Cambridge International A Level Biology Answers to self-assessment questions
below, there are two quadrats for each Table showing the critical values of rs at p =
species where this applies. 0.05 for different numbers of paired values
Next, calculate D (the difference between the
ranks). Then calculate D2 for each species (as Number of pairs of Critical value at p = 0.05
in the χ2 test this removes the negative signs). measurements (5%)
Calculate the sum (Σ) of D2. (You could type
the table into a spreadsheet and get it to do 5 1.00
all the calculations for you.)
6 0.89
% cover % cover Rank 7 0.79
C. V.
vulgaris myrtillus C. C. 8 0.74
vulgaris vulgaris
D D2 9 0.68
30 15 9 8 11 10 0.65
37 23 11 10 1 1 11 0.60
15 6 5.5 4 1.5 2.25 12 0.59
15 10 5.5 5.5 0 0 13 0.56
20 11 7 7 00 14 0.54
9 10 3 5.5 –2.5 6.25 15 0.52
3 3 1 2 –1 1 16 0.51
5 1 2 1 11 17 0.49
10 5 4 3 1 1 18 0.48
25 17 8 9 –1 1 19 0.45
35 30 10 11 –1 1 20 0.44
n = 11 ΣD2 = 30 0.36
15.5
If the value for rs is greater than the critical
The figures are now inserted into the above value, then you can reject the null hypothesis.
equation. If the value of rs is less than the critical value
6 × ΣD2 you can accept the null hypothesis.
rs = 1 – n3 – n In this case the value for rs is greater than
the critical value so the ecologist can
rs = 1 – 163 ×3 11 –5 .151 reject the null hypothesis and accept the
alternative hypothesis.
rs = 1 – 0.070 d There is a significant correlation between
the abundance (as measured by percentage
rs = 0.930 cover) of the two species on the moorland.
e There are two variables: soil moisture and
A correlation coefficient of +0.93 is very close to percentage cover (or some other measurement
+1, so we can conclude that there is a positive of abundance, such as species density)
correlation between the two species and that Random sampling or systematic sampling
the strength of the association is very high. could be used in this investigation. This is the
Now look up the Spearman’s rank coefficient method for random sampling:
in the table of critical values that correspond
to the number of pairs of measurements in
results table (there are 11).
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Cambridge International A Level Biology Answers to self-assessment questions
■■ mark out a large plot of the moorland with step 2: multiply each pair of values (for each
tapes at right angles to each other tree) together to give their product, xy.
step 3: calculate the sum of the products xy,
■■ use a random number generator to select to give Σxy
coordinates where the quadrat will be These steps are shown in the table.
placed in each quadrat
Mean width of crack in bark / mm Height x – x (x – x)2 Width y – y (y – y)2 product
Tree■■ record percentage cover of each species /m / cm xy
■■ take a soil sample using any suitable x y
method – such as using a soil auger, which 1 1.77 0.161667 0.026136 50 –0.08333 0.006944 88.5
is ‘screwed’ down into the soil and then
removed to extract a soil sample all the 2 1.65 0.041667 0.001736 28 –22.0833 487.6736 46.2
way down to a fixed depth
■■ take at least three samples of soil from 3 1.81 0.201667 0.040669 60 9.916667 98.34028 108.6
each quadrat (as replicates)
■■ weigh a small quantity of soil from each 4 0.89 –0.71833 0.516003 24 –26.0833 680.3403 21.36
sample
■■ dry in an oven or leave in the sun to dry 5 1.97 0.361667 0.130803 95 44.91667 2017.507 187.15
■■ continue to dry and weigh the soil until the
mass remains constant / there is no further 6 2.15 0.541667 0.293403 51 0.916667 0.840278 109.65
decrease in mass
■■ calculate the mass of water in the soil (wet 7 0.18 –1.42833 2.040136 2 –48.0833 2312.007 0.36
mass as extracted from the soil – dry mass
= mass of water / moisture content) 8 0.46 –1.14833 1.318669 15 –35.0833 1230.84 6.9
■■ calculate the percentage of the original soil
sample that was water: 9 2.11 0.501667 0.251669 69 18.91667 357.8403 145.59
decrease in mass (= mass of water) divided
by the original mass of the soil × 100 10 2.00 0.391667 0.153403 64 13.91667 193.6736 128
■■ draw a scattergraph of percentage cover
against percentage soil moisture 11 2.42 0.811667 0.658803 74 23.91667 572.0069 179.08
■■ calculate Spearman’s rank correlation
coefficient 12 1.89 0.281667 0.079336 69 18.91667 357.8403 130.41
x= Σ(x – x)2 = y = Σ(y – y)2= Σxy =
1.608333 5.510767 50.08333 8308.917 1151.8
s2 0.500979 s2 755.3561
9 a 100 sx 0.707799 sy 27.48374
90 rM=eaΣnxnys s–ax snnyxd y standard deviations can be
80 calculated with a calculator or the whole
70 calculation set out on a spreadsheet in the
60 same way as in the table.
50 Now insert the figures into the formula:
40 r = 1151.81 2– ×(1 02. 7×1 1 ×.6 2078. 4×8 50.08)
30 0.5 1.0 1.5 2.0 2.5 3.0 185.4563
20 Circumference of trees / m r = 234.1296
10
0
0.0
b The following steps are carried out: r = 0.79
step 1: calculate the mean and standard
deviation for each variable c This shows that there is a fairly strong
x = mean for height sx = standard deviation correlation, but is this significant? As with
for height Spearman’s rank, we can test this. First we
y = mean for width sy = standard deviation for need a null hypothesis that there is no linear
width correlation between the circumference of the
trees and the width of the cracks in the bark.
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Cambridge International A Level Biology Answers to self-assessment questions
This is the table to use if the student has b i taxon – a unit of classification, a category
made the hypothesis that there is a linear that indicates a rank or level in the
correlation between the two variables. classification system. Examples are
species, genus, family, etc.
Number of pairs of Critical value at p = 0.05
measurements (5%) ii hierarchical system – each taxon above the
level of species includes all the taxa below
5 0.88 it. For example, an order includes all the
families that have similar features.
6 0.81
7 0.75 11
8 0.71 Feature Domains
9 0.67 Bacteria Archaea Eukarya
10 0.63 Cell structure prokaryotic prokaryotic eukaryotic
(e.g. no
membrane–
bound
11 0.60 organelles)
12 0.58 Nucleus ✘ ✘ ✔
15 0.51 DNA circular circular linear
20 0.44 DNA with ✘ ✔ ✔
In this case there are 12 pairs of histones
measurements and the critical value is 0.58.
The value of r (0.79) is greater than this, so Plasmids present in present in present in very
the null hypothesis can be rejected. many some few, e.g. yeasts
d The student can reject the null hypothesis
and we can accept the alternative hypothesis Ribosomes all 70S all 70S all in the
that there is a correlation between the height cytosol are
of the trees and the width of the cracks in 80S (70S in
their bark. mitochondria
and
chloroplasts)
Peptidoglycan ✔ ✘ ✘
in cell wall
1 0 a Taxonomic rank African bush elephant Method of cell binary fission binary fission mitosis
division
domain Eukarya Organisation single–celled single–celled unicellular
or chains / or chains / / colonial /
kingdom Animalia groups of cells, groups of cells, multicellular
e.g. filaments e.g. filaments
phylum Chordata
class Mammalia
order Artiodactyla
family Giraffidae
genus Giraffa
species Giraffa camelopardalis
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1 2 a i Protoctista and Plantae source of potential medicines
economic reasons, e.g. for ecotourism
ii Protoctista, Fungi and Animalia to maintain, food webs / nutrient cycles
b to maintain ecosystems that provide
Features Kingdoms ‘services’ for humans
Protoctista Fungi Plantae Animalia aesthetic reasons, e.g. the intrinsic beauty of
the natural world
Type of body unicellular composed multicellular, multicellular,
and of hyphae; branching most have 15
multicellular yeasts are body a compact Some possible answers:
unicellular body pressure on national parks of many visitors
the need to provide facilities for visitors
Cell walls present in ✔ (made of ✔ (made of ✘ the need to restrict any damage done by
visitors
some species chitin) cellulose)
Cell vacuoles ✔ ✔ ✔ ✔ so some areas may be fenced off to prohibit
large large small entry, e.g. during breeding season for some
algae permanent permanent temporary animals / time of flowering for plants
have large vacuoles vacuoles e.g. populations of animals / plants must be kept
permanent lysosomes in check, e.g. to prevent overgrazing
vacuoles, and food
protozoans vacuoles
have small,
temporary
vacuoles park must provide habitats appropriate for
the species that live there
human activities, such as building /
Type of autotrophic heterotrophic autotrophic heterotrophic development / transport / farming, need to
nutrition and be controlled in some parks
heterotrophic
Cilia / some have ✘ some ✔
flagella cilia for gametes 1 6 a Zoos may not be able to provide the right
locomotion have flagella, ‘habitat’ suitable for breeding. Animals may
and feeding otherwise need particular factors in their environment
(e.g. Stentor) none have before their reproductive systems become
them
Motility some are ✘ ✘ ✔ able to produce sperm or eggs – for example,
(ability highly mobile, muscular to have plenty of space, or to have many
to move others are tissue others of their species around them. These
themselves) fixed to a factors affect their physiology and their
substrate
Nervous ✘ ✘ ✘ ✔ behaviour. Courtship may be difficult in the
conditions in a zoo. They may need particular
coordination
changes in day length or in food supply to
trigger hormonal changes associated with
1 3 a Viruses are acellular / they do not have a reproduction. These may not occur in the
cellular structure. latitude where the zoo is situated. Males and
females may be from very different social
b The type of nucleic acid, DNA or RNA and the groups and refuse to associate with each
number of strands in these molecules – 1 or 2.
1 4 a Some examples: other let alone mate.
pollution b Artificial insemination – semen is collected
deforestation
overgrazing from the male and stored in thin tubes
known as straws. These are injected into the
natural catastrophes (hurricanes, etc) vagina or uterus using a catheter around the
hunting / fishing
habitat destruction time of ovulation.
In vitro fertilisation (IVF) – sperm are collected
from male; eggs are collected from the ovary.
b source of alleles for breeding to improve Sperm and eggs are mixed together in a dish
agricultural species
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or tube to fertilise. The embryo is kept for a 1 8 a Many wild plants are threatened with habitat
few days in culture. loss and the effects of climate change. Seeds
Sperm bank – sperm are collected from can be kept for a long time – possibly until
males, mixed with an albumen solution and suitable places and conditions are available
put into thin tubes known as straws. These for the plants to be reintroduced. Seed
are kept at –196 oC until required. banks keep the genetic material that would
Embryo transfer – embryos formed during IVF otherwise be lost.
are inserted into the uterus for implantation
to take place. They could be inserted into b Seeds that can be stored in seed banks are
the female that donated the eggs or into orthodox seeds. Some plants do not produce
another female which has been treated with this type of seed. Their seeds are known as
hormones so is prepared for implantation recalcitrant. The plants concerned have to
(embedding in the endometrium). grow as plants in a gene bank, e.g. cocoa and
Surrogacy – one or more embryos are coconut.
placed into the uterus of a female who did 1 9 a Selection pressures in the natural habitat
not provide the female gamete. Pregnancy might include the ability of the adult plants
therefore occurs in a female who is to survive grazing, wide variations in rainfall
biologically unrelated to the embryo / fetus. or competition with other species. In the
This is to save the female who is the source of seed bank, none of these selection pressures
the eggs from the risks of pregnancy. would apply. In the seed bank, the greatest
‘Frozen zoo’ – a store of sperm, eggs and selection pressure will become the ability
embryos of animals, many / all of which are of the seeds to survive the conditions in
endangered species. Stored material can which they are stored for a long period of
be made available to zoos anywhere in the time. Seeds from store are germinated every
world. The frozen zoo is a store of genetic few years to check on their viability. Also
diversity to prevent inbreeding. some plants may be grown in protected
environment to set seeds to replenish the
c To prevent inbreeding. This reduces the stock in cold store.
chances of them becoming increasingly
homozygous over several generations. The b It is possible that the plants that grow from
health and fertility of animals is likely to the seeds that have been saved will not
decrease if they are inbred. For example, the have characteristics that will allow them
risk of developing genetic diseases caused by to survive the selection pressures they will
recessive alleles increases. encounter in their natural habitat. This could
reduce the chances of success in returning
17 Some possible answers: them to the wild.
Zoos provide a refuge for endangered c Providing enough space to grow all the plants
animals, when their habitat no longer exists necessary to have a collection of the genetic
or is too fragmented to support populations diversity in the species. Keeping plants
in the wild. They can keep animals in disease–free. There are problems involved
better heath than in the wild. They conduct in maintaining the collection in areas that
research on the best ways to breed them might be susceptible to natural catastrophes,
to increase the populations also ensuring such as hurricanes, severe storms, flooding
that health is maintained or improved by and drought.
restricting inbreeding. They allow people to
see animals that they would not otherwise
see (except on video or TV). Zoos provide an 2 0 a Some possible answers
important role in conservation education. Alien species may be carnivores so will prey
on many animals. They will compete with
existing predators in the ecosystem. They
may be herbivores in which case they will
Cambridge International AS and A Level Biology © Cambridge University Press 2014
Cambridge International A Level Biology Answers to self-assessment questions
compete with the existing herbivores in
the ecosystem. If they are plants, they will
compete with existing species for resources
such as light, minerals, water and space.
They may introduce disease that the existing
species have not experienced so have no
immunity. They will compete with native
species for space and breeding sites, etc.
They may change the environment so that
native species cannot survive.
b Find two areas: one invaded by an alien
species and one that has not. Carry out
random sampling. Count the number of
different species / make a species list. Record
the abundance of the different species.
Use an appropriate method, e.g. species
frequency, species density, percentage cover
or an abundance scale. Use Simpson’s Index
of Diversity to compare the different areas. If
this is repeated in several places, where the
alien species is present then Spearman’s rank
correlation coefficient could be calculated
to see if there is a correlation between the
abundance of the alien species present in the
areas sampled and the index of diversity for
each area.
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Answers to SAQs
Chapter 19 5 Suspect 3 – the bars in his blood sample exactly
match those found at the scene of the crime.
1 a A single stranded length of several nucleotides
at the end of a fragment of DNA that can form 6 A primer is a short length of DNA that
hydrogen bonds with an equivalent length attaches to one end of a single strand of DNA
of complementary nucleotides. Sticky ends and allows DNA polymerase to begin to make
are formed by restriction enzymes that make a complementary copy of the single template
staggered cuts across DNA. They can also be strand. A probe is also a short length of DNA
formed by adding a short section of single that attaches to one end of a single strand
stranded DNA to a blunt-ended fragment of of DNA. The probe, however, is labelled in
double stranded DNA. some way, often with the form of phosphorus
that emits beta radiation. The position of the
b A: 6 probe indicates the position of DNA.
B: 8
C: 1 7 a 256
D: 4 b There is no enzyme that will use an RNA
2 Advantages of using plasmids as vectors: template to make double stranded RNA.
■■ they exist naturally in bacteria, which are Instead, reverse transcriptase uses an RNA
able to take them up from their surroundings template to make single stranded DNA.
■■ they are small so that they are easy to use This DNA can then be replicated using DNA
■■ they can be produced artificially by polymerase and this can be used in PCR as
combining lengths of DNA from different shown in Figure 19.12. In this way, multiple
sources copies of cDNA can be made which hold the
■■ they are double stranded so genes from information in the original mRNA.
prokaryotes and from eukaryotes can be
inserted into them 8 a Gene probes from all of the genes in
■■ they replicate independently within the human genome can be placed on a
bacteria to clone any genes that are microarray. The mRNA from the cytoplasm
inserted into them of a cell is isolated and converted into single
■■ they can be transferred between different stranded DNA using reverse transcriptase
bacterial species. that uses DNA nucleotides that have
fluorescent tags. The quantity of this DNA
3 It depends where the gene for fluorescence is can be increased by PCR. Single stranded
inserted into the plasmid. It is possible that cDNA is then applied to the microarray. cDNA
some bacteria might have taken up plasmids hybridises with the appropriate gene probes
that do not contain the desired gene but do on the microarray and these are detected
still contain the fluorescence gene. when the microarray is scanned. Spots that
fluoresce indicate the genes that have been
4 Father and mother – HbAHbS / heterozygous transcribed to produce the mRNA that was
Children from left to right: isolated from the cytoplasm.
lane 4: HbAHbA (this child has normal
haemoglobin and is not a carrier of sickle cell b Mature red blood cells have no nucleus so
anaemia) transcription does not occur. However, there
lane 5: HbAHbS (this child does not have may still be some mRNA left in the cytoplasm
sickle cell anaemia, but is a carrier) especially if the red blood cell has only just
lane 6: HbSHbS (this is the child that has left the bone marrow where it was produced
sickle cell anaemia) from stem cells. Most of this mRNA will be for
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Cambridge International A Level Biology Answers to self-assessment questions
the α and β globin polypeptides for making
haemoglobin molecules.
9 Some possible answers are:
■■ genetic screening provides information
about the increased risk of people having
genetic conditions
■■ allows people to prepare for late onset
genetic conditions, such as Huntington’s
disease
■■ genetic tests can identify whether embryos
produced by IVF or embryos/fetuses
developing in the womb are going to
develop a genetic condition
■■ this identifies fetuses that will need early
treatment if they go to full term
■■ also allows parents to prepare for the birth
of a child who will need treatment for a
considerable time or even throughout life
■■ identifies carriers of genetic conditions
■■ helps to provide early diagnosis
■■ allows couples who are both carriers of a
genetic condition to make decisions about
starting a family / having more children /
seeking IVF and embryo biopsy.
10 The allele that causes cystic fibrosis is
recessive. The inserted allele would be
dominant, so would be able to make its
product and affect the phenotype.
1 1 a Without a promoter, a gene will not be
expressed, as no mRNA will be made from it.
b Plasmids and the bacterium, Agrobacterium
tumefaciens were used as vectors. The
plasmid inserted the required genes into A.
tumefaciens and this bacterium then inserted
them into the plant cells.
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Answers to SAQs
Chapter P2 5 x (x − x) (x − x)2
2.8 −0.22
1 Read the position of the meniscus at time 0, 3.1 0.047
and then again after a measured length of 2.9 0.08
time – say five minutes. The rate of movement 3.2 −0.12 0.007
of the meniscus will be proportional to the 2.9
rate of respiration of the yeast. 2.7 0.18 0.013
3.0 −0.12
2 All except c, the length of the capillary tubing. 2.8 −0.32 0.034
It is probably also unimportant to control the 2.9 −0.02
light intensity, f. 3.0 −0.22 0.013
3.2 −0.12
3 a Use a top pan balance to measure 1 g of 3.1 −0.02 0.100
glucose. Place in a volumetric flask with 3.0
some distilled water and dissolve fully. Make 3.2 0.18 0.000
up to 100 cm3 with distilled water. 3.0 0.08
3.1 −0.02 0.047
b Use a top pan balance to measure 2.5 g of 3.3 0.18
glucose. Place in a volumetric flask with 3.2 −0.02 0.013
some distilled water and dissolve fully. Make 2.9 0.08
up to 250 cm3 with distilled water. ∑x = 57.3 0.28 0.000
n = 19 0.18
c Calculate the relative molecular mass of x = 3.02 −0.12 0.034
glucose. This is (6 × 12) + (12 × 1) + (6 × 16)
= 72 + 12 + 96 = 180. 0.007
Use a top pan balance to measure 180 g
of glucose. Add this to a small amount of 0.000
distilled water in a volumetric flask. Add
distilled water to make the solution up to a 0.034
total volume of 1 dm3.
0.000
d Measure 0.5 dm3 of the 1 mol dm–3 solution
into a volumetric flask. Add water to make up 0.007
to 1 dm3.
0.081
4 They are all the same.
0.034
0.013
∑(x − x)2 = 0.48
n − 1 = 18
Σ (x – x)2 = 0.03
n–1
s = 0.16
6 s = 0.16
n = 19
s
SM = n
SM = 0.16
19
SM = 04..3166
SM = 0.04
7 No. The error bars overlap, showing that there
is no significant difference between the petal
lengths in the woodland and in the garden.
Cambridge International AS and A Level Biology © Cambridge University Press 2014
Cambridge International AS Level Biology Answers to end-of-chapter questions
Answers to EOCQs
Chapter 1 1 A; [1]
The mark schemes, suggested answers and 2 C; [1]
comments that appear in this CD-ROM were written
by the author(s). In examinations, the way marks 3 A; [1]
would be awarded to answers like these may be
different. 4
Notes about mark schemes Feature Light Electron
microscope microscope
A or accept indicates an alternative acceptable source of radiation light; electrons;
answer. wavelength of 400–700 nm; about 0.005 nm
R = reject. This indicates a possible answer that radiation used
should be rejected. maximum resolution 200 nm; 0.5 nm in
; The bold semi-colon indicates the award of 1 mark. practice
/ This indicates an alternative answer for the same
mark. The alternatives may be separated from the lenses glass electromagnets;
rest of the answer by commas.
( ) Text in brackets is not required for the mark. specimen living or non- non-living or
Underlining This is used to indicate essential living or dead; dead
word(s) that must be used to get the mark. stains
AW means ‘alternative wording’. It is used to image coloured dyes heavy metals;
indicate that a different wording is acceptable
provided the essential meaning is the same, and is coloured black and white;
used where students’ responses are likely to vary
more than usual. Award 1 mark for each correct entry in the table
AVP means ‘additional valid point’. This means (shown in italics). [8]
accept any additional points given by the student
that are not in the mark scheme, provided they are 5 nucleus; (smooth) endoplasmic reticulum;
relevant. But accept only as many additional points rough endoplasmic reticulum;
as indicated by the bold semi-colons, e.g. AVP;; 25 nm / larger / 80S ribosomes; linear / non-
means award a maximum of 2 extra marks. circular DNA;
ORA means ‘or reverse argument’ and is used when chromatin / chromosome(s);
the same idea could be expressed in the reverse lysosome(s); Golgi body; mitochondrion /
way. For example: ‘activity increases between pH mitochondria; microtubule(s);
2 and pH 5 ORA’ means accept ‘activity decreases centriole(s); centrosome; vacuole(s);
between pH 5 and pH 2’. microvillus / microvilli; cilium / cilia;
max. This indicates the maximum number of marks nucleolus / nucleoli; nuclear envelope;
that can be given. nuclear pore(s); AVP; [max. 10]
6 a magnification is the number of times larger
an image is compared with the real size of the
object; AW
resolution is the ability to distinguish
between two separate points / the greater
the resolution, the greater the detail that can
be seen; AW
a statement linking the terms, such as both
terms used with reference to microscopy; [3]
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Cambridge International AS Level Biology Answers to end-of-chapter questions
b light microscope uses light as a source of 7 In each case, one mark for any three
radiation; appropriate organelles.
electron microscope uses electrons as a
source of radiation; [2] a e.g. nucleolus, ribosome, centrioles,
centrosome, microtubule;;;
c both organelles / both found in eukaryotic cells;
nucleolus is located inside nucleus; b e.g. lysosomes, rough ER, smooth ER, Golgi
nucleus controls cell activity; body;;;
nucleolus makes ribosomes;
AVP;; e.g. nucleus surrounded by envelope, c nucleus, mitochondrion, chloroplast;;; [9]
no membrane round nucleolus [max. 4]
8 a Golgi body;
d chromatin and chromosomes both contain DNA b nucleolus;
(and protein / histones) / both found in nucleus; c ribosome;
chromatin is the loosely coiled form of d ER / rough ER;
chromosomes; e rough ER;
chromatin is the form that exists between cell f mitochondrion;
/ nuclear divisions; g nucleus;
chromosomes are formed just before / h chloroplast;
during, cell / nuclear division; [3] i centrosome (accept centriole);
j nucleus;
e an envelope consists of two membranes (one k membrane;
just inside / outside the other); l ribosome / microtubule; [12]
a membrane is a thin (partially permeable)
barrier found around cells and some 9 a 1 mark for each accurately measured
organelles; ‘observed size’ (to within ±2 mm) and 1 mark
example of at least one organelle surrounded for each accurately calculated ‘actual size’;;;;;;
by an envelope is given; 1 mark for applying the formula;
membranes found in / around all cells, 1 mark for measuring in mm and converting
envelopes only in eukaryotes; [max. 3] mm to µm for each calculation;
1 mark for rounding up actual size to no more
f both consist of flattened membrane-bound than one decimal place; [9]
sacs;
both found spreading through cytoplasm of b quality of drawing:
eukaryotic cells; sharp pencil used;
smooth ER lacks ribosomes, rough ER has more than half of available space used;
ribosomes on surface; clean, continuous lines / not sketchy;
one function of smooth ER given, e.g. makes interpretation of structures accurate;
lipids / steroids; representative parts of main organelles
rough ER transports proteins made by drawn, including those below for which label
ribosomes on its surface; [max. 4] marks are awarded; [5]
labels:
g prokaryotes have no nucleus, eukaryotes nucleus;
have nucleus; nuclear envelope;
prokaryotes are smaller / simpler; nuclear pore;
prokaryotes have few organelles, eukaryotes nucleolus;
have many organelles, some membrane- rough ER;
bound / compartmentalisation / more ribosome(s);
division of labour; mitochondrion;
one other important difference given / crista or cristae;
eukaryotes evolved from prokaryotes; [4] Golgi body;
[Total: 23] Golgi vesicle / secretory vesicle; [max. 9]
Cambridge International AS and A Level Biology © Cambridge University Press 2014
Cambridge International AS Level Biology Answers to end-of-chapter questions
c mitochondria will appear circular if they are
cut, in transverse section / across (the long
axis); AW [1]
d i A protein made on the ribosome is moving
into the rough ER;
B rough ER buds off small vesicles;
vesicles fuse to form the Golgi body;
(therefore) protein moves into Golgi body;
protein may be modified / processed inside
Golgi body;
C Golgi body buds off Golgi vesicles;
D Golgi vesicles travel to cell surface
membrane;
Golgi vesicle(s) fuses with cell surface
membrane;
protein / enzyme leaves cell;
exocytosis / secretion; [max. 8]
ii ribosome / messenger RNA; [1]
iii nuclear pore; [1]
iv ATP; [1]
[Total: 35]
1 0 a i 100 000 g
ii 1000 g
iii 10 000 g; [1]
b lysosomes are, similar in size to / slightly
smaller than, mitochondria;
therefore sediment at same / similar, g force
/ speed;
therefore contaminate mitochondrial
sample; AW
therefore cannot be sure whether effects
due to mitochondria or lysosomes in any
experiments; [4]
[Total: 5]
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Cambridge International AS Level Biology Answers to end-of-chapter questions
Answers to EOCQs
Chapter 2 1 D; [1]
The mark schemes, suggested answers and 2 C; [1]
comments that appear in this CD-ROM were written 3 B; [1]
by the author(s). In examinations, the way marks 4
would be awarded to answers like these may be
different.
Notes about mark schemes Globular protein,
e.g. haemoglobin
A or accept indicates an alternative acceptable Fibrous protein,
answer. e.g. collagen
R = reject. This indicates a possible answer that Monosaccharide
should be rejected. Disaccharide
; The bold semi-colon indicates the award of 1 mark. Glycogen
/ This indicates an alternative answer for the same Starch
mark. The alternatives may be separated from the Cellulose
rest of the answer by commas. Lipid
( ) Text in brackets is not required for the mark.
Underlining This is used to indicate essential Monomer ✘ ✘ ✔ ✘ ✘ ✘ ✘✘
word(s) that must be used to get the mark.
AW means ‘alternative wording’. It is used to Polymer ✔ ✔ ✘ ✘ ✔ ✔ ✔✘
indicate that a different wording is acceptable
provided the essential meaning is the same, and is Macromolecule ✔ ✔ ✘ ✘ ✔ ✔ ✔✘
used where students’ responses are likely to vary
more than usual. Polysaccharide ✘ ✘ ✘ ✘ ✔ ✔ ✔✘
AVP means ‘additional valid point’. This means
accept any additional points given by the student Contains subunits that form ✘ ✘ ✘ ✘ ✔ ✔ ✘✘
that are not in the mark scheme, provided they are branched chains
relevant. But accept only as many additional points
as indicated by the bold semi-colons, e.g. AVP;; Contains amino acids ✔ ✔ ✘ ✘ ✘ ✘ ✘✘
means award a maximum of 2 extra marks.
ORA means ‘or reverse argument’ and is used when Made from organic acids and ✘ ✘ ✘ ✘ ✘ ✘ ✘✔
the same idea could be expressed in the reverse glycerol
way. For example: ‘activity increases between pH
2 and pH 5 ORA’ means accept ‘activity decreases Contains glycosidic bonds ✘ ✘ ✘ ✔ ✔ ✔ ✔✘
between pH 5 and pH 2’.
max. This indicates the maximum number of marks Contains peptide bonds ✔ ✔ ✘ ✘ ✘ ✘ ✘✘
that can be given.
One of its main functions is to act ✘ ✘ ✘ accept ✔ ✔ ✘✔
as an energy store ✔ or ✘
Usually insoluble in water ✘ ✔ ✘ ✘ ✔ ✔ ✔✔
Usually has a structural function ✘ ✔ ✘ ✘ ✘ ✘ ✔✘
Can form helical or partly helical ✔ ✔✘ ✘ ✘ ✔ (see ✘ ✘
structures amylose)
Contains the elements carbon, ✘ ✘ ✔ ✔ ✔ ✔ ✔✔
hydrogen and oxygen only
[8]
5
Category Example
structural
collagen; keratin; AVP
enzyme e.g. elastin, viral coat
hormone; protein; [max. 2]
respiratory pigment / AW;
AVP e.g. amylase;
defensive
insulin
contractile / AW;
haemoglobin and
myoglobin
antibodies / fibrinogen /
AVP;
actin and myosin
storage casein / ovalbumin / AVP;
[8]
Cambridge International AS and A Level Biology © Cambridge University Press 2014
Cambridge International AS Level Biology Answers to end-of-chapter questions
6 dissolve easily in water; in the diagram above. The form used to make
sweet; the disaccharide is the beta form of galactose,
general formula (CH2O)n / contain the but students will not need to know this, other
elements carbon, hydrogen and oxygen / than for interest.
hydrogen and oxygen are present in ratio of
2 : 1; [3] e alpha glucose / α-glucose;
the –OH group on carbon atom 1 is below the
7 a lactose could be a source of energy; ring; [2]
it could be digested to, monosaccharides /
glucose and galactose, which could then be f carry out a Benedict’s test on both solutions;
used as building blocks for larger molecules; lactose would give a brick-red / brown
[2] precipitate, sucrose would not;
accept positive result for lactose, negative
b condensation; [1] result for sucrose [2]
[Total: 10]
c glycosidic bond; [1]
8 a i vi alternative answers
d 6CH2OH O OH on C atom 1 above CH3 O AA OH
5 ring (β-galactose) H CC H HO H iv
OH N H NC C N
4 H OH CH2 O
1 H H
H CC
DH H H H OH
32
v peptide bonds
H OH
C of COOH joined to N of NH2 for both peptide
galactose bonds;
(drawn according to convention) peptide bonds shown as C=O joined to –NH
(i.e. water has been eliminated);
H OH all three amino acids joined and in correct
sequence; accept even if errors in bonding [3]
3 2 ii primary structure; [1]
H iii water; [1]
H OH H iv ring drawn around –OH or whole R group
4
1 (–CH2OH) of serine; [1]
OH H v rings drawn around two peptide bonds
5 O OH
and bonds labelled appropriately; [1]
6CH2OH vi ring drawn around –NH group one side of a
galactose peptide bond and group labelled A; [1]
(molecule rotated 180° in order b held in place by hydrogen bonding;
to line up with α-glucose) secondary structures;
all the –NH and –C=O groups of, peptide
H 6CH2OH O H bonds / polypeptide backbone, are involved;
4 5 1 [3]
OH H OH c molecule made from repeating subunits;
subunits similar or identical to each other;
OH H giant molecule / macromolecule; [max. 2]
2
d i XXX, XXY, XYY, XYX, YYY, YYX, YXX, YXY; [1]
3 ii 23; [1]
H OH
[Total: 15]
α-glucose OH on C atom 1
below ring
glucose correctly drawn;
galactose correctly drawn; [2]
Carbon atoms need not be numbered. Note
that galactose will probably be drawn
‘upside down’ as in the disaccharide – the
conventional way of drawing it is also shown
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Cambridge International AS Level Biology Answers to end-of-chapter questions
9 a A identified as lipid, d carbon, hydrogen, oxygen, nitrogen, iron;;
B identified as phospholipid; [1] 2 marks for all five correct, 1 mark for four
correct, 0 marks for 3 or fewer correct [2]
b i junction between head and tail for all [Total: 10]
three tails is indicated on diagram;;
Allow 1 mark if only one or two junctions
indicated [2]
ii fatty acids;
glycerol; [2]
c head of phospholipid is labelled phosphate;
[1]
d i phospholipid / B; [1]
ii phosphate is, charged / polar / hydrophilic;
[1]
e lipid:
energy store / insulator / buoyancy / source
of metabolic water / any other suitable
example;
phospholipid:
any reference to the importance of
phospholipids in structure of membranes; [2]
[Total: 10]
1 0 a Collagen Haemoglobin
Globular or fibrous globular
fibrous?
Entirely entirely partly
or partly
helical?
Type of helix triple helix/ alpha
extended helix/
three-stranded
Prosthetic no yes
group
present?
Soluble in no/insoluble yes/soluble
water?
1 mark for each correct row. No half marks. [5]
b 1 mark for structural feature, 1 mark
for linking this feature to its function,
e.g. haemoglobin contains iron.
iron combines with oxygen; [2]
c molecule has more than one polypeptide
chain; [1]
R molecule has four polypeptide chains
Cambridge International AS and A Level Biology © Cambridge University Press 2014
Cambridge International AS Level Biology Answers to end-of-chapter questions
Answers to EOCQs
Chapter 3 1 B; [1]
The mark schemes, suggested answers and 2 D; [1]
comments that appear in this CD-ROM were written
by the author(s). In examinations, the way marks 3 D; [1]
would be awarded to answers like these may be
different. 4 C; [1]
Notes about mark schemes 5 straight line drawn from origin at zero to
show steepest gradient of curve; [1]
A or accept indicates an alternative acceptable
answer. 6 a maximum activity / optimum pH, is pH 5;
R = reject. This indicates a possible answer that activity gradually increases between pH 2
should be rejected. and pH 5, and decreases from pH 5 to pH 10;
; The bold semi-colon indicates the award of 1 mark. activity very low at pH 2 and pH 10; AW
/ This indicates an alternative answer for the same [max. 2]
mark. The alternatives may be separated from the
rest of the answer by commas. b pH is a measure of the hydrogen ion
( ) Text in brackets is not required for the mark. concentration;
Underlining This is used to indicate essential hydrogen ions are positively charged;
word(s) that must be used to get the mark. hydrogen ions can interact with the R groups
AW means ‘alternative wording’. It is used to of amino acids;
indicate that a different wording is acceptable affects ionic bonding / affects ionisation of
provided the essential meaning is the same, and is R groups;
used where students’ responses are likely to vary affects tertiary structure / affects 3D shape of
more than usual. enzyme;
AVP means ‘additional valid point’. This means therefore substrate may not fit active site
accept any additional points given by the student (as precisely); [max. 4]
that are not in the mark scheme, provided they are [Total: 6]
relevant. But accept only as many additional points
as indicated by the bold semi-colons, e.g. AVP;; 7 a optimum temperature; [1]
means award a maximum of 2 extra marks.
ORA means ‘or reverse argument’ and is used when b 37 °C; accept 40 °C [1]
the same idea could be expressed in the reverse
way. For example: ‘activity increases between pH c as temperature increases the kinetic energy
2 and pH 5 ORA’ means accept ‘activity decreases of the molecules increases;
between pH 5 and pH 2’. the rate of collision between substrate and,
max. This indicates the maximum number of marks enzyme / active site, increases;
that can be given. rate of reaction increases; [3]
d the enzyme is gradually being denatured;
when the rate is zero the enzyme is
completely denatured;
ORA enzyme loses tertiary structure;
substrate no longer fits into active site
/ active site loses its (specific) shape so
substrate does not fit;
AVP e.g.hydrogen bonds broken / increased
vibration of enzyme molecule; [max. 3]
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e the extra energy which must be given to the 1 0 a replication increases reliability; AW [1]
substrate;
before it can be converted into the product; b to act as a reference to show what happens if
[2] there is no denaturation; AW [1]
[Total: 10] c 40 °C is the optimum temperature for a
mammalian enzyme; [1]
8 a succinic acid; [1] d enzyme / amylase (molecules) diffuse(s) from
wells into the agar;
b malonic acid acts as a competitive inhibitor; enzyme / amylase digests the starch;
it has a similar shape / structure to succinic to maltose;
acid; forms rings / halos, of digested starch around
it therefore competes with succinic acid for a the wells;
place in the active site of the enzyme; [3] amount of digestion / rate of digestion, is
c i cysteine; [1] related to degree of denaturation of enzyme /
ii –SH groups form disulfide bridges; amylase; [max. 4]
used to determine tertiary structure; e the more enzyme / amylase added, the
greater the amount of digestion of starch
heavy metal would prevent formation of or
disulfide bridges; want results to be due to differences in
preheating times, not to differences in
could change shape of active site; amount of amylase / enzyme; AW [1]
heavy metal could affect shape either by f
binding directly in the active site, or by
binding at another site which then results Time (heated) at 60 °C / min Diameter of halo / mm
in change in shape of the active site;
substrate would not be able to fit into 0 24
active site; [max. 4]
iii (non-competitive) irreversible; [1] 1 19
[Total: 10] 5 10
9 a carry out Benedict’s test on solutions A, B 10 6
and C;
a positive result / brick-red precipitate will be 30 0
seen, with the glucose solution;
heat separate samples of the two remaining table drawn with ruled lines for border and
solutions, in boiling water bath / to high to separate columns and headings (ideally
temperature (e.g. 80 °C), for suitable time / at ruled lines between rows, but not essential
least two minutes (enzyme will be denatured); for mark);
for each heated solution, mix it with an correct headings to columns with units;
unheated sample of the other solution; first column is independent variable (Time
leave several minutes / suitable time (for heated at 60 °C);
reaction to take place); correct measurements of halos; [4]
carry out Benedict’s test on the two tubes; g measure the four halos and calculate the
only one will give a positive result (due to mean; [1]
presence of maltose) and this will be the one (any anomalous results should be ignored)
which contained the unheated enzyme;
Accept alternative wording for all steps in
the procedure, provided the same logical
sequence is described [max. 6]
b hydrolysis; [1]
[Total: 7]
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h 24 Diameter /mm
20
16
12
8
4
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30
Time at 60˚C /minutes
x-axis (horizontal axis) is labelled ‘Time 40 °C is a control (for reference to find out size
(heated) at 60 °C’, y-axis (vertical axis) is of halo with no denaturation);
labelled ‘Diameter’; at least five temperatures, e.g. 40, 60, 80, 100,
units given on axes, min / minutes and mm; 120 °C;
regular intervals on both axes (check that heat for suitable length of time (e.g. one
0, 1, 5, 10, 30 are not regularly spaced on hour, at least ten min);
x-axis); cool to room temp / 40 °C, add equal volumes
points plotted accurately; to wells in starch–agar plates, replicate wells
points joined with straight lines or in each plate (e.g. four), leave 24 hours,
smooth curve; [5] test for starch, measure diameters of halos;
[max. 5]
i enzyme was completely denatured after Background information: amylase enzymes
30 minutes; from the bacterium Bacillus licheniformis and
rate of denaturation was rapid at first and the fungus Aspergillus have been developed by
then gradually slowed down; biotechnology companies for use in industrial
data quoted; processes. For example, a bacterial amylase
enzyme loses tertiary structure; that functions in the range 90–110 °C has
substrate no longer fits into active site been developed and is used in beer brewing
/ active site loses its (specific) shape so and other processes, and a fungal amylase
substrate does not fit; that operates in the range 50–60 °C is used for
AVP e.g. hydrogen bonds broken / increased pastry baking and maltose syrup production.
vibration of enzyme molecule; [max. 4]
k pH;
j heat samples of mammalian, fungal and substrate concentration;
bacterial amylases at different temperatures; enzyme concentration; [3]
suitable range, e.g. between 40 °C and 120 °C; [Total: 30]
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11 a see Figure 3.14b. Award 1 mark for each
correct label;;; [3]
b inhibitor A had no effect on Vmax;
and increased Km; [2]
c inhibitor B decreased Vmax;
and had no effect on Km; [2]
d inhibitor A is competitive, B is non-
competitive;
A is competitive because:
it increased Km / did not affect Vmax;
decreased the affinity of the enzyme for its
substrate;
the substrate is competing with the inhibitor
for the active site;
the inhibition is overcome by increasing
substrate concentration; [max. 4]
Or
Alternative ways of explaining the same
marking points;
B is non-competitive because:
it did not affect Km/decreased Vmax;
did not affect the affinity of the enzyme for its
substrate;
the substrate is not competing with the
inhibitor for the active site;
the inhibition cannot be overcome by
increasing substrate concentration;
e i Z; [1]
ii Y; [1]
iii X; [1]
Reasons:
Accept any valid points up to a maximum of
2 marks for each inhibitor, for example:
ii the lines Y and Z cross the y-axis at the
same point, which is 1/Vmax;
therefore Vmax is the same for both;
line Y meets the x-axis at a less negative
value than line Z;
therefore Km is increased; [max. 2]
iii the lines X and Z cross the y-axis at the
same point, (which is –1/Km );
therefore both have the same Km;
line X crosses the y-axis higher than line Z,
so 1/Vmax has a higher value;
therefore Vmax has a lower value; [max. 2]
[Total: 18]
Cambridge International AS and A Level Biology © Cambridge University Press 2014
Cambridge International AS Level Biology Answers to end-of-chapter questions
Answers to EOCQs
Chapter 4 1 C; [1]
The mark schemes, suggested answers and 2 D; [1]
comments that appear in this CD-ROM were written
by the author(s). In examinations, the way marks 3 D; [1]
would be awarded to answers like these may be
different. 4 Information for constructing this table can be
found on pages 72–73.
Notes about mark schemes Suggested mark allocation:
phospholipids: [2]
A or accept indicates an alternative acceptable cholesterol: [4]
answer. glycolipids: [3]
R = reject. This indicates a possible answer that glycoproteins: [3)
should be rejected. proteins: [3]
; The bold semi-colon indicates the award of 1 mark. [Total: 15]
/ This indicates an alternative answer for the same
mark. The alternatives may be separated from the 5 a Information for answering this question can
rest of the answer by commas. be found on page 77 and in the answer to
( ) Text in brackets is not required for the mark. Question 4.8.
Underlining This is used to indicate essential Award 1 mark for each term correctly used [14]
word(s) that must be used to get the mark.
AW means ‘alternative wording’. It is used to b Information for answering this question can
indicate that a different wording is acceptable be found on page 00 and in the answer to
provided the essential meaning is the same, and is Question 4.8.
used where students’ responses are likely to vary Award 1 mark for each term correctly used [15]
more than usual. [Total: 29]
AVP means ‘additional valid point’. This means
accept any additional points given by the student 6 a A phosphate head (of phospholipid);
that are not in the mark scheme, provided they are B fatty acid tail(s) (of phospholipid);
relevant. But accept only as many additional points C phospholipid bilayer / membrane; [3]
as indicated by the bold semi-colons, e.g. AVP;;
means award a maximum of 2 extra marks. b Award max. of 2 marks: 2 or 3 correct answers
ORA means ‘or reverse argument’ and is used when 1 mark, 4 correct answers 2 marks
the same idea could be expressed in the reverse
way. For example: ‘activity increases between pH i hydrophilic
2 and pH 5 ORA’ means accept ‘activity decreases
between pH 5 and pH 2’. ii hydrophobic
max. This indicates the maximum number of marks
that can be given. iii hydrophobic
iv hydrophilic; [2]
c ions move by diffusion;
channel has shape which is specific for
particular ion;
channel is hydrophilic / water-filled / allows
movement of polar substance;
ions move down concentration gradient;
[max. 3]
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Cambridge International AS Level Biology Answers to end-of-chapter questions
d both intrinsic proteins; 8 Award 1 mark for each correct row
both have specific shape; [2]
Feature Cell wall Cell
e channel proteins have a fixed shape / carrier membrane
proteins have a variable shape; [1]
is the thickness µm nm
f width of C measured in mm; normally
mm converted to µm and µm converted to measured in nm
nm; or µm?
correct formula used magnification: M = I/A =
width of C/7 accept mm, µm or nm; location surround some surround all
correct answer in nm; [4] cells / not animal cells / may be
[Total: 15] cells / only outside found inside
/ surrounding cells cells
7
Uses Uses Specific Controllable chemical contains cellulose phospholipids,
Process energy proteins by cell composition in plants, protein,
peptidoglycans (sometimes)
diffusion ✘✘ ✘ ✘ accept any / murein in cholesterol
statements prokaryotes, (chitin
osmosis ✘✘✔ ✘ that serve to in fungi) / contains
distinguish a strengthening
facilitated ✘ ✔ ✔ ✘ between cell material / contains
diffusion wall and cell a polysaccharide
membranes. (or polysaccharide-
active ✔ ✔ ✔ ✔ Examples are like substance) AW
transport given.
endocytosis ✔ permeability freely permeable partially
and ✔ ✘ ✔ permeable
exocytosis
[20] function mechanical selective
NB: It could be argued that facilitated strength barrier AW
diffusion is controllable, because the number
of channel proteins in the membrane can fluid or rigid rigid fluid
affect the rate.
[6]
9 description:
rate of entry of water is rapid at first but
slows down gradually;
until rate is zero / no further entry of water
or water enters until water potential of
cell = water potential of pure water = 0
(= equilibrium);
exponential / not linear;
rate depends on / proportional to, difference
in water potential between cell and external
solution; [max. 3]
explanation:
water (always) moves from a region of higher
water potential to a region of lower water
potential;
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(in this case) by osmosis; NB: This is similar to the effect of substrate
through partially permeable cell surface concentration on rate of enzyme activity.
membrane of cell; [max. 5]
as cell fills with water, cell / protoplast [Total: 12]
expands and pressure (potential) increases; 1 1 a the living contents of a plant cell; [1]
until water potential of cell = zero / water b i at 90% = 22 kPa (accept 21 or 23 kPa), at
potential of pure water; 95% = 100 kPa, at 100% = 350 kPa; [1]
cell wall rigid / will not stretch (far), and
prevents entry of more water; ii change 90–95 % = 78 kPa (accept 77 or
cell is turgid; [max. 5] 79 kPa);
[Total: 8] change 95–100% = 250 kPa; [2]
1 0 a the greater the concentration difference, the iii as water enters the cell, the cell wall is
greater the rate of transport; [1] stretched / protoplast pushes against cell
wall;
b (net) diffusion and facilitated diffusion only
occur if there is a concentration, difference / cell wall is (relatively) rigid;
gradient, across the membrane
or water cannot be compressed;
at equilibrium / if no concentration
difference, there is no, net exchange / therefore pressure builds up more and
transport across membrane / rate of more rapidly (for given volume of water)
transport, is same in both directions; AW / small increase in amount of water has
active transport can occur even if no large effect on pressure; AW [max. 2]
concentration difference;
because molecules / ions are being pumped; (This could be compared with pumping up
AW [3] a bicycle tyre – pressure increases much
more rapidly for a given amount of air
c i active transport; [1] towards the end due to the elastic limit of
ii active transport depends on a supply of the tyre being reached.)
ATP; iv 350 kPa; [1]
provided by respiration; [2]
d graph for diffusion is linear / straight line c i zero (kPa); [1]
(with no maximum rate); ii 86%; [1]
purely physical process / not dependent
on transport proteins / channel or carrier iii incipient plasmolysis; [1]
proteins;
graph for facilitated diffusion is a curve with a iv water potential = solute potential; [1]
maximum rate; AW
facilitated diffusion depends on presence v ψ = ψs + ψp; [1]
of, transport proteins / channel or carrier vi the cell continues to lose water /
proteins;
as concentration increases, the receptor protoplast continues to shrink;
sites of these proteins become more and
more saturated / the more saturated these protoplast pulls away from cell wall =
plasmolysis;
shrinks until equilibrium is reached;
when water potential of cell = water
potential of outside solution;
solute potential gets lower / more
negative;
become, the less the effect of increasing because cell contents becoming more
concentration; concentrated; [max. 5]
rate reaches a maximum when all, transport
/ channel or carrier proteins, are working at d only a small amount of water is needed to
full capacity / when all receptor sites are, full bring about a large change in pressure;
/ saturated; because the cell wall is (relatively) rigid;
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this is not enough to significantly change the
concentration of the cell contents; AW [3]
[Total: 20]
1 2 a if it were diffusion, there would be (net)
movement of ions from a region of
higher concentration to a region of lower
concentration until equilibrium is reached
when concentration inside = concentration
outside; AW [1]
R because concentrations different inside
and outside
b active transport;
active transport involves pumping ions
against a concentration gradient; [2]
c if respiration is inhibited, no ATP is produced;
active transport uses ATP as energy source;
active transport stops;
diffusion continues;
ions move down concentration gradients by
diffusion until equilibrium reached; [max. 4]
[Total: 7]
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Cambridge International AS Level Biology Answers to end-of-chapter questions
Answers to EOCQs
Chapter 5 1 D; [1]
The mark schemes, suggested answers and 2 B; [1]
comments that appear in this CD-ROM were written
by the author(s). In examinations, the way marks 3 D; [1]
would be awarded to answers like these may be
different. 4 centrosome is a microtubule organising
centre;
Notes about mark schemes makes spindle during mitosis;
contains two centrioles;
A or accept indicates an alternative acceptable located just outside nucleus;
answer. centriole has 9 triplets of microtubules;
R = reject. This indicates a possible answer that no function (during mitosis);
should be rejected. centromere is region of a chromosome that
; The bold semi-colon indicates the award of 1 mark. holds two 2 chromatids together;
/ This indicates an alternative answer for the same point of attachment for microtubules during
mark. The alternatives may be separated from the mitosis;
rest of the answer by commas. point of attachment of chromatids to
( ) Text in brackets is not required for the mark. spindle;
Underlining This is used to indicate essential AVP; [max. 6]
word(s) that must be used to get the mark.
AW means ‘alternative wording’. It is used to 5 a A anaphase; B prophase; C metaphase; [3]
indicate that a different wording is acceptable b Information for this answer can be found in
provided the essential meaning is the same, and is
used where students’ responses are likely to vary Figure 5.10 on page 92.
more than usual. 1 mark for each correct summary. [3]
AVP means ‘additional valid point’. This means
accept any additional points given by the student [Total: 6]
that are not in the mark scheme, provided they are
relevant. But accept only as many additional points 6 a i metaphase; [1]
as indicated by the bold semi-colons, e.g. AVP;; ii prophase drawing shows two single
means award a maximum of 2 extra marks.
ORA means ‘or reverse argument’ and is used when chromosomes, each with a centromere
the same idea could be expressed in the reverse (not paired chromatids), ‘randomly’
way. For example: ‘activity increases between pH distributed, surrounded by cell surface
2 and pH 5 ORA’ means accept ‘activity decreases membrane but with no spindle; [1]
between pH 5 and pH 2’. b a long and a short chromatid, each with
max. This indicates the maximum number of marks a centromere, are shown inside each
that can be given. new nucleus; [1]
c six chromatids about half-way between
equator and each pole (12 chromatids in all);
two long, two short, two hooked in each
direction;
centromere leading for each chromatid; [3]
[Total: 6]
7 a microtubules are made out of tubulin
molecules;
the tubulin molecules stick together in a
particular pattern to form the microtubules,
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so the presence of colchicine would interfere 9 Award 1 mark for correct statement (True or
with this; AW [2] False). No explanation is required.
A True;
b spindle; Centrosomes replicate during interphase,
centrioles; [2] before M phase begins.
c (held up in) prophase; B True;
spindle cannot form (due to presence of
colchicine); Sister chromatids are formed by the
replication of DNA. Each contains one
so metaphase cannot occur; daughter DNA molecule identical to the
metaphase, normally follows prophase / is
next stage of mitosis; [max. 3] parent molecule.
[Total: 7] C False;
Microtubules extend from the kinetochore to
the nearest pole. The kinetochores in sister
8 a 8.7 cm = 87 mm = 87 000 µm; chromatids are connected to opposite poles.
(87 000 µm of DNA packed into 10 µm of D False;
chromosome) therefore packing ratio = This occurs during M phase during spindle
87 000 ÷ 10; manufacture (polymerisation) and chromatid
= 8700; [3] movement (depolymerisation).
b total length of chromosomes = 46 × 6 µm = E False;
276 µm; Kinetochores are found on chromatids.
1.8 m = (1800 mm =) 1800 000 µm; F False;
(1.8 m/1 800 000 µm of DNA packed into Telomeres are the caps at the ends of
276 µm of chromosomes) therefore packing chromosomes. Microtubules are attached at
ratio = 1 800 000 ÷ 276; the centromeres (kinetochores).
= 6522; [4] G True;
c 4.5; (50 ÷ 11) [1] Chromatids separate at the start of
d linker DNA is 53 × 0.34 nm in length; anaphase.
= 18.02 nm; accept 18 nm
therefore 50 + 18 = 68 nm of DNA in 11 + 18 [Total: 7]
= 29 nm of chromatin;
packing ratio = 68 ÷ 29;
= 2.34; accept 2.3 [5]
e 40; (40 000 ÷ 1000 remember that [1]
1 µm = 1000 nm)
f it suggests that any change in amino acid
composition would be harmful (so only
organisms with conserved molecules survive);
the way they function must therefore involve
nearly all their amino acids / they may have
a very precise shape which is vital for their
functioning; AW [2]
g i (1014 × 1.8 m = )1.8 × 1011 km; [1]
ii Distance to Sun and back
= 2 × 1.5 × 108 = 3 × 108 km;
1.8 × 1011 km ÷ 3 × 108 km
(= 1.8 × 103 ÷ 3 = 0.6 × 103);
= 600 times; [3]
[Total: 18]
Cambridge International AS and A Level Biology © Cambridge University Press 2014
Cambridge International AS Level Biology Answers to end-of-chapter questions
Answers to EOCQs
Chapter 6 1 B; [1]
The mark schemes, suggested answers and 2 C; [1]
comments that appear in this CD-ROM were written
by the author(s). In examinations, the way marks 3 D; [1]
would be awarded to answers like these may be
different. 4 a P adenine P cytosine P guanine
sugar sugar
Notes about mark schemes sugar
[1]
A or accept indicates an alternative acceptable b P thymine
answer. [1]
R = reject. This indicates a possible answer that sugar
should be rejected. [1]
; The bold semi-colon indicates the award of 1 mark. c P uracil
/ This indicates an alternative answer for the same
mark. The alternatives may be separated from the sugar
rest of the answer by commas.
( ) Text in brackets is not required for the mark. 5 nucleotide:
Underlining This is used to indicate essential a molecule made up of a pentose sugar, a
word(s) that must be used to get the mark. phosphate group and a nitrogenous base;
AW means ‘alternative wording’. It is used to nucleic acid:
indicate that a different wording is acceptable a polymer of nucleotides / a polynucleotide;
provided the essential meaning is the same, and is [2]
used where students’ responses are likely to vary
more than usual. 6 labels should include:
AVP means ‘additional valid point’. This means parent molecule;
accept any additional points given by the student daughter molecules;
that are not in the mark scheme, provided they are parent / old strand acts as template;
relevant. But accept only as many additional points new strands made from nucleotides binding to
as indicated by the bold semi-colons, e.g. AVP;; old strands by complementary base pairing; [4]
means award a maximum of 2 extra marks.
ORA means ‘or reverse argument’ and is used when 7 Met–Phe–Pro–Asp–[stop]; [2]
the same idea could be expressed in the reverse
way. For example: ‘activity increases between pH 8 tRNA DNA triplet from which mRNA
2 and pH 5 ORA’ means accept ‘activity decreases anticodon was transcribed
between pH 5 and pH 2’. mRNA
max. This indicates the maximum number of marks codon
that can be given.
UUA AAU AAT
UUG AAC AAC
CUU GAA GAA
CUC GAG GAG
CUA GAU GAT
CUG GAC GAC
[6]
Cambridge International AS and A Level Biology © Cambridge University Press 2014
Cambridge International AS Level Biology Answers to end-of-chapter questions
9 a gene mutation / substitution; [1] 1 2 a X mRNA;
Y ribosome;
b Val / valine; [1] Z (poly)peptide chain / chain of amino acids;
c sickle cell anaemia; [1] [3]
[Total: 3] b from left to right;
1 0 a many amino acids have more than one triplet increasing length of polypeptide chain; [2]
code; [Total: 5]
so sequence of amino acids is unchanged; [2]
b adding or deleting three nucleotides may add 1 3 a the DNA in the spleen and thymus of the
or remove the coding for one amino acid; same organism is the same;
this may not affect the final shape of the the same genes are present in both organs;
protein; [2]
adding or deleting one nucleotide affects the b the DNA in different species is different;
arrangement of all subsequent triplets; different genes are present; [2]
this ‘frameshift’ may alter the coding of c DNA has double helix / is double stranded;
all amino acids following the addition or the numbers of A and T, and of C and G, are
deletion; similar because A pairs with T and C pairs
a triplet may be altered to a stop signal; with G; [2]
[max. 4]
[Total: 6] d the DNA is single stranded;
no base pairing occurs; [2]
11 [Total: 8]
Transcription Translation
Site in cell where nucleus ribosome (in
process occurs cytoplasm)
Molecule used as a DNA (in cytoplasm)
template in process
Molecule used
as a template
in process DNA
mRNA
Molecule produced mRNA polypeptide /
by process protein
Component RNA amino acids
molecules nucleotides
(monomers) used in
process
Name one other type RNA tRNAs / enzyme
of molecule that is polymerase (peptidyl
essential for process transferase)
to occur / ribosomal
RNA (rRNA)
/ ribosomal
protein
Award 1 mark for each correct row [5]
Cambridge International AS and A Level Biology © Cambridge University Press 2014
Cambridge International AS Level Biology Answers to end-of-chapter questions
Answers to EOCQs
Chapter 7 1 A; [1]
The mark schemes, suggested answers and 2 B; [1]
comments that appear in this CD-ROM were written
by the author(s). In examinations, the way marks 3 C; [1]
would be awarded to answers like these may be
different. 4 a refer to pages 131–132 1 mark for each valid
point up to a max of 3 [3]
Notes about mark schemes
b refer to pages 129–131 1 mark for each valid
A or accept indicates an alternative acceptable point up to a max of 4 [4]
answer.
R = reject. This indicates a possible answer that c refer to pages 127–129 1 mark for each valid
should be rejected. point up to a max of 3 [3]
; The bold semi-colon indicates the award of 1 mark. [Total: 10]
/ This indicates an alternative answer for the same
mark. The alternatives may be separated from the 5 a i vessel elements;
rest of the answer by commas.
( ) Text in brackets is not required for the mark. tracheids;
Underlining This is used to indicate essential
word(s) that must be used to get the mark. parenchyma;
AW means ‘alternative wording’. It is used to
indicate that a different wording is acceptable fibres; [max. 3]
provided the essential meaning is the same, and is
used where students’ responses are likely to vary ii sieve (tube) elements;
more than usual.
AVP means ‘additional valid point’. This means companion cells;
accept any additional points given by the student
that are not in the mark scheme, provided they are parenchyma;
relevant. But accept only as many additional points
as indicated by the bold semi-colons, e.g. AVP;; fibres; [max. 3]
means award a maximum of 2 extra marks.
ORA means ‘or reverse argument’ and is used when b vessel element:
the same idea could be expressed in the reverse transport of water / support / transport of
way. For example: ‘activity increases between pH mineral ions;
2 and pH 5 ORA’ means accept ‘activity decreases tracheid:
between pH 5 and pH 2’. transport of water / support / transport of
max. This indicates the maximum number of marks mineral ions;
that can be given. sieve element:
transport of, sucrose / organic solutes;
companion cells:
loading / unloading, phloem (sieve element) /
forms functional unit with sieve element;
parenchyma:
storage / gas exchange;
fibres:
support / mechanical strength;
[max. 4]
[Total:10]
6 soil solution > root hair cell > xylem vessel
contents > mesophyll cell > dry atmospheric
air; [1]
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7 a the lower the relative humidity, the higher therefore water enters plant through roots;
the tension / the lower the hydrostatic time delay between rate of transpiration and
pressure, in the xylem; rate of water uptake is due to time taken
more evaporation from leaf (mesophyll cells) for effect of transpiration to be transmitted
when low relative humidity; through the plant; AW [max. 4]
results in lower water potential in leaf [Total: 10]
(mesophyll cells); 1 0 a to rinse out / remove, any existing K+ ions
therefore more water moves from xylem from the root;
(vessels to replace water lost from leaf ); present in the apoplast; [2]
down a water potential gradient;
sets up tension in the xylem vessels; [max. 4] b there is a rapid initial uptake of K+ ions;
lasts for about 10 minutes; accept 10–20
b lowest / most negative, hydrostatic pressure minutes;
is at the top of the tree; ORA the rate of uptake then slows down and
because water is being lost at the top of the remains steady / linear / continuous; [3]
tree;
this sets up a tension which is greatest at the c rapid uptake of K+ ions for first 10–20 minutes
top of the tree; / as Expt A for first 10–20 minutes;
there is a, hydrostatic pressure / tension, but not quite as rapid as Expt A / at 25 °C;
then no further uptake; [3]
gradient in the xylem vessels;
some pressure is (inevitably) lost on the way d initial rapid phase of uptake is due to uptake
down the tree; [max. 3] of K+ ions into the apoplast;
[Total: 7] by mass flow;
8 transpiration / loss of water vapour / loss of or by diffusion;
second phase / after 10–20 minutes due to
water by evaporation, from the leaves occurs active transport;
during the day;
because the stomata are open; which happens as soon as uptake starts
(0 minutes);
this results in tension in the xylem (vessels); at 0 °C enzyme activity is reduced / stops;
walls of xylem vessels are pulled slightly
inwards / vessels shrink slightly; AW therefore no active transport (because active
transport depends on respiration which is
overall effect is for diameter of whole trunk controlled by enzymes); [max. 5]
to, shrink / get smaller;
stomata close at night, so no transpiration at e no further uptake of K+; [1]
night; [max. 3] f inhibits respiration, therefore inhibits active
transport;
9 a the loss of water vapour; which is dependent on ATP produced during
from the leaves / from the surface of a plant; [2] respiration; [2]
b light intensity; [Total: 16]
temperature; [2] 1 1 a hydrogen ions are actively transported out of
c rate of water uptake shows the same pattern the, sieve element–companion cell; [1]
as rate of transpiration; AW
but there is a time delay, with changes in rate b there are more hydrogen ions / there is a
of transpiration occurring before changes in build-up of hydrogen ions, outside the sieve
water uptake; AW [2] element–companion cell units compared
with inside;
d transpiration causes water uptake; hydrogen ions are positively charged; [2]
loss of water (by transpiration) sets up a
water potential gradient in the plant; c ATP is needed for the active transport of
water potential in roots is lower than water hydrogen ions out of the tubes; [1]
potential in soil; [Total: 4]
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Cambridge International AS Level Biology Answers to end-of-chapter questions
1 2 a actual length = observed length/
magnification or A = I/M;
observed length of sieve element = 50 mm
(allow ±1 mm);
actual length = 50 mm/150 = 0.50 mm; accept
conversion of mm to µm: answer = 500 µm [3]
b i 1 metre = 1000 mm;
1000/0.50= 2000;
or
1 metre = 1 000 000 µm;
1 000 000/500 = 2000; [2]
ii to maintain the pressure gradient inside
the sieve tubes;
without the sieve plates the different
pressures at source and sink would
quickly equilibrate; [max. 1]
iii sieve pores; [1]
c (sieve element is 0.50 mm long)
(1 hour = 3600 seconds)
3600 seconds to travel 1 metre;
therefore: 0.50/1000 × 3600 seconds to travel
0.50 mm;
= 1.8 seconds (to one decimal place);
Accept 500 µm and 1 000 000 (µm) instead of
0.50 mm and 1000 (mm). [3]
[Total: 10]
13 a i when seed is forming / just after [1]
fertilisation;
ii germination; [1]
iii young immature leaf / leaf that is still
growing; [1]
iv mature photosynthesising leaf; [1]
v when food is being accumulated / when
storage organ is growing (in size) /
developing / end of plant’s growing season
/ just before winter; [1]
vi when plant starts to grow (using food from
the storage organ); [1]
b i to make starch;
respiration; [2]
ii to make cellulose;
respiration; [2]
[Total: 10]
Cambridge International AS and A Level Biology © Cambridge University Press 2014
Cambridge International AS Level Biology Answers to end-of-chapter questions
Answers to EOCQs
Chapter 8 1 C; [1]
The mark schemes, suggested answers and 2 C; [1]
comments that appear in this CD-ROM were written
by the author(s). In examinations, the way marks 3 D; [1]
would be awarded to answers like these may be
different. 4 B; [1]
Notes about mark schemes 5 Information for constructing this table can be
found on pages 158–163 [6]
A or accept indicates an alternative acceptable
answer. 6 Information for constructing this table can be
R = reject. This indicates a possible answer that found on pages 164–166 [6]
should be rejected.
; The bold semi-colon indicates the award of 1 mark. 7 the haemoglobin molecule is a protein with
/ This indicates an alternative answer for the same quaternary structure;
mark. The alternatives may be separated from the hydrogen bonds, ionic bonds and van der
rest of the answer by commas. Waals forces hold the protein in its three-
( ) Text in brackets is not required for the mark. dimensional shape;
Underlining This is used to indicate essential the primary structure of each polypeptide
word(s) that must be used to get the mark. chain determines how the chain will fold /
AW means ‘alternative wording’. It is used to where the bonds will form (thus determining
indicate that a different wording is acceptable its three dimensional shape);
provided the essential meaning is the same, and is the haemoglobin molecule has R groups
used where students’ responses are likely to vary with small charges on its outer surface
more than usual. (hydrophilic R groups), which help to make it
AVP means ‘additional valid point’. This means soluble in water;
accept any additional points given by the student this allows it to dissolve in the cytoplasm of a
that are not in the mark scheme, provided they are red blood cell;
relevant. But accept only as many additional points each haemoglobin molecule is made up of
as indicated by the bold semi-colons, e.g. AVP;; four polypeptide chains, each with a haem
means award a maximum of 2 extra marks. group at its centre;
ORA means ‘or reverse argument’ and is used when each haem group can bind reversibly with
the same idea could be expressed in the reverse one oxygen molecule;
way. For example: ‘activity increases between pH when one oxygen molecule binds with one
2 and pH 5 ORA’ means accept ‘activity decreases of the haem groups, it slightly changes the
between pH 5 and pH 2’. shape of the haemoglobin molecule;
max. This indicates the maximum number of marks so that it becomes easier for more oxygen
that can be given. molecules to bind with the other haem
groups; [max. 6]
8 a (the word ‘gradually’ is not correct)
the partial pressure of oxygen is high in the
lungs and low in muscle and does not change
gradually as the blood flows from the lungs
to the muscle, (because it is only when it gets
to the muscle that the blood is in contact
with anything that is using oxygen);
Cambridge International AS and A Level Biology © Cambridge University Press 2014
Cambridge International AS Level Biology Answers to end-of-chapter questions
while it is inside an artery, it remains fully which maintains diffusion gradient for CO2
oxygenated; to diffuse into the blood from respiring
the blood is only exposed to a low partial tissues; [max. 3]
pressure of oxygen once it enters a capillary
inside a respiring tissue, such as a muscle; d i 73%, 62%; [1]
capillary walls, unlike those of arteries, are
thin and easily permeable to oxygen; [max. 2] ii presence of carbon dioxide causes affinity
b (arteries do not pump blood) of haemoglobin for oxygen to decrease;
elastic artery walls enable the artery to
expand and recoil as pulses of high-pressure hydrogen ions (from the dissociation of
blood pass through; H2CO3) bind with haemoglobin;
recoil of the artery wall does help to give
cause change in shape of Hb molecule;
[max. 2]
iii Bohr effect; [1]
the blood a further ‘push’ in between these iv causes more release of oxygen (than if this
pulses; effect did not occur);
but this is not ‘pumping’ and is due only to
elasticity, not to muscle contraction; [max. 2] in respiring tissues;
c each haemoglobin molecule can combine where demand for oxygen is high / where
with eight oxygen atoms;
one red cell contains well over 200 million production of carbon dioxide is high; [3]
[Total: 16]
haemoglobin molecules; 1 0 a blood goes through heart twice on one
complete circuit of the body; [1]
d (red blood cells do have a large surface area,
but oxygen does not attach to their surface) b has more smooth muscle / elastic tissue;
the large surface area allows more oxygen to to withstand higher (blood) pressure;
diffuse in and out at any one time; to withstand fluctuating (blood) pressure;
[max. 2]
therefore increasing the rate at which the cell
can take up and release oxygen;
once inside the cell, the oxygen does not c to prevent blood flowing into the capillary
attach to its surface, but to the haemoglobin bed / to divert blood to other capillary beds;
molecules within its cytoplasm; [max. 2] [1]
[Total: 8] d permeable walls / reference to pores in walls;
allow water / dissolved ions / dissolved
substances (from plasma) to pass out;
9 a reference to diffusion; [max. 2] do not allow large protein molecules / cells to
down concentration gradient; pass out;
through the wall of a capillary; reference to greater hydrostatic pressure
inside capillary than in tissue fluid; [max. 3]
b lower pressure; e i (plasma contains) more proteins;
lower concentration of oxygen; has lower water potential;
lower concentration of glucose;
lower water potential; has lower, carbon dioxide / HCO3–
lower concentration of proteins / amino acids concentration;
/ fatty acids / other named nutrient;
higher concentration of urea; [max. 3] has greater glucose concentration;
has greater oxygen concentration; [max. 3]
c i carbonic anhydrase; [1] ii lymph; [1]
ii hydrogencarbonate ions diffuse out of red [Total: 11]
blood cells;
(hydrogencarbonate ions) are transported 1 1 a i about 0.75 seconds;
in solution in blood plasma; ii 60 ÷ 0.75 = 80 beats per minute;
conversion of CO2 to hydrogencarbonate
reduces concentration of CO2 in the blood;
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For b, c, d, e and f, see figure below.
16 semilunar semilunar
valves open valves close
14 aorta
12
Pressure / kPa 10 left
8 ventricle
6 atrioventricular atrioventricular right
valves close valves open ventricle
4
left
2 atrium
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3
Time / s
Stage atrial ventricular ventricular and atrial ventricular ventricular and
systole systole atrial diastole systole systole atrial diastole
1 2 a i right ventricle; impulses spread upwards through
pulmonary vein; [2] ventricle walls;
ii they open to allow blood to flow from atria causing ventricles to contract from bottom
to ventricles; upwards;
delay of 0.1 to 0.2 s after atrial walls; [max. 5]
they close during ventricular systole /
when ventricles contract; [Total: 13]
reference to closure being caused by
differences in pressure in atria and
ventricles; [max. 2]
b
Event during the cardiac cycle Number
atrioventricular (bicuspid) valve opens 6
ventricular systole 1;
semilunar (aortic) valve closes 5;
left ventricle and left atrium both relaxing 2;
semilunar (aortic) valve opens 4;
[4]
c SAN produces rhythmic pulses of electrical
activity;
which spread across the muscle in the atria;
causes muscle in atria to contract;
specialised tissue, in septum / near AVN,
slows spread / delays transfer to ventricles;
Purkyne tissue conducts impulses down
through septum;
Cambridge International AS and A Level Biology © Cambridge University Press 2014
Cambridge International AS Level Biology Answers to end-of-chapter questions
Answers to EOCQs
Chapter 9 1 A; [1]
The mark schemes, suggested answers and 2 D; [1]
comments that appear in this CD-ROM were written
by the author(s). In examinations, the way marks 3 A; [1]
would be awarded to answers like these may be
different. 4 C; [1]
Notes about mark schemes 5 C; [1]
A or accept indicates an alternative acceptable 6 B; [1]
answer.
R = reject. This indicates a possible answer that 7 a i A squamous epithelial cell;
should be rejected.
; The bold semi-colon indicates the award of 1 mark. B red blood cell;
/ This indicates an alternative answer for the same
mark. The alternatives may be separated from the C endothelial cell; [3]
rest of the answer by commas.
( ) Text in brackets is not required for the mark. ii D (blood) plasma; [1]
Underlining This is used to indicate essential
word(s) that must be used to get the mark. b Award 2 marks for the correct answer. If
AW means ‘alternative wording’. It is used to answer is incorrect, award 1 mark for the
indicate that a different wording is acceptable correct working.
provided the essential meaning is the same, and is distance of scale bar is 15 mm [2]
used where students’ responses are likely to vary 15 mm = 10 µm
more than usual. X–Y distance measured on page is 43 mm
AVP means ‘additional valid point’. This means 43 ÷ 15 × 10 = 29 µm
accept any additional points given by the student
that are not in the mark scheme, provided they are c (very) large number forming a large surface
relevant. But accept only as many additional points area;
as indicated by the bold semi-colons, e.g. AVP;; squamous epithelial cells are very thin to give
means award a maximum of 2 extra marks. short diffusion distance;
ORA means ‘or reverse argument’ and is used when surrounded by capillaries so well supplied
the same idea could be expressed in the reverse with blood;
way. For example: ‘activity increases between pH capillaries are very close so short diffusion
2 and pH 5 ORA’ means accept ‘activity decreases distance;
between pH 5 and pH 2’. well ventilated so air constantly refreshed;
max. This indicates the maximum number of marks maintains concentration gradients for oxygen
that can be given. and carbon dioxide; [max. 4]
[Total: 10]
8 a i P – ciliated epithelial cell;
R – goblet cell; [2]
ii S – cilium / cilia;
T – mitochondrion;
U – Golgi body;
W – nucleolus; [4]
b i T / mitochondria, provide, energy / ATP;
for movement of cilia; [2]
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Cambridge International AS Level Biology Answers to end-of-chapter questions
ii U / Golgi apparatus, packages proteins into b breathing / ventilation;
vesicles; introduces, fresh air / atmospheric air;
removes, stale air / air rich in carbon dioxide;
for secretion; [2] [3]
c Award 2 marks for the correct answer. If c increase
answer is incorrect, award 1 mark for the depth of breathing;
correct working. rate of breathing;
length of cell P on page = 73 mm = 73 000 µm
magnification = ×750 expansion of alveoli to give a larger surface
area;
actual length = 73 000 ÷ 750 = 97 µm [2] diameter of airways; [max. 3]
[Total: 10]
d cell P – cilia beat / move back and forth;
move mucus;
upwards / towards throat;
cell R – secretes mucus; 1 1 a tar stimulates, goblet cells / mucous glands,
to secrete more mucus;
mucus traps, dust / bacteria / viruses / pollen; paralyses / destroys, cilia;
prevents entry to, alveoli / gas exchange mucus not moved up the, bronchioles /
surface; [max. 4] bronchi / trachea / airways;
[Total: 16] mucus accumulates in the airways;
9 a Trachea Respiratory bacteria multiply within the airways;
bronchiole (leads to) chronic bronchitis;
Structure tar contains, carcinogens / named
carcinogen; e.g. benzpyrene
(tar) settles on bronchial, epithelial cells /
smooth muscle cells ✔ ✘
ciliated epithelial cells ✔ ✔ epithelium;
mutation(s) / change to DNA;
growth of tumour;
mucous glands ✔✘ bronchial carcinoma / lung cancer; [max. 8]
cartilage
✔ ✘ b nicotine:
increases heart rate;
elastic fibres ✔✔
Award 1 mark for each row [5] increases blood pressure;
increases chance of blood clotting /
b mucus secreted; promotes thrombosis;
by mucous glands (in the trachea) / goblet decreased flow of blood to, extremities /
cells (in trachea and bronchi); fingers / toes; [max. 3]
bacteria / viruses / pathogens, stick to mucus; carbon monoxide:
cilia move mucus, upwards / towards throat; combines (irreversibly) with haemoglobin;
mucus and pathogens swallowed; forms carboxyhaemoglobin;
destroyed by acid in stomach; reduces oxygen carrying capacity of,
macrophages / phagocytes, in the alveoli; haemoglobin / blood;
engulf and digest any pathogens; [max. 5] damages lining of arteries;
[Total: 10] promotes atherosclerosis; [max. 3]
1 0 a oxygen diffuses down its concentration [Total: 14]
gradient; 1 2 a fewer alveoli;
from alveolar air into red blood cells; larger air spaces;
carbon dioxide diffuses down its fewer capillaries;
concentration gradient; scar tissue in, bronchioles / bronchi;
from, red blood cells / plasma, to alveolar air; few / no, cilia;
across epithelial cells of alveolus and few / no, goblet cells;
endothelium of capillary; [max. 4] enlarged mucous glands;
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Cambridge International AS Level Biology Answers to end-of-chapter questions
enlarged smooth muscle;
may be pre-cancerous / cancerous cells;
tumour / bronchial carcinoma; [max. 4]
b i difficulty breathing / breathlessness;
wheezing;
tiredness;
not able to do (much) exercise; [4]
ii small(er) surface area for gas exchange;
less oxygen absorbed;
poor oxygenation of the blood;
bronchi / bronchioles / airways, blocked by
mucus;
increased resistance to flow of air; [max. 3]
[Total: 11]
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Cambridge International AS Level Biology Answers to end-of-chapter questions
Answers to EOCQs
Chapter 10 1 A; [1]
The mark schemes, suggested answers and 2 C; [1]
comments that appear in this CD-ROM were written
by the author(s). In examinations, the way marks 3 D; [1]
would be awarded to answers like these may be
different. 4 C; [1]
Notes about mark schemes 5 2, 4, 6, 3, 5, 1
1 mark for every 2 correct answers [3]
A or accept indicates an alternative acceptable
answer. 6 a unprotected sexual intercourse;
R = reject. This indicates a possible answer that sharing needles (between intravenous drug
should be rejected. users) / reuse of unsterilised needles;
; The bold semi-colon indicates the award of 1 mark. blood transfusion / blood products;
/ This indicates an alternative answer for the same (mother to child) across the placenta / at
mark. The alternatives may be separated from the birth / in breast milk; [max. 3]
rest of the answer by commas.
( ) Text in brackets is not required for the mark. b doctor’s / hospital, records;
Underlining This is used to indicate essential tests for HIV status (e.g. at antenatal clinics
word(s) that must be used to get the mark. for pregnant women);
AW means ‘alternative wording’. It is used to death certificates; [max. 3]
indicate that a different wording is acceptable
provided the essential meaning is the same, and is c determine how numbers of people infected
used where students’ responses are likely to vary are changing;
more than usual. see where medical resources should be
AVP means ‘additional valid point’. This means targeted;
accept any additional points given by the student e.g. drugs for treating HIV infection;
that are not in the mark scheme, provided they are monitor success of HIV/AIDS programmes:
relevant. But accept only as many additional points in reducing spread of HIV infection;
as indicated by the bold semi-colons, e.g. AVP;; in treating people who are HIV+ so they do
means award a maximum of 2 extra marks. not develop AIDS;
ORA means ‘or reverse argument’ and is used when to see if more education is required;
the same idea could be expressed in the reverse to provide support to national / regional
way. For example: ‘activity increases between pH health organisations; [max. 3]
2 and pH 5 ORA’ means accept ‘activity decreases
between pH 5 and pH 2’. d i 0.052 : 1; [1]
max. This indicates the maximum number of marks
that can be given. ii better health care in North America;
better diagnosis, so people who are HIV+
start treatment early;
more affluent countries, so drugs available
to (nearly) all people who are HIV+;
[3]
[Total: 13]
7 a female Anopheles (mosquito);
takes a blood meal from an infected person;
transfers parasite / pathogen / Plasmodium,
in saliva when takes a blood meal from an
uninfected person; [3]
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b distribution for one mark: ref to very high number of cases in Haiti;
malaria occurs between the tropics / tropics
and sub-tropics / in equatorial regions / South may depend on remoteness of regions
and Central America, Africa and S-E Asia; affected by cholera;
explanation for four marks:
Anopheles / mosquito / vector, distributed or ways in which, emergency supplies /
throughout tropical regions; personnel, can reach affected areas;
Plasmodium falciparum cannot complete its
life cycle within mosquitoes at temperatures ref to high case fatality rates in, Nigeria /
less than 20 °C; Cameroon;
Anopheles, needs high rainfall / high humidity
/ standing water; use of data to compare case fatality rates
Anopheles lays eggs in water; in individual country with global rate;
no transmission at high altitude; [max. 3]
too cold for parasite to complete life cycle;
no transmission in deserts; iii cholera is a serious disease;
no breeding grounds for mosquitoes; [max. 5]
death can occur very quickly after infection;
c Plasmodium is intracellular parasite;
inside, red blood cells / liver cells; spreads quickly in population (especially
antigenic concealment; after a disaster);
antibodies are ineffective;
short stage in plasma when antibodies are deaths are avoidable;
effective;
Plasmodium is eukaryotic; if ORT is available immediately;
has many, genes / antigens;
difficult to develop a vaccine; data is useful to predict, situations /
resistance to drugs used, to treat malaria / as places, where cholera may occur;
prophylactics;
example of drug; WHO can coordinate responses to
Anopheles / vector, breeds in small pools of outbreaks; [max. 3]
water;
difficult to control all breeding places; c i infected person travelled from an area with
mosquitoes come into contact with humans; an outbreak of cholera; [1]
mosquitoes become resistant to insecticides;
[max. 6] ii water supply is not contaminated with
[Total: 14] (human), sewage / faeces;
8 a bacteria pass out in faeces of infected piped water / water supply is treated to
person; kill bacteria;
carried in, water / food, consumed by
uninfected person; [2] V. cholera destroyed in sewage treatment;
[max. 2]
b i 2.22; [1] [Total: 12]
ii treatment for cholera involves supply of
9 a i Mycobacterium tuberculosis; accept
oral rehydration therapy; Mycobacterium bovis [1]
and provision of safe drinking water;
better response to emergencies (in some ii infected person, coughs / sneezes / spits;
countries); aerosol / droplets, containing bacteria,
effectiveness of response may depend on breathed in by uninfected person; [2]
number of cases; b i Sub-Saharan Africa;
South-East Asia;
countries of, old Soviet Union / Russian
Federation;
India / Pakistan / Afghanistan;
South America / Bolivia;
Papua New Guinea; [max. 3]
ii TB linked with HIV infection;
HIV weakens immune system;
TB is an opportunistic disease;
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as many people are infected although
show no symptoms;
transmission where there is, overcrowding
/ poor housing;
poverty;
poor ventilation of housing;
poor nutrition;
poor access to health care;
poorly organised treatment for people
with TB; [max. 4]
[Total: 10]
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Answers to EOCQs
Chapter 11 1 C; [1]
The mark schemes, suggested answers and 2 D; [1]
comments that appear in this CD-ROM were written
by the author(s). In examinations, the way marks 3 D; [1]
would be awarded to answers like these may be
different. 4 A; [1]
Notes about mark schemes 5 B; [1]
A or accept indicates an alternative acceptable 6 a i bone marrow; [1]
answer.
R = reject. This indicates a possible answer that ii mitosis; [1]
should be rejected.
; The bold semi-colon indicates the award of 1 mark. iii plasma cell; [1]
/ This indicates an alternative answer for the same
mark. The alternatives may be separated from the iv antibody; [1]
rest of the answer by commas.
( ) Text in brackets is not required for the mark. b i antigen refers to any substance that
Underlining This is used to indicate essential stimulates the production of antibodies;
word(s) that must be used to get the mark.
AW means ‘alternative wording’. It is used to antibodies are proteins produced by,
indicate that a different wording is acceptable plasma cells / (activated) B-lymphocytes;
provided the essential meaning is the same, and is
used where students’ responses are likely to vary each antibody is specific to an antigen; [3]
more than usual.
AVP means ‘additional valid point’. This means ii self refers to antigen(s) within a person’s
accept any additional points given by the student body (e.g. those of the ABO blood group
that are not in the mark scheme, provided they are system which they have);
relevant. But accept only as many additional points
as indicated by the bold semi-colons, e.g. AVP;; all the antigens that the immune system
means award a maximum of 2 extra marks. does not recognise as foreign; [max. 1]
ORA means ‘or reverse argument’ and is used when
the same idea could be expressed in the reverse non-self refers to antigen(s) that are not in
way. For example: ‘activity increases between pH a person’s body (e.g. those of, pathogens /
2 and pH 5 ORA’ means accept ‘activity decreases the ABO system, that they do not have);
between pH 5 and pH 2’.
max. This indicates the maximum number of marks all the antigens that the immune system
that can be given. recognises as foreign; [max. 1]
c memory cell;
remains in circulation / lymph system / body;
is specific to an antigen on tetanus bacteria;
responds quickly to another infection by
(same strain of) pathogen;
as there are a large number / is a large clone;
during (secondary / subsequent) immune
response;
differentiate into plasma cells;
to give large number of antibody molecules
in short space of time; [max. 3]
[Total: 12]
7 a phagocyte has:
a lobed nucleus;
lysosomes;
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cytoplasm with granules / granular c herd / mass, vaccination / immunity;
cytoplasm; [3] prevented spread through population;
surveillance for infected people;
b presentation of antigen(s) by, macrophages / very easy to identify infected people / no
other APCs; symptomless carriers;
some T-lymphocytes have receptors contact tracing to find people who may have
complementary to antigen; become infected;
these are selected; ring vaccination / vaccination of all people in
divide by mitosis; surrounding area;
helper T cells secrete cytokines; prevented spread from isolated infected
to activate B-lymphocytes; people;
to secrete antibodies; one dose of the vaccine was enough to give
killer T cells search for cells infected by, life-long immunity / no boosters required;
parasite / pathogen; vaccine contained ‘live’ virus;
destroy host cell (and pathogen); smallpox virus was stable / did not mutate;
prevent reproduction of pathogen; [max. 6] no antigenic variation;
c B-lymphocytes can be activated by presence same vaccine was used for whole programme
of, antigen / pathogen alone;
without involvement of macrophages; / did not need to be changed;
heat-stable / freeze-dried, vaccine;
B-lymphocytes differentiate into plasma suitable for hot countries / isolated areas /
cells;
secrete antibodies (T cells do not secrete rural areas;
bifurcated / steel, needle made vaccinating
antibodies); [max. 2] easy;
[Total: 11]
did not need to be done by health
professionals; [max. 6]
8 a i immunity is gained by the transfer of [Total: 16]
antibodies from another source;
no immune response within the body; 9 a transcription (of DNA);
antigen(s) / pathogen(s), have not entered translation (of RNA);
the body; [max. 2]
assembly of amino acids to make four
ii natural passive immunity: polypeptides;
assembly of polypeptides to make antibody
antibodies cross the placenta; molecule;
in breast milk / colostrum; [2] packaged in Golgi body into vesicles;
b i baby has passive immunity; exocytosis; [max. 4]
antibodies against measles antigens (from b i X = variable region / antigen-binding site;
mother) will interact with measles viruses /
antigens in vaccine; Y = constant region; [2]
so prevent an immune response; ii disulfide; [1]
therefore no memory cells will be formed; c variable region(s) are antigen-binding sites;
[max. 3] variable regions, are specific /
complementary, to antigen;
ii difficulty with timing first vaccination; variable region has different amino acid
sequences for different antigens;
many children are not vaccinated at 20 different amino acids can be arranged to
appropriate time; form different shapes;
disulfide bonds hold polypeptides together;
measles is highly infectious; hinge region allows flexibility in binding to
antigen;
vaccination programmes concentrated on
other infectious diseases which have more
severe effect, such as smallpox and polio;
[max. 3]
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constant region for binding to receptors on many lymphocytes (especially T cells) leave
phagocytes; [max. 4] lymph nodes and the spleen and circulate in
[Total: 11] the blood patrolling for bacteria;
Patient B is HIV positive:
1 0 a 2; 5; 3; 1; 4; [5] the numbers of white cells are not much less
b i binds only with tumour-associated antigen than the numbers shown in Table 11.1;
(TAG) glycoprotein in cancer cells; the number of helper T cells is much smaller
source of gamma rays concentrated by than the typical number (between 500 and
cancer; 1500 cells mm–3);
gamma rays can be detected from outside idea that helper T cells are central to the
the body; [3] immune response;
ii emits gamma rays, which pass through soft this person is at risk of developing many
tissue and so can be detected from outside opportunistic infections;
the body; Patient C has acute myeloid leukaemia:
weak gamma rays, reducing chance of this is a cancer of the stem cells that produce
mutation; neutrophils;
short half-life, so no continuing source of this explains the great increase in immature
gamma rays; neutrophils;
indium has no biological role; [max. 3] Patient D has acute lymphoblastic
iii add, a drug / an enzyme to activate a drug, leukaemia:
to kill cancer cells; much higher numbers of white blood cells;
in place of radioactive label; most of which are immature lymphocytes;
[2] produced by cancerous stem cells in bone
[Total: 13] marrow;
the development of cancerous stem cells
in the bone marrow has disrupted the
production of red blood cells so the numbers
11 a have decreased; [max. 8]
All white cells Neutrophils Lymphocytes
Red cells
Patient c i 1.1 million red blood cells mm–3;
typical range typical range typical range typical range
4000 white blood cells mm–3; [2]
A higher within higher higher higher higher higher higher
ii 23.9%;
B lower lower lower lower lower lower lower within 57.2%; [2]
C lower lower higher higher higher higher higher within d the patient may never have had such a blood
test before;
D lower lower higher higher lower lower higher higher there is no way to know if the numbers
of blood cells have changed as a result of
table constructed as above with headings for the infection;
columns and rows; wide ranges of results are normal;
typical and range comparisons included for there are a variety of causes for high or low
each group of cells; numbers of blood cells;
all entries in the table completed as above; AVP; [max. 3]
[3] [Total: 18]
b Patient A has a bacterial infection:
body releases neutrophils from stores in the
bone marrow;
high number includes those on their way
from bone marrow to sites of infection;
lymphocytes increase as a result of clonal
selection and clonal expansion;
Cambridge International AS and A Level Biology © Cambridge University Press 2014
Cambridge International AS Level Biology Answers to end-of-chapter questions
Answers to EOCQs
Chapter P1 1 C; [1]
The mark schemes, suggested answers and 2 C; [1]
comments that appear in this CD-ROM were written
by the author(s). In examinations, the way marks 3
would be awarded to answers like these may be
different. Investigation Independent Dependent Two important
variable variable control variables
Notes about mark schemes
The effect sucrose % plasmolysis ■■ source of onion
A or accept indicates an alternative acceptable cells
answer. of sucrose concentration of onion cells
R = reject. This indicates a possible answer that ■■ length of time of
should be rejected. concentration immersion
; The bold semi-colon indicates the award of 1 mark.
/ This indicates an alternative answer for the same on
mark. The alternatives may be separated from the
rest of the answer by commas. plasmolysis of
( ) Text in brackets is not required for the mark.
Underlining This is used to indicate essential onion cells
word(s) that must be used to get the mark.
AW means ‘alternative wording’. It is used to The effect of pH rate of activity ■■ concentration of
indicate that a different wording is acceptable pH on the rate
provided the essential meaning is the same, and is of activity of of amylase, amylase
used where students’ responses are likely to vary amylase
more than usual. measured as ■■ concentration of
AVP means ‘additional valid point’. This means e.g. the rate of substrate
accept any additional points given by the student disappearance
that are not in the mark scheme, provided they are of starch or rate ■■ temperature
relevant. But accept only as many additional points
as indicated by the bold semi-colons, e.g. AVP;; of appearance
means award a maximum of 2 extra marks.
ORA means ‘or reverse argument’ and is used when of maltose
the same idea could be expressed in the reverse
way. For example: ‘activity increases between pH The effect of temperature percentage of ■■ wind movement
2 and pH 5 ORA’ means accept ‘activity decreases temperature open stomata
between pH 5 and pH 2’. on the in leaf ■■ light
max. This indicates the maximum number of marks percentage of
that can be given. open stomata ■■ humidity
in a leaf
■■ pre-treatment of
leaf (i.e. anything
that might affect
its degree of
hydration)
■■ source of leaf
(same species,
age, position on
plant)
One mark for each box completed correctly
[9]
4 a A epidermis;
B cortex / parenchyma;
C phloem;
D endodermis; [4]
b i LP plan drawn with no cell detail;
xylem only drawn inside circle;
correct proportions;
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Cambridge International AS Level Biology Answers to end-of-chapter questions
lines continuous, not sketchy and sharp thin, clear, best-fit line drawn or points
pencil used; [4] joined with ruled lines – no extrapolation;
[4]
ii LP plan drawn showing vascular bundles
only and no cell detail; ii correct reading from graph, including unit
(mean number of bubbles per minute); [1]
sclerenchyma, xylem and phloem drawn in
outline; [2] iii accuracy:
iii no. of squares of graph paper covered by use water bath to change independent
lignified tissue in root counted; [1] variable;
control of significant named variable plus
iv no. of squares of graph paper covered by method of control (e.g. use same type of
whole root section counted; [1] yeast);
v % squares occupied by lignified tissue in use named apparatus (e.g. gas syringe) to
root calculated correctly from student’s
answers to iii and iv; (answer should be collect gas (for measurement of dependent
variable);
greater than for root) [1] reliability:
vi no. of squares covered by lignified tissue in increase number / range of temperatures;
stem counted;
no. of squares covered by whole stem repeat each temperature three times and
calculate mean; [max. 3]
counted;
% squares occupied by lignified tissue in b hypothesis is supported;
quote figures for change in mean number
stem calculated correctly; (answer should of bubbles between any two temperatures
be around 1%) [3]
between 15 °C and 40 °C;
vii reference to no data below 15 °C or above 40 °C;
stem needs more support than root; so cannot tell if hypothesis is also supported
outside this range; [max. 2]
because upright in air and needs support [Total: 10]
to prevent it falling over / collapsing; AW [2]
c roots subjected to tugging / pulling pressure 6 a x-axis is ‘Time / minutes’, y-axis is ‘Number of
from parts above ground; individuals’;
roots spread out, so like a series of guy ropes; scales on both axes with suitable range and
stem a single column; interval;
greater strength from a ring of rods than from all bars plotted accurately or points plotted
one central rod; accurately (using a cross or an encircled dot);
ring of rods provides greater resistance to all lines neat and thin, plus key; [4]
compression from above than a single central
rod; b on both days, minimum time taken is 35 min
AVP; accept any reasonable suggestion(s) and maximum time taken is 55 min;
which are based on different stresses to on both days, number of individuals is
which roots and stems are subjected. [max. 2] greatest near the centre of the range;
[Total: 20] on day 1, greatest number of individuals
take 45 minutes to digest starch, but on day
5 a i ‘Temperature / °C’ on x-axis and ‘Enzyme 2 greatest number of individuals take 40
activity / mean number of carbon dioxide minutes to digest starch;
bubbles released per minute’ on y-axis; mean time is greater on day 1 than on day 2;
[max. 3]
suitable scales on both axes – range from
10 or 15 to 40 on x-axis and 0 or 5 to 20 on c temperature may have been higher on day 2;
y-axis, in intervals of 2 or 5; animals on day 2 may have eaten recently
and so had more saliva / amylase in their
all points plotted accurately, using crosses mouths; [max. 1]
or encircled dots;
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Cambridge International AS Level Biology Answers to end-of-chapter questions
d use individuals of same age / mass /
body weight;
ensure pre-treatment is the same (e.g. food
given, environment);
use same volume of saliva;
use same volume and concentration
of starch;
keep temperature the same by using a
water bath; [max. 3]
[Total: 11]
Cambridge International AS and A Level Biology © Cambridge University Press 2014
Cambridge International A Level Biology Answers to end-of-chapter questions
Answers to EOCQs
Chapter 12 1 C; [1]
The mark schemes, suggested answers and 2 D; [1]
comments that appear in this CD-ROM were written
by the author(s). In examinations, the way marks 3 D; [1]
would be awarded to answers like these may be
different. 4 a energy currency:
immediate donor of energy to all energy-
Notes about mark schemes requiring reactions in a cell;
energy storage:
A or accept indicates an alternative acceptable short-term (glucose, sucrose) or long-term
answer. (starch, glycogen, triglyceride) store of
R = reject. This indicates a possible answer that chemical potential energy; [2]
should be rejected.
; The bold semi-colon indicates the award of 1 mark. b decarboxylation:
/ This indicates an alternative answer for the same a reaction in which carbon dioxide is
mark. The alternatives may be separated from the removed from a compound; [1]
rest of the answer by commas. dehydrogenation:
( ) Text in brackets is not required for the mark. a reaction in which hydrogen is removed
Underlining This is used to indicate essential from a compound; [1]
word(s) that must be used to get the mark. [Total: 4]
AW means ‘alternative wording’. It is used to
indicate that a different wording is acceptable 5 a NAD:
provided the essential meaning is the same, and is a hydrogen carrier molecule: it accepts a
used where students’ responses are likely to vary hydrogen from one reaction and donates it to
more than usual. another; [1]
AVP means ‘additional valid point’. This means
accept any additional points given by the student b coenzyme A:
that are not in the mark scheme, provided they are a carrier of an acetyl group from the link
relevant. But accept only as many additional points reaction to the Krebs cycle; [1]
as indicated by the bold semi-colons, e.g. AVP;;
means award a maximum of 2 extra marks. c oxygen:
ORA means ‘or reverse argument’ and is used when the final electron acceptor and hydrogen ion
the same idea could be expressed in the reverse acceptor in oxidative phosphorylation: the
way. For example: ‘activity increases between pH oxygen is reduced to water; [1]
2 and pH 5 ORA’ means accept ‘activity decreases [Total: 3]
between pH 5 and pH 2’.
max. This indicates the maximum number of marks 6 ATP ATP Net gain
that can be given. used produced in ATP
Feature
–2 +4 +2
glycolysis
0 00
link reaction
Krebs cycle 0 +2 +2
oxidative 0 +28 +28
phosphorylation
total –2 +34 +32
[5]
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7 a Lipid has more hydrogen atoms per molecule
than does carbohydrate;
most energy liberated in aerobic respiration
comes from the oxidation of hydrogen to
water; [2]
volume of carbon dioxide given out in unit time
b RQ = volume of oxygen taken in in unit time
or
R Q = mmoloelse/sm/moloelceucluienluseunsonifotictftaiotmirxmbeyogenenditoaxkiedneignivinen [2]
c Respiratory substrate RQ
carbohydrate 1.0
lipid 0.7
protein 0.9
[3]
17.5 cm3
d i RQ = 25 cm3
= 0.7; [2]
ii lipid; [1]
e C6H12O6 + 6O2 → 6CO2 + 6H2O + energy;
6CO2
6H2O = 1.0; [2]
[Total: 12]
8 1 air spaces;
2 aerenchyma;
3 roots;
4 oxygen;
5 anaerobic pathways;
6 ethanol / alcohol;
7 alcohol dehydrogenase; [7]
9 a provide hydrogen to reduce NAD and FAD;
reduced carriers pass to electron transport
chain;
provide energy for ATP synthesis in oxidative
phosphorylation;
refer to chemiosmosis; [max. 3]
b i increasing the concentration of aluminium
ions from 0 to 40 µmol increases rate of
fumarate production;
increases from 40 to 120 µmol have
little effect; [2]
ii aluminium binds to enzyme / refer to
cofactor; optimises shape of active site; [2]
[Total: 7]
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Answers to EOCQs
Chapter 13 1 B; [1]
The mark schemes, suggested answers and 2 C; [1]
comments that appear in this CD-ROM were written
by the author(s). In examinations, the way marks 3 A; [1]
would be awarded to answers like these may be
different. 4 a allows chlorophyll and other pigments to be
arranged into photosystems;
Notes about mark schemes provides large surface area for pigments;
increases efficiency of light harvesting;
A or accept indicates an alternative acceptable allows electron carriers to be arranged
answer. appropriately;
R = reject. This indicates a possible answer that provides structure for proton gradient for
should be rejected. chemiosmosis;
; The bold semi-colon indicates the award of 1 mark. anchors ATP synthase; [5]
/ This indicates an alternative answer for the same
mark. The alternatives may be separated from the b
rest of the answer by commas.
( ) Text in brackets is not required for the mark. Structural feature Shared by chloroplast and
Underlining This is used to indicate essential typical prokaryotic cell
word(s) that must be used to get the mark.
AW means ‘alternative wording’. It is used to circular DNA ✔
indicate that a different wording is acceptable
provided the essential meaning is the same, and is DNA combined with ✘
used where students’ responses are likely to vary structural protein to
more than usual. form chromosomes
AVP means ‘additional valid point’. This means
accept any additional points given by the student ribosomes about ✔
that are not in the mark scheme, provided they are 18 nm in diameter
relevant. But accept only as many additional points
as indicated by the bold semi-colons, e.g. AVP;; complex arrangement ✘
means award a maximum of 2 extra marks. of internal membranes
ORA means ‘or reverse argument’ and is used when
the same idea could be expressed in the reverse peptidoglycan wall ✘
way. For example: ‘activity increases between pH
2 and pH 5 ORA’ means accept ‘activity decreases size ranges overlap ✔
between pH 5 and pH 2’.
max. This indicates the maximum number of marks [6]
that can be given. [Total: 11]
5 a photolysis of water occurs in light;
H+ released;
accepted by DCPIP / methylene blue;
colourless when reduced;
shows ‘reducing power’ of chloroplasts;
[max. 4]
b NADP [1]
[Total: 5]
6 a cyclic photophosphorylation:
electron emitted by chlorophyll of
photosystem I returns to chlorophyll by a
series of carriers;
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non-cyclic photophosphorylation: photochemical reactions are not affected by
electron emitted by chlorophyll of temperature;
photosystem II does not return to that at low light intensities, light intensity is the
chlorophyll (but is absorbed by photosystem rate-limiting factor;
I and electron emitted by photosystem I is at high light intensities and low
absorbed by NADP); [2] temperatures, temperature is the rate-
b photophosphorylation: limiting factor; [max. 5]
synthesis of ATP using light energy in [Total: 10]
photosynthesis in a chloroplast;
oxidative phosphorylation: 9 a
synthesis of ATP using energy released from
the electron transport chain in aerobic Item Mesophyll Bundle
respiration in a mitochondrion; [2] PEP carboxylase cell sheath
c NAD: cell
hydrogen carrier in respiration; ✔
NADP: ✘
hydrogen carrier in photosynthesis; [2]
rubisco ✘✔
[Total: 6]
RuBP ✘✔
7 a and b enzymes of Calvin cycle ✘✔
CO2 (1C)
rubisco high concentration of oxygen ✔ ✘
active here
light-dependent reactions
✔ ✘
contact with air spaces ✔✘
RuBP (5C) GP/PGA (3C) [7]
b Photorespiration:
rubisco catalyses the combination of RuBP
and oxygen;
the result is an overall intake of oxygen and
loss of carbon dioxide; [2]
[Total: 9]
triose phosphate (3C) 1 0 a absorption spectrum:
a graph of the absorbance;
of different wavelengths of light by a
[Total: 5] compound;
8 a limiting factor: one factor, of many affecting action spectrum:
a process, that is nearest its lowest value and a graph of the rate of a process;
hence is rate-limiting; [1] e.g. photosynthesis at different wavelengths
of light; [4]
b light intensity; b number of bubbles shows rate of
light wavelength; photosynthesis;
concentration of carbon dioxide; rate similar at 450 nm (blue) and 650 nm (red);
temperature; [4] these are wavelengths that are absorbed by
chlorophyll;
c shows that there are two sets of reactions in rate, much lower / refer to figures, at 550 nm
photosynthesis; (green);
a light dependent photochemical stage; very little absorbed by any pigment; [max. 4]
a light independent temperature-dependent
stage; [Total: 8]
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Answers to EOCQs
Chapter 14 1 C; [1]
The mark schemes, suggested answers and 2 B; [1]
comments that appear in this CD-ROM were written
by the author(s). In examinations, the way marks 3 D; [1]
would be awarded to answers like these may be
different. 4 B; [1]
Notes about mark schemes 5 2, 7, 4, 6, 3, 1, 5
All correct = 4, subtract marks for mistakes.
A or accept indicates an alternative acceptable [4]
answer.
R = reject. This indicates a possible answer that 6 a excretion:
should be rejected. removal from the body;
; The bold semi-colon indicates the award of 1 mark. of waste products of metabolism;
/ This indicates an alternative answer for the same carbon dioxide / nitrogenous waste / urea /
mark. The alternatives may be separated from the uric acid / any other example;
rest of the answer by commas. substances in excess of requirements;
( ) Text in brackets is not required for the mark. water / salts / sodium ions / potassium ions /
Underlining This is used to indicate essential any other example; [max. 3]
word(s) that must be used to get the mark.
AW means ‘alternative wording’. It is used to b i A distal convoluted tubule;
indicate that a different wording is acceptable B Bowman’s capsule;
provided the essential meaning is the same, and is C glomerulus / capillary;
used where students’ responses are likely to vary D proximal convoluted tubule; [4]
more than usual.
AVP means ‘additional valid point’. This means ii cortex;
accept any additional points given by the student glomeruli / convoluted tubules, are only
that are not in the mark scheme, provided they are
relevant. But accept only as many additional points found in the cortex; [2]
as indicated by the bold semi-colons, e.g. AVP;;
means award a maximum of 2 extra marks. iii distance = 10 mm [2]
ORA means ‘or reverse argument’ and is used when = 10 000 µm [Total: 11]
the same idea could be expressed in the reverse a=c1t01u 80a00l0distance = 56 µm
way. For example: ‘activity increases between pH
2 and pH 5 ORA’ means accept ‘activity decreases 7 a hypothalamus; [1]
between pH 5 and pH 2’. b 1555 cm3 (or any answer within the range
max. This indicates the maximum number of marks
that can be given. 1150 to 1160 cm3 or equivalent in dm3); [1]
c any four from:
water was absorbed into the blood;
water increases the water potential of the
plasma;
any effect of an increase in water potential
of the plasma on, cells / tissues, e.g. water
enters cells by osmosis / cells will swell /
decreases efficiency of reactions inside cells /
cells may burst;
osmoreceptors detect increase in water
potential;
Cambridge International AS and A Level Biology © Cambridge University Press 2014
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Cambridge International AS and A Level Biology Coursebook
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Cambridge International AS and A Level Biology Coursebook
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