Answers to end-of-chapter exam-style questions
question answer
number
Movement in and out of cells (page 35)
marks
Characteristics and classification (page 19)
1
C
1
2
B
1
3
C
1
4
Group
Spider
Description of characteristics
Cells with a definite cell wall but
no chlorophyll
Produce spores and cells contain
chlorophyll
Two body parts and eight jointed
legs
Body is made of a single cell,
with a clear nucleus and
cytoplasm
Three body parts and six jointed
legs
Insect
Fungus
Fern
Protoctist
(a) a-3, b-5, c-1, d-2, e-4
(b) i. bacterial cells have no distinct (bound in a
membrane) nucleus
ii. viruses cannot reproduce outside the cells of their
host
4
A: Barbatula barbatula
B: Phoxinus phoxinus
C: Osmerus eperlanus
D: Thymallus thymallus
E: Barbus barbus
4
6
(a)
1 – has single bladders on its fronds
2 – has serrated edges to the fronds
3 – has pairs of bladders on its fronds
4 – has a single, unbranched frond
5 – has a finger-like branched frond
5
(b) Several possibilities – perhaps begin with frond or no
frond
i. turgid means fully inflated/supported by water
ii. water presses against inside of cellulose cell walls
iii. A
iv. B
v. the chip would become turgid as water enters the cells
down the water potential gradient
1
1
1
1
2
2
(a)
5
2
3
4
3
Kilometre – metre – millimetre – micrometre;
2
1000/50 = 20
2
(a) Differences: epithelial cell has no cellulose cell wall/
chloroplasts/central vacuole;
Similarities: both cells have cytoplasm/cell surface
membrane/nucleus
Adapted to their functions – palisade cell
photosynthesises, epithelial cell does not
(b) i. 50µm ii. 70µm iii. 35µm iv. 5µm v. 5µm
7
structure
Liver
cell
Palisade
cell
Cell surface membrane
Chloroplasts
Cytoplasm
√
x
√
√
√
√
Cellulose cell wall
Nucleus
Starch granule
Glycogen granule
Large, permanent vacuole
Mitochondrion
Ribosome on rough
endoplasmic reticulum
x
√
x
√
x
√
√
√
√
√
x
√
√
√
5
a-6, b-3, c-1, d-8, e-5, f-4, g-7, h-2
6
(a) i. D
ii. contains the genetic information/chromosomes/DNA
(b) i. C and E
ii. E
50
50
50
50
50
51
0.08
0.06
0.02
-0.02
-0.06
-0.10
54
53
51
49
47
46
2
2
1
(a) amino acids
(b) keratin/collagen/haemoglobin/amylase and many others
(c) liver; muscle; pancreas
(d) grind up seed in water to produce a suspension/solution;
add biuret solution; shake; check colour change from
blue to purple/violet; compare intensity of colour; wear
safety goggles
1
2
2
4
2
(a) a mean can be calculated, to minimise the effects of any
one result
(b) orange juice 1.9/2.4 × 10 = 7.9 g dm-3; grapefruit juice
1.9/3.1 × 10 = 6.1 g dm-3; apple juice 1.9/9.4 × 10 = 2.0
g dm-3;
3
(a) i. A – substrate B – enzyme C – product
ii. enzyme active site binds to substrates of
complementary shape; stress induced in substrates to
form product; product(s) leaves active site
iii. lock and key because the active site shape (lock) must
‘match’ the shape of the substrate (key)
3
3
(b) i. the shape of starch is complementary to the active site
of amylase but not of protease
ii. Human body temperature – 37oC – is optimum
temperature for these enzymes. Enzymes are not
damaged/denatured at this temperature but are at 73 oC.
1
4
(a) i. Table and ii. pie chart drawn.
Pie chart is more visually clear
(b) i. oxygen ii. Protein iii. Water iv. Fats v. nucleic
acid/DNA vi. Bones/teeth
-
5
(a) i. valid ii. Valid iii. Valid iv. Invalid
(b) to act as a control (i.e. water alone does not digest
starch)
(c) repeat and calculate mean value
(d) maltose/reducing sugar. Heat with Benedict’s Reagent –
positive test is a change from blue to brick-red.
3
4
Organelle – cell – tissue – organ – system – organism
Percentage
change
Enzymes and biological molecules (page 43)
Cells and organisation (page 28)
1
Mean length Mean length
at start
after
Suitable graph
(b) 0.25 mol. dm-3 as this is where there is no net entry or
loss of water by osmosis
(c) makes comparison easier, as original lengths might be
different
1
1
5
1
3
2
2
2
(c) i. this represents the optimum temperature – the
balance between high energy collisions between
substrates and enzymes, and damage/denaturation
4
of enzyme which destroys active site
(c) ii. check labels on axes (pH on x axis)/linear
4
axes/accuracy of plot/title
(c) iii. there is an increase in rate (from 4 to 30) between pH
3 and pH 7, then a decrease to 9 from pH 7 to pH 9
rate of reaction falls as enzyme is denatured (loss
of shape of active site) so fewer collisions between
enzymes and substrate molecules
5
10
Photosynthesis and plant nutrition: part 2 (page 60)
7
1
B
1
2
C
1
3
(a)
i. keep the plant for 48h in the dark
ii. So that the plant has no starch stored at the beginning
of the investigation
(b) by immersing the leaf in hot alcohol
(c) blue-black in green areas, but not in white areas or
under the black paper
1
1
(a) i. palisade layer 47/5 = 9.4 spongy layer 28/5 = 5.6
ii. upper layer of leaf more exposed to light so optimum
position for light absorption
(b) i. measured thickness = 80mm length of scale = 28mm
therefore actual thickness = 100 × 80/28 = 350µm
ii. leaf is thin/large surface area/stomata for absorption
of carbon dioxide
4
2
1
1
2
1
7
(a) A
(b) cell wall; chloroplast; cilium
1
3
8
a – E; b – A; c – D; d – C; e – B
4
9
(a) CELLS – TISSUES – PHLOEM/XYLEM/BLOOD –
ORGAN – SYSTEMS – EXCRETORY SYSTEM
(b) SPECIALIZED – RED BLOOD CELL – DIVISION OF
LABOUR – NERVOUS – ENDOCRINE
(c) PALISADE CELL – CHLOROPLASTS – LEAF –
EPIDERMIS - XYLEM
6
4
5
5
1
1
2
2
2
5
6
7
(a) i. vary distance of light source from plant
ii. Measure the length of oxygen bubble formed in a
fixed length of time
iii. Water bath will stabilise temperature,
hydrogencarbonate solution will maintain optimum
carbon dioxide concentration
(b) The limiting factor is the condition furthest from its
optimum value. This will control the rate of
photosynthesis. In this example, light intensity is the
limiting factor up to 5.0 arbitrary units, then another
limiting factor would prevent a further increase in rate.
2
2
(a) purple
(b) in E – in the dark – respiration exceeds photosynthesis
so carbon dioxide concentration rises and indicator
would be yellow
(c) in D there is no photosynthesis but the snail is respiring
so indicator would be yellow
(d) as controls, to show that no other factor is causing a
change in the indicator
2
2
(a) absence of nitrate causes poor protein synthesis so
limited growth
Absence of magnesium causes poor production of
chlorophyll so leaves are yellow/pale green
(b) as controls, to show that no other factor is affecting leaf
colour or growth
2
8
Bounty
Maltesers
Mars
Milky way
Minstrels
Snickers
Treets
Twix
4
4
2
3
(a) X – capillary/arteriole Y – lacteal Z – epithelium/
epithelial layer
(b) to squeeze gut contents along the intestine/peristalsis
(c) ileum/duodenum/small intestine
(d) amino acids; glucose; minerals such as iron; water
(e) diffusion moves solutes from a high concentration to
a lower concentration, e.g glucose from gut contents
into capillaries; active transport can move solutes
against a concentration gradient, using energy. Carriers
in the epithelium move amino acids in this way.
(a) i. bile
ii. Secreted by the liver and passes along bile duct into
the duodenum/small intestine (ileum)
iii. Enzymes in this region of the gut have an optimum
pH which is alkaline/bile contains hydrogencarbonate,
which is alkaline
iv. Bile emulsifies fats i.e. increases their surface area by
converting large globules into smaller ones. Greater
surface area aids digestion by lipase.
(b) the ‘use’ of the absorbed products of digestion e.g. the
synthesis of proteins from amino acids in the live
1
2
1
1
3
4
5
5
(a)
4
Enzyme
6
7
maltose
amino acids
fatty acids and
glycerol
(b) i. hepatic portal vein
ii. excess amino acids are deaminated to urea and
carbon dioxide or converted to glucose and other
products
1
3
(a) Caitlin – she has eight fillings but Hannah has only four
(b) There may be genetic variation between them;
malnutrition leading to Caitlin having weaker teeth
(c) This removes gender and age from the list of input
variables
(d) The cheek teeth crush food and so the sticky food spends
longer there, and can fit into crevices between the teeth
(e) They may not have had so many cavities, and there
would have been fewer teeth in the milk dentition
2
2
(a)
2
pH
5.0
5.5
6.0
6.5
7.0
7.5
8.0
9.0
Time
taken
20
15
8
4
1.25
1.25
3
8
0.07
0.13
0.25
0.8
0.8
0.33
0.13
Reaction 0.05
rate
(b) check axes (pH on x axis)/axes linear/labels on axes to
include units/accuracy of plot
(c) 7.0 – 7.5
(d) mouth/ileum – presence of sodium bicarbonate
(e) iodine test for starch/Benedicts test for reducing sugar
(f) temperature/concentration of enzyme/concentration of
starch solution
D
1
(a) A: 5.0; B: 0.2; C: 5.0; D: 0.4; E:0.25
(b) if stomata are open for carbon dioxide absorption the
plant will also lose water vapour by diffusion hence the
plant loses more water in high light intensity
(c) B will have a low diffusion gradient, E has low humidity
but stomata are closed in the dark
(d) C has low humidity as wind moves humid air away
from leaf, so high rate of water loss, but D also has low
humidity but the stomata are closed in the dark
6
3
(a) OSMOSIS – SUPPORT – SOLVENT/TRANSPORT
MEDIUM – PHOTOSYNTHESIS
4
(b) ROOTS – ROOT HAIRS – SURFACE AREA – IONS/
MINERALS – MAGNESIUM – NITRATE – ACTIVE
TRANSPORT
(c) XYLEM – PHLOEM - VASCULAR
7
(a) check height of bars/bars not touching/correct labels
on axes
(b) warm, bright, windy – all move humid air away and
keep stomata open
(c) check that bars are lower, as rolled leaves have a
reduced access to stomata and keep a humid layer close
to leaf surface
(d) evapotranspiration/evaporation
4
2
2
3
4
3
1
Circulation (page 108)
End products
amylase
protease
lipase
3
4
6
2
DIARRHOEA; ILEUM; COLON; CHOLERA;
ANTIBIOTICS
starch
protein
fat
1
1
3
4
Food
D
C
2
3
2
1
1
1
2
1
2
(b) i. check accuracy of grid from Paul’s notes
ii. Bread, biscuits
iii. Plant protein
iv. Loss of villi reduces surface area for absorption
4
1
1
1
1
2
5
1
1
2
2016
1816
1960
1775
2016
2476
2016
Transport in plants (page 94)
3
(a) i. cucumber
ii. (900) + (50/100 × 1600) + 10/100 × 3000) = 2000 kJ
2
818
(b) bar chart – bars should not be touching
(c) i. Bounty
ii. Treets
(d) 1180
(e) 196.7 minutes; used to maintain temperature/keep nervous system
working/digest food for example
(f) 39 minutes
(g) more – he could use some stored energy e.g. glycogen from
the liver
2
Human nutrition and health (p.83)
1
(a)
B
C
3
B
1
4
(a) A – left atrium; C – left ventricle; E – right atrium;
(b) valves prevent the backflow of blood from ventricles to
atria
(c) i. F
ii. D
3
1
5
(a) plasma= 55%, white cells = 1%, red cells = 44%;
(b) hormones/urea/carbon dioxide/minerals such as Na+
(c) i. B
ii. A
iii. C
iv. A
3
2
1
1
1
1
6
(a) it falls (from about 17 kPa to about 2 kPa)
(b) blood must be forced along the arteries to the respiring
tissues
(c) valves
(d) they are surrounded by sphincters (rings of muscle)
which can constrict or relax to regulate blood flow to
different tissues
(e) one cell thick, so short diffusion distance/have pores
which allow rapid transfer of solutes to tissues
1
1
7
(a) i. 5
ii. fewer than 2
(b) this has eliminated two other variables (gender and
smoking) which might influence the likelihood of
developing CHD
1
1
2
8
(a) first set of data as line graph, second set as bar chart
(check on bar chart that bars are not touching)
(b) i. higher than normal blood cholesterol
ii. 9.9 times more likely (3.2 × 3.1)
4
(a) smoking/lack of exercise
(b) high fat diet/too much salt
(c) males more likely to develop CHD
1
1
1
2
2
1
9
4
1
2
2
3
1
1
2
1
1
1
1
2
3
1
2
Blood and defence against disease (page 120)
2
1
B
1
2
D
1
3
(a) pathogen
(b) from a surface, such as a table top; from a cooking
implement, such as a knife
(c) by direct contact (contagion); by animal vectors, such as
a mosquito; in droplets in the air; via body fluids, such
as blood or semen
1
2
4
4
5
6
7
8
(a) skin/mucus secretion/lysozyme in tears
(b) in colostrum
(c) i. B lymphocytes produce plasma cells which release
specific antibodies; produce memory cells which
can release antibodies more quickly on subsequent
contact with the same antigen
ii. T HELPER cells coordinate action of lymphocytes
against pathogen
(d) method B provides a protective level of antibodies more
quickly
3
2
(a) urea, from liver to kidneys; carbon dioxide, from
respiring tissues to lungs
(b) 8000 – may rise due to infection
(c) 5 000 000: 8 000 = 625:1
(d) car exhaust; inefficient burning of gas in heaters;
cigarette smoke
(e) it falls, as red blood cells/haemoglobin bind to carbon
monoxide
(f) 120 days, the time taken to replace worn out red blood
cells
(g) liver contains a high concentration of haemoglobin,
which contains iron
4
(a) this represents the time taken for antigen to trigger
formation of specific antibody-producing cells
(b) the second response is faster/greater; because memory
cells are present to secrete the antibody
(c) antibody levels may decrease below the effective level;
the person may be exposed to the same antigen in the
future
2
(a) antigen X; it has a shape complementary to the arms of
the antibody
(b) i. the antibody only recognises one antigen; only has
the complementary shape to one antigen; the chicken
pox pathogen may have differently shaped antigens
on its surface
ii. antigens may resurface from another source (e.g.
another country); the pathogen may mutate so that it
cannot be recognised by the specific antibody
2
C
1
6
2
2
2
1
2
2
2
1
2
2
3
3
2
2
(a) graph plotted appropriately – time on x axis
(b) Alan – he can get by with fewer breaths during periods
of exercise
(c) 8:1 (4:0.5)
(d) 30 breaths per minute (each one takes 2 seconds)
(e) Increases both rate and depth of breathing
(f) Heart rate would increase to a maximum as exercise
levels increased
(g) More oxygen (and glucose) are delivered to the
respiring muscles
5
3
(a)
2
Rate
4
2
1
2
5
2
2
55 70 80 90 120 140 150 170
Output per beat 73 69 65 62 50
43
39
27
(b) accurate plot as directed
(c) i. rising to peak then falling,
ii. Output per beat falls as rate increases
(d) i. there is no benefit in terms of total cardiac output,
ii. Output per beat may be higher (heart muscle is more
elastic in trained athletes)
(e) it is less likely that heart muscle will become fatigued/
low rate means that there is further to go before the
peak rate is reached
(f) to supply more oxygen to be transported by the blood to
the respiring tissues
4
1
1
2
1
2
6
2
Gas exchange (page 138)
1
2
3
4
5
PULMONARY; RESPIRING; ENERGY; CARBON DIOX10
IDE; HYDROGENCARBONATE; CAPILLARIES; THIN;
SURFACE AREA; ALVEOLI; CARBON DIOXIDE; OXYGEN;
DIFFUSION; PULMONARY; LEFT ATRIUM
(a) – inhaled air, as it represents atmospheric air without
any oxygen removed
(b) Nitrogen is inert so there is no removal of this gas by
the cells of the body
(c) respiration
(d) water evaporates from the moist surfaces of the alveoli
2
(a) check labels on axes/accuracy of plot/correct scales/full
use of available grid/key to identify three different flasks
(b) respiration
(c) this is because germination requires water
(d) boiling kills the seeds/denatures respiratory enzymes
in the seeds
(e) microbes on the surfaces of the seeds are killed by the
disinfectant, so less respiration
(f) as a control, to show that only the living cells can respire
5
(a) The diaphragm
(b) They will inflate
(c) Exhalation/expiration
(d) No intercostal muscles shown
(e) 1 – rib; 2 – sternum; 3 – (external)intercostal muscle;
4 – backbone/vertebral column
1
1
1
1
4
(a) 45 (100 – 55)
(b) Either – the pie chart makes a visual comparison more
straightforward
(c) i. normal tissue has fewer large cavities/no ‘broken’
walls
ii. Reduced surface area of emphysematous lung less
able to absorb oxygen (also, walls are less elastic)
2
4
4
1
1
2
2
2
Excretion and homeostasis (page 150)
2
1
2
Breathing and exercise (page 133)
1
(a) graph plotted appropriately – temperature on x
(horizontal) axis
(b) to prevent entry of oxygen by diffusion
(c) (anaerobic) respiration
(d) glucose
alcohol + carbon dioxide (C6H12O6
2C2H5OH + 2CO2 would be chemical equation)
(e) the enzymes in the yeast cells are denatured, and so no
respiration is possible
(f) they would use up all of the sugar, and they would be
subjected to toxic concentrations of alcohol
1
1
1
7
1
2
2
2
1
(a) This is a condition in which the body temperature falls
significantly below its normal level
(b) Old people have less efficient circulation, and also their
temperature control systems work less well
(c) Children have a high surface area to volume ratio, so
they lose heat more quickly than they can produce it
1
(a) any change in body temperature is detected by a sensor,
and then corrective actions (shivering/sweating;
constriction of skin blood vessels/dilation of skin blood
vessels) return the temperature to normal body
temperature
(b) shivering is rapid muscle contraction. This generates
heat from respiration (and some friction)
(c) liver is the main metabolic centre of the body, so
generates a great deal of heat (especially from
respiration)
(d) in negative feedback a change from the optimum/norm
causes actions which correct (cancel out) the original
change
4
(a) Show artery connected to aorta, vein to vena cava and
tube carrying urine to the entry to the bladder
(b) Rejection occurs because the body detects that the organ
does not belong to the recipient, and the immune system
produces more cells which attack the transplant
i. 1500 cm3
ii. The kidney tubule can select how much water is
reabsorbed depending on blood water potential due to
water losses from sweating more or less.
2
(a) It passes by diffusion i.e. down a concentration gradient
into the dialysis fluid (the dialysis fluid has a very low
urea concentration)
(b) The transplanted kidney will carry out all of the
functions of a normal kidney, and will not run the risk
of infection that there is with an external connection to
a dialysis machine.
1
(a) i. A – kidney cortex B – medulla C – renal artery
D - renal vein E – ureter
ii. C delivers blood (containing high concentration of
urea) to the kidney and D takes blood (less urea) back
into the circulation
(b) Ammonia is very soluble, and so affects water balance.
Ammonia is also very toxic – it can denature blood
proteins
(c) i. protein molecules are too large to cross the
membrane in the Bowman’s Capsule
ii. These solutes are valuable and so are reabsorbed in
the kidney tubule
iii. Water is reabsorbed from the filtrate so the
concentration of urea and mineral ions rises
iv. Urea has been formed in the liver and so is present
in the blood arriving at the kidney
(a) i. so that thermometers recorded temperature of water
and not glass
ii. The aluminium foil around tube B reflects heat back
into the tube
(b) i. baby’s body generates heat; foil
blanket reflects heat back to baby’s body and so heat
loss is reduced
ii. Water evaporates from clothes; using heat from body;
body temperature falls; hypothermia possible
(c) sweat is evaporated from skin surface; heat is lost
(d) i. ADH changes the permeability of the collecting ducts
so that more water can be reabsorbed
ii. Negative feedback means that a change from the
norm (such as a drop in blood water) sets off
corrective measures to return the condition to normal
(a) the maintenance of a constant internal environment (by
a series of negative feedback mechanisms)
(b)
Name of organ
Factor controlled
Kidneys
Liver
Lungs
Blood urea concentration
Blood glucose concentration
Carbon dioxide concentration in
blood
Body temperature
Skin
1
1
1
2
2
2
2
2
2
5
2
2
2
1
2
2
1
2
2
2
3
3
2
1
2
(c) Body temperature rises……..temperature change
4
measured in hypothalamus…….motor neurones
increase activity of sweat glands…..sweat secreted………
sweat evaporates……body temperature returns to
normal
2
2
3
8
(a) 5000 (1994) – 1300 (1978) = 3700. 3700/1300 × 100 =
284%
(b) increasing gap between number waiting for transplants
and number of transplant tissues available
(c) lymphocytes/phagocytes
(d) recognise transplanted ‘foreign’ tissue and begin to at
tack/digest tissue surface
3
4
1
1
2
Coordination and response (page 164)
1
A
1
(a) Time on x axis; key/label for plots; activity periods
shown; accuracy of plot
(b) exercise raises blood glucose level, to provide glucose
for muscle respiration; insulin levels fall to limit removal
of blood glucose for storage as glycogen
(c) adrenaline – dilates blood vessels and increases rate/
depth of breathing; testosterone – increases aggression
(d) i. reduces effects of ‘rogue’ results;
ii. Eliminate starting blood glucose level as a variable
(e) half an hour
D
1
CORNEA LENS RETINA IRIS(DIAPHRAGM) PUPIL
RODS BLACK AND WHITE CONES COLOUR HIGH
RETINA INVERTED SMALLER OPTIC
INTEGRATION
6
(a)
2
(b) As line graph or bar chart. Number of cups drunk on
x-axis. Suitable scale. N.B. check that mean heart rate
data is to correct number of significant figures.
(c) Yes – heart rate increases as number of cups drunk
increases.
(d) People same age/gender/fitness. Same level of activity.
Coffee always made up of same concentration (of
caffeine). Coffee at same temperature.
4
5
(a) connector/relay neurone. Links sensory input with
motor output; allows modification of motor output by
‘higher’ centres of CNS.
(b) synapse
(c) i. electrochemical impulse
ii. Chemical/neurotransmitter
(d) spinal: knee jerk; hand withdrawal from heat,
cranial : blinking; salivation (normal); pupil reflex,
conditioned : salivation (substitute stimulus)
(e) i. no sensation
ii. still have sensation
(f) yes, until cut was complete a sensory input would
continue
(g) C – dendron (accept axon); D – myelin sheath;
E – cell body; F – axon; G – terminal dendrites.
No receptor on motor neurone, ends at effector, cell
body at CNS end not middle.
2
(a) i. A – NO; B – YES; C – NO; D – YES;
ii. Enzymes involved in hydrolysis of food stores and
in respiration will have an optimum temperature
closer to 30oC
(b) i. labels/units on axes; time on x axis; accuracy of plot;
good use of space;
ii. 19
iii. 30
iv. Hydrolysis of food stores to supply growth materials/
energy sources before photosynthesis begins
(a) Ciliary muscles/suspensory ligaments
(b) Independent is distance from eye, dependent is
thickness of lens
(c) Amount of light in room (could affect pupil size which
might have an effect on lens shape)/size and colour of
pencil (must make sure that the distance is the only
independent variable
(d) Repeat the measurements at each of the distances, and
take a mean of the measurements
2
2
(a) A – retina; B – iris; C – tear duct; D – choroid;
E – conjunctiva; F – sclera (all appropriately labelled on
drawing)
(b) i. The retina will not be damaged/bleached by the high
light intensity;
ii. Light – retina – brain – iris muscle
6
(a) 55 years
(b) Will be less able to accommodate to objects from
different distances, so some images on retina will be
blurred
(c) Once over 60, there is little change in the ability to alter
the shape of the lens
(d) Cornea – accommodation will not be so efficient
(e) Use model lenses, or lenses removed from animals
slaughtered for food. Amount of smoke reaching lens is
the independent variable, the measured hardness of the
lens is the dependent variable. A suitable control would
be a lens not subjected to cigarette smoke (to show that
it doesn’t harden without the smoke).
1
1
5
6
7
8
Number of cups drunk
0
1
2
3
4
5
Mean heart rate
73
75
79 79
83
89
Hormone
Effect
Adrenaline
Development of sperm
Insulin
Secreted if blood
glucose level falls
Increase in heart rate
Testosterone
Oestrogen
Glucagon
B
1
C
1
3
D
1
4
(a) A – ovary; B – pollen (grains); C – filament; D – anther; 7
E – stamen; F – style; G – stigma
3
(b) wind-pollinated, because the stamens and stigma are
exposed to the environment. Petals do not appear to be
brightly coloured.
3
6
1
1
1
5
3
(a) i. A ii. C iii. D
(b) show tube from stigma, growing down through style
and entering at micropyle
3
2
8
(a) transfer of pollen from the anther/stamen of one flower
to the stigma of another flower of the same species
(b) increases the possibility of cross-fertilisation/genetic
variation
2
1
2
2
2
1
2
2
1
2
Human reproduction (page 204)
1
2
1
C
1
2
A
1
3
(a) Fertilisation
(b) i. Semen (a body fluid) can contain HIV; semen can
enter vagina if no (condom) barrier
ii. HIV targets cells of the immune system; reduced
immunity increases risk of infection by other
organisms
(c) i. Gonorrhoea
ii. Pain when passing urine/creamy discharge from penis
or vagina; course of antibiotics
1
(a)17 days;
(b) menstruation
(c) i. oestrogen – stimulates development of secondary
sexual characteristics
ii. progesterone – prepares lining of uterus for
implantation and pregnancy
(d) increase in progesterone concentration after day 16
(e). decline in concentration after day 20-22
(f). thickening of lining; improved blood supply to lining
1
1
1
(a) FSH stimulates the production of ova in the ovaries/
stimulates the production of ova from the Graafian
Follicle
(b) i. ova cannot reach the most likely site of fertilisation in
the uterus
ii. FSH stimulates ovulation; ova collected from head of
oviduct; ova transferred manually to uterus;
fertilisation occurs in the uterus
2
(a) it is not losing heat to the environment/it has a high rate
of respiration, which generates heat
(b) i. Fat is an excellent insulator against heat loss;
ii. This prevents blood reaching the surface of the skin,
where heat can be lost to the surroundings
(c) i. change in temperature is detected by sensor –
thermostat controls action of heater – more/less heat is
released to return temperature to normal;
ii. Prevents heat loss to the environment/prevents baby
drying out/allows nurse/mother to view the newborn
baby
(d) i. carbohydrate – supplies energy/ protein – supplies raw
materials for growth;
ii. Antibodies – baby may have less immunity to child
hood diseases
1
1
2
3
4
4
5
6
Produced by the
pancreas as blood
glucose level increases
Deposition of fat in the
breasts
6
1
1
2
7
3
ADRENALINE – GLYCOGEN – GLUCOSE – OXYGEN –
DEEPER – FASTER – GUT – MUSCLES – PALES – DILATE
– STANDS UP – FIGHT - FLIGHT
4
2
7
4
4
2
(a) i. Y = (200+280+260+260+340+250+240+270+250+290)/
10 = 264; X = (400+350+420+610+640+600+340+460+
600+520)/10 = 494
ii. 494 – 264 = 230 mm
iii. soil water content/light availability/mineral (named)
availability
iv. new length would be less than 264, as new plant is
smaller and so can photosynthesise less to supply
material for growth of next runner
(b) runner must break/wither away
(c) asexual reproduction supplies very little variation i.e.
offspring are identical to parents
(d) i. photosynthesis – light energy converts carbon dioxide
and water to sugar and oxygen. The sugar condenses
to sucrose for transport.
ii. phloem
1
1
2
Male secretes more testosterone at puberty. Testosterone
leads to aggressive/territorial behaviour. As testosterone
concentration in blood rises, it inhibits production of
further testosterone. As a result, testosterone concentration
slowly falls (as it is removed from the blood) and so the
male does not become too aggressive.
2
1
1
1
1
a-3, b-4, c-1, d-5, e-2
2
2
2
Hormones (page 169)
1
2
Plant reproduction and growth (page 186)
2
3
4
4
4
2
2
1
2
1
1
1
2
2
4
1
1
4
2
2
2
7
(a) i. maternal – V; fetus – Y;
ii. T is the umbilical cord. It shrivels and the remains
fall off after birth.
iii. Fetus receives oxygen, soluble foods and antibodies,
fetus excretes carbon dioxide and urea
(b) oestrogen – maintains female secondary sexual char
acteristics, including development of mammary glands/
at birth makes uterus more sensitive to oxytocin (which
leads to contraction of uterine muscle); progesterone –
maintains the lining of the uterus and the blood supply
to the developing fetus/level falls in preparation for
birth
5
2
1
3
4
6
Inheritance (page 226)
1
a-2, b-5, c-4, d-1, e-3
4
heterozygote
genotype
dominant
GENE
MEIOSIS
HETEROZYGOUS
DIPLOID ZYGOTE
A O
RECESSIVE 5
B O
(a) parental genotypes I I × I I
Gametes IA IO
IB
IO
Offspring genotype IOIO
3
(b) heterozygous genotype IAIO or IBIO
Codominant alleles IA and IB
3
7
Phenotype blood group O (blood group A and blood group
B in parents would be acceptable answers)
5
6
1
ii. check axes – mass on x-axis, number in group on
y-axis. Bars touching.
(c) continuous
4
(a) artificial
(b) sow crop in water-deficient area; allow to grow; collect
seeds from those which grow well; sow next crop from
these seeds and repeat selection process
(c) i. identify gene which codes for the desired
characteristic; insert gene into vector – using
restriction enzyme to open vector DNA and DNA
ligase to stitch gene into vector DNA; use vector to
transfer gene into new host; grow host and test for
desired characteristic;
ii. May transfer gene into unsuitable species; modified
species may become a pest; licence holders may
control availability of modified variety
1
3
Some rats have a mutant gene which gives them resistance
to warfarin - this is natural variation in a population.
These rats are unaffected by the poison so can continue to
survive and reproduce when other rats die - survival of the
fittest.
Because these rats breed more successfully the gene for
resistance becomes more common.
3
3
3
(a) A carrier is heterozygous i.e. has the mutant allele but
does not express it in the phenotype e.g. Nn would be
a carrier
(b) Parental genotypes Nn Nn ; gametes N N n n;
offspring genotypes Nn Nn nn nn i.e. 50% chance
of a child being homozygous recessive.
1
(a) i. XhY
ii. XhXh
iii. XhX
(b) Queen Victoria – she must be heterozygous and Prince
Albert cannot be carrying the Xh allele or he would be
haemophiliac
(c) by mutation or by random fertilisation - Xh allele is still
within the family
(d) it must be recessive as Queen Victoria has the allele and
is not haemophiliac
1
1
1
3
Ecosystems, decay and cycles (page 258)
2
small bird
fox
3
caterpillar
aphids
8
5
1
1
2
1
2
4
5
1
0
1
4
5
9
11
5
1
1
(a) A – ovum/egg cell C – zygote E – cell in embryo
(b) B – fertilisation D – mitosis
F – differentiation/
specialisation
(c) G – muscle; H – pancreas (islets of Langerhans)
I – neurone/nerve cell
3
2
2
Number of individual nuts
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
The sum of the alleles on
the chromosomes in the
nucleus
A form of a gene that
codes for one of a pair of
contrasting characters
An allele which, if present, always has an effect
An organism with two
different alleles for a
particular characteristic
The sum of the measurable characteristics of an
organism
phenotype
4
(a) i. to eliminate ‘species’ as a variable – different species
may naturally have different sized seeds
ii. To avoid any bias in collecting the seeds
Mass of nut/g
Definition
allele
3
4
2
(b) i.
Term
2
(a) i. check that bars do not touch/blood group type on X
axis
ii. this form of variation – discontinuous – is a result of
GENES, so that there will always be a series of
discrete bars
(b) i. change in the type or amount of DNA
ii. ionising radiation /some chemicals e.g tar
in tobacco smoke
(a) (from top of web)
hawk
fungus
1
4
2
4
rabbit
green plant
Variation and evolution (page 238)
1
D
1
(b) the arrows represent the flow of energy
1
2
D
1
3
(a) The process in which the best adapted members of a
2
species survive, breed and pass on their genes to the
next generation
(b) i. overuse of antibiotics selects those cells which have
2
a natural resistance (NOT immunity!) to the
antibiotic so that these resistant cells can reproduce
and form a resistant population
ii. DDT killed most of the mosquitoes, but those with
2
natural resistance survived, reproduced and passed
on their gene for resistance to successive generations.
DDT resistant mosquitoes were then able to transmit
the malarial parasite (Plasmodium) and cause
malaria
(c) i. aphid/caterpillar/rabbit
ii. green plant
2
1
(d) fewer rabbits might mean less of the green plants are
eaten (so more food available for aphids/caterpillars),
and would mean that less food is available for foxes, so
the number of foxes would decline
2
2
(a) i. A – combustion B – respiration C – photosynthesis
D – nutrition/feeding
ii. cellulose
(b) increased combustion of fossil fuels increases carbon
dioxide concentration/removal of forests takes away the
trees which might otherwise reabsorb carbon dioxide
4
(a) i. haemoglobin can bind to and release oxygen. It is
responsible for the transport of oxygen from lungs to
respiring tissues
ii. iron
(b)
P
INIS
x
INIS
3
A – grass seed-mouse-weasel; B – oats-rabbit-flea; C – oak
tree-aphid-ladybird; D – cabbage-caterpillar-wasp parasite
4
4
(a) A – denitrifying B – nitrogen fixation C – nitrifying
D – nitrifying E - decay
(b) by the addition of inorganic fertilisers
5
5
(a) P – evaporation R – transpiration
(b) temperature – increases rate of evaporation from
internal leaf surfaces; wind speed – blows humid air
away from leaf; humidity – low humidity maintains
diffusion gradient for water vapour
(c) support of cells; raw material for photosynthesis
2
4
4
gametes
IN
F1
ININ
IS
IN
INIS
INIS
1
1
4
IS
ISIS
Probability that child will be heterozygous is ½ or 50%
(c) this advantage is noted in areas where the malarial
parasite is common. People who are heterozygous only
have mild anaemia, and are protected from malaria
since the malarial parasite cannot multiply in sickleshaped red blood cells.
3
5
1
2
1
2
6
Term
Ecology
Community
Habitat
Population
Environment
4
Definition
9
An area that can supply
food, shelter and a
breeding site
The study of living things in
relation to their
environment
The physical and biological
conditions that are present
in the place where an
organism lives
The living organisms of
different species which live
in a particular habitat
The number of individuals
of a particular species
present in a particular
habitat
Biotechnology (page 299)
1
(a) Penicillium (notatum)
(b) i. to provide optimum temperature/avoid denaturation ;
for the action of enzymes in the Penicillium;
ii. Respiration of the Penicillium generates heat; the cold
water jacket carries excess heat away
(c) to prevent contamination by other microbes; to avoid
competition for nutrients
(d) the common cold is a viral infection, and penicillin has
no effect on viruses; overuse of antibiotics may lead to
selection of antibiotic-resistant strains of bacteria.
1
2
2
2C2H5OH + CO2
(a) C6H12O6
(b) alcohol is a toxin; alcohol poisons the yeast
(c) the fermentation lock prevents the entry of oxygen from
the atmosphere
3
3
1
3
(a) beef has less carbohydrate/more fat/more cholesterol/
less fibre/more protein/more total energy
(b) mycoprotein has less fat/cholesterol so less likely to
cause CHD; mycoprotein has more fibre so less
likelihood of constipation/bowel cancer
2
a-2, b-4, c-1, d-5, e-3
Human impacts on ecosystems (page 288)
1
(a) i. flooding/earthquakes/forest fires
ii. Drought/salination of soil
(b) increase in population means more people to feed and,
perhaps, less land on which to grow food
(c) e.g. wheat: identify plants with more/larger seeds;
self-pollinate these plants; collect seeds; grow plants;
repeat process of selection/growth
(d) transfer of DNA from one organism to another
2
1
2
(a) industrial area – 27.8%; agricultural area – 5.2%
(b) dark form has higher survival potential when
camouflaged on soot-covered trees; can breed more
effectively; pass on genes for ‘camouflage’; successive
generations have altered frequency of dark form; has
lower survival potential when obvious on light coloured
trees; cannot breed effectively; does not pass on genes
for ‘camouflage’; successive generations have lowered
frequency of dark form
(c) less selective advantage in being dark, so frequency will
fall as breeding success is limited
(d) Biston (must have capital B)
2
4
(a) 8000km per year
(b) i. variety would fall
ii. More arable land would be available
(c) food may be more easily imported
(d) advantages – harvesting is easier, as all crop available at
the same time; all crop will need the same management
(e.g. timing of fertilisers) disadvantages – loss of
biodiversity (fewer habitats available); all crop
susceptible to the same pests/diseases
(e) A, C, D, H, I
(f) nitrate – required for protein synthesis; magnesium –
required to produce chlorophyll, which is necessary for
light absorption for photosynthesis
1
1
1
1
4
(a) some form of filtration
(b) hypothesis – pesticide will affect the numbers/
reproduction of insects; prediction – increased volume
of pesticide will reduce number of insects
(c) concentration of pesticide
(d) number of aphids alive after 24 hours
(e) they used the same species of plant and the same initial
number of aphids
(f) concentration on x axis; number alive on y axis; axes
labelled with units; clear plot using large proportion of
grid
(g) 22 g per 1000 dm3
(h) little effect until 15–20 g per 1000 dm3, and little
further effect after 30 g per 1000 dm3 so prediction only
partly supported
2
2
5
(a) washed from agricultural land, where they had been
sprayed to control pests on crop plants
(b) D – A – E – G – C – F - B
2
6
(a) bars not touching; bars drawn accurately; axes correct –
right way round and labelled
(b) 250
(c) December – less food available and low temperatures
increase food demands, and March – greater
competition for food in breeding season
(d) population = 100 × 80/25 = 320
(e) provide more breeding/feeding sites; control grey
squirrel population
4
(a) 53; 11; 13.8; 13.9; 8.3
(b) pollutant type on x axis; mass on y axis; units; bars not
touching
(c) e.g. nitrogen oxides ; human health – breathing issues;
environment – contributor to acid rain
2
4
2
3
4
7
8
(a) distance on x-axis, percentage oxygen concentration
on y-axis, number of organisms on y axis
(b) bacteria present in human sewage
(c) algae decline as bacteria rise because of competition for
nutrients and loss of light by shading
(d) adults lay eggs before they die, then eggs hatch
(e) oxygen concentration falls to minimum at 200m, then
rises again to 100% by 1000m
(f) oxygen concentration falls due to aerobic respiration by
bacteria then rises as bacterial population becomes more
dispersed/more oxygen available as algae begin to
photosynthesise again
2
2
2
2
2
Genetic engineering (page 305)
1
1
2
1
1
2
4
2
2
6
1
2
2
2
2
(a) algae
shrimps
fish
sea otters
(b) pesticides are absorbed by algae, then amplification
occurs so that otters receive highly concentrated
pesticide
6
1
1
2
2
2
2
1
(a) i. gene/DNA
1
ii. A surface antigen
(b) whole virus may remain active, and cause disease; whole 3
virus will have other antigens which may provoke severe
immune response
3
2 - 3 - (1) - 6 - 4 - (8) - 7 - 5
1
5
4
(a) i. (restriction) endonuclease;
ii. (DNA) ligase
(b) to act as a vector/to carry the inserted gene from one
organism to the host bacterium
(c) cheaper to produce/ human insulin so should not
provoke immune response/can be generated when
required
(d) engineered bacteria could escape from laboratory/
factory with unpredictable consequences; companies
may ‘own’ the new gene and its product, so not
available to people who need the product
5