Unit 2 - Biology Notes
, CELLULAR RESPIRATION
Why do we need ATP?
∗ For every major process such as:
1. Active transport
2. Catabolic reactions
3. Muscle contractions
4. Nervous system transmissions
Sources of ATP
Ways of Producing ATP (text page 95 96)
1. Substrate Level Phosphorylation (anaerobic)
In this process, the formation of ATP is coupled with an energy yielding process
(exergonic). More energy is released from the particular chemical reaction than is required
to bond the phosphate group onto the ADP. ATP is produced directly.
ADP + P + energy (from exergonic reaction) ATP
An example of this type occurs in glycolysis as well as lactic acid fermentation. It is the
fast and direct production of ATP.
1
,EFFECIENCY OF ANAEROBIC RESPIRATION:
1 mole of glucose = 686 kilocalories (x 4.18 to convert to kilojoules)
1 mole of glucose produces 2 moles of ATP (anaerobic)
Energy in 1 mole of ATP = 7.3 kilocalories ∴2 moles release 2 x 7.3 = 14.6 kilocalories
Efficiency = 14.6 (trapped) X 100 = 2%
686 (could be trapped)
∴ only 2% efficient
2. Chemiosmotic ATP Generation (Oxidative phosphorylation)
In this case, ATP is generated when protons move back through the membrane via protein
proton “pumps”. This is how most of the ATP in the cell is made.
There are 2 types of chemiosmotic phosphorylation depending on the source of the
electrons that drive the proton pumping system:
a) If the source of energy is from light, then this is called photosynthetic
photophosphorylation.
If the source of energy is from the oxidation of food, then this is referred to as oxidative
phosphorylation. The electrons in this case are extracted from redox reaction that involve the
breakdown of high energy molecules which we call food. This process is called cellular
respiration or, because it involves the use of oxygen, oxidative respiration. This method is
indirect and slower than substrate phosphorylation.
2
, EFFECIENCY OF AEROBIC RESPIRATION:
1 mole of glucose = 686 kilocalories
1 mole of glucose produces 36 moles of ATP
1 ATP = 7.3 kilocalories ∴ 36 ATP = 262.8 kilocalories
Efficiency = 262.8 X 100 = 38.3%
686
∴ 38.3% efficient that is good!
3
, CELLULAR RESPIRATION
Why do we need ATP?
∗ For every major process such as:
1. Active transport
2. Catabolic reactions
3. Muscle contractions
4. Nervous system transmissions
Sources of ATP
Ways of Producing ATP (text page 95 96)
1. Substrate Level Phosphorylation (anaerobic)
In this process, the formation of ATP is coupled with an energy yielding process
(exergonic). More energy is released from the particular chemical reaction than is required
to bond the phosphate group onto the ADP. ATP is produced directly.
ADP + P + energy (from exergonic reaction) ATP
An example of this type occurs in glycolysis as well as lactic acid fermentation. It is the
fast and direct production of ATP.
1
,EFFECIENCY OF ANAEROBIC RESPIRATION:
1 mole of glucose = 686 kilocalories (x 4.18 to convert to kilojoules)
1 mole of glucose produces 2 moles of ATP (anaerobic)
Energy in 1 mole of ATP = 7.3 kilocalories ∴2 moles release 2 x 7.3 = 14.6 kilocalories
Efficiency = 14.6 (trapped) X 100 = 2%
686 (could be trapped)
∴ only 2% efficient
2. Chemiosmotic ATP Generation (Oxidative phosphorylation)
In this case, ATP is generated when protons move back through the membrane via protein
proton “pumps”. This is how most of the ATP in the cell is made.
There are 2 types of chemiosmotic phosphorylation depending on the source of the
electrons that drive the proton pumping system:
a) If the source of energy is from light, then this is called photosynthetic
photophosphorylation.
If the source of energy is from the oxidation of food, then this is referred to as oxidative
phosphorylation. The electrons in this case are extracted from redox reaction that involve the
breakdown of high energy molecules which we call food. This process is called cellular
respiration or, because it involves the use of oxygen, oxidative respiration. This method is
indirect and slower than substrate phosphorylation.
2
, EFFECIENCY OF AEROBIC RESPIRATION:
1 mole of glucose = 686 kilocalories
1 mole of glucose produces 36 moles of ATP
1 ATP = 7.3 kilocalories ∴ 36 ATP = 262.8 kilocalories
Efficiency = 262.8 X 100 = 38.3%
686
∴ 38.3% efficient that is good!
3
