Biology 0610/41 October/November 2022: question paper and mark scheme
Cambridge IGCSE Biology (0610), Paper 4 Theory (Extended), October/November 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Fig. 1.1 is a side view of a human skull indicating the four types of teeth and the jaws.
upper jaw
incisors
canine premolar molars lower jaw
Fig. 1.1
(i) State the function of human teeth.
…… [1]
(ii) State the name of the visible outer layer of the teeth.
…… [1]
(iii) Explain the process of tooth decay in humans.
…… [3]
(b) Mammals can be classified according to the position and shape of their teeth.
Fig. 1.2 shows the skulls of seven mammals.
A
B
C
D
E
F
G
not to scale
Fig. 1.2
(i) Use the key to identify each species shown in Fig. 1.2.
Write the letter of each species (A to G) in the correct box in the key.
Key
1 (a) two or more different types of teeth go to 2
(b) fewer than two different types of teeth go to 3
Mark scheme
1(a)(i) mechanical / physical, digestion ; 1
1(a)(ii) enamel ; 1
1(a)(iii) any three from: 3
sugar / food / plaque, left on teeth ;
respiration (of sugar) by bacteria ;
produce (lactic) acid ;
dissolves, enamel / dentine / AW ;
AVP ; e.g. dentine is exposed / AW or
dentine, is softer / dissolves more rapidly (than enamel)
or ref. to, pulp / nerve endings, being exposed or (decay
reaches nerve endings) leading to pain
1(b)(i) 4 7 correct = 4 marks
5 or 6 correct = 3 marks
Orcinus orca E 3 or 4 correct = 2 marks
1 or 2 correct = 1 mark
Myrmecophaga tridactyla F
Cervus elephus G
Go to 5
Macropus rufus B
Equus ferus A
Lemur catta C
Pteropus niger D
;;;;
1(b)(ii) any two from: 2
double circulation ;
four-chambered heart ;
lungs / any named part of lungs ;
diaphragm ;
red blood cells without nuclei ;
AVP ;
1(b)(iii) vertebrates ; 1
(a) have wide gap between front and back teeth in both jaws go to 4
(b) have no wide gap between front and back teeth in both jaws go to 6
3 (a) all teeth of similar shape Orcinus orca
Myrmecophaga
(b) no teeth on either jaw tridactyla
4 (a) no incisors in upper jaw Cervus elephus
(b) incisors in both upper and lower jaw go to 5
5 (a) incisors on lower jaw longer than incisors on upper jaw Macropus rufus
(b) incisors on upper and lower jaw are similar in size Equus ferus
6 (a) incisors on lower jaw project forwards Lemur catta
(b) incisors on lower jaw do not project forwards Pteropus niger
[4]
(ii) Killer whales, Orcinus orca, are mammals.
State two internal features you would expect to find in a killer whale that you would not find in a fish.
1 ……
2 ……
[2]
(iii) State the name of the group of animals that includes mammals and fish.
…… [1]
[Total: 12]
2 Digestive enzymes catalyse the breakdown of large insoluble molecules.
(a) (i) Explain why it is important that large insoluble molecules are broken down by chemical digestion.
…… [2]
(ii) State the name of the substance that is the solvent for most molecules that have been digested by enzymes.
…… [1]
(b) The activity of two protease enzymes, A and B, was measured at different pHs. Both enzymes are found in the human alimentary canal.
The results are shown in Fig. 2.1.
900 140 enzyme A enzyme B
800
120
700
100
600
500 80 enzyme A activity enzyme B activity
/ U per mg 400 / U per mg
60
300
40
200
20
100
0 0
1 2 3 4 5 6 7 8 9 10 11 12 pH acid neutral alkaline
Fig. 2.1
Describe and explain the roles of the two protease enzymes, A and B, in the alimentary canal.
Use the information in Fig. 2.1 to support your answer.
…… [6]
(c) Maltase is a digestive enzyme that acts in the small intestine.
State the exact location of maltase in the small intestine.
…… [1]
[Total: 10]
Mark scheme
2(a)(i) any two from: 2
become soluble ;
for absorption ;
small enough, for diffusion / active transport ;
2(a)(ii) water ; 1
2(b) any six from: 6
1 protein → amino acids ; MP1 A breaks down proteins (to,
polypeptides / peptides)
2 ref to specificity ;
3 both enzymes are active between pH 3 and pH5 ; A at pH4
4 A is pepsin ;
5 optimum pH at 2 ;
6 enzyme shows, no activity / is denatured, from pH 5 ;
7 (functions) in stomach ;
8 where HCl is present / in acid conditions ;
9 B is trypsin ;
10 optimum pH at 10 ;
11 enzyme shows, no activity / is denatured, from pH 3 ;
12 (functions) in small intestine / secreted from pancreas ;
13 bile neutralises (stomach) acid / in alkaline conditions ; MP13 A pancreatic juice neutralises stomach acid
2(c) (membrane of) epithelium ; 1
(a) Fig. 3.1 shows some apparatus that was used to investigate water loss from a leafy shoot.
oil layer burette
water leafy shoot
jar cut end
water
balance
Fig. 3.1
(i) State the name of the process by which leafy shoots lose water.
…… [1]
(ii) Before the leafy shoot is inserted into the jar shown in Fig. 3.1, it must be recut under water.
Suggest why the end of the leafy shoot was cut under water.
…… [1]
(iii) State the purpose of the oil layer on top of the water in the burette.
…… [1]
(iv) Using the information in Fig. 3.1, describe one method that can be used to determine how much water is lost from the leafy shoot.
…… [1]
(b) The apparatus shown in Fig. 3.1 was used to investigate the effect of temperature on the rate of water loss in a species of plant. The results are shown in Fig. 3.2.
0.60
0.50
0.40 rate of water loss from leaves 0.30
/ g per m2 per s
0.20
0.10
0.00
25 27 29 31 33 35 37 39 41 43 45 temperature / °C
Fig. 3.2
(i) Using the information in Fig. 3.2, calculate how much water would be lost from 1 m2 of leaves in 12 hours if the plants were kept at 35 °C. Include the unit.
[3]
(ii) Using the information in Fig. 3.2, describe and explain the effect of increasing temperature on the rate of water loss in this species of plant.
…… [5]
(c) The apparatus shown in Fig. 3.1 can also be used to investigate the effects of changing humidity on water loss in plants.
(i) Suggest why the mass of water in the apparatus does not change when the leafy shoot is kept at 100% relative humidity.
…… [2]
(ii) Even at extremely low relative humidities the leafy shoot did not wilt.
Explain why the leafy shoot shown in Fig. 3.1 did not wilt.
…… [1]
(iii) The investigation on the effect of temperature was done at a relative humidity of 20%.
The investigation was repeated at a relative humidity of 80% and all other conditions were kept the same.
Predict how the water loss will differ from the trend shown in Fig. 3.2.
Sketch your prediction on Fig. 3.2.
[1]
[Total: 16]
Mark scheme
3(a)(i) transpiration ; 1
3(a)(ii) ensure continuous column of water / prevents air bubbles 1
/ prevents airlock ;
3(a)(iii) prevent, evaporation / condensation (from the top of the burette, 1
affecting the volume of water in the burette) / AW ;
3(a)(iv) measure the decrease in the volume of water (in burette over a 1
period of time) / record the decrease in mass (over time) ;
3(b)(i) 12 096 g (per m2) / 12.096 kg (per m2) ;;; 3 A 12 / 12.1 kg (per m2)
MP1 correct reading from graph 0.28 (g per m2 per s)
MP2 correct calculation
0.28 × 60 × 60 × 12 = 12 096 g
MP3 correct unit – g or kg ;
ecf for MP2 from incorrect MP1
3(b)(ii) any five from: 5
1 as temperature increases, (rate of) water loss increases ;
2 ref to steeper rate / greater loss of water, after 37 / 38 oC ;
3 any correct comparative data quote (with units at least
once) ;
4 water vapour lost, through stomata / between guard cells ;
5 evaporation from mesophyll into air spaces ;
6 (diffusion) down a, water potential, gradient ;
7 increasing temperature increases kinetic energy (of water
molecule) ;
8 faster (rate of) more, diffusion ;
9 stomata open wider / more stomata open in high(er)
temperatures ;
10 AVP ; e.g. transpiration / evaporation, cools the plant
3(c)(i) no diffusion (of water vapour) ; 2
(because) no water potential gradient / described ;
3(c)(ii) (it has a) continuous supply of water / AW ; 1
3(c)(iii) line drawn below original line ; 1
Chromosomes are made of DNA.
(a) Describe the structure of a DNA molecule.
…… [4]
(b) (i) Outline how antibiotic resistance develops in a population of bacteria.
…… [3]
(ii) Scientists use differences in antibiotic-resistance genes to distinguish between different strains of the bacterium, methicillin-resistant S. aureus (MRSA).
Suggest why scientists use differences in base sequences to classify the strains of MRSA rather than using other methods.
…… [1]
(iii) Explain why scientists are concerned that some strains of bacteria, such as S. aureus, have become resistant to antibiotics.
…… [2]
(c) (i) Describe how the use of antibiotics can be managed to reduce the development of resistant strains of bacteria.
…… [2]
(ii) Suggest why MRSA is unlikely to be transmitted from a mother to her unborn fetus.
…… [1]
(d) Many species of bacteria do not cause disease. Bacteria are very important in many biological processes.
State the names of three natural processes involving bacteria that are important to ecosystems.
1 ……
2 ……
3 ……
[3]
[Total: 16]
Mark scheme
4(a) any four from: 4 A as an annotated drawing
two strands ;
formed into (double) helix ;
ref. to four bases / A and T and C and G ;
pairing of, A with T / C with G ;
cross links between bases ;
4(b)(i) any three from: 3
1 variation (in the bacteria) ;
2 ref. to mutations / described ;
3 mutations, give bacteria an advantage ;
4 non-resistant bacteria, killed / inhibited / AW (by antibiotic) ;
5 competition with non-resistant bacteria ;
6 resistant bacteria reproduce ;
7 pass on, gene / allele (for resistance) ;
8 natural selection ;
9 AVP ; e.g. horizontal gene transmission / described as gene(s)
passed from one bacterium to another
4(b)(ii) any one from: 1
1 no difference in appearance (of different strains of
MRSA) / AW ;
2 no need to culture the bacteria ;
3 (gene sequencing is an) automated process / AW ;
/ easy to do / AW ;
4 only need small, number / quantity / amount (of bacteria) ;
5 more, accurate / precise (than traditional methods) / AW ;
6 explained in terms of comparing base sequences of
different, strains / AW ;
7 AVP ; e.g. (DNA sequences) within a species are very similar
4(b)(iii) any two from: 2
(resistant) bacteria will not be killed by antibiotics / AW ;
idea that no antibiotic of last resort / AW ;
(bacterial diseases) cannot be treated / AW ;
infection / disease / (resistant) bacteria, will spread ;
AVP ;; e.g. new antibiotics need to be developed
it takes time to make new antibiotics
4(c)(i) any two from: 2
antibiotics used only when essential ;
ensuring people, follow instructions / complete course of
antibiotics ;
do not prescribe antibiotics for, viral / fungal, infections ;
AVP ;; e.g. develop new antibiotics
4(c)(ii) placenta forms a barrier between mother and fetus / AW ; 1
4(d) any three from: 3
nitrogen fixation ;
decomposition / decay ;
nitrification ;
denitrification ;
deamination ;
fermentation ;
AVP ;; e.g. nitrogen cycle / carbon cycle / aerobic or anaerobic
respiration
Fires release carbon dioxide into the atmosphere.
(a) (i) State one other natural process that releases carbon dioxide into the atmosphere.
…… [1]
(ii) Carbon dioxide is a greenhouse gas.
State the name of one other greenhouse gas.
…… [1]
(b) Data scientists used satellite images to analyse the occurrence of fires globally, during a
14-year period. They tracked all fires that were larger than 0.21 km2 and therefore visible from space.
Table 5.1 summarises some of their data, categorising the fires by location. The locations include natural ecosystems and land that is managed by people. The expansion rate is the speed at which each fire becomes larger.
Table 5.1
estimated average estimated average estimated total expansion rate location of fire duration of fires number of fires of fires
/ days
/ km2 per day natural boreal forest 197 124 0.6 5.4
natural temperate forest 178 909 0.4 4.1 natural savannah
9 809 719 0.7 4.6
(grassland with few trees) managed land being
909 826 0.3 3.8 deforested managed agricultural land 1 631 918 0.3 3.4
(i) Using the information in Table 5.1, compare the data for the two managed locations with the data for the three natural locations.
…… [3]
(ii) Describe how the data in Table 5.1 could be used to estimate the total area that was burnt during the 14-year period, for each location.
…… [2]
(iii) Burning large areas of forest is a cause of habitat destruction.
Describe the possible consequences of habitat destruction.
…… [3]
[Total: 10]
Mark scheme
5(a)(i) any one from: 1
respiration ;
decomposition ;
volcanic eruptions ;
release of carbon dioxide from oceans (ocean-atmosphere
exchange) ;
5(a)(ii) methane / AVP ; 1
5(b)(i) any three from: 3
1 (total) number of fires in natural ecosystems is higher than
(total) number of fires in managed land
or
idea that the savannah (alone) has more fires than the
(total) fires in managed land ;
2 fires on managed land, have a lower expansion rate / spread
more slowly (than other fires) ;
3 fires on managed land, have a shorter duration / last less
time (than other fires) ;
4 more fires on land being deforested than in natural forest ;
5 comparative data quote (with units at least once) ; MP5 e.g. approx. 20% of all fires occur on land
managed by humans
6 AVP ;
5(b)(ii) multiply ; 2
number of fires by expansion rate by duration (for each row) ;
5(b)(iii) any three from: 3
1 reduction in population sizes of organisms ;
2 organisms / animals / plants, become endangered ;
3 organisms / animals / plants, become extinct ;
4 loss of biodiversity ;
5 loss of genetic diversity ;
6 ref. to negative effect on, food chains / food webs ;
7 ref. to negative effect on, nutrient / water, cycling ;
8 AVP ; e.g. migration of organisms / flooding / soil erosion /
leaching / desertification
(a) Fig. 6.1 shows part of the human gas exchange system.
K
A
J
B
H
G
F
E
D C
Fig. 6.1
(i) Table 6.1 shows the names of some parts of the human gas exchange system, their functions and the letters in Fig. 6.1 that identify the parts.
Complete Table 6.1.
Table 6.1
function name of the structure letter in Fig. 6.1
hairs in the nose A
prevents collapse of the airway
contracts to decrease the pressure in the thorax F
diaphragm
protects the lungs from mechanical damage
contains cilia to move mucus out of the airway
site of gas exchange alveoli
[7]
(ii) Describe and explain how the alveoli are adapted for gas exchange.
…… [3]
(b) (i) Explain the differences in composition between inspired and expired air.
…… [3]
(ii) Physical activity changes the concentration of carbon dioxide in the body.
State where this change is detected and how the body responds to the change.
…… [2]
(iii) State the name of a solution that can be used to test for the presence of carbon dioxide gas.
…… [1]
[Total: 16]
Mark scheme
6(a)(i) 7 one mark for each correct row
name of the letter on
function
structure Fig. 6.1
traps particles (before they enter the
hairs in the
airway) / mechanical barrier to A
nose
pathogens / AW
prevents collapse of the airway cartilage J/B
external
contracts to decrease the pressure in
intercostal F
the thorax
muscles
contracts to, reduce the pressure
diaphragm C
/ increase the volume, in the thorax
protects the lungs from mechanical
rib(cage) E
damage
trachea J
contain cilia to move mucus out of
/ bronchi /B
the airway
/ bronchiole /G
site of gas exchange / AW alveoli H
;;;;;;;
6(a)(ii) any three from: 3
thin / one cell thick ;
large surface area ;
(which) reduces diffusion distance / AW ;
good blood supply ;
AVP ; e.g. moist lining / presence of surfactant
6(b)(i) any three from: 3
less oxygen and more carbon dioxide in expired air (than
inspired air) ;
(as) oxygen is required for aerobic respiration ;
carbon dioxide is released by respiration ;
water vapour increases (as it is released by respiration) ;
AVP ;
6(b)(ii) detected: 2
either
(carbon dioxide is) in the blood
or
by the brain ;
response:
either
increases, rate / depth, of breathing
or
increases heart rate / AW ;
6(b)(iii) limewater / hydrogencarbonate indicator (solution) / AVP ; 1
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Biology 0610/41 for October/November 2022 is out of 80 marks, with 1 hour 15 minutes to answer it, across 6 questions.
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