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Respiration Flashcards

1
Q

Describe the importance of ATP in cells, giving 2 examples of processes in which it is used

A
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2
Q

(M/J 03) Compare the relative amounts of ATP produced by the substrate level phosphorylation and oxidative
phosphorylation when a molecule of glucose is completely oxidised. [2]

A
  1. Oxidative phosphorylation more than substrate level phosphorylation.
  2. 32 ATP per glucose in oxidative phosphorylation compared with 4 ATP per glucose in substrate level
    phosphorylation
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3
Q

(M/J 15 42) State 3 reasons why ATP is ideal as an energy currency in all living organisms. [3]
(O/N 14 43) Describe how the structure of ATP is related to its role as energy currency. [3]
(M/J 05) Explain how ATP is able to transfer energy in cells. [3]

A
  1. ATP is synthesised from ADP and Pi. Rate of interconversion of ATP and ADP is high.
  2. ATP is a small and soluble molecule which diffuses rapidly and can easily be transported around the cell.
  3. ATP is easily hydrolysed.
  4. On hydrolysis, energy released is 30.5 kJ mol–1.
  5. ATP links catabolic and anabolic reactions. It is a universal intermediate molecule between energy yielding
    and energy requiring reactions.
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4
Q

(O/N 08) ATP is described as having a universal role as the energy currency in all living organisms. Explain why it is
described in this way. [4]

A
  1. ATP is synthesised from ADP and Pi. Rate of interconversion of ATP and ADP is high.
  2. ATP is a small and soluble molecule which diffuses rapidly and can easily be transported around cell.
  3. ATP is easily hydrolysed.
  4. On hydrolysis, energy released is 30.5 kJ mol-1.
  5. ATP links catabolic and anabolic reactions. It is a universal intermediate molecule between energy yielding
    and energy requiring reactions.
  6. Found in all organisms.
  7. ATP is produced from a variety of reactions.
  8. The energy released on hydrolysis is used in processes/ reactions ie muscle contraction/ protein synthesis/
    DNA replication/ cell movement/ active transport.
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5
Q

(O/N 08) State precisely 2 places where ATP is synthesised in cells. [2]

A
  1. Inner mitochondrial membrane
  2. Grana/ thylakoids/ inner chloroplast membrane
  3. Cytoplasm
  4. Mitochondrial matrix
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6
Q

(M/J 13 42) The production of ATP by oxidative phosphorylation takes place in the electron transport chain in a
mitochondrion. State the part of the mitochondrion in which the electron transport chain is found. [1]

A
  1. Inner membrane/ cristae
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7
Q

Explain how ATP is synthesised in substrate-linked reactions in glycolysis and in the Krebs cycle.

A
  1. Energy is released by reorganising chemical bonds is used to make ATP (chemical potential energy).
  2. During glycolysis, 2 molecules of ATP are used and 4 molecules of ATP are made, making a net gain of 2 ATP
    per glucose.
  3. During Krebs cycle, 2 molecules of ATP are made.
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8
Q

O/N 14 43) All living organisms require a continuous supply of energy. Outline the need for energy in living
organisms. [2]

A
  1. Synthesis of complex substances or synthesis of named large molecule/ anabolic reactions.
  2. Transport of substances against concentration gradient/ active transport.
  3. Energy is needed for movement such as muscle contraction/ cilia movement/ locomotion.
  4. Energy is needed for bioluminescence/ electrical discharge/ temperature regulation.
    (Ignore ref. to energy currency)
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9
Q

(M/J 16 41) 1 (b) ATP provides an immediate energy source for metabolic processes such as anabolic reactions.
State two examples of anabolic reactions in a mammal that require ATP as an energy source. [2]

A

accept synthesise/ produce/ convert to for ‘make’ for all marking points
1. Make protein/ polypeptide/ peptides.
(Accept protein synthesis/ translation)
2. Make disaccharide/ oligosaccharide/ polysaccharide/ glycogen.
(Reject non-mammalian examples such as starch or cellulose)
3. Make triglycerides/ lipids/ phospholipids/ steroids/ cholesterol.
(Accept glycogenesis)
4. Make nucleotide/ polynucleotide/ nucleic acid/ DNA/ RNA.
(Accept transcription/ DNA replication)
5. Alternative Valid Point; e.g. named example of polymerisation/ condensation
(Accept phosphorylation example)
Generic terms like polymerisation, condensation and phosphorylation needed to be qualified with a specific
reaction example.

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10
Q
A

A – adenine R adenosine
B – ribose/ pentose

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11
Q

(O/N 14 41) State two ways in which the structure of ATP differs from the structure of an adenine nucleotide in a
DNA molecule. [2]

A
  1. ATP contains ribose (not deoxyribose).
  2. ATP has three phosphate groups (not one)
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12
Q
A
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13
Q

(b) Explain the consequences to a mitochondrion if the water potential of the liquid in the dishes is higher than the
water potential of the mitochondrial matrix. [2]

A
  1. Water enters the mitochondrial matrix
  2. by osmosis (down the water potential gradient).
  3. This results in the membranes rupturing and the bursting of mitochondrion.
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14
Q

(O/N 02) Role of NAD in respiration [3]

A
  1. NAD is a coenzyme.
  2. NAD becomes reduced in glycolysis, link reaction and Krebs cycle. NAD carries protons and electrons to ETC
    from glycolysis, link reaction and Krebs cycle.
  3. When NADH is reoxidised in the ETC, energy released is used to form ATP. 2.5 molecules of ATP are produced
    per reduced NAD.
  4. NAD is also involved in the oxidation of triose phosphate to pyruvate in glycolysis.
  5. Ethanal is reduced to ethanol by reduced NAD in alcohol fermentation and pyruvate is reduced by reduced
    NAD during lactate fermentation.
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15
Q

(M/J 16 41) 1 (d) Outline the roles of NAD in the cytoplasm of a cell. [2]

A
  1. Hydrogen carrier/ acceptor
    (Accept gets reduced or gains H/ H+ and electrons, Ignore donates, Reject H2/ hydrogen molecules)
  2. Acts as a coenzyme
    (Accept enables dehydrogenases to work)
  3. Allows glycolysis to continue during anaerobic respiration
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16
Q

(O/N 03) Explain why NAD cannot be regenerated from reduced NAD in mitochondria in the absence of oxygen. [3]

A
  1. Oxygen is the final hydrogen acceptor at end of electron transfer chain.
  2. In absence of oxygen electron transfer chain does not work.
    * Reduced NAD cannot be oxidized.
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17
Q

(M/J 10 42) Explain the role of reduced NAD in respiring yeast cells in the absence of oxygen. [4]

A
  1. Pyruvate is decarboxylated to ethanal using pyruvate decarboxylase.
  2. Ethanal reduced by reduced NAD to ethanol using ethanol dehydrogenase.
  3. Pyruvate acts as alternative hydrogen acceptor and NAD is regenerated.
    * This allows glycolysis to continue.
    * Prevents H+ from lowering pH.
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18
Q
A
  1. R - pyruvate
  2. S - carbon dioxide
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19
Q

(ii) explain what happens to the reduced NAD. [2]

A
  1. Hydrogen(s)/ protons and electrons are released
  2. at ETC for oxidative phosphorylation.
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20
Q

(O/N 17 42 7(b)(ii)) State the role of acetyl coenzyme A in respiration. [1]

A
  1. Carrier of 2C (unit)/ acetyl group/ acetate to the Krebs cycle/ oxaloacetate
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21
Q

(New syllabus) Outline the role of FAD in respiration.

A
  1. FAD is coenzyme.
  2. FAD becomes reduced in Krebs cycle. FAD carries protons and electrons to ETC from Krebs cycle.
  3. When FAD is reoxidised in the ETC, energy released is used to form ATP. 1.5 molecules of ATP is produced
    per FAD.
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22
Q

(New syllabus) Outline the role of coenzyme A (CoA) in respiration.

A
  1. In link reaction, pyruvate is decarboxylated, dehydrogenated and combined with CoA to give acetyl CoA.
  2. CoA acts as a carrier of acetyl group to the Krebs cycle.
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23
Q
A
  1. Both have ribose (sugars) (Reject ribulose)
  2. ATP has 1 ribose/ pentose/ sugar, NAD has 2 (Ignore ref. to additional)
  3. Hexose both have adenine/ purine (base) (Ignore adenosine)
  4. NAD has nicotinamide/ pyrimidine (base)
  5. ATP has 3 phosphates, NAD has 2
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24
Q
A
  1. Three phosphates
  2. Contains ribose / pentose
  3. Contains adenine
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25
(ii) Describe the role of coenzyme A in respiration. [3]
1. Coenzyme A combines with acetyl group/ acetate. 2. Coenzyme A is used in link reaction. 3. Coenzyme A delivers acetyl group/ acetate to the Krebs cycle 4. The acetyl group/ acetate combines with oxaloacetate.
26
1. Muscle/ liver cell
27
ii) State the mechanism by which glucose enters the cell. [1]
1. Facilitated diffusion
28
(iii) Name the type of reaction occurring at F and the type of reaction occurring at G. [2]
F – condensation/ polymerisation/ anabolic/ glycogenesis/ dephosphorylation G – hydrolysis/ catabolic/ glycogenolysis/ phosphorylation
29
(iv) With reference to Fig. 1.2, suggest one example of an ‘other metabolic pathway’ for phosphorylated glucose. [1]
1. Glycolysis/ respiration/ lipid synthesis
30
(M/J 06) Explain why the hexose is converted to substance hexose bisphosphate [2]
1. Hexose is energy rich. 2. However, it does not react easily. 3. Phosphorylation provides activation energy. 4. Activation process is required for the subsequent splitting of fructose 1,6-bisphosphate. 5. Phosphorylation of glucose also maintains concentration gradient of glucose inside and outside cell.
31
(O/N 11 41) Pyruvate can enter in mitochondrion when oxygen is present. Describe what happens to pyruvate in a yeast cell when oxygen is not present. [4] (M/J 06) Briefly describe what happens to pyruvate if yeast is deprived of oxygen. [4]
1. It does not enter Krebs cycle. 2. Pyruvate is decarboxylated to ethanal, with enzyme pyruvate decarboxylase. 3. Reduced NAD does not enter ETC. 4. Ethanal is reduced by reduced NAD to ethanol, with the enzyme alcohol dehydrogenase. 5. The reaction is irreversible.
32
(M/J 11 41) State 1 use of glucose within the liver cell. [1]
1. Converted to glycogen/ lipid 2. Used in glycolysis/ respiration
33
(M/J 11 41) Glucose is phosphorylated at the start of glycolysis in the muscle cell. Suggest why this phosphorylated glucose does not diffuse out of the cell into the surrounding tissue fluid. [2]
1. Phosphorylated glucose cannot pass through phospholipid bilayer. 2. It is too big to fit through the glucose's protein channel. 3. There is no specific transport protein which allows it to cross the phospholipid bilayer. 4. It might be used up as soon as it is made.
34
(O/N 11 41) Explain why glucose needs to be converted to hexose bisphosphate. [2]
1. This provides activation energy for it to split Activation process is required for the subsequent splitting of hexose bisphosphate.
35
(i) State the precise location of glycolysis in the cell. [1]
1. cytoplasm
36
state the steps where phosphorylation occurs:
Steps 1 and 3
37
state the step where oxidation occurs:
Step 5
38
name the type of reaction by which ATP is made during step 5:
substrate-linked phosphorylation
39
(b) Some cancer cells have different metabolic requirements from normal cells. These cancer cells obtain most of their ATP from glycolysis, even if oxygen is available. State how the glucose and oxygen requirements of these cancer cells differ from normal cells. [2]
1. Cancer cells need more glucose. 2. Cancer cells need less oxygen. 3. Cancer cells get little energy per glucose molecule.
40
41
(b) Uncontrolled cell division can lead to a cancerous tumour. Many cancer cells break down the amino acid glutamine and convert it to a 5-carbon intermediate compound, which is shown in Fig. 6.1. Suggest how the breakdown of glutamine can lead to the production of ATP in a cancer cell, other than that directly produced during step 3. [2]
1. Reduced NADs are passed to ETC 2. for oxidative phosphorylation.
42
1. Dehydrogenase enzymes provide H+ and electrons to form reduced NAD/ reduced FAD, which are then passed to ETC for oxidative phosphorylation 2. Electrons and protons are accepted by oxygen to form water. 3. This reaction is catalysed by cytochrome oxidase.
43
Suggest an explanation for this effect. [2]
1. Initial step rise up to 40 micromol Al 2. 2 paired figures 3. Ref Plateau above 80 micromol Al 1. Initially Al is activator/ cofactor/ coenzyme and changes the shape of the active site to fit the substrate 2. Enzyme/ substrate limiting after 40 micromol
44
Decarboxylation
Removal of carbon dioxide/ carboxyl group [1]
45
Dehydrogenation
Removal of hydrogen [1]
46
(M/J 10 43) State where the reduced NAD molecules are reoxidised and describe what happens to the hydrogen atoms. [5]
1. Reduced NAD are reoxidised to NAD in the ETC in the inner mitochondrial membrane by dehydrogenase enzymes and this releases hydrogen. 2. Hydrogen then splits into protons and electrons. 3. Electrons flow down ETC and the energy released is used to pump protons across inner membrane into intermembrane space, creating a proton gradient. 4. H+ diffuses down the concentration gradient through stalk particle, synthesising ATP from ADP and Pi 5. Oxygen acts as final acceptor of protons and electrons to form water. This reaction is catalysed by cytochrome oxidase.
47
(M/J 16 41) 1 (c) Name the type of chemical reaction by which ATP is made during the Krebs cycle. [1]
1. substrate-linked/ substrate-level phosphorylation (Ignore condensation reaction) Few candidates stated that substrate-linked phosphorylation makes ATP during the Krebs’ cycle. Non-scoring answers included ‘oxidative phosphorylation’ or just ‘phosphorylation’.
48
(O/N 17 42) 7 (a) In respiration, most ATP is synthesised during oxidative phosphorylation. Some ATP is made by substrate-linked reactions in glycolysis and Krebs cycle. Describe how ATP is made by substrate-linked reactions. [2]
1. Inorganic phosphate is added to ADP/ ADP + Pi 2. Phosphate is donated by a phosphorylated compound such as phosphoenolpyruvate (PEP). Give named phosphorylated compound
49
(b) Lipids can be metabolised to provide ATP.
* The enzyme lipase hydrolyses lipids to glycerol and fatty acids. * The hydrocarbon chain of the fatty acid breaks down into smaller, 2C compounds. * Each 2C compound reacts with coenzyme A to form acetyl coenzyme A.
50
(i) Name the covalent bond in lipids that is hydrolysed by lipase. [1]
1. Ester
51
(O/N 17 41) Name the enzyme used to produce ATP in chemiosmosis. [1]
1. ATP synth(et)ase
52
(M/J 13 42) The production of ATP by oxidative phosphorylation takes place in the electron transport chain in a mitochondrion. State the part of the mitochondrion in which the electron transport chain is found. [1]
1. Inner membrane / cristae
53
(M/J 05) State how ATP is synthesised in mitochondria. [4]
1. ATP is synthesised in mitochondria through substrate level phosphorylation and oxidative phosphorylation. 2. Oxidative phosphorylation is explained below in the theory of chemiosmosis. 3. Reduced NADs are reoxidised to NAD in the ETC located at the inner mitochondrial membrane by dehydrogenase enzymes and this releases hydrogen. 4. Hydrogen then splits into protons and electrons. 5. Electrons flow down ETC and the energy released is used to pump protons across inner membrane from the matrix into intermembrane space, creating a proton gradient. 6. H+ diffuses down the concentration gradient through stalk particle, synthesising ATP from ADP and Pi 7. Oxygen acts as final acceptor of protons and electrons to form water. This reaction is catalysed by cytochrome oxidase.
54
(M/J 10 41) (M/J 11 42) Outline the process of oxidative phosphorylation. [5]
1. Reduced NAD are reoxidised to NAD in the ETC located at the inner mitochondrial membrane by dehydrogenase enzymes and this releases hydrogen. 2. Hydrogen then splits into protons and electrons. 3. Electrons flow down ETC and the energy released is used to pump protons across the inner membrane from the matrix into intermembrane space, creating a proton gradient. 4. H+ diffuses down the concentration gradient through stalk particle, synthesising ATP from ADP and Pi 5. Oxygen acts as final acceptor of protons and electrons to form water. This is catalysed by the enzyme cytochrome oxidase.
55
(M/J 13 42) Describe briefly where the electrons that are passed along the electron transport chain come from. [3]
1. The electron comes from the hydrogen atom (Reject H+/H2) 2. which is produced through the oxidation of reduced NAD 3. during Krebs cycle/ link reaction/ glycolysis 4. in the matrix of mitochondrion (for Krebs cycle and link reaction)/ in cytoplasm (for glycolysis). The matrix of the mitochondrion is the site of the link reaction and the Krebs cycle, and contains the enzymes for these reactions.
56
(O/N 05) Explain how NAD is regenerated. 3m
1. Reduced NAD are reoxidised to NAD in the ETC located at the inner mitochondrial membrane by dehydrogenase enzymes and this releases hydrogen.
57
(O/N 17 41) State the specific role of oxygen in the mitochondrion. [1]
1. Acts as the final electron (and proton) acceptor in the ETC.
58
(M/J 13 42) Describe the role of oxygen in the process of oxidative phosphorylation. [2]
1. Oxygen acts as final acceptor of protons and electrons at the end ETC. 2. Oxygen combines with H+ to form water. 3. Since reduced NAD is oxidised to NAD, NAD is regenerated and can be reduced again. 4. ETC can continue to work as electrons can keep flowing along the ETC.
59
(M/J 04) Explain how the lack of oxygen will affect the respiratory processes in the mitochondria. [3] (M/J 03) Explain why oxidative phosphorylation is not possible in the absence of oxygen. [3]
1. No oxygen to combine with hydrogen at the end of the ETC 2. Without oxygen, reduced NAD cannot be oxidized and Krebs cycle stops 3. ETC cannot function as there is no electron flow. 4. No H+ gradient is produced and oxidative phosphorylation cannot take place. 5. No ATP is synthesized through chemiosmosis.
60
(M/J 13 41) Running requires rapid use of ATP by muscle cells in the legs and heart of a lizard. With reference to the events occurring inside a mitochondrion, explain why a faster use of ATP requires a greater rate of oxygen consumption. [4]
1. ATP is made in the electron transport chain/ by oxidative phosphorylation 2. The energy released from the transfer of electron (between electron carriers) is used to pump protons across inner membrane from the matrix into intermembrane space, creating a proton gradient. 3. H+ diffuses down the concentration gradient through stalk particle, synthesising ATP from ADP and Pi. 4. Oxygen acts as final acceptor of protons and electrons to form water. This reaction is catalysed by cytochrome oxidase. 5. If less oxygen (consumed/available) then fewer electrons are transferred along the chain.
61
(O/N 15 43) 1 (b) Dinitrophenol (DNP) is a compound used as a herbicide. DNP inhibits respiration by interfering with the formation of the proton gradient between mitochondrial membranes. When DNP was added to isolated mitochondria the following observations were made: * fewer ATP molecules were produced * more heat energy was released * the uptake of oxygen remained constant. Suggest explanations for these observations. [3]
fewer ATP molecules produced 1. No/ fewer protons/ H+ move through ATP synth(et)ase/ stalked particles or less steep proton/ H+ gradient (Ignore chemiosmosis) more heat energy released 2. H+ gradient/ electron flow/ ETC energy converted to heat/ thermal energy constant oxygen uptake 3. ETC still works/ oxygen acts as final electron acceptor (Ignore oxidative phosphorylation still works)
62
1. Nuclei 2. Ribosome
63
(M/J 10 41) Explain why carbon dioxide is produced when mitochondria are incubated with pyruvate but not when they are incubated with glucose. [3
1. Glycolysis does not occur in mitochondrion/ only occurs in cytosol or cytoplasm. 2. Pyruvate is produced in glycolysis. 3. Pyruvate can enter mitochondrion/ glucose cannot enter mitochondrion. 4. Carbon dioxide is produced/ decarboxylation in Krebs/ link reaction.
64
.(M/J 10 41) Explain why in the presence of cyanide, lactate is produced but carbon dioxide is not. [3]
1. Cyanide is a non-competitive enzyme inhibitor and inhibits cytochrome oxidase. 2. As a result, reduced NAD is not oxidised and Krebs cycle stops. 3. In anaerobic respiration pyruvate is converted to lactate by lactate dehydrogenase after glycolysis. 4. Pyruvate acts as alternative H acceptor and NAD is regenerated. * This allows glycolysis to continue.
65
66
67
(O/N 02) State stage of respiration for each of the structure below. [2]
1. Matrix – Krebs cycle 2. Cristae/inner mitochondrial membrane – Oxidative phosphorylation
68
(O/N 02) Describe how the structure of a mitochondrion is adapted to carry out these two processes. [3]
1. Membranes separate matrix of mitochondrion from rest of cytoplasm – allows different pH between cytoplasm and the matrix 2. Inner membrane folded to form cristae – large internal surface area 3. Inner membrane allows attachment of stalk particles, which act as channels for H+/ ATP synthesis 4. Linear arrangement of carriers in the ETC allows for greater efficiency in ATP synthesis 5. Matrix contains enzymes for Krebs cycle
69
(M/J 18 43) 7 (b) Explain why it is an advantage to the cell for the inner membrane of the mitochondrion to be folded. [2]
1. Increase in surface area 2. More carriers/ ATP synthase/ ETCs 3. More ATP is produced
70
71
(M/J 04) Describe the 2 ways in which the structure of the mitochondrion is adapted for oxidative phosphorylation. [4]
1. Intermembrane space * allows accumulation of H+ 2. Folded inner membrane/ cristae * increases surface area available 3. Impermeability of inner membrane to H+ * maintains H+ gradient/ H+ only go through channels 4. Stalked particles * act as channels for H+/ ATP synthesis 5. Linear arrangement of ETC on inner membrane * allows for greater efficiency in ATP synthesis
72
(O/N 17 41) Describe the role of the inner mitochondrial membrane (crista) in chemiosmosis. [4]
1. Site of electron transport chain. 2. Protons are pumped across the inner membrane from the matrix into intermembrane space, 3. creating a proton gradient. 4. Protons/ H+ diffuse to matrix 5. through stalked particles/ ATP synth(et)ase. 6. ADP + Pi → ATP 7. Oxidative phosphorylation takes place.
73
(M/J 14 41) Suggest the functions of the DNA and ribosomes in a mitochondrion. [3]
1. DNA for transcription of mRNA/ codes for mRNA. 2. Ribosomes for translation and 3. synthesis of respiratory enzymes/ named enzyme/ inner membrane proteins.
74
(O/N 13 41) Describe the structure of a mitochondrion and outline its function in a plant cell. [8]
1. A mitochondrion is 0.5–1.0 µm in diameter. 2. Each mitochondrion is surrounded by an envelope of 2 phospholipid membranes. 3. The outer membrane is smooth, but the inner in much folded inwards to form cristae, giving the inner membrane a large surface area. 4. The inner membrane is studded with tiny spheres, about 9nm in diameter, which are attached to the inner membrane by stalks. The spheres are the enzyme ATP synthase. 5. The inner membrane is the site of electron transport chain and contains carrier proteins necessary for this. 6. The intermembrane space allows accumulation of H+ for ATP production during oxidative phosphorylation. 7. The matrix of the mitochondrion is the site of the link reaction and the Krebs cycle, and contains the enzymes for these reactions. 8. It also contains 70s ribosomes and several identical copies of looped mitochondrial DNA.
75
76
(O/N 13 41) Describe the consequences for the cell of the following statements. * Each cell has only a very small quantity of ATP in it at any one time. * The molecules, ATP, ADP (adenosine diphosphate) or AMP (adenosine monophosphate) rarely pass through the cell surface membrane. [2]
1. Cell uses ATP as source of energy. 2. ATP is broken down to release energy. 3. Cell regenerates ATP 4. from ADP and Pi. 5. ADP is synthesised from AMP in the cell.
77
1. The mean ratio for group A is higher than B. 2. Group A mice have larger inner membrane in their mitochondria than group B, with more ETCs/ cytochromes/ ATP synth(et)ase/ stalked particles. 3. This enables more H+ to diffuse down the concentration gradient through stalked particles. 4. There is thus more oxidative phosphorylation, resulting in more ATP being produced. 5. Group A mice’s muscles can contract for a longer period of time without getting tired.
78
79
80
(ii) Suggest reasons for the difference in mean number of mitochondria per µm3 between the fat cell and the heart cell. [2]
1. Mitochondria/ respiration produce(s) ATP. 2. Heart/ cardiac muscle cell is more active OR needs more energy/ ATP. 3. Heart/ cardiac muscle cell contracts.
81
(iii) Suggest why it is difficult to compare the ability of the three types of cell to respire aerobically, based only on the mean number of mitochondria per cell. [1]
1. Mitochondria vary in size/ in surface area of inner membrane.
82
W - ethanal (or acetaldehyde/ C2H4O) X - carbon dioxide/ CO2 Y - reduced NAD (or NADH/ NADH2/ NADH+ + H+)
83
(O/N 03) Suggest why the rate of glycolysis increases significantly when yeast cells switch from aerobic to anaerobic respiration. [2]
1. Aerobic respiration produces more ATP than anaerobic respiration 2. to produce the same amount of ATP more glucose broken down in glycolysis. 3. Glycolysis is the only part of respiration used/ no ETC or oxidative phosphorylation.
84
(M/J 18 43) 8 (c) Muscle cells sometimes have to carry out respiration in anaerobic conditions. Describe how respiration occurs in anaerobic conditions in a muscle cell and state why it is important that this process occurs. [4]
description 1. Pyruvate is reduced by reduced NAD that is produced from glycolysis and is converted to lactate by lactate dehydrogenase. 2. The conversion is reversible. importance 3. This allows glycolysis to continue. 4. Anaerobic respiration produces a small amount of ATP.
85
(O/N 04) Describe how the lactate that appears in the blood is formed. [3]
1. After glycolysis pyruvate is converted to lactate by lactate dehydrogenase. 2. In shortage of oxygen to muscles, pyruvate acts as hydrogen acceptor. NAD is regenerated. * This allows glycolysis to continue.
86
(O/N 04) Outline how blood lactate is linked to oxygen debt. [3]
1. Lactate must be oxidised. 2. The extra oxygen required is the oxygen debt.
87
(O/N 04) Suggest why the buildup of lactate occurs at a higher workload in the distance runner. [1]
1. More anaerobic respiration/ insufficient oxygen supply.
88
(O/N 11 43) Describe how anaerobic respiration in mammalian cells differ from anaerobic respiration in yeast cells. [3] (M/J 10 43) Describe how the production of lactate in muscle tissue differs from anaerobic respiration in yeast. [3]
1. Lactate is produced. (No point awarded for this in 2010) 2. No decarboxylation/ carbon dioxide released. 3. The enzyme involved is lactate dehydrogenase. 4. Pyruvate is converted into lactate in a single step, 5. and the conversion is reversible.
89
M/J 11 41) Suggest why anaerobic respiration is said to be less efficient than aerobic respiration. [2]
Anaerobic 1. Less ATP/ only 2 ATP produced per mol glucose 2. Lactate still contains energy 3. Only glycolysis involved/ stages other than glycolysis are not involved 4. Not sustainable/ cannot go on indefinitely Anaerobic respiration * Glycolysis Aerobic respiration * Glycolysis * Krebs cycle * Oxidative phosphorylation
90
(O/N 14 41) Explain why less ATP can be synthesised from the same mass of glucose in anaerobic respiration than in aerobic respiration. [3] (O/N 11 43) Explain why anaerobic respiration results in a small yield of ATP compared with aerobic respiration. [3] In anaerobic respiration
1. Only glycolysis occurs/ Krebs cycle stops/ link reaction stops 2. Glucose not fully broken down/ (pyruvate/ lactate/ ethanol) still contains energy * Pyruvate does not enter mitochondrion 3. No oxygen so no final electron acceptor in ETC 4. ETC cannot function as there is no electron flow. 5. No H+ gradient is produced and oxidative phosphorylation cannot take place.
91
(O/N 15 43) 1(a)(ii) (ii) State two differences between anaerobic respiration in yeast cells and anaerobic respiration in human muscles cells. [2
In yeast cells – or reverse argument for muscle cells 1. Ethanol is produced as opposed to lactate/ lactic acid. 2. Carbon dioxide is released. 3. Different dehydrogenases are involved/ reduction of ethanal instead of pyruvate. 4. Pyruvate is converted into ethanol in two steps and the conversion is irreversible. 5. Two steps/ two enzymes are involved/ decarboxylation/ ref. to (pyruvate) decarboxylase/ CO2 production. Alternatively, reverse argument for muscle cells In muscle cells 1. Lactate is produced. 2. No decarboxylation/ no carbon dioxide is released. 3. The enzyme involved is lactate dehydrogenase. 4. Pyruvate is converted into lactate in a single step and the conversion is reversible.
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93
1. Ethanol evaporated. 2. Other microorganism metabolised ethanol.
94
1. At high temperatures, reactions/enzyme activity/metabolism is faster 2. because molecules/enzymes/substrates have more kinetic energy 3. and hence collisions become more frequent. 4. Therefore respiration/Krebs cycle/electron transport chain/production of reduced NAD takes place at a faster rate. 5. Idea of increase in rate of anabolic reactions (requiring more ATP).
95
1. Oxygen consumed = oxygen inhaled – oxygen exhaled 2. Measure oxygen consumption at rest (x) and after exercise stops (y) 3. Extra oxygen consumed/oxygen debt = y – x 4. Measure mass of lizard
96
1. Less oxygen debt for Varanus 2. Difference is greater at higher temperatures 3. Any two comparative figures at one temperature including units A 102.0 cm3 O2 kg-1 at 30°C and 40°C
97
(iii) Varanus is a fast-moving carnivore. Sauromalus is a slow-moving herbivore. Explain how the results in Table 4.1 indicate that Varanus is well-adapted for its mode of life. [3]
1. Varanus uses less anaerobic/more aerobic respiration when running. 2. More ATP is produced per glucose molecule. 3. Varanus is able to run for long time and 4. has a good chance of catching prey.
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(iv) Most lizards, including Sauromalus, have very simple lungs with no alveoli. Varanus, however, has lungs that are more like those of mammals, containing large numbers of air sacs similar to the alveoli of human lungs. Suggest how this difference could account for the differences in the oxygen debts of Sauromalus and Varanus shown in Table 4.1. [2]
assume Varanus throughout 1. Larger surface area in lungs for gas exchange. 2. More oxygen absorbed into blood (per unit time)/faster rate of gas exchange. 3. More oxygen supplied to muscles (so oxygen debt lower).
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(O/N 13 43) (O/N 16 41) 8 (ii) Outline the role of oxygen in aerobic respiration. [3]
1. In the process of oxidative phosphorylation, 2. oxygen acts as final acceptor of protons and electrons at the end of ETC. 3. Oxygen combines with H+ to form water. 4. ETC can continue to work as electrons can keep flowing. 5. Oxygen increases ATP production 6. as in the absence of oxygen only glycolysis continues.
100
(M/J 14 41) Some parasitic worms, such as tapeworms, live in a mammalian gut where there is no oxygen. Suggest how a tapeworm produces ATP in this environment. [5]
1. Through anaerobic respiration. 2. Triose phosphate is oxidised to pyruvate in glycolysis. 3. Pyruvate is then reduced by reduced NAD that is produced from glycolysis and is converted to lactate by lactate dehydrogenase. 4. NAD is regenerated, and 5. this allows glycolysis to continue. 6. There is a production of 4 ATP per glucose, with a net gain of 2 ATP per glucose.
101
(O/N 14 43) Thermus thermophilus is a bacterium found in geothermal environments, such as hot springs. The bacterium respires aerobically, even though at high temperatures the solubility of oxygen in water is low. (i) Explain how aerobic respiration may be affected by a decrease in oxygen availability. [2]
1. Less/ decreased aerobic respiration 2. Oxygen is the final electron acceptor/ needed for ETC 3. Oxidative phosphorylation is decreased/ chemiosmosis is decreased 4. Regeneration of NAD/ Kreb’s cycle/ link reaction is decreased 5. ATP synthesis decreases/ ATP synthetase activity is decreased
102
(ii) One strain of T. thermophilus, HB8, has an enzyme, nitrate reductase, which allows nitrate to be used as the final electron acceptor in the electron transport chain (ETC). Suggest an advantage to the bacterium of this adaptation. [1]
1. More ATP produced (for population growth)
103
(i) Compare the growth of the two strains of bacteria in aerobic and anaerobic conditions in separate cultures. [2]
1. HB8 always does better than mutant HB8. 2. HB8 and mutant HB8 both do better in aerobic than in anaerobic conditions. 3. data quote to support: for mp1 [950 × 106 per cm3 v 900 × 106 per cm3] and [490 × 106 per cm3 v 410 × 106 per cm3] or manipulated figures for mp2 [950 × 106 per cm3 v 490 × 106 per cm3] and [900 × 106 per cm3 v 410 × 106 per cm3] or manipulated figures [max 2]
104
(ii) Compare the growth of the two strains of bacteria in aerobic and anaerobic conditions in mixed cultures. [2]
1. Both grow better in aerobic compared to anaerobic. 2. There is significant difference found in mutant HB8 (aerobic compared to anaerobic). 3. data quote to support for mp1 [880 × 106 per cm3 v 460 × 106 per cm3] and [840 × 106 per cm3 v 50 × 106 per cm3] or manipulated figures for mp2 [840 × 106 per cm3 v 50 × 106 per cm3] or [460 × 106 per cm3 v 50 × 106 per cm3] or manipulated figures
105
(iii) Suggest an explanation for the results shown in flask 6. [1]
1. HB8 is a better competitor than mutant HB8. 2. In mutant HB8 activity of enzyme/ nitrate reductase is reduced.
106
(O/N 05) State what is meant by the term respiratory quotient (RQ). [1] (M/J 15 42) State how RQ is calculated. [1]
107
(M/J 15 42) Give the typical RQ values obtained from the respiration of carbohydrates and lipids. [2]
1. Carbohydrates – 1.0 2. Lipids – 0.7
108
(M/J 15 42) Suggest what happens to RQ value when respiration in yeast becomes anaerobic. [1]
1. Becomes greater than 1
109
O/N 05) Explain why carbohydrates release half as much energy per unit mass as fats and oils. [2]
1. Less C-C bonds 2. Less C-H bonds 3. More oxygen A O R O2
110
(O/N 06) Explain the relative energy values of carbohydrate and lipid as respiratory substrates. [3]
Carbohydrates 1. Less reduced/ less hydrogen/ less C-H bonds 2. for aerobic respiration/ ETC/ NAD/ ATP. 3. Less energy released 4. per unit mass/ mole. 5. Carbohydrate has lower energy density.
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1. RQ remains stable between 3 °C and 10 °C 2. Rise between 10 °C and 20 °C 3. 0.74 to 0.76/ 0.8 4. Sharp rise between 25 °C and 27 °C/ after 25 °C 5. 0.8 to 0.91/ peaks at 0.91 6. At low temperatures hamster used lipids. 7. More heat is generated from lipid respiration. 8. At higher temperatures more carbohydrates are used
112
(O/N 06) State a circumstance under which RQ value would rise over 1.0. [1]
1. Anaerobic respiration/ conversion of carbohydrate to fats as animal hibernates.
113
1. RQ value falls steeply initially/ 40-80 min. 2. Then very little change. 3. Sugar/ carbohydrate is metabolised at the start. 4. Then fat is metabolised 5. due to fasting/ carbohydrate running out
114
(M/J 09) Hummingbirds regulate their body temperature whereas butterflies do not regulate their body temperature. Explain briefly the effect of an increase in temperature on the rate of respiration of a butterfly. [2]
1. Increased in rate of respiration. 2. Kinetic energy increases/ more enzyme-substrate complexes/ enzyme activity increases. 3. Too high a rise in temperature leads to denaturation of enzymes.
115
1. Palmitic acid has more C-H bonds 2. per mole. 3. Hydrogen is needed for ATP production/ chemiosmosis/ oxidative phosphorylation.
116
(ii) Describe the circumstances in which alanine and lactate are used as respiratory substrates.
Alanine - starvation/ lack of fat or carbohydrate. Lactate - after anaerobic respiration.
117
1. Lipid releases most energy 2. because it has more C-H bonds 3. per unit mass. 4. More hydrogen for ATP production/ oxidative phosphorylation/ chemiosmosis.
118
(ii) Suggest why more water is produced from the metabolism of lipid than from the other two substrates.
1. More C-H bonds, more hydrogens available to reduce oxygen to water.
119
(O/N 14 41) Explain why more ATP can be synthesised in aerobic respiration from one gram of lipid than from one gram of glucose. [3]
1. Lipid contains more C-H bonds per unit mass. 2. More reduced NAD/ FAD produced. 3. More electrons passed along ETC. 4. More hydrogen for ATP production/oxidative phosphorylation/chemiosmosis.
120
1. Lipids contain more C-H bonds per unit mass. 2. More reduced NAD/FAD produced. 3. More electrons passed along ETC, more ATP produced per unit mass. 4. More hydrogen for ATP production/oxidative phosphorylation/chemiosmosis. 5. Fats only broken down aerobically.
121
122
123
124
(O/N 02) Describe how photophosphorylation differs from oxidative phosphorylation [3]
1. Light is involved 2. Occurs in chloroplast/ thylakoid 3. On thylakoid membranes 4. Photophosphorylation can be cyclic or non-cyclic 5. Photolysis of water produces oxygen Oxidative phosphorylation 1. Light not involved 2. Oxygen final hydrogen acceptor/ oxygen not evolved
125
(O/N 05) State how the formation of ATP in Krebs cycle differs from the formation of ATP in oxidative phosphorylation. [1]
1. Substrate level phosphorylation 2. No ETC 3. No proton gradient involved 4. No ATP synthase
126
127
128
O/N 16 43) 8(c) In anaerobic conditions, the pyruvate formed in glycolysis is converted to ethanol in yeast cells and to lactate in mammalian tissue. Compare the pathways by which pyruvate is converted to ethanol or to lactate. [5]
129
130
(i) State the role of potassium hydroxide solution in the use of a respirometer. [1]
1. Absorbs carbon dioxide
131
(c) Respirometers, as shown in Fig. 8.1, were used to investigate the effect of temperature on the rate of respiration of germinating pea seeds. Four respirometers, A, B, C and D were set up: * A and B in a water-bath maintained at 10 °C. * C and D in a water-bath maintained at 25 °C. * A and C each contained 30 germinating pea seeds. * B and D each contained glass beads with a total volume equivalent to 30 pea seeds. * The respirometers were left in the water-baths for 10 minutes. * In each respirometer the position of the coloured liquid in the graduated tube was then marked (time 0 minutes). * After 5 minutes the distance moved by the coloured liquid was measured. * The volume of oxygen taken up was calculated for each respirometer. * This was repeated after 10, 15 and 20 minutes.
1. Equilibration / acclimatising / adjusting
132
ii) Suggest why respirometers B and D were used in this investigation. [2]
1. To act as a control. 2. Control eliminates effects of variables other than the independent variable/ temperature. 3. Changes in A and C are due to seeds/ respiration.
133
(iii) Calculate the rate of oxygen uptake in cm3 per minute for respirometer C between 5 and 20 minutes. Give your answer to two significant figures. Show your working. [2]
134
(iv) Explain why there is an increased rate of respiration of germinating pea seeds between 10°C and 25°C. [2] at 25°C (ora for 10°C)
1. Enzymes are involved. 2. There would be an increase in kinetic energy and 3. consequently more enzyme-substrate complexes formed.
135
v) Suggest why carrying out the experiment with germinating seeds at 50 °C could result in a lower rate of respiration than at 25 °C. [2]
1. Enzymes would be denatured. 2. There would be a change in the active site or the 3D shape of the enzyme.
136
1. Potassium hydroxide/ sodium hydroxide (solution)
137
(ii) The respirometer can be used to measure the effect of temperature on the rate of respiration of organisms. Suggest one factor that would need to be taken into account when using woodlice rather than germinating seeds. [1]
1. Must not raise temperature too high for animals 2. Problem of the woodlice moving around
138
iii) As respiration takes place, oxygen is used by the woodlice and the coloured liquid moves down the graduated tube. Name the stage of aerobic respiration where oxygen is used. [1]
1. Oxidative phosphorylation
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(ii) Explain the difference in the rates of oxygen uptake at 15 °C and 25 °C. [3] at 25 °C (ora for 15 °C)
1. The higher temperature (and hence increased kinetic energy) results in enzymes/ substrates/ molecules moving faster. 2. More collisions between enzyme and substrate. 3. More ESCs are formed. 4. Q10 = 2
141
(O/N 10 41) Using examples, outline the need for energy in living organisms. [9]
1. ATP acts as a universal energy currency. 2. Light energy is needed for photosynthesis. 3. ATP is used for the conversion of GP to TP. 4. ATP is used to regenerate RuBP. 5. Energy is needed for anabolic reactions. 6. Examples of anabolic reactions are protein synthesis/ starch formation/ triglyceride formation. 7. Activation energy 8. is needed to activate glucose in glycolysis. 9. Energy is needed in active transport. 10. such as when the sodium/ potassium pump is working 11. Energy is needed for movement/ locomotion, 12. for example, muscle contraction/ cilia beating. 13. Energy is needed for endocytosis/ exocytosis/ pinocytosis/ bulk transport. 14. Energy is needed for temperature regulation.
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(O/N 17 42) 9 (a) Describe how the structure of a mitochondrion is related to its function. [8]
1. Each mitochondrion is surrounded by an envelope of 2 phospholipid membranes. 2. The outer membrane is smooth, but the inner in much folded inwards to form cristae, giving the inner membrane a large surface area. 3. The inner membrane is studded with tiny spheres, about 9nm in diameter, which are attached to the inner membrane by stalks. The spheres are the enzyme ATP synthase. 4. The inner membrane is the site of electron transport chain and contains carrier proteins necessary for this. 5. The space between the two membranes of the envelope usually has a lower pH than the matrix of the mitochondrion. 6. The matrix of the mitochondrion is the site of the link reaction and the Krebs cycle, and contains the enzymes for these reactions.
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(M/J 2018 41) 9 (a) Describe the features of ATP that make it suitable as the universal energy currency. [6]
1. ATP is a short-term store of energy, 2. where the energy is derived from food/ respiration/ photophosphorylation/ chemiosmosis. 3. ATP transfers energy to other molecules in cellular processes. 4. ATP is found in all organisms. 5. ATP is synthesised from ADP and Pi. Rate of interconversion of ATP and ADP is high. 6. ATP is a small and soluble molecule which diffuses rapidly and can easily be transported around cell. 7. ATP is easily hydrolysed. On hydrolysis, energy released is 30.5 kJ mol-1. 8. ATP links catabolic and anabolic reactions. It is a universal intermediate molecule between energy yielding and energy requiring reactions. The energy released on hydrolysis is used in: (max 2m) 9. active transport/ action potential/ electrical discharge 10. muscle contraction 11. anabolic reactions/ condensation reactions/ transcription/ translation/ DNA replication/ Calvin cycle/ phosphorylation reactions 12. exocytosis/ endocytosis/ intracellular transport 13. bioluminescence
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(O/N 09 41) Describe the structure of ATP and its universal role as the energy currency in all living organisms. [8]
1. A nucleotide with 3 phosphate groups, an organic nitrogenous base (adenine) and a pentose sugar (ribose). 2. ATP is synthesised from ADP and Pi. Rate of interconversion of ATP and ADP is high. 3. ATP is synthesised through oxidative phosphorylation in mitochondria/ chloroplasts with the enzyme ATP synthase. 4. ATP is easily hydrolysed. On hydrolysis, energy released is 30.5 kJ mol-1. 5. ATP links catabolic and anabolic reactions. It is a universal intermediate molecule between energy yielding and energy requiring reactions. 6. The energy released on hydrolysis is used in processes/ reactions ie muscle contraction/ protein synthesis/ DNA replication/ cell movement/ active transport.
145
(M/J 12 42) Explain the role of ATP in active transport of ions and in named anabolic reactions. [7]
1. ATP provides energy for active transport and anabolic reactions. Active transport 1. Requires energy because movement of molecule or ion against concentration gradient. 2. Carrier protein in membrane binds to specific ion and the protein changes shape to transport the molecule or ion across the membrane Anabolic reactions 1. Synthesis of complex substances from simpler ones. 2. Formation of glycosidic bonds between monosaccharides to form starch. 3. Formation of ester bonds between fatty acids and glycerol to form triglyceride. 4. Formation of peptide bonds between amino acids to form protein. 5. Accept other named polymer from suitable monomer.
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(O/N 03) Explain the role of NAD in cellular respiration. [3]
1. NAD is a coenzyme. 2. NAD becomes reduced in glycolysis, link reaction and Krebs cycle. 3. NAD carries protons and electrons to ETC from glycolysis, link reaction and Krebs cycle. 4. When NADH is reoxidised in the ETC, energy released is used to form ATP. 5. 2.5 molecules of ATP are produced per reduced NAD.
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(O/N 04) Explain the role of NAD in aerobic respiration. [6]
1. NAD is a coenzyme for dehydrogenase. 2. NAD becomes reduced in glycolysis and Krebs cycle. 3. NAD carries protons and electrons to ETC from glycolysis, and Krebs cycle. 4. When NADH is reoxidised in the ETC, energy released is used to form ATP. 5. 2.5 molecules of ATP is produced per reduced NAD
148
(M/J 08) Explain the roles of NAD is anaerobic respiration in both plants and animals [6]
In cytoplasm 1. NAD, becomes reduced by accepting H 2. during glycolysis. In plants 1. Pyruvate is converted into ethanal. 2. Ethanal is reduced 3. by reduced NAD 4. to form ethanol. In animals 1. Pyruvate is converted to lactate 2. by reduced NAD 3. in muscles. 4. This allows glycolysis to continue.
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(O/N 18 42) 9 (a) Explain why carbohydrates, lipids and proteins have different relative energy values as substrates in respiration in aerobic conditions. [6]
1. Different substrates have different numbers of hydrogen atoms/ C-H bonds. 2. Lipids have relatively more hydrogen atoms/ C-H bonds than carbohydrates or proteins. 3. These hydrogen atoms/ C-H bonds are located in fatty acid (tails of lipids). 4. Breakdown/ oxidation of substrate provides hydrogen atoms 5. for reduction of NAD/ FAD. 6. Reduced NAD/ FAD provides/ releases hydrogen to ETC. 7. The hydrogen in turn dissociates into protons and electrons. 8. The energy released used to set up proton gradient. 9. Chemiosmosis results in ATP production. 10. Thus, there is more ATP/ energy from lipids per unit mass than carbohydrates/ proteins or lipids more energy dense/ have higher relative energy value.
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(O/N 10 41) Explain the different energy values of carbohydrates, lipid and protein as respiratory substrates. [6]
1. Idea lipid > protein > carbohydrate/ AW; A lipid has more energy than either protein or carbohydrate. 2. Comparative figure 39.14, 17.0, 15.8 3. kJ/ g; per unit mass. 4. More hydrogen atoms in a molecule means more energy could be yielded. 5. Lipid have more hydrogen atoms/ C-H bonds. 6. Most energy comes from oxidation of hydrogen to water using reduced NAD/ FAD in ETC. 7. Chemiosmosis results in ATP production.
151
(O/N 18 42) 9 (b) Define the term respiratory quotient (RQ) and describe how you would carry out an investigation to determine the RQ of germinating barley seeds. [9]
152
O/N 09 41)(M/J 16 42 9(a)) Describe the process of glycolysis. [7]
1. Glucose is phosphorylated by addition of 2 P molecules which come from the hydrolysis of 2 ATP to 2 ADP and fructose bisphosphate is formed. 2. Phosphorylation of glucose results in its activation. 3. Fructose bisphosphate (6C) then splits R sugar splitting 4. into 2 molecules of triose phosphate (3C). 5. Dehydrogenation occurs. 2 reduced NAD are formed, with 1 reduced NAD formed for the oxidation of a single TP to pyruvate 6. 4 ATPs are produced with a net gain of 2 ATPs. 7. 2 pyruvates are produced. 8. Reduced NAD are passed to inner mitochondrial membrane for oxidative phosphorylation/ redox.
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(O/N 04) (O/N 09 42) (M/J 16 42 9(a)) Outline the main features of the Krebs cycle. [9]
1. Acetyl CoA combines with oxaloacetate (4C) 2. to form citrate (6C). 3. The citrate is decarboxylated and dehydrogenated in a series of steps, to yield carbon dioxide and hydrogens which are accepted by the carriers NAD and FAD. 4. It is a substrate level phosphorylation which involves a series of steps 5. and enzyme catalysed reactions . 6. At the end of the cycle, oxaloacetate is regenerated. 7. Occurs in mitochondrial matrix.
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(M/J 08) Describe the process of oxidative phosphorylation in mitochondrion. [9] (M/J 16 41 9 (a)) Outline how ATP is synthesised by oxidative phosphorylation. [8]
1. Reduced NAD and FAD are passed to the ETC which is located at the inner mitochondrial membrane to be oxidised by dehydrogenase enzymes and this releases hydrogen. 2. Hydrogen splits into electrons and protons. 3. Electrons flow down ETC as it is passed from one carrier to the next, moving downhill in energy terms. 4. The energy released is used to pump protons across inner membrane from the matrix into intermembrane space, creating a proton gradient. 5. H+ diffuses down the concentration gradient through stalk particle, synthesising ATP from ADP and Pi. 6. Oxygen acts as final acceptor of protons and electrons to form water. 7. This is oxidative phosphorylation explained in the theory of chemiosmosis.
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(M/J 2018 41) 9 (b) Outline respiration in anaerobic conditions in mammalian cells and describe how it differs from respiration in anaerobic conditions in yeast cells. [9] (O/N 10 43) Outline anaerobic respiration in mammalian cells and describe how it differs from anaerobic respiration in yeast cells. [7]
1. Pyruvate cannot enter mitochondrion/ remains in cytoplasm. [no mark] 2. It becomes the alternative hydrogen acceptor. 3. Pyruvate is reduced by reduced NAD that is produced from glycolysis and 4. is converted to lactate by lactate dehydrogenase. 5. This allows glycolysis to continue. 6. There is no decarboxylation/ CO2 removed. 7. Pyruvate is converted into lactate in a single step and the conversion is reversible. 8. Lactate could be converted back to pyruvate by oxidation. 9. The liver oxidizes some (20%) of the incoming lactate to carbon dioxide and water via aerobic respiration when oxygen becomes available again. The oxygen needed to allow this removal of lactate is called the oxygen debt. 10. Ethanol is not produced in anaerobic respiration in mammalian cells.
156
(M/J 12 42) Outline the process of anaerobic respiration in both mammal and yeast cells. [8]
1. Reduced NAD is produced by glycolysis. 2. Small amount of ATP is produced by glycolysis. In yeast cells 1. Pyruvate is converted to ethanal. 2. Carbon dioxide is released/ decarboxylation occurs. 3. Ethanal is reduced by reduced NAD to ethanol, with the enzyme alcohol dehydrogenase. 4. The reaction is irreversible. In mammalian cells 1. Pyruvate is converted to lactate 2. by reduced NAD 3. in muscles. 4. Pyruvate is converted into lactate in a single step and the conversion is reversible.
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(M/J 16 41) 9(b) Describe respiration in yeast cells in anaerobic conditions. [7]
1. Pyruvate is formed by glycolysis. 2. Reduced NAD is produced in glycolysis. 3. Pyruvate is converted to ethanal. 4. Carbon dioxide is released/ decarboxylation occurs. 5. Ethanal is reduced by reduced NAD to ethanol, with the enzyme alcohol dehydrogenase. 6. NAD is regenerated and can now accept hydrogen atoms so glycolysis can continue.
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