4....ʘ but if you coupled it with atp hydrolysis ∆go=-4 (it is large amount), much more glucose...

RAHAF Muwalla

Rahaf Muwalla

4.

FAISAL

RAHAF+Leen Attar

1 | P a g e L E E N A T T A R & R A H A F M U W A L L A

FREE ENERGY:

The direction in a chemical reaction proceeds are

determined by the degree to which two factors

change during the reaction.

These are enthalpy ∆H,( a measure of the change

*∆+ in heat content of the reactants and products)

and entropy ∆S,( a measure of the change in

randomness or disorder of the reactants and products).

** Enthalpy and entropy can be used to define a third

which predicts the direction free energy (G),quantity,

in which a reaction will spontaneously proceed.

.STANDARD FREE ENERGY CHANGE ∆Go, AND EQUILIBRIUM CONSTANT Keq

∆Go: The free energy at specific condition, when reactants and products are at a concentration of 1 mol/L.

Keq: is obtained by dividing [products] to [reactants] when the reaction reaches equilibrium. (Equilibrium does not mean that there is no reaction, but means that both reactions occur at the same rate).

Keq= [product]/[reactant] At equilibrium ∆G = 0

((The ratio of product to the reactant is CONSTANT, is not equal.))

ʘ We can apply the equation >>

∆G= ∆Go +RT ln[product]/[reactant]

(At equilibrium the ∆G=0, so the ratio of product to reactant is constant value (Keq).)

Free energy (G), enthalpy (H), and

entropy (S).

T is the absolute temperature in

Kelvin (K), where K = °C + 273


ʘ So, we can replace these in general equation

0=Go +RT ln Keq (we can arrange the equation)) -> ∆Go = - RT ln Keq

ʘ Keq is constant in any reaction/ T is also constant / R is gas constant(1.987 cal/mol K)

So, ∆Go for any reaction is constant value. ((For any reaction it has its own ∆Go value))

ʘ ∆Go is a special condition not at equilibrium, it is special at standard conditions.

Also. At standard condition we can apply ->> ∆G= ∆Go +RT 2.3 log 1

Log 1=0 ∆G= ∆Go ( a special condition, when The ratio of [product] to [reactant] =1 )

**************

ʘ NOW, let’s apply this concept to simple chemical reaction:

Isomerization reaction, which means to convert glucose 6-phosphate to fructose 6-

phosphate.

* Glucose (aldose) and fructose (ketose) are isomers.*

At equilibrium condition, the concentration

of glucose-6-phosphate is equal twice the concentration

of fructose-6-phosphate. ((The ratio is 2:1)).

ʘ We can apply it at the equation:

G= ∆Go + RT 2.3 log (0.33/ 0.66) ∆

Go +1.987 *(25 + 273) * log (0.33/ 0.66) ∆0=

∆Go = +.4 Kcal/mol (=400cal/mol)

and this is the constant value for this reaction. (∆G=0), so the reaction goes equilibrium.


At Non- equilibrium condition ( not standard condition):

Go= + 0.4 kcal/mol) ∆(

∆G= ∆Go +RT 2.3 log 0.09/0.9

∆G = - 0.96 (minus sign means that the

reaction goes FORWARD.

_______________________________________________________________________

At standard condition:

∆G= ∆Go + RT 2.3 log 1/1

∆G= ∆Go= +.4Kcal/mol.(positive sign means the

Reaction goes BACKWARD.

SUMMARY>> Free energy change (∆G) of a reaction depends on the concentration of

reactant and product. For the conversion of glucose 6-phosphate to fructose 6-

G is ∆ ,r than 1is large to product reactantthe ratio of phosphate, ∆G is negative when

positive and equal to ∆Go under standard conditions, and ∆G is ZERO at equilibrium.

Remember>> The standard free energy change, ∆G0, is so called because it is equal to

the free energy change, ∆G, under standard conditions (that is, when reactants and

products are at 1 mol/l concentrations)


** ʘ ∆G0 and reaction direction: Under standard conditions, ∆G0 can be used to

predict the direction of the reaction proceeds because, under these conditions, ∆G0 is

equal to ∆G.

****************

ʘ Relationship between ∆G0 and Keq

at equilibrium:

ʘ Notice that keq increase 10 times but

∆Go just twice !

Because it is log function

( each unit in log function equals 10 time )

ʘ keq < 1

here if keq very small

more reactant than product

here the reaction is almost impossible to

is very high 0G∆occur.


Exergonic reactions in Biochemistry:

ʘ Means that ∆G is negative ( the product has a lower free energy than the substrate),

then the reaction is spontaneous (loss energy from the system).

ʘ Complex structures converted to simple structures.

Examples>> Proteins → amino acids (that happens during digestion/ or degraded

protein in the cell)

ʘ Exergonic reaction means that the reaction will occur at very high rate ?

NO, because it requires enzyme.

Ex: if you have protein solution will be remain protein solution because digestion

require enzymes!

Starch → n glucose

glucose + O2 → CO2 + H2O

So, Hydrolysis reactions : Breaking bonds by adding H2O.

Decarboxylation reactions (release of CO2 )

pyruvate ( C3 ) → acetyl- (C2) +CO2

ʘ These two reactions are always exergonic reactions.

Oxidation with O2 (O2 very strong oxidant).

ʘ Now let’s talk about endergonic reactions, you can imagine

that the exergonic reaction all what they need is enzymes

then the reaction can occur. But how can endergonic reactions

occur?

Endergonic reaction can be driven by an exergonic reaction if the two reactions can be coupled.

For

categorization,

you look at the

reaction then

decide:

- complex

reaction

converted to

simple one >>

EXERGOANIC

reaction.

- simple reaction

converted to

complex one>>

ENDERGONIC


ʘ Example on coupling reaction, hypothetical reaction:

A → B ∆G = + 5 kcal/mol (endergonic reaction)(even if you have the enzyme, the

reaction cannot occur)

C → D ∆G = - 9 kcal/mol (exergonic reaction) (you must have an enzyme so the

reaction occurs)

ʘ When the two are coupled:

A + C → B + D ∆G = - 4 kcal/mol

A + C → I → B + D (for this reaction to

proceed we need enzyme, when there

is no enzyme, it will not be converted).

So, there is no enzyme allows A to convert

to B without coupling the two reactions.

ʘ How the coupling occurs?

Through common intermediate, the enzyme has to bind A and C together and the

intermediate will occur and after this B and D will produce.

ʘ Presence of enzyme plus common intermediate allows the second reaction to drive

the first reaction.

ʘ coupling means they can’t separate from each other, there should be an enzyme to

allow conversion AB ONLY if CD*

ʘ In the picture above, red will not go up unless green goes down, by intermediate.

ʘ If C converted to D by enzyme the energy will release as heat as a byproduct only, but

if we coupled the two reactions the energy will be transfer from catabolism to

anabolism.

Green can go down slow, because it’s down sloop

(exergonic)

Raising Red box needs energy (endergonic), so they are

coupled by a rob.

RED WON’T GO UP, UNLESS GREEN GOES DOWN, by

intermediate.


ʘ Coupling allow the specific reaction to occur with another reaction.

ʘ Endergonic and exergonic reactions are INDIRECTLY coupled by ATP cycle.

ATP + H2O → ADP + Pi ∆G0 = - 7.3 (ATP hydrolysis, exergonic reaction) (needs

enzyme).

ADP + Pi → ATP + H2O ∆G0= 7.3 (same value with positive number) (endergonic)

ʘ If A → B is coupled with ATP hydrolysis, enzyme takes A , takes ATP and allow the

conversion A → B ( ∆G = 5-7.3 = -2.3 )

ʘ And again, the ATP formation is coupled again with conversion of C to D

(∆G = -9 + 7.3 = -1.7)

ʘ So, energy which released from

exergonic reaction will captured as a

chemical energy in the product of the

endergonic reaction and the difference is

heat (is a byproduct).

ʘ Most of the reactions in our body released

heat.


ʘ What is the structure of

ATP?

The carbon number 1 in ribose

binds with adenine by N-

glycosidic bond.

And carbon number 5 with

phosphate by ester bond.

ʘ Anhydride bond is known as high energy phosphate bond

ʘ What’s make hydrolysis of ATP produce a high amount of energy (7.3kcal/mole)?

ʘ By looking at the

structure of ATP and

to the structure to

the product (ADP+P)

There are two reasons:

1- In the ATP we have three negative charges and when phosphate is released negative

charges will be separated away from each other, so the products are more stable than

the reactant.

2- This terminal phosphate forms many form resonances as you see in the picture

above.

ʘ As conclusion, the products are more stable than the reactant.


ʘ If we convert adenosine diphosphate ADP to (AMP+P) same amount of

energy will be released (7.3).

ʘ NOTE: Usually what it used of energy production is ATP, so even the

hydrolysis of ADP to AMP is produces same amount of energy, but usually

the enzymes don’t use ADP energy to convert it to AMP.

ʘ Conversion of AMP to (adenosine +p) (hydrolysis of ester bond will not

produce energy)

****************

ʘ You have six phosphorylated compounds in the table above.

ATP in the middle, is it advantage to have a value for the hydrolysis that is

intermediate? Why the ATP is the suitable for carrying energy and not phosphoenol

pyruvate for example?

If it is phosphoenol pyruvate, it will be synthesize from pyruvate by too much energy!

And this is the problem

But by being in the middle it can be formed by transfer of phosphate group from the

compound above it.

For example, glucose-6-phosphate can be synthesized by transfer of phosphate group

from ATP.


ʘ If we take this example :

Glucose + phosphate Glucose-6-phosphate + water

The ∆Go is 3.3 as you see at the table above, how it can be synthesized?

ʘCan you put (glucose + p) with proper enzyme to have it occurs?

this case. ined formphosphate will be -6-amount of glucose elttil No, very

ʘ But if you coupled it with ATP hydrolysis ∆Go=-4 (it is large amount), much more

glucose 6-phosphate will be form than glucose.

ʘ So, coupling it with ATP hydrolysis has a great advantage.

ʘ Notice that ATP is not a long term storage of energy, every time there is hydrolysis

and by coupling with amino acid catabolism. Look at the picture below

************************

ʘ How energy is stored in the cell?

If you say as a form of ATP this is false because ATP use just for transfer of energy.


C6H11O6 + 6O2 6CO2 + 6H2O + 36ATP

ʘ You can take any tissue and calculate how many oxygens are consumed, for example

in the brain approximate is 3.4mole/day, if I divide this value on 6 I can know the moles

of glucose.

ʘ The most of oxygen consumption is used by the brain (3.4mole/day)and liver

(3.6mole/day)

ʘ If the total moles ATP is consumed per day equal to 90.6, how many kilograms are

consumed daily ? (1mole = 500grams)

90.6*500=45000 grams = 45kilograms

ʘ and all of ATP in our cells are 50 gram

ʘ So, the turnover is very rapid and the amount of ATP present in the cell is sufficient

for less than one minute, if there is no ATP continuous production ATP will finish in less

than one minute! So, it is not storage form, it is for immediate use.

****************


ʘ There are other nucleotides, they are the same, all have the same amount of energy,

all synthesize from ATP.

ʘ If GTP is required as source of energy in some reaction, we must take it from ATP

indirectly. AND if the reaction produces GTP you cannot use it as such(as we said that

the enzymes use ATP as a source of energy not GTP even they produce the same

amount of energy), it has to be converted into ATP as a reaction in the picture above.

ʘ To make polysaccharide like glycogen from glucose, how it is synthesize?

Is there an enzyme add one glucose?

Yes, but what is the form of glucose that is suitable for addition?

It is UDP glucose. It is suitable for synthesizing polysaccharide, disaccharide,

oligosaccharide.

GOOD LUCK

4....ʘ but if you coupled it with atp hydrolysis ∆go=-4 (it is large amount), much more glucose...

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