Principles of Biochemistry 2 |Entropy| Class Notes |HarvardX

What is life?

Carbon is the base of life. But organisms are not only a piece of carbon.
Living life constantly carries a series of energy exchange processes.

digraph F { rankdir = DU;

subgraph cluster_1 {
    rank = UD
    minerals [label = "minerals", shape = box, fillcolor = "#FF0000" ];
    O2 [label = "O2", shape = diamond, color = green, size = 2];
    CO2 [label = "CO2", shape = diamond, color = green, size = 2];
    organics   [label = "organics"      , shape = box, color = coral];

}

subgraph cluster_0 {
    Food [label = "Food", shape = box, color = deepskyblue1];
    Sun [label = "Sun", shape = box, color = deepskyblue1];
}
subgraph cluster_2 {
  Metabolism [label = "Metabolism", shape = circle, color = pink, size = 3];
  Catabolism [label = "Use for variety bio-activities"]
  Anabolism [label = "Build body component"]
  Metabolism -> Anabolism [label = Anabolism];
  Metabolism -> Catabolism [label = Catabolism];
}

subgraph cluster_0 {
	style=filled;
	color=lightgrey;
	node [style=filled,color=white];
	label = "Potential Energy";
}


subgraph cluster_1 {
	node [style=filled];
	label = "Surrounding Energy";
	color=blue
}

subgraph cluster_2{
  node [style=filled];
	label = "Living Cell";
	color= red
}
Food -> Metabolism [ label = "Catabolism" ];
Sun ->  Metabolism;
minerals -> Metabolism;
organics -> Metabolism;
O2 -> Metabolism;
CO2 -> Metabolism;

}

Types of Metabolism

digraph F {

NSunlight [label = “Not Sunlight”];
Organisms -> Sunlight -> Phototrophs;
Organisms -> NSunlight -> Chemotrophs;
Phototrophs -> Food;
Phototrophs -> CO2;
Chemotrophs -> Food -> Heterotrophs;
Chemotrophs -> CO2 -> Autotrophs;

subgraph cluster_0{
node [style=filled];
label = “Energy source”;
color= blue
Sunlight;
NSunlight;
Phototrophs;
Chemotrophs;
}

subgraph cluster_1{
node [style=filled];
label = “Carbon Source”;
color= pink;
Food;
CO2;
}

subgraph cluster_2{
node [style=filled];
label = “Metabolic type”;
color= red;
Autotrophs;
Heterotrophs;
}

{rank=“same”; Heterotrophs; Autotrophs}
Heterotrophs -> Autotrophs [label = CO2; color= red];
Heterotrophs -> H2O -> Autotrophs [ color= red];
Autotrophs -> O2 -> Heterotrophs [ color= blue];
Autotrophs -> organics -> Heterotrophs [ color= blue];
}

Entropy and reaction

Gibb’s free energy equation

$$
G = H - TS
$$
G: Free engergy
H: Enthalpy
T: Temperature
S: Entropy

Enthalpy (H): Total energy of the system (Energy in bonds)
Entropy (S): Quantitative expression for the randomness of the system; (the disorder of the system, the amount of energy that cannot do work)
Temperature (T): Temperature; Degrees Kelvin
$TS$: Increased temperature intensifies random molecular motion, leading to increased disorder.
Free Energy (G): The only portion of the energy that is able to do work.

In a cell that carries reactions, the amount of energy lost as heat, which contributes to random motion, increased entropy, and the reaction was irreversible.

Spontaneous Reaction

In a spontaneous reaction, the free energy is decreased,
therefor:
$\Delta G < 0$
Because:
$\Delta G = \Delta H - T\Delta S$
As a result:
$\Delta H < T \Delta S$

So,
$\Delta H$ is negative: $(-)\Delta H$, the heat is release, it is a exothermic reaction.
$\Delta H$ is positive: $(+)\Delta H$, the heat is absorbed, it is a endothermic reaction.

The parameters of the reaction:

• Spontaneous or not;
• Equilibrium constant;
• Directionality;
• Velocity;

Equilibrium

1. A reaction can occur spontaneously only if $\Delta G$ is negative. (exergonic)
2. A system is at equilibrium and no net change can take place if $\Delta G = 0$
3. A reaction can not occur spontaneously, only input of free energy to lead it happens when the $\Delta G <0$ . (endergonic)

Equilibrium constant
$S \rightleftharpoons P$

The concentration of S and P constantly changes.

This ration is called: Equilibrium constant ($K_{eq}$)
$K_{eq} = \frac{[P]_ {eq} } { [S]_{eq}}$

Significant: most chemical reactions are reversible. By knowing the $K_{eq}$ and the initial the concentration, we can predict the direction of reaction.

Calculating Exp. 1:

Dihydroxyacetone phosphate (DHAP)
Glyceraldehyde 3-phosphate (G3P)

$$DHAP \rightleftharpoons G3P$$

$K_{eq} = \frac{[G3P]}{[DHAP]}$
$\ \ \ \ \ \ \ = 0.475$

Stabdard free energy change ($\Delta G^{\circ}$):
$$\Delta G^{\circ} = -RTln(K_{eq})$$

R: gas constant
T: temperature expressed in degrees Kelvin.

So, a standard condition is a condition that the reaction proceeds at a constant temperature-- T equals 298 Kelvin or $25^{\circ}$C:

$$T = 298K = 25^{\circ}C$$
Reactants and products = $1M$
$R = 1.98 \times 10^{-3} kcal^1 mol^{-1} deg^{-1}$

$\Delta G^{\circ} = -RTln(K_{eq})$
$\ \ \ \ \ \ \ \ = - (1.98 \times 10^{-3}) \times 298 \times ln(0.0475)$
$\ \ \ \ \ \ \ \ = 1.80 kcal/mol$

So, the $\Delta G^{\circ}$ is positive.

But the thing that determines the property spontaneously is the free energy change. It gives a function
$$\Delta G = \Delta G^{\circ}+ RTln(K)$$

Here, K is the actual ratio of glyceraldehyde 3-phosphate and dihydroxyacetone phosphate concentrations in the cell. In this example, the concentration of the DHAP is $2 \times 10^{-4}M$ and the G3P’s concentration is $3 \times 10^{-6}M$ (the initial of the concentration).
Under this condition, We now have:

$\Delta G = \Delta G^{\circ}+ RTln(K)$
$\ \ \ \ \ \ \ = 1.80 + RTln(\frac{[G3P]}{[DHAP]})$
$\ \ \ \ \ \ \ = 1.80 + (1.98 \times 10^{-3}) \times 298 \times ln(\frac{3 \times 10^{-6}}{2 \times 10^{-4}})$
$\ \ \ \ \ \ \ = - 0.7 kcal/mol$

As a result, the $\Delta G < 0$
That’s means, it is a spontaneous reaction!

Delta G

We already know that:
$\Delta G = \Delta G^{\circ}+ RTln(K)$

When there are no works is down, then, $\Delta G = 0$:
$0 = \Delta G^{\circ} + RTln(K_{eq})$
$\Delta G^{\circ} = - RTln(K_{eq})$

Calculating Exp. 2:

In the following reaction:
$S \longrightarrow P $
$K_{eq}=4$
Initial concentration of P and S: $[P]= 10M;[S]=1M$
The forward reaction is ?:

• Standard Calculation:
  $\Delta G = \Delta G^{\circ} + RTln(K)$
  $\Delta G = - RTln(K_{eq}) + RTln(K)$
  $\Delta G = RT(RTln(K) - ln(K_{eq}))$
  $\Delta G = RT(ln(10) - ln(K_{4}))$
  $\Delta G = 1.98 \times 10^{-3} \times 298 \times (ln(10) - ln(4))$
  $\Delta G = 0.5406482kcal/mol$

• or can avoid the calculation:
  $\Delta G = \Delta G^{\circ} + RTln(K)$
  $\Delta G = - RTln(K_{eq}) + RTln(K)$
  $\Delta G = RT(ln(K) - ln(K_{eq}))$
  $\Delta G = RT(ln(10) - ln(K_{4}))$
  $\because RT > 0$
  $\because ln(10) - ln(K_{4}) > 0$
  $\therefore RT \times ( ln(10) - ln(K_{4}) ) > 0$
  $\therefore \Delta G > 0$

• Another way:
  $\because K_{eq} = \frac{[P_{eq}]}{[S_{eq}]} = 4$
  $\because K = \frac{P}{S} = 10$
  $\therefore K_{eq} < K$
  $\therefore ln(K) - ln(K_{eq}) > 0$
  $\because RT > 0$
  $\therefore RT \times ( ln(10) - ln(K_{4}) ) > 0$
  $\therefore \Delta G >0$

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Cover: Walkowski, Slawomir & Szymas, Janusz. (2011). Quality evaluation of virtual slides using methods based on comparing common image areas. Diagnostic pathology. 6 Suppl 1. S14. 10.1186/1746-1596-6-S1-S14.