Biology 0610/42 October/November 2019: question paper and mark scheme
Cambridge IGCSE Biology (0610), Paper 4 Theory (Extended), October/November 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Fig. 1.1 shows four arthropods.
A
B
C D
not to scale
Fig. 1.1
(i) State two features, visible in Fig. 1.1, that are common to all arthropods.
1 ……
2 ……
[2]
(ii) Fig. 1.2 is a dichotomous key for the arthropods shown in Fig. 1.1.
Complete Fig. 1.2 by writing suitable statements in:
• box 2 to identify species B
• box 3 to separate species C and A.
START
box 1 yes animal has more than ten pairs of legs D
no
box 2 yes
B
no
box 3 yes
A
no
C
Fig. 1.2
[2]
(b) The Hawaiian happy-face spider, Theridion grallator, is found on several of the Hawaiian islands. Some of the spiders have a very distinctive pattern on their bodies as shown in
Fig. 1.3.
Fig. 1.3
(i) State one feature, visible in Fig. 1.3, that identifies T. grallator as an arachnid.
…… [1]
(ii) Scientists think that the pattern on the bodies of the spiders is an adaptive feature.
Explain the term adaptive feature with reference to this pattern.
…… [3]
(c) DNA can be extracted from the webs of spiders. This DNA can be used to identify the species of spider that made the web, and the species of prey caught in the web.
Explain how DNA extracted from spider webs can be used to identify different species.
…… [2]
(d) Many species of spider are thought to be endangered.
State three reasons why animal species such as spiders become endangered.
1 ……
2 ……
3 ……
[3]
[Total: 13]
Mark scheme
1(a)(i) exoskeleton / AW ; 2
jointed / segmented, limbs / legs / appendages / AW ;
pairs of, limbs / legs / appendages / AW ;
segmented (body) / AW ;
bilateral body symmetry ;
1(a)(ii) Box 2: any one from: 2
animal has, 3 pairs of legs / 6 legs / less 4 pairs of legs / less than 8
legs ;
wings ;
head, thorax, abdomen / body in three sections ;
no, pincers / claws / carapace ;
Box 3: any one from:
(animal has) claws / pincers, of different sizes / AW ;
eyes on stalks ;
smooth, carapace / body / ‘shell’ ;
body, has five sides / is angular ;
hairs / bristles / AW, on, legs / claws ;
1(b)(i) 4 pairs of legs / 8 legs ; 1
fused head and thorax / cephalothorax / two sections or parts of the
body ;
(pedi)palps / described ;
AVP ;
1(b)(ii) inherited feature / controlled by gene(s) / allele(s) / DNA ; 3
functional / AW, feature ;
increase, fitness / chances of survival ;
increases chance of reproducing (to leave offspring) ;
any suitable example visible in T. grallator ;
e.g. camouflage / warning signal / scares predators / lure for
prey / sexual attractant
1(c) identification can be done using: 2
base, sequences / order / pattern, in DNA / genes ;
each species, has unique / AW, DNA / genes ;
idea that compare with, reference DNA / base sequences / genes,
of known species ;
idea that if a match with DNA from known species then DNA is from
that species OR closely related species have fewer differences in
their, base sequences / DNA / genes ;
AVP ; e.g. any technique involved in DNA analysis
1(d) hunting / collecting / AW ; 3
pollution ;
pesticide(s) / insecticide / (chemical) spray(s) that kill
animals / poisons ;
loss of habitat / any example ;
disease ;
increased competition / described ;
increase in predator(s) ;
climate change / any example ;
lack of food / loss of prey species ;
AVP ;
Fig. 2.1 shows part of a cross-section of the stem of a young sunflower plant.
area Y
Fig. 2.1
(a) Draw a circle around one vascular bundle on Fig. 2.1.
Label the xylem in the vascular bundle with the letter X. [2]
(b) Explain how the cells in area Y are able to support the stem so that it stays upright.
…… [3]
(c) Researchers used carbon dioxide that contained a traceable source of carbon (13C) to investigate translocation of sucrose from the leaves of bean plants, Phaseolus vulgaris.
Fig. 2.2 shows that glucose produced in photosynthesis is converted to sucrose for translocation.
carbon dioxide glucose sucrose
Fig. 2.2
Researchers selected four plants, Q, R, S and T, which had leaves that were of similar sizes.
The leaves on the four plants were supplied with 13CO2.
After the leaves had started to make sucrose, the researchers cut away a ring of tissue in different places as shown in Fig. 2.3. The rings of tissue that were removed from plants R, S and T contained the phloem.
Q R S T
13CO 13CO 13CO 13CO
2 2 2 2
Key: the positions on the stems where rings of tissue containing phloem were removed.
Fig. 2.3
The quantities of sucrose containing 13C in the shoot tips and in the roots were determined.
The results are shown in Table 2.1.
Table 2.1
quantity of sucrose containing 13C / arbitrary units plant shoot tip root
Q 3.24 0.94
R 0.00 0.44
S 4.14 0.00
T 0.00 0.00
Describe and explain the effect of removing the phloem on the translocation of sucrose in plants Q, R, S and T.
…… [5]
[Total: 10]
Mark scheme
2(a) any shape drawn that includes one whole vascular bundle including 2
all of the xylem and phloem with or without sclerenchyma ;
label line from X to xylem in any of the vascular bundles ;
2(b) cell vacuoles / cells, contain (much) water / have high water 3
potential ;
water absorbed, by osmosis / down water potential gradient ;
cells, are turgid / have a turgor pressure ;
cell contents / vacuole / cell membrane, pushes out (against cell
wall) ;
cell wall, does not stretch / is inelastic / is rigid ;
AVP ; e.g. cells are tightly packed / AW
2(c) Q sucrose / 13C, is in shoot and root ; 5
T no, sucrose / 13C, in shoot or root ;
R sucrose / 13C, in root only / (in root but) not in shoot ;
S sucrose / 13C, in shoot only / (in shoot but) not in root ;
idea that no transport of, sucrose / 13C, where phloem is removed ;
phloem transports (sucrose) in both directions ;
leaf is source / carbon (dioxide) is fixed in leaf / sucrose is made in
leaf ;
roots / shoots, are sink(s) / described ;
e.g. of descriptions respired / stored as starch / converted to another
(named) compound ;
AVP ; e.g. Q is a control
Cells in the lining of the stomach secrete gastric juice, which contains hydrochloric acid and pepsin.
(a) (i) State two roles of hydrochloric acid in the stomach.
1 ……
2 ……
[2]
(ii) Describe the function of pepsin.
…… [2]
(b) Define the term assimilation.
…… [2]
(c) There are stem cells in the epithelial tissue that forms the lining of the stomach.
Explain why these stem cells are necessary.
…… [2]
(d) The epithelial cells of the small intestine have many microvilli.
State the role of the microvilli.
…… [2]
(e) Lactobacillus is a type of bacterium. A study was carried out to investigate the ability of
Lactobacillus to attach to the epithelial cells that form the lining of the small intestine.
Researchers added Lactobacillus bacteria to epithelial cells that had been grown in Petri dishes.
Every 15 minutes, the researchers estimated the average number of bacteria that were attached to the epithelial cells in the Petri dishes.
The results are shown in Fig. 3.1.
140
130
120
110
100
90
80 average number of bacteria
70 attached to epithelial cells
60
50
40
30
20
10
0
0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 time / minutes
Fig. 3.1
Calculate the percentage increase in the average number of bacteria attached to epithelial cells from 45 minutes to 75 minutes.
average number of bacteria at 45 minutes …… .
average number of bacteria at 75 minutes …… .
Give your answer to the nearest whole number.
Space for working.
…… %
[3]
[Total: 13]
Mark scheme
3(a)(i) provides, suitable / optimum, pH for (correct named) enzyme 2
action ;
activates, enzyme / pepsin ;
kills / AW, bacteria / viruses / pathogens / microbes ;
AVP ;
3(a)(ii) (catalyses) breaks down / (chemically) digests, of protein ; 2
to amino acids ;
3(b) movement of digested food molecules into cells ; 2
food molecules become part of cells ;
3(c) (stem cells) divide by mitosis ; 2
form (named) specialised cells (in stomach) ;
to replace cells (in the lining of stomach) ;
idea that cells are worn away from the surface of the stomach ;
for repair of any damage to tissues ;
3(d) increase / large, (surface) area ; 2
for absorption (of named substances) / described;
AVP ;
3(e) 54(%) ;;; 3 one mark for correct readings (78 minutes and 120
minutes)
one mark for correct calculation
one mark for whole number correctly rounded
(a) Explain how the human gas exchange system is protected against pathogens.
…… [4]
(b) Tuberculosis (TB) is a transmissible disease caused by a bacterial pathogen. The spread of this disease can be controlled by vaccination.
Explain how vaccination provides a defence against transmissible diseases.
…… [4]
(c) TB is a disease that can be treated with antibiotics. HIV infection cannot be treated with antibiotics.
Explain why viral diseases cannot be treated with antibiotics.
…… [2]
(d) Lung cancer is a disease that is strongly linked with smoking tobacco.
Fig. 4.1 shows some data about smoking and lung cancer in country A between 1900 and
2020 (2020 data has been estimated):
• percentage of the male population that smoke tobacco
• percentage of the female population that smoke tobacco
• number of deaths in males from lung cancer per 100 000 of the male population
• number of deaths in females from lung cancer per 100 000 of the female population.
percentage of deaths from population that lung cancer per smoke tobacco 100 000
100 100
90 90
80 80
70 70
60 60
50 50
40 40
30 30
20 20
10 10
0 0
1900 1920 1940 1960 1980 2000 2020
Key: year
percentage of population that smoke tobacco deaths from lung cancer per 100 000 male female male female
Fig. 4.1
(i) Describe the differences between the percentages of males and females in country A that smoke as shown in Fig. 4.1.
…… [3]
(ii) Fig. 4.2 shows the same smoking statistics for country B between 1950 and 2020.
percentage of deaths from population that lung cancer per smoke tobacco 100 000
100 100
90 90
80 80
70 70
60 60
50 50
40 40
30 30
20 20
10 10
0 0
1900 1920 1940 1960 1980 2000 2020
Key: year
percentage of population that smoke tobacco deaths from lung cancer per 100 000 male female male female
Fig. 4.2
Discuss the evidence for and against this statement:
‘The changes in the percentage of the population that smoke tobacco, and the deaths from lung cancer per 100 000, follow the same trends in males and in females.’
Use the information for country A in Fig. 4.1 and country B in Fig. 4.2 in your answer.
…… [6]
[Total: 19]
Mark scheme
4(a) ref. to, mechanical / chemical, barriers ; 4
nasal hairs, trap / filter / AW, pathogens ;
mucus traps pathogens / pathogens stick to mucus ;
mucus, produced / secreted, by goblet cells ;
cilia move mucus (upwards / towards mouth / away from alveoli) ;
coughing / sneezing / swallowing ;
phagocytes / phagocytosis / described ;
4(b) introduces harmless form of pathogen / AW ; 4
ref. to antigen(s) ;
stimulates an immune response ;
ref to active immunity ;
lymphocytes produce antibodies ;
(lymphocytes develop into) memory cells ;
memory cells, remain in the body / give long-term immunity / can
produce antibodies (at a later time) ;
respond quickly when an infection (of the same pathogen)
occurs / before symptoms occur ;
AVP ;
4(c) idea that viruses have no antibiotic targets ; 2
viruses, are not alive / are not living / cannot be killed / not cells ;
no cell membrane ;
no cell wall ;
no protein synthesis / no ribosomes ;
no metabolism / do not respire ;
AVP ; e.g. viruses are inside (host) cells
4(d)(i) percentage of males smoking increases faster than females ; ora 3
more men smoked than women / higher percentage of men smoked
(over all the time) ; ora
peak for percentage of men smoking occurred earlier ; ora
decrease in percentage of men smoking is greater than in females ;
ora
difference between percentages smoking decreases after peaks ;
any comparative use of percentages ;
4(d)(ii) for from A 6
percentage of men and women that smoke increases and
decreases ;
numbers of deaths from lung cancer in both groups increases and
decreases ;
both show lag between peak smoking and peak deaths from lung
cancer ;
country A shows, same time gap / 40-year gap, between peaks for
smoking and deaths from smoking ;
for from B
percentage of men and women who smoke has decreased
(overall) ;
numbers of deaths from lung cancer in both groups increase ;
against from B
percentage female smokers, fluctuates / AW, but no fluctuation in
men ;
numbers of deaths in men increase (continually), whereas deaths
in females does not ;
country B may show the same but no data to be sure / no data from
before 1950 ;
any data quote - percentage / number and approximate year and
units ;
Grass plants are wind-pollinated. Fig. 5.1 shows a flower from a species of grass plant.
A
B
D
C
Fig. 5.1
(a) (i) Describe and explain how the features of the flower shown in Fig. 5.1 are adaptations for wind-pollination.
…… [3]
(ii) State one letter in Fig. 5.1 that identifies a structure where meiosis occurs.
…… [1]
(b) Wheat has flowers similar to those in Fig. 5.1, but reproduces by self-pollination.
Explain the consequences of self-pollination for a population of plants.
…… [4]
(c) Wheat is an important crop plant in many countries. Wheat can be infected by diseases leading to total loss of the crop which results in famine.
Outline factors, other than plant diseases, that can cause famines.
…… [3]
(d) Black stem rust is a disease of wheat that is caused by a fungus.
Plant breeders used two varieties of wheat to produce a variety of wheat that is both rust-resistant and has a high yield.
Fig. 5.2 shows the breeding programme that was used.
step 1 rust-resistant variety of × variety of wheat with a high yield wheat with a low yield
step 2 new variety 1 × variety of wheat with a high yield
step 3 selected rust-resistant offspring × variety of wheat with a high yield
repeat step 3 for several generations
rust-resistant variety with a high yield
Fig. 5.2
(i) Suggest how plant breeders make sure that the plants that they use for step 3 are rust-resistant.
…… [2]
(ii) Suggest why step 3 is repeated for many generations before the new rust-resistant variety is made available for farmers to grow.
…… [1]
(e) Wheat plants are monocotyledons.
State one feature of monocotyledons that can be used to distinguish them from dicotyledons.
…… [1]
[Total: 15]
Mark scheme
5(a)(i) anthers / stamens / filaments / stigma, hang / AW, outside (the 3
flower) ;
large, anthers / C, produce large quantities of pollen ;
idea that anthers / C, easily release pollen ;
‘feathery’ / AW, stigma / B ;
stigma has, feathery surface / large surface area, to catch pollen ;
bracts / A, are, small / inconspicuous ; A petals
5(a)(ii) C; 1
5(b) idea that pollination / fertilisation, always going to happen / AW ; 4
no agent of pollination needed ;
little wastage of pollen ;
reduced / little, variation / diversity ;
increased competition between plants (as have the same
adaptation) ;
increase chance of, genetic / inherited, disease ;
all plants adapted to same, conditions / environment ;
little ability to adapt to changing conditions / little ability to evolve ;
an infectious disease can kill all of the population / all plants more
susceptible to the same disease ;
risk of extinction ;
AVP ;
5(c) drought / lack of rain / lack of (irrigation) water ; 3
flooding ;
fire ;
tsunamis / cyclones / hurricanes ;
earthquakes / volcanic eruptions ;
plagues of, animals / insect pests ;
pests / diseases, of stored food / livestock ;
conflict / war ;
rising prices of food ;
poverty ;
unequal distribution of food ;
growing, non-food crops / biomass for fuels / crops for export ;
increase in population / migration of people ;
soil degradation / soil erosion / desertification / salination of
soils / loss of soil fertility / barren land / AW ;
5(d)(i) expose the plants to (stem), rust / fungus ; 2
find out if plants have DNA for rust resistance ;
only use those plants that show no symptoms / AW ;
5(d)(ii) to increase the, numbers of plants / population ; 1
incorporate more genes from the high yield variety ;
to maximise the yield of wheat from individual plants ;
check that the plants keep their resistance to rust ;
check the plants grow well in field conditions ;
make sure the plants are, pure-breeding / homozygous ;
5(e) one from: 1
one, cotyledon / embryonic leaf / seed leaf
oblong leaves / narrow leaves / straight leaves
parallel-veined leaves / straight veins
(named) flower parts in multiples of three
fibrous roots / adventitious roots
scattered vascular bundles in stem
AVP ;
Carbon dioxide forms approximately 0.04% of the atmosphere.
Fig. 6.1 shows part of the carbon cycle.
carbon dioxide
A B
C
feeding death plants animals decomposers
death
D
coal, oil, gas
Fig. 6.1
(a) Complete Table 6.1 by naming the processes labelled A to D in Fig. 6.1.
Table 6.1
letter on name of the process in the carbon cycle
Fig. 6.1
A
B
C
D
[4]
(b) Complete the sentences by writing the missing words in the spaces.
Carbon dioxide is a greenhouse gas. Greenhouse gases trap ……
in the atmosphere. Water vapour is the most common greenhouse gas in the atmosphere.
Another greenhouse gas is methane which is released from ……
and …… . Carbon dioxide and methane released from human
activities over the past 200 years have contributed to the ……
greenhouse effect.
Other atmospheric pollutants can cause acid rain. One of these is
…… which is a waste gas from some power stations.
One source of pollution in both aquatic and terrestrial ecosystems is
…… plastics. [6]
[Total: 10]
Mark scheme
6(a) 4
letter on Fig. 6.1 name of the process in the carbon cycle
A photosynthesis ;
B combustion ;
C respiration ;
D fossilisation ;
6(b) gap 1: heat OR long-wave / infra-red, radiation ; 6
gaps 2 and 3, any two from: ;;
paddy fields / rice farming
(named) animals / livestock
decay / decomposition (by bacteria)
rubbish tips / landfill
sewage / dung / faeces
(natural) gas extraction / fracking
melting tundra
waterlogged soil / swamp / marsh
biomass burning / forest fires / peat fires
gap 4: enhanced ;
gap 5: sulfur dioxide / sulfur trioxide / nitrogen oxide(s) ;
gap 6: non-biodegradable / micro- / non-recyclable / single-use ;
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