257Answers

Answerstype of organism concentration of DDT

in cells (ppm)

plant planktonanimal planktonminnowgrebe

0.040.160.966.00

Table A.1.1 (2)

c) (i) The higher up (further along) the position of the organism in the food chain, the higher the concentration of DDT in its cells.

(ii) Grebe. Its cells contained the highest concentration of poisonous DDT. (3)

d) Because it is a poisonous substance that does not break down into harmless substances in the environment. (1)

e) A food chain takes the form of a pyramid of numbers. The animal at the top of the pyramid eats many of the organisms beneath it. If these contain molecules of non-biodegradable DDT, an accumulation of this poison takes place in the cells of the final consumer. (3)

f) (i) The grebe eats animals which have already accumulated many molecules of pesticide in their cells; the moorhen eats plants which have only taken in a few molecules of pesticide.

(ii) Unlike grebes, gannets are sea birds and the animals that they eat live in the sea. The sea is not as severely affected as freshwater environments which receive DDT directly from nearby farmland. (4)

Section 2

1 a) To move the animal forward. (1)b) It is able to feed on organic substances present in the

water. (1)c) It will be able to make food by photosynthesis in light.

(1)d) Chloroplast. (1)e) Cell wall. (1)f) Euglena contradicts this idea because it has features

typical of both plants and animals. (1)

2 10 micrometres. (1)

3 a) 1 5 C, 2 5 A, 3 5 B. (2)b) The cylinder pinned to cork 1 increased in size so it

must have been in tube C where it gained water by osmosis from a region of HWC. The cylinder pinned to cork 2 decreased in size and became soft so it must have been in tube A where it lost water by osmosis to the very concentrated sugar solution (LWC). The cylinder pinned to cork 3 remained unchanged

Section 1

1 a) 7.1 ppm. (1)b) Loch 4. (1)c) Species H. (1)d) Loch 3. (1)e) 6. (1)f) (i) The number of species of snail decreases as the

concentration of calcium in the water decreases. (ii) As the concentration of calcium decreases, less

is available for use to make shells and therefore fewer and fewer species of snail can survive. (2)

2 a) The total number of spiders decreases as the distance from the ground increases. (1)

b) Species Y. (1)c) Moving upwards from site 1 to site 4, the relative

number of species X increases and Y decreases. (1)d) Most of species X will gather at end B and most of

species Y will gather at end A. End B is dry and like sample site 4 where most of X were found; end A is moist and like site 1 where most of Y were found. (2)

3 a) To get rid of fur and bones of prey that they are unable to digest. (1)

b) See figure A.1.1. (3)c) wheat → field mouse → barn owl (1)d) No. Almost all of the prey animals in the table are

small mammals. (1)e) The barn owls help to keep the crops clear of field

mice. (1)

field mouse

smallbird

vole

shrew

Figure A1.1

4 1 5 B, 2 5 E, 3 5 C, 4 5 A, 5 5 D. (2)

5 a) plant plankton → animal plankton → minnow → grebe (1)

b) See table A.1.1.

258 Answers

so it must have been in tube B where the water concentration of the dilute sugar solution was equal to that of the cells. (3)

4 0.115g. (1)

5 When Paramecium is in distilled water, the difference in water concentration (concentration gradient) between inside and outside the animal is very high and the animal gains much water by osmosis. The contractile vacuoles have to work hard to remove all this excess water so the number of pulsations is found to be at the highest level.

On the other hand, when Paramecium is in 1% salt solution, there is only a slight difference in water concentration between inside and outside and the animal gains little water. A few pulsations are enough to remove this. (4)

6 a) C. (1)b) 13 hours. (1)c) 65%. (1)

7 a) By removing a few cells from a living organism and growing them in a nutrient solution. (2)

b) 17 hours and 15 minutes. (1)c) The rate of division increases. (1)d) 64. (1)e) (i) B and C.

(ii) C.(iii) A and D. (3)

f) Nutrition. Food gives the cells the energy and building materials that they need to grow and divide. (2)

8 a) Use an equal volume of flour suspension and an equal number of pieces of snail’s gut in each tube. (2)

b) A. (1)c) A. (1)d) It is possible that the fresh snail’s gut in A contains an

enzyme that digested starch to sugar. It is possible that no breakdown of starch took place in B because the enzyme had been destroyed by boiling. (2)

9 a) 27°C. (1)b) 28cm3. (1)c) 39°C. (1)d) 3.5 times. (1)e) 12°C. (1)

Section 3

1 176 million metric tonnes. (1)

2 a) Beans and rice. (1)b) Sweet corn and rice. (1)c) Cabbage. (1)d) Green peppers. (1)e) 3. (1)f) 5. (1)g) 1.7%. (1)h) 1.1%. (1)i) 20g. (1)

3 a) 21. (1)b) See table A.3.1.

pupil type of plant

floweringplant

seaweed moss fern evergreentree

fungus

1 6 2 3 1 0 0

2 3 0 0 0 1 0

3 10 2 1 0 0 0

4 0 1 0 3 0 2

5 1 0 4 0 0 1

total 20 5 8 4 1 3

Table A.3.1 (4)

4 See figure A.3.1. (4)

4 species of Geranium

white flower coloured flower

red flower purple flower

hairy stem non-hairy stem

Veined Cranesbill

Shining Cranesbill

Bloody Cranesbill Wood Cranesbill

Figure A3.1

5 20%. (1)

6 Make the mass of cotton wool and the number of seeds equal in both flasks and give the one on the right an airtight stopper. (3)

7 Light intensity and temperature. (2)

8 a) 2. (1)b) Light. (1)c) D. (1)d) 4. (1)

9 a) Pink to blue. (1)b) The paper strip may have picked up water vapour

from the air in the room while it was being taken across the room. (1)

c) Place the desiccator beside the plant before taking out

259Answers

the paper strip so that the paper strip is exposed to the air in the room for the least possible time. (1)

Section 4

1 a) 10%. (1)b) 60g. (1)

2 Food eaten by a herbivore is ground up between enamel ridges on the herbivore’s molar teeth by a side to side movement. On the other hand food eaten by an omnivore is crushed between rounded cusps on the omnivore’s molar teeth by an up and down movement. (4)

3 a) (i) The period of time would be longer. (ii) The period of time would be shorter. (2)

b) Volume of starch ‘solution’. (1)

4 9 weeks. (1)

5 a) See figure A.4.1. (4)b) 300g. (1)c) 4 weeks. (1)d) 1650g. (1)e) 1000g. (1)f) 550g. (1)

2500

2000

1500

1000

500

0

mas

s of

fetu

s (g

)

12 16 20 24 28 32 36age of fetus (weeks)

Figure A4.1

6 a) 4.6kg. (1)b) 5.6kg. (1)c) 2.4kg. (1)d) 8.6kg. (1)e) 5.0kg. (1)

7 a) Insufficient insulin being made by the body. (1)b) There would be a higher concentration of glucose

in the blood of the untreated sufferer of diabetes mellitus. (1)

c) Hunger, thirst and excessive urination accompanied by loss of weight and strength. (3)

d) The doctor could test the person’s urine for the presence of glucose. (1)

e) Strict diet and injections of insulin. (2)f) When the kidney’s filtration units become damaged. (1)g) It allows earlier detection of protein in urine and

earlier application of preventative measures. (2)

8 a) 50cm3/30min. (1)b) By 400cm3/30min. (1)c) 2 hours. (1)d) Two different people were used so this introduced

a second variable factor. This could be overcome by doing all the tests on the one person. (2)

e) Unlike tap water, 0.96% salt solution has no significant effect on urine output. (1)

9 a) The two pieces of damp blotting paper are unequal in size making the dark side of the plastic tube damper than the light side. Heat from the bright lamp is heating up the light side of the plastic tube more than the dark side. (2)

b) Make the two pieces of damp blotting paper equal in size. Move the bright lamp to a central position so that any heat that it gives out affects both sides equally. (2)

10 a) X 5 owl, Y 5 badger, Z 5 butterfly. (1)b) Robin, squirrel and butterfly are diurnal. Owl, badger

and moth are nocturnal. (1)

Section 5

1 a) S. (1)b) Q. (1)c) R. (1)

2 There would be nothing left of the bone. Its flexible material would be removed by heating it in the oven and its hard mineral material would be removed by placing it acid. (3)

3 a) 7. (1)b) 4. (1)

4 a) (i) 0 metres. (ii) 5300 metres. (2)

b) (i) 4.2 million.(ii) 6.1 million. (2)

c) (i) Day 78.(ii) 6.4 million. (2)

d) (i) They are directly related. As one increases so does the other.

(ii) Red blood cells pick up oxygen in the lungs and carry it round the body for use in respiration. When the air is thinner, less oxygen is gained per breath. Therefore the body makes extra red blood cells to increase oxygen uptake and ensure survival. (3)

e) 66 days. (1)f) They would have a higher number of red blood cells

per mm3 of blood. (1)g) Graph B lags behind graph A. It takes a few days for

the body to respond to the thinner air at high altitudes and start making extra red blood cells. (2)

260 Answers

5 The volume of lime water should be equal in both tubes. The length of tubing immersed in the lime water should be equal in both tubes. (2)

6 a) The distance and the number of tapping attempts varied between the two parts of the experiment. (2)

b) Get the second boy to stand two paces away in both parts of the experiment and allow him ten tapping attempts both times. (2)

7 1 5 d), 2 5 f), 3 5 b), 4 5 e), 5 5 c), 6 5 a). (1)

8 As age increases, MOI (and fitness) decreases. Non athletes have a lower MOI (and are less fit) than athletes. (2)

9 a) See figure A.5.1. (4)b) 10–20 min. (1)c) 10mg/100cm3. (1)d) 100mg/100cm3. (1)e) The athlete had stopped exercising. (1)f) 30 min. (1)

100

80

60

40

20

0

lact

ic a

cid

con

cent

ratio

n of

blo

od (m

g/10

0cm

3 )

0 10 20 30 40 50 60 70 80 90 100

time (min)

Figure A5.1

Section 6

1 a) See figure A.6.1. (3)b) Continuous. (1)c) 8. (1)d) (i) 41–45.

(ii) 26–30. (2)

12

11

10

9

8

7

6

5

4

3

2

1

026-30 31–35 36-40 41-45 46-50 51-55 56-60 61-65

range of ray floret number

1 2 5 12 23 32 40 44

3 6 13 24 33 41 45

4 7 14 25 34 42

10 17 28 37

11 18 29 38

19 30 39

20 31

21

22

8 15 26 35 43

9 16 27 36

num

ber

of d

aise

s

Figure A6.1

2 a) See figure A.6.2. (3)b) Mother and grandfather/grandmother/son/daughter. (1)c) (i) Mother.

(ii) Father. (2)d) See figure A.6.2. 1 in 2 (50%) chance. (2)e) See figure A.6.2. 1 in 4 (25%) chance. (2)

1 1 1

grandfatherroller

fathernon roller

motherroller

grandmotherroller

son daughter

rr

rr

RR

Rr Rr Rr

Rr Rr

Figure A6.2

3 a) Black and red. (1)b) Black. (1)c) See figure A.6.3. (3)d) No. Although they are black, they also possess the

allele for red coat which they could pass on to some of their offspring. (2)

parents

gametes

F1

BB bb

all B all b

all Bb

Figure A6.3

4 a) Gg and gg. (2)b) 1Gg:1gg. (2)c) GG. (1)

5 a) It has brought about an increase in annual milk yield. (1)b) 3349 litres. (1)c) 4.8%. (1)

6 a) From each generation of pigs, they chose those animals that were slowest moving, with fat bodies carrying much meat and with a docile nature. They allowed only these animals to breed and produce the next generation. They repeated this procedure for many generations over a very long period of time. (3)

b) The modern pig’s body carries much more meat that can be used for human consumption. The modern pig is more docile and easier to tend on the farm. (2)

261Answers

7 a) See figure A.6.4. (4)b) (i) 4.4%.

(ii) 8.4%.(iii) 3 times.(iv) 30. (4)

4020 30

20

19

18

17

16

15

14

13

12

11

10

9

8

7

6

5

4

per

cent

age

pro

tein

pre

sent

in g

rain

s

number of generations of artificial selection

0 10 50

strain A

strain B

Figure A6.4

Section 7

1 a) Proteus. (1)b) It is not motile and it is able to grow on sucrose,

lactose and citrate. (1)

2 a) Temperature. (1)b) 6. (1)c) Volume of glucose solution. Presence of live yeast cells.

(2)d) Concentration of glucose solution. (1)e) 10°C 5 E, 20°C 5 A, 30°C 5 F, 40°C 5 B,

50°C 5 D, 60°C 5 C. (5)

3 a) A. (1)b) Before pipe A at sample point 1, the unpolluted water

contained a very low number of bacteria, a very high concentration of oxygen and many fish. After pipe A at sample point 2, the water contained a very high number of bacteria, a very low oxygen concentration and no fish. This indicates that the water had been polluted with untreated sewage which provided food for a population boom of bacteria. These in turn exhausted the river water’s oxygen supply leading to the death of the fish. (4)

4 Both methods of biological treatment of sewage involve the conversion of sewage effluent to ‘purified’ effluent by the action of bacteria which need oxygen.

During biological filtration the effluent is sprayed onto a bed of stones coated with the microbes. As the effluent trickles down between the stones, the microbes act on it using oxygen from the air spaces between the stones. However in the activated sludge process, the bacteria are added to a tank of effluent and the oxygen that they need to act on the effluent is supplied by blowing compressed air through the tank. (4)

5 a) 2.41. (1)b) 1946–50. (1)c) 1956–60. The annual number of cases of the disease

per 1000 dropped significantly. (2)d) It increased. (1)e) The earthquakes may have caused damage to the city’s

water mains pipes allowing the spread of the disease-causing microbes from the sewage system into the drinking water. (2)

f) (i) 0.61. (ii) 25%. (2)

g) (i) 11000. (ii) 660. (2)

6 a) Y. (1)b) X. Since it works at 30°C (and pH 7) this suggests that

it contains a ‘normal’ enzyme that digests protein. (2)c) Z. It has failed to work at the ‘low’ temperatures of

30°C and 60°C suggesting that it does not contain a digestive enzyme and needs a higher temperature to work. (2)

7 a) X. (1)b) Y. (1) c) Z. (1)d) X. The curved (concave) part of colony X that faces Y

shows it is sensitive to Y and unable to grow near it. The bulging (convex) part of colony X that faces Z shows that it is resistant to Z and is able to grow near it. (3)