from last time… pressure = force/area two major pressures in plants

from last time…

Pressure = force/area

Two major pressures in plants.

1. The positive pressure (turgor) inside living cells and that’s required for cell and tissue growth.

2. The negative pressure (tension) that exists in the cells of the xylem of transpiring plants.

In general we’ll use units of pressure to express theenergy status of water, the “water potential” .

How is pressure like energy/volume?

Nm2

multiply by unity as m/m

N x m m2 m

N . m = force x distance = energy m3 volume volumeSo, we can use units of pressure to express the

energy status of water. We’ll see that water tendsto move from areas of higher to lower energy/vol, or pressure.

force/area = energy/volume?

The gas constant, R

Remember PV = nRT?

R, the “gas constant” makes the relationship amongP, V, n, and T work.

R shows up in lots of energy equations.

Values and units for R8.314 J mol-1 K-1

8.314 m3 Pa mol-1 K-1

We’ll use R a lot!

a bit more review….

energy must equalvolume x pressure

How to put some numbers to all the energyexpended in doing the work of life.

•Chemical reactions - synthesizing compounds, degrading others

•Solute transport - maintaining concentration differences across membranes

•Maintaining electrical potentials and movingions across charged membranes.

How can we understand when these processes require energy and how much?

Bioenergetics and Free Energy

Free Energy, G, is the energy available to do work.G is the change or difference (in G during a process or reaction.

G equations help us understand whether a reaction:1) yields energy and can happen spontaneously (G < 0),

2) requires energy input to occur (G > 0),

3) or is at equilibrium (G = 0).

We will use G equations for understanding bioenergetics of chemical reactions and the transport of charged and uncharged solutes.

the free energy equations give usvalues of energy per mole, J mol-1

1. G of a chemical reaction General equations:

1. G = G0 + 2.3 RT log (K)1. or

G = G0 + RT ln (K)

G0 is the standard free energy change, defined for standardized conditions. It allows comparisons of G of different reactions. K is the equilibrium constantK = ([product]*[product]) ([reactant]*[reactant])

So, G = G0 + 2.3 RT log (K) can be written as:

G = G0 + 2.3 RT log ([product]*[product]) ([reactant]*[reactant])

Example using a very important reaction

G for ATP hydrolysis: ATP ADP + Pi

G0 = - 33kJ mol-1

In typical cellular conditionsG = –50 kJ mol-1 to –65 kJ mol-1,

The reaction releases energy and can happen spontaneously.

Compare with ATP synthesis: ADP + PI ATP

G > 0, requires energy, is not spontaneous.

2. Solute transportTransport is from C1 to C2

[C1] [C2]------------->

There are 3 possibilities:

1. [C1] < [C2], so log [C2]/[C1] > 0, so G > 0

2. [C1] > [C2], so log [C2]/[C1] < 0, so G < 0

3. [C1] = [C2] so log [C2]/[C1] = 0, so G = 0 Which happens spontaneously?

The standard equation: G = 2.3 RT log [C2]/[C1]

2. Solute transport

G = 2.3 RT log [C2]/[C1] transport is from C1 to C2

[C1] > [C2] means that log [C2]/[C1] < 0so G < 0

This can happen spontaneously, without energy input

C1 C2

------------->

2. Solute transport

G = 2.3 RT log [C2]/[C1]

transport is from C1 to C2

[C2] = [C1] means log[C2]/[C1] = 0

So, G = 0, equilibriumC1 C2

<---------->

Numerical example

[C2] = 100mM, [C1] = 10mM 37 0C

How much energy to transport a mole of C?

C1 C2

------------->

G = 2.3 RT log[C2]/[C1]

Dimensional analysis - do the units make sense?

G R T log[C2]/[C1]

Units: J mol-1 = J mol-1 K-1 K mol l-1/mol l-1

J mol-1 = J mol-1

G = 2.3 RT log[C2]/[C1]

G = (2.3)(8.314 J mol-1 K-1)(310 0K) log(100/10)

= 2.3 x 8.314 x 310 x 1 J mol-1

= 5928 J mol-1 or 5.928 kJ mol-1

Energy is required to move a solute “up” a concentration gradient

Now fill in the numbers

C1 C2

------------->

3. G for ion transport: ions are charged solutes

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.

Fig. 6.4

K+

NO3-

Ca+2

SO4-2

Which ion requires the mostenergy to move acrossthe membrane, assuming thesame concentration gradientfor all four?

Biological membranes are electrically polarized, like a battery.

3. G for ion transport

G = zF Em

z is charge on the ion: K+ = +1, NO3

- = -1, Ca+2 = +2, SO4-2 = -2

other molecules have not net charge

F is Faraday’s constant = 9.65 x 104 J vol-1 mol-1

Em is membrane potential, volts

NO3-

Example: uptake of NO3- against -0.15 volt potentialG = zF Em

G = (-1)x9.65x104 J Volt-1 mol-1 x (-0.15Volt) = 1.45 x 104 J mol-1 = 14.5 kJ mol-1

4. Movement along electrical and concentration gradients

G = zF Em + 2.3 RT log(C2/C1)

note rearrangment as Em = -2.3 RT/zF log(C2/C1)

R = 8.314 J mol-1 K-1

z is charge of soluteF is Faraday’s constant = 9.65 x 104 J vol-1 mol-1 Em is membrane potential, volts

Enzyme kineticsWhat are enzymes?

What kinds of molecules are they made of?

What do they do to reaction rates?

How do they work?

What’s the world’s most abundant enzyme?

What conditions affect the rate of enzyme-catalyzed reactions?

reaction rate, V(moles of productper second)

Substrate concentration, S(moles/liter)

Vmax

V

S (substrate concentration)

Vmax

1/2 Vmax

Km

Vmax x S Km + SV =

Michaelis-Menten equation

Enzyme specificity is not perfect,other molecules can compete for the active site.

“competitive inhibition”

Enzyme structure can be modified by other molecules,reducing enzyme activity.

“non competitive inhibition”

Competitive inhibitor increases Km and does not affect Vmax, but a higher [S] is required to reach Vmax.Mechanisms1. Competitive inhibitor binds at same active site as

substrate, making less enzyme available to catalyze E+S reaction.

2. Competitive inhibitor binds at another site on enzyme, causing a conformational change in active site that reduces affinity for the primary substrate. “allosteric inhibitor”.

Substrate concentration

With competitiveinhibitor

No inhibitor

Conditions affecting enzyme activity.

1. pHEnzymes have an optimum pH at which activity is maximum,with sharp declines in activity at lower and higher pH.

pH affects enzyme activity by altering ionization state of active site or byaffecting the 3-D conformation of the active site.

2. TemperatureEnzyme activity has an optimum temperature, with sharp declines in activity at lower and higher pH.

Reaction rates increase with temperature because enzymes and reactantsare moving faster and have higher probability of encountering one another.

Enzyme activity decreases at temperatures high enough to cause“denaturation”, the unfolding of protein structure and loss of proper conformation for catalysis.