2020 H2 A level Paper 3

Section A

Answer all questions.

1 Animals that possess hair are classified within the phylum Chordata and class Mammalia.

(a) Define biological classification and explain how classification relates to phylogeny.[3]
1. Biological classification groups organisms based on overall morphological similarities and
usually does not consider their evolutionary history;

2. It uses a naming system where each organism is given a binomial name* and grouped
into a domain, kingdom, phylum, class, order, family, genus and species in a hierarchical
manner;

3. Phylogeny also groups organisms but they are grouped based on their evolutionary history
/ ancestor-descendent relationships using molecular data and;

4. It does not rank organisms but instead assigns each organism a position on a
phylogenetic tree* based on its evolutionary relationship with other organisms on the
tree;

(b) (i) Epithelial stem cells and melanocyte stem cells are examples of a particular type of
stem cell.

State two key features of this type of stem cell. [2]

1. They are multipotent* stem cells that can differentiate into a limited range of
specialized cell types in the presence of the correct molecular signals;

2. They are able to undergo extensive proliferation and self-renewal by mitosis* in the
presence of the correct molecular signals and hence give rise to a constant pool of
cells;

(ii) Discuss how the presence of stem cells within the skin and the presence of hairs
covering the skin influence the harmful effects of high levels of ultraviolet light on the
health of mammals. [4]

1. The presence of hairs covering the skin play a protective role as they reduce
exposure of the skin by blocking the harmful ultraviolet radiation;
2. Since stem cells are near the skin surface, the high levels of ultra violet light can
damage the DNA is stem cells as;
3. the ionizing radiation can cause the DNA is the stem cells in the skin to mutate and
if these mutations are in genes that regulate cell cycle checkpoints/proto oncogenes/
tumour suppressor genes;
4. it could lead to uncontrolled cell division/ tumour formation and increase the chances
of an individual getting cancer;
5. especially because stem cells already have telomerase activity;

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(c) (i) The shape of MC1R is characteristic of a particular group of transmembrane receptor
proteins.

Name this group of proteins.[1]

7 pass transmembrane G-protein coupled receptor *

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(ii) In cats with agouti hairs, ASIP is secreted at intervals during hair growth.

With reference to Table 1.1 and the information provided, explain how this interrupted
pattern of ASIP secretion leads to striped agouti hairs in cats.[4]

1. During intervals when ASIP is not secreted, α-MSH binds to MC1R (a GPCR),
causing the MC1R to undergo a conformation change in its intracellular domain
allowing G-protein to bind to it; and

2. G-protein is activated when it displaces its attached GDP for GTP;

3. Activated G-protein will translocate along membrane and bind to enzyme

adenylyl cyclase and phosphorylate it, thus activating it;

4. Adenylyl cyclase will catalyze conversion of ATP to cAMP, which will trigger a
phosphorylation cascade that will cause black eumelanin pigment to be
produced, making the hair black;

5. However, during intervals when ASIP is secreted, ASIP will prevent α-MSH
from binding to MC1R/ ASIP will preferentially bind to MC1R;

6. Thus MC1R will not undergo a conformational change that allows G-protein to
bind to it and so the G protein will not be activated, adenylyl cyclase will not
activated, cAMP is not produced and so yellow phaeomelanin pigment is
produced making the hair yellow;

Hence the interrupted pattern of ASIP secretion, will lead to the striped yellow and
black agouti hairs in cats.

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Table 1.2

(d) (i) Use Table 1.2 to identify:

a recessive allele: 2

a dominant allele : 15

an allele containing a frameshift mutation 2 [3]

(ii) With reference to Table 1.1, Table 1.2 and the information provided, explain why melanic
(black) individuals of F. catus have hairs that do not contain any phaeomelanin. [4]

1. Melanic F.catus has 2 copies of 2 and hence has 2 copies of the mutated ASIP
alleles;
2. A 2 base pair deletion in the gene will result in a frame shift in the DNA resulting in
a change the sequence of codons* in the mRNA sequence;
3. which will cause a change in the sequence of amino acids in the polypeptide which
in turn will affect the R group interactions between the amino acids in the tertiary
structure of the ASIP protein, resulting in a change in its 3D conformation;
A: truncated non-functional polypeptide
4. As a result, the non-functional ASIP (ligand) cannot bind to the MC1R (receptor) on
the melanocyte cell as it is no longer complementary* in shape to it;
5. Thus and cAMP (second messenger) production will continue and black eumelanin
is produced in the melanocyte cell;

Thus the melanic individuals have hairs that do not contain any phaeomelanin;

(iii) Scientists have concluded that the black fur colour phenotype found in several cat
species has evolved more than once.

Use information in Table 1.2 to justify this conclusion.[1]

Since cats with the same phenotype, that is, having black fur, have 2 different
mutations in 2 different gene loci, either in the ASIP locus or the MC1R locus, it shows
that black fur colour phenotype in cat species evolved more than once;

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(e) (i) Explain how temperature variation of the skin contributes to the phenotype of the cat in
Fig. 1.1.[2]

1. Since the ears, face, paws and tail are the cooler parts of the body of the cat, the
temperature sensitive tyrosinase enzyme does not denature in those areas and
hence black pigment can be produced in the hairs in those areas;
2. The other parts of the body are warmer above 30°C and the temperature sensitive
tyrosinase enzyme starts to denature, lowering enzyme activity and hence reducing
the production of pigment in the hairs there;

(ii) Draw a genetic diagram to show the predicted coat colours of the offspring of this
cross.

Insert the symbols A and a for the alleles of the ASIP gene and the symbols H and h
for the alleles of the tyrosinase gene in the key to symbols table, before completing the
genetic diagram. [4]

key to symbols
∆2 allele of the ASIP gene a

normal allele of the ASIP gene A

Himalayan allele of the tyrosinase gene h

non-Himalayan allele of the tyrosinase gene H

genetic diagram showing predicted coat colours:

Parental phenotype Himalayan cat with X Agouti, Non-Himalayan cat
black extremities

Parental genotype aahh AaHh

Gametes ah AH Ah aH ah

Gametes ah
AH Ah aH

ah AaHh Aahh aaHh aahh

Agouti, Non- Himalayan cat Himalayan cat

Black, Non-
Himalayan cat with Agouti with black
Himalayan cat
extremities extremities

Offspring genotype: AaHh Aahh aaHh aahh

Offspring phenotype: Agouti, Non- Himalayan cat Black Non- Himalayan cat
Himalayan cat with Agouti Himalayan cat with black
extremities extremities
Offspring phenotypic: 1 : 1 : 1 : 1
ratio

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1m for correct parental phenotype and genotype ;

1m for correct gametes which are circled ;
1m for correct offspring genotype and phenotype ;
1m for correct phenotypic ratio ;

(iii) Put a tick (✓) in one box to indicate whether or not this cross is a test cross.

Give a reason for your answer.[1]

test cross ✓ not a test cross

reason : The heterozygous cat was crossed with a homozygous recessive cat with
both the mutant ASIP and heat sensitive tyrosinase gene loci

(iv) Name a statistical test that would allow you to test the assumptions you made in your
genetic diagram by comparing the observed results of many crosses of this type with
your expected results.[1]

Chi-square test

[Total: 30]

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2 (a) Describe and explain the relationship shown in Fig. 2.1 between the number of cases of
measles and the percentage uptake of vaccine.[4]

1. From 1969 to 1996, as the percentage of uptake of vaccine increases from 32% to 94%,
the number of measles cases decreases from 320 000 to 0 cases;
2. After 1996, as the percentage of uptake of vaccine remained high above 80%, the number
of measles cases were 0;
3. Antigen presenting cells (APCs) macrophages / dendritic cells (A: antigen processing cells)
take up the measles vaccine by phagocytosis, process antigen and present it as a
peptide:MHC complex;
4. APC activates naïve T cell which will undergo clonal expansion and differentiation to form
helper T cells, cytotoxic T cells and memory T cells;
5. T helper cells* bind to secrete cytokines that activate specific naïve B cells* to undergo
clonal expansion and differentiation and form antibody-secreting plasma cells and memory
B cells;
6. Memory B and T cells* when exposed to virus, will recognize it and mount a faster and
stronger and longer lasting secondary immune response;
7. Specific antigen binding site of antibody binds to specific epitope on antigen of
incoming virus;
8. Prevent/block viral glycoproteins from binding to host cell receptors, resulting in
neutralization*;
9. Prevents attachment and subsequent infection of cells, thus reducing the number of
cases of infection;

(b) If the percentage of people who are immune to a disease exceeds the herd immunity
threshold, the disease can no longer persist in the population. Assuming 100% efficacy of the
vaccine, the herd immunity threshold is calculated as:

100 × (1 − 1/𝑅𝑜 )

(i) Calculate the herd immunity threshold for measles, if Ro = 11.

herd immunity threshold = …………………………………..% [1]

90.9 or 91%

(ii) Use Fig. 2.1 and your answer to (b)(i) to identify the years when the herd immunity
threshold was achieved or exceeded in this part of the United Kingdom.[1]

from 1992 to 1999 inclusive

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(c) The mutation rate in the measles virus haemagglutinin gene is estimated to be 6.0 x 10-4
substitutions per nucleotide per year. The mutation rate in human nuclear DNA is estimated
to be 2.5 x 10-8 substitutions per nucleotide per generation time of 20 years.

(i) Suggest reasons for the difference in mutation rate of the measles virus
haemagglutinin gene and the mutation rate of human nuclear DNA.[2]
1. Measle virus has a single-stranded RNA genome unlike human DNA that is
double stranded;
2. No backup copy to do correction and hence has a higher mutation rate;
OR
3. (RNA-dependent/ viral) RNA polymerase lacks proof-reading ability unlike human
RNA polymerase that has proof reading ability;
4. more errors occur in viral genome replication;
OR
5. RNA genome more reactive (due to the 2’OH group) than DNA genome; R:
unstable
6. Higher frequency of mutations of the RNA* genome leading to the changes in the
RNA sequence
Note: It is important to make comparative statements. Use of words like more and
higher are important

(ii) Explain why knowledge of mutation rates is useful in reconstructing phylogenies.[2]

1. Rate of mutation/time taken for mutation to accumulate in virus types is constant;
2. from the number of nucleotide differences between different virus strains, it is
possible to distinguish between different virus strains /infer evolutionary
relationships/degree of genetic divergence between different virus strains
or
infer common ancestor of virus strains;
3. from the number of nucleotide differences, it is possible to infer the time when a
certain strain virus emerged or share common ancestor;
Reject time of speciation because viruses are not able to form new species.
A: if not specifically reference to virus

[Total: 10]

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3 (a) Name these two products and explain how the presence of each of these could help coral
growth.[4]

1. Oxygen* is produced by the algae during the light dependent reaction of

photosynthesis;
2. It is used by the corals for aerobic respiration as the final electron acceptor* of the
electron transport chain* during oxidative phosphorylation*;
3. Glucose* is produced by algae during the light independent reaction of
photosynthesis;
4. It is the main substrate of respiration by the corals;
5. To produce energy in the form of ATP* for metabolism and growth;

(b) (i) Calculate the overall mean percentage changes in coral cover per year in area 1 and
area 2 from 1985 to 2012.

State which area shows the greater overall mean percentage change in coral cover
per year.

overall mean percentage change in area 1 = (+2.07 – 0.77 – 1.05 – 0.36)% = -0.11%

overall mean percentage change in area 2 = (+2.34 – 1.59 – 1.75 – 0.04)% = -1.04%

overall mean percentage

change in coral cover per year in area 1 = …-0.11%………..

overall mean percentage

change in coral cover per year in area 2 = ……-1.04%...…...

area with greater overall mean

percentage change in coral cover per year = …area 2…......[2]

(ii) Use the data in Table 3.1 to evaluate whether or not the changes in coral cover in
these two areas of the Great Barrier Reef from 1985 to 2012 can be attributed to
global climate change.[4]

Decrease can be attributed to climate change because

1. Coral bleaching is a result of increase in sea water temperatures which cause

zooxanthellae to be expelled from corals, eventually leading to death of coral;
2. Mean percentage decrease due to coral bleaching was 0.36% and 0.04% in
area 1 and area 2 respectively;
3. Tropical storms are more severe due to increased temperatures from climate
change;
4. Mean percentage decrease due to tropical storms 1.05% and 1.75% in area 1
and area 2 respectively;

Decrease should not be attributed to climate change because

1. There were only 2 sites studied, both in the Great Barrier Reef;
2. Changes may be due to localised environmental conditions and not global
climate change;
3. Data is represented as a mean value over 27 years, so unable to see changes
on a yearly basis to make an informed conclusion;
[Total:10]

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Section B
Answer one question in this section.

4 (a) Outline the structural features and genomic organisation of bacteria. [15]
(A) Structural features of bacteria
1. Bacteria cells have simple internal structures with no membrane-bound
organelles*.
2. The bacterial chromosome is a double-stranded, circular DNA molecule*;
3. that makes up a dense region within the cell called the nucleoid* region;
4. Some bacteria cells may contain extrachromosomal double-stranded small
circular DNA molecules called plasmids*;
5. It contains 70S ribosomes* needed for protein synthesis;
6. Nutrients and chemical reserves may be stored in the cytoplasm in the form of
granules, e.g. glycogen granules, lipid granules, etc;
7. The bacterial cytosol is enclosed by cell surface membrane* made up of a
phospholipid bilayer* with proteins embedded in carrying out specific functions
such as transport, ATP synthesis, etc;
8. The peptidoglycan cell wall* confers rigidity to the bacterial cell and protects it
from osmotic lysis;
9. Gram-positive bacteria have a cell wall with a thick peptidoglycan layer;
10. Gram-negative bacteria have a cell wall with a thin peptidoglycan layer;
11. Some bacteria may have a layer of polysaccharides known as glycocalyx to the
exterior of the cell wall;
12. If the glycocalyx layer is a distinct layer, it is known as a capsule;
13. If the glycocalyx layer is a diffused mass, it is known as a slime layer;
14. Some bacteria may have short, bristle-like fibres extending from the cell surface
called fimbriae*;
15. Fimbriae may be evenly distributed over the entire cell surface or at the poles of
the cells;
16. Other bacterial appendages include pili*, which are longer, hollow hair-like
structures found on the cell surface;
17. They occur in fewer numbers compared to fimbriae and may have specialized
functions such as the sex pilus during conjugation;
18. Some bacteria cells may have a long, hollow cylindrical protein thread called the
flagella*;
19. Flagella are usually used by bacteria cells for motility as it helps propel the
bacteria by rotation;

(B) Genomic organization of bacteria

20. The bacterial chromosome is small in size about 104 – 107 base pairs;
21. It comprises of a single double-stranded, circular DNA molecule*;
22. The circular DNA forms looped domains by associating with a small amount of
histone-like proteins, followed by further supercoiling to form highly condensed
DNA;
23. that makes up a dense region within the cell called the nucleoid* region;
24. The genome is simple with a single origin of replication*;
25. where genes with related functions or belonging to the same metabolic pathway
are organized together in an operon* under the control of a single promoter*;
26. Bacterium being prokaryotic, has no intron sequences in its chromosome;
27. Some bacteria cells may contain extrachromosomal double-stranded small
circular DNA molecules called plasmids*;
28. There may be more than 1 copy of the same plasmid in the bacterial cell;
29. The plasmid may contain genes which may confer advantages for the bacteria
living in stressful environments, e.g. antibiotic resistance genes;

QWC: Address both (A) and (B)

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(b) Compare how genetic variation arises in prokaryotes and eukaryotes. [10]

Point of prokaryote eukaryote

comparison

1. Method of 1A. Mainly by horizontal gene 1B. Mainly by meiosis and

obtaining transfer; sexual reproduction;
genetic
variation

2. Methods of 2A1. Conjugation whereby the 2B1. Meiosis: During

genetic F plasmids to transferred from crossing over between
variation host to the recipient via a non-sister chromatids of
mating bridge; homologous
chromosomes* in prophase
1 results in new
combinations of alleles on
2A2. Specialised chromatids. (& eventually a
transduction* whereby a variety of offspring)
temperate phage takes up an
adjacent bacterial DNA of the 2B2. Independent
provirus when it undergoes the assortment of homologous
lytic phase and phage infects chromosomes at the
another bacteria. This is metaphase plate & their
transferred to another bacteria subsequent separation
via homologous recombination; during metaphase I &
anaphase I respectively &

2A3. Transformation* whereby 2B2. Random orientation of

fragments of naked DNA are non-identical sister
taken in by a competent chromatids of each
bacteria cell and introduced into chromosome at the
the bacterial chromosome by metaphase plate & their
crossing over of homologous subsequent separation
regions; during metaphase II and
anaphase II respectively

2B3. This results in gametes

with different combinations
of maternal & paternal
chromosomes. (& eventually
in a variety of offspring)

3. Random fusion of
gamete occurs during sexual
reproduction/fertilisation
results in offspring with a
variety of genotypes &
possibly phenotypes (&
hence a variety of
offspring);

4. Mutations As there is only 1 chromosome, As there are more than 1

there will not be mutations that chromosome, non-
involve more than 1 disjunction, translocation
chromosome, such as non- and polyploidy can occur;

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disjunction, polyploidy and

translocation

Similarities:
5. Mutations are a source of genetic variation for both prokaryotes and eukaryotes;
6. Mutations such as point mutations and gene mutations such as inversion,
substitution, deletion and addition can occur in both;
7. Role of mutations in both can be significant as novel phenotypes can be selected for
during natural selection hence increasing the frequency of alleles coding for the
favourable phenotype in the subsequent generation;

8. QWC: Mention both similarities and differences.

[Total: 25]

5 (a) Outline the processes that occur in aerobic respiration in eukaryotic cells. [15]
1. Aerobic respiration occurs when organic food substances such as
carbohydrates, fats and proteins are oxidised to provide energy, protons and
electrons that will ultimately be used to generate ATP;
2. Oxygen is required for the complete oxidation of these organic food substances
and ATP is produced in the process;

(A) Glycolysis
3. Glycolysis* occurs in the cytoplasm of the eukaryotic cell;
4. Phosphorylation of glucose involves the initial investment of 2 ATP molecules
which phosphorylate the two ends of glucose molecule forming fructose-1,6-
bisphosphate*;
5. Phosphofructokinase* is the enzyme involved in the second phosphorylation
step;
6. The phosphorylated 6C sugar splits into two 3C sugar phosphates;
7. Each fructose molecule ultimately gives rise to two molecules of
glyceraldehyde-3- phosphate (G3P)*;
8. Glyceraldehyde-3-phosphate undergoes oxidation by dehydrogenation* to
form 1,3-bisphosphoglycerate;
9. NAD+* is reduced to NADH* in the process;
10. 2 ATP is produced by substrate level phosphorylation* of 1,3-
bisphosphoglycerate forming pyruvate*;
11. Since each glyceraldehyde-3-phosphate yields 2 ATP and 1 NADH, each
glucose molecule yields 2 ATP, 2 pyruvate and 2 NADH taking into account the
2 ATP as an initial investment;

(B) (B) Link reaction

12. Link reaction* occurs in the matrix* of the mitochondria;
13. Pyruvate formed from glycolysis enters the mitochondrial matrix via transport
proteins;
14. Pyruvate undergoes oxidative decarboxylation* where it combines with
coenzyme A to form acetyl CoA(2C)*;
15. Carbon dioxide* is released and NAD+* is reduced to NADH* in the process;
16. Each glucose molecule produces 2 pyruvate which undergoes the link reaction
to produce 2 acetyl CoA, 2 CO2 and 2 NADH;

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(C) (C) Krebs cycle

17. Krebs cycle* occurs in the matrix* of the mitochondria and when oxygen is
present;
18. Acetyl CoA (2C) combines with oxaloacetate (4C)* to form citrate (6C)*;
19. Citrate is decarboxylated and dehydrogenated to form α-ketoglutarate (5C)*
and NADH.
20. Each decarboxylation step results in a loss of carbon in the form of a carbon
dioxide;
21. Regeneration of oxaloacetate (4C) involves one decarboxylation step and three
dehydrogenation steps to yield 2 NADH, 1 FADH2 and 1 CO2;
22. High energy electrons originally from the glucose molecule have now been
transferred to electron carriers NAD+ and FAD;
23. NAD+ + 2H+ + 2e- → NADH + H+ (or reduced NAD)
FAD + 2H+ + 2e- → FADH2 (or reduced FAD)
24. 1 ATP* is also produced through substrate-level phosphorylation* during this
regeneration process;
25. Altogether 1 molecule of glucose will yield 6 NADH, 2 FADH2 & 2 ATP through
the Krebs cycle. The coenzymes with their reducing power will next be
transported to the electron transport chain where the bulk of ATP is generated.

(D) Oxidative phosphorylation

• Oxidative phosphorylation occurs on the folds of the inner membrane of
mitochondria = cristae*;
• NADH donates electrons to first electron carrier of the electron transport
chain* found on the cristae;
• The electron carriers alternate between reduced and oxidized states as they
accept and donate electrons;
• Each successive carrier has a higher electronegativity;
• Each FADH2 donates electrons to the chain at a lower level and hence
generates 2 ATP compared to 3 ATP generated by NADH;
• Electron transport through the series of carriers is coupled to the active
pumping of H+ into the intermembrane space. This generates a proton
gradient*/ proton motive force* across the mitochondria membrane;
26. The proton motive force is used to phosphorylate ADP. As protons diffuse back
into the matrix via the ATP synthase complex, ADP is phosphorylated to ATP*.
27. This process is known as chemiosmosis*;
• and produces 90% of the ATP during aerobic respiration.
• With oxygen* as the final electron acceptor* having the highest
electronegativity;
• 2e- +2H+ + 1/2O2 → H2O.

QWC: Must have at least two points from (A), (B), (C) and (D).

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(b) Compare how ATP is generated in chloroplasts and mitochondria. [10]

Differences:
ATP generation in ATP generation in
Feature of comparison
chloroplasts mitochondria
1. Process(es) in which Substrate-level
ATP is generated Photophosphorylation* phosphorylation* (Krebs
in the light-dependent cycle) and oxidative
reaction phosphorylation* in
aerobic respiration
2. Site of ATP formation Takes place in/at the Takes place in/at the
stroma*/thylakoid matrix*/intermembrane
membrane* of chloroplast space* of mitochondria
3. Source of energy Energy for synthesis of Energy for the synthesis
ATP comes ultimately of ATP comes from the
from light; oxidation of glucose which
stores chemical energy;
4. Source of electrons Water is the electron [oxidative phosphorylation
which travels down donor in the non-cyclic only]: NADH and FADH2
the electron transport pathway while are the electron donors to
chain Photosystem I* is the the first electron carrier of
electron donor in the ETC;
cyclic pathway;
5. Location of proton Protons are pumped from [oxidative phosphorylation
reservoir which stroma across the only]: Protons are pumped
supplies protons that thylakoid membrane, into from mitochondrial matrix
diffuse down ATP the thylakoid space* to across the inner
synthase establish a proton membrane, into the
gradient for ATP intermembrane space*
synthesis; to establish a proton
gradient for ATP
synthesis;
6. Direction of proton Protons diffuse from Protons diffuse from
flow through ATP thylakoid space* to intermembrane space*
synthase stroma* to matrix*
7. Type of phosphate Inorganic phosphate [substrate-level
phosphorylation only]:
phosphate derived from
respiratory
intermediates/substrate
8. By-product Splitting of water [oxidative phosphorylation
produces oxygen as by- only]: Water is produced
product during non-cyclic as by-product when
pathway; oxygen combines electron
and proton at the end of
ETC;
9. Energy conversion Light energy is converted Chemical energy (from
to chemical energy in the glucose) is converted to
process; chemical energy (in the
form of ATP) in the
process;

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Similarities:
1. Both involve the transfer of electrons down electron carriers* of increasing
electronegativity in an electron transport chain*;
2. In both cases, energy release from the transfer of electrons down the electron
transport chain* is coupled to the pumping* of H+;
3. Both involve the generation of a proton gradient* across a membrane;
4. Both involve facilitated diffusion* of H+ through the hydrophilic pore* of ATP
synthase* down its proton gradient;
5. In both, ATP* is produced from ADP* and inorganic phosphate/Pi * via
chemiosmosis*;

QWC: essays need to include at least one similarity and one difference.

[Total: 25]

2020 H2 Biology A level Paper 3 [Turn over