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1
Q

What is a homologous pair of chromosomes?
[1 mark]

A

(Two chromosomes that) carry the same
genes;

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2
Q

Give two ways in which the arrangement of prokaryotic DNA is different from the
arrangement of the human DNA in Figure 1.
[2 marks]

A

(Prokaryotic DNA) is
1. Circular (as opposed to linear);
2. Not associated with proteins/histones ;
3. Only one molecule/piece of DNA
OR
present as plasmids;

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3
Q

When preparing the cells for observation the scientist placed them in a solution that
had a slightly higher (less negative) water potential than the cytoplasm. This did not
cause the cells to burst but moved the chromosomes further apart in order to reduce
the overlapping of the chromosomes when observed with

A
  1. Water moves into the cells/cytoplasm by
    osmosis;
  2. Cell/cytoplasm gets bigger;
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4
Q

The dark stain used on the chromosomes binds more to some areas of the
chromosomes than others, giving the chromosomes a striped appearance.
Suggest one way the structure of the chromosome could differ along its length to
result in the stain binding more in some areas.
an optical microscope.
Suggest how this procedure moved the chromosomes apart.
[2 marks]

A

Differences in base sequences
OR
Differences in histones/interaction with
histones
OR
Differences in condensation/(super)coiling;

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5
Q

What is meant by ‘species richness’?
[1 mark]

A

(A measure of) the number of (different) species in
a community;

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6
Q

Three of the bee species collected in the farmland areas were Peponapis pruinosa,
Andrena chlorogaster and Andrena piperi.
What do these names suggest about the evolutionary relationships between these
bee species? Explain your answer.
[2 marks]

A
  1. A. chlorogaster and A. piperi are more
    closely related (to each other than to
    P. pruinosa);
  2. Because they are in the same genus;
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7
Q

Formation of an enzyme-substrate complex increases the rate of reaction.
Explain how.
[2 marks]

A
  1. Reduces activation energy;
  2. Due to bending bonds
    OR
    Without enzyme, very few substrates have
    sufficient energy for reaction;
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8
Q

Lyxose binds to the enzyme.
Suggest a reason for the difference in the results shown in Figure 5 with and without
lyxose.
[3 marks]

A
  1. (Binding) alters the tertiary structure
    of the enzyme ;
  2. (This causes) active site to change
    (shape);
  3. (So) More (successful) E-S
    complexes form (per minute)
    OR
    E-S complexes form more quickly
    OR
    Further lowers activation energy;
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9
Q

A change from Glu to Lys at amino acid 300 had no effect on the rate of reaction
catalysed by the enzyme. The same change at amino acid 279 significantly reduced
the rate of reaction catalysed by the enzyme.
Use all the information and your knowledge of protein structure to suggest reasons for
the differences between the effects of these two changes.
[3 marks]

A
  1. (Both) negatively charged to positively charged
    change in amino acid;
  2. Change at amino acid 300 does not change the
    shape of the active site
    OR
    Change at amino acid 300 does not change the
    tertiary structure
    OR
    Change at amino acid 300 results in a similar
    tertiary structure;
  3. Amino acid 279 may have been involved in a
    (ionic, disulfide or hydrogen) bond and so the
    shape of the active site changes
    OR
    Amino acid 279 may have been involved in a
    (ionic, disulfide or hydrogen) bond and so the
    tertiary structure changed;
    OR
    Amino acid 279 may be in the active site and
    be required for binding the substrate;
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10
Q

The scientists tested their null hypothesis using the chi-squared statistical test.
After 1 cycle their calculated chi-squared value was 350
The critical value at P=0.05 is 3.841
What does this result suggest about the difference between the observed and
expected results and what can the scientists therefore conclude?
[2 marks]

A
  1. There is a less than 0.05/5% probability
    that the difference(s) (between observed
    and expected) occurred by chance;
  2. Calculated value is greater than critical
    value so the null hypothesis can be
    rejected;
  3. (The scientists can conclude that) the
    proportion of plants that produce 2n
    gametes does change from one breeding
    cycle to the next;
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11
Q

When a person is bitten by a venomous snake, the snake injects a toxin into the
person. Antivenom is injected as treatment. Antivenom contains antibodies against
the snake toxin. This treatment is an example of passive immunity.
Explain how the treatment with antivenom works and why it is essential to use passive
immunity, rather than active immunity.
[2 marks]

A
  1. (Antivenom/Passive immunity) antibodies
    bind to the toxin/venom/antigen and
    (causes) its destruction;
  2. Active immunity would be too slow/slower
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12
Q

A mixture of venoms from several snakes of the same species is used.
Suggest why.
[2 marks]

A
  1. May be different form of antigen/toxin
    (within one species)
    OR
    Snakes (within one species) may have
    different mutations/alleles;
  2. Different antibodies (needed in the
    antivenom)
    OR
    (Several) antibodies complementary (to
    several antigens);
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13
Q

During vaccination, each animal is initially injected with a small volume of venom.
Two weeks later, it is injected with a larger volume of venom.
Use your knowledge of the humoral immune response to explain this vaccination
programme.
[3 marks]

A
  1. B cells specific to the venom reproduce by
    mitosis;
  2. (B cells produce) plasma cells and memory
    cells;
  3. The second dose produces antibodies (in
    secondary immune response) in higher
    concentration and quickly
    OR
    The first dose must be small so the animal
    is not killed;
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14
Q

Describe the role of two named enzymes in the process of semi-conservative
replication of DNA.
[3 marks]

A
  1. (DNA) helicase causes breaking of hydrogen/H
    bonds (between DNA strands);
  2. DNA polymerase joins the (DNA) nucleotides;
  3. Forming phosphodiester bonds;
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15
Q

Describe the gross structure of the human gas exchange system and how we breathe
in and out.
[6 marks]

A
  1. Named structures – trachea, bronchi, bronchioles,
    alveoli;
  2. Above structures named in correct order
    OR
    Above structures labelled in correct positions on a
    diagram;
  3. Breathing in – diaphragm contracts and external
    intercostal muscles contract;
  4. (Causes) volume increase and pressure decrease
    in thoracic cavity (to below atmospheric, resulting
    in air moving in);
  5. Breathing out - Diaphragm relaxes and internal
    intercostal muscles contract;
  6. (Causes) volume decrease and pressure increase
    in thoracic cavity (to above atmospheric, resulting
    in air moving out)
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16
Q

Mucus produced by epithelial cells in the human gas exchange system contains
triglycerides and phospholipids.
Compare and contrast the structure and properties of triglycerides and phospholipids.
[5 marks]

A
  1. Both contain ester bonds (between glycerol and
    fatty acid);
  2. Both contain glycerol;
  3. Fatty acids on both may be saturated or
    unsaturated;
  4. Both are insoluble in water;
  5. Both contain C, H and O but phospholipids also
    contain P;
  6. Triglyceride has three fatty acids and
    phospholipid has two fatty acids plus phosphate
    group;
  7. Triglycerides are hydrophobic/non-polar and
    phospholipids have hydrophilic and
    hydrophobic region;
  8. Phospholipids form monolayer (on
    surface)/micelle/bilayer (in water) but
    triglycerides don’t;
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17
Q

Mucus also contains glycoproteins. One of these glycoproteins is a polypeptide with
the sugar, lactose, attached.
Describe how lactose is formed and where in the cell it would be attached to a
polypeptide to form a glycoprotein.
[4 marks]

A
  1. Glucose and galactose;
  2. Joined by condensation (reaction);
  3. Joined by glycosidic bond;
  4. Added to polypeptide in Golgi (apparatus)
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18
Q

Describe how a non-competitive inhibitor can reduce the rate of an
enzyme-controlled reaction.
[3 marks]

A
  1. Attaches to the enzyme at a site other than the
    active site;
  2. Changes (shape of) the active site
    OR
    Changes tertiary structure (of enzyme);
  3. (So active site and substrate) no longer
    complementary so less/no substrate can fit/bind;
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19
Q

No large lipid droplets are visible with the optical microscope in the samples from
suspension A.
Explain why.
[2 marks]

A
  1. Emulsification;
  2. (Cannot be seen) due to resolution (of optical
    microscope);
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20
Q

Give two ways the students would have ensured their index of diversity was
representative of each habitat.
[2 marks]

A
  1. Random samples;
  2. Large number (of samples)
    OR
    (Continue sampling) until stable running mean;
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21
Q

The scientists calculated a P value of 0.03 when testing their null hypothesis.
What can you conclude from this result? Explain your answer.
[3 marks]

A
  1. Probability that difference (in frequency of
    births above 4500 g) is due to chance is
    less than 0.05
    OR
    Probability that difference (in frequency of
    births above 4500 g) is due to chance is
    0.03;
  2. Reject null hypothesis;
  3. Presence of KIR2DS1/allele does
    (significantly) affect the frequency of high
    birth mass;
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22
Q

Describe the structure of the human immunodeficiency virus (HIV).
[4 marks]

A
  1. RNA (as genetic material);
  2. Reverse transcriptase;
  3. (Protein) capsomeres/capsid;
  4. (Phospho)lipid (viral) envelope
    OR
    Envelope made of membrane;
  5. Attachment proteins;
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23
Q

The scientists determined the percentage of heart cells undergoing DNA replication
by using a chemical called BrdU. Cells use BrdU instead of nucleotides containing
thymine during DNA replication.
Describe how BrdU would be incorporated into new DNA during semi-conservative
replication.
[5 marks]

A
  1. DNA helicase;
  2. Breaks hydrogen bonds (between 2 DNA
    strands);
  3. BrdU complementary to adenine (on template
    strand)
    OR
    BrdU forms hydrogen bonds with adenine (on
    template strand);
  4. DNA polymerase joins (adjacent) nucleotides (to
    incorporate BrdU into the new DNA strand);
  5. Phosphodiester bonds form (between
    nucleotides);
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24
Q

Cells with BrdU in their DNA are detected using an anti-BrdU antibody with an enzyme
attached.
Use your knowledge of the ELISA test to suggest and explain how the scientists
identified the cells that have BrdU in their DNA.
[3 marks]

A
  1. Add antibody (anti-BrdU with enzyme attached)
    to cells/DNA
    OR
    Add cells/DNA to antibody (anti-BrdU with
    enzyme attached);
  2. Wash (cells/DNA) to remove excess/unattached
    antibody
    OR
    Wash (immobilised antibody) to remove
    excess/unattached cells/DNA;
  3. Add substrate to cause colour change;
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25
Q

Unlike plants, Ulva lactuca does not have xylem tissue.
Suggest how Ulva lactuca is able to survive without xylem tissue.
[1 mark]

A

Short diffusion pathway (to cells)
OR
It has a surface permeable (to water/ions into cells);

26
Q

Ulva prolifera also produces haploid, mobile single cells that can fuse to form a zygote.
Suggest and explain one reason why successful reproduction between Ulva prolifera
and Ulva lactuca does not happen.
[2 marks]

A
  1. They are different species;
  2. (So) if fused together they would not
    produce fertile offspring
    OR
    (So) they have named characteristics that
    means they are reproductively isolated;
27
Q

Use your knowledge of gas exchange in leaves to explain why plants grown in soil
with very little water grow only slowly.
[2 marks]

A
  1. Stomata close;
  2. Less carbon dioxide (uptake) for less
    photosynthesis/glucose production;
28
Q

The data in Table 7 show differences between the oxyhaemoglobin dissociation curve
for a mouse and the oxyhaemoglobin dissociation curve for a horse.
Suggest how these differences allow the mouse to have a higher metabolic rate than
the horse.
[2 marks]

A
  1. Mouse haemoglobin/Hb has a lower
    affinity for oxygen
    OR
    For the same pO2 the mouse
    haemoglobin/Hb is less saturated
    OR
    At oxygen concentrations found in tissue
    mouse haemoglobin/Hb is less saturated;
  2. More oxygen can be
    dissociated/released/unloaded (for
    metabolic reactions/respiration);
29
Q

Mammals such as a mouse and a horse are able to maintain a constant body
temperature.
Use your knowledge of surface area to volume ratio to explain the higher metabolic
rate of a mouse compared to a horse.
[3 marks]

A

Mouse
1. (Smaller so) larger surface area to volume
ratio;
2. More/faster heat loss (per gram/in relation
to body size);
3. (Faster rate of) respiration/metabolism
releases heat;

30
Q

Explain five properties that make water important for organisms.
[5 marks]

A
  1. A metabolite in condensation/hydrolysis/
    photosynthesis/respiration;
  2. A solvent so (metabolic) reactions can occur
    OR
    A solvent so allowing transport of substances;
  3. High heat capacity so buffers changes in
    temperature;
  4. Large latent heat of vaporisation so provides a
    cooling effect (through evaporation);
  5. Cohesion (between water molecules) so
    supports columns of water (in plants);
  6. Cohesion (between water molecules) so
    produces surface tension supporting (small)
    organisms;
31
Q

Describe the biochemical tests you would use to confirm the presence of lipid,
non-reducing sugar and amylase in a sample.
[5 marks]

A

Lipid
1. Add ethanol/alcohol then add water and
shake/mix
OR
Add ethanol/alcohol and shake/mix then
pour into/add water;
2. White/milky emulsion
OR
emulsion test turns white/milky;
Non-reducing sugar
3. Do Benedict’s test and stays blue/negative;
4. Boil with acid then neutralise with alkali;
5. Heat with Benedict’s and becomes
red/orange (precipitate);
Amylase
6. Add biuret (reagent) and becomes
purple/violet/mauve/lilac;
7. Add starch, (leave for a time), test for
reducing sugar/absence of starch;

32
Q

Describe the chemical reactions involved in the conversion of polymers to monomers
and monomers to polymers.
Give two named examples of polymers and their associated monomers to illustrate
your answer.
[5 marks]

A
  1. A condensation reaction joins monomers
    together and forms a (chemical) bond and
    releases water;
  2. A hydrolysis reaction breaks a (chemical)
    bond between monomers and uses water;
  3. A suitable example of polymers and the
    monomers from which they are made;
  4. A second suitable example of polymers and
    the monomers from which they are made;
  5. Reference to a correct bond within a named
    polymer;
33
Q

The action of the carrier protein X in Figure 1 is linked to a membrane-bound ATP
hydrolase enzyme.
Explain the function of this ATP hydrolase.
[2 marks]

A
  1. (ATP to ADP + Pi ) Releases energy;
  2. (energy) allows ions to be moved against a
    concentration gradient
    OR
    (energy) allows active transport of ions;
34
Q

The movement of Na+ out of the cell allows the absorption of glucose into the
cell lining the ileum.
Explain how.
[2 marks]

A
  1. (Maintains/generates) a
    concentration/diffusion gradient for Na+
    (from
    ileum into cell);
  2. Na+ moving (in) by facilitated diffusion,
    brings glucose with it
    OR
    Na+ moving (in) by co-transport, brings
    glucose with it;
35
Q

Describe and explain two features you would expect to find in a cell specialised for
absorption.
[2 marks]

A
  1. Folded membrane/microvilli so large surface
    area (for absorption);
  2. Large number of
    co-transport/carrier/channel proteins so fast
    rate (of absorption)
    OR
    Large number of co-transport/carrier proteins
    for active transport
    OR
    Large number of
    co-transport/carrier/channel proteins for
    facilitated diffusion;
  3. Large number of mitochondria so make
    (more) ATP (by respiration)
    OR
    Large number of mitochondria for aerobic
    respiration
    OR
    Large number of mitochondria to release
    energy for active transport;
  4. Membrane-bound (digestive) enzymes so
    maintains concentration gradient (for fast
    absorption);
36
Q

Figure 2 shows the SGLT1 polypeptide with NH2 at one end and COOH at the
other end.
Describe how amino acids join to form a polypeptide so there is always NH2 at
one end and COOH at the other end.
You may use a diagram in your answer.
[2 marks]

A
  1. One amine/NH2 group joins to a
    carboxyl/COOH group to form a peptide
    bond;
  2. (So in chain) there is a free amine/NH2 group
    at one end and a free carboxyl/COOH group
    at the other
    OR
    Each amino acid is orientated in the same
    direction in the chain;
37
Q

Describe the role of micelles in the absorption of fats into the cells lining the ileum.
[3 marks]

A
  1. Micelles include bile salts and fatty acids;
  2. Make the fatty acids (more) soluble in water;
  3. Bring/release/carry fatty acids to cell/lining (of the
    ileum);
  4. Maintain high(er) concentration of fatty acids to
    cell/lining (of the ileum);
  5. Fatty acids (absorbed) by diffusion;
38
Q

Describe the role of DNA polymerase in the semi-conservative replication of DNA.
[2 marks]

A
  1. Joins (adjacent DNA) nucleotides;
  2. (Catalyses) condensation (reactions);
  3. (Catalyses formation of) phosphodiester bonds
    (between adjacent nucleotides);
39
Q

Cyclin D stimulates the phosphorylation of DNA polymerase, which activates the
DNA polymerase.
Describe how an enzyme can be phosphorylated.
[2 marks]

A
  1. Attachment/association of (inorganic) phosphate
    (to the enzyme);
  2. (Released from) hydrolysis of ATP
    OR
    (Released from) ATP to ADP + Pi;
40
Q

Particulate matter is solid particles and liquid particles suspended in air. Polluted air
contains more particulate matter than clean air.
A high concentration of particulate matter results in the death of some
alveolar epithelium cells. If alveolar epithelium cells die inside the human body they
are replaced by non-specialised, thickened tissue.
Explain why death of alveolar epithelium cells reduces gas exchange in human lungs.
[3 marks]

A
  1. Reduced surface area;
  2. Increased distance for diffusion;
  3. Reduced rate of gas exchange;
41
Q

Alpha-gal is a disaccharide found in red meat.
Alpha-gal is made of two galactose molecules. Galactose has the chemical formula
C6H12O6
Give the chemical formula for the disaccharide, alpha-gal, and describe how it is
formed from two galactose molecules.
[2 marks]

A
  1. C12H22O11;
  2. Condensation reaction
    OR
    With a glycosidic bond;
42
Q

A tick is a small animal that bites humans and feeds on their blood. This results in
proteins from the tick saliva entering the human body.
Scientists have suggested one hypothesis for the allergic reaction to alpha-gal in
red meat. They think that an earlier immune response to a tick bite can cause a
person to have an allergic reaction to alpha-gal in red meat.
Suggest how one antibody can be specific to tick protein and to alpha-gal.
[2 marks]

A
  1. (Part of tick protein and alpha-gal) have a similar
    shape/structure;
  2. Antibody is complementary to both (tick protein
    and alpha-gal)
    OR
    Antigen-binding site is complementary to both
    (tick protein and alpha-gal)
    OR
    Antibody can form antigen-antibody complex with
    both (tick protein and alpha-gal);
43
Q

Complete Table 2 to show three differences between DNA in the nucleus of a
plant cell and DNA in a prokaryotic cell.
[3 marks]

A

Plant v prokaryote
1. (Associated with) histones/proteins v no
histones/proteins;
2. Linear v circular;
3. No plasmids v plasmids;
4. Introns v no introns;
5. Long(er) v short(er);

44
Q

Scientists investigated the genetic diversity between several species of sweet potato.
They studied non-coding multiple repeats of base sequences.
Define ‘non-coding base sequences’ and describe where the non-coding multiple
repeats are positioned in the genome.
[2 marks]

A
  1. DNA that does not code for protein/polypeptides
    OR
    DNA that does not code for (sequences of)
    amino acids
    OR
    DNA that does not code for tRNA/rRNA;
  2. (Positioned) between genes;
45
Q

Describe how mRNA is formed by transcription in eukaryotes.
[5 marks]

A
  1. Hydrogen bonds (between DNA bases) break;
  2. (Only) one DNA strand acts as a template;
  3. (Free) RNA nucleotides align by complementary
    base pairing;
  4. (In RNA) Uracil base pairs with adenine (on
    DNA)
    OR
    (In RNA) Uracil is used in place of thymine;
  5. RNA polymerase joins (adjacent RNA)
    nucleotides;
  6. (By) phosphodiester bonds (between adjacent
    nucleotides);
  7. Pre-mRNA is spliced (to form mRNA)
    OR
    Introns are removed (to form mRNA);
46
Q

Describe how a polypeptide is formed by translation of mRNA.
[6 marks]

A
  1. (mRNA attaches) to ribosomes
    OR
    (mRNA attaches) to rough endoplasmic
    reticulum;
  2. (tRNA) anticodons (bind to) complementary
    (mRNA) codons;
  3. tRNA brings a specific amino acid;
  4. Amino acids join by peptide bonds;
  5. (Amino acids join together) with the use of ATP;
  6. tRNA released (after amino acid joined to
    polypeptide);
  7. The ribosome moves along the mRNA to form
    the polypeptide;
47
Q

Define ‘gene mutation’ and explain how a gene mutation can have:
* no effect on an individual
* a positive effect on an individual.
[4 marks]

A

(Definition of gene mutation)
1. Change in the base/nucleotide (sequence of
chromosomes/DNA);
2. Results in the formation of new allele;
(Has no effect because)
3. Genetic code is degenerate (so amino acid
sequence may not change);
OR
Mutation is in an intron (so amino acid sequence
may not change);
4. Does change amino acid but no effect on tertiary
structure;
5. (New allele) is recessive so does not influence
phenotype;
(Has positive effect because)
6. Results in change in polypeptide that positively
changes the properties (of the protein)
OR
Results in change in polypeptide that positively
changes a named protein;
7. May result in increased reproductive success
OR
May result in increased survival (chances);

48
Q

Describe the induced-fit model of enzyme action and how an enzyme acts as a
catalyst.
[3 marks]

A
  1. Substrate binds to the active site/enzyme
    OR
    Enzyme-substrate complex forms;
  2. Active site changes shape (slightly) so it is
    complementary to substrate
    OR
    Active site changes shape (slightly) so
    distorting/breaking/forming bonds in the
    substrate;
  3. Reduces activation energy;
49
Q

Explain the advantage for larger animals of having a specialised system that
facilitates oxygen uptake.
[2 marks]

A
  1. Large(r) organisms have a small(er) surface
    area:volume (ratio);
    OR
    Small(er) organisms have a large(r) surface
    area:volume (ratio);
  2. Overcomes long diffusion pathway
    OR
    Faster diffusion;
50
Q

Describe how one amino acid is added to a polypeptide that is being formed at a
ribosome during translation.
[3 marks]

A
  1. tRNA brings specific amino acid (to ribosome);
  2. Anticodon (on tRNA) binds to codon (on
    mRNA);
  3. Amino acids join by condensation reaction
    (using ATP)
    OR
    Amino acids join to form a peptide bond (using
    ATP);
51
Q

Suggest two ways the student could improve the quality of his scientific drawing of the
blood vessels in this dissection.
[2 marks]

A

Describe two precautions the student should take when clearing away after the
dissection.
[2 marks]

52
Q

Describe how a sample of chloroplasts could be isolated from leaves.
[4 marks]

A
  1. Break open cells/tissue and filter
    OR
    Grind/blend cells/tissue/leaves and filter;
  2. In cold, same water potential/concentration, pH
    controlled solution;
  3. Centrifuge/spin and remove nuclei/cell debris;
  4. (Centrifuge/spin) at high(er) speed, chloroplasts
    settle out;
53
Q

Some proteins found inside the chloroplast are synthesised inside the chloroplast.
Give one feature of the chloroplast that allows protein to be synthesised inside the
chloroplast and describe one difference between this feature in the chloroplast and
similar features in the rest of the cell.
[2 marks]

A

Mark in pairs, 1 and 2 OR 3 and 4
1. DNA;
2. Is not associated with protein/histones but
nuclear DNA is
OR
Is circular but nuclear DNA is linear
OR
Is shorter than nuclear DNA;
3. Ribosomes;
4. Are smaller than cytoplasmic ribosomes;

54
Q

Name the three phases of mitosis shown by C, D and E on Figure 7.
Describe the role of the spindle fibres and the behaviour of the chromosomes during
each of these phases.
[5 marks]

A
  1. C = prophase and
    D = metaphase and
    E = anaphase;
  2. (In) prophase, chromosomes condense;
  3. (In) prophase OR metaphase, centromeres
    attach to spindle fibres;
  4. (In) metaphase, chromosomes/pairs of
    chromatids at equator/centre of spindle/cell;
  5. (In) anaphase, centromeres divide;
  6. (In) anaphase, chromatids (from each pair)
    pulled to (opposite) poles/ends (of cell);
  7. (In) prophase/metaphase/anaphase, spindle
    fibres shorten;
55
Q

Use Figure 6 to suggest why iron-deficient plants have a reduced growth rate.
[3 marks]

A
  1. Less (thylakoid) membrane
    OR
    Fewer/smaller grana;
  2. Smaller surface area (of membrane in
    chloroplast)/less chlorophyll;
  3. (Less chlorophyll so) reduced light absorption;
  4. (So) slower rate of photosynthesis;
56
Q

Use your knowledge of phagocytosis to describe how an ADC enters and kills the
tumour cell.
[3 marks]

A
  1. Cell ingests/engulfs the antibody/ADC
    OR
    Cell membrane surrounds the antibody/ADC (to
    take it inside the cell);
  2. Lysosomes fuse with vesicle/phagosome
    (containing ADC);
  3. Lysozymes breakdown/digest the antibody/ADC
    to release the drug;
57
Q

Some of the antigens found on the surface of tumour cells are also found on the
surface of healthy human cells.
Use this information to explain why treatment with an ADC often causes side effects.
[2 marks]

A
  1. ADC will bind to non-tumour/healthy cells;
  2. Cause death/damage of non-tumour/healthy
    cells
    OR

Cause damage to other organs/systems;

58
Q

Describe how a triglyceride molecule is formed.
[3 marks]

A
  1. One glycerol and three fatty acids;
  2. Condensation (reactions) and removal of
    three molecules of water;
  3. Ester bond(s) (formed);
59
Q

The scientists used a data logger to measure the length of the root rather than a ruler.
Suggest one reason why they used a data logger and explain why this was important
in this investigation.
[1 mark]

A
  1. To increase accuracy/resolution because
    differences/lengths are small;
  2. To increase accuracy because reduces risk of
    human error;
  3. To increase accuracy because roots are less
    (likely to be) damaged;
60
Q

Describe the structure of DNA.
[5 marks]

A
  1. Polymer of nucleotides;
  2. Each nucleotide formed from deoxyribose, a
    phosphate (group) and an organic/nitrogenous base;
  3. Phosphodiester bonds (between nucleotides);
  4. Double helix/2 strands held by hydrogen bonds;
  5. (Hydrogen bonds/pairing) between adenine, thymine
    and cytosine, guanine;
61
Q

Name and describe five ways substances can move across the cell-surface
membrane into a cell.
[5 marks]

A
  1. (Simple) diffusion of small/non-polar molecules down a
    concentration gradient;
  2. Facilitated diffusion down a concentration gradient via
    protein carrier/channel;
  3. Osmosis of water down a water potential gradient;
  4. Active transport against a concentration gradient via
    protein carrier using ATP;
  5. Co-transport of 2 different substances using a carrier
    protein;

Biology Past Paper Questions - KEYWORDS (3 decks)

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