OCR A Biology
2.1
History of the light microscope and development of cell theory
1 idea that cell is unit of life (1) / many organisms unicellular (1) / (most) cells are too small to see
without microscope (1) / cell components / organelles, are even smaller (1) / idea that need to see
organelles to determine function (1)
2 Prior to the mid-19th century microscopes were of too low a magnification (1) to see and identify
cells (1) and cell components (1).
Sample preparation
1 a so light can shine through it (1) / details can be seen (1)
b reduce / prevent, diffraction between liquid and glass (1) / prevent / reduce distortion of
image (1)
c reduce / prevent air bubbles being trapped (1)
Using staining
1 Gram negative have thinner cell wall (1) / penicillin disrupts cell wall formation (1) / less cell wall
formation (in gram negative) / membrane (around gram negative) prevents entry of penicillin (1)
2 Avoid skin / eye contact (1) / wear gloves / goggles (1)
Less is more
1 a shading / label lines not touching relevant object (1) label lines not parallel with top of
page (1) no magnification stated (1)
Summary questions
1 Both plant and animal tissue is composed of cells (1); cells are the basic unit of all life (1); cells
only develop from existing cells (1)
2 Staining provides contrast (1) / different structures/organelles absorb stain differently allowing
identification (1).
3 Objective lens and eyepiece lens (1); objective lens magnifies the specimen (1); eyepiece lens
magnifies image (from objective lens) (1); higher magnification (produced than with just one
lens) (1)
4 a i 0 × 10 = 100 (1)
ii 10 × 40 = 400 (1)
b
area of field of view =   1000 2 = 3.141 × 10 μm
6
average area of cell =   30 = 2.827 × 10 μm
2
3
3.141 × 10 6
= 1111.11 = 1111 whole cells
2.827 × 10 3
2
2
(1)
(1)
(1)
2.2
Using a graticule to calibrate a light microscope
1 graticule / stage micrometre, eyepiece graticule (1)
2 20 divisions of the eyepiece graticule were equivalent to 9.5 micrometre divisions (1); 1
micrometre division = 10 µm (1); 20 graticule units = 95 µm (1); so 1 graticule unit = 95/20 =
4.75 µm (1); calibration factor of the ×10 lens = 4.75.
3 a 14 (1)
b
Results
Diameter of pollen grain /
divisions
Diameter of pollen grains /
µm
c
d
1
2
3
11
16
14
11 × 4.75 = 52.25
16 × 4.75 = 76.00
14 × 4.75 = 66.50
52.25 + 76.0 + 66.5 = 194.75/3 = 64.9 µm
Scale on eyepiece graticule is always the same (1); but magnification of other lenses changes
(1); need to calibrate eyepiece graticule for each lens to know actual measurements
represented by eyepiece graticule at different magnifications (1); necessary to calculate real
size of objects seen.
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OCR A Biology
Summary Questions
1 simplifies calculation (1) / reduces errors (1)
2 3846 × 10 × 1000 × 1000 (1) / = 3.846 × 1010 (1)
3 Contrast is difference in colour/shade between two objects (1) Resolution is the smallest distance
between two objects that can still be seen as separate (1)
4 Approximately × 366, when the diameter is 22 mm (2).
5 diffraction happens when light passes through structures (1) / light waves spread out (1) /
(light waves) overlap (1) / individual objects do not appear separate (1) / causes blurring (1)
6 (eyepiece graticule is) arbitrary scale / calibrated for each lens (1) / using stage micrometer (1)
2.3
Sample preparation for electron microscopes
fixation stabilise sample / prevents decomposition (1) / dehydration prevent vaporisation of water in
vacuum (1) vaporisation would damage sample (1) / embedding allows thin slices to be obtained (1)
staining with heavy metals creates contrast (1) in electron beams (1)
Scientific drawings from electron micrographs
a
is best (1) no shading (1) / label lines parallel with top of page (1)
Identifying artefacts
Evidence supports artefact theory (1) / not present normally (1) / idea that antibiotics responsible for
appearance (1).
Fluorescent tags
a ability to see individual objects as separate (1)
b resolution the same (1) / resolution limited by wavelength of light (1) / fluorescence is
light emitted (1) / super resolved fluorescent microscopy has higher resolution (1)
Atomic force microscopy
1 image not formed by light (1) / (image formed by) deflections of, tip / probe (1) / (as tip / probe)
moves across surface of specimen (1)
2 higher resolution (than electron microscope) (1) / magnification depends on resolution (1)
3 (AFM) only scans surface (1) / idea that cannot see into cells (1) / idea that need to see how
organelles are related to understand function (1)
Super resolved fluorescence microscopy
1 a specimen preparation kills cells (1); detail e.g., fixation (1)
b single molecules can fluoresce (1); multiple images obtained (1); different molecules fluoresce
in each image (1); images superimposed (1); idea of individual molecules seen in relation to
each other interacting (1)
Summary questions
1 Electron microscopes use electrons instead of light and electrons have a shorter wavelength
than light (1) which produces images with a higher resolution (1).
2 a An artefact is a visible object (1) or distorted cell structure (1) present in an electron
micrograph (or other micrograph) due to the sample preparation process (1).
b more sample preparation (in electron microscopy) (1) / (leads to) more damage to specimen
(1) / damage results in artefacts (1)
3 a Left: transmission electron microscope Right: scanning electron microscope (1)
b organelles visible in a (1) / a has greater magnification (1) b shows surface detail (1)
c TEM advantages greater, magnification / resolution (1) / more detail (1) TEM disadvantages
2D image (1) / very thin specimens needed (1) more preparation so more artefacts (1) SEM
advantages specimens do not need to be thin (1) 3D image (1) SEM disadvantages lower,
magnification / resolution (1) (max 6)
4 a Answers a emission of light (1); (that has been) absorbed (1)
b increase intensity (of light) (1)
c scattered light / light from outside the focal plane (1); is eliminated (1); reduces blurring /
increases resolution (1)
d idea of light penetration (of sample) is limited (1)
2.4
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OCR A Biology
Cell movement
1 microtubules (and microfilaments) are involved (1) / undergo polymerisation and hydrolysis (1) /
intermediate fibres are not involved (1) (have) role in cell stability (1)
Summary Questions
1 Lysosomes are specialised vesicles (1) that contain hydrolytic enzymes (1) for breaking down
waste material. The membrane that forms lysosomes has an important role in compartmentalising
these enzymes away from cell structures that could be damaged by activity of the enzyme (1).
2 Incompatible reactions / catabolic and anabolic reactions require different conditions / damage
due to hydrolytic enzymes (3) three named examples (e.g. nucleus, vesicle, lysosome,
mitochondrion, Golgi body, endoplasmic reticulum, chloroplast) (1).
3 Rough ER has ribosomes attached and smooth ER does not have ribosomes attached (1); rough
ER protein synthesis (and modification) (1); smooth ER lipid synthesis (1).
4 The cytoskeleton has three components: microfilaments (1) are contractile fibres made of actin
that bring about cell contraction during cytokinesis (1); microtubules (1) are formed from the
cylindrical protein tubulin and form scaffold like structures used both in the movement of
organelles and vesicles and as spindle fibres in the segregation of chromosomes/chromatids in
cell division (1); intermediate fibres give mechanical strength to cells (1).
5 a 7 × 107 × 0.34 × 10−9 (1) / = 2.38 × 10−2 × 46 (1) 1.09 m (1)
b coiled / wrapped (1) / around histones (1) / further coiling (1) / formation of chromatin (1)
6 microfilaments composed of actin (1) / (actin is) contractile (1) / microtubules composed of tubulin
(1) / (tubulin) polymerises (1) / (contraction and polymerisation lead to) change in length of
filaments (1) / change in length (of filaments) results in movement of cell (1) / intermediate fibres
have fixed length (1) / for stability (1)
2.5
Summary Questions
1 cell wall (1) / chloroplast (1) / plant cell (1) / presence of, chloroplast / cell wall (1).
2 a plant cell walls contain cellulose (1)
b prevent cells bursting (1) / allows turgidity (1) / idea that keep plants upright (1).
3 both have three named organelles (e.g. nucleus, cell surface membrane, mitochondria,
ribosomes, Golgi body, endoplasmic reticulum) (1) / only plants have two named organelles (e.g.
chloroplasts, cell wall, large (central) vacuoles) (1) / centrioles present in animal cells but not
flowering plants (1).
2.6
Endosymbiosis
1 mitochondria / chloroplasts, are (about) the same size as bacteria (1) / have a double membrane
(1) / (second membrane) acquired upon entry to cell (1) / contain DNA (1) / necessary for protein
synthesis (1) / (and) replication (1)
Prokaryotic cell study
1 a correctly drawn scientific diagram (1) / showing cell wall (1) / DNA / chromosome (1) /
cytoplasm (1)
b idea of many more structures (1) / membrane bound organelles (1) / three named structures
(e.g. nucleus, mitochondria, endoplasmic reticulum, Golgi body) (1)
Summary Questions
1 prokaryotic cells: no nucleus / no membrane bound organelles, e.g. mitochondria / smaller / 70s
ribosomes / plasmid / extra chromosomal DNA / peptidoglycan / murein cell wall. (Any 3). Accept
reverse arguments for eukaryotic cells.
2 Prokaryotic cells have ribosomes (1), which are needed for protein synthesis (1). Ribosomes are
not membrane bound (1).
3 Eukaryotic cells do not have peptidoglycan (1) cell walls (1) and these antibiotics do not damage
any other cell components (1) named example (e.g., nucleus, ribosomes, mitochondria) (1)
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2 Basic components
of living systems
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
1
D;
1
2 (a) (i)
X endoplasmic reticulum / ribosome;
Y golgi, apparatus / body;
Z vesicle;
1
1
1
3
2 (a) (ii)
eukaryotic; presence of, nucleus / membrane
bound organelle / named (membrane bound
organelle);
carbohydrases / hydrolytic;
2
nucleus; production of mRNA / transcription;
ribosome; protein synthesis / translation; golgi
apparatus; protein modification / alternative
wording (AW); vesicle; transported to cell
surface membrane; exocytosis / described;
vesicle (containing protein); fuses with cellsurface membrane; ATP / energy, required;
protein released (from cell);
vesicles have role in, transport / secretion;
vacuoles have role in storage; vacuoles contain
mostly water; vacuoles are larger (than vesicles
and lysosomes); lysosomes contain hydrolytic
enzymes; lysosomes have role in digestion;
15 mm=15 000 µm
15 000 / 2.6
Magnification × 5769
ability to see two adjacent points as separate
points; higher resolution means more detail can
be seen;
degree by which the size of an image is larger
than the size of the actual object;
magnification is increased without increase in
resolution so no more detail is visible;
objective lens magnifies image of object;
eyepiece lens magnifies image again;
dry mount; sample sectioned / AW; cover slip
placed over sample; wet mount; sample
suspended in liquid; cover slip is placed on at
an angle; squash slide; wet mount is pressed;
smear slide; liquid sample is smeared to
produce thin coating;
object visible in field of view; produced during
preparation of sample; e.g air bubble;
advantages greater magnification; greater
resolution; disadvantages sample must be
dead; expensive, set up / sample preparation;
large; many artefacts; technical skills required;
black and white images;
laser; point illumination / described;
fluorescence (from components labelled with
dye); light from specimen, travels through
pinhole / is filtered; (only) light from close to
5
2 (b) (i)
2 (b) (ii)
2 (c)
2 (d)
3 (a)
3 (b) (i)
3 (b) (ii)
3 (b) (iii)
3 (c)
3 (d)
4 (a)
4 (b)
4 (c)
© Oxford University Press 2015
Guidance
1
Max 5
3
4
Max 4, 1 mark per organelle extra
mark for extra detail.
2
2
1
1
4
6
2 marks per technique
2
4
4
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2 Basic components
of living systems
Answers to practice questions
OCR Biology A
5 (a)
5 (b)
6
focal plane is detected; unwanted radiation
does not go through pinhole;
present; membrane bound and non-membrane
bound; cellulose or chitin if present; small / 70s;
present;
cell components; organelles / named examples
(at least three);
prokaryotic genes arranged into operons;
number of genes controlled together; reduces
space needed for control elements; single
chromosome; eukaryotic more complex DNA
packing; supercoiled DNA; DNA wrapped
around histones; formation of multiple
chromosomes;
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5
2
6
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OCR A Biology
3.1
Summary Questions
1 Atoms form bonds with each other when pairs of electrons are shared (1) according to the bonding
rules (1).
2 A cation is an ion with a net positive charge (1), i.e. it has lost one or more electrons (1). An
anion is an ion with a net negative charge (1), i.e. it has gained one or more electrons (1).
3 Water – one oxygen atom binds to two hydrogen atoms (1), oxygen can form two bonds, each
hydrogen can only form one bond (1). Carbon dioxide – one carbon atom and two oxygen atoms
(1), carbon can form four bonds, each oxygen atom can form two bonds, therefore carbon forms a
double bond with each oxygen atom (1).
4 a X ray diffraction does not involves lenses (1) / electron microscope uses electromagnetic lenses
(1) / beams focused in electron microscopy to produce image (1).
b Cells are larger than ribosomes (1) / cells are larger than, half the wavelength / resolution limit, of
light (1) / electron microscopes have greater resolution (than light microscopes)
(1) / idea that molecules are smaller than resolution limit of light and larger than resolution limit of
electron beam (1).
3.2
Summary Questions
1 Oxygen and hydrogen share electrons unequally when they bond. Oxygen, has a greater share/ is
more negative (1). Hydrogen, has a smaller share/is more positive (1). The more negative oxygen
atom is attracted to the more positive hydrogen atom (1).
2 Water is composed of hydrogen and oxygen atoms and bonds between oxygen and hydrogen
involve unequal sharing of electrons (1) in bonds resulting in the oxygen atom being more negative
and the hydrogen atoms being more positive (1).
3 Liquid so transport medium (1); polar solvent (1); (many) biological molecules / examples
(e.g., enzymes, glucose), are polar (1); ions are charged (1); coolant so (relatively) resistant to
temperature change (1).
4 (water is) liquid (1) / allows movement of substrates and enzymes (1) / idea that this is necessary
for reactions to occur (1) / (water is) a polar solvent (1) / substrates / enzymes / products, are, polar /
ionic (1) / (water is) substrate for (some) reactions (1).
3.3
Summary Questions
1 Hydroxyl group on carbon 1 is in a different position (1); in alpha glucose it is below the ring in beta
glucose it is above the ring (1).
2 Bond formed between two glucose molecules (1) – hydroxyl group of carbon 1 on one molecule
(1) and carbon 4 (1) on the other interact in a condensation reaction/removal of water molecule (1) to
form an ‘oxygen bridge’.
3 Cellulose is straight chain molecule (1) with many hydrogen bonds between individual chains (1)
and staggered ends (1). This confers strength to the fibres (1).
4 In beta glucose the hydroxyl group at carbon 1 is above the ring (1) so alternate glucose molecules
must rotate 180 degrees (1) so the hydroxyl groups on carbon 1 and carbon 4 are close enough to
react (1) condensation reaction (1) forming a glycosidic bond (1). The rotation of molecules produces
a straight chain molecule (1) – cellulose.
3.4
Quantitative methods to determine concentration
Colorimetry
1 100% – transmission % = absorbance % (1)
2 (to) maximise absorption (1); complementary colour / red for Benedict’s solution (1)
3 use distilled water (1); set colorimeter to 100% (1)
4 unreacted Benedict’s solution (1); supernatant (1)
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OCR A Biology
5 correct axes (1); correct plots (1); line of best fit (1)
6 concentration of glucose at 44% absorbance (1); units (1)
Biosensors
1 Biological detector (1); presence of (toxic) gas causes a change (1); distress of bird is display (1);
canary in a cage is a biosensor (1); disadvantage not specific to one gas (1); ethical considerations of
causing harm to an animal (1).
Summary Questions
1 (enzymes have) active site (1); (active site) specific (1); to, substance / molecule, testing for (1)
2 Reducing sugars react with copper ions in Benedict’s reagent resulting in the addition of electrons to
blue Cu2+ ions (1), reducing them to Cu+ ions which form a brick red precipitate (1).
3 In an iodine test a purple/black colour indicates the presence of starch (1). Starch is a product of
photosynthesis (1). The test shows that starch is produce when light is available to the plant, but not
when the plant is kept in the dark (1).
4 Reagent strips are quantitative (1); they can be used to estimate the concentration (1) of reducing
sugar (glucose) (1) in the blood. They are simple to use and interpret (1).
3.5
Fats in our diet
1 hydrogenation / addition of hydrogen (1); removes double bonds in fatty acids (1); closer packing of
molecules (1)
2 unsaturated and saturated fat have high energy content (1); excess energy intake leads to obesity
(1)
Summary Questions
1 Oils are (usually) unsaturated (1); unsaturated fatty acids contain double bond(s) (1); molecules
cannot pack closely (1); fats are usually saturated so fatty acids have no double bonds (1).
2 Hydroxyl group from glycerol (1); hydroxyl group from fatty acid (1); condensation reaction (forms
ester bond) (1); idea that hydrolysis is reverse of the process described (1).
3aA
b Both have phosphate group attached to glycerol (1); both have fatty acid (tail) (1); cross links
between fatty acid tails in A (1) ; no oxygen attached by double bond (on A) / ester bond not present
(on A) (1).
c Cross links (1); stabilise membrane (1)
4 procedure for emulsion test statement 2, sample / lipid, dissolved in ethanol (1); water is mixed with
ethanol (and lipid) solution (1); statement 3, idea that water displaces lipid from ethanol forming
suspension (1); statement 1, (suspension forms because) lipids not soluble in water (1)
3.6
Separating amino acids using thin layer chromatography
1 a to prevent contaminating stationary phase (1); idea of biological material (on skin) (1) b testing
unknown compounds (1); not known whether, polar / non-polar (1); idea that the different solvents will
dissolve both polar and non-polar compounds (1)
c so the concentrated spots were not covered (1)
d (so) air inside jar is saturated with (solvent) (1); prevents evaporation of solvents (1)
2 a from bottom to top glycine, proline, phenylalanine (1)
b from bottom to top alanine, methionine (4)
Identification of proteins
1 mauve / lilac / purple (1)
2 no peptide bonds present (as no protein) (1); test is negative (1); solution (remains) blue (1); as
copper sulfate solution is blue (1)
3 Biuret test identifies peptide bonds (1); degree of colour change dependent on number of peptide
bonds (1); different proteins have different numbers of peptide bonds (1); idea that different degrees
of colour change could indicate different proteins not different quantities of protein (1).
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OCR A Biology
Summary Questions
1 Diagram showing the amine group (1); carboxylic acid group (1); and variable group (1); in correct
positions. See Figure 1 in Topic 3.6, Structure of proteins.
2 Condensation reaction (1) between amine group of one amino acid (1) and carboxylic acid group of
another (1), forming a water molecule.
3a
b Oxygen is relatively negative and hydrogen (attached to nitrogen) is relatively positive (1); oxygen
and hydrogen are attracted to each other (1).
c Secondary structures are (simple) repeating structures (1); globular protein / haemoglobin, has a
tertiary structure (1); tertiary structure is formed from complex folding of secondary structure (1).
4 R groups on amino acids interact (1); tertiary structure – interactions within a protein molecule (1)
determines shape of molecule (1); quaternary structure – interactions between protein molecules (1);
holds molecules together (1). Both involve the same interactions (1), i.e., hydrogen bonds, ionic
bonds, disulfide bonds, and hydrophobic and hydrophilic interactions.
3.7
The structure of fibrous proteins – elastin and collagen
1 strength, non-elastic
2 (collagen is a) large molecule so unlikely to enter skin (collagen), has a complex structure, idea of
individual components arranged in hierarchical structure, idea that new molecules would not
incorporate into existing collagen
3B
Summary Questions
1 Conjugated proteins contain a non-protein group (1) called a prosthetic group (1), simple proteins
do not (1).
2 insulin globular protein (1); soluble (1); specific shape (1); binds to receptor (1); chemical
messenger / described (1); keratin fibrous protein (1); strong (1); structural function / example (e.g.
hair, nails)
3 globular proteins hydrophobic R groups, in the centre (of the molecule) not in contact with water (1);
hydrophilic R groups, on the outside (of the molecule) / in contact with water (1); hydrophobic R
groups are repelled by water / hydrophilic R groups are attracted to water (1); fibrous proteins have R
groups on the outside of the molecule (1)
4 similarities globular (protein) (1); alpha helices (1); prosthetic group (1); hydrophobic R groups
positioned towards the centre (of the molecule) (1); differences single polypeptide not four
polypeptides / myoglobin tertiary not quaternary (1); no beta chains (1)
3.8
DNA extraction
1 reduce activity of enzymes (1); reduce breakdown of DNA (1)
2 disrupts membrane structure (1); phospholipids form suspension in aqueous solution (1)
Summary Questions
1 DNA nucleotide – deoxyribose sugar, thymine base (1); RNA nucleotide – ribose sugar, uracil base
(1).
2 A pyrimidine base always pairs with a purine base (1). Adenine and thymine/uracil always hydrogen
bond together (1) and cytosine and guanine always hydrogen bond together (1).
3 Polymer so contains a lot of information (1); idea that base sequence is used as a code (1); double
stranded so molecule is stable (1); double stranded so accurate replication (1).
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OCR A Biology
4 Adenine always base pairs with thymine, so same amount of thymine (17%) (1). Adenine and
thymine together is 34%, so cytosine and guanine must be 100 – 34 = 66% (1). Cytosine always base
pairs with guanine so cytosine amount equals guanine, 66/2 = 33. Therefore cytosine 33% and
guanine 33% (1).
3.9
Continuous and discontinuous replication
1 Continuous replication – DNA polymerase binds to the end of a strand, free DNA nucleotides added
without any breaks; discontinuous replication – DNA polymerase cannot bind to the end of a strand,
free DNA nucleotides added in sections, sections then joined.
2 Enzymes are (substrate) specific (1); DNA polymerase catalyses the joining of nucleotides (1);
nucleotides have a different shape to Okazaki fragments (1).
Summary Questions
1 Semi-conservative means ‘half the same’ (1). When DNA replicates the double helix unwinds
into two separate strands (1); free nucleotides pair with their complementary bases (1); two new
molecules of DNA are produced (1); each with one old strand and one new strand (1).
2 The triplet code is a particular sequence of three bases (1); that codes for a specific amino acid (1).
3 A mutation in the DNA changes the triplet code (1); meaning different amino acids are incorporated
into the protein/enzyme (1); that the DNA codes for. If such a change affects the precise structure of
the active site (1); a substrate may not be able to bind (1), rendering the enzyme non-functional.
4 The triplet code of DNA is degenerative (1), there are 64 different triplets/codons but only 20 amino
acids (1); therefore an amino acid can be coded for by more than one codon (1); so more opportunity
for differences in DNA sequence than amino acid sequence (1).
3.10
Summary Questions
1 From column left to right: UAC CGG AGU GCA
2 mRNA – copies gene from DNA (1), takes copy to ribosome (1); tRNA – brings amino acid to
ribosome (1); rRNA – formation of ribosome (1)
3 a catalyse the formation of bond between two amino acids (1); peptide bond (1)
b bind to tRNA (1); complementary base pairing (1)
c free floating ribosomes produce proteins for use in, cell / cytoplasm (1); bound ribosomes produce
proteins for export from the cell (1)
4 a role of protein dependent on structure (1); shape / 3D structure, dependent on primary structure /
sequence of amino acids (1); base triplets / codons, on mRNA, code for amino acids (1); introns
would code for, unnecessary amino acids / stop signal (1); codons could cause frameshift (1)
b different proteins produced from one gene (1)
c idea that originally functional gene(s) (1); mutation/s (1); base sequence/s changed (1); no longer
code for (useful) amino acids (1)
3.11
Summary Questions
1 sugar/ribose sugar (1); joined to a base/adenine (1); and to three phosphates (1)
2 It is present in all cells (1); it is present in all organisms (1). It releases energy in, small/ manageable
quantities (1).
3 (fat is) long term energy store (1); idea that fat is stable molecule and ATP is unstable molecule (1);
fat has other uses (1); e.g., insulation (1)
4 a bond formation releases energy (1); bond uses energy (1)
b ATP provides energy for, reactions / processes (1); ATP is present in all living organisms (1); idea
that there is no other equivalent molecule (1); therefore statement is valid (1).
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3 Biological molecules
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
1 (a)
two or more atoms; connected by (chemical)
bonds;
unequal distribution of electrons (across bond /
between atoms); (leading to) relatively, positive
/ negative areas (of molecule);
(water is) liquid; (water is a) polar solvent; e.g.
blood and glucose; sap and sucrose;
2
filter precipitate; dry; weigh; more precipitate
means more glucose; plot calibration curve; use
mass of precipitate from unknown solution
concentration to estimate concentration;
 ;  -; - -;
5
hydrophilic means attracted to water; phosphate
heads are hydrophilic; hydrophobic means
repels water; fatty acid tails are hydrophobic;
dissolve sample in alcohol; mix solution with
water; shake; cloudy solution is a positive result;
non-protein; tightly bound to protein; required
for function of protein; metal ion / iron ion, in
haem;
A phosphate group; B ribose / pentose / 5C
sugar; C nitrogenous base; D phosphodiester
bond;
many monomers joined; monomers are
nucleotides;
RNA is single stranded / DNA is double
stranded; RNA has uracil / DNA has thymine;
RNA is shorter;
DNA is double strand held together by hydrogen
bonds between bases; adenine binds to
thymine; cytosine binds to guanine; all bases,
form part of base pair / are bound to another
base;
incorrect base, inserted / deleted / substituted;
(leads to) change in sequence of bases;
mutation;
as rate of cell division increases the rate or
ribosome synthesis increases; linear
relationship; figures quoted; idea that trend
more closely followed at higher values;
mRNA binds to ribosome; tRNA binds to
ribosome; idea that tRNAs positioned to allow
binding of amino acids; enzyme that catalyses
formation of peptide bond is present in
ribosome;
as cell number doubles ribosome synthesis
doubles; figures quoted; new cells are
(genetically) identical; protein requirements the
same for each new cell;
A transcription; B translation;
4
arrows show information exchange; protein
does not code for anything; DNA cannot leave
3
1 (b)
1 (c)
2
3
4 (a)
4 (b)
5
6 (a)
6 (b)
6 (c)
6 (d) (i)
6 (d) (ii)
7 (a)
7 (b)
7 (c)
8 (a)
8 (b)
© Oxford University Press 2015
Guidance
2
4
Blood and sap are liquid which is
required for movement; glucose
and sucrose are polar molecules;
Max 5
3
4
4
4
2
2
4
3
4
3
4
2
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3 Biological molecules
Answers to practice questions
OCR Biology A
8 (c)
9
10
nucleus; DNA needs mRNA intermediate to
code for proteins;
RNA virus; RNA copied into DNA by reverse
transcriptase;
structure DNA nucleotide deoxyribose sugar /
four different bases ATP ribose sugar / one type
of base Both three phosphates / one base;
function DNA nucleotide part of genetic code;
ATP energy transfer;
1 hydrogen bond represented as,
horizontal / vertical, dashed line between O on
one
molecule and H on the adjacent molecule ;
2 hydrogen / H, bond label
(on any drawn bond between 2 molecules) ;
+
3 (delta positive)  on each drawn H
and (delta negative) (2)  on each drawn O ;
2
5
3 marks for structure and 2 marks
for function
3
1 DO NOT CREDIT if >1 H bond
is drawn between the same two
molecules 3 if both molecules
+
drawn,  and  on all atoms.
ACCEPT d (lower case) for 
11 (a)
ice floats
(ice less dense because) molecules spread out;
molecules form, crystal structure / lattice / AW ;
ice forms insulating layer / clearly described ;
water (below ice), does not freeze / still liquid /
remains water / kept at higher temperature ;
P1
P2
P3
P4
If possible, leave them off. If not,
yes, we should explain that they
are marks related to property
linked to survival.
organisms do not freeze ;
animals / organisms, can still, swim / move ;
allows, currents / nutrients, to circulate ;
S1
S2
S3
P3 e.g. acts as a barrier to the
cold
solubility
ions / named ion, polar / charged ;
ions /named ion, attracted to / bind to / interact
with, water;
P5
P6
(named) organisms / plants / animals,
uptake / AW, minerals / named mineral /
nutrients ;
correct use of named, mineral / nutrient, in
organism ;
S4
S1 DO NOT ACCEPT die
(because ‘survival’ stated in
stem)
S5
S4 ACCEPT obtain / enters /
goes in / gets
temperature stability
many / stable, (hydrogen) bonds between
molecules ;
at lot of energy to, force apart molecules / break
bonds ;
© Oxford University Press 2015
P7
S5 needs to be more specific
than ‘for growth /
metabolism’ suitable examples
include but are not
limited to: nitrates for amino acids
/ protein / (named)
nucleic acid / phosphate for ATP /
phospholipids /
plasma membrane / magnesium
for chlorophyll etc
P8
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3 Biological molecules
Answers to practice questions
OCR Biology A
high (specific) heat capacity ;
P9
temperature does not change much /
small variation in temperature ;
S6
effect of temperature on , enzymes / metabolic
rate ;
gases remain soluble ;
S7
Award once in any section hydrogen bonds ;
H
S8
8 max
P7 Many hydrogen bonds
between molecules = 2 marks
(gets P7 and H)
P8 ACCEPT heat as alternative
to energy
P9 DO NOT CREDIT latent heat
capacity
S6 could refer to organisms or
surrounding water
ACCEPT stays cool in summer /
stays warm in winter
DO NOT CREDIT constant alone
S7 ACCEPT any reference to
temperature affecting
enzyme activity / metabolic rate
DO NOT CREDIT if in incorrect
context
(e.g. they are strong bonds)
11 (b)
hydrolysis / hydrolytic ;
hydrophilic ;
© Oxford University Press 2015
2
ACCEPT phonetic spelling
throughout
IGNORE head
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OCR A Biology
4.1
Summary Questions
1 a protein (1)
b amino acids (1)
c specific, 3D shape / tertiary structure (1); (formation of) active site (1); binds to substrate(s) (1);
catalyses reaction (1)
2 Catabolism is breaking down of molecules (1); anabolism is building of molecules (1); reactions
involve breaking down and building of molecules (1); idea of metabolism is sum of all reactions (1).
3 a simple / easy to understand (1); representation (1).
b Both models substrate interacts with R groups in active site (binds) (1); (leading to) bond strain in
substrate molecule (1); Lock and key substrate is complementary to active site (of enzyme)
(1); Induced fit active site is flexible (1); (active site) changes shape as substrate binds, closer fit
between active site and substrate.
4 a energy required (1); to start reaction (1)
b bonds are broken (1); in substrate (1); energy required (1); energy (of system (increases) (1)
c Idea of improved technology (1); idea of continually investigated (1); more evidence (1); more
accurate representation (1).
4.2
Enzymes in action
Extremities are at, same temperature as rest of the body / higher temperature than normal, in womb
(1); enzyme / tyrosinase, not denatured (1); pigment is broken down (1).
Investigations into the effects of different factors on enzyme activity
1 easy to obtain (1); contains catalase (1) 2H2O2 ➝ 2H2O + O2
2 reactant (1); products (1)
3 volume of oxygen released increased (1); figures quoted (1); enzyme / catalase, catalysed reaction
(1)
4 Enzymes are proteins (1); (boiling) denatures protein (1); tertiary structure (of protein) changed
(1); active site no longer complementary to substrate (1); fewer enzyme-substrate complexes formed
(1); decreased reaction rate (1).
5 Same apparatus (1); range of hydrogen peroxide concentrations (1); temperature kept constant (1);
ruler used to measure change in water height (1); readings taken at set time intervals (1);
radius/diameter, of test-tube used to calculate volume (of gas collected) (1), repeats (1).
6 a independent variable – concentration (hydrogen peroxide) dependent variable – volume of gas
collected (1) b controlled variables – concentration of hydrogen peroxide (1); volume of hydrogen
peroxide (1); mass of liver tissue (1); surface area of liver tissue (1).
7 only independent variable / temperature, is only factor that is changed (1)
8 a reading that, lies outside the normal range / does not follow trend (1)
b 65 test 2 100 at 20 seconds
9 reliability / identify anomalies (1)
10 correct graph – axes including units (1); plots (1); lines of best fit (1); use of graph paper (1)
11 (Initial) rate of reaction higher with higher concentration (of substrate) (1); more substrate present
(1); more enzyme-substrate complexes formed (per unit time) (1); shown by steeper line (1); rate of
reaction, slows down / plateaus, as substrate used up (1); lower plateau at lower substrate
concentration (1).
12 results are valid because – reliability is good as repeats are (relatively) consistent (1); expected
trend observed (1); only one anomaly (1); many intermediates (1) results may not be valid because –
only three different concentrations (of substrate) used (1); temperature, may have changed / not
controlled (1); pH, may have changed / not controlled (1); may have been timing errors (1).
Serial dilutions
−3
Serial dilution (1); described e.g., 1 ml of stock solution and 9 ml of distilled water (1); 2 mmol dm
−3
−3
solution (1); repeat with diluted solution (s) (1); 0.2 mmol dm and 0.02 mmol dm and 0.002 m mol
−3
dm (1)
Summary Questions
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OCR A Biology
1 R-group interactions are disrupted (1); change in tertiary structure (1); change in 3D shape of active
site preventing binding with substrate (1).
2 Curve A (1): low/acidic pH is optimum pH and the stomach contains acid/has a low pH (1).
3 Bacterial enzymes have high optimum temperatures, human body temperature is lower (1).
Enzymes will have low activity (1) and bacteria will not thrive (1).
4 (at low temperatures) Kinetic energy is low (1); substrates / enzymes, move slowly (1); (so) fewer
collisions (1); collisions have less energy (1); increased flexibility of active site (1); increases chances
of successful collision (1).
4.3
Summary Questions
1 A non-competitive inhibitor binds to an enzyme away from the active site (1) at an allosteric site (1),
which has a different shape than the active site (1).
2 Inhibitor will always be present (1); some enzymes always inhibited (1).
3 End-product inhibition regulates rate of reaction (1); concentrations of substrate and product
determine reaction rate (1); (so must be) competitive (1); substrate concentration has no effect in noncompetitive inhibition (1); e.g. ATP and PFK in respiration (1).
4 Ethanol has similar shape to ethylene glycol (1); (ethanol) binds to active site of enzyme which
breaks down ethylene glycol (1); competitive inhibition (1); less ethylene glycol broken down (1); more
(ethylene glycol) leaves body unchanged (1); fewer toxic effects (1).
4.4
Enzyme activation and the blood clotting-mechanism
Enzymes responsible for blood clotting are present as precursors / described (1); e.g., Factor X,
prothrombin (1); prevents clotting unless required (1).
Summary Questions
1 Transfer, atoms / groups, between reactions (1); form part of active site (1).
2 Coenzymes bind loosely to enzymes (1); e.g. NAD (1); prosthetic groups are a permanent feature of
/ bind tightly to, proteins / enzymes (1); e.g. iron ion in haemoglobin (1).
3 Presence of cofactor (1); e.g., vitamin K and Factor X (1); change in tertiary structure / described
(1); e.g. (activated) factor X catalyses the breaking of bonds in prothrombin (1); forming thrombin (1);
thrombin catalyses the conversion of fibrinogen to fibrin (1).
© Oxford University Press 2015
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4 Enzymes
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
1
Statement 3;
1
2 (a)
5
2 (b) (i)
A / rate of reaction increasing active sites
available; increasing number of collisions
between active sites and substrates; more
enzyme substrate complexes forming;
B / reaction rate constant all active sites
occupied by substrate; idea of substrate
queuing up for active sites; Vmax;
enzymes;
2 (b) (ii)
break bond; involving the addition of water;
2
2 (b) (iii)
enzymes specific; amylose is hydrolysed to
maltose; active site of maltase is
complementary to maltose;
maltose hydrolysed / digested, by enzymes;
enzymes are proteins; stomach contents
contain proteases; proteases hydrolyse
proteins;
stomach contents are acidic; optimum pH of,
enzymes / named, is at higher pH; (pancreatic)
enzymes not denatured;
reaction is at maximum rate; all active sites
occupied; enzyme activity is maximum;
amino acids;
3
malonate is similar shape to substrate /
succinate; part of / group on, malonate the
same as on succinate; malonate, fits into / is
complementary to, active site of enzyme;
prevents substrate binding; competitive
inhibition;
B;
4
idea of control mechanism; need to be
reversible so more product can be made, when
needed; with competitive inhibitor, more
substrate would overcome inhibition; (leading
to) excess product;
prevent build-up of, excess / unnecessary
products; maybe toxic; prevent waste of
resources / energy; ensure sufficient levels of
required products;
take samples at a range of times / AW ;
same volumes (of solutions) added / removed
(each time) ;
heat with, Benedict’s (solution) / CuSO4 and
NaOH ;
(use of ) excess Benedict’s ;
changes to, green / yellow / orange / brown /
(brick) red ;
remove precipitate / obtain filtrate ;
colorimeter ;
calibrate / zero, using, a blank / water /
(unreacted) Benedict’s ;
3
Reaction rate of B eventually
equals A.
Max 3
3
Max 3
1
B2
B2 must be in context of
Benedict’s test rather than
reaction mixture
B3 DO NOT CREDIT boil / warm
B3 DO NOT CREDIT if
Benedict’s added to the
mixture at the beginning
C6 CREDIT description of
method
e.g. filtering / centrifuging /
decanting
8 IGNORE ‘control’
2 (b) (iv)
2 (c)
2 (d)
3 (a)
3 (b)
3 (c)
3 (d) (i)
3 (d) (ii)
4
© Oxford University Press 2015
Guidance
Maximum 3 for A or B.
1
4
3
3
1
1
B3
B4
B5
C6
C7
8
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4 Enzymes
Answers to practice questions
OCR Biology A
use (red / orange) filter ;
reading of, transmission / absorbance
OR
mass of precipitate ;
more transmission / less absorbance, of filtrate,
OR
greater mass ppt, = more maltose present ; ora
using, standard / known, concentrations (of
maltose) ;
(obtain) calibration curve ;
plot, transmission / absorbance / mass of ppt,
against (reducing sugar) concentration ;
use graph to read off concentration of maltose /
AW ;
9
T10
11
12
13
14
15
9 DO NOT CREDIT if colour of
filter is incorrect
T10 ACCEPT ‘measure how
much light, does / does
not, pass through’
11 if unfiltered Benedict’s /
precipitate is clearly
indicated as being present in
sample, ACCEPT ‘less
transmission / more absorbance,
= more maltose present’
11 DO NOT CREDIT if precipitate
is added to colorimeter
12 CREDIT ‘serial dilutions’
6 max
QWC – correct sequence ;
5 (a)
5 (b)
1 of mps B2 to B5, then mp C6 or
C7, then mp T10
increases / greater / faster ;
1
1 ACCEPT any time between
reaction completed in / plateaus after /
2
3.45 and 3.55 min.
concentration is 100% after, 3.5 minutes ;
3 two maltose concentrations (+
figures with units to support mp 1 ;
3
or – chloride) for a given time or
two times (+ or – chloride) for
2 max
given maltose concentration.
3 ACCEPT calculated difference
3 DO NOT CREDIT if ‘%’ and
‘min.’ not given
3 ACCEPT any concentration
within ± 1 % and time within ±
0.05 min.
Presence or
The percentage concentration of maltose (%)
absence of
present every half a minute
chloride ions 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
min min min min min min min min min
Chloride ions 0
24
54
70
80
88
95
100 100
present
Chloride ions 0
12
20
29
36
40
45
48
50
absent
Difference in 0
12
24
41
44
48
50
52
50
maltose
concentration
when
chloride ions
are either
present
or absent
Allow a + /- 1% for any concentration of maltose and a +/- 2% for the difference in maltose
concentrations
temperature ;
1
Mark the first three answers only
pH ;
2
regardless of
enzyme / amylase / chloride, concentration ;
3
which line they are on
substrate / starch / amylose, concentration ;
4
DO NOT CREDIT refs to, time
constant / regular, stirring ;
5
3 IGNORE ‘amount’ or ‘volume’
(fixed) volume of solution
6
3 DO NOT CREDIT
(removed each time for sampling) ;
3 max
‘concentration’ unqualified
4 IGNORE ‘amount’ or ‘volume’
4 DO NOT CREDIT
‘concentration’ unqualified
© Oxford University Press 2015
1
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OCR A Biology
5.1
Summary Questions
1 Membranes form cells and separate areas within cells (1); isolating each area from its external
environment (1).
2 Intrinsic protein – embedded in both sides of the bilayer (1); for example a channel protein or
carrier protein (1). Extrinsic protein – embedded in one side of the bilayer (1); for example a
glycoprotein or enzyme (1).
3 Lipid soluble molecules can pass through membranes (1); (by) simple diffusion (1); (so) diffuse
quickly through (whole) body (1).
4 Process occur within/across, membranes (1); process is enzyme controlled (1); folding gives
increased surface area (1); (so) more enzymes (1); increased rate of reaction(s) (1); and therefore an
increased rate of ATP production (1).
5.2
Investigating membrane permeability
1 a to remove all surface pigment released from damaged cells (1)
b to allow the mixture to equilibrate (1)
c repeats for reliability (1)
d because the pigment is red (1)
2 more pigment molecules absorb more light (1); light transmitted decreases (1)
3 the membrane was disrupted between 40 and 50°C (1)
4 same procedure except temperature constant (1); different (organic) solvents used (1);
e.g., ethanol (1)
Summary Questions
1 water is a polar solvent (1); phospholipids will not dissolve in water (1)
2 use of colorimeter (1); detail (e.g., use of filter) (1); range of readings taken (at different
temperatures) (1); graph (1)
3 Alcohol is lipid soluble and dissolves in membrane bilayer (1); this disrupts the bilayer and
stops/reduces transport of materials (1); preventing the normal functioning (1); and may cause cell
death. The liver is particularly affected due to its role in filtering substances from the blood (1); this
may ultimately be fatal if the liver function is destroyed (1); or if impulse transmission is depressed,
prevent involuntary reflexes (1); such as breathing and the gag reflex (which prevents choking).
© Oxford University Press 2015
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OCR A Biology
5.3
Rate of diffusion and surface area
1
Cube
size
(cm)
Surface
area
2
(cm )
Volume
3
(cm )
Surface
area /
volume
Diffusion
distance
(cm)
Rate of
diffusion
using
distance
(cm / min)
Rate of
diffusion
using
volume
3
(cm / min)
Rate of
diffusion
using
volume
per
3
64 cm
agar
28.8
57.6
111.2
4×4×4
96
64
1.5
0.3
0.03
28.8
2×2×2
24
8
3.0
0.3
0.03
7.2
1×1×1
6
1
6.0
0.3
0.03
1.8
2 sodium hydroxide solution has diffused into the agar (1); sodium hydroxide is an alkali (so
phenolphthalein indicator turns pink) (1)
3 larger width of pink colour in smaller blocks (1); smaller blocks have larger surface area to volume
ratio (1); (so) sodium hydroxide has diffused further into the blocks (1)
Investigations into the factors affecting diffusion rates in model cells
1 qualitative detects the presence of, reducing sugar / glucose (1); quantitative colour change is
estimate of concentration (of reducing sugar / glucose) (1);
2 tied dialysis tubing is, simplified / practical, representation of real cell (1); (tied dialysis tubing) has
the same properties as membrane (of cell) (1); demonstrates diffusion across (cell) membrane (1);
3 a cell membranes more complex (1); ORA (or reverse argument) (cell membranes) have, carrier
proteins / channel proteins (1); ORA active transport and diffusion across (cell) membranes (1); (cell
membranes) have hydrophobic core (1); ORA (cell membrane permeability) determined by size and,
polarity / charge (1); ORA
b (dialysis) tubing permeability based on pore size (1); dialysis tubing is not a barrier to (small) ions
(1); phospholipid bilayer (of cell membrane) is barrier to ions (1); ions diffuse through channel proteins
(in membrane) (1)
Summary Questions
1 Increased temperature increases the kinetic energy of particles (1); causing the particles to move at
increased speed (1).
2 Increased surface area (1); and reduced thickness (1).
3 Diffusion is described as passive because it does not require an external (metabolic) energy source
(1); diffusion relies on the energy from the natural random movement of particles (1).
5.4
Summary Questions
1 Diffusion is always a passive process, it does not require a metabolic energy source (1); in
facilitated diffusion a channel/co-transport protein aids diffusion (1).
2 Active transport requires metabolic energy in the form of ATP (1); produced in the mitochondria (1);
so these cells have more mitochondria.
3 Plants need mineral ions (1); concentration of mineral ions higher in root hair cells (than soil
solution) (1); mineral ions will diffuse out of root hair cells (1); energy required to move mineral ions
against concentration gradient (1).
5.5
Osmosis investigations
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OCR A Biology
1
Sugar
concentration
–3
(mol dm )
0.0
0.1
0.3
0.5
0.7
Original
mass (g)
Final mass
(g)
Difference in mass
(g)
% mass
change
3.0
4.0
1.0
33
3.0
4.1
1.1
37
3.3
4.2
1.1
33
3.0
3.5
0.5
17
3.2
3.6
0.4
13
2.9
3.3
0.4
14
3.0
3.0
0
0
2.9
3.0
0.1
3
3.2
3.2
0
0
3.2
2.8
–0.4
–13
3.0
2.6
–0.4
–13
3.1
2.7
–0.4
–13
3.1
2.2
–0.9
–29
3.3
2.4
–0.9
–27
3.0
2.0
–0.1
–33
Mean mass %
mass change
34
15
1
–13
–30
2 axes x sugar concentration with units and y mean mass change with units (1); plots (1); use of
graph paper (1)
3 increase in mass decreases as sugar concentration increases (1); mass increases as sugar
concentration of solution, is lower / has higher water potential, than tissue (1); water moves into tissue
by osmosis (1); mass decrease as concentration of sugar solution, is higher / has lower water
potential, than tissue (1); water moves out of tissue (by osmosis) (1)
4 three repeats (for each reading) increases reliability (1); no anomalous results suggest good
reliability (1); five concentrations used so range covered (1); results are valid (for potato) (1); number
of intermediates could be increased around estimated water potential (1); other plant tissue could be
used (1)
Summary Questions
1 One arrow from pure water to dilute solution (1); one arrow from dilute solution to concentrated
solution (1).
2 The water potential of pure water is zero (1); addition of solute decreases water potential (1);
therefore all solutions have negative water potential.
3 Where the graph line crosses the x-axis/where mass change is 0% (1); this is the isotonic value.
4 Electrolytes/solutes/minerals are necessary for many body processes (1) and help prevent excess
water loss by osmosis (1); to help maintain correct fluid balance for reactions (1).
5 Pine kernel tissue has the highest solute concentration/lowest water potential (1); as it does not
reach isotonic state even at the highest sodium chloride concentration (1); pine kernels (sometimes
called pine nuts) are the seeds of pine trees. Seeds store nutrients (1); for the seedling that will grow
and they have low water content (1); while dormant, requiring uptake of water to germinate.
© Oxford University Press 2015
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5 Plasma membranes
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
1
D;
1
2 (a) (i)
protein without, prosthetic / non-protein, group;
2
2 (a) (ii)
primary; secondary; tertiary / quaternary;
3
1 mark per protein structure
2 (b)
glycoproteins; cell, recognition / identification /
signalling / adhesion / receptors; channel
proteins; transport by facilitated diffusion; carrier
proteins; active transport;
use beetroot; contains betalain dye; cut beetroot
into small cubes; place in water at different
temperatures; (at least) 5 different
temperatures; sample water after set time ;
use filter in colorimeter; zero colorimeter ;
(for each sample) record, absorbance /
transmission;
use the same beetroot; same area of beetroot;
use cubes of the same size; same volume of
water;
repeat (at least) three times at each;
temperature; identify anomalies; calculate
mean;
alcohol disrupts cell membrane; of bacteria;
(helps) prevent infection;
positive correlation; as lipid solubility increased
so did permeability; quoted figures;
no correlation; molecule size does not affect
permeability; lipid solubility is more important
than size;
only certain ions / molecules can cross cell
membranes; ORA (or reverse argument)
depends on, size / charge / presence of
transport proteins; semi -permeable does not
give the idea of selectivity; ORA
should be, cell surface / plasma, membrane;
5
Max 2 per protein type.
6
Max 6
water molecules are small; moving fast;
between phospholipid molecules; through
transient gaps;
1 active transport requires ATP ;
at low temperatures:
2 (molecules have) little kinetic energy ;
3 (therefore) less, respiration / ATP made ;
4 less active transport or less, movement /
loading, of sugars into sieve tube (element) ;
5 less, osmosis / movement of water, into sieve
tube (element) ;
6 low (hydrostatic) pressure created ;
as temperature increases:
7 (molecules have) more kinetic energy ;
8 (therefore) more, respiration / ATP made ;
9 more active transport or more, movement /
loading, of sugars into sieve tube (element) ;
10 more , osmosis / movement of water, into
sieve tube (element) ;
2
3 (a)
3 (b) (i)
3 (b) (ii)
3 (c)
4 (a)
4 (b)
5 (a) (i)
5 (a) (ii)
5 (b)
6
© Oxford University Press 2015
Guidance
2
2
3
2
3
2
1
Max 2
1 ACCEPT loading / uptake for
transport
3 IGNORE no respiration / no
ATP made / no loading of
sucrose
4 ACCEPT slow active transport /
slow loading
9 ACCEPT faster active transport
/ faster loading
12 DO NOT CREDIT cells
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5 Plasma membranes
Answers to practice questions
OCR Biology A
7
11 higher / more (hydrostatic) pressure created;
12 at high temperature (plant), enzymes /
proteins, denatured ;
phospholipids; form bilayer; (bilayer) has
hydrophobic core; prevents diffusion of, polar
molecules / ions; idea that barrier separates
different areas; (bilayer) allows diffusion of nonpolar molecules; glycoproteins; membrane
stability; cell recognition; cholesterol; regulates
membrane fluidity;
© Oxford University Press 2015
3 max
6
denatured
12 CREDIT change to tertiary
structure, damage to proteins
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OCR A Biology
6.1
Cell cycle regulation and cancer
Similarities bind to enzymes (1); not at the active site (1); cell regulation (1); specific (1). Differences
inhibitors stop the activity of enzymes (1); cyclins activate enzymes (1).
Summary Questions
1 Mitosis is the process of replicating and diving the genome (1); cytokinesis is the physical division of
the cell (1).
2 DNA has been checked for errors (1); change in sequence of bases is a mutation (1); (leads to)
change in amino acid sequence (1); function of protein dependent on, 3D shape / tertiary structure
(1); tertiary structure dependent on primary structure (1); primary structure is sequence
of amino acids (1).
3 a Mutations occur during DNA replication (1); indefinite replication, increases chances of mutation /
accumulation of mutations (1); increased chance of harmful mutation (1).
b indefinite replication (1); cancer / formation of tumour (1).
4 a DNA is double stranded (1) b 3 × 109 / 50 (1); 6 × 107 (1)
c many origins of replication (1); idea of simultaneous replication of different lengths (1)
d (prokaryotic) genome, is shorter / has fewer genes / has no introns (1)
6.2
Summary Questions
1 Chromosomes only become visible under the microscope during mitosis/meiosis (1). DNA needs to
replicate for cell division. Chromosomes consist of two sister chromatids, which are identical copies of
DNA (1).
2 So that each daughter cell has identical DNA after mitosis/cell division (1); and correct number of
chromosomes (1) i.e., diploid after mitosis and haploid after meiosis.
3 Animal cells cleavage furrow forms around middle of cell (1); furrow pulls inwards and fuses (1);
plants cells furrow cannot form due to cell wall (1); vesicle assemble across centre of cell and fuse (1)
4 Prophase – 92 chromosomes have replicated G1 – zero replication has not occurred yet
5 Plant root tips continually grow at regions called meristems (1). Meristems are a good source of
cells for studying mitosis as they are constantly diving (1). Plant cells are easy to obtain (1) and
prepare for microscopy (1).
6.3
Summary Questions
1 a Meiosis I / the first division is a reduction division as each daughter cell is haploid (1).
b Gametes are the sex cells and two sex cell (one from each parent) must combine to produce a
diploid offspring (1); therefore gametes must contain only half the number of chromosomes/DNA, i.e.,
be haploid, otherwise with each new generation the number of chromosomes would increase (1).
2 A pair of same chromosomes, one from each parent (1); which have the same genes but can have
different alleles of each gene (1).
3 Anther(s) from a flower should be used (1); prepare a squash slide (1); use stain (1); observe using
microscope (1).
4 Crossing over – homologous chromosomes pair up (in prophase), non-sister chromatids entangle
(chiasmata) (1); and exchange genes/ alleles when they pull apart (at anaphase) (1). This produces
new combinations of alleles. Independent assortment – pairs of homologous chromosomes (meiosis
I)/chromosomes (meiosis II) line up on the equator (at metaphase) and each (pair/chromosome)
orientates independently (1); before being separated to opposite poles of the cell (at anaphase) (1).
This produces new combinations of alleles.
5a
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(1 mark for each allele pair)
b Creating different allele pairs during meiosis is an important source of genetic variation (1); in a
population. Genetic variation is important for the process of natural selection (1); giving individuals in
a population characteristics/traits that might confer an advantage (1); in changing environment (1); for
example pathogen resistance. If there was no genetic variation in a population, the entire population
would be vulnerable to such an external factor and there would be no opportunity for adaptation (1).
6.4
Summary Questions
1 Squamous (1); flattened cells provide thin surface (1) e.g. (alveoli) in lungs (1); diffusion of gases
(1); ciliated (1); (have cilia), for movement, of cell / liquid outside cell (1); e.g. trachea (1); movement
of mucus (1).
2 Any two appropriate examples with detail of structure related to function. (2 marks each).
3 A tissue is a collection of cells (1); that work together (1); an organ is a collection of tissues
(1); that work together (1).
4 Digestive system is a group of organs working together carry out a function (1); pancreas produces
digestive enzymes (1); stomach contains acid for digesting food (1); liver produces bile to aid
digestion of fats (1); small intestine digests and absorbs soluble food (1); large intestine absorbs
water from undigested food, producing faeces (1).
6.5
Gene therapy using stem cells
1 Bone marrow contains stem cells (1); stem cells can differentiate (1); into T cells (1); immune
system is functional (1).
2 Tissue (from) donor is not a good match (1); rejection (1); transplanted cells destroyed (1); patients
own cells used in gene therapy (1); no chance of rejection (1).
3 Another gene damaged during process (1); mutation of (another gene) (1); (lead to) uncontrolled
cell division (1).
Plant stem cells and medicines
Medicines are (often) derived from plants (1); many plants destroyed (in production of medicine) (1);
using stem cells reduces number of plants destroyed (1).
Summary Questions
1 Pluripotent – stem cells that can form all tissue types but not whole organisms (1). Only present in
embryos (1). Multipotent – stem cells that can only form a range of cells within a certain type of tissue
(1). For example, bone marrow is multipotent (or any appropriate answer) (1).
2 Shoot tips / root tips (1); (meristematic tissue contains) dividing cells (1); (leading to) growth (1); new
cells / stem cells can differentiate (1); (leading to) specialisation (1).
3 Embryos left over from fertility treatment (1); discarded anyway (1); embryos now created (to supply
stem cells) (1); embryos then destroyed (1); religious objections (1); life begins at conception (1);
embryo has rights (1); ownership of genetic material (1); (incurable) diseases cured (1); improved
quality of life (1).
4 a stem cells, divide and specialise (1); damaged tissue replaced (1)
b Progress of Parkinson’s disease delayed by drugs (1); symptoms of Alzheimer’s disease reduced
using drugs (1); drugs are only short term measure (1); possible side effects of drugs (1); idea of stem
cell therapy will lead to repair of tissue so, long term / permanent treatment (of both) (1); no / few, side
effects (1).
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6
Cell division
Answers to practice questions
OCR Biology A
Question Answer
number
Marks Guidance
1
D;
1
2 (a)
2
2 (e)
idea of specialised structures develop; carry out
specific functions;
have cilia; cilia can move; fluids can be moved
past the cell; e.g. mucus;
cells are individual; epithelium is a tissue; group of
cells; working together ;
cilia (on ciliated cells); waft / described, fertilised
egg;
D, B, A, E, C;
3 (a)
Q, T, P, R ; ; ; ;
4
3 (b) (i)
growth of cell / growth of organelles / increase
number of organelles / synthesis of proteins ;
1
3 (b) (ii)
mutation / faulty DNA produced / error in copying ;
daughter cells will not receive identical genetic
information ;
proteins / (daughter) cells, not made / do not
function ;
haploid / half genetic information / chromosome
number is n ;
genetic information not identical / produces
genetically different cells ;
4 cells produced ;
2
2
4 (b) (ii)
(three days = 72 hrs) 72 hours – 36 hours = 36
4
hours; (36 hours divided 8) = 4 2 = 32;
two sets of chromosomes; one set from each
parent;
gamete;
4 (c)
Do not pair up ; n ; 2n ; 2 ; 2 ; 1 ;  ;
4
1 mark per correct row
4 (d) (i)
two homologous chromosomes circled ;
1
ACCEPT one circle around both
chromosomes or two circles
The two chromosomes must be
of same length
2 (b)
2 (c)
2 (d)
3 (c)
4 (a)
4 (b) (i)
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3
4
2
1
2 max
Allocate marks for the following
pairs: S – Q Q – T T – P P - R
DO NOT ACCEPT ‘growth’
unqualified
DO NOT ACCEPT refs to DNA
replication
IGNORE ref. to respiration
ACCEPT named steps in protein
synthesis
ACCEPT ‘daughter cells will not
be clones’
ACCEPT ‘proteins / daughter
cells function differently’
ACCEPT use of comparative
chromosome numbers as
example
DO NOT ACCEPT identical / not
identical without ‘genetic’
DO NOT ACCEPT smaller cells
2
1
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6
Cell division
Answers to practice questions
OCR Biology A
4 (d) (ii)
three chromosomes, one from each pair ;
chromosomes drawn as one bar ;
2
Chromosomes should be of
different lengths however if two
are of similar length, look for
different centromere position to
award mark
ACCEPT
DO NOT CREDIT two joined
together at centromere
4 (e)
5 (a)
5 (b)
6 (a)
6 (b)
6 (c)
6 (d)
bacteria are prokaryotes / do not have a nucleus;
single chromosome; need homologous
chromosome(s) for meiosis;
15/16 × 100; 93.8%;
3
cell divides at the right time; cell does not divide
too often; cell is the right size ;
DNA has been checked; chromosomes are in the
right position;
Immature cell; undifferentiated cell; capable of
division;
embryo; umbilical cord; adult; meristematic tissue;
3
3
Max 3
ability to differentiate into different cell types ;
totipotent; can differentiate into any type of cell;
pluripotent; can form all tissue types (but not whole
organisms); multipotent; can only form a range of
cells within a certain type of tissue;
Heart disease; repair of muscle tissue in the heart
is damaged as result of a heart attack; Diabetes;
replace (beta) cells destroyed by the immune
system; Parkinson's disease; replace dopamine
producing cells in the brain; Alzheimer's disease;
replace brain cells are killed by abnormal proteins;
genetic disease; reverse previously untreatable
birth defects;
5
Max 5
6
2 marks per example
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2
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Module 2 Foundations in biology
Application and Extension
OCR Biology A
Question
number
Answer
Marks
Application
1 (a)
Non-polar macromolecules containing carbon
hydrogen and oxygen – most common form are
triglycerides, glycerol with three fatty acids
attached.
1 (b)
1 (c)
The membranes
Shows up the phospholipid bilayer structure of
the membrane.
2
Fatty acids are long chain carboxylic acids –
they all have long chain of carbon atoms with
hydrogen atoms attached, with a –COOH group
on one end. Monounsaturated fatty acids have
a single double bond in the carbon chain.
Polyunsaturated fatty acids have several double
bonds between carbon atoms. Saturated fatty
acids have no double bonds between the
carbon atoms of the hydrocarbon chain
(daigrams useful in all of these).
3 (a)
The people eating the Mediterranean diets,
whether with extra nuts or extra olive oil, had a
significantly reduced risk of dying from a heart
attack, stroke or cardiovascular disease over
the time of the study. The longer the study
continued the stronger the effect – after five
years around 6.5% people on low fat diet died,
compared to 4.5% on the Mediterranean diets.
Any other sensible points
3 (b)
How many people dropped out, balance of men
and women in each study, level of compliance
with the diet, is the same effect seen with
people who are not recognised as at risk of
heart disease, any other sensible suggestions
3 (c)
Look for evidence of students looking at several
different sources, the use of relevant and up-todate research, and clarity of explanations and
writing for the audience.
Extension 1
(a)
Protein released from the cardiac muscle cells
when they are damaged during a heart attack.
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Module 2 Foundations in biology
Answers to practice questions
OCR Biology A
Proteins are macromolecules made up of long
chains of amino acids. They have various levels
of structure primary the order of amino acids,
secondary the folding of the amino acid chain
into helices or pleated sheets (based on
hydrogen bonding), tertiary and quaternary
structure the 3-D structure of the single
helix/sheet or several chains joined (based on
hydrogen bonding, ionic bonding, and disulfide
bonds). If more of the heart muscle is damaged,
the higher the levels of cardiac troponins are in
the blood, so they act as indicators of both the
occurrence and the severity of a heart attack.
Don’t appear for several hours after the event.
1 (b)
Enzyme – protein which catalyses a reaction, in
this case between creatine and
phosphocreatine, an energy store in the muscle.
Enzymes are specific to a reaction or type of
reaction because of the shape of the active site
in the 3-D structure of the protein. Creatine
kinase is released by damaged muscles e.g., in
a heart attack but also severe bruising and
other conditions so it is not a specific marker for
heart atttack but it can be indicative with other
symptoms.
1 (c)
Myoglobin is a conjugated protein – a protein
attached to a non-protein prostheic group, in the
case of myoglobin this is iron. Molecule with a
high affinity for oxygen found in muscle tissue.
Released rapidly from the damaged heart
muscle after a heart attack – useful for early
diagnosis.
1 (d)
Peptides are chains of amino acids. There is a
peptide hormone produced by the heart muscle
that reduces blood volume and so the blood
pressure. Studies suggest that raised levels of
thiese peptides suggest a poor outcome for a
patient with a heart attack.
Should have section showing biochemistry of
carbohydrates and importance as energy supply
2
Similar section on poster for lipids
Cytology of adipose tissue and fat cells,
adaptations, numbers
Any other interesting material included should
be credited – for example discussion of
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Module 2 Foundations in biology
Answers to practice questions
OCR Biology A
amounts of food needed and exercise
Clear links to biochemistry and cell biology
required.
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Module 2 practice questions
Answers
OCR Biology A
Question Answer
number
Marks
1 (a)
Cation ion with positive charge; anion ion with negative
charge; ion in solution;
3
1 (b)
Ion
Function
calcium ions
component of teeth and bones
2+
(Ca )
phosphate ions
component of bones, ATP
3–
(PO4 )
+
sodium ions (Na ) role in nervous system
potassium ions
role in nervous system
+
(K )
ammonium ions
nitrogen source, removal of
+
(NH4 )
nitrogen
DNA polymerase; Okazaki fragments; DNA strand; DNA
ligase;
5
2 (b) (i)
Simple outline of process; representation;
2
2 (b) (ii)
A;
1
2 (b) (iii)
Fewer errors; original strand used as template; errors can
be corrected; idea that errors would not be as obvious if
both strands were new;
Increases initially; to a peak / then decreases; numbers
quoted;
Increase due to enzyme activation; copper is a cofactor at
low concentration; decrease because copper is an inhibitor
at high concentration; high concentrations of copper,
denature / change the 3D shape of enzyme;
Activity of enzyme reduced due to denaturation; at
increased temperature, bonds / named, broken; 3D shape
of active site changed; substrate no longer complementary
to active site;
B; horizontal label lines; label lines end at part of drawing
being labelled; heading;
Diameters of at least two ribosomes measured; mean
calculated; answer divided by 362 500;
Electron; transmission;
4
mRNA binds to ribosome; tRNA brings specific amino acid;
complementary anticodons bind to codons; peptide bonds
form between adjacent amino acids;
4
2 (a)
3 (a) (i)
3 a (ii)
3b
4a
4 b (i)
4 b (ii)
4 b (iii)
5a
Guidance
4
3
4
4
4
3
2
3
description
letter
an animal cell that
has been placed in
distilled water
an animal cell that
has been placed in
a concentrated sugar
solution
a plant cell that has
been placed in
distilled
N;
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L;
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Module 2 practice questions
Answers
OCR Biology A
5 (b)
5 (c)
water
a plant cell that has
M
been placed in a
concentrated sugar
solution
water moves out of cell;
by osmosis;
cell has higher / greater / less negative water potential (than
surrounding solution) / ORA;
(water moves) down water potential gradient/from high to
low water potential;
small, non-polar substances:
diffuse (through membrane / phospholipid bilayer);
large substances:
(using), transport / carrier, proteins;
endocytosis / phagocytosis / described;
polar substances:
through, pore / channel, proteins;
(using), transport / carrier, proteins;
general – must be used in correct context, each once only
ref to facilitated diffusion;
ref to active transport / use of ATP;
3 max
5 max
QWC – technical terms spelled AND used in correct
context;
6 (a) (i)
6 (a) (ii)
6 (b) (i)
6 (b) (ii)
use starch solution; constant concentration and volume;
iodine solution; blue/black, indicates the presence of starch
or yellow/brown indicates the absence of starch;
colorimeter; correct filter; (colorimeter) calibrated; constant
volume of lead nitrate solution; lead nitrate solutions
prepared by serial dilution; lead nitrate solution and starch
solution mixed for set time;
19.7;
4 max
idea that result was very different from other two results;
anomalous result;
correct axes including units; plots correct; line of best fit;
use of graph paper;
2
© Oxford University Press 2015
note: this is explain not
describe ACCEPT Ψ
for water potential must
be comparative – DO
NOT ACCEPT high
alone
DO NOT ACCEPT
across or along water
potential gradient
DO NOT ACCEPT ref
to water concentration
anywhere
IGNORE ref to solute
potentials
ACCEPT diffusion /
diffuses
ACCEPT protein pump
DO NOT ACCEPT
channel proteins here
ACCEPT pinocytosis
apply only to large /
polar substances
apply only to large /
polar substances
DO NOT ACCEPT ref
to active transport with
channel proteins
(three from:
phospholipid / bilayer /
diffusion / facilitated
diffusion / active
transport / transport
protein / carrier protein
/ channel protein /
pinocytosis /
endocytosis /
phagocytosis) if protein
spelled incorrectly
throughout, only
penalise once
1
3 max
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Module 2 practice questions
Answers
OCR Biology A
6 (c) (i)
(mean) transmission of light decreased as concentration of
lead nitrate increased; non-linear relationship / described;
figures quoted; increased inhibition decreased, breakdown /
hydrolysis of starch; more starch present resulted in
decreased transmission (of light);
increasing concentrations of inhibitor lead nitrate decreased
activity of enzyme amylase;
correct for the concentrations (of lead nitrate) tested; for the
enzyme amylase;
incorrect as not all concentrations tested; only one enzyme
tested;
valid results follow expected trend; repeats are similar; no
anomalous results;
not valid for all concentrations of inhibitor; different
enzymes;
use different inhibitor; use different enzyme; more
intermediates; more repeats;
stem / undifferentiated;
(bone) marrow;
differentiate;
meristem(atic) / cambium;
4 max
7 b (i)
idea of: create flow of water / move water;
1
7 (b) (ii)
strain / filter (the water) OR trap particles;
to catch food (particles);
1 max
6 (c) (ii)
6 (d)
6 (e)
6 (f)
7 (a)
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1
4
2 max for each
4 max
1 max
4
Mark the first answer
for each prompt line. If
the answer is
correct and an
additional answer is
given that is incorrect
or
contradicts the correct
answer then = 0 marks
ACCEPT totipotent /
pluripotent
IGNORE unspecialised
(as specialised in the
passage)
IGNORE specialise as
given in the passage
ACCEPT callus
Mark the first answer
only. If the answer is
correct and an
additional answer is
given that is incorrect
or contradicts the
correct answer then = 0
marks
DO NOT CREDIT ref to
movement of, organism
/ cell
IGNORE ref to liquid /
food particles
Mark the first answer
only. If the answer is
correct and an
additional answer is
given that is incorrect
or contradicts the
correct answer then = 0
marks
IGNORE trap
substances unqualified
ACCEPT named
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Module 2 practice questions
Answers
OCR Biology A
7 (c)
Xylem:
consists of vessels;
4 max
one cell specialisation described;
transpiration stream OR
movement of water / minerals;
phloem:
sieve tube element(s) and companion cell(s);
one cell specialisation described;
translocation OR
transports, sucrose / assimilates / products of
photosynthesis / amino acids;
AVP;
8 (a)
5.6; –1.4;
2
8 (b)
correct axes including units; plots correct; line of best fit;
use of graph paper;
as the sucrose concentration increased the final diameter
–3
decreased; 0.05 and 0.10 mol dm solutions caused
diameter (of cells) to increase and 0.20, 0.40 and 0.80
–3
mol dm caused diameter to decrease; figures quoted
(including units); solutions with lower concentration had
higher water potential than cytoplasm; so water moved into
cells by osmosis; solutions with higher concentration had
lower water potential than cytoplasm; so water moved out
of cells by osmosis;
4
8 (c)
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suitable food particles,
e.g. bacteria
IGNORE ref to
preventing infection /
catching pathogens
IGNORE ref to
nutrients unqualified as
these are dissolved
IGNORE ref to catching
dust
ACCEPT cells joined
end to end
ACCEPT continuous
column / tube
e.g. wall water proof /
wall lignified / no end
walls /
(bordered) pits / hollow
/ no organelles / no cell
contents
IGNORE dead
IGNORE transpiration
unqualified
ACCEPT sieve element
/ sieve tube, and
companion cell
e.g. sieve plates
(between phloem
elements)
no nucleus / few
organelles, in sieve
tube (elements)
little cytoplasm in sieve
tube (elements)
many plasmodesmata
many mitochondria /
dense cytoplasm, in
companion cells
ACCEPT sugar
IGNORE load / unload
sugars alone
in either xylem or
phloem
ref to fibres
ref to, packing cells /
parenchyma cells
4 max
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Module 2 practice questions
Answers
OCR Biology A
8 (d)
red blood cells;
1
9 (a)
correct axes including units; plots correct; line of best fit;
use of graph paper;
tangent drawn; gradient correctly calculated; (approx.) 0.7;
mm/min;
rate of diffusion decreases as distance (from well)
increases; figures quoted; idea that rate of diffusion is
higher when concentration difference/gradient is
greater/steeper; concentration gradient decreases as
distance (from well) increases;
±0.5 mm;
4
9 (b)
9 (c)
9 (d) (i)
9 (d) (ii)
4
4 max
1
4
9 (d) (iii)
0.5/14; ×100; 3.57; %;
using a ruler with smaller divisions;
1
9 (d) (iv)
repeated the experiment; (at least) twice;
2
9 (e)
4 max
10 (c)
valid because trend was as expected; percentage error was
low; no points far from trend; not valid because no repeats;
do not know, how reliable the results are / if there are any
anomalous points;
as the pH moves away from neutral the activity of the
enzyme, amylase will decrease;
independent variable pH; dependent variable time (taken
for starch to be, broken down / hydrolysed);
concentration / volume of starch; concentration / volume of
enzyme; temperature;
pH
Time for
Mean
complete
time /
hydrolysis / minut
minute
e
1st run
2nd
3rd run
run
5
11
10
8
9.6
6
7
6
7
6.6
7
3
4
3
3.3
8
4
5
6
5.0
9
10
9
10
9.7
10 (d)
repeats; identified anomalous result(s);
2
10 (e)
support hypothesis; enzyme activity highest at pH 7 / lowest
at high and low pH; figures quoted; idea that highest activity
may be between pH 6 and 7 or pH 7 and 8 which would not
support the hypothesis;
4
10 (a)
10 (b) (i)
10 (b) (ii)
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2
2
4
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OCR A Biology
7.1
Summary Questions
1 Metabolic activity relatively low (1); so relatively little oxygen needed or carbon dioxide produced
(1). SA : V is large (1); so diffusion distances small (1).
2 Large SA for exchange to overcome limitations of SA : V ratio of larger organisms (1); thin layers so
distances substances have to diffuse short, making the process fast and efficient (1); good blood
supply so substances constantly delivered to and removed from exchange surface which maintains
steep concentration gradient for diffusion (1); ventilation (for gaseous systems) maintains
concentration gradients and makes process more efficient (1).
3 Radius 2 au = 3:2 Radius 6 au = 1:2
(1 for correct ratio, 1 for correct workings in each case)
The SA : V ratio of smaller animal is three times bigger than that of larger animal, this illustrates how
the SA : V ratios of larger animals are much smaller than those of smaller animals (1); as a result they
need specialised exchange systems to get enough oxygen in, or carbon dioxide out of the system (1).
7.2
Attacking asthma
1 Give immediate relief from symptoms (1); attach to receptors on membranes of smooth muscle cells
(1); relax smooth muscles (1); dilate airways (1).
2 Steroids taken every day (1); reduce sensitivity of lining of airways to asthma triggers (1); reduce
likelihood of an attack (1).
The first breath
1 Before first breath tissue has never been extended (1); drawing air in baby has to overcome the
elastic recoil of the lungs (1); and the adhesion of the surfaces (1); 15–20 times more effort than next
breath (1); lung surfactant stops alveoli collapsing and surfaces sticking together (1); means
subsequent breaths easier (1).
Summary Questions
1 a Nose: large SA with good blood supply warms the air to body temperature (1); hairy lining
secretes mucus which traps dust and bacteria, protecting delicate lung tissue from irritation and
infection (1); moist surfaces increase humidity of incoming air, reducing evaporation from exchange
surfaces (1); produces air at similar temperature and humidity to air already in lungs (1). (max 3)
b Trachea: wide tube, supported by incomplete rings of strong, flexible cartilage that stop tube
collapsing (1); rings incomplete so food moves easily down oesophagus behind trachea (1); lined with
ciliated epithelium with goblet cells between epithelial cells (1); goblet cells secrete mucus to trap dust
and bacteria (1); cilia beat and move mucus and trapped particles away from lungs to throat to be
swallowed and digested. (max 3)
c Bronchioles: Small tubes spreading into both lungs (1); the smaller bronchioles (diameter 1mm or
less) have no cartilage rings. The walls contain smooth muscle which contracts to close up
bronchioles and relaxes to dilate them, changing the amount of air entering the lungs (1); lined with
thin layer of flattened epithelium, making some gaseous exchange possible (1). (max 3)
2
2 a large SA of ~50–75 m for gaseous exchange (1); thin layers so short diffusion distances (1); good
blood supply with large capillary network supplying alveoli bringing carbon dioxide and picking up
oxygen, maintains steep concentration gradient for carbon dioxide and oxygen between air in alveoli
and blood in capillaries (1); good ventilation as breathing moves air in and out of alveoli, helping
maintain steep diffusion gradients for oxygen and carbon dioxide between blood and air in lungs (1).
b Alveolar structure breaks down giving air sacs with much bigger radii (1); this reduces surface to
volume ratio which makes them much less effective for gaseous exchange (1), e.g., (Students can
choose any radius they like – it will have a noticeable effect)
3 Trachea lined with ciliated epithelium with goblet cells that secrete mucus (1); mucus traps dust and
bacteria (1); cilia beat to move mucus and trapped particles away from lungs to throat to be
swallowed and digested (1); in smokers cilia anaesthetised so do not beat (1); mucus with its load of
bacteria and dust moves down into the lungs (1); more pathogens reach lungs so smokers more likely
to get infections of breathing system than non-smokers with active cilia (1).
7.3
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OCR A Biology
Summary Questions
1 Record number of breaths for a timed period and repeat (1); calculate means of results under
different conditions (1); use spirometer to observe breathing rate (1); any other sensible suggestion.
2 Ventilation rate is tidal volume of air breathed in at each breath, multiplied by number of breaths per
minute (breathing rate) (1); units are cm3 or litres per minute.
VR = TV × bpm (1)
Oxygen uptake closely related to ventilation rate, the more air is moved into the lungs, the more
oxygen can be taken up by haemoglobin in blood (1); so as ventilation rate increases oxygen uptake
also increases (1).
3 VR = TV × BR dog under stress so pants, breathes rapidly, but breathes shallowly so although
breathing rate increases tidal volume falls (1); so the ventilation rate stays the same (1).
3
4 a VR = TV × BR (1) so VR/BR = TV (1); TV = 45000/30 = 1500 cm . (2)
3
Normal TV = 500 cm . (1) So during strenuous exercise it is 3 × higher (1)
b VR = TV × BR (1); so with infection VR = 300 × 25 = 7500 (2)
Normal 500 × 18 = 9000 (1)
9000 – 7500 = 1500 (1)
1500/9000 × 100 = 16.7% fall in ventilation rate (1)
7.4
Discontinuous gas exchange cycles in insects
Look for thought and ingenuity on the part of the student along with recognition of the difficulties in
these types of investigation and the need for safe and ethical handling of insects (up to 5 marks for
each experiment suggested).
Dissecting, examining, and drawing gaseous exchange systems
1 Easy to make alterations and corrections (1); won’t run if it gets damp/splashed (1)
2 For each diagram give marks for use of pencil; size; accuracy; and quality of drawing; accuracy of
labelling.
The histology of exchange surfaces
1 For each diagram give marks for use of pencil, size, accuracy, and quality of drawing, accuracy of
labelling (3).
2 Show up more detail than can be seen with the naked eye (1); can use stains to show up specific
aspects of tissues or organs (1).
Summary Questions
1 In air gill filaments all stick together (1); SA for gas exchange is greatly reduced and so fish dies
from lack of oxygen (1).
2 Table should compare key differences between humans, insects, and bony fish. For example the
main organ of gaseous exchange, entry into exchange system, main site of gaseous exchange, and
how ventilation occurs (5 marks).
3 Fluid towards end of tracheole limits penetration of air for diffusion (1); when energy demands high
lactic acid build up in tissues, water moves out of tracheoles by osmosis, exposing more surface area
for gaseous exchange (1); tracheal system can be mechanically ventilated with air actively pumped
into system by muscular pumping movements of thorax and/or the abdomen (1);movements change
volume of body, changing pressure in tracheae and tracheoles so air drawn into trachea and
tracheoles, or forced out, as pressure changes, making gaseous exchange more efficient (1). Some
very active insects have collapsible enlarged tracheae or air sacs which act as air reservoirs, used to
increase amount of air moved through gas exchange system (1); they are usually inflated and
deflated by ventilating movements of thorax and abdomen (1). (max 6)
4 Gills have large stacks of gill filaments carrying gill lamellae that have large surface area (1); good
blood supply (1); and thin layers (1); needed for successful gaseous exchange. Constant flow of water
maintained over gills so best possible diffusion gradient for the respiratory gases (1); tips of gill
filaments overlap – increasing resistance to flow of water, slowing it down for more effective gaseous
exchange (1); water and blood flow in opposite directions. Countercurrent exchange system
maximises the potential exchange of gases (1). An annotated diagram/diagrams to make any or all of
these points would be acceptable.
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7 Exchange surfaces
and breathing
Practice questions
OCR Biology A
Question Answer
number
Marks
1
D;
1
2 (a)
fastest rate of air flow from lungs; indication of lung
function;
always higher for men; same pattern for men and
women as age increases; (for both) initial increase in
lung function; (for both) to a peak and then
decreases; quote figures;
line of the same shape but below bottom line (for
men); asthma would lead to reduction in lung
function; at all ages; (unlikely to be seen) as would
normally be treated;
for inhalation diaphragm contracts; requires, ATP /
energy; for exhalation elastic recoil of lungs;
large surface area of lungs; ventilation/AW, maintains
concentration gradient; blood flow maintains
concentration gradient; thin alveolar wall;
oxygen concentration lower in water than air; ORA
operculum; ram ventilation / described; counter
current mechanism / described;
chitin;
2
composed of monomers / repeating units; Nacetylglucosamine is the monomer;
spiracles; trachea; lined within chitin to keep them
open; tracheoles; run between /next to, individual
cells; gaseous exchange between tracheoles and
cells by diffusion;
both held open by rigid material; to maintain air flow;
chitin in insects; cartilage in mammals;
3
3
5.75 – 2.75 = 3000 cm ; 2.75 – 2.25 = 0.50 cm ;
2
2 (b)
2 (c)
2 (d)
3 (a)
3 (b)
4 (a)
4 (b)
4 (c)
4 (d)
5 (a)
5 (b)
5 (c)
4
Guidance
Max 4
3
3
1 mark per point. Max 3
3
Max 3
3
1
4
Max 4
2
2
2
Shorter red arrow = expiratory reserve volume
Longer black arrow = residual volume
tidal volume is higher during pregnancy; figures
quote; expiratory reserve volume lower; more oxygen
required for fetus; total lung volume / residual
volume, decreased, due to space occupied by fetus;
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6
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OCR A Biology
8.1
Summary Questions
1 Transports requirements for metabolism, e.g., oxygen, food molecules, to cells (1); removes waste
products of metabolism from cells and carries them to excretory organs (1); transports materials made
in one place to another place where they are needed (1).
2 Unicellular organisms have large SA : V ratio so diffusion distances small and metabolic demands
low so diffusion can supply and remove substances quickly and efficiently enough (2). Multicellular
organisms have small SA : V ratio, so long diffusion distances. Metabolic demands are high – diffusion
alone can no longer supply all needs quickly and efficiently enough (2).
3 Similarities: liquid transport medium (1); vessels to transport the medium (1); pumping mechanism
to move transport fluid around system (1).
Differences: open has few vessels; closed has transport medium (blood) enclosed in vessels
(1). In open transport medium is pumped into body cavity (haemocoel) under low pressure; in closed
heart pumps blood around body under pressure (1). In open, transport medium is in direct contact
with body cells; in closed transport medium has no direct contact with body cells (1). In open transport
medium returns to heart through open ended vessel; in closed blood flows relatively fast and returns
to heart all within vessels (1) (max 6)
4 Land predators top land predators hunt so need ability to move in fast bursts (1); they grow large
and maintain own body temperature (1); need to support body against gravity (1); they may be
pregnant and so have to support needs of growing fetus as well as own body needs (1) (max 3
marks) high metabolic rate (1); they need a very efficient circulatory system supply. Double circulatory
system supplies blood to lungs to be oxygenated and then returns it to heart to be pumped around
body (1); so tissues receive a high level of oxygen and high levels of carbon dioxide
can be removed. (1).
Aquatic predators such as pike need to hunt so also need efficient circulatory system (1); their single
system less efficient than a double system (1); but bony fish have operculum so continuous flow of
water over gills to oxygenate blood (1); countercurrent flow allows efficient oxygen uptake (1); they do
not maintain their own body temperature and are supported by water (1); so demands of tissues much
lower than those of an animal like a fox (1); so single circulation is adequate to supply their needs.
8.2
Collagen, elastin, and aortic aneurysms
1 A higher proportion of collagen: elastin increases likelihood that blood vessels will develop an
aneurysm. (2); blood vessel less elastic (1); so less able to withstand surges of blood in aorta (1);
more likely to stretch and bulge permanently (1).
2 Reduce high blood pressure (1); regular screening of aorta for signs of aneurysm developing (2).
Summary Questions
1 Arterial blood under pressure from pumping of blood and elastic recoil of artery walls, so no
tendency for it to flow backwards (1). After passing through capillary beds blood in veins under much
lower pressure, there is no pumping from heart and little elastic recoil in veins so blood might flow
backwards (1); as it moves back towards heart against gravity. Valves prevent this happening – they
open as blood flows towards heart and close if it flows in opposite direction (1).
2 Arterioles have more smooth muscle and less elastin in walls than arteries, as they have little pulse
surge (1); smooth muscle means they can constrict or dilate to control flow of blood into individual
organs by preventing blood flowing into a capillary bed (vasoconstriction) or allowing it to flow
(vasodilation) (1).
3 a Diagrams are helpful in describing and comparing structures
Large veins have thin walls as don't have to withstand high pressures of arterial system (1); large
lumen as they contain large volume of blood (1); smooth muscle in veins contracts/ relaxes allowing
constriction/dilation to change amount and pressure of blood (1); walls contain collagen and relatively
little elastic fibre, so there is a limit to amount of blood that can flow through them (1); wide lumen and
smooth lining mean blood flows easily. (3 – for any three relevant points).
Medium sized veins have similar structures and function to large veins but also have valves, which
prevent backflow and move it through the venous system to largest veins and so back to heart (1).
Venules less structure in walls than veins; very thin walls with a little smooth muscle to allow blood to
flow onto into veins; venules do not have valves so cannot control blood flow. (max 2)
b In all other areas of adult body veins and venules carry deoxygenated blood back from body to
heart (1). In lungs they carry oxygenated blood from lungs back to heart (1).
© Oxford University Press 2015
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OCR A Biology
8.3
Summary Questions
1 Transport of oxygen (1); and carbon dioxide (1); (to and from respiring cells, respectively); transport
of digested food from intestine to cells (1); transport of nitrogenous wastes from tissues to excretory
organs (1); transport of: hormones (1); platelets (for clotting) (1); and antibodies (1); immune response
(1); maintaining constant body temperature (1); and pH (1). (max 4)
2 a Platelets fragments of large cells called megakaryocytes (1); found in red bone marrow (1).
b Carried around in body in circulatory system (1); involved in blood clotting mechanism (1); which
prevents blood loss after injury.
c 250 000 (1)/5257000 (1) ×100= 4.76% (1).
3 Plasma: straw coloured liquid which contains water, dissolved glucose and amino acids, mineral
ions, hormones and large plasma proteins including albumin (important for maintaining osmotic
potential of blood), fibrinogen (important in blood clotting) and globulins (involved in transport and
immune system). (3 for any three points)
Tissue fluid: Liquid contains same constituents as plasma except the plasma proteins – so no
albumin, fibrinogen or globulins (1).
Lymph: Liquid similar to tissue fluid but with less oxygen and digested food and more carbon dioxide
and waste (it has been past the cells) and more fatty acids from small intestine. (2)
4 Hydrostatic pressure is the pressure from heart beat forcing liquid out through junctions of capillary,
which at arterial end of capillaries is 3.2kPa (1); oncotic pressure is result of water potential in
capillary from plasma proteins moving water into capillary which is −2kPa (1). At arterial end of
capillary hydrostatic pressure higher than oncotic pressure, water is forced out of capillary and forms
tissue fluid (1); as blood moves along capillary more fluid moves out and residual force from heart
beat is lost. By venous end of capillaries hydrostatic pressure has fallen to 0.5kPa (1); plasma
proteins don’t leave capillary as they can’t pass through loose junctions so oncotic pressure is still
−2kPa (1); as a result water now moves back into capillary by osmosis and by end of the capillary
network around 90% of tissue fluid is back in capillaries again (1).
8.4
Summary Questions
1 Biconcave shape gives large surface area for gaseous exchange (1); and makes it possible to move
through capillaries (1); erythrocytes contain oxygen carrying pigment haemoglobin (1); mature
erythrocytes have no nucleus so more room for maximum amount of haemoglobin (1); contains
enzyme carbonic anhydrase involved in carriage of carbon dioxide in blood (1). (max 3)
2a
(1 mark correct axes; 1 mark correct graph lines)
b Foetal haemoglobin has higher affinity for oxygen than maternal haemoglobin (1); so foetal blood
takes oxygen from maternal blood (1); enables foetus to survive and grow (1).
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OCR A Biology
3a
(1 mark correct axes; 1 mark correct graph lines)
b Myoglobin in muscles has higher oxygen affinity than haemoglobin in blood (1); so muscles can
take oxygen from haemoglobin in blood, enabling muscles to get extra oxygen when they are
contracting during exercise (1).
4 Flow chart should clearly shows main stages of carbon dioxide transport. It should not include
plasma transport as specifically states red blood cells (6).
8.5
Dissecting a heart
Atria not very clearly displayed – either lost at butchers OR very small compared with ventricles so not
easy to see; lack of major blood vessels e.g.. pulmonary vessels, aorta – lost with removal of atria;
blood vessels not as clear as on Figure 1 and 2, valves not as clearly defined as in Figure 2, difficult
to pick up heart wall from ventricular lining; Figure 1 and 2 schematic designed to show the principles
of the structures and how they are related to each other and to their function – NOT an accurate
anatomical representation; any other sensible point. (max 4)
A hole in the heart
Small hole means blood flows past, and two sides of heart remain effectively separate (2)
Large hole means deoxygenated blood from right side of heart mixes with oxygenated blood from left
side of heart so blood does not carry enough oxygen to tissues (3).
Blood pressure
Weak heart does not beat strongly so blood leaves heart at a lower pressure than in a healthy person
– giving low blood pressure (3); damaged, closed or less elastic vessels – lumen is narrower so blood
flowing through is under higher pressure (same amount of blood flowing through a smaller space) (3).
Summary Questions
1 Heart cardiac muscles needs good supply of oxygen and glucose to contract with a regular rhythm
(1); coronary arteries supply blood carrying glucose and oxygen to heart (1); healthy coronary arteries
provide good supply of blood to heart muscle so it can continue to beat (1).
2 a First heart sound – blood hitting against atrioventricular valves (1). Second heart sound: sound of
semilunar valves as they close to prevent a backflow of blood (1).
b Pressure difference between atria and ventricles as atria empty and ventricles s start to contract
means blood is forced against the atrio-ventricular valves which close to prevent backflow of blood
into the atria (3); pressure difference between blood in artery and ventricles as they empty means
blood hits semilunar valves which are closed to prevent backflow of blood into heart (3).
3 a Bradycardia is the slowing of the heart (1); when animals dive they need to conserve their oxygen
and food to last for whole dive (1); they undergo bradycardia as part of slowing down metabolism to
enable them to stay under water as long as possible (1).
b Tachycardia is speeding up of the heart (1); at altitude there is less oxygen available in air – this
means there is less oxygen available in blood (1); heart speeds up to compensate and
carry more oxygen to tissues, even if it isn’t
effective because of low oxygen atmosphere (1).
4 a (i) Approx 63 bpm (2) (ii) Approx 48 bpm (2) (iii) Approx 107 bpm (2)
b Normal 0 (1); 23% decrease (1); 69% increase (1).
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8 Transport in animals
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
Guidance
1
squamous epithelium
short(er) diffusion, distance / path ;
large number of alveoli
large(r) surface area ;
good blood supply
high / large / steep, concentration gradient
OR
removes oxygen (from lung surface) / brings
carbon
dioxide (to lung surface);
good ventilation
high / large / steep, concentration gradient
OR
supplies oxygen (to alveoli) / removes carbon
dioxide
(from alveoli) ;
large surface area; alveoli and lamellae; thin
diffusion distance; one cell thick epithelium
(alveoli and lamellae); movement of blood (in
capillaries);
gills, external / extend into water and lungs,
internal / air enters lungs; flow of water over gills
is in the opposite direction to blood flow /
countercurrent; (this) maximises concentration
gradient; lower concentration of oxygen in water
(compared to air);
Arterial: 13
Venous: –7
drains into lymphatic system; eventually
returned to blood; (if this did not happen) fluid
would accumulate; oedema / swelling of tissue;
diffusion is necessary for movement, into / out
of, cells or movement, into /out of, transport
medium; diffusion distance has to be short;
transport system reduces diffusion distance;
open circulatory system; few vessels; open
body cavities; transport medium pumped
straight from heart to body cavity; open ended
vessels;
prevent backflow of blood;
4
ACCEPT reduced / shorter
diffusion distance
ACCEPT thin diffusion barrier
IGNORE thin diffusion pathway
blood under pressure; from contraction of
ventricle; does not pool in vessels;
contains a non-protein; haem group; porphyrin
+
and Fe ;
co-operative binding (of oxygen to
haemoglobin); first oxygen cannot bind easily;
this changes shape of haemoglobin; following
oxygens bind more easily; first oxygen hardest
to remove from oxyhaemoglobin;
carbaminohaemoglobin;
2
myoglobin has a higher affinity for oxygen;
higher saturation at the same partial pressure of
oxygen;
2
2 (a)
2 (b)
3 (a) (i)
3 (a) (ii)
3 (b)
3 (c)
4 (a)
4 (b)
4 (c) (i)
4 (c) (ii)
4 (d)
4 (e) (i)
© Oxford University Press 2015
ACCEPT increases surface area
IGNORE SA : Vol ratio
ACCEPT maintains / creates
concentration gradient
IGNORE ref diffusion gradient
4
ACCEPT maintains / creates
concentration gradient
IGNORE ref diffusion gradient
IGNORE ref to air
Max 4
3
Max 3
2
4
3
Max 3
4
Max 4
1
Max 2
2
4
Max 4
1
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8 Transport in animals
Answers to practice questions
OCR Biology A
4 (e) (ii)
4 (e) (iii)
5 (a)
5 (b)
5 (c)
it only has one haem group; no co-operative
binding;
sigmoidal; in between the curve for myoglobin
and haemoglobin;
protein;
2
increases rate of reaction; lowers activation
energy (of reaction);
carbon dioxide reacts with water; forms
carbonic acid; carbonic acid dissociates;
hydrogen and hydrogen carbonate, ions;
2
© Oxford University Press 2015
2
2
3
Max 3
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OCR A Biology
9.1
Observing xylem vessels in living plant stems
1 Less detailed, only gives position of xylem, not other tissues, but shows the movement of water
happening in living tissue.
2 Can only see xylem – cannot be modified to show phloem. To get good sections where xylem
vessels can be clearly seen is largely dependent on how sharp blade is and having a steady hand.
If longitudinal sections are cut in wrong place in stems then you won’t see any xylem.
Summary Questions
1 Too big for diffusion alone to supply needs as SA : V ratio too small for diffusion to be effective
means of transport (1); transport system required for transporting oxygen and glucose for respiration
(1); waste product removal (1); water and mineral ions from roots to all the cells (1); hormones made
in one part of a plant to the areas where they have an effect (1).
2 In plants no heart to act as central pump, whereas many multicellular animals have heart (1); in
plants one type of vessel is made of dead cells – all animal vessels made of living tissue (1); in plants
there are two different transport systems carrying different materials – animals have different types of
vessels but the same transport medium in both (1).S
3 Stem – vascular bundles around outside (1); helps give strength and support to structure (1);
roots – vascular bundles in centre (1); to help give strength against tugging forces when plant blown
by wind (1); leaves – large central vein containing vascular tissue (1) gives supports to broad
structure of leaf (1).
4 Similarities both transport materials around plant; both made up of cells joined end to end forming
long, hollow structures, any other sensible points.
Differences Xylem largely non-living tissue, phloem living; xylem transports water, mineral
ions, and supports plant, phloem transports organic solutes around plant from leaves; in xylem flow of
material from roots to shoots and leaves, in phloem flow of material up and down; xylem cell walls
lignified, phloem not; xylem have wide lumen; mature phloem cells have no nucleus; other cells
associated with xylem in herbaceous dicots include xylem parenchyma and xylem fibres, equivalent in
phloem include fibres and scleroids. Any other sensible points. (max 6)
9.2
Summary Questions
1 Very small to penetrate soil particles (1); large SA : V ratio for water absorption (1); thin layers for
ease of diffusion (1); high solute concentration in cytoplasm gives low water potential so water moves
in from soil by osmosis (1).
2 Symplast pathway: relies on osmosis as water moves through cell membranes and cytoplasm (1).
Water moving in from soil by osmosis into root hair cell raises water potential compared to next cell so
water moves again by osmosis (1). Active transport of ions needed to move water from endodermis to
xylem by osmosis (1). Apoplast pathway: water moves through cellulose cell walls by cohesive forces
between water molecules and as result of transpiration pull up xylem (1); moves into symplast
pathway in endodermis as a result of Casparian strip (1); and needs active pumping of ions into
xylems followed by osmosis (1) before water moves back to apoplast pathway in xylem.
3 a Temperature increase increases root pressure BUT although increase in temperature increases
rate of chemical reactions (1) it also increases rate of passive processes such as diffusion and
osmosis (1).
b Active transport needs energy in form of ATP (1); cyanide poisons mitochondria where cellular
respiration takes place so cyanide prevents ATP formation (1); Oxygen and respiratory substrates
needed for cellular respiration and ATP production (1). Factors which interfere with ATP production
also interfere with development of root pressure (1); suggesting that development of root pressure is
an active process requiring ATP (1).
9.3
Measuring transpiration
1 5 cm (1)
2 Water vapour lost from plants through open stomata in transpiration (1); most water taken up by
plant is used for transpiration (1); as the rate of water uptake slows down significantly with vaseline on
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OCR A Biology
lower surface this suggests that most stomata are on under surface of leaf (1); and once covered very
little water vapour could be lost (1); so very little water taken up by shoot (1).
Summary Questions
1 Transpiration: evaporation of water from surface of a leaf (1). Transpiration stream: flow of water
moved up from soil into root hair and through root cortex by osmosis, into xylem and up through stem
by cohesion of water molecules, cross leaf cells by osmosis and out of leaf by evaporation and
diffusion (1).
2 Root pressure: active movement of solutes followed by passive movement of water by osmosis,
which gives a positive pressure forcing water up the xylem (1).
Transpiration pull: pulling of a constant stream of water molecules up xylem held together by cohesive
forces as a result of evaporation of water from surface of spongy mesophyll cells in leaf (1); a passive
process and a negative pressure (1).
3 All conditions except air movements kept the same (1); sensible suggestions as to how to
investigate effect of air movements, e.g., use of fan for set time (1); fan placed at different distances
from potometer (1); control readings in still air allow plant recovery time between different distances
(1), more than one reading for each distance (1) etc. Total of 4 marks from these or any other
sensible suggestions.
4 a 0.01 cm/s (1); dark so no photosynthesis (1); and most stomata closed (1); little gas exchange and
so little water lost by transpiration (1); so little water movement up stem to replace it (1).
b 0.14 cm/s (1); photosynthesis taking place (1); quite a few stomata open for gas exchange (1); so
water lost in transpiration (1); moves up stem to replace it (1).
c 0.2 cm/s (1); bright light – maximum photosynthesis (1); so many or all stomata open for maximum
gas exchange (1); lots of water lost by transpiration (1); and lots taken up to replace it (1).
d 0.3 cm/s (1); bright light – maximum photosynthesis so many or all stomata open for maximum gas
exchange so more water lost in transpiration (1); when windy water vapour moves away from leaf as
soon as it appears reducing humidity(1); and increasing diffusion gradient for water out of leaf (1);
even more water lost in transpiration (1); so more needed to replace it giving rapid
uptake of water (1).
5 a Water molecules evaporate from surface of mesophyll cells into air spaces in leaf and move out of
stomata into surrounding air by diffusion down concentration gradient (1); loss of water by evaporation
from mesophyll cell lowers water potential of cell (1); water moves into cell from adjacent cell by
osmosis, along apoplast and symplast pathways (1); repeated across leaf to xylem. Water moves out
of xylem by osmosis into cells of leaf (1); water molecules form adhesive bonds with walls of narrow
xylem vessels and also form hydrogen bonds and so tend to stick together – cohesion (1); as a result
of these cohesive forces, along with adhesion to walls of xylem, water is pulled up xylem in
continuous stream to replace water lost by evaporation, this is the transpiration pull (1); transpiration
pull results in tension up xylem which helps to move water across roots from soil, movement is partly
by osmosis (change in water potential in cells across root) and partly by cohesion through apoplast
pathway (1).
b (i) Cohesive forces in xylem cause negative pressure (1); that draws tissues in and reduces
diameter of tree (1); in daytime when maximum photosynthesis and so maximum transpiration is
taking place (1); minimum transpiration takes place during night so less tension in the xylem tissue
and it expands (1).
(ii) Tension and negative pressure in xylem means air may be pulled into xylem when stem is cut (1);
this breaks cohesive stream of water molecules (1); and prevents more water moving up stem (1); as
a result flowers droop and die very quickly, as air bubble is pulled up the stem (1).
9.4
Summary Questions
1 Source: supplier of carbohydrates needed by cells of plant (1); e.g., leaves, stems, storage organs,
seeds (1); sink: area of plant that needs assimilates in phloem sap (1), e.g., roots, meristems,
developing fruit, seeds, and storage organs (1).
2 Aphids pierce plant tissues and push their stylet directly into phloem vessels (1); if aphid removed,
leaving stylet in place, pressure in phloem vessel continues to force sap out of stylet (1); this can be
to show presence and concentration of assimilates in phloem (1); pressure in phloem (1) and how
these things change with manipulation of other factors such as light intensity (1).
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OCR A Biology
3 a Sucrose moves from cells to companion cells and sieve tube elements by diffusion along a
concentration gradient, is moved into companion cells and sieve tube elements by an active process
(1); hydrogen ions (H+) actively pumped out of companion cell into surrounding tissue using ATP (1);
they return into companion cell down a concentration gradient via a co-transport protein carrying
sucrose as well (1); build-up of sucrose in companion cell and sieve tube element means water
moves in by osmosis causing a build-up of hydrostatic pressure (1). Water carrying assimilates moves
into tubes of sieve elements, reducing pressure in companion cells and moving up or down the plant
by mass flow (1). Sucrose moves out of phloem into cells which need it by diffusion, followed by water
by osmosis.
b Companion cells have membrane folding to give large surface area for transport of sucrose (1); also
have many mitochondria to produce ATP needed for active transport (1); if mitochondria in companion
cells are poisoned, translocation in phloem stops (1); the pH of the companion cells is higher (more
alkaline) than the surrounding cells supporting idea of hydrogen ion pump (1); the flow of sugars in
the phloem is about 10 000 times faster than it would be by diffusion alone, suggesting an active
process is driving mass flow (1).
9.5
Investigating stomatal numbers
1 Expect few stomata in xerophytes (1); and expect them to be hidden in pits, or grooves on leaf (1);
covered in hairs (1); inside rolled leaves (1); etc to maintain a still, moist microclimate (1); and reduce
loss of water by transpiration (1).
2 The leaves are reduced to spines, so few stomata and difficult to find (1); the stomata on the stems
are protected by spines (1).
Summary Questions
1 Students can choose any three from: thick waxy cuticle, sunken stomata, reduced leaves, hairy
leaves, curled leaves, succulent leaves. NOT leaf loss as that is not a structural adaptation of the leaf.
Students should give a clear and full explanation of the relationship between the structural adaptation
and the function in conserving water. (1 mark for each)
2 Students can choose any three adaptations and give a clear and full explanation of the relationship
between the structural adaptation and the function in conserving water. (1 mark for each)
3 Dry conditions: conflict between need to open stomata for gaseous exchange and loss of water by
evaporation from open stomata. Hot dry air so evaporation will take place fast. Ground dry or frozen
so little water available. Tissues need water to maintain turgor and carry out photosynthesis. Any
other sensible points. (up to 2 marks)
In water: waterlogging of tissues – no access to oxygen. Sinking – need to be near the surface to get
light for photosynthesis. Slow diffusion of oxygen in from water. Any other sensible points.
(up to 2 marks)
4 Characteristics should be well argued and make sense in terms of either withstanding drought or
withstanding flooding. They should be able to explain the advantages of the characteristics they have
chosen and why they think they would be particularly useful to the crop.
Example for drought: Hairy leaves (1): these trap a microclimate around the stomata and reduce
water loss by transpiration. Only suitable for crops where the leaves aren’t eaten.
Example for flooding: Resistance to rotting: (1) Plants in flood conditions often rot so resistance to
fungi or bacteria which cause rot would be very useful.
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9 Transport in plants
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
1 (a)
transport system in plants; composed of xylem
and phloem;
2
1 (b)
1 (c)
2 (a) (i)
2 (a) (ii)
2 (b) (i)
2 (b) (ii)
2
bulge above the ring; translocation past ring
prevented; sucrose accumulates; lowers water
potential; water moves into area by osmosis;
loss of water vapour; from aerial parts of plant;
4
photosynthesis; requires carbon dioxide;
stomata must be open; water vapour diffuses
out;
potometer;
4
measures water uptake; water may be used in
photosynthesis ;
2
2 (c) (i)
2 (c) (iii)
Treatment
0
1
Light
0
3
0.11
6
0.11
9
0.11
12
0.11
15
0.11
18
0.11
21
0.11
24
0.11
27
0.018
30
0.018
assume constant thickness of leaves; measure
mass of section of leave with known surface
area; leaf mass multiplied by this figure equals
surface area of leaf;
independent variable light intensity or wind
speed or humidity; dependent variable water
loss; controlled variables (when not
independent variable) light intensity / wind
speed / humidity / temperature;;
© Oxford University Press 2015
Max 4
2
2
Water loss
ml
2 (c) (ii)
Guidance
1 mark for using the surface area
2
to calculate water loss per m and
1 mark for getting all the
calculations correct.
3
4
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9 Transport in plants
Answers to practice questions
OCR Biology A
2 (c) (iv)
axes correct; units; plots; line of best fit;
4
2 (c) (v)
fan 18, fan 30 and mist 18, mist 27; (these)
results, do not follow trend / very different from
expected;
rate of water loss is measure of transpiration
rate; increased concentration of water vapour
outside leaves decreases water vapour
potential gradient; ORA rate of water loss higher
with fan than (just) light because air movement
increased water vapour potential gradient; rate
of water loss decreased with mist because
water droplets (in mist) reduced water vapour
potential gradient;
A osmosis; down water potential gradient;
B osmosis; down water potential gradient;
C mass flow; down pressure gradient ;
D diffusion; down water (vapour) potential
gradient;
passive movement from high to low
concentration;
2
10 × 10 × 6 = 600 mm and 10 × 10 × 10 =
3
1000 mm ; 600/1000 = 0.6;
as (surface area to volume) ratio decreased the
rate of diffusion decreases / proportional;
figures quoted;
large plants have large leaves; SA/volume ratio
increases; increases, rate of transpiration /
water loss; reference to presence of
adaptations;
divided length of side by time taken ;
2
3 (d) (ii)
idea that student used whole length of side,
rather than half length ;
1
4 (a)
plant adapted to survive in dry environments;
1
4 (b)
reduced size of leaves / needles; chloroplasts in
stem; thick cuticle; stores water in stem /
succulent;
active transport requires, energy / ATP; is
movement against concentration gradient;
facilitated transport is passive; is movement
down concentration gradient;
movement is only upwards in xylem; movement
in any direction in phloem; between source and
sink;
1 phloem must be under pressure; presence of
pressure gradient;
2 gradients of organic solvents will mean water
potential gradients are present; movement of
water by osmosis;
3 must be an active process; photosynthesis
leads to the production of ATP / respiratory
substrate / glucose;
3
2 (c) (vi)
2 (d)
3 (a)
3 (b)
3 (c) (i)
3 (c) (ii)
3 (d) (i)
5 (a)
5 (b)
5 (c)
© Oxford University Press 2015
Max 2
4
6
Max 6 is best.
1
2
2
2
Max 2
1
IGNORE divide mm by s (units
alone too vague)
ACCEPT needs to divide answer
by 2 / distance has to be to
centre of cube rather than whole
length of side / assumed
diffusion occurs (across whole
cube) from one side
Max 3
3
2
6
2 marks per point
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9 Transport in plants
Answers to practice questions
OCR Biology A
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Module 3 Exchange and transport
Application and Extension
OCR Biology A
Question
number
Answer
Marks
Application
1
Plants have mechanisms for blocking damaged
phloem vessels and preventing the loss of sap.
Different proteins coagulate and block the
phloem vessels so the sap cannot come out of
the cut ends. As a result, the rest of the plant
continues to transport substances and survives.
2
Aphids transport sugar-rich phloem sap up very
narrow, delicate stylets. If the proteins in the
phloem sap coagulated in the stylet, or around
the entrance to the stylet, it would be blocked
and the insect would be unable to transport the
food from the plant into its gut so it would
starve.
3
Read around the resources and plan carefully
what to include and what is not necessary
Guidance
Produce 6–10 slides – no more
Clear, well labelled diagrams can carry a lot of
information.
Make every word count
Extension
1
Key points should be included about blood
clotting – rapid cascade, involves proteins such
as prothrombin and fibrinogen, involves calcium
ions to enable proteins to form clot.
Flow chart, for example:
Human blood: damaged tissues release
platelets  platelets release enzyme
thromboplastin  thromboplastin catalyses the
conversion of protein prothrombin to enzyme
thrombin in presence of calcium ions 
thrombin acts on fibrinogen converting it to fibrin
 fibrin forms a mesh of insoluble fibres 
Platelets and blood cells get caught in the mesh
to form a clot
Plant phloem systems:
Sieve element damage plastids e.g.,
chloroplasts, amyloplasts burst open  plastid
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Module 3 Exchange and transport
Answers to practice questions
OCR Biology A
contents coagulate to block phloem tube
Sieve element damage  Endoplasmic
reticulum proteins become separated
coagulate to block phloem
Sieve element damage  calcium flows into
sieve element from adjacent cells  reacts with
forisomes forisomes expand and block sieve
element
Similarities: rapid response, proteins heavily
involved in response, involves calcium ions in
some stages of process, involves
coagulation/clotting, any other sensible points
Differences: clotting cascade single complex
mechanism which prevents blood loss from
damaged vessels, plants have a number of
different responses which all help to block
damaged phloem vessels, not all plant systems
depend on calcium ions, any other sensible
points.
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Module 3 practice questions
Answers
OCR Biology A
Question
number
Answer
Marks
1 (a) (i)
Loss of water vapour; from the aerial parts of a
plant;
2
1 (a) (ii)
Stomata are open during transpiration; carbon
dioxide diffuses in; for photosynthesis;
photosynthesis necessary for growth;
Er has lowest rate; Er and Lp have highest rate;
Er and Pt have lower rate than Lp;
figures/numbers quoted;
4
Er and Lp; highest transpiration rate; highest
growth rate in winter; more water removed from
soil;
A: left semi-lunar; B: right semi-lunar; C:
tricuspid / left atrioventricular; D: bicuspid / right
atrioventricular;
Valve prevents backflow of blood; A from aorta
to left ventricle; B from pulmonary artery to right
ventricle; C from left atrium to left ventricle; D
from right atrium to right ventricle;
When open, unhealthy valves result in greater
aperture; edges of unhealthy valves are wavy;
when closed unhealthy valves do not close
completely; so blood could leak through;
Backflow of blood; from, atria to, veins;
ventricles to atria; aorta / pulmonary artery, to
ventricles; reduced pressure increase; reduced
blood flow;
Rate of diffusion of carbon dioxide constant /
fluctuates, in control; decreases and then
increases in stressed; figures quote; water
potential constant in control; increases then
decreases in stressed; figures quote;
Decrease in rate of carbon dioxide diffusion due
to stomatal closure; so transpiration rate
decreases; less water vapour lost; so water
potential inside leaf increases;
Comparison; to observe the effect of water
stress;
X symplast; Y apoplast;
3
A: epidermis; B: Casparian strip; C:
endodermis;
Casparian strip is impermeable to water; so
water moves from apoplast pathway; to
symplast pathway;
Elongated extension; increased surface area;
increased number of, carrier proteins / channel
proteins; increased rate of, active transport /
diffusion / osmosis;
Maximum volume of air that can be moved in
and out of lungs in one breath;
(vital capacity is) tidal volume + inspiratory
reserve volume + expiratory reserve volume;
3
1 (b) (i)
1 (b) (ii)
2 (a) (i)
2 (a) (ii)
2 b (i)
2 (b) (ii)
3 (a)
3 (b)
3 (c)
4 (a) (i)
4 (a) (ii)
4 (a) (iii)
4 (b)
5 (a) (i)
© Oxford University Press 2015
Guidance
4
4
5
4
5
4 max
4
2
2
3
4
1
ACCEPT VC = IRV + TV + ERV
ACCEPT VC = TLC - RV
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Module 3 practice questions
Answers
OCR Biology A
5 (a) (ii)
5 (b) (i)
5 (b) (ii)
6 (a)
6 (b)
6 (c)
7 (a)
7 (b)
8
9 (a)
9 (b)
(vital capacity is) total lung capacity – residual
volume;
Volume of air that can be expired in the first
second of forced expiration;
80;
1
2
DO NOT CREDIT the volume,
exhaled / expired, in, a single
/ one, breath
Correct answer = 2 marks
If final answer is incorrect (not
rounded or incorrectly rounded)
or missing allow 1 mark for 4.5
÷5.6
One mark for each CORRECT
row
IGNORE  in COPD column for
patient D
Diagnosis
Asthma
COPD

C

D

E
Cardiac output: volume up blood pumped; from
the left ventricle / around the circulatory system,
in one minute ;; stroke volume: volume of blood
pumped from the left ventricle; in one
contraction ;; heart rate
number of heart beats / contractions, in one
minute;
stroke volume × heart rate = cardiac output;
3
A ventricle walls relaxed / ventricular diastole;
blood flows into ventricles; from atria;
atrioventricular valves are open; B atrial walls
contract / atrial systole; blood is forced into
ventricles (from atria); C ventricle walls contract
/ ventricular systole; blood is forced into aorta
and pulmonary artery; semi-lunar valves are
open and atrioventricular valves are closed;
waterproof; prevent loss of water; strength;
prevent collapse of vessel / keep vessel open;
spiral; allows flexibility;
6 max
myoglobin = 63 – 65
and
haemoglobin = 23 – 25;
large / active, organisms have high(er), demand
for oxygen / need;
to remove CO2;
small(er), surface area to volume ratio / SA:V /
surface area: volume;
surface area too small / distance too large /
diffusion takes too long (to supply needs);
create / maintain, (steep), diffusion /
concentration, gradient;
1
IGNORE units
2 max
ACCEPT ORA throughout
IGNORE ref to nutrients
ACCEPT diffusion too slow
look for reason why diffusion not
good enough
3 max
could give mark in any row as an
additional mark – but only once
DO NOT ACCEPT any vague
reference to ‘gases’ throughout
ACCEPT short diffusion distance
here even if given above
ACCEPT breathing in and out /
Patient
epithelium
short (diffusion)
distance;
capillaries
delivers carbon
dioxide (to be
removed from
blood) / carries
oxygen away (from
alveoli);
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1
4
2
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Module 3 practice questions
Answers
OCR Biology A
short (diffusion)
distance;
diaphragm /
intercostal muscles
9 (c)
ventilation / supply of
oxygen (to alveoli) /
removal of carbon
dioxide (from alveoli);
diaphragm (contracts / flattens and) moves
downwards;
intercostal muscles contract to move ribs, up /
out;
increase volume of thorax;
reduce pressure inside thorax;
to below atmospheric pressure / creates
pressure gradient / AW;
4 max
9 (d) (i)
a clear X placed on any part of trace where line
is sloping down;
1
9 (d) (ii)
3 dm ;
1
10 (a) (i)
X AV valves; Y semi-lunar valves;
2
10 (a) (ii)
prevent backflow of blood; to atria (from
ventricles);
left ventricle increased force of contraction;
ORA increased pressure; ORA to pump blood
around the body; right ventricle blood pumped
to lungs; prevent damage to capillaries;
idea of no recognisable pattern in b; ventricular
systole not happening in b; higher rate of
contractions in b;
cardiac arrest change in flow of impulses from,
SAN / AVN; arrhythmia / abnormal heart
rhythm; heart stops beating; heart attack
blockage of (coronary) artery; blood flow to
heart muscle, reduced / stopped; (section of)
heart muscle dies; heart (normally) continues
beating;
2
10 (a) (iii)
10 (b)
10 (c)
3
© Oxford University Press 2015
IGNORE ref to internal / external
ACCEPT increase volume of
lungs / chest
ACCEPT decrease pressure in
lungs / chest
must ensure the pressure
gradient is in correct
direction – lower in lungs
ACCEPT label line with X
DO NOT ALLOW X on tip of crest
/ trough
correct units must be given
ACCEPT litres
5
3
6
4 max for cardiac arrest or heart
attack.
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OCR A Biology
10.1
The development of classification systems
The earliest classification systems were based on visual similarities between organisms. Advances in
our understanding of the biological or genetic make-up of organisms has provided evidence for how
organisms are linked. This information informs revisions to the system of classification which is used.
Summary Questions
1 Any 2 appropriate reasons. For example: Enables scientists to share information / makes
communication easy (1); provides information about an organism, based on members of the same
group (1); allows accurate identification of an organism (1).
2 Ligers cannot reproduce to produce more ligers therefore they are not a species (1). Both lions and
tigers reproduce to produce fertile offspring, therefore they are species (1).
3 a Phylum (1); b Erithacus (1); c rubecula (1)
4 Both parents are members of the same genus (Rubus) (1) but different species (ursinus and idaeus)
(1). Two different species cannot produce fertile offspring / according to the taxonomic classification
system the loganberry should not be fertile (1).
10.2
Summary Questions
1 Any two from: Plants have chloroplasts / chlorophyll, whereas fungi do not (1); plants are
autotrophs, whereas fungi are heterotrophs (1); fungi may be unicellular, plants are always
multicellular (1); fungi store food as glycogen, whereas plants store food as starch (1); plant cell walls
are composed of cellulose, whereas fungi cell walls are composed of chitin (1) Or other suitable
example.
2 a Prokaryote (1), b Fungi (1), c Protoctista (1)
3 Any three from: Advances in biological techniques have identifies large differences in composition
(1); ribosomes/rRNA differ (1); cell walls differ – peptidoglycan not found in archaea (1); old
classification does not show correct phylogeny (1).
4 Any six from: Living organisms classified into two kingdoms based on major differences in
characteristics (1) for example, those that moved and ate (animals) and those that didn’t (plants) (1);
scientific advances/use of microscope allowed smaller details to be observed (1); organisms divided
into five kingdoms (1); Plants, animals, fungi, protoctista, prokaryotes (1); Advances in science
allowed DNA and proteins to be studied (1); Provided evidence for evolutionary relationships (1);
Three domain system proposed (1); Relevant scientists mentioned (Linnaeus, Whittaker, Woese) (1).
10.3
How do you interpret phylogenetic trees?
Summary Questions
1 Historical classification systems based on physical characteristics / niche occupancy, whereas
phylogeny based on evolutionary relationships (1).
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2 It takes into account evolutionary relationships that might not be obvious by just looking at
characteristics (1); it forms a continuous tree so organisms do not have to be forced into groups
(1); is not hierarchical therefore different groups on the tree are represented according to their
evolutionary position – and can thus be compared (1).
3 a snakes (1)
b They have become extinct / not present in the world today (1); they are placed along the timeline at
the point they existed in time (1).
c Bird and crocodile branches are closer together than bird and turtle (1); birds shared a common
ancestor with dinosaurs. This organism shared a common ancestor with the crocodiles (the common
ancestor with turtles is much further back in history) (1).
10.4
Evolutionary embryology
Very unlikely that an embryo could form a fossil, as it is made up only of soft tissue (only in rare
cases, such as encasement in tar or resin, are bodily tissues preserved).
Summary Questions
1 A diagram used to show evolutionary relationships between organisms (1); the closer the branches
of the tree the closer the evolutionary relationships (1).
2 Advantages (2): for example – radioisotopes can be used to date fossils / changes can be tracked
over time / chronological order apparent in rock strata. Disadvantages (2): for example – many
organisms decompose quickly before they have a chance to fossilise/destroyed by volcanoes/
destroyed by earthquakes.
3 1 mark for scientist, 1 mark for contribution (max 6)
Lyell – suggested that fossils were actually evidence of animals that had lived millions of years ago.
Hutton – proposed theory of uniformitarianism.
Darwin – came up with theory of evolution by natural selection through observations in Galapagos
islands / jointly published theory.
Wallace – came up with theory of evolution by natural selection in Borneo/jointly published theory.
4 Any three from: Study of similarities and differences in proteins and nucleic acid/DNA of an
organism (1); changes in highly conserved molecules can help identify evolutionary links (1); such as
cytochrome C / ribosomal RNA (1); species that are closely related have the most similar DNA and
proteins /distantly related have far fewer similarities. (1)
10.5
Studying variation in identical twins
1 They have no genetic variation therefore all variation is the result of the environment.
2 They have been exposed to greater amount on environmental influences.
3 Eye colour is determined solely by genes as both identical, ear piercing determined solely by
environment as they have opposite results, mass and height controlled by a combination of both
genetic and environmental factors as results vary, mass controlled more by the environment than
height as greater variation shown.
Summary Questions
1 Interspecific variation is differences between individuals of different species whereas intraspecific is
differences between individuals of the same species (1).
2 a Any two suitable examples, e.g., scar, tattoo, dyed hair (1)
b Any two suitable examples, e.g., eye colour, blood group, lobed, or lobeless ears (1)
3 Caused by a combination of genetics and the environment (1). Genes determine the natural colour
of hair and texture, e.g. curly/straight (1). Environment affects final appearance, e.g. if hair is cut,
dyed, or lightened by sunlight (1).
4 Any 3 from: Individuals produced by asexual reproduction are clones/genetically identical to parents
(1); no fertilisation so no mixture of genetic material (1); meiosis does not take place/ no production of
gametes (1); DNA can only be altered as a result of mutation (1).
10.6
Flagella length variation in Salmonella
1 2.4 μm
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2 0.49
3 1.91—2.89 μm (68% of population will fall within standard deviation of the mean).
4 Continuous variation, the length of flagella can take any value within a range.
Summary Questions
1 Continuous – b, c (1). Discontinuous – a, d (1)
2 Characteristics which show discontinuous variation are purely controlled by genetics/no
environmental influence (except scars/tattoos just environment) (1). Normally controlled by a single
gene (1). Characteristics which show continuous variation are controlled by a combination of genetic
and environmental causes (1). Controlled by a number of genes/ polygenes (1).
3 Any two from: The values of a characteristic which shows discontinuous variation fall into discrete
categories (1) if a mean is calculated it may produce a value which does not fit into a category (1)
many of the characteristics do not have a numerical value (1).
4 Any four from: Continuous variation (1) as controlled by both genetic and environmental causes (1).
Normal distribution (1) very few rabbits would be extremely large or extremely small (1) most would be
within one standard deviation of the mean (1).
5a
Diameter of
stem / mm
Rank
Number of
thorns per
unit length
1
1
8
2
2
11
3
3
9
5
4
12
8
5
12
10
6
27
11
7
23
14
8
30
(2) – 1 mark for each ranked column correct
∑ 𝒅𝟐 = 4.5 (1)
𝑟 = 𝟏−
𝟔 ∑ 𝒅𝟐
𝒏(𝒏𝟐 −𝟏)
Rank
1
3
2
4.5
4.5
7
6
8
(1)
r = 1 – (6 x 4.5) / (8 × 63) (1)
r = 0.946 (1)
b df=6 (1); = 0.946 > 0.881 therefore >99% confidence. Therefore there is very little (<0.1%)
likelihood that the null hypothesis is true (1).
10.7
Classification of giant pandas
1 Giant pandas have: a large body mass, shaggy fur, a pseudo-thumb, the ability to climb
2 Student’s own answer. Any conclusion drawn should be consistent with the evidence available: for
example, a student may conclude that the giant panda is a panda, due to the similarities in its teeth,
snout and paws to the red panda.
3 The molecular sequence of a particular molecule is compared, by looking at the order of DNA bases
or at the order of amino acids in a protein. Species that are closely related have similar DNA and
proteins, whereas those that are distantly related have far fewer similarities.
Summary Questions
1 Anatomical – camouflage, sharp canine teeth (1); Physiological – melanin production, production of
toxins (1); Behavioural – migration, courtship dance (1).
2 Analogous structures are structures that have adapted to perform the same function but have a
different origin whereas homologous structures appear superficially different but have the same
underlying structure (1).
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3 a Insect and bird wing – both have evolved to fly to escape predators/hunt for food (1).
4 2 marks for named adaptation and suitable explanation. 2 marks for correctly naming the adaptation
as behavioural/anatomical/physiological.
5 They have analogous structures – anatomical features that perform the same function in different
organisms, but have a different origin (1).
Any two from: Both burrow through soft soil to find insects (1). Both have a streamlined body shape,
and modified forelimbs for digging (1). Both have velvety fur which allows smooth movement through
the soil (1).A
10.8
Anolis lizards
1 When few individuals of a species colonise a new area their offspring initially experience loss in
genetic variability, resulting in individuals that are physically and genetically different from their source
population (1).
2 Have no lizard populations.
3 Area covered in scrub/short vegetation, short hind limbs provide stability to walk along narrow
perches.
4 Any four from: released pairs of lizards on islands with no lizards but the same vegetation;
measured genetic variation; measured hind leg length; hind leg length shortened over time providing
evidence for natural selection; founder effect would produce random leg length.
5 Decreases fitness of population and their ability to survive and reproduce, as a result of less
variation, increased chance of recessive disorders.
Summary Questions
1 Any three from: availability of light / water / nutrients / carbon dioxide / space, risk of being eaten,
disease, ability to cross-pollinate, or other suitable example (1).
2 Variations exist within a population (1); those with the best characteristics survive AND reproduce
(1); characteristics are passed onto their offspring through genes (1).
3 A mutation occurred / existed in the mosquitos DNA which made them DDT resistant (1) these
organisms survived exposure to DDT and reproduced (1); mutation which caused resistance is
passed onto their offspring (1); frequency of the DDT-resistant allele increases in the population (1).
4 Any six from (or other appropriate examples):
Flavobacterium digests nylon waste (1); positive – used to clean up factory waste (1); bacteria e.g.
MRSA – antibiotic resistance (1); negative – no longer killed using current medical treatment (1);
sheep blowfly – insecticide resistant (1); negative – no longer killed by insecticide so increased sheep
death (1).
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10 Classification and evolution
Answers to practice questions
OCR Biology A
Question Answer
number
Marks Guidance
1
B
1
2 (a)
Echiniscus ; order ; phylum ; Animalia ; Eukaryota
5
2 (b)
1 (phylogeny) is evolutionary relationships (between
organisms) ;
Max 3
2 (phylogeny is study of) closeness of (evolutionary)
relationships ;
3 Phylogeny is basis of / used in, natural / scientific /
modern, classification ;
4 Idea that the closer the (evolutionary or genetic)
relationship the closer the (taxonomic) grouping ;
5 Correct use of example ;
2 (c)
Too small to see ;
2
(unable to see them) until invention of microscope /
development of suitable viewing apparatus / AW ‘
1 IGNORE ‘evolution’ without
further qualification
1 & 2 phylogeny is the
closest of evolutionary
relationships = 2 marks
1 ACCEPT phylogeny is
evolutionary history
3 ACCEPT new
3 IGNORE related to
classification
4 ACCEPT ref to recent
common ancestors as AW for
close relationship
4 ACCEPT named taxonomic
group for ‘grouping’
4 ACCEPT ‘if DNA is very
different then the group is not
the same’
5 e.g. gorillas and
chimpanzees (closely
grouped)
‘can only be seen under
microscope’ = 1 mark (mp1)
IGNORE ‘can’t see it’ without
the idea of size, e.g. can’t
see it clearly = 0 marks,
Can’t see its features = 0
marks
ACCEPT implication of being
too small to see, e.g. ‘you
need a microscope to see
them’ = mp1
‘people couldn’t see them in
the past because we didn’t
have microscopes’ = 2 marks
(mp1 and mp2)
IGNORE type of microscope
if stated
ACCEPT ‘magnifying glass’
Only 0.3 mm in length ;
ACCEPT ± 0.1 mm
3 (a)
Kingdom
Membranebound
organelles
Cell wall
Types of
nutrition
6
Heterotrophic
and
autotrophic
© Oxford University Press 2015
Mark the first answer in
each box f the answer is
correct and an additional
answer is given that is
correct or contradicts the
correct
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10 Classification and evolution
Answers to practice questions
OCR Biology A
Protoctists/
protoctista
IGNORE case of initial ‘P’
ACCEPT ‘’ or ‘yes’
present
Plant(s)/
Plantae
(present
and
made
of)
cellulose
IGNORE case of initial ‘P’
ACCEPT ‘’ or ‘yes’
present
3 (b)
fungi;
1
3 (c)
Assume answers refer to 3 domain classification
unless otherwise stated
based on (differences in) , DNA / RNA / nucleic acids /
polynucleotides ;
idea that more accurately reflects origins (of,
prokaryotes / eukaryotes) ;
(domain) divides / AW , prokaryotes ; ora
idea that domain reflects differences / AW , between
(eu)bacteria and archaea ;
example of two differences to support point 3 or 4 ;
(domain) groups / AW , eukaryotes together ; ora
idea that domain reflects the fact that there are
similarities between eukaryotic kingdoms ;
example of two or more similarities to support point 6
or 7 ;
1
2
3
4
5
6
7
8
3 max
© Oxford University Press 2015
Mark the first answer. If the
answer is correct and an
additional answer is given
that is incorrect or contradicts
the correct answer then = 0
marks
ALLOW fungus/ fungal/
fungae
IGNORE case of initial ‘f’
CREDIT Latin forms of
domain names throughout
IGNORE case of initial letter
1 CREDIT in the context of
an example
3 ‘prokaryotes are split into
groups because bacteria and
archaea are different’ = 2
marks (mp 3 and 4)
4 ACCEPT phonetic spellings
of ‘archaea’
4 ACCEPT ‘archaebacteria’
4 IGNORE multiple examples
for this mp, must be a
general statement
5 IGNORE if mp 3 or 4 not
awarded
5 e.g. (differences between)
cell wall / cell membrane /
flagella / (named) RNA
enzymes / ATPase / proteins
bound to genetic material /
DNA replication /
transcription etc
6 IGNORE as part of a list of
domains. Answer must state
that eukaryotes have been
placed in the same group.
6 ‘eukaryotes are placed in
the same group because they
have similarities’ = 2 marks
(mp 6 and 7)
6 IGNORE ‘are similar’
7 IGNORE multiple examples
for this mp, must be a
general statement
8 IGNORE if mp 6 or 7 not
awarded
8 e.g. all eukaryotes have,
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10 Classification and evolution
Answers to practice questions
OCR Biology A
nuclei / membrane bound
organelles / 80S ribosomes /
large cell size / linear DNA /
chromosomes / histones etc.
4 (a)
4 (b)
4 (c)
(a)
behavioural adaptation – the way an organism acts;
physiological adaptation – process which takes place
inside an organism
Any three from:
Curled leaves – to minimise the surface area of leaf
exposed to the wind
Hairs on the leaf inner surface – to trap moist air close
to leaves, reducing the diffusion gradient
Sunken stomata – making them less likely to open and
lose water
Thick waxy cuticle – reducing water loss through
evaporation
Unrelated species which show similar traits;
which have evolved separately, but under similar
evolutionary pressures
natural / directional , selection ;
mutation ;
(mutation / genetic variation, is) random / due to
chance / spontaneous / pre-existing ;
selection pressure is lack of / competition for , food /
prey ;
individuals with mutation(s) / allele(s) / gene(s) (for
echolocation) , survive ; ora
(echolocation) allele(s) / gene(s) / mutation(s) ,
passed on ( to next generation) ;
over many generations frequency of , echolocation /
allele / characteristic , increases ;
1
1
1
1
1
3 max
1
1
1
2
3
4
5
6
7
4 max
1
(b) (i)
Pipistrellus ;
(b) (ii)
similar / same, (body) mass ;
similar wingspan ;
`similar / same, colour ;
all characteristics , similar / same, except echolocation
/ wingspan ;
© Oxford University Press 2015
1 max
2 DO NOT CREDIT if implied
as a consequence of
selection
pressure
4 ACCEPT ‘selection
pressure is ability to hunt’
4 ACCEPT ‘selective
pressure’
5 IGNORE refs to breeding /
reproduction
5 ACCEPT ‘individuals that
can echolocate survive’ ora
5 DO NOT CREDIT if answer
implies that echolocation is a
learned behaviour
6 IGNORE ‘genetic trait(s)’
7 Answers must imply
multiple generations
7 ACCEPT ‘over time’ as an
alternative to ‘over many
generations’ but must be
further qualified
Mark the first answer. If the
answer is correct and an
additional answer is given
that is incorrect or contradicts
the correct answer then = 0
marks
IGNORE case of initial letter
‘P’
DO NOT CREDIT if species
name given as well
Mark the first answer. If the
answer is correct and an
additional answer is given
that is incorrect or contradicts
the correct answer then = 0
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10 Classification and evolution
Answers to practice questions
OCR Biology A
previously unable to measure echolocation
(frequency);
(b) (i )
(inter)breed / AW ;
determine if offspring are fertile ;
if offspring are infertile / no offspring produced,
then different species ; ora
(c)
Most marks (apart from C2, C5 and D5) are stand
alone and do not need to be linked to context.
However, max 5 if any statements are mismatched.
continuous ;
(continuous / AW , is) effect of , many genes /
polygenic / genes and environment / genetic and
environmental / environment ;
quantitative ;
there is a range / any value is possible / intermediate
values / no distinct groups / AW ;
example to illustrate any C marking point ;
discontinuous ;
(effect of) one / few, genes ;
little / no, environmental effect ;
discrete categories / no intermediates / AW ;
example to illustrate any D marking point ;
© Oxford University Press 2015
2 max
C1
C2
C3
C4
C5
D1
D2
D3
D4
D5
marks
IGNORE ‘similar appearance’
ACCEPT ‘both 5.5 g’
IGNORE ‘same’
ACCEPT ‘almost the same’
or ‘small difference’ or ref to
figures
ACCEPT ‘both (medium to
dark) brown’
ACCEPT ‘mate’ / ‘reproduce’
CREDIT ‘observe to see if
populations are
reproductively
isolated’ as resitting A2
candidate might consider
phylogenetic species
definition
This mark is for assessing
the fertility of the offspring
'if they belong to the same
species they will be able to
breed with each other and
produce fertile offspring' = 2
st
marks (1 and 3rd)
For example ‘ some variation
is controlled by only one
gene this variation will have
intermediates’
AWARD D2 and C4 but max
5 for the whole question and
DO NOT AWARD QWC and
put CON in the margin
C2 IGNORE alleles
C2 IGNORE example of
environmental factor, e.g.diet
C2 Must be linked to context
of continuous variation
C3 No ora for discontinuous
C5 must be linked to another
C mark
CREDIT only , body mass /
wingspan / colour / range of
pitch within species
D2 ACCEPT ‘there is a gene
for pitch’ or ‘there are highpitched and low-pitched
alleles’
D2 ACCEPT any suggestion
of a low number of genes
D2 IGNORE ‘variation is
genetic’
D3 ACCEPT ‘only influences
by genes’ / AW
D3 IGNORE unqualified refs
to genes
D4 ACCEPT ‘set groups’
D5 Must be linked to another
D mark
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10 Classification and evolution
Answers to practice questions
OCR Biology A
7 max
© Oxford University Press 2015
D5 CREDIT only these
examples:
low-pitched or high-pitched /
pitch variation between
species / sex / no bat call
between 47 and 52 Hz
D5 IGNORE ‘colour’ as an
example to support a D mark
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OCR A Biology
11.1
Summary Questions
1 Species richness – the number of different species living in a specific area (1).
Species evenness – the number of individuals within the species living in a community (1).
2 Habitat biodiversity: desert – low; coastline – high (1). Suitable habitat examples given (1). Species
biodiversity: desert – low; coastline – high (1).
3 Answer must include: (max 4) Greater genetic variation / wider range of alleles (1); therefore
increased likelihood some organisms are suited to a habitat change (1). Plus up to two from: some
organisms may be suited to different habitats (1); therefore more areas may be colonised by the
species (1); some individuals will be resistant to a new disease (1); therefore lower probability of all
organisms being killed by the disease (1); some organisms will be better adapted to avoid new or
adapted predators / catch prey (1); therefore less chance of being eaten / starvation (1); other suitable
example with consequence (2).
11.2
Summary Questions
1 Random – all organisms have an equal chance of selection; non-random – different organisms have
higher / lower probabilities of being selected (1).
2 Use random sampling (1); removes sampling bias (1); use as large a sample size as possible
(1); removes the effects of chance (1).
3 a Systematic sampling (1); abiotic conditions vary as you travel downstream affecting the type and
abundance of organisms present (1).
b Random sampling (1); any one from: environment easy to study/fairly uniform (1); reduces sample
bias/increases reliability (1).
11.3
Belt transect in a National Park
1 Mark out line across sampling region (1); quadrats are placed at fixed intervals (1m apart), and used
to measure percentage cover in that area (1).
2 Students were trying to investigate specifically how the distribution of plant species varied according
to land use. Random sampling would not sample required region in a systematic manner (1).
3
Correct axes, labels, and suitable scale (1); correct plot and line (1)
4 Any 4 from: Bare ground is the path (1); path is 3 / 4m wide / is found between positions 3 / 4 and 6
(1); where the ground is trampled, grass is the dominant species (1); there appears to be a direct
correlation between the maximum height of vegetation and the percentage cover of grass (1);
untrampled regions have greater species biodiversity (1); larger species are found away from the path
region (1); other relevant points from the data (1)
5 Suggested abiotic factor e.g. soil pH (1); explanation of how the factor could be measured at each
station along the transect in a systematic manner e.g. measured using a pH meter positioned at the
centre of each quadrat position (1)
© Oxford University Press 2015
This resource sheet may have been changed from the original.
OCR A Biology
6 Only one set of data was collected along the line (1) which may not be representative of the whole
region (1); not all samples sum to 100% percentage cover (1) therefore errors were made /some
unidentified species were not recorded (1); therefore the conclusions drawn are tentative (1)
Summary Questions
1 a pooter (1) b sweep net (1) c pH probe (1) d quadrat (1).
2 Any two from: likely to have higher resolution; rapid changes can be monitored; less possibility for
human error; data can be stored on a computer (2).
3 At least one advantage and disadvantage should be included for each sampling measure.
Population density – advantages: accurate; disadvantages: time consuming, can only be used when
individual members of a species can be identified (1) Frequency – advantages: rapid, can be used
when individual members of a species cannot be identified; disadvantages: only gives an approximate
result (1) Percentage cover – advantages: allows a lot of data to be collected quickly; disadvantages:
only gives an approximate result; least precise sampling technique (1).
11.4
Summary Questions
1 Any three from: large number of successful species; ecosystem fairly stable; many ecological
niches available; environment unlikely to be hostile; complex food webs exist (3 max).
2 a Pond B because it has the higher value of Simpson’s Index of Biodiversity (1).
b Pond A was more polluted. It has the lower value of Simpson’s Index of Biodiversity (1).
Fewer species can survive (1) the harsher environmental conditions (1).
3 N = 28 (1)
Organism
𝑛𝑛 2
� �
𝑁𝑁
Bird’s-foot trefoil
0.01
Crested Dog’s-tail
0.03
Meadow buttercup
0.10
Oxeye daisy
0.06
Rough hawkbit
0.01
Smaller cat’s-tail
0.01
2
∑(n/N) = 0.21 (1)
Simpson’s Index of Diversity = 0.79 (1)
11.5
Summary Questions
1 Any two from: number of alleles in a population must increase (1); mutations can create new alleles
(1); gene flow can introduce new alleles into population (1).
2 More genetically biodiverse a species is the greater variation in DNA/number of alleles present (1);
species more likely to survive a change to the environment (1) as there is a higher probability that
some members of the species will have the allele to survive the change and reproduce (1).
© Oxford University Press 2015
This resource sheet may have been changed from the original.
OCR A Biology
3 Species A 0.48/48% of genes were polymorphic (1). Species B 0.6/60% of genes were polymorphic
(1). Species B is more genetically diverse as it has a higher percentage pf polymorphic genes (1)
therefore more alleles are present (1).
11.6
Loss of biodiversity in the UK
1 Economic gain (1)
2 Twice the proportion of chalk grassland (compared to lowland mixed woodland) (1).
3 300 000 hectares (1)
4 Any two from: Replanting hedgerows (1); provides an additional habitat allowing more species to be
supported (1). Mixed planting/crop rotation (1); greater range of plant species enabling more animal
species to be supported (1); reduced/no chemical use (1); pests/weeds not destroyed which provide
food source for a range of species (1). Any other suitable reason and explanation.
Summary Questions
1 a Only one type of plant is grown (1); this will be the food source for only one / a few species of
animal (1); or other relevant reason, correctly justified.
b Deforestation / removal of habitat / reduction in plant diversity (1); leading to reduction in animal
diversity because more limited range of foods / shelter available (1); or other relevant reason,
correctly justified.
c Pesticides remove pest species (1); which reduces the food source for the animals that live off
these species (1); or other relevant reason, correctly justified.
2 Any two from: Deforestation removes the food source for a / many species, which causes starvation
for the predator species which feed on them (1); loss of habitat removes the shelter used by
organisms, causing local extinction / migration (1); migration of species can lead to reduction in
species diversity in adjacent areas (1).
3 Any three from (effect and explanation required for each mark): Polar ice caps melting / reducing in
size leads to reduction in size of ice sheets, removing the habitat of the native species (1); rising sea
levels leads to loss of habitats in low lying land / salination of rivers further upstream, affecting the
species which inhabit these areas (1); higher temperatures can lead to drought in a region, affecting
non-drought resistant species (1); insect life cycles and populations may change, spreading diseases
to new regions, affecting the animal organisms which live in these areas (1). Other suitable
suggestion with explanation of its effect on a population.
11.7
Keystone species
1 Species that are essential for maintaining diversity – they have a disproportionately large effect on
their environment relative to their abundance (1).
2 Predators keep the populations of their prey at a consistent level, thus allowing for the balanced
coexistence of other species (1).
3 Up to two marks for stating how a reduction in the population of prairie dogs would lead to a
reduction in the population of purple coneflowers; up to two marks for a linking explanation of the
effect. Less aeration of the soil (1) leading to fewer decomposers redistributing nutrients (1).
Less animal droppings / movement of soil (1) less redistribution of nutrients (1).
Water being lost (by evaporation) / not channelled into the water table (1) lack of sufficient water for
growth / photosynthesis (1).
Summary Questions
1 Aesthetic – how the region appears; economic – how the region can provide an income; ecological
– the species that can be supported in the region (1).
2 Any two sensible suggestions, for example: All living organisms have right to survive and live in the
way they have become adapted (1); habitat and biodiversity loss prevents many organisms living
where they should (1); moral responsibility to conserve for future generations (1).
© Oxford University Press 2015
This resource sheet may have been changed from the original.
OCR A Biology
3 a Only two potato varieties plants in Ireland so little genetic diversity (1); when new disease
introduced (Phytophthora infestans) no potato had resistance to disease (1).
b Plant wider range of crops to increase genetic biodiversity within populations (1); if new
disease/climate change/pests introduced some will have resistance/ ability to tolerate changing
conditions, so some crops will survive (1).
11.8
Summary Questions
1 Captive breeding – animals reared and bred in human-controlled environments (1). Seed banks –
collections of seeds are stored (at low temperatures) for future use (1). Botanic gardens – collections
of plants are grown in controlled environments (1).
2 Up to two advantages, and two disadvantages:
Advantages – allows for an individual of a species to survive; maintains or increases endangered
populations (1); allows for reintegration of a species into its natural habitat (1)
Disadvantages – leads to loss of genetic diversity (1); leads to organisms not learning behaviours of
their wild counterparts (1); can decrease the disease resistance of a population (1); genetic problems
can occur in offspring due to in-breeding (1).
3 Any four from: controlled grazing allows species time to recover, rather removing them entirely from
habitat (1); restricting human access to prevent poaching / avoid plant species being trampled (1);
feeding animals to ensure native populations able to survive to reproductive age (1); culling/removal
of invasive species to ensure native species able to access resources / to remove competition for
resources (1); halting succession ensuring the habitat remains in its current state, so appropriate
resources are available for current species (1).
4 Any four from: Landowners/countries have economic/cultural reasons for exploiting natural
resources (1); exploitation leads to loss of biodiversity (1); (financial) incentives often needed to
replace income exploiting a resource would provide; animals do not respect nation’s boundaries (1);
agreements between nations needed to manage endangered populations and help limit trade of
controlled/protected species (1); to preserve number of species in natural habitat (1).
© Oxford University Press 2015
This resource sheet may have been changed from the original.
Chapter 11
Practice questions
OCR Biology A
Question
number
Answer
Marks
1
B
1
2
B
1
3
Term
a. Biodiversity
Description
The variety of living
organisms present in
an area.
Genetic biodiversity
b. The variety of
genes which make
up a species
c. Random sample
A sample produced
without bias; each
individual has an
equal likelihood of
selection.
Abiotic factor
d. Non-living
component of a
habitat
e. Systematic
A sampling technique
sampling
where different areas
within a habitat are
identified and
sampled separately;
for example, by using
a line transect.
Collecting a number of representative organisms
5
Any two from:
to reduce time required to collect / count organisms;
to minimise costs;
some organisms cannot be easily located / collected;
organisms move in / out of habitat; other appropriate suggestion
Random – all individuals have an equal chance of selection; nonrandom – some element of bias affects the likelihood of an
individual’s selection;
random advantage – more representative of the population /
sample produced is unbiased;
non-random advantage – allows easier comparison between
abiotic factors / changing abiotic conditions can be studied;
Any six from:
Up to two animal sampling techniques
e.g. Use of sweep nets to catch flying insects / insects in tall
grass;
Up to two plant sampling techniques
e.g. Use of quadrats to measure percentage cover of different
plant species;
Use of random / non-random sampling approach;
with justification;
Collection of several samples with use of averaging;
with justification;
Measurement of a named abiotic factor;
in relation to plant / animal abundance;
Correct use of term species richness / species evenness;
1
4 (a) (i)
4 (a) (ii)
4 (b)
4 (c)
© Oxford University Press 2014
Guidance
1
3
2
2
1
1
1
1
1
1
1
6 max
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Chapter 11
Practice questions
OCR Biology A
5 (a)
41 667
2
5 (b)
1 Part of ecosystem / habitat for other organisms;
2 Part of food, chain / web
3 wood useful for specific purpose
4 (potential) source of medicine
5 genetic resource
6 aesthetic value / give pleasure / beautiful trees
7 ethical reason / moral responsibility
8 resource for (non-medical) scientific research
3 max
5 (c) (i)
Not in, natural / normal, habitat / environment;
5 (c) (ii)
1 most plants produce an excess
2 (so) can be collected (from wild) without damaging (wild) ,
plants / organisms / population / habitat;
3 take up little space; ora
4 able to store, large numbers / more species ; ora
5 easy / cheaper, to transport / AW ; ora
6 idea of remaining viable for long periods; ora
7 less susceptible to, disease / pests / environmental change ; ora
© Oxford University Press 2014
Award 2 marks for a
correct answer,
even if no working
shown.
ALLOW 1 mark for
41 666 666; 41
666.7, 41 666.67,
41 666.667, 41 700,
41 666, 41668 or 42
000.
If the answer is
incorrect ALLOW 1
mark for
(2500x100)/6
Mark the first three
reasons regardless
of lines
1 IGNORE
biodiversity
2 ACCEPT food
source
2 IGNORE home
3 e.g. making,
fences / furniture /
boundary marker
5 ACCEPT
description or
example but must
refer to genes
6 ACCEPT tourism
7 ACCEPT idea that
they have a right to
exist
7 DO NOT CREDIT
‘playing God’
1
4 max
5 ACCEPT can
easily be sent where
wanted
6 Answers must
have some
reference to
survival, not just
‘can be stored for a
long time’
7 IGNORE
recovery/survival,
from disease
7 CREDIT answers
that describe
(greater) disease
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Chapter 11
Practice questions
OCR Biology A
5 (c) (iii)
1 (maintain/increase) genetic variation / gene pool;
2 reduced chance of (future), disease / environmental change,
affecting (whole) population;
3 reduces chance of interbreeding;
4 maintain, geographical variation / varieties / races / strains /
subspecies ;
3 max
6 (a) (i)
3
species
Dog’s
mercury
Wild
strawberr
y
Common
avens
Wood
sorrel
Number
of
individual
s (n)
40
n/N
(n/N)2
0.40
0.16
13
0.13
0.02
43
0.43
0.19
4
0.04
0.01
N = 100
2=
Σ(n/N)
0.38
1-(
2
Σ(n/N) )
= 0.62
;;;
6 (a) (ii)
species richness
number of species (in an area / habitat) ;
species evenness
number of / how many, individuals there are of,
each / every, species (in an area / habitat) ;
© Oxford University Press 2014
resistance as a
property of the
seeds themselves
Or that the seed
bank is a (more)
protected
environment for the
seeds
IGNORE cheaper
unqualified
1 ACCEPT different
alleles
1 DO NOT CREDIT
different genes
2 ACCEPT ‘so if
one dies from a
disease some might
survive’
2 ACCEPT ‘to get
some plants that are
resistant to different
diseases’
4 IGNORE variation
unqualified
Award 3 marks for
the correct answer
(0.62 to 2 dp)
If answer is
incorrect:
IGNORE numbers
in first 4 rows
‘N = 100’ = 1 mark
2
Σ(n/N)
ALLOW ecf for
correct calculation
from candidate’s
incorrect N
value
2
1-(Σ(n/N) )
ALLOW ecf for
correct calculation
from candidate’s
2
Σ(n/N) value
Answer must be
given to 4 dp for ecf
2
IGNORE organisms
/ abundance /
quantity / variety
DO NOT CREDIT
amount
ACCEPT
‘organisms’ as AW
for individuals
CREDIT relative
abundance of
(each) species /
population size of
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Chapter 11
Practice questions
OCR Biology A
6 (a) (iii)
(habitat) dominated by, one / few / AW, species ;
change in one species , likely to affect whole
habitat / AW ;
community / ecosystem / habitat / area , is
unstable / not able to withstand change /
easily damaged ;
2
6 (b)
1 idea of random sampling ;
2 standardisation of technique ;
3 use of, key/identification chart ;
4 survey at different , times of year / season ;
5 include , trees / species larger than quadrat ;
2
7 (a)
idea that:
1 not all , areas explored / species yet discovered ;
2 microscopic / small / nocturnal / camouflaged , species difficult
to see ;
3 sampling might miss rare species ;
4 organisms mistakenly identified as one species may actually be
two (or more) species ;
5 concept of species is difficult to define ;
2
7 (b) (i)
1 both / assessed and threatened , show increase ;
2 number of assessed (species) , always / AW , higher (than
threatened species) ; ora
3 idea of: widening gap between assessed (species) and
threatened (species) / higher rate of increase for assessed
species ;
4 between 2000 and 2002 / in first two years , both / assessed
and threatened , were level / AW ;
3
© Oxford University Press 2014
each species
IGNORE relative
abundance of, a /
one, species
DO NOT CREDIT
amount
ACCEPT high
number of one
species
IGNORE
environment /
biodiversity as AW
for community
IGNORE the
community / AW will
be damaged
IGNORE prompt
lines and mark as
prose
1 ACCEPT
description of
randomisation
method
2 ACCEPT
description of
standardisation
method
2 ACCEPT count
the same way each
time
4 IGNORE ‘repeat’
unqualified
4 IGNORE different
times of day /
different times
CREDIT any valid
point where seen
1 ACCEPT ‘not all
species have been
identified (yet)’
1 IGNORE ‘yet to be
named’
1 IGNORE refs to
speciation
1, 2, 3 ACCEPT
‘organism’ as AW
for species as it is
an
‘idea that’ marking
point
Marking points 1-5
must be stated in
words, not implied
by
figures
1 IGNORE both are
similar shape
unqualified
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Chapter 11
Practice questions
OCR Biology A
5 after 2004 , both / assessed and threatened , have, reduced
rate of increase / slower increase / AW ;
6 figures to support any above statement ;
7 (b) (ii)
Year
Total
number
of
species
Total
species
threatened
2000
2001
2002
2003
2004
2005
2006
2007
2008
2009
2010
16500
16500
16500
22000
38000
38500
40000
41500
45000
47500
57500
Accept
+/- 500
11500
11500
11500
12500
15500
15500
16500
16500
17000
17500
18500
accept
+/- 500
© Oxford University Press 2014
Increase Increase
in total
in number
number
of species
of
threatened
species
since
since
2000
2000
0
0
0
0
5500
1000
21500
4000
22000
4000
23500
5000
25000
5000
28500
5500
31000
6000
41000
7000
accept
accept
+/- 1000 +/- 1000
Acceptable 2
range for
% or total
65-75
65-75
65-75
53-60
39-43
38-42
40-43
38-41
36-39
35-38
31-33
1 ACCEPT general
statement or
referring to given
time
period
1 ACCEPT
assessed and
threatened show
positive
correlation
4 IGNORE ‘at the
start’ answers must
mention years
5 IGNORE ‘between
2004 and 2005’
answers must imply
whole of
subsequent time
period
6 figures must
support another
point that has been
credited
6 Answers must
quote numbers for
total assessed
species
and for threatened
species along with
two years
6 ACCEPT
calculated
comparisons
Examples of
acceptable figure
quotes to support
each point
mp1 “between 2000
and 2009 total
assessed species
increase by 31000
and threatened
species increase
from 11500 to
17500”
mp2 “in 2004 total
assessed species
was 38000 and
threatened was
15500”
mp3 “in 2000 there
were 5000 more
assessed species
than threatened, in
2006 the gap was
23500”
mp4 “between 2000
and 2002 assessed
This resource sheet may have been changed from the original.
www.oxfordsecondary.co.uk/acknowledgements
Chapter 11
Practice questions
OCR Biology A
7 (b) (iii)
1 (total species assessed is increasing because) ,
a idea of more sampling / exploration (leads to more
species identified)
or
b improved identification , techniques / described ;
2 (threatened species is increasing because) ,
a loss of habitat
or
b climate change
or
c increased human population
or
d idea of interspecific competition from introduced species
or
e idea that some of the newly-identified species are likely to be
threatened ;
3 (there is a widening gap between total and threatened species
because) ,
a new species tend to be discovered in areas where humans
don’t live so they are not threatened
or
b conservation techniques are working / AW ;
2
7 (c)
range / number , of habitats / ecosystems ;
genetic variation (within species) ;
CITES 2 max
C1 regulate / monitor / prevent , trade in , selected / certain /
endangered , species
C2 ensure (international) trade does not endanger , wild
populations / AW ;
C3 prohibit (commercial) trade in wild plants ;
C4 allow trade in , artificially propagated plants / AW ;
C5 allow (some) trade in less endangered , wild species /
organisms / animals and plants ;
Rio Convention 2 max
R1 sustainable use of , organisms / habitats / ecosystems ;
R2 share genetic resources ;
R3 share access to , scientific knowledge / technology ;
1
7 (d)
© Oxford University Press 2014
4
species were 16500
and threatened
were 11500”
mp5 “in the 4 years
before 2004, total
species rose by
21500 and
threatened by 4000.
In the 4 subsequent
years total assessed
rose by13000 and
threatened rose by
1500.”
1 IGNORE refs to
speciation as time
frame too short
1 eg DNA
fingerprinting
1 IGNORE study if
used in the context
of species that have
already been
identified
IGNORE idea of
conservation not
working
IGNORE refs to
hunting
IGNORE
‘competition from
newly discovered
species’ as
this implies that the
candidate thinks the
species was not
present until it was
discovered
e.g ‘as more
species are
discovered, the
number of
threatened species
will go up’
CREDIT only these
answers
If correct points
included under the
wrong headings
then award
max 1 for that
convention
ACCEPT suitable
synonyms for trade
throughout, e.g.
‘buying and
selling’
C1 ACCEPT ref to
products from
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Chapter 11
Practice questions
OCR Biology A
R4 idea of promoting (named) ex situ conservation
method(s) ;
R5 idea of raising profile of (biodiversity) with , governments /
public bodies / general public ;
R6 idea of international cooperation (on biodiversity issues) ;
© Oxford University Press 2014
endangered
species, e.g.
leopard skin
C1 ACCEPT ‘illegal’
as AW for ‘selected
/ AW’
C2 DO NOT
AWARD if ‘all trade
in wild plants’ stated
R1 ACCEPT
example e.g.
replanting trees /
fishing quotas /
large
mesh size
R2 AWARD in
context of access to
or benefits from
genetic resources
R4 e.g. ‘set up seed
banks’ / ‘captive
breeding
programmes’
R4 IGNORE ‘zoos’
unqualified
R4 IGNORE ‘in situ’
R5 ACCEPT ‘take
biodiversity into
account during
planning processes’
R5 ACCEPT
‘informing people
that it is their duty to
consider biodiversity
This resource sheet may have been changed from the original.
www.oxfordsecondary.co.uk/acknowledgements
OCR A Biology
12.1
Summary Questions
1 a Disease that can be passed from one organism to another (1).
b Pie chart should show: 9% injuries (1); 23% communicable diseases (1); and 9% injuries (1).
2 Table should summarise key comparisons between the four. For example, whether they are
prokaryote/eukaryote, whether they have plasmids or not, whether they are heterotroph/autotroph,
whether they are beneficial/pathogenic/neutral, comparison of cellular structure, whether they have a
nucleus or not. Extra credit for students who remember 70s and 80s ribosomes for
prokaryotes/eukaryotes.
3 13 000 000 is 23% of all deaths 68% of all deaths are caused by non-communicable diseases (1)
68% is 13 000 000/23 × 68 (1) = 36–37 million (1); then approximate because working with
approximate numbers (1).
4 Take over cell metabolism, viral genetic material inserted into host DNA (2); take over cell and
digest contents (e.g., some Protista) (1); completely digest living cells and destroy them (e.g., fungi)
(1); produce toxins which poison or damage host cells, some toxins break down cell membranes or
inactivate enzymes or prevent cell division (e.g., most bacteria) (2).
5 a Viruses insert genetic material into host DNA (1); and take over cell metabolism to make new
viruses before breaking out of cell (1); protists take over cells and feed on cell contents (1); and divide
before breaking out of the cell (1).
b Viruses only active when inside a host cell (1); have little structure and take over whole host cell (1).
12.2
The threat to English Oak trees
Acute oak decline caused by bacterium found on or in oak jewel beetles showing they may transfer
disease (2); trees become infected in regions where no oak jewel beetles so they are not
vectors/cause of disease or not the only vector/cause of disease (2). Any other sensible suggestions.
Banana diseases and food security
1 30 000 000 000 kg bananas (1)
2 Increased malnutrition as people deprived of their main staple food (1); increased disease and
death as people less able to resist disease due to malnutrition (1); shortages of other foods as people
try to buy other staples (1); increased food prices as a result of short supplies driving up demand (1)
any other sensible point
3 a lack of awareness of causes of disease (1); lack of biocontrol on farms (1); contamination
between farms (1) (max 2)
b Cloned Cavendish plants so if one plant susceptible to disease, they all will be and therefore whole
plantations can be wiped out (1).
TB, cows, and badgers
1 Wild animals such as badgers and possums which are infected with TB regularly use same
pastures and so bacterium is around in the grass (2).
2 It isn’t possible to tell if animal is infected with TB or has been vaccinated so would be impossible to
protect consumers against infected milk products etc (2).
3 Any sensible suggestions for example: People's perception of badgers are different to that of cattle;
some people put a higher value on wildlife than the health of farm animals and people; some people
feel that there are more humane alternatives that would protect both badgers and cattle (max 2).
4 Catching the animals to deliver vaccine would be difficult and stressful for animals and people;
would never know what proportion of the population was vaccinated; difficult to tell if an animal is
vaccinated or not; if vaccine in food difficult to quantify how many badgers eat food and impossible to
control dose. Any other sensible point (4).
5 Look for clear evidence of good research skills, balanced approach, citing sources etc.
Zoonotic influenza
1 Viruses not affected by antibiotics so cannot be cured (1).
2 Enables scientists to identify cause of outbreak (1); track its spread (1); and helps in the
development of vaccines (1); and (if bacterial) medicines to treat the disease (1).
a They were much younger than usual flu victims – up to 80% were under 65, whereas in normal flu
outbreaks, around 90% of deaths are in people over 65.
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OCR A Biology
b H1N1 was a new strain of flu which crossed species barrier from animals to people. It has
happened before – older people may have met a similar virus earlier in their lives whereas people
under 65 had not encountered a similar virus before and so it was extremely damaging to them (3).
Identifying pathogens
1 Appropriate treatment can be used (1); and appropriate steps taken to prevent the spread of the
pathogen (1).
2 Benefits: Relatively cheap (1); available in all hospitals (1); special stains show up classes of
organisms easily (1). Limitations: can take some time for culture to grow (1); some organisms e.g.
viruses can be very difficult to culture (1); light microscope limited e.g. viruses not visible (1).
3 Look for evidence of good research skills, clear explanations, citing of sources etc.
Summary Questions
1 Bacterial: Ring rot (caused by Clavibacter michiganense) affects potatoes, aubergine, and
tomatoes. TB (Mycobacterium tuberculosis and M. bovis) affects humans, cows, badgers, deer.
Bacterial meningitis affects humans. Viral: Tobacco mosaic virus affects tobacco plants and 150 other
species. HIV/aids affects humans and some apes. Influenza (Orthomyxoviridaespp affects mammals
including humans, pigs, and birds. Protist: Potato blight (Phytophthera infestans) affects potatoes and
tomatoes. Fungal: Black sigatoka (Mycosphaerella fijensis) affects bananas and plantains. Ring worm
(Trichopyton verrucosum) affects cattle (other spp. Affects most animals including people). Athlete’s
foot (Tinia pedia) affects human feet (max 6).
2 Show evidence of understanding of the ways in which bacteria attack animal and plant populations
and the different effects they have. For example: ring rot (plants) and TB (animal), any diseases may
be chosen as long as bacterial (1); both caused by bacteria, both cause tissue destruction, both
remain infective in the environment (3); animal disease can be cured by antibiotics/prevented by
vaccination, no treatment or vaccine for plant disease (2).
3 Students should show awareness of the variety of ways in which the diseases they have chosen
can be spread for example: For animals direct transmission from one animal to another via direct
contact, inoculation, and ingestion and indirect transmission e.g., droplet infection, fomites vectors
etc. For plants direct transmission plant to plant and indirect transmission including soil contamination
and different types of vectors e.g., wind, water, animals, humans. 3 for any three correct animal
methods, 3 for any three correct plant methods including up to 2 for specific types of vectors.
12.3
Preventing the spread of communicable diseases in humans
1 Removes and destroys pathogens so not transmitted through direct contact, ingestion,
or leaving on fomites.
2 Living close together increases droplet infection risk and contagion (2); poor sanitation increases
risk of contagion, contaminated water and food, and vectors (2); and poor nutrition means people are
more vulnerable to infection (1); all increase the risk of the spread of communicable diseases so
improving them lowers the risk (1).
3 Mosquitos breed in water (1); mosquitoes carry malaria (1); any waste container which holds water
provides a breeding space for mosquitos (1); and so increases malaria (1); removing the waste
reduces breeding opportunities for mosquitos(1); and so reduces the incidence of malaria (1).
4 Any sensible well-structured answer using variety of sources.
Preventing the spread of communicable diseases in plants
Points could include: If pathogen getting to plant prevented there will be no disease so methods of
reducing spread of pathogens would be key (e.g., human hygiene not spreading spores, disposing
diseased plant tissue carefully, removing all traces of damaged plant tissue, insect vector control, any
other sensible points); breed plants that are not susceptible to infection or disease resistant; if
conditions favourable for healthy plant growth plants will increase disease resistance (so good
management e.g., soil fertilising, pest control); need to balance all three elements to avoid disease.
Summary Questions
1 Direct pathogen is spread directly from one organism to another (1). Indirect – the pathogen is
spread from one organism to another through another medium, e.g. the air, a vector (1). Show
awareness of the differences between organisms that can move around and organisms that cannot.
2 Similarities: being crowded close together increases risk of direct and indirect transmission.
Weakened individuals more at risk of infection. Damage to protective outer layers can allow
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OCR A Biology
pathogens in (2). Differences: Animals actively exchange body fluids (sex, kissing, bites) plants don’t.
Animals transfer food and drink into body through mouth which plants don't (2).
3 Show awareness of the differences between organisms that can move around and organisms that
cannot. Similarities: animals, wind and water can act as vectors, fomite, e.g. bedding, sacks,
machinery can carry disease from one individual to another, soil contamination is common indirect
method of disease spread in plants and can affect animals too. Differences: droplet infection from
coughs and sneezes doesn’t affect plants BUT droplets and splashes from one leaf to another can do.
4 Treating people to reduce pool of infection using medicines against disease (1); or using vaccines
(no really effective ones developed yet); destroying mosquitoes that spread the disease (insecticide
sprays on water and homes) (1); preventing mosquitoes breeding (1), draining swamps (1), removing
waste filled with water (1); preventing mosquitoes reaching people (mosquito nets over beds, screens
at doors and windows) (1) (max 5).
12.4
Summary Questions
1 Receptors respond to molecules from pathogens (1); or to chemicals produced by the plant cell wall
when it is attacked (1); these attach to receptors, stimulating the release of signalling molecules to
switch on genes in the nucleus, triggering cellular responses (1).
2 Diagram of table to include: Production of defensive chemicals, e.g., insect repellants, insecticides,
antibacterial compounds including antibiotics, antifungal compounds, anti-oomycetes, general toxins
(2); physical defences, e.g.; callose barriers immediately, callose and lignin deposition in cell walls
longer term, callose blocking sieve plates to prevent spreading through the phloem, callose deposited
in plasmodesmata to prevent spread of pathogens from one cell to another (2); sending alarm signals
to uninfected cells so they can put defences in place (2).
3 Show evidence of careful research from reputable sources and reference their sources (6).
12.5
Summary Questions
1
Adaptation
How it prevents entry of pathogens
skin
impermeable barrier between air/water and inside the body
sebum
chemical produced by skin which inhibits growth of pathogens
mucus
traps pathogens in nose/trachea etc uses lysosomes to destroy bacteria/fungal spores,
phagocytes to engulf and digest pathogens
tears
lysozymes break down pathogens
urine
lysozymes break down pathogens
stomach acid
destroys most bacteria and fungal spores entering the gut
expulsive reflexes
coughing/sneezing expel pathogens from gas exchange system
vomiting and diarrhoea expel pathogens from digestive system
blood clotting
protects again the entry of pathogens through broken skin
2 Localised inflammatory response to pathogens at site of wound, mast cells activated and release
histamines and cytokines (1); histamines cause vasodilation causing localised heat and redness (1);
raised temperature helps prevent pathogens reproducing; histamines make blood vessels leaky
forcing tissue fluid out, causing oedema and pain (1); cytokines attract phagocytes to site which
phagocytose pathogens; accumulation of dead phagocytes and pathogens forms visible pus layer (1).
3 a Pathogens produce chemicals which attract phagocytes (1); phagocytes recognise nonhuman proteins
in pathogen (1); this is not a response to a specific type of pathogen, simply to a cell or organism which is
‘not self’(1); phagocyte engulfs the pathogen and encloses it in a vacuole called a phagosome (1);
phagosome combines with a lysosome to form a phagolysosome (1); enzymes from lysosome digest and
destroy pathogen (1).
b Cytokines act as cell signalling molecules (1); that stimulate phagocytes to move to a site of infection or
inflammation (1). Opsonins bind to pathogens (1); and tag them so they are more easily recognised by
phagocytes (1); because phagocytes have receptors on their membranes which bind to common opsonins,
e.g., antibodies (1).
© Oxford University Press 2015
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OCR A Biology
12.6
Summary Questions
1 immunoglobulins, Y shaped glycoproteins that bind to specific antigens on pathogens/toxins/foreign
cells. Specific antibody for every antigen. Made up of two heavy and two light polypeptide chains with
an active site made up of 110 amino acids which fits the antigen (1). Work by binding to antigen
forming antigen-antibody complex which is then either engulfed by phagocytes or simply cannot
function as a pathogen anymore (1).
2 Similarities: Both T and B cells form clones of active cells (1); both form memory cells which mean
that when they meet a pathogen a second time there is a rapid response, destroying the pathogen
before it can cause disease (1). Differences: T cells stimulate B cells (1); T cells destroy pathogens
directly (1); B cells produce antibodies which act as opsonins stimulating phagocytes to engulf
pathogens (1); T cells also regulate immune response so it stops once a pathogen is removed and
doesn’t turn against body cells (1).
3 In autoimmune disease immune system stops recognising ‘self’ and starts attacking healthy cells.
Immunosuppressant drugs reduce activity of immune system (1); preventing/reducing destruction of
healthy tissue BUT susceptibility to infection increases (1); as immune system less effective at
recognising pathogens (1).
4 Humoral immune system responds to antigens outside of cells (1); bacterial and fungal cells present
in body have antigens to which humoral system can respond (1); system makes antibodies to
bacterial and fungal surface antigens, forms antigen-antibody complexes so macrophages readily
engulf pathogen (1). Cell-mediated system responds to changes in cells (1). Viruses get into body
cells and take over cell metabolism (1) – not so obvious in blood presenting antigens e.g., bacteria.
However, cell-mediated response detects changes in-virus infected cells and killer T cells attack and
destroy them.
12.7
Case study: Influenza
1 Vaccination (if available); isolating infected people as soon as any symptoms appear; preventing
travel into and out of infected countries (1) any other sensible point.
Summary Questions
1 Flow diagram that covers every step of how artificial active immunity is induced and is clear and
easy to follow as well as accurate and informative. (4 marks)
2 Any four sensible reasons including: Some diseases so severe that patient killed before body can
develop antibodies (1); some people do not have children vaccinated against diseases (1); if child
immunocompromised, has a comorbidity, or neglected/malnourished will be more vulnerable to
infections (1); bacterial disease may be resistant to current antibiotics (max 4).
3 a day 56 (1) b approximately 8600(1) c approximately 800 (1) d 86% (1)
4 a Most people vaccinated, so if bitten they are given course of injections to deliver antibodies
(produced in another animal) directly into blood stream (1); antibodies form antigen–antibody complex
with rabies virus, allowing phagocytes to destroy pathogen (an example of artificial passive immunity).
b Epidemic occurs when communicable disease spreads rapidly to a lot of people at either local/
national level (1); in vaccination, immune system stimulated to make antibodies to a pathogen by
exposure to safe form of an antigen injected into blood stream (1); if an epidemic begins to build,
mass vaccination (1); can protect people in the community by building immunity to infecting pathogen
and prevent pathogen spreading disease into the wider population (1).
5 Health departments increase public health awareness decreasing levels of communicable disease
(1); chlorinated water reduces water-borne infections and reduces number of deaths (1); public
hygiene will mean fewer rats etc to act as vectors of disease (e.g., plague) (1); penicillin reduces
deaths from bacterial diseases but does not affect viral diseases such as flu and polio (2); vaccines
such as polio reduced deaths from infectious diseases where a vaccine has been developed (1); Any
other sensible pint could be substituted for one of these.
© Oxford University Press 2015
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12 Communicable diseases
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
1
C;
1
2 (a)
microorganism / example; that causes
disease;
(leaf) epidermis; cuticle; closure of stomata;
2
no circulatory system; no, B / T,
lymphocytes; no memory cells; no
antibodies;
number of cases and number of deaths
increase over the time period; rate of
increase in both increases; rate of increase
in number of deaths is lower; ORA figures
quoted;
infectious disease is reason for increase in
both; increasing number of cases reported
is reason for increasing difference; idea of
aid / volunteers / treatment centres,
improved so proportion of people dying is
decreasing;
nucleic acid / RNA / DNA; protein coat /
capsid / capsomeres; (few) enzymes;
acellular / described; idea that cell is
fundamental unit of life;
4
drug / vaccine, development is expensive;
developing countries cannot afford the cost;
developed countries not affected; outbreaks
are rare; outbreaks affect (relatively) few
people; idea of balance cost of
development against development of other,
vaccines / drugs, which many benefit large
numbers of people;
decrease in number of cases of HIV; idea
of more awareness so precautions taken;
decrease in number of cases of TB; (due
to) improved, treatment / drugs; (due to)
decrease in number of cases HIV, TB is
opportunistic infection; decrease in number
of HIV status unknown; (due to) better
health care / diagnosis ;
figures quote with units;
idea of infection which develops when;
immune system weakened; due to
malnutrition / AW; primary infection;
TB airborne droplets and HIV bodily fluids;
TB
coughing / sneezing; overcrowded living
conditions / poor ventilation;
HIV
5
2 (b)
2 (c)
3 (a) (i)
3 (a) (ii)
3 (b)
3 (c)
4 (a)
4 (b)
4 (c)
© Oxford University Press 2014
Guidance
3
4
3
4
5
Max 5
3
4
Max 2 for either
This resource sheet may have been changed from the original.
www.oxfordsecondary.co.uk/acknowledgements
12 Communicable diseases
Answers to practice questions
OCR Biology A
sexual contact / injecting with used needles
/ blood transfusions / breast feeding;
promiscuity / drug use / contaminated blood
products;
© Oxford University Press 2014
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www.oxfordsecondary.co.uk/acknowledgements
Module 4 Biodiversity,
evolution, and disease
Application and Extension
OCR Biology A
Question Answer
number
Application
1 (a)
1 (b)
MarksGuidance
A disease that can be passed from one organism to another,
either directly from one organism to another by some
mechanism or indirectly e.g., via wind, water or an animal
Any sensible points, for example:
Spirometra
TB
HIV/AIDS
erinaceieuropaei
Communicable
disease
Communicable
disease
Communicable
disease
Can affect the
brain and other
tissues
Can affect the
brain and
other tissues
Can affect the
brain and
other tissues
Tapeworm
Bacteria
Virus
Rare in people,
found in
crustacea,
reptiles,
amphibian, and
mammals such
as cats and
dogs
Common in
people, also in
other
mammals
including
cows,
badgers, and
deer
Common in
people, rare in
other species
except for
some apes.
Probably
treatable by
anti-helminthic
medicine
Treatable by
antibiotics
No vaccine
Vaccine
Infection by
eating raw
infected meat,
eye contact with
raw infected
tissues
1 (c)
Droplet
infection
Cannot be
cured – can be
managed
using antivirals etc
No vaccine
Direct
transmission
via body fluids
A localised non-specific response to the presence of
pathogens – mast cells activated if tissue damaged release
histamines and cytokines, histamines increase blood flow
and cause oedema (swelling) and pain, cytokines attract
phagocytes to destroy a pathogen.
Suggestion: As tapeworm moved across brain, localised
inflammatory response would affect the working of that area
© Oxford University Press 2015
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www.oxfordsecondary.co.uk/acknowledgements
Module 4 Biodiversity,
evolution, and disease
Answers to practice questions
OCR Biology A
of the brain and cause changing symptoms.
2 (a)
Examination of observable features including under a
microscope
Comparision of proteins e.g., by chromatography, gel
electrophoresis
DAN (genome) sequencing
2 (b)
Natural selection: organisms within a species show variation in
their characteristics. This variation is caused by differences in
the alleles of their genes – genetic variation. Mutation also
introduces variation, creating a new version of an allele.
Predation, competition (for mates and resources) and disease
cause a struggle for survival. The organisms whose
characteristics are best adapted to these selection pressures
have an increased chance of surviving and reproducing. Less
well adapted organisms die or fail to reproduce. This process is
known as ‘survival of the fittest’.
Genes from the successful organisms are passed onto their
offspring in the next generation. This means the offspring are
likely to possess the characteristics that made their parents
successful. This process is repeated. Over time, the number of
individuals with the advantageous adaptation increases.
Therefore the frequency of the allele which codes for this
particular characteristic increases in the population.
Evolution: This takes place when natural selection takes
place in two different populations, separated by geography,
niche etc. The selection takes the populations in different
directions until eventually they cannot interbreed successfully
and new species have been formed. The formation of new
species by natural selection is evolution.
2 (c)
The parasite has a number of different hosts. The more hosts
it can attack the more likely it is to survive. It will need
different proteins to survive in the different hosts. A large
genome means it can code for a wide range of different
proteins at different stages in its life cycle, and is therefore a
useful adaptation to the parasitic way of life.
(d)
Spirometra erinaceieuropaei DNA sequencing allows
comparison with other species and similar specimens for
identification, enables identification of pathogen. Allows
identification of vulnerabilities to drugs
Wider points – any sensible suggestions e.g., classification –
allows identification of new species, identification of
evolutionary links by similarities and differences in DNA
sequences, intra-specific differences and genetic diversity
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Module 4 Biodiversity,
evolution, and disease
Answers to practice questions
OCR Biology A
can be measured
Medicine – real time identification of pathogens in outbreak,
monitoring spread of an outbreak, development of vaccines
and medicines, identification of strains resistant to medicine
etc.
Extension
1
Clear description given of the expected specific immune
response – a flow chart or diagram is a useful way to do this.
2
Evidence of research into the ways in which parasites avoid
rejection/destruction by their host organism. For example,
gross anatomy – hooks, suckers etc which allow the parasite
to position itself in the host. Some are physiological, for
example, reproductive strategies (used by Spirometra
erinaceieuropaei). Many are biochemical – this is the way
Spirometra erinaceieuropaei and other tapeworms suppress
the host immune system to prevent it recognising and
attacking them, for example, by producing oxylipins
(oxygenated forms of fatty acids) and eicosanoids such as
prostaglandins which interact with the host immune system
and, for example, suppress phagocytosis and cytokine
actions, block enzyme action in host cells, and interfere with
the immune system cascades.
3
One or two page report on chosen biochemical response,
with clear explanations and including diagrams if they are
helpful. Evidence of research from multiple sources
(including scientific papers).
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Module 4 practice questions
Answers
OCR Biology A
Question
number
Answer
Marks
1 (a) (i)
Discontinuous;
1
1 (a) (ii)
Environmental influence because it can be
learned; identical twins have the same genes;
so could be continuous; discrete characteristic;
idea that unlikely to have environmental
influence as done without thought; identical
twins share trait; (mainly) genetic;
Organisms that have similar characteristics; and
can interbreed to produce fertile offspring;
Mutation produce new alleles; (so) new
characteristics; (characteristics) may be
advantageous; different selection pressure(s) /
new niche; change in allele frequency;
Environmental factor; that leads to the, death /
reduced reproductive rate, of some organisms
(in a population);
Positive influence directional selection;
divergence of organisms; negative influence
idea of insufficient organisms able to survive;
Species evenness / described; species richness
/ described;
Deforestation reduces biodiversity; loss of
habitats; monoculture reduces biodiversity; idea
of growing one species of crop over large areas;
fertilisers reduce biodiversity if specific to one
plant / reference to algal bloom; increase
biodiversity if growth of all plants increased;
culling of predators increase biodiversity;
number of prey species increase;
Decreased number of species; figures quote;
decreased Simpson’s index of diversity; figures
quote;
Biodiversity; number of species in an area;
number of individuals within each species;
Increased number of species at higher
elevations; so increased biodiversity at higher
elevations; biodiversity stays about the same at
medium elevations; reduced biodiversity at
lower elevations;
Immune system; antigen; mitosis; clone; killer;
1 (b)
2 (a) (i)
2 (a) (ii)
2 (b) (i)
2 (b) (ii)
3 (a)
3 (b)
4 (a) (i)
4 (a) (ii)
4 (b)
5
6 (a)
6 (b)
Time scale is not specified; many generations;
example, e.g. humans change as they grow /
develop; idea of permanent change in
characteristics of population; no reference to
genetics;
Common ancestor on mainland; Finches flew to
different islands; isolated; no gene flow;
different, environments / niches /examples;
different selection pressures; descent with
modification / evolution; Finches on different
islands now have different, characteristics /
© Oxford University Press 2015
Guidance
4
3
2
4 max
2
3
2
8
4
3
4
5
4
6 max
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Module 4 practice questions
Answers
OCR Biology A
examples;
3
8
Preservation maintain / keep, environment the
same; conservation environment changes;
(changes) managed; (changes) sustainable;
In situ conserving species in their natural
habitat; ex situ conserving species in captive
breeding programs; zoos; gene banks;
A;
9 (a)
A continuous; B discontinuous;
2
9 (b)
Change in environment / new selection
pressure; some individuals will be better suited;
due to variation; (these individuals) survive and
reproduce; pass on alleles for advantageous
traits to offspring; change in allele frequency;
Random; mutation; new allele combinations;
new gene products;
Idea of different ways to measure success;
population number; humans are not the most
abundant species; length of time species exists;
there are many species (past / present) that
have lasted longer than humans; AVP; e.g.
moral arguments
Antigen / microorganism , attracted to
phagocyte; phagocyte (membrane) invaginates
/ AW; (antigen / microorganism) engulfed /
described; formation of phagosome;
(phagosome) fuses with lysosome; hydrolytic
enzymes (in lysosome); breakdown, antigen /
microorganism;
Microorganism / example; causes, disease /
harm;
Influenza caused by virus; antibiotics kill
bacteria; antibiotics do not effect viruses;
Prevent secondary infections; caused by
bacteria; opportunistic;
Weaken / attenuate, virus; (virus) no longer
causes disease; inactivate virus; use antigens
from, virus / bacteria;
Variable regions; bind to antigens; constant
region; bind to phagocyte; hinge region; bind to
more than one antigen; agglutination; disulfide
bridges; for strength;
Evolutionary; history / AW;
4 max
Divided into different levels; becoming, more /
less, inclusive / exclusive;
Classification; placing into groups / AW;
2
Placing into groups; based on common
evolutionary history;
Kingdom five divisions; eukaryotic group
divided; based on physical characteristics;
Domain three divisions; prokaryotic divided;
based on, molecular / physiological /
embryological / behavioural, characteristics;
evolution; change in characteristics of an
organism over time; random mutations;
2
7 (a)
7 (b)
9 (c)
9 (d)
10 (a)
10 (b)
10 (c)
10 (d)
10 (e)
10 (f)
11 (a) (i)
11 (a) (ii)
11 (a) (iii)
11 (a) (iv)
11 (b)
11 (c)
© Oxford University Press 2015
4
1
4
6
5 max
2
3
3
4
6
2
2
5
3 max for kingdom or domain.
6
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Module 4 practice questions
Answers
OCR Biology A
12 (b)
variation within species; change in environment;
selection pressure; organisms with
advantageous traits survive and reproduce;
ORA alleles (for advantageous trait) passed to
offspring; change in allele frequency;
bone marrow;
plasma (cell) / effector (cell);
antibody / immunoglobulin / IgG;
Antigens;
12 (c) (i)
P;
1
12 (c) (ii)
provide, long-term immunity / immunological
memory;
remain in, body / blood / lymph nodes, after
infection / AW;
produce, plasma cells / secondary response /
faster response, (when re-infected);
no Z in blood until day, 5 / 6;
concentration rises quickly / AW;
maximum concentration, at day 20 (A 19 – 21) /
of 7.5 au (A 7 – 8 au);
then falls slowly;
to about 2 au at day 60 (A 59-60);
1 max
DO NOT CREDIT ‘to remember
the antigen /
pathogen’
3
natural passive
immunity
natural active
immunity
artificial passive
immunity
artificial active
immunity
1
CREDIT rapidly / steeply / steep
gradient / faster
(than fall)
CREDIT maximum concentration
if described as
‘increase followed by decrease’
but must have
units
CREDIT ‘40 days after peak’
instead of 60 days
DO NOT CREDIT ‘units’ for au
No mark if more than one box
ticked
Crossed out tick to be treated as
‘blank’ if there is
another tick
If crossed out tick is the only
response then mark
this tick
12 (a)
12 (d) (i)
12 (d) (ii)
3
1
;
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OCR A Biology
13.1
Summary questions
1a Water potential in cells (1) b Blood pH (1) or any other sensible suggestions
2 Cell releases a chemical (1); which has an effect on a target cell (1).
3 Any six from: Organism needs to respond to internal/external changes for survival (1); occurs by
electrical impulses/nervous system in animals (1); chemicals/hormones/hormonal system in
plants/animals (1); / named example of hormone or chemical (1); different cells rely on others for
materials/removal of waste (1); named example e.g., glucose/oxygen (1); different organs work
together to ensure homeostasis (1); named example e.g., brain and skin in temperature control (1);
cells communicate through cell signalling (1); named example of where this occurs (1).
13.2
Multiple sclerosis
1 0.15% (1)
2 A disease causing the body’s immune system to attack its own cells (1)
3 Speeds up transmission of nerve impulses (1)
4 Light energy detected by sensory receptor/rod/cone cell in eye (1); impulse does not reach
CNS/brain or impulse received too slowly (1); due to sensory neurone axon being broken down (1).
Summary questions
1 Sensory neurones transmit impulses to the CNS / from receptor, motor neurones transmit impulses
away from the CNS / to an effector (1).
2 One mark for correctly drawn and labelled structure: dendrites – drawn as multiple short protrusions
from cell body (1); cell body containing a nucleus (1); long single axon (1); myelin sheath with nodes
of Ranvier (1).
3 Any four from: Axon of a myelinated neurone is covered in myelin (1); myelin is an electrical
insulator (1); the sheath is formed by Schwann cells growing around the axon several times (1); there
are gaps in the myelin sheath known as nodes of Ranvier (1); electrical impulse moves in a series of
‘jumps’ from one node to the next/saltatory conduction (1); impulse transmitted much faster than
along an unmyelinated axon (1).
13.3
Summary questions
1 Detect stimuli (1); convert energy into a nervous impulse (1)
2 Light energy is converted into a nervous impulse/action potential (1)
3 Any six from: When you touch the pin it exerts mechanical pressure on your skin (1); Pacinian
corpuscle found within skin detects pressure (1); pressure changes shape of Pacinian corpuscle (1);
stretch-mediated sodium channel in neuronal membrane stretches (1); channel widens (1); sodium
ions diffuse into membrane (1); membrane is depolarised/generator potential created (1); generator
potential creates an action potential (1); action potential transmitted along neurones to CNS/brain (1).
13.4
Measuring action potentials
1 Depolarisation/potential across membrane goes from negative to positive (1)
2 Any 4 from: At 5.5 ms, the inside of the membrane is negatively charged (–70 mV) with respect to
the outside of the membrane (1); because there is a higher concentration of Na+/sodium ions outside
the membrane (1); stimulus causes (voltage-gated) sodium ion channels to open (1); Na+ ions diffuse
into the neurone (1); down concentration / electrochemical gradient (1); this increases the potential of
the inside of the membrane / makes the inside of the membrane more positive (1) to +44 mV.
3 Time between action potentials = 5 ms (1)
Frequency = 1 ÷ time = 1 ÷ 0.005 / 1000 ms in 1 s, so 1000 ÷ 5 (1)
Frequency = 200 action potentials per second / 200 Hz (1)
Summary questions
1 The larger the stimulus the more frequent the nerve impulses/action potentials (1)
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OCR A Biology
2 Depolarisation – voltage/potential difference becomes more positive; repolarisation –
voltage/potential difference becomes more negative (1); hyperpolarisation – when potential difference
is lower than resting potential (1).
3 The axon could be immediately depolarised after an action potential (1); therefore the action
potential could travel backwards / in both directions / not reach target cell (1).
4 Any four from: Sodium potassium pump actively transports sodium and potassium ions (1); three
sodium ions moved out (1); two potassium ions moved in (1); potassium ions can diffuse out through
open potassium ion channels / sodium ions cannot diffuse in as sodium ion channels closed (1); more
positive ions outside than inside the axon / outside of the cell is more positively charged (1); cell/axon
membrane is polarised / there is a difference in voltage across it (1).
5 Any six from: Heat energy acts a stimulus (1); stimulus causes (voltage–gated) sodium ion channels
to open (1); sodium ions diffuse into axon (1); down electrochemical gradient (1); inside of axon
becomes less negative/more positive/depolarises (1); an action potential is triggered (1); along a
sensory neurone (1); to the CNS/Brain (1).
13.5
Effects of drugs on synapses
1 An inhibitory drug results in less action potentials being created / a stimulatory drug creates more
action potentials (1)
2 Any four from: Amphetamines stimulate the release of more neurotransmitter (1); from the
presynaptic neurone (1); therefore it will bind to more receptors/more receptors activated (1); on the
postsynaptic membrane (1); threshold is reached (1); more action potentials generated in the post
synaptic neurone (1); greater response to stimulus (1); neurotransmitter can only be released from the
pre-synaptic membrane (1).
3 Alcohol binds to GABAA receptors (1); this changes shape of receptor allowing more
neutrotransmitter (GABA) to bind (1); this increases activity of the neurotransmitter (1); GABA
decreases the activity of the brain by preventing nervous transmission (1).
4 Any six from: Serotonin would normally be taken back into the presynaptic neurone to be used
again (1); prozac would prevent this occurring resulting in higher levels of serotonin in synapse (1);
more serotonin can bind to neurotransmitter receptors (1); on postsynaptic membrane (1);
depolarisation threshold reached (1); more action potentials (1); nervous system more stimulated
reducing depression symptoms (1).
Summary questions
1 Junction between one neurone and another (or neurone and effector cell) (1)
2 Neurotransmitter receptors are only present on the postsynaptic membrane (1); so can only cause
depolarisation of this membrane resulting in action potential (1).
3 Similarity – both result in a build-up of neurotransmitter in the synapse / both result in an action
potential being triggered (1); difference – spatial is a result of neurotransmitter being released from
many neurones into the same synapse, temporal is a result of small amounts of neurotransmitter
being released from a neurone several times in a short period (1).
4 Any six from: The action potential reaches the end of the pre-synaptic/motor neurone (1);
depolarisation of the pre-synaptic membrane causes calcium ion channels to open (1); calcium ions
diffuse into the presynaptic knob (1); synaptic vesicles fuse with the presynaptic membrane and
release neurotransmitter / exocytosis takes place (1); neurotransmitter diffuses across the synaptic
cleft /binds with receptor on the postsynaptic membrane (1); sodium ion channels open and sodium
ions diffuse into the post-synaptic neurone (1); threshold potential reached / depolarisation triggers an
action potential (1); muscle cell contracts (1).
13.6
Summary questions
1 The CNS is the brain and the spinal cord, the PNS is all other neurones (1)
2 Somatic – c throwing a ball, d walking (1) Autonomic – a pupil dilation, b blood pressure (1)
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OCR A Biology
3 Reasonable suggestion (1); with explanation of the context is relevant (1); and why the ability to
control the function is important (1); for example: breathing can be controlled (1); so that a person can
hold their breath underwater (1); as it is not possible to breathe normally in this environment (1).
13.7
Summary questions
1 Anterior pituitary produces hormones, posterior pituitary stores and secretes hormones produced by
the hypothalamus (1)
2 Sound receptor in left ear  sensory neurone  temporal lobe in right cerebral hemisphere (3)
3 Cerebellum (1); because this region of the brain is responsible for muscular movement, body
posture, and balance (1).
13.8
Measuring reflect reaction time
1 Any two from: height ruler is dropped from / same object being dropped / volume of caffeinecontaining drink consumed / time of testing after caffeine consumption / other appropriate named
variable (1).
2 The brain has been involved in the decision-making process of how the body will respond to the
situation (1)
3 Any four from: Reaction time should be tested before a caffeine drink is given (1); this should be
repeated many times / with many people to obtain reliable results (1); caffeine drink of known
concentration is given to subjects (1); after a fixed time period, reaction time test is repeated in an
identical way to the initial test (1); a range (ideally 5+) of caffeine concentrations should be
investigated (1); reference to control group / use of placebo caffeine drink (1).
4 Any two from: Use of a placebo is designed to highlight bias in results (1); as the subject believes
they have received a stimulus which would affect the outcome of the test (1); placebo or control group
provides baseline against which result changes can be measured (1); to ensure results are due to the
independent variable, rather than a learned behaviour or response (1).
Summary questions
1 (pressure) receptor in skin  sensory neurone  relay neurone  motor neurone  (extensor)
muscle (1)
2 a and d (1)
3 Relevant situation named e.g., hand placed over burning material, reflex arc formed does not
include the brain / it is an involuntary response (1) which decreases the time taken for the body to
react to the situation (1).
4 Researcher should take into account: Health benefits / risks (1); the drug administered could have
an effect on the subject’s body (1); which could be positive or negative (1). Ethical considerations (1);
when employing a placebo, people may believe that they have received a drug which has a beneficial
effect which may or may not be the case (1); exemplification on the use of a placebo – for example,
an experimental drug may be able to cure a condition; use of a placebo means that some of the
sample group would not receive the drug (1) (max 4).
13.9
Drawing a labelled diagram of a sarcomere
These are areas where actin and myosin filaments to not overlap.
Slow-twitch and fast-twitch muscle fibres
1 Any three from: Slow twitch muscle fibres would appear redder in colour (1); due to more
myoglobin/blood vessels (1); there would be more blood capillaries (1); and more mitochondria (1); as
these cells carry out aerobic respiration (1).
2 Any four from (accept converse): Marathon runners will have a higher proportion of slow twitch
fibres (1); as these fibres provide contraction over long periods of time (1); without tiring (1); marathon
runners mainly use aerobic respiration during exercise (1); which provides the energy for slow-twitch
muscles (1).
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OCR A Biology
Summary questions
1 Muscles are made up of bundles of muscle fibres (1); each fibre is made up of bundles of myofibrils
(1); each myofibril is made up of protein filaments / actin and myosin (1).
2 A. Z-line (1) B. H-zone (1) C. Light band / I-band (1) D. Dark band / A-band (1)
3 Any four from (at least one must be a similarity): Similarities: Both involuntary / unconscious control
(1); both uninucleated / one nucleus per cell (1). Differences: Cardiac striated, involuntary not (1);
Cardiac – fibres branched, involuntary – fibres spindle shaped (1); involuntary are slower to contract
(1); involuntary can contract for longer (1).
4 a Longitudinal – sample taken in the orientation of the muscle fibre(s), Transverse – sample taken
across the muscle fibre(s) (1).
4 b Any four from: The small dots represent the actin filaments, and the large dots the myosin
filaments (1); the last diagram represents the dark band (1); in this region the actin and myosin
filaments overlap (1); the second diagram represents the H-zone (1); here only myosin filaments are
present (1); the first diagram represents the light band (1); here only actin filaments are present (1).
13.10
Monitoring muscle activity with sensors
1 Any four from: Select a named repetitive exercise for a specified muscle (1); connect EMG sensors
to the area of muscle to be studied (1); collect data from the subject at rest – this will provide
comparative baseline data (1); carry out named repetitive exercise over designated period of
monitoring (1); record EMG data (1); using a data logging interface and computer software (1); repeat
experiment at least three times to provide reliable data set (1).
2 Any three from: EMG trace compared to resting trace (1); initial period of activity within EMG trace
as muscle contracts/is activated (1); as muscle becomes fatigued, electrical activity to the muscle
increases / as the muscle becomes fatigued, the amplitude of the EMG trace increases / frequency of
signal decreases / signal becomes more erratic (1); time taken between initial stimulation and EMG
trace change measures the time taken to when / point at which muscle first becomes fatigued (1).
Summary questions
1 Any two from: In a contracted muscle the light band is narrower (1); Z lines closer together (1); Hzone becomes narrower (1).
2 Any three from: During a sprint, muscles work so hard oxygen cannot be replaced as quickly as it is
used up (1); aerobic respiration alone can’t be used / anaerobic respiration required (1); creatine
phosphate is a source of phosphate (1); the more creatine phosphate, the more ADP can be
phosphorylated (1); muscles can perform at maximum rate for longer (1).
3 ATP is needed to break bond/cross-bridge between actin and myosin (1); if no ATP available actin
remains bonded to myosin (1); filaments remain in contracted state / filaments can’t slide back to
original position (1).
4 Any four from: Muscles do not contract as much (1); as fewer calcium ions released into
sarcoplasm/from sarcoplasmic reticulum (1); fewer calcium ions bind to troponin (1); less troponin
moves/pulls tropomyosin (1); myosin-actin binding sites remain blocked (1); myosin cannot bind to
actin to pull filament (1).
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13 Neuronal communication
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
1 (a) (i)
1 (a) (ii)
1(b) (i)
Actinomyosin
Summation
Between 1.2 and 2.0 µm, tension higher as
length longer; non-linear change / described;
reference to peak; between 2.2 and 3.6 µm
tension lower as length shorter; figures quoted;
1
1
4 max
1(b) (ii)
2.0–2.2; µm;
2
1(b) (iii)
As length of sarcomere increases overlap
decreases between actin and myosin; fewer
cross bridges form during contraction; reduced
power stroke; as length of sarcomere decreases
reduced, movement / sliding, of filaments;
reduced contraction of muscle;
Myelin sheath
5
Reduced speed of impulse transmission; loss of
insulation along neurone; idea that sodium ions
can diffuse (into axon or dendron) at points
other than Nodes of Ranvier; shorter local
circuits;
Damage to sensory neurones affects vision and
sensation in limbs; damage to motor neurones
effects muscle contraction; (so) muscles of
bladder and digestive system; damage to
neurones present in central nervous system
effect balance and co-ordination;
A vesicle (containing neurotransmitter); F
neurotransmitter; G pre-synaptic membrane; H
post-synaptic membrane;
Exocytosis
4
Serotonin would remain bound to receptor (on
post-synaptic membrane); repeated action
potentials / continual stimulation;
P is complementary shape to serotonin
transporter; P binds to transporter; less / no,
serotonin taken into synaptic knob; increased
levels of serotonin in synapse;
Cocaine is not selective; levels of, other
neurotransmitters / named, increased; more /
increased, side effects; e.g., blood pressure;
Reduced blood flow; reduced formation of
tissue fluid;
Increases / produces, contrast; more detail
visible;
fMRI scan more detail; range of colours;
differences in blood flow visible; ORA
MRI scans advantages more detail of soft
tissue; safer / ionising radiation not used;
disadvantages more expensive; take longer;
less detail of bones claustrophobia; ORA
Treatment has to start quickly; CT scans still
provide some detail;
2
2 (a) (i)
2 (a) (ii)
2 (b)
3 (a) (i)
3 (a) (ii)
3 (b) (i)
3 (b) (ii)
3 (c)
4 (a)
4 (b)
4 (c)
4 (d)
4 (e)
© Oxford University Press 2015
Guidance
1
4
4
1
4
4
2
2
3
4 max
2
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OCR A Biology
14.1
Summary questions
1 Produce and secrete hormones (1); named example of gland and corresponding hormone e.g.,
adrenal gland and adrenaline (1).
2 Stimulus detected → gland stimulated → hormone secreted → travels in blood (plasma) → target
cell → binds to receptor/triggers response (2 max).
3 Neuronal communication (1); faster response needed to prevent damage / electrical impulses travel
faster than hormones in blood stream (1).
4 Detected by sympathetic nervous system (1); triggers the adrenal glands to secrete hormones (1);
from the adrenal medulla (1); named hormone – adrenaline / noradrenaline (1); effect of hormone on
body – raises heart rate / increases blood glucose concentration / widens air passages / narrows
blood vessels in non-essential organs / raises blood pressure (1); explanation of how hormonal effect
increases chance of survival e.g., increased rate of respiration / muscles are able to work harder /
longer (1).
14.2
Histology of the pancreas
Look for a clear well labelled scientific drawing, including scale and key features labelled.
Summary questions
1 Hormones are produced by endocrine tissue whereas substances such as enzymes are produced
by exocrine tissues (1).
2 Any two from: Lighter stained / appear whiter than surrounding tissue (1); larger cluster of cells (1);
when viewed under high power an acinus is visible / small groups of cells are visible surrounding a
duct (1).
3 Brown cells are beta cells (1); which produce insulin (1); found in islets of Langerhans (1); three
islets visible in slide (1); islets are endocrine tissue / secrete hormones (1); surrounding (blue) tissue
is exocrine / acini (1); which is exocrine / produces and secretes digestive enzymes (1) (max 6).
14.3
Summary questions
1 Negative feedback is feedback which causes any corrective measures to be reduced or switched
off, allowing the system to return to its original (normal) level (1).
2 Glucagon is responsible for increasing blood glucose concentration (1); binds to receptors on liver
cells (1); causes glycogenolysis (1).
3 Any four from: Glucose enters the cell via a glucose transporter (1); Glucose is metabolised / ATP
generated (1); ATP binds to potassium channels / potassium channels close (1); Cell becomes
depolarised (1); Calcium ion channels open (1); Calcium ions cause vesicles to release insulin /
exocytosis of secretory vesicles (1).
4 Any six from: Cause blood glucose concentration level to rise (1); rise in blood glucose
concentration is detected by β cells in islet of Langerhans (1); β cells release insulin into blood (1);
insulin binds to glycoprotein receptors / receptors on cell surface membrane (1); cells take up / absorb
more glucose (1); cell respiration rate increases (1); increases glycogenesis / conversion of glucose
into glycogen (1); results in a decrease in blood glucose concentration (1).
14.4
Type 2 diabetes
Look for evidence of a well thought out discussion amongst students – no right or wrong answers.
Students are thinking about their own opinions and perceptions about health, lifestyle, and healthcare.
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OCR A Biology
Summary questions
1 Award one mark for each correct row (max 3)
Type 1
Cause
inability to produce insulin
When does it develop
Period of development
childhood
fast
Type 2
no insulin produced or
cells not responding
properly to insulin
later in life
slow
2 Answer must include (max 3): Blood glucose concentration varies throughout the day / as a result
of diet and exercise (1); correct example given e.g., increased exercise will cause blood glucose
concentration to decrease / eating carbohydrates will cause blood glucose concentration to increase
(1). Plus any one of the following: knowing the exact blood glucose concentration ensures the correct
dose of insulin is given (1); if a sufferer gives too low a dose they will remain hyperglycaemic, this can
result in death if left untreated (1); if a sufferer gives too high a dose they can become hypoglycaemic.
This can result in a coma/unconsciousness (1). Any other sensible suggestion.
3 Any three from: They are suffering from type 2 diabetes (1); their body cells do not respond to
insulin/not sensitive to insulin/cell surface receptors do not bind to insulin (1); cells do not take up
enough glucose from the blood (1); so blood glucose concentration remains high (1).
4 One mark awarded for each correctly stated advantage or disadvantage. Must include one
advantage and disadvantage for each treatment. Acceptable answers given here, but allow other
relevant answers. (6 marks max)
Insulin: Advantages – readily available, dose can be altered easily, cheap. Disadvantages – reliant on
drugs, side effects, person has to inject themselves.
Transplant: Advantages – permanent cure, no reliance on drugs. Disadvantages – could be rejected,
person’s immune system is permanently repressed, not enough donor organs available, person
becomes susceptible to infections.
Stem cells: Advantages – permanent cure, no reliance on drugs, no risk of rejection, stem cells readily
available. Disadvantages – technology not ready yet, embryos have to be destroyed, risk of tumours
developing.
14.5
Summary questions
1 Any two named responses with appropriate explanation (2 max). For example: Heart rate increase –
to pump oxygenated blood around the body (1); pupil dilation – to take in as much light as possible for
better vision (1).
2 Less blood in the skin to keep it warm, hence the skin feels cold (1); as the blood has been
redirected to muscles to aid movement/ability to run (1).
3 Any six from: Danger detected by the autonomic nervous system (1); hypothalamus triggers
sympathetic nervous system (1); nervous impulse triggers release of hormones from adrenal medulla
(1); adrenaline causes glycogenolysis in liver cells (1); credit detail of second messenger model (1);
increased blood glucose used for respiration/create energy for muscle contraction (1); other
appropriate hormone and response (1); pituitary gland stimulates the adrenal-cortical system (1);
hormones release from adrenal cortex (1).
14.6
Monitoring heart rate
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OCR A Biology
1 Any four from: Resting pulse rate measured (1); by taking an average of several results (1); use of
data logging sensor e.g., pulse monitor to collect data (1); standardised exercise stated e.g., each
student performs 100 star jumps (1); pulse rate monitored during exercise (1); students allowed
resting time (e.g., 2 minutes) (1); pulse rate taken again (1); entire experiment repeated several times
(1).
2 The students needed a baseline measurement (1).
3 a 65.6 bpm (1)
3 b 5.79 (2); (1) for correct substitution.
4 a 𝑥𝑥̅1 = 65.6; 𝑥𝑥̅2 = 91.3 (1) Percentage increase = 39.2% (1)
b
Resting heart
–1
rate / min
Rank
Exercise
heart
–1
rate / min
Rank
d
d
57
2
84
1.5
0.5
0.25
64
4.5
89
4
0.5
0.25
54
1
84
1.5
–0.5
0.25
78
10
101
10
0
0
74
9
98
9
0
0
68
7
92
6
1
1
65
6
90
5
1
1
60
3
86
3
0
0
64
4.5
95
8
–3.5
12.25
72
8
94
7
1
1
2
(1) For each ranked column – max (2)
d = 16 (1); rs = 1 –
2
6∑ d 2
n(n 2 – 1)
(1); rs = 1 – ((6 × 16) / (10*99)) (1); rs = 0.903 (1).
c df = 8 (1); r = 0.902 >> 0.794 therefore p<0.01, so 99% confidence (1).
Summary questions
1 One mark for each correct row (4)
Stimulus
Receptor
high blood pressure
low blood pressure
low blood CO2
concentration
high blood CO2
concentration
baroreceptor
baroreceptor
chemoreceptor
Nervous system
involved
parasympathetic
sympathetic
parasympathetic
Effect on heart
rate
decrease
increase
decrease
chemoreceptor
sympathetic
increase
2 Any two from: Stress (1); leads to release of adrenaline / noradrenaline (1); which increases the
frequency of impulses from SAN (1).
3 a Increased respiration leads to increased production of CO2 (1); CO2 dissolves in the blood to form
carbonic acid, which lowers blood pH (1).
3 b Any three from: Chemoreceptors detect decrease in blood pH (1); frequency of impulses sent to
medulla decrease (1); decrease in impulses sent to SAN (1); SAN decreases heart rate (1).
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14 Hormonal communication
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
1 (a) (i)
Insulin
1
1 (a) (ii)
Different 3D shape due to mutation
1
1 (b)
Increased conversion of glycogen to glucose /
glycogenolysis; decreased use of glucose in
respiration; increased rate of, gluconeogenesis /
described;
3
1 (c) (i)
Increased; linearly; figures quote;
3
1 (c) (ii)
3
3 a (i)
(Relatively) constant / minor fluctuations; figures
quote; reference to dip at 90 mins;
correlation because prevalence of diabetes
increase as mean body mass increases; cause
not explained;
overall prevalence of diabetes increases as
mean body mass increases; increase in
prevalence fluctuates; figures quote;
weight loss – glucose absorbed from diet is lost
in urine; as lack of insulin reduces uptake into
cells; increased urination – increased glucose
concentration in filtrate lowers water potential;
less water reabsorbed from filtrate; dehydration
– increased water loss from body;
energy surplus would mean increased ATP
present; ATP synthase is enzyme required for
ATP synthesis; reduced levels of ATP would
mean increased sensitivity;
Tissue
3 a (ii)
Islets of Langerhans
1
3 a (iii)
4
3 (c) (i)
Both regulate blood glucose concentration;
detect blood glucose concentration; beta cells
release insulin when blood glucose
concentration is high; alpha cells release
glucagon when blood glucose concentration is
low;
red blood cells transport oxyhaemoglobin;
deliver oxygen; oxygen required for respiration;
ATP required for, production/release, of,
hormone/named;
Acinus / acini, cell;
3 (c) (ii)
Release digestive enzymes;
1
3 (c) (iii)
Cells / A form duct; for secretions; enzymes into
small intestines; exocrine gland; alpha / beta,
cells secrete into blood;
Repeated; identify anomalies; calculate mean;
5
Cotton wool placed on microscope slide; to
reduce movement of Daphnia; reduce counting
3
2a
2b
2c
2d
3b
4 (a) (i)
4 (a) (ii)
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Guidance
2
3
5
3
1
3 max
1
3
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14 Hormonal communication
Answers to practice questions
OCR Biology A
errors;
4 (a) (iii)
4 (b)
4c
4 (d)
4 (e)
Drops of pond water placed on the slide;
Daphnia do not dry out; no cover slip; maintain
oxygen for Daphnia;
Correct axes; correct plots; line of best fit;
2 max
Heart rate decreases as concentration
decreases; figures quote; linear;
Spread of data about the mean;
3
t = 3.38; one degree of freedom and p = 0.05;
less than critical value; no significant difference;
4
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1
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15.1
Summary questions
1 a 1 mark for each correct receptor and what it detects (3 max). For example: Pacinian corpuscle
/mechanoreceptor detects changes in pressure. Photoreceptors detect changes in light,
chemoreceptors detect chemical changes e.g., pH. Thermoreceptors detect temperature changes
b 1 mark for each correct effector and what it does (2 max). For example: Muscles – move limbs,
squeeze gut, squeeze chemicals from glands. Glands – secrete hormones and enzymes
2 a The maintenance of a dynamic equilibrium (1); within narrow ranges in the body (1).
b The body needs sensory receptors to monitor changes in the internal environment (1); effectors to
respond to those changes (1); and restore the original balance (1).
3 In a negative feedback system, when a change takes place systems in the body act to return the
situation to normal (1); – they inhibit the change (1); in a positive feedback system, when a change
takes place systems in the body act to reinforce the change (1). In homeostasis, the body seeks to
maintain a dynamic equilibrium (1); if there is a change, the need is to inhibit it and return things to the
original state (1); this is possible with negative feedback systems but not with positive feedback
systems (1).
15.2
The Namaqua chameleon – a highly adapted ectotherm
Adaptation
black in morning/on side exposed to Sun
pale grey away from Sun/at certain times of day
orientation of body sideways to Sun
How it helps animal warm up or cool down
absorbs more energy so can get warmer
reflects light so reduces heat loss
maximise surface area available to absorb
sunlight
increased heart rate in cool morning
increases metabolic rate, increasing temperature
inflation/deflation of body
maximises/reduces surface area exposed to Sun
presses body to sand
allows animal to warm up by conduction
heart rate slows down
reduces metabolic rate and so less warming
panting during day / mouth open
cools by evaporation of water from surfaces and
radiation from tissues
holds away from desert surface
to allow cooling into air by conduction, radiation,
etc
Summary questions
1 a Animals that use heat from their surroundings to warm their bodies (1); so their core body
temperature is heavily dependent on their environment (1).
b Lizard, locust, any other suitable examples (1).
2 a When an ectotherm such as a lizard basks in the Sun (1); it gains heat by radiation from the
sunlight (1) (or any other suitable example).
b When an ectotherm such as a lizard presses against the hot earth (1); it gains heat by conduction
(1) (or any other suitable example)
c When an ectotherm such as a lizard stands up as high as it can off the ground (1); it will lose heat
by convection currents in the air around it (1) (or any other suitable example)
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d When an ectotherm such as a lizard wallows in mud or water (1); it loses heat by evaporation of the
water from the surface of the skin (1) (or any other suitable example).
3 a Black colour – absorbs more heat by radiation (1); on exposed rocks – maximum radiation from
Sun to get warm and raise body temperature (1); alter position and posture to maximise heat
o
absorption by radiation and conduction until reach around 36 C then alter position to maintain
temperature but avoid overheating (1); need to reach body temperature of around 36°C in order to be
able to move quickly when diving (1); and to make sure their body temperature does not decrease too
much when diving (1) (max 5).
b For 6 marks all 4 statements must be addressed. Length of dive depends on temperature of animal
before the dive starts and the size of the animal (1); heat lost to the water by conduction (1); and
convection (1); bigger animals have smaller surface area : volume ratio (1); and so lose heat more
slowly (1); water temperature is lower than iguanas core temperature so heat is quickly transferred to
the water reducing the core temperature (1); slow and clumsy because body temperature dropped too
low for effective movement and control (1); until they have done more basking and gained heat again
(1) (max 6).
15.3
Summary questions
1 Reactions of respiration etc are controlled by enzymes (1); and have an optimum temperature (1); if
temperature of an organism is too low – reactions are very slow (1); so there is not enough energy for
muscle contraction for movement etc (1); if temperature is too high, enzymes are denatured (1); cells
of the body die which can lead to death (1) (any other sensible suggestions).
2 Ectotherms do not sweat (1); they can cool down using evaporation of water only if they wallow or
submerge in water (1); when they emerge from the water or mud they cool down as the water
evaporates from their skin surface (1); endotherms sweat as the core body temperature increases (1);
the evaporation of the water in sweat lowers the skin surface temperature (1); which in turn lowers
temperature of blood by conduction (1).
3 Peripheral temperature receptors are in the skin and detect changes in surface temperature (1);
receptors in the hypothalamus detect blood temperature within the body (1); the peripheral receptors
respond to environmental stimuli while receptors in the hypothalamus respond directly to changes in
core temperature (1).
4 a Pale colours reflect light (1); and therefore heat (1); so they reduce the amount of heat absorbed
from sunlight (1).
b Dark colours absorb more heat (1); therefore help to increase the body temperature (1).
c Cold places tend to have a lot of ice and snow (1); dark colours show up against the white (1);
makes animals very visible to predators (1); increased risk of being eaten is a stronger evolutionary
driver than advantage of heat absorption (1).
15.4
Looking at liver cells
Look for pencil drawings with clear accurate labels, genuine drawings from slides not copies of
diagrams, accurate magnification etc.
Cirrhosis of the liver
1 Hepatocytes divide rapidly (1); and liver is a very fast growing organ (1); so if the organ is not too
damaged it can regenerate over a few months (1).
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2 Alcoholic drinks contain the addictive drug ethanol (1); so people become addicted to the drug and
cannot function properly without it (1); look for evidence of sensible discussion of how to help those
affected, for example, use of medication to help treat withdrawal, support such as counselling or
alcoholics anonymous (2).
3 Any six from: Tiredness and weakness (1); inability to make urea which affects kidneys and may
retain fluid (1); weight loss due to inability to break down excess amino acids and use them (1);
inability to make bile leads to difficulty digesting fats properly causing weight loss (1); lack of balance
of proteins and other chemicals affects brain and behaviour (1); become jaundiced (yellow) as
breakdown products of red blood cells build up (1); inability to break down toxins properly so
symptoms of urea poisoning (1); any sensible or researched points.
4 Look for balanced arguments – if they haven’t stopped drinking will damage new liver, without
transplant will die, if still drinking heavily may forget to take immunosuppressant tablets, transplant
can offer a new start and a new chance of life, other sensible suggestions (6 max).
Summary questions
1 Excretion is the removal of the waste products of metabolism from the body (1); including carbon
dioxide, urea, and bile (2); defecation is largely the removal of undigested food, dead cells, and
bacteria from the body (1); but some excretion takes place as bile produced from the breakdown of
haemoglobin from old red blood cells (1); is removed from the body in the faeces (1).
2 a Type of cell that makes up most of the liver (1); and carries out most of the homeostatic functions
in the liver (1).
b Large nuclei – indicates active DNA transcription (1); Lots of Golgi apparatus indicates high level of
protein packaging and modification (1); likely to make many proteins that are modified for different
uses (1); many mitochondria hepatocytes very active (1); high production of ATP for cell reactions (1)
(max 3 one from each feature).
3 a Clear scientific drawing showing liver, correct labelling should include key features such as
Kupffer cells and hepatocytes.
b Any 6 from: Blood from hepatic portal vein brings products of digestion and cell metabolism to the
liver (1); blood from hepatic artery brings oxygenated blood to liver (1); blood from the HPV and HA
combine in the sinusoids, providing raw materials and oxygen for hepatocytes (1); hepatocytes line
sinusoids and absorb digested food, toxins, and oxygen from blood, also break down toxins, convert
glucose to glycogen, deaminate amino acids (1); Kupffer cells act as macrophages and engulf and
digest foreign cells and debris (1); bile secreted into canaliculi as haemoglobin broken down in
hepatocytes (1); hepatic vein removes deoxygenated blood carrying products of detoxification e.g.,
urea away from the liver (1).
4 a Ethanol from alcoholic drinks is absorbed and concentrated in the liver (1); to be detoxified (1);
hepatocytes detoxify ethanol using the enzyme alcohol dehydrogenase (1); forming ethanal (1); that is
further modified to form ethanoate (1). So an excess of toxic ethanol would affect the cells of the liver
first (1).
4 b Ethanoate produced as ethanol is detoxified in the liver (1); can be fed into the pathway
synthesising fatty acids (1); and so lipids (1); if there is a lot of alcohol a lot of fat is likely to build up in
the hepatocytes as a result of the detoxification process (1).
15.5
Investigating the kidneys
1 Regions of the kidney much less clear on the dissection than the diagram (1); not all areas can be
seen on a dissection (1); dissection shows what it really looks like, whereas diagram makes it much
easier to understand the areas involved in the way the kidney works (1). Any other sensible points.
2 Second point is subjective – but look for sensible and logical justification to award marks. For
example, Figure 7 shows the Bowman’s capsules and coiled blood vessels also shows how closely
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packed the tubules are around the capsules. Figure 8 shows the arrangement of cells and the circular
cross section of the tubules very clearly – allows you to see how you might get countercurrent
multipliers etc.
How long is your loop of Henle?
1 Expect bar or pie charts but any effective way will do (2).
2 Expect longer loops as camel makes more concentrated urine compared to plasma (1); to do this a
steep diffusion gradient is needed across the medulla (1); and this is built up along the loop of Henle,
so long loop needed to get the diffusion gradient to move water out along the length of the collecting
duct as the urine passes through the medulla (1).
3 Porpoise in sea water has effectively a relatively dry environment – plenty of water but also has salt
to deal with and excrete through the kidney (1); therefore would expect it to have similar length loops
of Henle and similar reabsorption of water possible from the collecting ducts across the loop of Henle
(1).
4 Movement of sodium and chloride ions out of the loop needed to produce steep gradients in the
medulla (1); depends on active transport out of the ascending limb (1); many mitochondria produce
the ATP needed (1); many cristae mean large surface area inside mitochondria for enzymes of
respiration so particularly effective at producing ATP (1). This is used in active transport to move
mineral ions out of the ascending limb to create a very steep concentration gradient to move water out
of the collecting duct by diffusion (1).
Summary questions
1 Layer of fat (1); to protect against mechanical damage (1); good blood supply (1); to ensure
diffusion gradients maintained (1); different sized blood vessels going into and out of the glomerulus
(1); to give high blood pressure for ultrafiltration (1).
2 a 180 l (1)
b 1–2 litres of urine produced in 24 hours (1); mean urine production in 24 hours is around 1.5 litres
(1); therefore percentage filtrate lost is 1.5/180 × 100 (1) = 0.83% (1).
3 a The forcing of liquid out from the blood capillaries (1); into the Bowman’s capsule (1); due to the
high blood pressure in the glomerulus (1); resulting from the fact that the arteriole feeding into the
knot of capillaries is wider than the arteriole leaving the glomerulus (1).
b Water, urea, glucose, amino acids, mineral ions (2) – the same as the blood without the red blood
cells and the large plasma proteins (2).
c Podocytes provide an extra filter (1); preventing any large proteins forced out of capillaries from
getting into filtrate in tubule (1); if the podocytes are damaged by infection, the structure will break
down (1); and proteins can get through into the kidney tubules (1); they are not reabsorbed in the
tubules (1) so will be lost in the urine (1).
4 a Mean volume of urine produced per day = 1.5 l (1); median amount of urea in urine over 24 hours
is (0.43 + 0.72)/2 = 0.575 moles (1); so mean concentration is 0.575/1.51 = 0.38 moles / l (1).
b There is around 1000 times more urea in the urine than in the blood (1).
c (i) 1 mark for each suitable example: Amount of protein eaten, amount of exercise done, blood
pressure levels, activity of the kidneys. Any other sensible suggestion. 3 max.
(ii) Because urea is constantly removed from the blood by the kidneys (1).
5 a Active transport of substances back into capillaries (1); water follows by osmosis (1); through
permeable walls (1); up to 80% of the filtrate reabsorbed (1).
b Water leaves by osmosis into concentrated tissue of medulla (1); sodium and potassium ions move
into the tubule by diffusion down concentration gradients (1); from the tissue fluid of the medulla (1);
as concentration of tissue increases through medulla, diffusion gradients maintained all the way along
the descending limb (1); so contains very concentrated solution at hairpin (1).
c Chloride ions removed from filtrate into tissue fluid of medulla by active transport (1); sodium ions
follow down electrochemical gradient (1); but walls of ascending limb are impermeable to water (1); so
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water cannot follow by osmosis (1); results in very high concentration of mineral ions in the tissue fluid
of the medulla (1).
d Permeability of the walls of the distal tubule varies with the levels of ADH (1); sodium ions may be
actively pumped out (1); with chloride ions following down an electrochemical gradient (1); further
concentrating the medulla; water may leave by osmosis concentrating the urine (1); if the walls of the
tubule are permeable in response to ADH (1); other substances may be actively secreted into the
distal tubule concentrating the urine (1).
e Permeability of the collecting duct to water is also controlled by the level of ADH (1); if walls are
permeable water moves out by osmosis into the concentrated tissue fluid of the renal medulla (1);
urine becomes more concentrated (1); water can be removed by osmosis along the length of the
collecting duct (1); as the concentration of the tissue fluid of the medulla increases from cortex to
pyramids maintaining a concentration gradient (1); this produces urine very hypertonic to blood (1).
15.6
ADH, water balance, and blood pressure
Severe sickness and diarrhoea – big loss of fluid from the digestive system (1); little or no water
absorbed by the body from the gut (1); concentration of solutes increases – osmoreceptors in
hypothalamus cause release of ADH (1); which makes the distal convoluted tubule and collecting duct
very permeable to water (1); water is reabsorbed into the blood and so little or no urine is produced
(1); 20% blood loss – large decrease in blood pressure detected by baroreceptors causes an increase
in release of ADH from pituitary (1); kidneys respond to reduce water loss from the body by makings
distal convoluted tubule and collecting duct very permeable to water. Water is reabsorbed into the
blood and so little or no urine is produced (6 max).
Summary questions
1 In cool weather you sweat less (1); so less water lost to keep cool (1).
2 Variety of ways students might do this but answers must include: stimulus of increased ADH from
pituitary (1);  binds to receptors on outside of tubule cell membranes (1);  triggers second
messenger (cAMP) in cell (1);  triggers movement of vesicles to membrane (1);  vesicles fuse
with membrane and insert protein based water channels (1); water moves out of tubules back into
blood by osmosis (1).
3 a Production of large quantities of very dilute urine (1); dehydration of body tissues (1); osmotic
effects on cells (1); constant thirst (1). Any other sensible suggestion
b Kidney tubules impermeable to water (1); protein-based water channels remain in vesicles (1); no
ADH means no second messengers (1); so no cascade to place water channels in membrane (1);
therefore no water moves out of distal convoluted tubule or collecting duct by osmosis (1); all fine
control over osmoregulation lost (1).
c Suggest managing water intake to conditions and activities (1); so don’t need any fine control of
water in body (1); severe – artificial ADH to replace the natural hormone (1).
15.7
Summary questions
1 Urine samples easy to get (1); non-invasive (1); urine contains drugs or metabolites excreted from
the kidney and stored in bladder so concentrated and can show up some time after drug use (they are
removed relatively rapidly from the blood) (1).
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2 a Allows confirmation tests to be done (1); if immunoassay indicates an illegal substance, a second
urine sample taken at exactly the same time is available (1); allows further testing using a different
technique – usually gas chromatography or mass spectrometry (1); to confirm the initial findings (1).
b Drugs can be used to improve performance/build muscles/increase RBC count etc during training
and then stopped long enough before a major competition to disappear from the system (1); while the
anatomical and physiological benefits remain (1). Random testing reduces likelihood that athlete will
use illegal substances as they don’t know when they will be tested, also more likely to be caught if
they are cheating (1).
3 a Similar – based on monoclonal antibodies and an immunoassay (1); different – in testing for illegal
drugs urine is divided into two samples (1); if immunoassay is positive gas chromatography/mass
spectrometry used to confirm, not done in a pregnancy test (1). b If test is done on day period is
missed woman might be pregnant but not have produced enough hCG to produce a strong enough
colour response to show positive (1); a few days later the hormone levels will have increased and the
second test will show positive if the woman is pregnant (2).
15.8
Summary questions
1 a Kidneys remove toxins and maintain water and electrolyte balance (1); if they fail, toxins are not
removed and may cause damage (1); if water or electrolyte balance is too far away from normal cells
may suffer osmotic damage (1).
b Diffusion (1)
2 Any 6 from: Blood leaves from an artery (1);  anticlotting chemicals added (1);  flows into
machine (1);  passes between semipermeable membrane surrounded by dialysis (1); fluid flowing in
opposite direction to blood (1);  excess mineral ions and urea diffuse out of blood down
concentration gradients into dialysis fluid (1) blood returned to patient through vein (1).
3 a Live organ taken straight from donor to recipient (1); alive all the time so no risk of tissue damage
(1); tissue match likely to be close (1) (max 2)
b Family members not necessarily close tissue match (1); living donors donating to strangers rare (1);
involves risk of surgery for donor (1); and risk if they ever have kidney problems as they are left with
only one functioning kidney (1). Any other sensible points. (max 2)
4 a Important to set up diffusion gradients between blood and dialysis fluid so excess salt, minerals,
urea, etc move out of the blood into the fluid across the dialysis membrane (1); glucose, salt, and
minerals present in plasma so want levels to equilibrate with dialysis fluid (1); hence normal plasma
levels in dialysis fluid (1); as much urea as possible needed to be removed so the steepest possible
diffusion gradient is required to ensure urea moves out of blood (1).
b Maximises exchange by maintaining the best possible diffusion gradients (1); flowing in opposite
directions maintains a countercurrent exchange, where the exchange continues all the way along the
membrane because there is always a gradient between the plasma and the dialysis fluid (1); (diagram
may be used to help explanation). Constant circulation of dialysis fluid also helps maintain maximum
diffusion gradients (1).
c Any urea formed, salt taken in etc, will pass into blood and be removed during dialysis session (1);
however as dialysis session progresses, excess protein will not be processed in body in time to be
converted into urea and removed before dialysis session ends (2).
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15 Homeostasis
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
1 (a) (i)
A Anaerobic respiration;
C Glycogenesis;
Glycogen;
2
1 (a) (ii)
Guidance
1
1 (a) (iii)
Liver;
1
1 (b)
Homeostasis is the maintenance of constant
internal environment; lactate lowers pH;
removal of lactate maintains (normal) pH;
glucose production maintains blood glucose
concentration;
High concentrations of lactic acid are toxic; so
must not accumulate; lactic acid is energy
resource; recycled; produced when energy
demand is high;
A hepatocyte; B hepatic artery / bile duct;
4
C deamination; D carbon dioxide; E urea; F
water;
Region P – cortex;
Structure Q – ureter;
Structure R – pelvis;
Structure S – pyramid;
Structure T – renal vein/renal artery;
4
1 (c)
2 (a)
2 (b)
3 (a)
3 b (i)
3 b (ii)
c
d (i)
d (ii)
Afferent arteriole; efferent arteriole; glomerulus;
Bowman’s capsule;
Afferent arteriole has larger diameter than
efferent arteriole; high(er) hydrostatic pressure
in capillary network (glomerulus); basement
membrane between, endothelium/capillary wall,
and epithelium (wall of Bowman’s capsule);
filtration of blood to form filtrate;
1 (large) protein / amino acids, present;
2 blood (cells) present;
3 glucose present;
4 more water present / more dilute;
5 more, ions / salts / electrolytes , present;
6 (more) vitamins present;
7 cells/large proteins, unable to pass through
basement membrane;
8 filtrate formed from water and solutes;
protein / polypeptide;
1 the ions (in solution) are too large to pass
through the channel
© Oxford University Press 2015
5
2
5
4
4
4
Mark as prose - award marks
wherever they occur
1 ACCEPT more, protein / amino
acids, ACCEPT appropriately
named protein e.g. albumin /
antibodies / immunoglobulins
3 DO NOT CREDIT more
glucose
1
Mark the first answer. If the
answer is correct and
an additional answer is given that
is incorrect or
contradicts the correct answer
then = 0 marks
IGNORE alpha helix / intrinsic /
transmembrane
DO NOT CREDIT glycoprotein
Mark the first two suggestions
1 ACCEPT gap / hole for channel
2
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15 Homeostasis
Answers to practice questions
OCR Biology A
OR the channel is too narrow for the
ions (in solution) to pass through;
2 shapes not compatible;
3 idea that positive charge (in the channel)
repels the (positively charged) ions;
4 (a)
4 (b)
4 (c)
4 (d)
4 (e) (i)
4 (e) (ii)
4 (f)
3 DO NOT CREDIT repels and/or
attracts
(initially) metabolic rate decrease as the
ambient temperature increases; (after plateau/at
higher ambient temperature) metabolic rate
increases as ambient temperature increases;
rate of increase is slower than rate of decrease;
hypothalamus;
3
increased rate of metabolism leads to increased
rate of respiration; idea that respiration is not
100% efficient; energy is released as heat;
respiration depends on enzymes; activity of
enzymes if increased with increasing
temperature; increased activity of enzymes
increases rate of respiration;
brown adipose cell more mitochondria; larger
nucleus; small droplets of fat; ORA
mitochondria are site of link reaction/Krebs
cycle/oxidative phosphorylation; more energy
released from respiratory substrates; small
droplets of fat have greater SA/vol ratio
increasing rate of breakdown;
energy released due to electron transfer; is not
used to synthesise ATP;
3
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3
3
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16.1
Investigating the effect of hormones on plant growth
1 Advantage: You can identify the effect of different concentrations of hormone on plant growth,
helping you build a model of the concentrations involved in the plant (1). Any other sensible
suggestion. Disadvantage: You might use concentrations that are nothing like those involved in the
plant cells – the concentrations effective in the cells might be so low that it is difficult to work at those
levels experimentally. Any other sensible suggestions.
2 Significant increase means the increase in fresh mass was statistically significant – in other words
the increase in fresh mass observed is very unlikely to have arisen by pure chance (1); to calculate
whether a difference is significant you need to apply the most appropriate statistical test. With the
information given in this case it is most likely that the Student’s t-test will be the test to use, to
compare means (1); applying the formula for the test gives a test statistic result, this is then compared
with the critical value (1); with the relevant confidence level (95% / p = 0.05) and the correct degrees
of freedom given the details of the experiment (1); if the test statistic is equal to or greater than the
critical value, the difference is reported as being significant (1).
Summary questions
1 Plants are multicellular and often large so need coordination (1); plants don’t appear to have
nervous systems so no electrical coordination system (1); chemicals can be carried in plant transport
systems and move from cell to cell to coordinate responses (1).
2 a Any three hormones and functions from Table 1 – must have correct hormone and function for the
mark (3).
b Analogous to animal hormones (1); involved in coordination and control of the plant (1); made in
one place (1); and carried through the transport system to another region where they have an effect
(1) (max 3)
3 a There are a variety of examples students could choose. One example is given here to show the
level of detail required for the marks: Auxin produced in tip of growing shoot stimulates growth in
some regions of the plant and inhibits growth in others – the apical shoot grows and lateral shoots are
inhibited (1); if tip of leading apical shoot is removed, growth in that shoot slows as stimulation of
auxin removed (1); lateral shoots grow faster as auxin inhibition removed (1); replace auxin artificially
on leading shoot (1); and apical shoot stimulated and grows fast again (1); while lateral shoots
inhibited again and growth slows (1).
b One chemical can exert control in different ways in different parts of the plant, allowing for complex
coordination without production of huge numbers of different chemicals (2).
4 Current model: Auxin molecules bind to specific receptor sites in the plant cell membrane, activating
process which pumps hydrogen ions into cell wall spaces, lowering pH to about 5, the optimum pH for
the enzymes that break down bonds between cellulose microfibrils, so they slide past each other
easily and the walls remains very flexible and plastic allowing cells to stretch and grow (2); graph
shows decrease in pH of cell wall from almost 6 to below 5 after the application of auxin (1); this is
followed by increased rate of shoot elongation from 1–2 microns per minute (1); to approximately 6
microns per minute (1); appears to confirm both change in pH and resulting increase in stretching of
cell wall and growth (1).
16.2
Summary questions
1 They are rooted to the ground so cannot move their bodies (1); therefore very important that they
are sensitive so they can grow in the right direction and make the best of the circumstances where
they have germinated (1).
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OCR A Biology
2 The amount of photosynthesis that can take place decreases as day length is reduced and
temperatures fall (1); so the amount of glucose produced by photosynthesis falls (1); the amount of
glucose needed for respiration to maintain leaves through the winter (1); and produce chemicals to
prevent freezing damage increases (1); it becomes more efficient to lose the leaves and become
dormant until the days lengthen and temperatures increase again (1).
3 Falling light levels  decreased concentration of auxin leaves produce hormone ethene 
ethene initiates gene switching in abscission zone at base of leaf stalk  gene switching causes
production of new enzymes  new enzymes digest and weaken cell walls in outer layer of abscission
zone (separation layer)  vascular bundles sealed off, fatty material deposited in cells on stem side
of separation layer  layer forms protective scar when lead falls preventing pathogen entry  cells in
separation zone respond to hormonal cues by retaining water and swelling putting more strain on
outer layer  further abiotic factors finish process  strain is too much and leaf separates from plant
leaving neat waterproof scar (max 6).
4 a Chemicals such as abscisic acid (1); trigger gene switching (1); so plants make chemicals such as
sugars or proteins (1); which lower the freezing point of the cytoplasm (1); or protect the cells against
damage by ice crystals if they do freeze (1).
b Water in intercellular spaces freezes and energy released raises temperature of cells (2); solute
concentration in cytoplasm/vacuoles maintains lower freezing points (1).
c Protection mechanisms include leaf loss, production of chemicals to act as antifreeze or ice
protection depend on gene switching, and production of new compounds in response to triggers of
day length and temperature which take place over several weeks (2); sudden early frost – cold
protection mechanisms not in place so cells damaged and plant may be killed (1); late frost – reverse
gene switching has occurred as plants move into summer mode (1); so levels of protection have fallen
(1); and can’t respond to freezing temperatures so cells destroyed (1).
5 Water stress induced experimentally by withholding water from plants (1); ABA content of leaves
increases shortly after water stress induced (1); stomatal resistance increases (stomata close) as
ABA levels increase (2); as soil rehydrates, ABA levels decrease rapidly in response (1); as ABA
levels drop, stomatal resistance decreases as stomata reopen (2).
16.3
Mimosa pudica – nerves and muscles in plants
Initial folding fast to avoid herbivory – if it was slow it wouldn’t frighten an animal away or make an
insect fall off the leaf to have a selection value must be fast (1). In survival terms, it is worth
expending ATP to save leaf (1). Recovery: if it recovers too fast herbivores may return and eat leaf
again, no need for rapid recovery so can rely on concentration gradients, etc. to restore original levels
and expend as little ATP as possible (1). Any other sensible points.
Summary questions
1 Herbivory is the process by which herbivores eat plants (1).
2 Any two can be chosen 1 mark for correct chemical, 1 mark for defensive role in plant, 1 mark for
human uses. For example tannins (1); bitter taste puts animals off eating leaves OR toxic to insects
(1); flavour tea and red wine (1). Alkaloids (1); affect metabolism, often poison animals OR prevent
germination in plants OR stop roots of neighbouring plants spreading OR taste bitter (1); insecticide
terpenoids; toxic to insects OR repel insects; used as insect repellent (1).
3 a A pheromone is a chemical made by an organism which affects the social behaviour of other
members of the same species (1); plants are not mobile and do not behave socially so pheromones
are not entirely appropriate (1).
b They must travel through the air inside or outside of the plant to carry their message (1); so they
must be in the form of a gas or vapour (1).
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OCR A Biology
c Any suitable examples chosen by student – the discussion about pheromones must include
recognition that pheromones act between members of the same species to affect social behaviour (3
marks max for each example). For example, pheromone produced by maple tree when attacked by
insects that is absorbed by leaves on other branches and nearby trees (1); that make protective
chemicals such as callose (1); pheromone because acts between members of the same species to
affect their behaviour (1); OR not a pheromone as does not affect behaviour, only biochemistry. Apple
tree attacked by spider mites produce chemicals that attract predatory mites (1); that come and
destroy the spider mites attacking the tree (1); NOT a pheromone as affects the social behaviour of a
different species (the predatory mites) not the same species (apple trees).
16.4
Practical investigations into phototropisms
Any sensible points raised by students, for example, immediate pictorial record (1); able to make a
series of photographs to record changes (1); time-lapse photography to show movement (1); have
camera always same distance away and so make accurate measurements of growth movements (1);
easy to save record, combine images and graphs. Disadvantages: Not everyone has access to smart
phones/tablets, in experimental setting potential to get things wet/drop them etc (1); have to
remember to always position in exactly same place at same angle or can introduce anomalies/false
evidence (1); can lose data/hard delete by accident (1); if lose or break phone or tablet all data lost (1)
any other sensible points (max 6).
Practical investigations into geotropisms
Plants respond to unilateral light by growing towards it (1); in dark or all round light they do not grow
towards light specifically so can assume any responses will be due to gravity (1). Any other sensible
suggestion.
Summary questions
1 a Plant growth responses to a directional stimulus (1)
b Phototropisms: Plant responses to unilateral light (1); Geotropisms: Plant responses to gravity (1).
c Shoots positively phototropic and negatively geotropic (1); roots negatively phototropic and
positively geotropic (1).
2 Variety of ways students might do this but answers must include: Unilateral light causes lateral
movement of auxin (1); across from the illuminated side to the shaded side (1);  low levels of auxins
reach elongating cells on lit side and higher than usual levels of auxin reach elongating cells on shady
side of shoot (1);  Reduced elongation and growth on lit side of shoot, extra elongation and growth
on shady side of shoot (1);  shoot bends as it grows to face the light as a result of asymmetric
growth (1);  auxin distribution becomes even again so shoot grows straight towards light (1).
3 Water soluble substances will pass through gelatin as it is made up of water (1); butter is a fat so
water soluble substances will not pass through it, but fat soluble substances will (1); gelatine allows
auxin to pass through it and so the response to unilateral light is maintained (1); butter prevents auxin
moving through it so the response to unilateral light is lost (1); indicating that auxin is water soluble
and not fat soluble (1).
4 Look for students commenting on some or all of the following (max 6): the importance of space flight
in the investigation of geotropism should be mentioned for full marks (1); the role of the root cap (1);
gravity perceiving cells (1); amyloplasts and sedimentation under gravity (1); role of calcium ions (1);
work in space to remove gravity (1); possibility of magnetic gradients (1); role of auxin (1).
16.5
Summary questions
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OCR A Biology
1 Ethene involved in natural fruit ripening (1); used to ripen fruit such as bananas, mangos, tomatoes
etc. at desired time (1); cytokinins and gibberellins can be used to delay fruit ripening (2).
2 Ripe fruit is easily damaged in transport (1); once fruit is ripe it has a finite life before it goes off (1);
if fruit is transported unripe and hard (1); much less likely to be damaged and doesn’t start to go off
(1); controlled ripening when needed gives uniform product (1); and minimises waste (1).
3 Should include: Auxins – rooting powder, selective weedkillers, development of seedless fruit, leaf
fall, (1.5 marks for at least 2 suggestions). Ethene – controlled ripening, fruit dropping, leaf fall (1.5
marks for at least 2 suggestions). Gibberellins: delays fruit ripening, increases fruit size, speeds up
the brewing process (1.5 marks for at least two suggestions). Cytokinins: prevent ageing of ripened
fruit and lettuces etc., and in micropropagation to control tissue development (1.5 marks for at least
two suggestions) (max 6).
–1
4 Ethene levels rise at rate of approximately 5 µl / kg h from day 2 to day 4 (1); and then remain at a
–1
steady level (1); carbon dioxide levels increase at a rate of approximately 15 ml / kg h from day 2 to
day 4 (1); as ethene levels rise (1); as the ethene levels plateau, the carbon dioxide levels decrease
–1
steadily (1) at a rate of around 5ml / kg h for 5 days(1).
b Initial rise in ethene levels triggers start of ripening process (1); many different reactions take place
to bring about fruit ripening (1); requires a lot of energy so respiration rates increase (1); and so
carbon dioxide levels increase dramatically reflecting increased cellular respiration (1); raised ethene
levels maintain ripening process (1); but reactions gradually slow down as aspects of ripening are
completed so demands on respiration are less so amount of carbon dioxide produced is reduced (1).
Any other sensible scenario can be given some credit. Link between metabolic increase in response
to ethene, reflected by increased levels of carbon dioxide must be included.
c Ripening is the result of many enzyme-controlled reactions(1); at lower temperatures reactions
occur more slowly (1); so ripening occurs more slowly even with ethene present (1).
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16 Plant responses
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
Guidance
1 (a)
1 to cope with changing conditions / AW;
2 avoid abiotic stress;
3 to maximise photosynthesis
or
to obtain more, light / water / minerals; ora
4 avoid, herbivory / grazing;
5 to ensure, germination in suitable conditions / pollination /
seed set / seed dispersal;
2 max
1 (b) (i)
1 in water / in
A / with no abscisic acid,
germination increases as conc. GA increases;
2 when abscisic acid present / in
B, no germination;
–3
3 maximum germination 90% with 5 mol dm GA,
in water / without abscisic acid;
4 2 comparative figures (x and y refs. plus units);
5 GA concentration increases,
logarithmically / by a factor of 10, on x axis;
6 10 times more GA gives,
3 (conc 0.05 to 0.5) / 0.5 (conc 0.5 to 5),
times more germination;
4 max
1 Looking for a
general statement
DO NOT CREDIT
“adapt to change”
3 CREDIT named
elements / ions
IGNORE nutrients
4 methods of
preventing grazing
could include
producing more
toxins / more spines /
encouraging stinging
ants
IGNORE predation
5 DO NOT CREDIT
‘maximise
reproduction’
without further
qualification
2 DO NOT CREDIT
‘inhibits germination’
(as this is a
conclusion not a
description)
3 ACCEPT 91% (±
2%) for 90%
4 EITHER compare
A and B at the same
GA conc
OR two points on
same line
with units for both
GA
A
B
conc (%)
(%)
(mol
3
dm )
0
10 ± 0
2
0.05 22 ± 0
2
0.5
66 ± 0
2
5
91 ± 0
2
1 (b) (ii)
gibberellin hydrolysis of starch; glucose used for respiration;
ATP/energy, required for germination; abscisic acid inhibits
cell growth so germination;
1 so temperature doesn’t affect results /
so only desired variable(s) changed /
to show just the effect of plant hormones;
2 since temperature affects enzyme activity;
4
1 (b) (iii)
© Oxford University Press 2015
2 max
1 ACCEPT fair test
IGNORE to control
temperature /
temperature is a
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16 Plant responses
Answers to practice questions
OCR Biology A
3 suitable / optimum, temperature for (lettuce) germination;
1 (b) (iv)
1 volumes of liquid(s);
2 ABA concentration;
3 oxygen availability;
4 age of seeds;
5 previous storage of seeds / viability idea;
6 genotype / variety, of seeds;
7 size / type of, petri dish / filter paper;
8 length of time experiment left for (before recording
results);
9 space between seeds;
10 AVP;
3 max
1 (c)
1 seedless, fruits / grapes;
2 weed killers;
3 rooting powder / to grow cuttings / used in tissue culture;
4 control fruit ripening;
5 controls fruit drop;
6 restrict hedge growth;
7 preserve, cut flowers / green vegetables;
8 specific example of improved fruit quality;
9 producing malt / in brewing;
10 AVP;
11 AVP;
2 max
© Oxford University Press 2015
limiting factor /
temperature is a
controlled variable
2 CREDIT “optimum
temperature for
enzyme activity” or
“this is the
temperature when
enzymes work best”
3 ACCEPT ‘these’
seeds
Mark the FIRST
suggestion on each
numbered line
DO NOT CREDIT
conc, GA / giberrellin
(as this is the
independent variable)
IGNORE number of
seeds (as given in the
question)
1 DO NOT CREDIT
amounts / levels
CREDIT volume of,
water / GA / ABA
3 IGNORE carbon
dioxide
6 CREDIT “from
same batch of seeds”
or
“seeds from same
plant”
10 e.g.
light qualified
(duration / intensity /
wavelength)
use of distilled water
all lids, off / on
Mark the FIRST TWO
suggestions
IGNORE the names
of plant growth
regulators
4 could be used to
speed up or slow
down
8 e.g.
longer stalks on
grapes
longer apples
10 & 11 e.g.
promoting sexual
maturity in conifers
promoting latex flow
in rubber plants
promoting sexual
maturity in female
cucumber plants
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16 Plant responses
Answers to practice questions
OCR Biology A
2 (a)
auxin / IAA;
(positive) phototropism;
plants / shoots, bend towards light;
etiolation / plants grow taller;
climbing plants climb, up / over, other plants;
(positive) thigmotropism / sense of touch;
grow roots towards, water / minerals;
allelopathy / description;
2 (b)
Auxins, apical dominance / tropisms / example described;
gibberellins, growth / germination / flowering; ethylene,
ripening / aging / abscission; abscisic acid, dormancy;
Not all plant hormones are responsible for plant growth;
idea that hormones are chemical messengers; (role in) cell
communication; reference to target cells; wide range of
effects; long-term effects;
3 (a)
3 (b)
3 (c)
4 (a) (i)
4 (a) (ii)
4 (b)
4 (c)
4 max
longer nodes in sugar
cane
restricting growth in,
chrysanthemums /
other e.g.
IGNORE other named
hormones
IGNORE apical
dominance
DO NOT ACCEPT
phototrophic /
thigmotrophic (but
penalise once)
IGNORE move, grow
IGNORE nutrients
4
5
4
Both plant hormones; auxins apical dominance / growth of
lateral buds suppressed; delays abscission; promotes cell
elongation; cytokinins fruit growth; promotes abscission;
promotes cell division;
Control plants leaf, number and surface area, relatively
constant; stressed plants leaf, number and surface area,
decreased with time; figures quote;
Growth / development, reduced; lack of water caused
stomatal closure; reduced carbon dioxide uptake; reduced
photosynthesis;
Reduced auxin concentration due to lower light intensity (in
the autumn); stimulates the production of ethene; ethene
switches on genes; in cells of abscission zone; enzymes
produced; (enzymes) digest cell walls (in cells of abscission
zone);
Both result in loss of leaves; due to action of hormones;
different hormones; abscisic acid in autumn leaves; ethene
in water stressed leaves;
© Oxford University Press 2015
6
3
3
5
5
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OCR A Biology
17.1
Energy transfer
1
Bond
Number
C=O
O—H
C—C
C—H
C—O
O=O
12
12
5
7
7
6
Bond energy
(kJ / mol)
803
464
347
414
358
495
Total (kJ / mol)
9636
5568
1735
2898
2506
2970
2 Total energy required to break the bonds in six carbon dioxide molecules = 9636 kJ / mol (1); total
energy required to break the bonds in six water molecules = 5568 kJ / mol (1); Total energy used =
15 204 kJ / mol (1); total energy released when all the bonds in one glucose molecule are formed =
9459 kJ / mol (1); total energy released when the bonds in six oxygen molecules are formed = 2970
kJ / mol (1); total energy released = 12 429 kJ / mol; energy required to make up difference = 2775
kJ / mol (1). 3 2775 kJ / mol
4 1163/2775 × 100 = 42%
Summary questions
1 Energy cannot be created (or destroyed) energy is transferred (1); ATP is produced (1).
2 ATP is not very stable (1); easy to remove phosphate group (1); organic molecules are (more)
stable (1); idea that organic molecules are more energy dense (1).
3 In photosynthesis light energy is converted into chemical energy (1); inorganic molecules are
converted into organic molecules (1); water and carbon dioxide are converted to glucose (and
oxygen) (1); respiration uses oxygen produced in photosynthesis (1); organic molecules are broken
down into inorganic molecules (1); energy released is used to synthesise ATP (1); carbon dioxide
produced is used in respiration (1) (max 5).
17.2
Summary questions
1 Universal energy currency (1); energy transfer is, quick/immediate (1); energy is in, small/usable,
quantities (1); (energy transfer) is quick, (energy transfer) in quantities that can be used; ATP can be
resynthesised (1) (max 3).
2 Impermeable to, ions/protons (1); idea that there can be different concentrations of protons on each
side of a membrane (1); contains, embedded / integral, proteins (1); e.g., ATP synthase (1); enzyme
responsible for synthesis of ATP (1).
3 Facilitated diffusion (1); ATP synthase provides hydrophilic channel for diffusion of protons (1);
catalyses the synthesis of ATP (1); lowers activation energy (1).
4 Reversible reaction (1); idea that the products formed from reaction in one direction are the
substrates of reaction in reverse direction (1); active site is complementary to both substrates and
products (1); so substrates and products can bind to same active site (1); direction of reaction
depends on relative concentrations (of substrates/products) on either side of reaction (1).
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OCR A Biology
17.3
Plants use sunscreen as well
1 a Anthocyanins are red pigment (1); produced when light intensity and sugar concentration is high
(1); apples have high sugar concentration; high intensity light affects one side more than the other.
b Produced by enzymes (1); enzyme activity is temperature dependent (1).
2 Enzymes involved in metabolic pathway (1); 3D shape of enzyme is different at different
temperatures (1); shape of active site changes (1); enzymes activated/inactivated.
Investigating photosynthetic pigments
1 Small strip of TLC plate (1); grind leaves with propanone/organic solvent (1); apply drop of extract to
TLC strip near to one end (1); repeat until concentrated spot produced (1); place TLC strip in test tube
(1); put solvent into test tube so level of solvent below spot (1); ensure strip not touching sides of tube
(1); leave solvent to run (up strip) (1); remove TLC strip before solvent reaches top of strip.
2 Carotenoids 0.1 (1); Chlorophyll b 0.38 (1); Chlorophyll a 0.54 (1); Pheophytin 0.64 (1); Carotene
0.96 (1)
Photorespiration
1 Less carbon dioxide fixed (1); less, organic molecules / named examples, synthesised (1); reduced
yield (1).
2 Photosynthesis produces oxygen (1); idea that little / no, oxygen present in atmosphere when plants
began to evolve (1).
Summary questions
1 Synthesis of ATP/addition of phosphate group to ADP (1); using energy from light (1)
2 Pigments absorb light (1); of specific wavelength/colour (1); absorbed light energy provides energy
for photosynthesis (1), more light absorbed leads to higher rate of photosynthesis (1).
3 Green light is reflected/not absorbed (1); no energy for light-dependent stage (1); no ATP and
reduced NADP for light-independent stage (1).
4 Inorganic carbon (1); added to organic molecule (1)
5 a Calvin cycle happens during the light as well (1)
b Calvin cycle requires ATP (1); and reduced NADP (1); supplied from light-dependent stage (1)
6 ATP produced and reduced NADP not produced (1); electrons not required from PSII/photolysis (1);
less ATP used in Calvin cycle (1); more ATP available for other metabolic processes (1).
7 10 molecules of TP every six turns of cycle (1); using ATP (1); 6 molecules of RuBP formed (1);
reference to (3/5) carbon shuffle (1); reference to individual steps/named (1); light (1); enzymes (1);
coenzymes/named (1); ATP (1); carbon dioxide does not combine directly with water (1).
17.4
Investigating the factors that affect the rate of photosynthesis
1 After point A change in dissolved oxygen % (y axis) divided by change in time min (x axis) (1);
based on the following values (sensible answer but others accepted) 70/80 = 0.88 % / min (1).
2 Rate of photosynthesis increased (1); higher than rate of respiration (1)
3 Sodium hydrogen carbonate solutions (1); different carbon dioxide concentrations (1); place flask in
water baths at different temperatures (1); place lamp at different distances to vary light intensity.
4 Number of bubbles easier to count (1); ORA volume / concentration, of oxygen (obtained using a
data logger) more precise than number of bubbles (1); (this) increases validity.
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OCR A Biology
Artificial photosynthesis, a win-win solution
Pigment/cell, to absorb light energy (1); catalyst to split water (1); enzyme to reduce carbon dioxide
(1); any other sensible suggestion
Different types of photosynthesis
1 Cactus swells (1); storing water in wet periods (1); leaves are spines (1); small surface area (1);
spines collect water (from atmosphere) (1); extensive and shallow root system (1); open stomata at
night (1); cactus / CAM plants, use PEP carboxylase to fix carbon dioxide (1); carbon dioxide stored
until daytime (1); limited quantity stored (1); limited photosynthesis so limited growth (1).
2 Carbon dioxide required for photosynthesis (1); carbon dioxide enters plant through stomata (1); no,
organic molecules / named examples, so no respiratory substrate (1); essential reactions / named
examples, cannot take place (1).
Summary questions
1 Factor which limits rate of a process (1); e.g., light (1) in photosynthesis (1).
2 Oxygen is used in respiration (1); so not all oxygen produced is released (1).
3 Limiting factors light, temperature, and carbon dioxide concentration (1); light is required for lightdependent stage (1); glasshouses are transparent / built from glass (1); carbon dioxide is required for
light-independent stage (1) ventilation of glasshouse so concentration of carbon dioxide does not fall
(1); temperature affects reactions in light-independent stage (and light-dependent stage) (1); use of
heater (1); fossil fuel burners also release carbon dioxide (1); idea that each factor is maintained at a
level that maximises the rate of photosynthesis but minimises waste (1).
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17 Energy for Biological processes
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
1 (a)
Limited space due to glasshouse size; idea of
temperature control, e.g. too hot in the summer;
ventilation due to plants being enclosed;
Rate of, photosynthesis and respiration,
increase as temperature increases; rate of
photosynthesis higher than rate of respiration at
lower temperatures; rate of photosynthesis
levels out at higher temperatures; due to,
limiting factor / named; figures quote; enzyme
controlled reactions in photosynthesis and
respiration; enzyme activity effected by
temperature;
Assimilates / named, required for, growth /
development; idea that rate of photosynthesis
must be higher than rate of respiration for net
production of assimilates; temperatures need to
be lower to increase saleability of tomatoes;
tomatoes would be, smaller / less sweet, if
grown at higher temperatures;
3
3
2 (a)
Maintenance of optimum, temperature / carbon
dioxide concentration; expensive to heat or cool
/ ventilate; idea that cost of maintaining
maximum rate of photosynthesis would reduce
profits;
A
2 (b) (i)
Different positions on plant / described;
2
2 b (ii)
A
1
2 b (iii)
Cells large number of chloroplasts; leaves large
surface area to absorb maximum light; large
number of stomata for maximum diffusion of
carbon dioxide; adaptations to high
temperatures; whole plants taller/broader;
124 (%) / 123.7 (%);;
5
Benefit
allows entry of more CO2;
Explanation
(CO2) for , light-independent reaction / Calvin
cycle
or
light-dependent reaction is taking place quickly /
reduced NADP building up / ATP building up
or
CO2 not as limiting (than when there are fewer
stomata)
or
idea that increases access to air spaces
for distribution of CO2;
OR
benefit
reduces transpiration;
2
1 (b) (i)
1 (b) (ii)
1 (b) (iii)
3 (a)
3 (b)
© Oxford University Press 2015
Guidance
5 max
4
1
2
Read through complete answer.
Award 2 marks if a benefit and
explanation are correctly linked.
If benefit and explanation are not
correctly linked:
Award Max 1 for either a benefit
or an explanation.
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17 Energy for Biological processes
Answers to practice questions
OCR Biology A
3 (c)
4 (a)
4 (b) (i)
explanation
idea of stomata sheltered from , air currents /
heat
(when on lower surface)
or
idea that diffusion shells maintained;
Equal sample size for sun and shade leaves /
increase sample size of shade leaves /
greater numbers of sun and shade leaves;
measure thickness of cuticle /
make cuticle observations quantitative;
record range / calculate SD / calculate SE /
(named) statistical analysis;
record data on leaf,
length / width / area / colour / chlorophyll
content;
record data on ,
size of stomata / stomatal count on upper
surface;
define what is a sun or shade leaf /
measure light levels to classify type of leaf;
repeat / replicate , the (whole) experiment /
using other plants of the same species;
Oxygen
1 oxygen only produced in one (named) stage
of
photosynthesis;
2 oxygen produced might be used for
respiration;
carbon dioxide
3 CO2 only used in one (named) stage of
photosynthesis;
4 CO2 produced during respiration might be
used for ,
photosynthesis /
light independent reaction /
Calvin cycle;
5 O2 / CO2 / both , could be an underestimate
or
represents net production (O2)
or
represents net use (CO2);
Light intensity;
2 max
DO NOT CREDIT refs to
controlling temperature
or light or wind or time
2 max
1 CREDIT for O2 ‘only measures
the rate of the
light dependent stage /
photolysis’
3 CREDIT for CO2 ‘only
measures the rate of the
Calvin cycle’
5 ACCEPT a description e.g.
‘measurement is
less than expected because
not all the oxygen produced
can be measured’
(but not if expressed in terms of
terms of experimental error –
e.g. dissolves in the water)
IGNORE refs to reliability /
accuracy
1
4 (b) (ii)
Carbon dioxide concentration / partial pressure
of CO2 /
temperature;
AVP;
1
4 (b) (iii)
(aerobic / anaerobic) respiration;
1
© Oxford University Press 2015
DO NOT CREDIT ‘high’ or ‘low’,
as these indicate
situations rather than factors
eg stomatal density
stomatal size
chlorophyll concentration
number of chloroplasts
enzyme turnover rate
IGNORE (temporary) changes in
stomatal ,
opening / closing
IGNORE ref to water availability
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17 Energy for Biological processes
Answers to practice questions
OCR Biology A
4 (b) (iv)
4 (c) (i)
4 (c) (ii)
1 at 0 , respiration only / no photosynthesis;
between 0 and X
2 idea that (rate of) respiration is greater than
(rate of) photosynthesis;
at X
3 idea that (rate of) respiration equals (rate of)
photosynthesis / at compensation point;
after X
4 idea that (rate of) photosynthesis is greater
than (rate of) respiration;
3 max
+
Reduced NADP / NADPH / NADPH2 / NADPH ;
ATP; oxygen;
1 prevents photophosphorylation;
2 cyclic and non-cyclic;
3 no / less , ATP / reduced NADP , for , lightindependent stage / Calvin cycle / GP to TP;
4 no (named) substrate made for respiration;
© Oxford University Press 2015
Assume that candidate is
answering in the same order as
the bullet points, unless otherwise
indicated. IGNORE
photorespiration throughout
CREDIT ‘Calvin cycle’ for
‘photosynthesis’ throughout For
mps 2, 3 & 4 must include clear
ref. to both respiration and
photosynthesis
2 DO NOT CREDIT no
photosynthesis
3
2 max
3 ‘no ATP for photosynthesis’ is
not quite enough DO NOT
CREDIT (oxidised) NADP
4 substrate eg glucose / starch /
carbohydrate / sucrose / sugars
IGNORE triose phosphate / food /
nutrients
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OCR A Biology
18.1
Summary questions
1 Dehydrogenation – removal of hydrogen from triose phosphate molecules to form pyruvate and
reduction of NAD / formation of reduced NAD (1); phosphorylation – addition of phosphate group to a
glucose molecule forming hexose bisphosphate (1); (both) catalysed by enzymes (1).
2 NAD accepts hydrogen (atom) and is reduced (1) during the formation of pyruvate (1); supplies
hydrogen to enzyme involved in later stage of respiration (1).
3 Addition of phosphate group (1); to ADP (1); or formation of ATP (using phosphate) from another
molecule (1).
4 Dehydrogenation – hydrogen removed in breakdown of glucose (1); hydrogen required at a later
stage (1). Phosphorylation – addition of phosphate groups destabilises (large) molecules/glucose (1);
leads to breakdown of glucose (1); and synthesis of ATP (1).
18.2
Summary questions
1 Hydrogen is also removed (1); removal of hydrogen oxidises pyruvate (1).
2 Acetyl group (1); carbon dioxide (1)
3 Pyruvate (1); acetyl CoA (1); reduced NAD (1);
4 Enzymes required are in cytoplasm/ORA (1); glucose molecule too large to move into
mitochondrion (1); no transport proteins for pyruvate (1); mitochondria not originally present in
(eukaryotic) cells (1).
18.3
Summary questions
1 ATP – three phosphate groups (1); one ribose (1); one nitrogenous base (1); NAD – two phosphate
groups (1); two riboses (1); two nitrogenous bases (1) (max 3 comparisons)
2 Idea that it is used to link reactions (1); idea that energy is released as a result of the activity of one
enzyme and used by another enzyme (1).
3 Students answers may vary but must include: glucose to triose phosphate (1); triose phosphate to
pyruvate (1); addition of two ATP (1); production of four ATP and two reduced NAD (1) (2 max).
4 One per turn (1) two in total (1).
5 Hydrogen needs to be removed for cycle to continue (1); hydrogen removed using NAD/FAD and
reduced (1), then NAD/FAD are oxidised at electron transport chain (1); oxygen required for electron
transport (1).
6 Enzymes are specific (1); active site complementary to substrate (1); different steps have different
substrates (1); different steps require different enzymes (1); different enzymes (may) require different
coenzymes (1); only one step in cycle has enzyme which requires FAD coenzyme (1).
18.4
Summary questions
1 Actively pumped to increase concentration gradient (1); energy required as moving from low to high
concentration (1); membrane impermeable to ions so ions diffuse down concentration gradient (1);
ATP synthase provides hydrophilic channel (1).
2 Reduced NAD releases electrons to carriers at the start of the ETC (1); reduced FAD releases
electrons to carriers after the start of the ETC (1); with FAD electrons transported a shorter distance
(1); so fewer protons are actively transported (1).
3 Stops flow of electrons (1); stops active transport of protons (1); proton gradient not formed (1);
(less) ATP synthesised; so less energy available for (vital) metabolic processes (1).
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OCR A Biology
4 ATP synthase is not actually part of the electron transport chain – agree (1); not an electron carrier
(1).
Oxygen is required for the transfer of electrons along the electron transport chain – agree (1); oxygen
is final electron acceptor, required for electron transport (1) Hydrogen ions return to the matrix by
facilitated diffusion – agree (1); diffuse through hydrophilic channels (of ATP synthase) (1).
18.5
Investigation into respiration rates in yeast
1 Vacuum flask to control the temperature (1); paraffin prevents oxygen entering the solution (1);
ensures respiration is anaerobic (1).
2 Tangent drawn a steepest part of curve (1); change in CO2 ppm (y axis) divided by change in time s
(x axis) (1); based on the following values (sensible answer but others accepted) 1500/250 = 6 ppm / s
(1).
3 Respiratory substrate used up (1); accumulation of metabolic waste/alcohol (1); toxic to yeast (1)
4 Layer of liquid paraffin removed (1); (then) investigation carried out as for anaerobic respiration (1)
Small-scale and large-scale adaptations to low oxygen environments
1 Flexible rib cage; idea that lungs can collapse under high pressure (1); air compressed (1);
maintaining concentration gradients (1); exhalation before inhalation (1); increase proportion of air
exchanged (1); larger lungs would increase buoyancy (1); more energy would be used during dives
(1).
2 Streamlined (1); heart rate slowed (1); reduced energy requirements (1); blowhole on top of the
head (1); large breath (when surface) (1); larger red blood cells (1); more haemoglobin (1); more
blood (1); faster oxygen transport (1); more myoglobin (1); increased oxygen storage (1).
Summary questions
1 Yeast cells normally respire aerobically (1); can respire anaerobically when required (1)
2 Electron transport chains present in some types of anaerobic respiration (1); aerobic respiration
always includes presence of electron transport chains (1); no electron transport chains present in
lactate fermentation (1).
3 Increase in lactic acid leads to decrease in pH (1), muscle contraction depends on protein (1) e.g.,
enzymes, contractile proteins (1); decreasing pH denatures protein (1); protein no longer functional
(1).
4 Red blood cells adapted to carry oxygen (1); lack of mitochondria means more space for
haemoglobin (1); increased oxygen transport (1); lactic acid not produced in cardiac muscle (1);
enzymes not denatured, no fatigue (1); blood flow to rest of organism not interrupted (1).
18.6
Calculating the respiratory quotient
1 At rest 10/10.5 = 0.95 (1); in flight 113.6/160 = 0.71 (1).
2 Respiratory substrate changes (1); (from) mainly carbohydrate to carbohydrate and lipid (1)
Low carbohydrate diets
1 Statement is accurate triglycerides are hydrolysed to fatty acids (and glycerol) (1); fatty acids,
broken down/oxidised, to acetyl groups / acetyl-coA (1); acetyl group transferred to, oxaloacetate /
Krebs cycle (1); carbohydrates / pyruvate, maintain oxaloacetate level / keep Krebs cycle going (1);
statement is inaccurate not all fats are triglycerides (1).
2 Benefits weight loss (1); (relatively) fast (1); diabetes risk reduced (1); drawbacks intake of, healthy
foods/fruit/vegetables, reduced (1); increased, gluconeogenesis / described (1); increased lipid
metabolism (1), risk of ketosis (1); increased protein metabolism (1); risk of, liver/kidney, damage (1);
muscle wastage/described (1).
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OCR A Biology
Practical investigations into the factors affecting rate of respiration using respirometers
1
Temperature Respirometer Reading at
Reading
Difference
Corrected
Rate of
3
3
°C
start (cm )
after 20
(cm )
difference
oxygen
3
minutes
(cm )
uptake
3
3
–1
(cm )
(cm min )
A
0.93
0.74
0.19
0.16
0.008
15
B
0.93
0.86
0.07
0.04
0.002
C
0.91
0.88
0.03
A
0.94
0.63
0.31
0.27
0.014
25
B
0.93
0.84
0.09
0.05
0.003
C
0.95
0.91
0.04
C rows are beads only, used to get the corrected differences.
2 To make the volume of, contents / peas, the same (1); volume of peas in A is greater than volume
of peas in B (1); peas in A have absorbed more water (1); without beads there would be more,
air/oxygen (1).
3 Find difference in volume between soaked peas and dry peas (1); difference represents the volume
of glass beads required (1); calculate volume of one bead to determine number of beads required (1).
4 Increased kinetic energy (1); of enzymes / named, involved (1).
5 Reactions, require aqueous medium, take place in water (1); so enzymes and substrates can collide
(1); soaked seeds need more energy (1); ORA for named process e.g., protein synthesis (1).
Summary questions
1 Triglyceride is broken down into fatty acids and glycerol (1); fatty acids undergo beta oxidation
forming acetyl groups (1); acetyl groups are taken into Krebs cycle by coenzyme A (1); glycerol is
converted to pyruvate, which undergoes oxidative phosphorylation (1).
2 Both measure oxygen uptake/carbon dioxide release (1); so rate of respiration (1); respirometer is
modified spirometer/(usually) used for smaller organisms (1).
3 A – carbohydrate/C6H12O6 = 12/24 = 50% (1)
B – amino acid/C2NO2H5 = 5/10 = 50% (1)
C – fatty acid/C18O2H36 = 36/56 = 64% (1)
Highest proportion of hydrogen in fatty acid because it has more C–H bonds (1); lipids have highest
energy value (1); equal proportion of hydrogen in carbohydrate and amino acid so carbohydrate and
protein energy values are (almost) the same (1).
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18 Respiration
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
1 (a) (i)
Addition of (inorganic) phosphate
1
1 (a) (ii)
3 max
2 (a) (i)
Similarities ATP produced; phosphate group
transferred; differences OP requires oxygen;
electron transport chain involved in OP; SLP
occurs during, glycolysis / Krebs cycle;
NADH 10/4;
FADH 6/4; = 1.5
NADH = 2.5 and FADH = 1.5;
12 NADH (glycolysis 2, link reaction 2, Krebs
cycle 8) + 2 FADH (Krebs cycle 2);
(12 × 10) + (2 × 6) = 132;
132/4 = 33;
glycolysis / glycolytic pathway;
2 (a) (ii)
cytoplasm;
1
2 (a) (iii)
ATP;
NAD;
pyruvate;
(pyruvate / F) converted to lactate;
F / pyruvate , accepts hydrogen (atoms);
hydrogen from , reduced NAD / reduced E;
(catalysed by) lactate dehydrogenase;
no, oxygen / O2 , to act as (final),
hydrogen / electron, acceptor;
(so) link reaction / Krebs cycle / ETC, cannot
take place;
NAD / E, regenerated / recycled / able to be reused;
allows glycolysis to continue / pyruvate
continues to be made;
limited / small amount of / some, ATP can be
produced;
physical (probably from diagrams)
1 large nostrils (open) to take in air;
2 (when submerged) nostrils close / nose closes
, to , keep air in / stop air from escaping;
3 lungs / airways , have high (vital) capacity;
links to respiration
4 idea that seal , has low(er) metabolic rate /
has low(er) respiratory rate / has low(er) energy
requirements / uses (relatively) little ATP;
5 able to respire anaerobically for a long time /
more glycolysis;
6 large supplies of NAD (to accept H);
7 (this) prevents , build-up of lactate / lowering
of pH;
8 idea that (seal) tolerates lactate / removes
lactate quickly;
9 idea that (seal) tolerates high CO2
concentration;
10 idea that (seal) tolerates low pH / has more
pH buffers; synoptic / inference
11 idea that blood diverted from certain regions
3
1 (b) (i)
1 (b) (ii)
2 (b)
2 (c)
© Oxford University Press 2015
Guidance
3
3
1
5
3 max
1 ACCEPT oxygen
2ACCEPT oxygen IGNORE ref to
keeping water out
3 ACCEPT deep / barrel / large ,
chest IGNORE big lungs CREDIT
large lung volume / takes in large
volume of oxygen / larger
numbers of alveoli / larger
(exchange) surface area /
increased number of capillaries
4 e.g. (streamlined, less
resistance so) uses less energy
(insulated so retain heat so) uses
less energy
(buoyant so) less energy required
(small flippers so less surface
area of extremity so loses less
heat so) uses less energy
5 ‘anaerobic’ needs time ref
7 ACCEPT lactic acid
8 ACCEPT lactic acid
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18 Respiration
Answers to practice questions
OCR Biology A
3 (b) (i)
/ certain regions have reduced metabolic
activity;
12 idea that has plenty of , haemoglobin / red
blood cells / myoglobin (as oxygen source);
13 idea that haemoglobin has a higher affinity
for oxygen / dissociates less readily
/dissociation curve shifted to left;
X adenine;
Y ribose;
Z (tri / 3) phosphate(s);
transfers energy / energy ‘currency’ / releases
energy /
universal energy molecule / energy
intermediate /
(immediate) source of energy;
phosphate(s) can be removed by hydrolysis;
–1
to , release / provide , 30 kJ (mol ) energy;
(energy released for) metabolism /
appropriate named reaction /
appropriate reaction described;
ADP can attach a phosphate (forming ATP)
during ,
respiration / photosynthesis;
energy released in ,
small ‘packets’ (to prevent cell damage) /
suitable quantity;
crista;
3 (b) (ii)
chemiosmosis / oxidative phosphorylation;
1
3 (c)
substrate respired changes over time;
initially respires (mostly) , glucose /
carbohydrate;
lower / decrease in / 0.75 , RQ indicates (more)
, fat / lipid , as substrate
or
as time goes by (more) lipid is respired; glucose
/ carbohydrate , used up / decreases (over
time);
protein not likely to be used as substrate /
protein only used as a last resort;
3 max
3 (a) (i)
3 (a) (ii)
© Oxford University Press 2015
11 DO NOT CREDIT zero
respiration rate
3
3
1
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Module 5 practice questions
Answers
OCR Biology A
Question
number
Answer
Marks
1 (a)
e.g. Protein synthesis; mitosis;
2
1 (b)
2
1 (c) (i)
ADP required to synthesis ATP; ADP not
supplied from cytoplasm;
Protein;
1 (c) (ii)
Complete set of DNA; of an organism;
2
1 (c) (iii)
Reduced energy demand; growth /
development, no longer occurring;
Electron transport, stopped / reduced; no /
fewer, proteins pumped; less ATP synthesised;
Protons are charged; hydrophobic core /
described; barrier to ions; channel protein
provides hydrophilic channel; carrier protein
moves protons across using energy;
Facilitated diffusion of protons; reduces proton
gradient; no / less, ATP synthesised; reference
to uncoupling;
Glycolysis; detail; Link reaction; detail; substrate
level phosphorylation;
Rate of basal respiration = rate of overall
respiration (initial) – rate of non-mitochondrial
respiration; rate of ATP linked respiration = rate
of basal respiration – (rate of proton leak
respiration + rate of non-mitochondrial
respiration); rate of proton leak respiration =
rate of overall respiration (after oligomycin) –
rate of non-mitochondrial respiration; maximal
respiratory capacity = rate of overall respiration
(after FCCP) – rate of non-mitochondrial
respiration; mitochondrial reserve capacity =
maximal respiratory capacity – rate of basal
respiration;
ATP-linked respiration; they have the same
effect as oligomycin;
ATP levels fall; linearly; level of rigor mortis
increases; figures quote;
Respiration stopped; no more ATP synthesised;
2
Residual ATP present; ATP is hydrolysed;
energy used for muscle contraction;
Limited ATP; cross bridges not broken; myosin
head not repositioned;
pH decreases due to increase in lactic acid;
anaerobic respiration;
Breakdown / hydrolysis, of proteins; due to
enzymes; cross bridges broken; no ATP
synthesised;
Temperature; effects activity of enzymes;
muscle mass; more muscles leads to greater
degree of rigor mortis; gender; effects muscle
mass;
A myosin; B myosin head; C actin;
3
1 (d)
1 e) (i)
1 (e) (ii)
1 (f)
1 (g)
1 (h)
2 (a) (i)
2 (a) (ii)
2 (a) (iii)
2 (a) (iv)
2 (b)
2 (c) (i)
2 (c) (ii)
2 (d) (i)
© Oxford University Press 2015
Guidance
1
2
4 max
3 max
5
5
2
4
2
3
2
4
4 max
3
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Module 5 practice questions
Answers
OCR Biology A
2 (d) (ii)
ATP;
1
2 (d) (iii)
A moves to the left;
1
2 (d) (iv)
Cofactor; changes 3D shape of, enzyme / active
site; activates enzyme;
Myosin head attaches and rotates; detaches;
repeats;
Amino acid sequence / primary structure;
different sequence in each; different R group
interactions;
Energy (from ATP) changes structure of
dyneins / kinesins; Energy (from ATP) used to
break cross bridges; dyneins / kinesins, do not
form cross bridges, ORA dyneins / kinesins,
result in movement within cell; ORA
Mitosis / meiosis; cell movement;
3
Lipid structure so hydrophobic; moves across
cell membrane by simple diffusion; hydrophobic
core is not a barrier; secondary messengers
needed for hydrophilic hormones;
Plasma is composed mainly of water; lipid
soluble hormones are not soluble in plasma;
(some) proteins are soluble in plasma;
Consumption of oxygen increases with
temperature; in all cases; figure quote;
3
Germinating corn 1.6/20 = 0.08 cm /min; non3
germinating 0.2/20 = 0.01 cm /min;
Yes oxygen consumption increases; respiration
uses oxygen; no seeds not germinating so no
respiration; change in oxygen concentration due
to experimental error;
(the factor) which is at its, least favourable /
nearest its minimum, value;
(the factor which) if increased would speed up
process;
factor which limits the rate of a, reaction /
process;
4
guard cells lose turgidity / AW;
closure of stomata;
restricts carbon dioxide entry (through stomata);
R stops carbon dioxide
entry
ref. to ABA and stomatal closure;
shade plant more photosynthesis at low light
intensities / ora;
sun plant more photosynthesis at high light
intensities / ora;
shade plant reaches, compensation point / net
rate of photosynthesis, at
low light intensity / ora;
shade plants , plateau / max photosynthetic
rate, at low light intensities /
ora;
fewer, cells / named cells/ less, biomass /
named tissue (present per leaf
surface area); fewer mitochondria;
2 max
2 (d) (v)
2 (e)
2 (f) (i)
2 (f) (ii)
3 (a)
3 (b)
4 (a)
4 (b)
4 (c)
5 (a) (i)
5 (a) (ii)
5 (b)
5 (c) (i)
© Oxford University Press 2015
3
3
4
2
2 max
3
2
4
1 max
2 max
1
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Module 5 practice questions
Answers
OCR Biology A
5 (c) (ii)
5 (d) (i)
5 (d) (ii)
(at low light intensity) rate of photosynthesis /
primary productivity, low;
small amount of sugars formed / AW; R food,
nutrients
sugars / photosynthetic products used in
respiration;
idea of low rate of respiration means products
used slowly / ora;
–2 –1
Group 1 21 mg (CO2 ) cm h ;
–2 –1
Group 3 7.5 mg (CO2 ) cm h ;
No units max 1
If units shown once and figures correct 2 marks
detect light intensity;
ref to PGRs;
PGR works via genetic control;
different, growth rate / pattern /appearance;
example of phenotypic
feature
© Oxford University Press 2015
3 max
2
3 max
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OCR A Biology
19.1
Effects of mutation
1 Amorphic mutation results in loss of functional protein (1); gene/allele, codes for enzyme (1);
enzyme no longer catalyses reaction (1); so recessive allele (1)
Sickle-cell anaemia
1 Heterozygotes have one normal and one mutant allele (for haemoglobin) (1); normally healthy (1);
except in low oxygen concentrations (1); part of malarial parasite life cycle is in red blood cells (1); red
blood cells of heterozygotes are, sickled/destroyed, when parasites enter (1); infection reduced (1);
idea that being heterozygous is a clear advantage in an area where malaria is present and a
disadvantage in an area where malaria is absent (1).
2 Change in primary structure (1); change in secondary and tertiary structure (1); change in 3D shape
(1); change in function (1); reduced oxygen carrying ability (1).
Summary questions
1 Ability to digest lactose is a beneficial characteristic (1); drinking milk prevented starvation (1);
reduced osteoporosis (1); directional selection (1).
2 Mutation is random (1); majority of DNA is non-coding; mutations more likely to occur in non-coding
regions (1); mutations in non-coding regions do not affect phenotype (1);
3 Majority of mutations are silent (1); idea that random change to protein structure is more likely to
reduce function (1); idea that beneficial mutations increase chances of survival if environment
changes (1); surviving organisms reproduce and pass new alleles to offspring (1); leading to evolution
(1).
19.2
1a enzyme coded for by lac operon enables lactose to enter bacteria (1); lactose binds to repressor
protein; (repressor) protein changes shape (1); transcription no longer blocked (1); enzymes needed
to metabolise lactose are synthesised (1) (3 max).
b β-galactosidase catalyses the hydrolysis of lactose (1); to galactose and lactose (1); lactose
permease enables the entry of lactose into cells (1).
2 Tryptophan binds to repressor protein (1); shape of repressor protein changes (1); repressor protein
binds to promoter (1); blocks RNA polymerase from binding (1); transcription prevented (1); of genes
coding for enzymes responsible for tryptophan synthesis (1) (5 max).
3 Cofactors bind to proteins that regulate transcription (1); changes binding of proteins to control
elements (1); rate of transcription changed (1); RNA polymerase activated (1).
19.3
Ontology doesn’t mimic phylogeny
Hox genes regulate the expression of other genes (1); increased transcription of genes responsible
for tumour development (1); decreased expression of tumour suppressor genes (1).
Summary questions
1 Bilateral symmetry is along (single plane through) central axis (1); e.g., two arms, two legs (1); radial
symmetry is along a plane at any angle through central axis (1); e.g., tentacles around central axis
(1).
2 Isolate (Pax6) gene from one species (1); detail e.g., PCR, restriction enzymes (1); test in different,
tissue/species (1); example of positive result e.g., eyes develop on legs (1); switch gene off early in
development and eyes will not develop (1); DNA sequencing and compare genes from different
species (1) (max 5).
3 Statement is valid (1); Hox genes are one form of homeobox gene (1); present in vertebrates (1); in
Hox clusters (1); other forms of homeobox gene present in other clusters (1).
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19 Genetics of living systems
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
1 (a)
Herbicide, is selective agent / exerts selective
pressure;
natural selection;
resistants have, selective advantage / AW; e.g.
competitive advantage
resistants survive / susceptibles die;
(more) resistants (reproduce and) pass,
mutation / allele / trait, to offspring;
increasing frequency of, mutation / allele (in
population); R gene
(common weed so) sprayed in many different
places / AW;
(common weed so) large number of seeds /
rapid spread / many generations in a year;
(weed so likely to have) good dispersal
mechanism / described;
ref to, large number of loops of DNA in
chloroplasts / large number of chloroplasts, so
greater chance of, replication error / mutation;
Substitution;
no missing base pair / 88 + 35 = 123bp / 88 +
35 + 154 = 123 + 154 bp
(277bp);
(mutation) removes / no, target site;
for the restriction enzyme;
so 123 bp fragment is not cut in two (88 + 35);
Mutation A ecf from (i)
5
1 (b) (i)
1 (b) (ii)
1 (b) (iii)
Guidance
2 max
2 max
2
ref to (mutation) altering DNA triplet code;
(so) codes for different amino acid;
(so) primary / tertiary, structure of protein
different;
or codes for premature stop triplet;
incomplete / no, protein produced;
AVP;
2 (a)
effect in terms of gene product / protein /
enzyme / transcription factor /AW
(altered) protein no longer, binds / inactivated
by, triazine / AW; A fits
(altered) protein no longer allows triazine
through membrane;
(altered) protein inactivates triazine;
(altered) enzyme breaks down triazine; A
denatures
AVP; e.g. ref to different metabolic pathway (so
triazine not effective)
Autosomal / not sex-linked, because
approximately equal numbers of male (5) and
female (4) sufferers;
appears in every generation / no alternate
generation pattern;
male passes trait to son;
Dominant / cannot be recessive, because
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3 max
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19 Genetics of living systems
Answers to practice questions
OCR Biology A
2 (b) (i)
2 (b) (ii)
2 (c) (i)
2 (c) (ii)
3 (a)
3 (b)
3 (c)
3 (d)
4 (a)
4 (b)
4 (c)
4 (d)
4 (e)
only inherited from sufferer; ORA
approximately equal numbers of sufferers and
non-sufferers in each
generation/ora expect smaller numbers if
recessive;
e.g. 2 sufferers and 2 non-sufferers in offspring
of sufferers in generation 2
;
e.g. 3 sufferers and 2 non-sufferers in offspring
of sufferers in generation 3
;
cuts DNA into, pieces / fragments / AW; at
specific sites; close to, but not in stutter; detail
of site; e.g. 4-6 base pairs, palindromic
negatively charged; detail; e.g. phosphate
groups attracted to anode; smaller fragments
travel further (towards anode) / ORA; smaller
fragments have less impedance / AW / ORA;
inherited from three different individuals/ each
parent (unaffected parents
of A, B and C);
length of stutter increases in each generation/ C
longer than A/different
sizes of stutter;
homeotic / regulatory, (gene);
contains, 180 bp / homeobox, sequence;
that codes for homeodomain (on protein);
(gene product) binds to DNA;
initiates transcription / switch genes, on / off;
control of, development / body plan;
4
5
1
1
2 max
these genes very important;
mutation would, have big effects / alter body
plan;
many other genes would be affected / knock-on
effects;
mutation likely to be, lethal / selected against;
protein synthesis / transcription and translation;
respiration; DNA replication; mitosis;
cytokinesis; apoptosis; differentiation / gene
switching;
fungi / plants;
Increase expression of gene; bind transcription
factors / aid binding of RNA polymerase to
promotor;
Section of non-coding DNA.
2 max
Change in single, base / nucleotide; e.g.
substitution;
Regulatory gene; involved in body development;
hedgehog is a mammal;
Idea that rate of transcription determines
enzyme production; enzymes required for
protein synthesis; enzymes required for
respiration; energy / proteins, required for
growth;
2
© Oxford University Press 2015
2 max
IGNORE hox
CREDIT controls gene
expression, ref. transcription
factor(s)
ACCEPT description, eg polarity,
segmentation, position of
limbs
ACCEPT example, eg no arms
CREDIT selected against in
context of survival, not
reproduction DO NOT CREDIT
ora, not beneficial so not selected
for
IGNORE growth ACCEPT
programmed cell death
1
2
1
2 max
3 max
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OCR A Biology
20.1
Summary questions
1 Homozygous – contains identical alleles of a gene. Heterozygous – contains different alleles of a
gene (1).
2 Phenotype – the displayed characteristics of the oak tree e.g., how tall it grows (1); Genotype – the
alleles present for each characteristic (1).
3 Continuous – a characteristic which can take any value within a range (1); Discontinuous – a
characteristic which can take only specific values (1); continuous – affected by environmental and
genetic factors (1); discontinuous – affected by only genetic factors (1); example of continuous
variation with explanation (1); e.g., height, as genes affect potential height but diet can limit this
potential. Example given of discontinuous variation with explanation (1); e.g., ability to roll tongue, as
people do or do not have this ability – it cannot be learnt.
20.2
Summary questions
1 Codominant alleles are equally dominant (1); a heterozygous individual would display a phenotype
caused by the expression of both alleles (1).
O O
2 Correctly stated mother’s genotype: mother is homozygous / I I (1); correctly stated father’s
A B
A B O O
genotype: father is heterozygous / I I (1); correct gametes shown using standard notation I , I , I , I
A O B O
(1); correct possible offspring shown via a Punnet Square diagram or equivalent: I I , I I (1);
A O
B O
statement of offspring genotype: 50% of offspring would be I I , 50% would be I I (1); statement of
offspring phenotype: 50% of offspring would be blood group A, 50% blood group B (1).
3 Mother is a carrier / is heterozygous / has one copy of the faulty allele (1); father has the normal
C
c
(dominant) allele (1); correct gametes shown using standard notation e.g., X (normal allele), X
C C
(faulty allele) (1); correct possible offspring shown via a Punnet Square diagram or equivalent: X X ,
C c
C
c
C
c
X X , X Y, X Y (1); statement that male offspring could be X Y or X Y (1); statement that 50% of
possible male offspring would suffer colour blindness (1).
20.3
Summary questions
1 Monogenetic inheritance is the study of the inheritance of one gene, whereas dihybrid inheritance is
the study of the inheritance of two genes (1).
2 One mark for a correctly stated reason with explanation (max 2). For example, genes are linked –
cannot be inherited independently (1); sample size is too small (1); unlikely to produce exact ratio
from small sample size, due to random nature of each individual outcome, crossing over of
chromosomes (1); new allele combinations created (1).
3 Correctly stated genotype of yellow round seed producing pea plant: YyRr (1); correctly stated
genotype of green wrinkled seed producing pea plant: yyrr (1); correct gametes shown using standard
notation YR yR Yr yr (from yellow round parent), yr (from green wrinkled parent) (1); correct possible
offspring shown via a Punnet Square diagram or equivalent: YyRr Yyrr yyRr yyrr (1); genotypes
linked to correct phenotype: YyRr - yellow round Yyrr – yellow wrinkled, yyRr – green round, yyrr –
green wrinkled (1); statement of expected ratio of offspring phenotype: 1:1:1:1 (1).
20.4
Corn and the chi-squared test
1 9:3:3:1 (1)
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2 1 mark for each correct row
Grain
Observed
phenotype
number
Purple and
smooth
Purple and
shrunken
Yellow and
smooth
Yellow and
shrunken
Total number
Observed ratio
Expected ratio
216
10.3
9
79
3.8
3
65
3.1
3
21
1
1
381
Expected
number
381 × 9/16 =
214
381 × 3/16 =
71
381 × 3/16 =
71
381 × 1/16 =
24
Chi-squared
value:
(Observed –
2
expected) /
expected
2
(216 – 214) /
214 = 0.019
2
(79 –71) / 71
= 0.901
2
(65 – 71) / 71
= 0.507
2
(21 – 24) / 24
= 0.375
1.50
3 Degrees of freedom = 4 – 1 (1); locate the number on row three in column 0.05 probability (1);
critical value stated (1).
4 Probability value stated (1); difference (between expected and observed results) is not due to
chance (1); significant difference (1); so genes are linked (1).
Summary questions
1 Comparison (of data) (1); (two) sets of data / observed and expected, data (1); significant difference
/ differences due to chance alone (1).
2 Dominant epistasis (1); black allele / E, is dominant to, chestnut allele / e (1); grey allele / G, is
dominant to g (1); G (allele) is epistatic to E (allele) (1); E (allele) is hypostatic to G (allele) (1).
3 a 5.2 genetic map units (1).
b Cross Z,z with X,x and Z,z with Y,y (1); idea that if Z locus closer to Y than X then order is X, Y, Z or
X, Z, Y (1); idea that if Z is more than 5.2 GMU from Y then order is X, Y, Z (1) idea that if Z is less
than 5.2 GMU from X then order is X, Z, Y (1).
c Genes further apart more likely to cross more than once (1); (this will) cancel previous cross (1); so
incorrect recombination frequency (1).
20.5
Summary questions
1 Evolution is change in allele frequency (1); alleles do not change within organism (1); single
organisms do not adapt (1).
2 The Hardy–Weinberg principle assumes a stable, isolated breeding population of diploid organisms
with random mating, no mutations, and no selection pressure (1). Because people choose their
partners or have them chosen for them by their parents, mating is not random (1). Any conclusions
based on Hardy–Weinberg principles will not, therefore, be valid, because the equilibrium only holds
in ideal conditions (1).
b idea that individuals do not reproduce with visibly similar organisms (1).
2
3 Frequency of homozygous recessive ll (q ) =
2
10
90
= 0.11 = 11% (1)
q = 0.11
q = √0.11 = 0.33 (1)
p + q = 1 (1)
p=1–q
p = 1 – 0.33 = 0.67 (1)
2
So frequency of homozygous dominant LL = p = 0.672 = 0.45 or 45% (1)
Frequency of heterozygote genotype Ll = 2pq = 2 × 0.67 × 0.33 = 0.44 or 44% (1)
So expected frequencies:
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homozygous recessive = 11%
homozygous dominant = 45%
heterozygotes = 44%
4 Genetic drift (1); small population (1); mutation (in allele) (1); large impact on future generations (1)
5 Sexual reproduction in eukaryotes and asexual reproduction in prokaryotes (1); (involves) meiosis
(1); crossing over (1); idea that non-coding regions of DNA reduce disruption of genes (during
crossing over) (1).
20.6
Pedigree dogs and the ethics of artificial selection
Idea of both unnatural (1); religious objections (to both) (1); idea of problems due to reduced genetic
diversity (1); idea of animal welfare issues (1); show dogs produced for pleasure (1); plants/animals,
greater yields of food (1); reduce food shortage (in some countries) (1). Any other sensible points
(max 5).
Summary questions
1 Pre-zygotic reproductive isolation before fertilisation stage (1); e.g., geographical, behavioural,
seasonal, anatomical (1); post-zygotic reproductive isolation after fertilisation (1); e.g., zygote not
viable, offspring infertile (1).
2 a Outbreeding (1)
b Introduces new alleles (1); idea that (new alleles) can be beneficial (1); idea of increased (genetic)
variation (1); (population) more able to cope with changes in environment / new selection pressures
(1); number of homozygous recessive loci reduced (over time) (1) (max 3).
3 (Isolated) organisms exposed to different environments (1); (so) different selection pressures (1);
idea that variation means that a wide range of alleles is present (1); (in different environments)
different alleles selected for (1); idea that different changes in allele frequency leads to formation of
different species (1).
4 Sympatric speciation occurs within same habitat (1); selection pressures the same for all (1);
allopatric speciation occurs within isolated groups (1); isolated groups (may have) different selection
pressures (1); (some) gene flow in sympatric speciation and no gene flow in allopatric speciation (1);
organisms not isolated in sympatric speciation so gene flow is still possible (1) (max 5).
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20 Patterns of inheritance
and variation
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
1 (a) (i)
(dominant) epistasis;
B epistatic A hypostatic;
prevents transcription of A;
product of B binds to, promoter / AW, of A;
prevents translation of A mRNA;
product of B binds to, mRNA / ribosome;
product of B inhibits enzyme encoded by A / B
codes for enzyme which breaks down pigment;
Parental phenotypes (AABB) white (aabb)
white; (genotypes given in question)
gametes AB ab;
F1 genotype and phenotype all AaBb white;
F1 gametes AB Ab aB ab; A from P. square
F2 genotypes all correct;; delete 1 for each of
first two mistakes
F2 phenotypes correctly related to genotypes; A
key
F2 ratio 13 white : 3 red;
(see Punnett square)
Genes linked / AW;
ref. locus involved in production of toxin;
resistance inherited with, allele A / allele b;
close together (on same chromosome / in same
linkage group);
few without resistance from crossing over;
Crossing over has occurred;
in, meiosis I / prophase I;
exchange of (part of) non-sister chromatids;
diagram;
mutation;
Green-based is dominant (to uniform colour) / G
is dominant to g / ora re recessive;
Red (fruit) is dominant to orange (fruit) / R is
dominant to r / ora re recessive;
ggrr / rrgg;
3 max
4 max
2 (d) (i)
Parental phenotypes green-based red x uniform
orange;
parental genotypes
GgRr
ggrr;
gametes GR
Gr gR gr
gr;
offspring genotypes and phenotypes
GgRr green-based red
Ggrr green-based orange
ggRr uniform red
ggrr uniform orange;;
3;
2 (d) (ii)
> 0.1/ greater than 0.1;
1
2 (d) (iii)
Difference from expectation is not
significant/statistically different;
above (critical) value (0.05) / AW;
result due to chance;
prediction correct / null hypothesis should be
accepted;
3 max
1 (a) (ii)
1 (b) (i)
1 (b) (ii)
2 (a) (i)
2 (a) (ii)
2 (b)
2 (c)
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Guidance
8
2 max
2 max
1
1
1
1
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20 Patterns of inheritance
and variation
Answers to practice questions
OCR Biology A
2 (e)
3 (a) (i)
3 (a) (ii)
3 (b)
3 (c)
loci (apparently), assort independently / not
linked;
Loci (too) far apart (for linkage to be detected);
ref. recombinants;
crossing over has occurred;
detail crossing over;
[in prophase I of meiosis / exchange of (part of)
non-sister
chromatids] ignore chiasmata
has occurred, twice / even number of times;
(therefore) restoring loci to parental
combinations;
diagram;
cross 1: species 1 - n=7; species 2 - n=7
cross 2: emmer wheat - n=14; goat grass - n= 7;
Chromosome number has doubled;
ref. polyploidy;
nuclear division but no cell division;
failure of spindle in mitosis;
non-disjunction;
Different numbers (42 & 14) of chromosomes;
different numbers (21 & 7) of chromosomes in
gametes;
chromosomes cannot pair;
ref., synapsis / homologous pairs, in meiosis;
meiosis fails; [R ‘cannot occur’]
hybrid/new plant sterile; R not viable
Act as gene banks;
source of genetic diversity; (a) maintain genetic
diversity
store of alleles;
for future use;
in selective breeding of wheat;
to restore alleles lost in selective breeding/
counter effect of inbreeding/genetic erosion;
in case different traits needed/ changed
consumer demand;
in case climate change;
e.g. global warming / temperature rise;
e.g. drier conditions; not environmental change
in case new, pathogen / disease;
in case new pest;
may have as yet unrecognized trait;
AVP; e.g. detail / i.e. disease resistance to new
disease
AVP; e.g. genetic engineering
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3 max
2
2 max
2 max
8 max
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OCR A Biology
21.1
Separation of nucleic acid fragments by electrophoresis
Proteins are polymers of amino acids, which are different sizes and are charged molecules (1); as a
result they would move through gel during electrophoresis under influence of an electric current in
same way as nucleotides do – and they would similarly be separated by size and charge (1).
Pitfalls of profiling
1 11 loci was more than the earlier DNA tests and was affordable (1); producing a DNA profile now is
increasingly faster and less expensive so a higher number of loci that are theoretically more sensitive
and with less chance of misidentification is now affordable (1).
2 Siblings are different as a result of random gene assortment in gamete formation and random
fusing of gametes (1); they share up to 100% of their DNA (1); although normally have different
mixture of alleles, introns, etc (1); but by chance some siblings will have much more genetic similarity
than others and could be seen as the same individual using 17 loci (1); cousins are less closely
related, sharing at most 50% of their DNA, so chances of them having the same DNA profile is much
reduced (1).
Summary questions
1 An intron is a large, non-coding (1); region of DNA (1); that is removed from messenger RNA before
it is translated into a polypeptide chain (1).
2 Polymerase chain reaction (PCR) is a process by which a small piece of DNA is amplified/replicated
many times (1); need relatively large DNA sample for DNA profile/sequencing (1); in forensic
cases/criminal investigations samples available are often extremely small (1); PCR amplifies samples
so DNA profiling can be carried out (1).
3 Benefits: can be used in criminal cases to prove guilt or innocence (1); tiny amounts of DNA can be
used (1); DNA lasts a long time so ‘cold’ cases can be revived by DNA evidence (1); can be used to
prove paternity / to prove or disprove familial relationships in immigration cases (1); can be used to
identify species / can be used to find evolutionary relationships (1). Limitations: can be too dependent
on it and ignore other evidence in criminal cases (1); DNA profiles can be done at different levels and
mistakes can be made (1); contamination of samples with DNA from other organisms (1). Any other
sensible points (6 max, up to 3 marks for benefits and up to 3 marks for limitations).
4 Variety of ways students might do this but must include: extract DNA from sample  PCR to amplify
DNA if necessary  use restriction endonucleases to cut DNA sample into fragments at recognition
sites  place fragments in wells of agarose gel electrophoresis plate with buffer, identifying fragments
 pass electric current through plate so fragments move dependent on size and charge  transfer
gel plate to alkaline buffer to denature DNA fragments  carry out Southern blotting by placing nylon
filter of nitrocellulose paper over gel and using absorbent material to draw fluid through leaving DNA
fragments on filter  fix fragments using UV light  add excess of radioactive or fluorescent gene
probes to hybridise with fragments  wash off excess probes  use X-ray photography or UV light to
show up bands of DNA profile (max 6).
21.2
Summary questions
1 DNA for sequencing is mixed with a primer, DNA polymerase, excess of normal nucleotides (bases
A, T, C, and G) and terminator bases, each with a coloured fluorescent tag (1);  at optimum
temperatures, DNA polymerase builds chains. Whenever a terminator base is incorporated, the chain
is terminated and no more bases can be added (1);  process is repeated until all possible chains
are created (1);  chains separated out by gel electrophoresis; (1)  lasers used to read fluorescent
tags and learn order of bases in complementary DNA strand, from this original strand can be
deciphered (1).
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2a Original bacterial genome – around 548 days, high-throughput methods – under 1 day. Difference
is 547 days (1).
b In original techniques each stage was carried out by hand in the laboratory (1); modern techniques
mainly carried out in machines, many DNA fragments processed at once, so much faster and more
efficient (1).
3a Bases that when added to a DNA chain during DNA synthesis terminate it / no more nucleotides
can be added (1).
b Using terminator bases all possible length DNA fragments are synthesised (1); having different
coloured fluorescent tags attached to four different terminator bases (1); makes it possible to work out
sequence of original DNA (1); once chains have been separated using gel electrophoresis (1); highthroughput sequencing is much more complex and rapid (1), but still relies on terminator bases to
terminate chains in final stages (1).
21.3
Synthetic life
Engineering new organisms to be used in biotechnology (1); engineering healthy genes to cure
genetic diseases (1); engineering new viruses to destroy disease causing bacteria – any other
sensible suggestions.
Infection outbreak – DNA Sequencing and clinical intervention
Benefits include: Ability to tell if an infection is bacterial or viral, and therefore whether to use
antibiotics or not (1); ability to tell what type of bacterium is causing an infection and therefore use the
right antibiotic to cure the infection and minimise risk of resistance developing (1); ability to pick out
hospital based or community based infections (1); ability to develop right vaccine for new strain of flu
etc (1); any other sensible points. Limits include: mainly cost (1); size of machines and expertise
needed to prepare samples and interpret results (1).
Summary questions
1 Bioinformatics allows scientists to analyse large amounts of data generated during sequencing of
billions of base pairs in genomes (1); display the data in ways that make sense and help identify
patterns (1). Computational biology takes these results and uses them to build up models (1); e.g., of
the spread of disease, the evolutionary relationships between organisms, the inheritance of antibiotic
resistance in bacteria (1); and use them to model possibilities in different circumstances (1).
2a Each strain of a pathogen has a slightly different genome and so can be accurately identified by
DNA sequencing (1); ability to identify a particular strain of a pathogen means country/place of origin
can be identified (for treatment, quarantine, etc.) (1); or individuals with disease can be identified (for
isolation, treatment, etc.) (1); spread of a strain of disease can be tracked and transmission methods
understood to prevent further spread (1).
b Traditionally species identification was done by observation of anatomical and physiological
features (1); with DNA sequencing genome similarities are examined and comparisons made to
standard species genome (1); much more accurate but harder in the field (1); in understanding
evolutionary relationships DNA sequencing looks at the difference in number of mutations between
species (1); and by calculating average mutation rate (1); you can calculate when two species
diverged (1).
3 Human genome sequencing shows there are 20–25 000 coding genes (1); proteomics suggests
there are between 250 000 to 1 000 000 proteins (1); technology used to sequence DNA can also
sequence amino acids in proteins – this is not always what you would predict from bases in DNA (1);
some genes can code for up to 1000 different proteins (1); introns and some exons may be removed
before mRNA lines up on ribosomes (1); spliceosomes combine exons in different orders to give
different versions of mRNA (1); which code for different amino acid sequences and different proteins
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OCR A Biology
(1); some proteins are modified by other proteins once they are formed (1); may remain intact,
shortened or lengthened which creates other proteins (1).
21.4
Summary questions
1 Genetic engineering is the practical technique of isolating genes for desirable characteristics in one
organism (1); and placing them into another organism (1); using a suitable vector (1).
2 Restriction endonucleases cut a required gene from DNA of an organism at specific sites within
DNA molecule (1); may leave uneven ends and extended regions of unpaired bases are called sticky
ends (1); used for inserting into DNA of a vector (1). Reverse transcriptase enzymes create DNA from
mRNA used to make a particular protein (1); reverse of normal DNA transcription (1); synthesised
DNA usually inserted into a vector same way as method using restriction endonucleases (1).
3 Plasmid cut open using same restriction endonucleases as used to excise the gene to be used (1);
so sticky ends of DNA fragment/gene and cut plasmid are complementary/match (1); sticky ends lined
up and DNA ligase joins them by forming phosphodiester bonds between the two strands of DNA (1);
transgenic plasmids taken up by bacteria as a result of calcium ions and temperature making
membrane permeable to plasmid or by electroporation when electric current makes membrane
porous to plasmids (1); plasmids have gene for antibiotic resistance, so bacteria that take up
engineered plasmids can be identified (1); plasmid also contains marker gene engineered in (site
where plasmid is cut open) to show engineering has worked – usually fluorescence or enzyme which
changes colour of medium – required gene is inserted in middle of marker – so bacteria showing
marker trait are not successfully engineered (1).
4 a and b Flow diagrams should cover all main points. Each diagram can be awarded a maximum of
6 marks. Diagrams should include the points of difference in the processes, 4 marks should be
reserved for the following points – without these a candidate cannot achieve full marks: infection of
plant cell with bacteria containing engineered plasmid (1); callus formation (1); and growing on of
cloned cells to form many individual transgenic plants (1); production of functional transgenic
bacterium after plasmid is reabsorbed into host bacterial cell (1).
21.5
Summary questions
1 Genetically engineered microorganisms have been used safely for many years (1); genetically
engineered microorganisms produce many beneficial materials from insulin to antibiotics so benefits
are very tangible (1); little empathy for microorganisms and no welfare issues (1). Any other sensible
points (max 3).
2 1 mark for each correct row
Somatic cell gene therapy
Germ line cell therapy
replacement of mutant/faulty gene in body cells replacement of mutant/faulty gene in egg,
with healthy/normal allele
sperm or early embryo with healthy/normal
allele
must be done in many cells
only needs to be done in one or two cells but
required for every child born to an affected
individual
have to get healthy allele into nucleus of cells
have to get healthy allele into nucleus of cells
gene needs to function long-term
gene needs to function long-term
healthy allele not passed on to offspring
healthy allele will function in individual and will
be passed on to their children
does not cure the genetic disease
cures the genetic disease
3a Many potential benefits for example, pest-resistant GM crop varieties (1); soya beans reduce
amount of pesticide spraying needed (1); protects other insects in the environment and helps poor
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OCR A Biology
farmers who cannot afford to spray regularly (1); yield can be increased giving more food (1); weed
killer resistant soya beans allow farmers to spray to get rid of weeds without affecting crop which then
gets all the resources and so has bigger yield (1); other plants are engineered specifically for high
yield (1); crops can grow in wider range of conditions/survive adverse conditions e.g., scuba rice
developed by IRRI to help rice farmers cope with extreme weather events (1); extended shelf-life of
some GM crops reduces food waste – either less food has to be produced or there is more food to go
around (1); nutritional value of crops can be increased – enhanced levels of vitamin A in golden rice,
enhanced protein, or carbohydrate in increased yields (1); plants used to grow medicines – vaccines
in bananas, tomatoes, etc., antibiotics (1); disease resistant varieties can be reduced which reduces
crop losses and provides more food. Any other sensible suggestions (1 mark for each potential
benefit, max 6).
b Evidence of balanced discussion with different opinions considered. For example: Non-pest insects
might be damaged by toxins in GM plants – for example Bt protein in modified soya plants could
affect larvae of other moths and butterflies (1); antibiotic genes could spread from marker genes into
wild populations and spread antibiotic resistance (1); transferred genes might mutate (1); biodiversity
could be reduced if GM crops are herbicide resistant (1); people might be allergic to proteins grown in
GM crops, e.g., Bt protein (1); insects may become resistant to pesticides in plant tissues. Two or
more insecticide genes may be used to reduce chances of resistance developing (1).
4 Copy of human gene coding for relevant/desired protein is isolated or synthesised, introduced into
genetic material of fertilised cow, sheep, or goat egg (1); promoter sequence added to ensure gene is
expressed only in mammary glands (1); fertilised, (female) transgenic embryo is returned to mother to
grow to birth. When mature transgenic animal conceives and gives birth, it produces milk containing
desired human protein to be harvested (1); well-being of transgenic animals appears unaffected –
simply make an extra protein in their milk (1); enormous benefits for human patients receiving
transgenic proteins/pharmaceuticals, however there is ethical consideration of whether human genes
should be put into animals (1); mother animal has to undergo fertility treatment to produce eggs and
then embryos are placed in surrogate animals, which have to be medically-prepared (1); success rate
of inserting human gene is low, so ova/embryos are wasted and destroyed (1); process involves germ
line cell manipulation so genes are passed on in perpetuity (1). Any other sensible points (6 max).
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21 Manipulating genomes
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
1 (a) (i)
Gel electrophoresis;
1
1 (a) (ii)
Placed (in a well) towards the negative end of
the gel; move to the right;
2
1 (a) (iii)
DNA is negatively charged;
1
1 (a) (iv)
(gel) slows rate of movement (of fragments); so
increase separation;
DNA probe; nucleotide sequence with,
fluorescent / radioactive, label; complementary
to required fragment; probe binds to required
fragment;
Small quantity of DNA obtained; PCR increases
quantity of DNA;
(DNA) primers binding to (5´ end) of DNA; so
DNA polymerase can bind; for replication;
Break hydrogen bonds (between strands); to
separate double stranded DNA; for semiconservative replication;
Both involve the breaking of bonds; due to
increased temperature; protein, involves
reforming of bonds; DNA, only hydrogen bonds;
protein, hydrogen bonds and, other bonds /
named; protein, change in, 3D shape
/described; DNA, separation of strands forming
double helix;
Gel electrophoresis, polymerase chain reaction,
gene sequencing;;
Bioinformatics allows scientists to analyse large
quantities of data; DNA sequencing involves
large quantities of data; DNA sequencing
provides evidence for epigenetics; possible role
of repetitive sequences in epigenetics;
Reasonable suggested order of importance;
explanation of importance of each principle (5);
comment on validity of placing principles like
these in order of importance;
Gel electrophoresis separates different sized
fragments of DNA; repetitive sequences are
different lengths;
Polymerase chain reaction;
2
Restriction enzymes; enzymes are specific;
different sequences of nucleotides have
different 3D shapes;
One / few, possibilities in paternity testing; lower
degree of match can rule subjects out; wider
range of possibilities in forensic testing;
Suspect 3
3
Homozygous loci will have identical alleles; with
identical lengths; producing a single band;
Both increasing; industrialised countries, rate of
increase is decreasing; ORA figures quote;
3
1 (b)
1 (c)
1 (d) (i)
1 (e) (i)
1 (e) (ii)
2 (a) (i)
2 (a) (ii)
2 (b)
3 (a) (i)
3 (a) (ii)
3 (a) (iii)
3 (a) (iv)
3 (b) (i)
3 (b) (ii)
4 (a) (i)
© Oxford University Press 2015
Guidance
4
2
2 max
3
5 max
2
4
7
2
1
3
1
3
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21 Manipulating genomes
Answers to practice questions
OCR Biology A
4 (a) (ii)
4 (b)
5 (a)
5 (b) (i)
5 (b) (ii)
5 (b) (iii)
Increase in intensive farming in industrialised
countries; more opposition in industrialised
countries; food shortages in developing
countries;
Difficult to stop spread of genetically modified
plants; idea of ‘super weeds’; affect food, chains
/ webs; reduce biodiversity; cause, allergy /
disease; consumers may not be aware of
modifications;
Somatic uses of body cells; temporary; not
passed on to offspring; germline use of,
gametes / embryos; permanent; passed onto
offspring;
Mitochondria involved in the production of ATP;
large number of mitochondria present (in these
cells);
Nucleus from affected egg transferred to
enucleated normal egg; idea of three parents;
reference to ethical issues;
Yes nuclei of gametes not modified;
mitochondrial DNA distinct from DNA in
nucleus; DNA not actually modified but
exchanged; no changes inherited; present in all
cells of offspring;
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3
6
4
2
3
5
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OCR A Biology
22.1
Practical cloning
1 Use a non-flowering stem – all plant resources available for growing new roots (1); make an oblique
cut in stem – maximises surface area available for rooting powder/new root development (1); use
hormone rooting powder – scientists unsure whether effect is the hormone directly or anti-fungal
action but seems to increase success rate (1); reduce leaves to two or four – minimises loss of water
by transpiration whilst maintaining photosynthesis (1); keep cutting well-watered – reduces water
stress (1); covering with plastic bag – keeps air humid and reduces water loss by transpiration.
2 Cuttings are genetically identical (1); so differences in rate of growth, time to flowering etc. down to
differences in conditions they are given (1).
Summary questions
1 Organ which contains stored food from photosynthesis e.g., potato (1); cloning – new bud/plants
may arise from the organ identical to original plant (1); allows plant to survive adverse conditions (1);
and produce a new shoot (1); using energy from food store (1).
2 One mark for each advantage or disadvantage, max 4. Must provide at least one advantage and
disadvantage. Cuttings – genetically identical to parent so likely to produce good crops (1); often
shorter time from planting to crop (1); reliable (1); don’t have to buy in (1); can use own plants (1).
Seeds – have genetic variation so more variability in quality of crop (1); but are more likely to
withstand disease of changes in circumstances (1); take time and right conditions to germinate and
grow to maturity (1); in some cases can collect seed and reuse for next planting but don’t always get
the same quality (1). Any other sensible suggestions.
3 Genetically identical because parts of the same plant (1); but eventually form will depend on
growing conditions – levels of light, water, temperature etc. (1); identical suggests appearance is the
same (1); cloned plants may not be identical because a mutation may take place in stem cells of
meristems (1); changing pattern of growth in the plant (1).
22.2
Yes, we have no bananas
1 All bananas are clones because they are sterile – so disease affected all of them (1).
2 Micropropagation allows disease-free Cavendish bananas to be produced in large numbers (1);
however, clones would still be all same strain so would be vulnerable to the same disease (1); but
micropropagation allows for mass production of genetically modified Cavendish plantlets immune to a
particular disease e.g., Black sigatoka (1). Banana explants could be used where parts of the plant
are transferred to a nutrient medium.
3 Any sensible points for example, from early humans taking plantlets that form at bottom of mutated
bananas (1); a technique used for centuries (1); culture methods remained same for a long time (1);
explanation of why asexual propagation needed (1); until introduction of micropropagation –
advantages it offers (1).
Summary questions
1 Variety of ways students might do this but answers must include: Tissue selection (1);  tissue
sterilisation (1);  tissue proliferation (callus formation) (1);  transfer of cells from callus to new
medium (1); for shoot and root development (plantlet formation) (1); plantlets potted into compost to
form small plants (1);  plants hardened off to be planted out and form mature plants (1).
2 Look for scientific explanation of whichever features chosen (6 marks max. 1 mark for each correct
feature, 2 marks for explanation of each feature). For example: Reliable way of increasing numbers of
endangered/rare plants. Will help towards maintaining biodiversity of ecosystems. Allows for rapid
production of large numbers of crops, very valuable for countries where there are food shortages.
3 a Advantages relatively easy (1); relatively cheap and readily available (1); history of use (1).
Disadvantages any disease in parent plant transferred with cutting etc. (1); limit to number of new
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plants that can be formed so cannot keep up with demand if there is a major threat to crop (1); still
produces clones (1). Any other sensible suggestions. Max 6, one point for each argument. Student
must provide at least two arguments for and against.
b Advantages can produce disease-free plants (1); can produce plants engineered to be resistant to
disease (1); can produce almost limitless numbers of plants fast (1). Disadvantages relatively
expensive (1); needs some infrastructure (1); still produces clones (1). Any other sensible
suggestions. Max 6, one point for each argument. Must provide at least two arguments for and
against.
22.3
Cloning humans
1 DNA sequencing – if child is a clone of an adult they will have identical DNA apart from
mitochondrial DNA which will come from egg donor (1).
2 Cloning primates – even primitive primates is very difficult because spindle mechanism needed for
cell division is very close to nucleus in primate cells and so often destroyed or damaged when
nucleus is removed (1); and synchronisation between stage of embryo and state of maternal
reproductive organs has to be finely attuned (1); no DNA evidence of cloned humans (1); ethical
objections among many people, unlikely to get a team together to do the research etc (1). Any other
sensible points.
Summary questions
1 Natural twinning early embryo splits (1); and two fetuses go on to develop (1); from the two halves
of divided embryo (1). Artificial twinning split in early embryo is produced manually (1); number of
identical embryos may be replaced in surrogate mothers (1); to produce a number of identical high
quality animals (1).
2 Observing births and recoding twin births when animals appear the same (1); genetic testing of any
twin cattle of the same gender (1).
3a Both processes involve removing eggs from an animal (1); both involve surrogate parents (1); both
potentially produce a number of genetically identical organisms (1). Any other sensible suggestion (3
max).
b In twinning either gametes meet outside the body (1); and early embryo develops before being split
(1); or early embryos flushed from the mother (1); egg cell contributes all maternal DNA (1); embryos
produced from gametes (1); embryos genetically related to two parents (1).
4 Outline of process: Egg cells retrieved are enucleated so only contribute mitochondrial DNA (1);
nucleus comes from an adult (somatic) cell of the animal to be cloned, not from gametes (1); an
electric shock is needed to stimulate development (1); no male gamete involved (1); embryos
genetically related only to adult cell nucleus donor (1). Any other sensible points. Look for evidence of
research and understanding by students. Look for evaluation of the evidence – indications of time
involved and cost balanced against potential benefits etc. Max 6, max 2 for decision and max 4 for
description of process.
22.4
Extra microorganisms
1 Reliable source of chymosin not dependent on numbers of animals slaughtered for meat (1); pure
enzyme – no contamination with animal products therefore less allergy risk (1); vegetarians can eat
cheese without ethical problems (1). Any other sensible points.
Summary questions
1 Points can include: no welfare issues (1); big range of microorganisms available to suit different
reactions (1); can genetically manipulate microorganisms to carry out required synthesis reactions
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e.g., human proteins (1); microorganisms have a very short life-cycle and rapid growth rate so can
produce large manufacturing volumes in short time space (1); nutrient requirements of
microorganisms often relatively cheap (1); can be genetically manipulated some microorganisms can
be modified to utilise materials which would otherwise be wasted (1); microorganisms produce own
enzymes so catalyse the reactions (1); processes using microorganisms use relatively low
temperatures and pressures (1). Any other sensible points, max 6.
2 Baking: Mixed with sugar and water (1); respires aerobically (1); carbon dioxide produced used to
make bread rise (1); yeast killed by heating as bread cooks (1); takes a couple of hours (1). Brewing:
mixed with malted barley and hot water (1); respiration (fermentation ) continues for days in anaerobic
conditions (1); ethanol produced as waste product (1); yeast eventually inhibited (not killed) by rising
pH (1); build-up of ethanol and lack of oxygen (1) (max 6, up to 3 marks for each process).
3a To destroy bacteria that would make it go bad rapidly (1); or cause diseases such as TB (1)
b The fat droplets are spread evenly through milk so cream doesn’t separate out (1); and creates a
uniform product (1).
c Cheese whole milk used (1); bacteria used to separate the curds from the whey (1); changing
texture, and bacteria ripen or mature the cheese in controlled slow reactions at low temperatures to
change taste (1); out-compete bacteria that would make the cheese go bad (1); takes weeks, months
or years (1); can last for years (1). Yoghurt skimmed milk powder added to milk to enrich it (1);
°
specific bacteria added and incubated at 45 C for 4–5 hours to produce extracellular polymers that
give the texture to yoghurt (1); lasts 2–3 weeks in a fridge (1) (1 mark per difference, 2 marks max).
22.5
Plants and bioremediation
1 No naturally occurring microorganisms that can accumulate heavy metals (1); interfere with
biological pathways, so genetic modification unable to find a way to use them as yet (1); plants have
cell walls, lignified areas, xylem etc that are not metabolically active or dead, where heavy metal ions
can be accumulated without doing any damage (1). Any other sensible point.
Summary questions
1 Fungi produce penicillin naturally (1); bacteria genetically engineered/modified to produce human
insulin (1).
2 Use of microorganisms or plants to break down pollutants and contaminants in soil or water (1);
often carried out on site (1); because area of contamination may be very large so not practical to
remove contaminated material/too expensive to remove contaminated material (1); organisms
involved in bioremediation grown and break down contaminants in situ (1); living organisms so they
grow and spread (1); may be harvested and contaminants retrieved (1). Any other sensible points.
3 Gene for making human insulin cut out of human genome (1); using restriction endonucleases (1);
 inserted into a bacterial plasmid using DNA ligase (1);  plasmid containing human insulin gene
inserted into bacteria through transformation e.g., electroporation (1);  bacteria grown in a fermenter
(1);  mixture removed and penicillin extracted in process called downstream processing (1).
22.6
Investigating factors which affect the growth of microorganisms – serial dilutions and bacterial
counting
3
a Allowing for the fact that only 0.1 cm is plated you can think of the effective dilutions as follows (if
3
–6
–7
–8
–5
–6
–7
you continue to work in cm ): A = 10 B = 10 C = 10 . Also credit A = 10 B = 10 C = 10 as
these are the absolute dilutions of samples.
–3
8
b A number of bacteria cm = 500 × 10 000 000 = 500 000 000 or 5.00 ×10
–3
8
B number of bacteria cm = 52 × 100 000 000 = 520 000 000 or 5.2 × 10
–3
8
C number of bacteria cm = 4 × 1 000 000 000 = 400 000 000 or 4 × 10
c Credit students choosing to take the mean of the results for A, B and C:
–3
(A + B + C) /3 = 4 733 333 333 or 4.73 × 108 bacteria cm in the original sample. However, also
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credit students who rely on the lowest dilution which gives countable results as the most accurate
guide to the true number of bacteria in the original sample. This means, since the plates are not
8
8
shown, that either plate A (giving an answer of 5.0 ×10 ) or plate B (5.2 × 10 ) could be the best
–3
answer for the true mean number of bacteria cm . Plate A if the colonies are small enough that few
or none will have merged (causing undercounting) or plate B if merging of colonies in plate A gives a
risk of undercounting. Additional credit should be given to students who make a distinction between
the number of bacteria as stated in the question and the number of viable bacteria or colony forming
units.
Summary questions
1 Both provide nutrients, suitable pH, moisture etc (1); both need to be maintained at optimum
temperature for growth (1); both must be kept sterile until inoculated with microorganisms (1); both
can be shaken at intervals to aerate it (1); agar plates remain closed once made up (1); broth is mixed
with known volumes of culture medium (1); agar plates inoculated using sterile wire loop and culture
medium (1); numbers in broth counted using turbidity, serial dilutions, and microscope graticules (1);
numbers on agar calculated using colony counting (1) (6 max, must be at least two similarities and
differences).
2 In large closed culture nothing gets in or out (1); initially, growth can be at theoretical rate as no
factors are limiting (1); as culture continues, numbers increase, food and oxygen are used up and
waste products build up often affecting pH (1); microorganisms run out of food or oxygen, are
inactivated by pH changes affecting enzymes or poisoned by waste products (1); so whilst theoretical
growth curve is exponential, real growth curve reaches a peak, plateaus, and declines (1).
3 a Vinegar is ethanoic acid therefore has a low pH (1); and inhibits bacterial growth (1)
b As temperatures fall bacteria growth slows but does not stop (1); so in fridge bacteria grow slowly
and eventually destroy food (1).
c Reactions in bacteria and fungi that act as decomposers affected by temperature (1); in August
temperatures relatively high so decomposition reactions relatively fast. In December, the
temperatures are much cooler so slower reactions in decomposers and rotting slower (1).
4 Could be done in a number of ways (max 6) e.g., set up series of cultures identical in every way
EXCEPT for factor you are investigating e.g., temperature, presence of a nutrient or antibiotic etc 
culture for several days  for each culture carry out a serial dilution  plate each dilution, making
streaks on agar plate with sterile inoculating loop and label carefully to show which original culture it
has come from and what dilution factor is repeat for all original cultures  culture agar plates for 5
°
days at 20 C for each factor find plate with number of bacterial colonies that can be counted. Work
out original number of bacteria using dilution factor  compare results across different cultures to see
effect on bacterial growth of factor being investigated.
22.7
Summary questions
1 Continuous processes run continuously once fermentation is started (1); sterile nutrient medium
added continuously once culture is growing exponentially (1); culture broth continually removed so
product can be processed and culture volume remains the same (1). Batch process everything added
at beginning in fixed volume of medium (1); nutrients used up and microorganisms, products, and
waste products build up (1); may be stationary phase when secondary metabolites formed, process
stopped, products extracted, reactor cleaned, and new process begun (1).
2 Any three from (max 2 per point): Temperature if temperature too low microorganisms will not grow
quickly enough, too high and enzymes will start to denature and microorganisms are inhibited or
destroyed. Bioreactors often have a heating and/or a cooling system linked to temperature sensors
and a negative feedback system to maintain optimum conditions. Nutrients if microorganisms use up
food supply they will start to die off so need a mechanism to keep food supplied, nutrient medium can
be added in controlled amounts to broth when probes or sample tests indicate that levels are
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decreasing to be mixed in using stirrers/paddles as will not spread through fast enough by diffusion
alone. Oxygen if microorganisms use up oxygen they will start to die off so need a mechanism to
keep nutrient medium oxygenated, oxygen is bubbled through broth when probes or sample tests
indicate that levels are dropping to be mixed in using stirrers/paddles as will not spread through fast
enough by diffusion alone. pH if waste products of microorganisms e.g., carbon dioxide build up then
pH of mixture will decrease. Change in pH can affect enzyme action and stop growth, buffers are
added to mixture and stirred in or alkaline solution added to maintain optimum pH.
3a Penicillin (1)
b Reaction continues after the maximum cell mass has been reached (1) – when food products would
be harvested (1); secondary metabolite is labelled (1).
c Graph should show process stopping once maximum cell mass is reached (1); could show removal
of cell mass (1); and addition of more nutrient medium and repeat of the exponential growth stage (1).
22.8
Immobilised enzymes in medicine
1 Advantage enzyme specificity means can get a very reliable response to one specific molecule in
the blood or urine (1). Any other sensible suggestion. Possible disadvantage enzymes usually work
best at a very specific pH and pH of urine can vary considerably which might limit use of biosensors
for chemicals in urine. Any other sensible suggestion.
2 Immobilising enzymes makes them less exposed to environment changes (1); buffers to maintain
steady pH could be included in immobilising structure (1); enzymes could be encapsulated in
membranes which only allow text molecules to pass through. Any other sensible point linking
immobilisation to protection from pH or other changes.
Summary questions
1 a Enzymes attached to an inert support system (1); over which the substrate passes and is
converted to product (1).
b More efficient (1); more specific; can optimise conditions for specific enzyme (1); less downstream
processing (1).
c Can be reused (1); easily separated from reactants and products (1); more reliable as control over
process (1); greater temperature tolerance(1).
2 Surface immobilisation – absorption onto inorganic carriers (1); covalent or ionic bonding onto
inorganic carriers (1); entrapment in a matrix (1); entrapment in membrane bound microcapsules (1).
3a Enzymes are accessible to substrates (1); allow continuous production by a continuous flow of
medium over the enzyme (1); conditions can be very tightly controlled over the enzyme beds (1);
changes in pH and temperature have less effect (1); any other sensible suggestions (3 max).
b Immobilising an enzyme may affect its ability to catalyse a reaction (1); diffusion of substrate to and
from active site of enzyme can be inhibited (1); by immobilising matrix or capsule and so slow reaction
(1); in surface immobilisation enzymes may be lost from matrix relatively easily (1). Any other sensible
points (4 max).
4 Look for evidence of research and of student having identified a suitable reaction catalysed by
immobilised enzymes – either one mentioned in the text or something different. Whichever reaction is
chosen the student should: give the reaction chosen (1); give the enzyme involved (1); explain how
the enzyme is immobilised (1 or 2 – one for simple statement of method, 2 for more explanation);
discuss the importance of the process (1); give the advantages of using immobilised enzymes over
other ways of catalysing the process (1 or 2 one for simple statement, 2 for more reasons or deeper
discussion) (6 max).
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22 Cloning and biotechnology
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
1 (a)
Reproductive cloning fertilised egg placed into
surrogate; cloned animal allowed to develop;
therapeutic cloning fertilised eggs / embryos,
remain in lab; e.g. stem cells collected;
Positive treatment of genetic disease; treatment
of infertility; medical research; negative idea of
unforeseen problems, e.g. shortened life span,
genetic disorders; religious objections;
emotional problems for, clone / parents;
reduction of genetic diversity;
Nucleus removed from adult (somatic cell);
nucleus removed from egg cell of different
animal; nucleus from somatic cell placed into
(enucleated) egg cell; electric shock to stimulate
division; embryo placed into surrogate;
3
Mitochondria in enucleated egg cell; contains
DNA;
Production of multiple embryos from one
embryo; increase success rate; reduce costs /
increase inefficiency;
Explant tissue removed, from, plant / animal;
transferred to nutrient medium as starting
culture; callus undifferentiated, mass of cells;
idea of large number of cells that can be split
into many cultures;
e.g. auxins / gibberellins; used in different
proportions; cell division; growth of roots;
growth of shoots;
Continuous; nutrients / examples, supplied;
products removed;
Feed valve for nutrient supply; harvest valve for
product removal; sensors to monitor, conditions
/ named; so, conditions / named, can be
maintained at optimum;
Advantages conditions maintained at optimum;
do not need to be cleaned as regularly; smaller
space; lower staff costs disadvantages
expensive; skilled workers required; not suitable
for all products;
Against cloned cows (may) suffer; example;
may affect products; many people think, cloning
should only be used in certain circumstances /
never used; for some people, think cloning is ok
or have no opinion; cloning may help solve food
shortages;
1 low(er) / less, energy (than beef);
2 useful for, slimming / weight control / AW;
3 low(er) / less, (total) fat;
4 (very) low / (much) less, saturated fat;
5 lower, cholesterol
OR
lower risk of, (coronary) heart disease / CHD /
cardio-vascular diseases / heart attack /
2
1 (b)
1 (c) (i)
1 (c) (ii)
1 (d)
2 (a)
2 (b)
3 (a)
3 (b)
3 (c)
4
5
© Oxford University Press 2015
Guidance
4 max
3 max
3
4
5
3
4
4
6
7 max
CREDIT ora for beef throughout.
IGNORE use of figures alone
when awarding mps 1, 3, 6, 7, 9
– look for descriptive statement,
e.g.
’12 g of protein’ = no mark
‘only 12 g protein’ = 1 mark
2 ACCEPT preventing obesity
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22 Cloning and biotechnology
Answers to practice questions
OCR Biology A
cardiac arrest /
myocardial infarction / MI / angina /
atherosclerosis / atheroma /
stroke / hypertension;
6 contains carbohydrate / AW;
7low(er) / less, iron content;
8 (increased risk of) anaemia / fewer RBCs /
less haemoglobin /
reduced oxygen carrying capacity of blood;
9 low(er) / less, protein;
10 (mycoprotein provides) more balanced diet;
11 need larger intake to meet requirements /
AW;
6 (a) (i)
6 a (ii)
6b
6c
ACCEPT ‘less energy to burn off
during exercise’
DO NOT CREDIT ‘burn off’
unqualified
6 ACCEPT ‘more carbohydrate
than beef’
IGNORE ‘carbs’
8 IGNORE answers phrased in
terms of role of iron alone
e.g. ‘haemoglobin contains iron’
=0
Answers must show
consequence of deficiency
e.g. ‘less haemoglobin’ = 1
Graticule total count; bacteria not necessarily
alive; dilution plating viable count; only live
bacteria can form visible colonies;
Colorimetry
4
Prevent contamination; possible pathogens;
inaccurate results;
7;
43; 43 / 1.26, x 1 000 000; 3.41 x 10
3
6;
1
3
6d
Using 63; 63 x 100 000; 6.3 x 10
3
7a
Gel entrapment; covalent / cross, linkage;
adsorption to inert surface; contained within
membrane;
Optimum temperature higher for immobilised
enzymes; ORA immobilisation provides
protection; rate of reaction lower for immobilized
enzymes; enzyme not in direct contact with
substrate; (substrate) takes time to diffuse to
active site; figures quote;
4
7b
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5 max
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OCR A Biology
23.1
Summary questions
1 Ecosystem whose biotic and/or abiotic factor(s) are constantly changing (1); all ecosystems are
dynamic (1).
2 Biotic populations of (named) organisms (1); competition between (named organisms) for a
resource (1). Abiotic any two from (for example): temperature, light intensity, dissolved oxygen
concentration, pH (2).
3 Animals able to migrate, e.g., towards warmer/cooler temperatures, to search for alternative food
supplies (1); animals able to move, e.g., to find shelter, to search out water supplies (1); many
animals able to regulate their internal temperature, so less affected by temperature changes (1);
many animals able to survive using a range of food sources (1). Any other sensible reasons.
23.2
Monitoring biomass during conservation
1 Secondary consumers (1)
2 a 86–87 mm (1)
2 b 25 mm (1)
2 c Sheyma island as larger urchins present (1); no predators present so have reached maturity (1)
3 After oil spill the kelp biomass will increase (1); as sea urchin population will have reduced (1); a
measured decrease in kelp biomass indicates recovery from spill, as kelp will be eaten by increasing
population of sea urchins (1).
Summary questions
1 Water removed from one organism by heating it in an oven at 80°C (1); the ‘dry’ organism is then
weighed to measure its mass in grams (1); this figure is multiplied by number of organisms present in
given area (1).
2 Through agriculture number of trophic levels in food chain is minimised (1); to maximise energy /
biomass transfer to humans (1).
3 6300/42000 (1) × 100 = 15% (1); potential gain in biomass of an organism is reduced due to energy
losses from the organism (1).
4 Energy is lost from an organism because (any three from): not all of an organism may be eaten,
parts of an organism may be indigestible, some energy is transferred to the environment through
metabolic heat, some energy is lost through excretion (3 max).
23.3
Reward and punishment
Any six from: Mutualistic relationships mean that both organisms benefit (1); Rhizobium fixes nitrogen
gas into ammonia (1); plant gains amino acids which use to form proteins (1); bacteria gain
carbohydrates which they use as an energy source (1); produced by the plant during photosynthesis
(1); animals gain nitrogen and carbon from eating plants (1); natural selection of most productive
bacteria ensures nitrogen fixation occurs at maximum rate (1).
Summary questions
1 Detritivores break down organic matter into small pieces, providing a larger surface area for
decomposers to act on (1); detritivores carry out internal digestion whereas decomposers carry out
external digestion (1).
2 Respiration carbon dioxide is released as a waste product of respiration and from the bodies of
dead organisms through respiration of decomposers (1). Combustion burning of forests/fossil fuels
releasing carbon dioxide as a waste product (1). Deforestation fewer trees and plants mean less
carbon dioxide is removed from the atmosphere (1).
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3 When a decomposer obtains its energy from dead organisms (1); by the process of external
digestion/process described (1).
4 Any two from: Ice core samples are collected from glaciers which have existed for many
(tens/hundreds of) thousands of years (1); air bubbles trapped within these glaciers are representative
of the atmosphere at a point in history (1); gas analysis of trapped air bubbles reveals atmospheric
composition at this point in history (1).
5 Max 1 mark for each correctly named bacterium in each process. Nitrogen-fixing bacteria
(Azotobacter/Rhizobium (1);) convert gaseous nitrogen into ammonia/organic compounds (1).
Nitrifying bacteria (Nitrosomonas (1);) oxidise ammonium compounds into nitrites (1); (Other) nitrifying
bacteria (Nitrobacter (1);) oxidise nitrites into nitrates (1); denitrifying bacteria convert nitrates into
nitrogen gas (1); ammonification is carried out by bacteria that convert nitrogen-containing molecules
into ammonium salts (1).
23.4
Conservation
1 Cows trample larger plants (e.g., bracken, gorse) (1); new growth is nibbled preventing plants
becoming established/larger (1); tree saplings are destroyed / removed / eaten (1); removal of
leguminous plants / gorse maintains low soil nutrient level (1); therefore, succession to woodland /
scrubland is prevented.
Succession on a sand dune
1 Primary succession
2 a Sea couch grass / other relevant example
b Any two from: deep tap roots to obtain available moisture (1); low profile to avoid strong wind (1);
waxy leaves to retain moisture (1); (high) salt tolerance (1); or suitable features explained of chosen
example.
3 Marram grass stabilises the dune by trapping sand in its roots (1); / nutrients are added to sand as
marram grasses die (1).
4 Environment becomes less alkaline (1); more (non-salt) water is present (1); presence of nutrients
increase soil depth / humus increases (1); plants generally experience less windy conditions (1); any
other suitable explained factor.
5 They are out-competed for factors such as light / water / space.
Summary questions
1 Climax community is the final stage of succession whereas plagioclimax community is the final
stage of succession as deflected by humans or some other outside factor (1).
2 Primary succession occurs on bare rock/sand without soil (1); whereas secondary succession takes
place where soil is present but the area contains no plants or animals (1).
3 Abiotic conditions/non-living environment becomes less hostile as soil forms when organisms
decay/ soil becomes more nutrient-rich/ soil retains more water (1); increased biomass supported (1).
4 Biodiversity increases (1); because range of habitats increase / number of food sources increases
(1); climax community develops (1); dominant species outcompete others for light / space / other
relevant example (1).
5 Any six from: Named example e.g., conservation of smooth snake at Studland Heath, Dorset (1);
sampling technique(s) used to monitor population size of endangered species (1); scientific
knowledge is required to understand a species’ habit requirements (1); scientific knowledge used to
produce coherent plan of action to conserve population of the species (1); human interference in an
ecosystem can stop succession (1); through land management / relevant example (1); this maintains
biotic and abiotic factors in favour of the endangered species (1); allowing the endangered species’
numbers to increase / preventing conditions favouring a competitor species (1).
23.5
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Monitoring biodiversity in the Sonoran desert
1 For example: plants random sampling using quadrats / belt transect (1); animals direct observation /
pitfall traps / filmed samples taken over time (1).
2
2 Figures for animal abundance per 50 m are not integer values, therefore must derive from a
number of observations (1)
3 (3 marks)
4 Positive correlation (1); the greater the plant abundance, the greater the animal abundance (1)
5 0.986 (1)
6 df = 13 (1); correlation of 0.986 >> 0.654 at p=0.01 (1); therefore we can say that the conclusion is
>99% certain (1); therefore, the conclusion formed is: In the Sonoran desert, the greater the plant
abundance, the greater the animal abundance (1).
Summary questions
1 Abundance is the total number of organisms whereas distribution refers to where individuals are
found (1).
2a Pooter/pitfall trap (1).
b 20 × 15 / 2 (1) = 150 (1)
3a Removes sample bias / makes sample more reliable (1).
b Allows scientist to study how differing abiotic factors may affect the distribution of a species (1).
4 Doesn’t take into account births/deaths (1); immigration / emigration (1); mark may rub off (1); mark
may make animals more visible to predators (1); animals don’t redistribute evenly (1); samples taken
may not be representative of the whole population (1); or other acceptable suggestions.
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23 Ecosystems
Answers to practice questions
OCR Biology A
Question Answer
number
Marks
1 (a) (i)
Secondary succession, does not begin with
bare ground.
To change the habitat; allowing different
species to colonise;
Pioneer species; formation of soil;
1
Outcompeted; by grasses and shrubs and
trees;
Climax community
2
4 max
2 (a) (ii)
Live increases and then plateaus; detritus
decreases and then plateaus; total
decreases, increases and then plateaus;
soil increases and then plateaus; figures
quote;
Soil would increase from zero; live would
start above zero; the total would not drop;
detritus would not decrease as much;
Increase in net primary productivity; to a
peak / then decrease; figures quote;
Increase in biomass; as succession
progresses; climax community is stable so
energy is no longer accumulated;
Position in food, chain / web at which an
organism feeds
A 2; B 1;
2 (a) (iii)
Biomass; e.g. more aphids than trees;
2
2 (a) (iv)
Along the side; from top to bottom;
2
3 (a)
5
3 (d)
Light intensity; carbon dioxide
concentration; nutrient availability;
competition; reference to pests;
Representative; capture as many
organisms as possible; mark captured
organisms; harmless method; release
individuals; recapture;
Marking may harm organism; capturing
organism may cause harm; marks may be
lost; marks may change interactions of
organisms; some organisms may be easier
to capture / organisms may learn to avoid
capture; ORA
208 / 10 357 = 24 064 / n; 24 064 / (208 /
10 357) = 1 197 214 91;
More accurate; decreases random error;
3 (e)
Increase time before recapture;
1
4 (a)
climate - tropical versus temperate
tropical has …
higher temperature / hotter;
more (sun)light / days longer;
photosynthesis faster;
idea that
4 max
1 (a) (ii)
1 (b) (i)
1 (b) (ii)
1 (b) (iii)
1 (c) (i)
1 (c) (ii)
1 (d) (i)
1 (d) (ii)
2 (a) (i)
3 (b) (i)
3 (b) (ii)
3 (c)
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Guidance
2
2
1
4
3
3
1
2
3 max
5
2
2
CREDIT reverse arguments for
temperate
tropical temperate
temperature
higher
lower
light intensity
more
less
photosynthesis more
less
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23 Ecosystems
Answers to practice questions
OCR Biology A
4 (b)
more storage of , organic molecules /
biomass / energy
or
more formation of , organic molecules /
biomass;
AVP;
vegetation - woodland or rainforest versus
grassland(s)
woodland or forest has …
idea of greater complexity / greater
biodiversity /
more niches;
competition for space less limiting;
AVP;
(bomb) calorimeter;
detail of technique;
detail of , measurement / analysis;
biomass made
more
less
eg less seasonal change
faster , mineral cycling / decomposition
CREDIT reverse arguments for
grassland
wood
grassland
complexity more
less
competition less
more
eg greater , humidity / shelter
2 max
eg known / dry , mass of (organic
material)
(material) burnt in oxygen
eg temperature rise of water measured
known volume of water
calculation described / converted to kJ
4 (c) (i)
(perch) 22; (cow) 1;
2
4 (c) (ii)
higher in bobcat / lower in cow;
for bobcat
more (energy) absorbed; ora
less (energy / waste) egested; ora
correct comparative figs. quoted from table;
meat more digestible; ora
mainly protein and fat;
contains no cellulose; ora
grasshopper;
idea of high conversion to biomass figure;
idea of herbivore / primary consumer /
low(er) trophic level than perch;
idea of more food available;
idea of one stage of energy loss in food
chain not two /
more energy passes through food chain (to
humans);
3
4 (c) (iii)
© Oxford University Press 2015
DO NOT CREDIT figs alone
IGNORE refs to grasshopper and perch
ALLOW ecf if cow calculated as > 6 in
(i)
bobcat 83(%) and cow 40(%)
(absorbed)
or
bobcat 17(%) and cow 60(%) (egested)
4 max
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OCR A Biology
24.1
Human population growth
1 Any two from: Food availability (famine), disease, climate, war, or other suitable example
9
9
9
9
9
2 Population change = 6.8 × 10 – 3 × 10 = 3.8 × 10 (1); population growth = 3.8 × 10 / 3 × 10 (1)
= 127% (1)
3 Suitable suggestion (1); explanation of effect on birth or death rate (1); link to population size (1);
for example: some religions discourage the use of contraception (1); this would lead to an increase in
the birth rate (1); increasing population size (1).
4 a Figure 5(a) has large numbers of women of child-bearing age in comparison to older people (1);
therefore, there is a large birth rate / population will increase over time (1). Figure 5(b) has a small
number of young people, and relatively large number of old people (1); therefore, relatively few
women of child bearing age are available to reproduce / there is a small birth rate / the death rate due
to an ageing population will be relatively high (1).
4 b Sketch or description of pyramid shape, wider at base than top (1); approximately linear decrease
in population sizes at different ages / consistent birth and death rates / relatively consistent numbers
of younger ages, with a decrease in numbers of older people (1).
Summary questions
1 Any three from (for example): increase in quantity or quality of food supply (1); availability of (clean)
water (1); lack of predators (1); favourable environmental conditions (1); availability of light (1).
2 The population doubles in size / increases by a fixed proportion (1); each time a fixed interval of
time elapses (1).
3 Any six from: Graph x-axis labelled time and y-axis labelled number of duckweed/population size
(1); graph: initial period of slow growth (1); few individuals to reproduce and increase population size /
birth rate exceeds death rate (1); graph: period of rapid growth (1); significant numbers of individuals
reproducing / birth rate significantly exceeds death rate (1); graph: period of approximately constant
population size (1); carrying capacity correctly labelled (1); graph: small fluctuations around the
carrying capacity (1); birth rate and death rate are approximately equal (1); biotic or abiotic factors
may (temporarily) affect the birth or death rate (1).
24.2
Summary questions
1 Interspecific competition is when members of different species compete for the same resource
whereas intraspecific competition is when members of the same species compete for a resource (1).
2 Initially resources plentiful so population increases in size (1); resources now limited as more
organisms compete for resource, resulting in population decrease (1); reduced population results in
less competition so organisms survive and reproduce, resulting in increased population (1).
3 a Intraspecific competition (1)
3 b The individuals which grow fastest / are better adapted will survive (1); as saplings grow taller
they will reduce light / other named resource for other saplings (1); this will result in death of other
saplings / only few dominant oak trees will survive (1).
24.3
Canadian lynx and snowshoe hare
1 The number of furs collected was a representative sample of the population (1).
2 a 1865 (1)
2 b Plausible reason e.g., snowshoe hares had an abundance of food (1); increasing survival and
reproductive rates (1).
3 a Figure between 1917 and 1919 (1)
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OCR A Biology
3 b Plausible suggestion e.g., prey population was also very low, so only enough food available for
very small population of lynx (1).
4 (Approximately) every 10 years (1)
5 Any four from: Hare population initially increases due to low numbers being eaten by lynx (1); large
hare population now provides more food for lynx (1); so more lynx survive and reproduce (1);
increasing lynx population (1); more lynx increase predation of hares / more hares die than are born
(1); so hare population decreases (1); no longer enough food to sustain lynx population (1); increasing
lynx death rate so lynx population decreases, cycle begins again (1).
Summary questions
1 Predator population would also fall as it has no food (1)
2 Any two from: The prey is not only food source (1); therefore can get energy by eating another
animal, so a decrease in this particular prey will not necessarily cause a decrease in the predator
population (1); predators contract a fatal disease (1); prey population can increase dramatically (1);
natural disaster (1); destroys both populations (1); seasonal fluctuations e.g., larger plant growth in
summer (1); increases prey food source so numbers climb (1). Any other suitable factor (1); with
explanation (1).
3 Graph drawn with axis labelled population size / number of organisms (y-axis) and time (x-axis) (1);
Predator (Didinium) and prey (Paramecium) population curves drawn, with predator curve mirroring
prey curve after a short (approximately consistent) time delay (1); two further marks for correct
annotations (2); for example, aphid population grows as plenty of food available and very few
predators.
24.4
Conservation of gray bats
1 Conservation exploitation of land surrounding caves was controlled (1). Preservation caves were
gated, thus preventing human access (1).
2 Advantages gray bat population rises, so species prevented from extinction (1); biodiversity is
maintained, which may help maintain balance in affected ecosystems (1); people’s awareness of
conservation issues is heightened (1); may have a wider effect in terms of promoting a more
sustainable lifestyle for those who were in receipt of education programme (1). Disadvantages
resources which could be used to support other organisms are diverted (1); ethical issue to decide
which species should be conserved, and therefore which species cannot be (1); human use of space
is prevented by gating the caves (1); this could, for example, mean preventing people access from
viewing an area of natural beauty (1); human exploitation of lands surrounding caves is restricted (1);
could mean a natural resource is not available for use, or that agricultural practices are not as efficient
as they could be (1).
Summary questions
1 Conservation active management of an ecosystem (1). Preservation protection of an ecosystem
from interference so that it remains in its original state (1).
2 a Suitable example (1); for example, to support a sustainable timber industry
2 b Suitable example (1); for example, to provide an environment for a human population to relax /
exercise
2 c Suitable example (1); for example, all animals have a right to life
3 Suitable conservation example (1); suitable explanation (1); for example, controlled grazing (1); to
prevent succession (1); suitable preservation example (1); suitable explanation (1); for example,
barring of human / animal access using fencing (1); to prevent trampling of rare plant species (1).
24.5
Overfishing of North Sea cod
1 1972 (1)
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OCR A Biology
2 Too few adult cod were left to breed and maintain cod numbers (1)
3 Number of fish allowed to be caught in an area limited (1); maintains fish stocks at a level (that
allows sufficient breeding to occur) to ensure new fish are produced at same rate as those which are
removed from the sea.
4 Immature fish grow and reproduce, allowing the population to recover (1).
5 Scientists monitor population sizes of specific fish species in different areas (1); if the population of
a species becomes too low to become sustainable action is suggested (1); governmental
organisations use scientific data to inform decision making (1); examples of actions which may be
taken include the introduction of fishing quotas (1); bans on catching specific species (1); limits on
fishing times (1); or bans on landing immature fish are introduced (1); this allows time for a fish
population to recover to sustainable levels (1); it is important that decisions are only taken on the
basis of accurate, impartial scientific evidence. Through the use of such data, fish populations are
able to be maintained and hence continue to meet the human demand for food (1) (max 6).
Summary questions
1 A renewable resource – one which is being economically exploited in a way that it will not diminish
or run out (1).
2 Any two from: To preserve the environment (1); to conserve / make available resources for future
generations (1); to maintain biodiversity in an area (1); to allow LEDCs to develop by exploiting their
natural resources (1); to create a more even balance in the consumption of resources between
LEDCs and MEDCs (1).
3 Any two from: Selective cutting is used, so only the largest trees are felled thus maintaining a
forest’s biodiversity (1); felled trees are replaced to maintain tree stocks / biodiversity / soil minerals /
nutrient cycles (1); new trees are planted an optimal distance apart to reduce competition (for light /
space / nutrients) (1); pest pathogens are treated / prevented to reduce plant loss (1); any other
relevant suggestion, explained (1).
4 In one area trees cut close to ground (1); timber is used for fuel / fencing / building / other
appropriate use (1); new shoots appear which are allowed to grow (1); another area of land is then
coppiced (1); when initial trees reach a certain height they are then coppiced and cycle begins again
(1).
24.6
Summary questions
1 Savannah / open grassland – open grasslands with occasional shrub and trees (1)
2 Tourism / ecotourism and farming / grazing (1)
3 Any four from: Tribes of region were traditionally semi-nomadic / equivalent (1); tribes of Masai have
been restricted to using certain defined areas / use areas around the game reserve (1); larger
population density in these regions (1); land farmed more widely / more land used for agriculture (1);
land farmed more intensively / greater reliance on fertilisers (1).
4 Answers must refer to an effect and its impact. At least one advantage and one disadvantage
should be included to achieve full marks. Any six from: People bring economic input into region (1);
which creates jobs / enables region to invest in infrastructure / technology / transport links / other
relevant example (1); visitors see local tribes (1); allows for continuation of traditional culture pastimes
/ educates visitors about their traditional way of life (1); research is carried out in the Masai Mara (1);
providing employment / improved facilities / better conservation approaches / ensuring maintenance
of biodiversity in the region (1); large visitor numbers requires infrastructure (1); can damage
environment / uses natural resources / affects local habitat (1); visitors / safaris require transportation
(1); leads to soil erosion / scares animals / increases pollution (1); other relevant example (1) with
effect (1).
24.7
Summary questions
1 Mixture of fertile agricultural land, with densely forested regions (1); region is hot and humid in the
summer months / subjected to monsoons (in rainy season) (1).
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OCR A Biology
2 Any two from: growing crops / agriculture (1); forestry / felling timber for building products / burning
as fuel (1); tourism (1).
3 Any four from: Local community forestry groups have been established (1); to set harvesting rules /
set agreed prices for timber products / reinvest profits for the benefit of the local population / enable
small businesses to gain FSC certification (1). Sustainable forestry practice has: Increased forested
area / forested density (1); therefore larger region available to support biodiverse ecosystem (1); /
improved water management / provided improved economic income to the region (1). Sustainable
agriculture has: Prevented further intensive agriculture in region (1); therefore retained biodiversity of
less cultivated regions (1); / improved irrigation of land / encouraged multiple cropping / planting of
leguminous crops / encouraged planting of disease / climatic / biotic-factor resistant crops / improved
fertilisation of land (1).
24.8
Summary questions
1 A region of spongy, waterlogged land containing decomposing vegetation (1).
2 Any two from: maintain water levels by reducing drainage from land / blocking draining ditches (1);
removal of tree seedlings from area to prevent water removal (1); / using controlled grazing to
maintain biodiversity of peatland surface (1).
3 To gain full credit, answers should refer to uses of peat itself and changes in surrounding land
usage which have an impact on the peatlands. Peat is used for fuel / as a gardeners’ soil additive (1);
therefore peat has been removed from the land to supply human demands (1); intensive farming /
irrigation / drainage of land / afforestation / other relevant effect has removed water from the wetlands
(1); drying out peatland regions thus leading to degradation or destruction of the peatlands (1).
24.9
Summary questions
1 Any three relevant suggestions, for example, to view rare or endangered species, to experience the
landscape, for recreational opportunities (1).
2 a Islands have never been connected to mainland / no large natural predators exist / almost
complete absence of mammals / other relevant feature (1).
b Extremely cold / dry / extended periods of sunlight or darkness / almost entirely covered by ice
sheet / other relevant feature (1).
3 Named region e.g., Antarctica (1). Suggestion of human activity e.g., whaling (1). Description of
effect on flora or fauna e.g., reduction in population of fin whales (1). Explanation of how human
activity has been controlled e.g., worldwide ban on whaling (1).
4 Named region e.g., Snowdonia National Park – no mark. Answers must be relevant to this region.
Answers should refer to at least one social and one environmental impact. Max 6.
Social (max 4): Visitors bring money (1); which creates jobs (1); enhances local population (1);
reduces rural depopulation (1); allows rural regions to be economically viable (1); enable rural areas
to invest in infrastructure (1); technology (1); supports local businesses (1); transport links (1); other
relevant example (1); (leisure) facilities created for visitors (1); also benefit local people (out of
season) / other relevant example (1); many visitors (1); can cause congestion / overcrowding /
disturbs tranquillity / can bring invasive species / transmit disease / other relevant example (1).
Environmental (max 4): Visitors bring money (1); which can be reinvested in conservation projects /
other relevant example (1); visitors raise profile of a region (1); to enable environmental campaigns to
gain publicity / other relevant example (1); many visitors (1); cause erosion of paths / trampling of flora
/ poaching / removing species from habitat / other relevant example (1).
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24 Populations and sustainability
Answers to practice questions
OCR Biology A
Question
number
Answer
Marks
1 (a)
A coppicing; B pollarding;
2
1 (b)
Trees not felled; continue to grow; all timber
removed regrows;
2
1 (c)
Not all habitats destroyed; e.g. birds; soil
structure maintained; maintains growth of
plants;
3 max
2 (a)
3
3 (a) (i)
Flights to more destinations; cheaper flights;
increase in disposable income;
Increased; large increase from 1992 / 3 to 2002
/3; figures quote;
Advantages increased awareness of ecological
problems; more money into the area;
disadvantages increased waste; increased
pollution; increased damage to ecosystems;
Idea of increased awareness of Antarctica due
to media coverage;
Tour operator more likely to see benefits;
tourists more likely to see benefits; travellers
more aware of threats than tour operators;
scientists have become more aware of the
threats and less likely to see benefits;
Good population has stabilised over the last 20
years; reason for decline has been removed;
not good numbers not increasing;
succession; IGNORE primary / secondary
3 (a) (ii)
mineral content; acidity / pH; water depth;
2
3 (b)
similarity chlorophyll breaks down / leaves
change colour;
differences (bog) minerals stay in plant / (forest)
minerals in soil; ORA
decomposers / fungi / bacteria, not, present /
active in bog; ORA
decomposers / named decomposers, not,
present / active; waterlogging reduces, air /
oxygen; acidity / low pH, stops (decay)
enzymes working;
bog / habitat / ecosystem, takes a long time to
form / hard to replace; loss of, biodiversity / rare
species; ACCEPT peat bogs maintain
biodiversity
3
2 (b) (i)
2 (b) (ii)
2 (b) (iii)
2 (c)
2 (d)
3 (c)
3 (d)
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Guidance
3
5
1
4
3
1
2
2
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Module 6 practice questions
Answers
OCR Biology A
Question
number
Answer
Marks
1 (a)
Weight has reduced; figures quote; length has
reduced; figures quote;
Smallest fish removed mean weight increased;
largest fish removed mean weight decreased;
fish randomly removed mean weight, fluctuated
/ stayed about the same; figures quote;
Smallest / largest, fish removed changed allele
frequency; larger frequency of, small / large,
alleles in next generation; directional selection;
random removal did not change allele
frequency; selection not directional;
Artificial selection; scientists are selection
pressure;
Genome complete set of DNA; of an organism;
operon more than one gene; switched, on / off,
together; expression switching on of a gene;
transcription initiated;
Regulatory genes products control the
expression of other genes; code for, repressor /
enhancer, proteins; products used elsewhere in
the, cell / organism; e.g. enzymes;
Both one regulatory gene; have operator and
promoter; repressor protein binds to operator;
and blocks promotor; lac has three structural
genes and trp has five structural genes;
Tryptophan binds to repressor protein;
repressor protein then binds to operator;
transcription halted; binding of, lactose /
tryptophan, changes 3D shape of repressor
protein;
Eukaryotes have nucleus; ORA DNA free in
cytoplasm in Prokaryotes; idea that transcription
and translation occur in different areas in
Eukaryotes; ORA mRNA has to leave nucleus
to attach to ribosome in Eukaryotes; proteins
contained in vesicles in Eukaryotes; Eukaryotes
have Golgi body; modification of proteins occurs
in Golgi body;
Species evenness is a measure of the number
of individuals in each species; species richness
is the number of different species;
Species richness is the same; species
evenness is higher at site 2; figures quote;
Foredune = 0.55; mature dune = 0.88;
4
More succession has occurred in mature dune;
closer to climax community; greater number of
species present/more individuals in each
species present;
Economical discovery of new drugs; local
communities may depend on forest; ecotourism;
ecological maintenance of biodiversity; ethical
idea of human responsibility; effect on local
communities; aesthetic ecotourism;
3
1 (b) (i)
1 (b) (ii)
1 (c)
2 (a)
2 (b)
2 (c) (i)
2 (c) (ii)
3
4 (a)
4 (b)
4 (c) (i)
4 (c) (ii)
5
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Guidance
4
5
2
6
4
5
4
7
2
3
2
7
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Module 6 practice questions
Answers
OCR Biology A
6 (a)
6 (b)
6 (c)
6 (d)
6 (e)
7 (a)
7 (b)
7 (c) (i)
Require different conditions / AW;
control agents might work / feed on each other
and not pest / AW;
idea of competition;
idea of limited resources;
further detail; e.g. M. calliginosus eats larvae
and so other two would not
be able to survive
AVP; e.g. ref. to cost and management of
releasing all three at same time
Parasitic wasps might be slower to work / AW;
parasitic wasp eggs might take a while to hatch;
parasitic wasps must fly and locate prey so
therefore slower to act / AW;
AVP;
Environmental benefits / AW;
(idea of) bioaccumulation;
pesticides might kill biological control agent;
pesticides might kill beneficial insects / non
target species;
pesticides (might kill pest) and remove food for
biological control agent /
disrupt food chains;
idea of pest resistance (over time); immunity
health benefits;
AVP; e.g. to demonstrate good practice
(weeks 1 and 2) no difference in aphid
population size; presence of natural predators
reduces growth of aphid population; ora
presence of natural predators aphid population
reaches plateau; relevant data quote; absence
of natural predators aphid population peaks
(and declines); relevant data quote;
ref. to cost;
pressure from agricultural companies;
pressure from supermarkets / no aphids allowed
on food / AW;
biocontrol needs to be reintroduced / labour
intensive;
ease of application
biocontrol slow to act;
lack of knowledge for biocontrol of all pests;
biocontrol cannot be used in large fields;
AVP;
(composition of ecosystem) changes; over time;
4 max
Greater number of individuals within each
species; greater number of different species;
more species per unit area; idea of increased
mass of living material; increased number of
heterotrophs; e.g. predators;
instability of, mud / soil;
waterlogged / low oxygen levels in soil / low
nitrate;
varying salinity / high salt concentration / low
water potential;
idea of periodic, desiccation / submergence;
ref to wave action on plants;
6
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2 max
4 max
3 max
3 max
2
2 max
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Module 6 practice questions
Answers
OCR Biology A
7(c) (ii)
Similarity – same, sequence / zonation / pattern,
as move up shore;
Salicornia always at lowest elevation / first in
succession;
Juncus always at highest elevation / final plant
in succession;
all plants found between 4 and 34 cm above
sea level / ora;
species have similar spread on both marshes
(23/24 cm);
elevation range of Salicornia does not overlap
with any other species /ora;
Salicornia only species with non-overlapping
ranges on A and B;
max 1
2 max
Difference – plants found at higher elevations in
salt marsh B / ora;
7 (c) (iii)
7 (c) (iv)
7 (d) (i)
7 (d) (ii)
7 (e)
8 (a) (i)
Limonium found at higher elevation than
Sarcocornia in A / ora;
species mean soil elevations greater range in A
than B (19/15);
max 1
Salicornia;
stabilise soil / develop soil structure;
increase humus content;
raise soil levels;
aerate soil;
decrease salinity / change pH;
provide food;
shelter / form microhabitat;
nitrogen fixation / increase nitrates / increase
minerals;
two from Spartina, Limonium and Sarcocornia;
mark first two answers
light / minerals / named mineral / carbon dioxide
/ pollinators / space;
R water, nutrients. mark first answer in list
transect;
continuous / belt / interrupted / line;
random placement of transect / AW;
use quadrats / point quadrat; R quadrant
measurement of elevation for each quadrat;
percentage cover within quadrat / number of
hits with point
quadrat;
use ACFOR scale;
reliability – large number of quadrats / repeat
transect;
AVP; e.g. use of keys, reference to safety,
calculate mean percentage cover.
sprayed / added to crops (to kill insect pests);
idea of persisting, in / on, plant crop or passed
onto animals (food chain);
ref. to leaching / run off, into water courses /
AW;
water to water treatment plant / used as drinking
water / used as water source;
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1
2 max
1
1
4 max
max 3 if no transect
2 max
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Module 6 practice questions
Answers
OCR Biology A
8 (a) (ii)
ref. to affecting nervous system / nerve agents;
ACh not broken down / synaptic transmission
impaired / AW;
continuous stimulation of post-synaptic
membranes;
AVP; e.g. detail of inhibition
partially permeable membrane;
allows organophosphate to diffuse through;
biological recognition layer;
(contains) immobilised acetylcholinesterase;
organophosphate binds to acetylcholinesterase;
(because of) complementary shape to active
site / binds elsewhere on enzyme;
A ref. to action of competitive / non-competitive
inhibitor ref. to adding acetylcholine (to obtain
product / acetate / choline and
acetate);
(organophosphate binding) product formation
inhibited;
ref. to higher concentration of organophosphate
linked to greater inhibition / less product;
acetate / H+ / choline, detected by transducer /
electrode; converted to electric signal /
impulse / current;
2 max
8 (a) (iv)
any three valid, e.g.
sensitive / detection of small concentrations;
accurate;
quantitative;
rapid results;
portable / can test in the field;
continuous monitoring;
(very) specific;
reusable;
ease of use;
small sample only required;
3 max
8 (b) (i)
ref. to specificity (of immobilised, molecule /
enzyme);
glyphosate, wrong shape / not complementary
shape / doesn’t fit (e.g. at active site) /
different enzyme needed / AW; binding does
not occur;
require enzyme(s) that are, inhibited by /
complementary to, glyphosate;
Advantages
increased yield;
reduces competition / AW;
makes glyphosate selective / AW;
can use weed killer when crops are growing (for
maximum effect);
AVP; e.g. only killing weeds, not crop plants /
crop plants do not die 1
max
Disadvantages
ref. to unknown harmful effects to humans; idea
of ‘superweeds’ / gene transfer;
may encourage overuse of glyphosate;
ref. to effects of above e.g. bioaccumulation,
pollution;
reduction of biodiversity;
AVP; e.g. may affect crop quality
2 max
8 (a) (iii)
8 (b) (ii)
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5 max
allow mps from suitable labelled /
annotated diagram
max 4 if principle of inhibition not
noted (e.g. organophosphate in
river
water sample and acetylcholine
added as substrate)
2 max
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Module 6 practice questions
Answers
OCR Biology A
8 (c)
(both) catalyse / speed up, reactions;
restriction endonuclease enzyme
to cleave / cut out / desired gene / AW;
ref. to recognition of specific sequences;
ref. to ‘sticky ends’ / staggered cuts;
to cleave / cut plasmid;
for gene insertion; 3 max
DNA ligase
seal sugar-phosphate backbone;
formation of phosphodiester bonds;
(desired) gene spliced / sealed / AW, into
vector / plasmid; forms recombinant DNA
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4 max
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Synoptic concepts
Answers
OCR Biology A
Question
number
Answer
Marks
1 (a) (i)
Amino acids;
1
1 (a) (ii)
Proteins;
1
1 (a) (iii)
Peptide (bonds);
1
1 (b)
Stimulate / inhibit; action potential; across
synapse / described;
Produced by fermentation; using bacteria; large
scale;
Growth rate higher in fed-batch (1); ORA rate of
glutamate production higher in fed-batch (1);
ORA figures quote (1); difference in growth rate
higher than difference in glutamate
concentration (1); ORA reference to similar
trends (1)
Batch all nutrients added at the start; fed-batch
nutrient added later in process; idea of more
control with fed-batch; ORA
Downstream processing;
3
Patients divided into two groups; one group
treated with placebo; patients do not know
whether they are taking placebo; doctors do not
know which patients are taking placebo;
For improves flavour of food; idea of naturally
occurring so not toxic; trials have not shown any
harmful effects; against some people are
sensitive to MSG; idea that use in large
quantities may lead to problems;
(Both) high glutamate content; enhance flavour
of food;
A glycolysis; B Calvin cycle / light-independent
reaction; C Krebs cycle;
synthesis / building up / making more complex
molecule from smaller units; e.g. from diagram
– glucose to starch / cellulose / polysaccharide,
amino acids to proteins, glycerol and fatty acids
to lipids, triose phosphate to RuBP / hexose
phosphate, NH3 + α ketoglutaric to glutamic
acid;
cellulose for cell wall; RuBP carboxylated / AW;
glycerate phosphate to triose phosphate; two
triose phosphates to hexose phosphate; beta
glucose; polymerised / condensation / glyosidic
bonds; phospholipids for cell membranes; triose
phosphate and glycerol; link reaction; acetyl,
groups / coA to fatty acids; add phosphate;
proteins, for cytoplasm / membranes; αketoglutaric acid to glutamic acid / amino acid;
add ammonia; transamination; protein synthesis
/ translation;
Energy required; to synthesise chemicals;
4
1 (c)
1 (d) (i)
1 (d) (ii)
1 (d) (iii)
1 (e) (i)
1 (e) (ii)
1 (e) (iii)
2a
2b
2c
2d
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Guidance
3
4 max
3
1
5
2
3
2
9 max
2
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Synoptic concepts
Answers
OCR Biology A
2e
3 (a)
3 (b) (i)
3 (b) (ii)
3 (c) (i)
3 (c) (ii)
3 (d) (i)
3 (d) (ii)
3 (d) (iii)
3 (e) (i)
3 (e) (ii)
3 (e) (iii)
3 (f)
4 a (i)
4 a (ii)
4 a (iii)
4 a (iv)
4b
4c
light intensity correct references to number of
chloroplasts; thickness of leaves; shape of
leaves; temperature correct references to
optimum temperature of enzymes; water
availability correct references to thickness of
cuticle; surface area of leaves; curling of leaves;
sunken stomata; wind correct references to
roots; loss of leaves;
Innate, genetic; stereotypical; no learning
involved; does not change;
Higher in larger axons; highest in mammals;
lowest in regular axons in insects; figures quote;
Larger diameter means greater surface area;
greater surface area means more channels for
diffusion (of sodium ions) so faster rate of
diffusion; myelination insulates axon except at
nodes (of Ranvier); leads to longer local circuits;
Sodium channels open before potassium
channels; figures quote; so sodium ions diffuse
into axon; resulting in depolarisation of
membrane; and formation of local circuit with
next section of membrane;
Membranes more permeable to potassium ions;
idea that potassium ions leak out of membrane
at greater rate than sodium ions leak into
membrane; so more positive on the outside of
the membrane;
Tetrodotoxin has complementary shape to
sodium ion channels; but not potassium ion
channels;
Diaphragm could not contract (1); so volume of
lungs could not be increased (1); so air could
not be drawn in (1); so no oxygen for respiration
(1)
Different shaped sodium ion channels;
6 max
Change in, base / nucleotide, sequence; in
DNA; e.g. substitution;
Change in primary structure of channel protein;
change in R group interactions; change in 3D
shape; tetrodotoxin no longer able to bind;
Fish with mutation more likely to survive; and
reproduce; pass (mutated) alleles onto
offspring; increase in allele frequency;
Convergent evolution;
2 max
Respiration synthesises ATP; no oxygen supply
for respiration;
ATP concentration decreases as time
increases; figures quote; degree of rigor mortis
increases as time increases; reference to delay;
No ATP; ATP required to break (Actinomyosin)
cross bridges; muscles remain contracted;
ATP not required for contraction; anaerobic
respiration; idea of residual nervous activity;
Lactic acid; produced by anaerobic respiration;
2
Decomposition; enzymes breakdown proteins;
enzymes breakdown Actinomyosin bridges;
4
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2 max
4
4
5
3
2
3 max
1
4
4
1
4
3
3
2
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Synoptic concepts
Answers
OCR Biology A
muscles no longer contracted;
4 d (i)
A myosin; B myosin head; C actin;
3
4 d (ii)
ATP;
1
4 d (iii)
A moves to the left;
1
4 d (iv)
Cofactor; changes 3D shape of, enzyme / active
site; activates enzyme;
Myosin head attaches and rotates; detaches;
repeats;
Amino acid sequence / primary structure;
different sequence in each; different R group
interactions;
Energy (from ATP) changes structure of
dyneins / kinesins; energy (from ATP) used to
break cross bridges; dyneins / kinesins, do not
form cross bridges; ORA dyneins / kinesins,
result in movement within cell; ORA
Mitosis / meiosis; cell movement;
3
4 d (v)
4e
4 f (i)
4 f (ii)
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3
3
4
2
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