Transcript Document

Wed. Sept 5
Metabolism -- ~P Homeostasis
The Ce ntral Role of ATP in Ce llular Ene rge tics
Storage Compounds
(Carbohy drates, Fats)
Catabolic
Pathways
(Breakdown of Complex
Molecules, Aerobic and
Anaerobic ~P
production)
(SUPPLY)
(Spontane ous)
Complex Molecules(poteins, lipids, complex carbs,
protein complexes
OR
Movement and transportrelated processes requiring
enery such as muscle contraction and activ e transport)
ATP
ADP + Pi
Anabolic
Pathways
and
Movement
(DEM AND)
(NonSpontane ous )
knp
Waste Products
(Mainly CO2
and Water)
Raw Materials (Amino Acids,Simple Carbs, etc; also
protein conf ormationsin the case of mov ement
The Measurement of
Metabolism
We will not directly follow these processes any more
than a chemist does who measures the heat evolved or
taken in by a process.
Methods all involve assumptions -- in a way it is like
measuring the economy and no one measure is perfect.
Phosphagen Cycle -Phosphagen Buffer During
High Demand
creatine kinase
C~P + ADP ----------------------> ATP + Creatine
demand reactions
Pi + ADP <----------------------- ATP
This process is controlled by the amount of enzyme
present (CK) and thermodynamically (amts. of
reactants/products)
Replenishment of
Phosphagen
creatine kinase
Creatine + ATP ----------------------> C~P + ADP
aerobic metabolism
ATP <----------------------- Pi + ADP
The Krebs Cycle
Take Home:
•Occurs in the
mitochondrial
matrix
•Rate is very
dependent on the
ratio of oxidized to
reduced coenzyme
•Feed ins -- 2 C
frags.. from
carbohydrate and
fat metabolism;
•
Amino acid
frags in other
places
Energy Schematic of the ETS
Take Home
•In inner
membrane
•Will maintain a
favorable
steady-state
ratio of oxidized
to reduced
coenzyme if
sufficient O2 is
present to
accept each
electron
produced in the
Krebs and other
reactions.
Mitochondrial Overall
Energy Schematic
Glycolysis
Energy
Schematic
Aerobic
Glycolysis
Overview
Energetics of Aerobic
Glycolysis
The overall reaction (path does not matter):
C6H12O6 + 6 O2 ----> 6 H2O + 6 CO2 - 2872 KJ
A distinct stoichiometry exists between all members
of this process.
This means that if we know carbohydrate is the fuel, then
if we measure the change in one component of this
reaction, we know the changes in all others.
How About Lipid?
Palmitic acid + 23 O2 --> 16 CO2 + 16 H2O - 9795 kJ
Once again, if we know that lipid (palmitic acid)
is the fuel, we can measure one factor in the
reaction and find all the others.
How do we know the fuel
being used?
You cannot tell simply from what the animal eats.
There is a certain ratio of CO2 production to O2
consumption exists for different fuels. For carbs:
C6H12O6 + 6 O2 ----> 6 H2O + 6 CO2 - 2872 KJ
Ratio is 1:1
The Ratio is Different In
Palmitic Acid
Palmitic acid + 23 O2 --> 16 CO2 + 16 H2O - 9795 kJ
16/23 = 0.7
The Respiratory Quotient (RQ)

RQ 

V CO 2

V O2

M CO 2

M O2
C6H12O6 + 6 O2 ----> 6 H2O + 6 CO2 - 2872 KJ
Palmitic acid + 23 O2 --> 16 CO2 + 16 H2O - 9795 kJ

RQ
0.71 (fat)
0.8 (protein)
0.85 mixed metabolism
(typical in mammals)
0.9
1.0 (carbohydrate)
Energy per Liter of O2
Consumed in Kcal
and (KJ)
4.69 (19.6)
4.8 (20.1)
4.86 (20.3)
Energy per Liter of CO 2
Released in Kcal
and (KJ)
6.61 (27.6)
6.00 (25.1)
5.72 (23.9)
4.92 (20.6)
5.05 (21.1)
5.47 (22.9)
5.05 (21.1)
How the Table Values For Energy Equivalence
Were Obtained – An example
C6 H12O6 +6O2 Ä 6CO2 + 6H 2O - 2872kJ
-2872 kJ
Energy Equivalent of 1 LO2 =
6 mols O2 *22.4 L O2 mol-1