Chapter 18 Respiration Flashcards
Describe the structure of mitochondria.
Mitochondria has a double membrane. There is the outer membrane, which has a role of compartmentalisation. The inner membrane is highly folded to form a structure called cristae. The cristae contains proteins for the electron transport chain. As the cristae is folded, this also means it has a large surface area for oxidative phosphorylation. The area of space between the outer and inner membrane is called the intermembrane space. The space within the cristae is called the matrix, which contains enzymes and mitochondrial DNA.
State the stages of aerobic respiration.
Glycolysis.
Link Reaction.
Krebs cycle.
Electron transport chain.
State the stages of anaerobic respiration in mammals and yeast.
In mammals> glycolysis which produces lactate.
In yeast> Glycolysis which produces ethanol and carbon dioxide (fermentation).
Where does the first stage of respiration take place and is oxygen needed?
Glycolysis takes place in the cytoplasm of the cell and no oxygen is needed.
In glycolysis, glucose is first phosphorylated. What does it require and what is produced through phosphorylation?
To phosphorylate glucose, 2 ATP molecules are required (which turns into 2 ADP), and it produces hexose bisphosphate.
In glycolysis, hexose bisphosphate splits to form 2 triose phosphate. After this, how is each triose phosphate converted to triose bisphosphate?
Each triose phosphate is phosphorylated again to form 2 triose bisphosphate. This phosphorylation does not require ATP molecules but is brought about by inorganic phosphate ions in the cytoplasm.
In glycolysis, how is triose bisphosphate converted to pyruvate?
In each compound of triose phosphate, there are 2 phosphate ions present. Each triose bisphosphate gets their phosphate ions (2 Pi) taken from them by 2 ADP molecules. These 2 ADP molecules are converted to 2 ATP molecules through the taking of phosphate ions from triose bisphosphate. The taking of 2 phosphate ions converts triose bisphosphate to pyruvate.
Additionally, each triose bisphosphate is oxidised by the removal of hydrogen atoms (dehydrogenation). The removal of these hydrogen atoms are donated to the coenzyme NAD to form reduced NAD (NADH).
The conversion of triose bisphosphate to pyruvate is described as substrate level phosphorylation. Why is this?
In the conversion of triose bisphosphate to pyruvate, 4 ATP molecules are formed (2 from each triose bisphosphate molecule). It is substrate level phosphorylation as this is the production of ATP without the use of an electron transport chain and ATP synthase. ADP is phosphorylated by an unstable intermediate to form ATP.
What are the products of the glycolysis stage in respiration?
2 pyruvate molecules.
Net gain of 2 ATP molecules.
2 reduced NAD molecules.
Where does the Link reaction take place?
In the mitochondrial matrix.
How does products of glycolysis go from the cytoplasm into the mitochondrial matrix?
The products, pyruvate, enters the mitochondrial matrix through active transport via specific carrier proteins in the mitochondrial membrane.
In order to form the acetyl group from pyruvate in the Link reaction, the pyruvate is said to have undergone ‘oxidative decarboxylation’. Why is this?
‘Decarboxylation’- Removal of carbon dioxide from pyruvate (a carbon is removed to give a 2-carbon molecule- the acetyl).
‘Oxidative’- Pyruvate loses a hydrogen, and hence, is oxidised. This hydrogen is accepted by NAD to form NADH.
What is the final product of the Link reaction?
How does the intermediate, the acetyl group, form this?
The final product of the Link reaction is acetylCoenzyme A. It is formed when the acetyl group bonds with Coenzyme A, to temporarily stabilise the structure so acetylCoenzyme A can carry the acetyl group to the next stage, the Kreb’s cycle.
The first intermediate of the Kreb’s cycle is citrate (citric acid). How is it formed?
The acetyl group, from acetylCoenzyme A of the Link reaction, is donated into the Kreb’s cycle. It reacts with a 4-carbon molecule, oxaloacetate, to form citric acid (which is 6 carbon).
After citrate is formed in the Kreb’s cycle, the citrate forms a 5 carbon intermediate. This 5 carbon intermediate then goes to form a 4 carbon intermediate. Through what processes are these intermediates formed?
To form the 5 carbon intermediate and 4 carbon intermediate, dehydrogenation and decarboxylation occurs.
Dehydrogenation- The intermediate loses a hydrogen atom, in which NAD accepts to form NADH.
Decarboxylation- The intermediate loses a carbon dioxide.