∆ However, species with either H2O or H+ requires consideration. For A + B C + D + nH2O [A][B] O][C][D][H RTln-G n 2o =∆ [A][B] [C][D] RTln-KRTln-G eq o =′=′∆ This is because water is at unity. Water is 55.5 M and for 1 mol of H2O formed: O]RTln[HGG 2 oo n+∆=′∆ -1oo molkJ9.96GG •+∆=′∆
  • 36. For: A + B C + HD H+ + D- Where a proton is in the equation K [HD] ]][D[H K -+ = ( ) o ooo ]Hln[RT K ][H-1 lnRT-GG + + +∆=′∆ M10][H -7 o =+ This is only valid for
  • 37. Coupled Reactions A + B C + D ∆G1 (1) D + E F +G ∆G2 (2) 0G1 ≥∆If reaction 1 will not occur as written. However, if 2G∆ is sufficiently exergonic so 0GG 21 ≤∆+∆ Then the combined reactions will be favorable through the common intermediate D A + B + E C + F + G ∆G3 0GGG 213 ≤∆+∆=∆
  • 38. As long as the overall pathway is exergonic, it will operate in a forward manner. Thus, the free energy of ATP hydrolysis, a highly exergonic reaction, is harnessed to drive many otherwise endergonic biological processes to completion!!
  • 39. Life obeys the Laws of Thermodynamics • Living organisms are open systems • Living things maintain a steady state • Enzymes catalyze biological reactions
  • 40. Water and elements in life processes • Lecture 3 1/22/2003

Editor's Notes

  1. Draw blackboard account of this