Chapter 18- Respiration

Question 1

What biological processes require ATP?

Answer 1

• active transport
• endocytosis
• exocytosis
• synthesis of large molecules such as proteins, enzymes and antibodies
• DNA replication
• cell division
• muscle contraction
• activation of chemicals
• metabolism

Question 2

The need for cellular respiration

Answer 2

➜ Maintaining body temp
➜ Movement (mechanical contraction of muscles or cellular movement of chromosomes)
➜ Anabolic reactions (synthesis of DNA)
➜ Transporting substances across membrane (Exocytosis of digested bacteria from WBC)

Question 3

Source of energy for Photo

Answer 3

➜ light energy from sun transformed into chemical potential during synthesis of carbohydrates
➜ carbohydrates then used in synthesis of ATP or are combined and modified to form all usable organic molecules essential for metabolic processes

remember autotrophs make own food

6 CO2 + 6 H20 ➜ C6H1206 + 6 O2

Question 4

Source of energy for resp

Answer 4

➜ from the breakdown of organic molceules
➜ involves transfer of chemical potential energy to a usable energy form from nutrients

heterotrophs eat food

C6H1206 + 6 O2 → 6 CO2 + 6 H20 + 2870kJ

Question 5

Mitochondrion

Answer 5

➜ rod shaped organelles and are 0.5 -1.0µm in diameter
➜ site of aerobic resp in eukaryotic cells
➜ synthesize ATP

Question 6

Structure of mitochondrian

Answer 6

➜ 2 phospholipid membranes:
Outer membrane is smooth and permeable to small molecules
➜ The inner membrane:
↳ folded to form cristae
↳ less permeable
↳ site of e- transport chain
↳ location of ATP synthase enzymes
➜ The intermembrane space:
↳ low pH due to high conc of protons
↳ conc grad across inner membrane formed during oxidative phos
➜ The matrix
↳ aq solution in inner membane of mito
↳ contains ribosomes, enymes and mito DNA

Question 7

Adaptations of Mitochondrian

Answer 7

➜ Large SA due to cristae - membrane can hold many e- transport chain proteins and ATP synthase enzymes
➜ More active cell types = larger mito with longer/tighter packed cristae to SA large

Question 8

4 stages of aerobic resp

has a greater yield tha anaerobic

Answer 8

➜ glycolysis
➜ link reaction
➜ krebs cycle
➜ oxidative phosphorylation

Question 9

Glycolysis

Answer 9

➜ takes place in cytoplasm
➜ traps glucose in cell and phosphorylated by 2 ATP to make fructos bisphosphate (6C)
➜ Lysis occurs where Fructose bisphosphate splits into 2 molecules of triose phosphate (3C)
➜ H removed from each triose phosphate and transferred to NAD to form 2 reduced NAD
➜ phosphates are transferred from intermediate substrate molecules to form 4 ATP
➜ pyruvate is produced at the end

Products: 2 pyruvate (3C), net 2 ATP, 2 reduced NAD (NADH)

Question 10

Link reaction

Answer 10

➜ when oxygen is available pyruvate (3C) enters mitochondrial matrix
➜ pyruvate moves across double membrane of mito via active transport (requires transport protein and smol amount of ATP)
➜ pyruvate is then oxidised (when NAD is reduced to NADH) by enzymes to produce acetate (2C) & CO2
➜ Coenzyme A (ribose and adenine and a vitamin) binds to acetate to form acetyl CoA

Products: Acetyl CoA, CO2, NADH

Question 11

Krebs cycle

Answer 11

➜ matrix
➜ Acetyl CoA (2C) enters cycle and combines with (4C) oxaloacetate to form citrate (6C)
➜ CoA (ribose and adenine and a vitamin) is released in this reaction
➜ citrate goes through decarboxylation where it releases CO2 as waste and product has 5C
➜ NAD is reduced
➜ The 5C loses another CO2 forming a 4C and NADH produced again
➜ ADP is then phophorylated to make ATP and FAD is reduced
➜ Another NADH formed
➜ This produces oxaloecetate which restarts the cycle again

Products: 2CO2, 3NADH, FADH2, ATP

Process occurs twicer per glucose so Products doubled

Question 12

Chemiosmotic theory

Oxidative phosphorylation

Answer 12

➜ chemiosmotic theory:
↳ energy from e- passed through a chain of protein membranes (e- transport chain) is used to pump protons up conc grad into intermembrane space
↳ H move via faciliated diffusion through a channel in ATP synthase into matrix
↳ energy of H allows phosphorylation of ADP into ATP by ATP synthase

Question 13

Oxidative phosphorylation

Answer 13

➜ inner mito membrane - forms many ATP and water
➜ H donated by reduced NAD and reduced FAD from krebs cycle
↳ H split into H+ and e-
↳ high energy electrons enter e- transport chain and release eergy as they move through
↳ released energy is used to transport protons across inner mito membrane from matric into intermembrane space
↳ conc grad of protons formed between intermembrane and marix
↳ protons return to matrix via faciliated diffusion through ATP synthase (channel proein)
↳ movement of protons provide energy for ATP synthesis
↳ Oxygen accepts electrons and combines with protons to form water
↳ process forms many ATP

Question 14

Coenzymes

Answer 14

➜ CoA = has a nucleoside(ribose and a purine/pyrimidine base) and a vitamin
➜ NAD & FAD are hydrogen carriers
➜ 2 NADH in glycolysis & Link reaction but 6 in krebs
➜ 2FADH2 from krebs
➜ Remove H = oxidised and Add H = reduced

Question 15

Consequences of no O2 or little O2

(Anaerobic resp)

THIS IS AN EXAM Q BTW

Has a lower yield as glucose is only partially oxidised

Answer 15

➜ No final acceptor of e- from chain
➜ e- chain dont function
➜ no more ATP via oxidative phosphorylation
➜ NADH and FADH2 aren’t oxidised by e- carrier
➜ no NAD or FAD available for dehydrogenation in krebs cycle
➜ krebs stop

Question 16

Ethanol fermentation

Yeast

Answer 16

➜ glycolysis goes on as normal
➜ pyruvate is decarboxylated to ethanAL which produced CO2
➜ NADH transfers H to ethanAL to form ethanOL
➜ ethanOL is a waste product (can’t be further metabolised duh its alcohOL)

Question 17

Lactate fermentation

Answer 17

➜ NADH transfers H to pyruvate so it is reduced to lactate by enzyme dehydrogenase
➜ Lactate can be metabolised.

Question 18

Metabolization of lactate

Answer 18

2 options:
➜ can be oxidised back to pyruvate
↳ then channelled into Krebs for ATP production
↳ this requires extra O2 which is known as oxygen debt
OR
➜ converted into glycogen
↳ stored in liver

Question 19

Respiratory substances

Carbohydrate

Answer 19

15.8 kJ g⁻¹

Question 20

Respiratory substances

Protein

Answer 20

17.0 kJ g⁻¹

Question 21

Respiratory substances

Lipid

Answer 21

39.4 kJ g⁻¹

Question 22

Molecular composition

Answer 22

➜ molecule with higher H content will result in a greater proton grad allowing of more ATP formation
➜ Lipid has the most H so has the highest energy val

Question 23

RQ

Answer 23

CO2/O2

Question 24

Respirometer

Answer 24

➜ used to measure and investigate the rate of oxygen consumption during aerobic respiration in organisms
∘ Set up respirometer - soda lime in control and experimental tube to absorb CO2 and glass beads in control tube and a respiring organism in the other tube (beads and organism must be same mass)
∘ test tubes are connected by a manometer which contains a coloured liquid
∘ run the experiment with both tubes in a temp controlled water bath
∘ use manometer reading to calc change in gas vol in a given time
∘ reset the apapratus by allowing air to reenter tubes via screw cap and reset manometer fluid via syringe
∘ change temp of water btah and allow tubes to acclimate then close screw clip
∘ use manometer reading to calc change in gas vol in same given time as before
∘ repeat at diff temps

πr²h = vol of O2 consumption (use diameter of tube and distance fluid moved)