BAHAN KULIAH BIOKIMIA POWER POINT BAGIAN 1 /BIOCHEMISTRY POWER POINT LECTURES PART 1 | Karya Tulis Ilmiah

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.

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

Thermodynamics and chemical

equilibria

• Lecture 2 1/15/2003

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?

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

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.

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)



U = U

_final

- U

_initial

= 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

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

.

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

Enthalpy (H)

When considering only pressure/volume work

H = qp - PV + PV = qp



H =

q

p

when other work is 0

qp is heat transferred at constant pressure.

In biological systems the differences between

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

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?

• 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.

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

W_L = 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)!

-(N

L!

N!

W

L



N L

W

2

N = 4 particles, 2 orientations or 2

4

= 16 different

arrangements of distinguishable particles

1 { 4 { 6* { 4 {

1 {

}

N+1

different states

The state that is the most probable is one with the highest value of W_L= 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

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

10

W



22 22

10 7

5x10

10

W



x

LnW

k

S



b

Each molecule has an inherent amount of energy which drives it to the most probable state or maximum disorder.

k_b 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)

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

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?

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

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

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

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 __

G

b

-G

a

-G

d

G

c

G







RTln[A]

G

-G

a oa



b a d c o

[B]

[A]

[D]

[C]

RTln

G

Now if we are at equilibrium or G = 0 Then b a d c o

[B]

[A]

[D]

[C]

RTln

G

0

G











b a d c o

[B]

[A]

[D]

[C]

RTln

G







eq

o

RTlnK

G







So what does Go really mean?

K

_eq

Go

G

b a

d c

o

[B] [A]

[D] [C]

RTln G

G  

If K_eq = 1 then G = 0

Go equates to how far K

K

_eq

can vary from 10

6

to 10

-6

or more!!!

Go is a method to calculate two reactions whose K

eq’s are different

However

The initial products and reactants maybe far from their equilibrium concentrations

so

b a

d c

[B]

[A]

[D]

[C]

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

Fortunately, this does not happen in nature.

R

S

T

1

R

H

-lnK

o o 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

S

o



T 1 = Intercept

K_{eq 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













G

o

G

o_f

(products)

-

G

o_f

(reactants

)

The f = for formation.

By convention, the free energy of the elements is taken as zero at 25 oC

The free energies of any compound can be measured as a sum of components from the free energies of

Standard State for Biochemistry

Unit Activity 25 oC

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

eq

K





G



o



G



Most times



G



o





G

o

However, species with either H₂O or H+ requires consideration.

For A + B C + D + nH₂O

[A][B] O] [C][D][H RTln -G n 2 o   [A][B] [C][D] RTln - K RTln -

Go  _eq 



This is because water is at unity. Water is 55.5 M and for 1 mol of H₂O formed:

O]

RTln[H

G

G



o





o



n

₂



-1 o

o

G

9.96

kJ

mol

For:

A + B C + HD

H

+

+ D

-Where a proton is in the equation

K

[HD]

]

][D

[H

K

-







o o o

o

RT

ln[

H

]

K

]

[H

-1

ln

RT

-

G

G

 











M

10

]

Coupled Reactions

A + B C + D



G

₁

(1)

D + E F +G



G

₂

(2)

0 G₁  

If reaction 1 will not occur as written.

However, if



G

₂ is sufficiently exergonic so



G

₁





G

₂



0

Then the combined reactions will be favorable through the common intermediate D

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

Life obeys the Laws of Thermodynamics

Water and elements in life processes