Biology 9700/22 February/March 2023: question paper and mark scheme
Cambridge AS & A Level Biology (9700), Paper 2 AS Level Structured Questions, February/March 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Table 1.1 lists cell structures that can be found in eukaryotic cells or prokaryotic cells. Some of these cell structures can be found in both types of cell.
Complete the table using a tick (3) to show that the cell structure can be present in a particular type of cell and a cross (✗) to show that the cell structure cannot be present.
Put a tick or a cross in every box.
The top row has been completed for you.
Table 1.1
cell structure eukaryotic cells prokaryotic cells
nucleus 3 ✗
Golgi body
circular DNA
70S ribosome
[2]
(b) All cells have a cell surface membrane. Fig. 1.1 shows a transmission electron micrograph of part of two adjacent animal cells, cell 1 and cell 2.
cell 1
R
cell 2
×300 000
Fig. 1.1
In the space provided, draw a diagram of the region in the box labelled R on Fig. 1.1. Your diagram should show the four dark lines.
Label the diagram to identify what is shown by the dark lines and each of the three spaces between them.
space for diagram:
[3]
(c) Mitogens are short chains of amino acids that function as cell-signalling molecules. Mitogens are released from secretory cells and travel in the blood to target cells, where the mitogens bind to cell surface receptors. The target cells respond by progressing from the G1 phase to the S phase of the mitotic cell cycle.
(i) Outline what happens in the G1 phase and S phase of the mitotic cell cycle.
G1 phase ……
S phase ……
[2]
(ii) As a result of mutation, the production and release of mitogens into the blood can be greatly increased.
Suggest a possible consequence for target cells of increased concentrations of mitogens in the blood.
…… [1]
[Total: 8]
Mark scheme
1(a) 2
cell eukaryotic prokaryotic
structure cells cells
nucleus ✓
Golgi body ✓
circular DNA ✓ ✓
70S ribosome ✓ ✓
one mark for each correct column ; ;
1(b) diagram showing: 3
dark line(s) labelled as phosphate heads ;
A phospholipid heads
clear area(s) between pairs of dark lines labelled as, fatty acid tails / hydrocarbon chains / hydrophobic core / AW ;
R if pointing to intercellular space
clear area between the two cell surface membranes labelled as, interstitial fluid / tissue fluid / extracellular
matrix / intercellular space ;
A intercellular, area / region
1(c)(i) G1 phase: 2
any one from:
RNA, synthesised / transcribed / translated ;
proteins / enzymes, synthesised ;
increasing quantity of organelles ;
increase in volume of cytoplasm ;
A growth
(at end of G1 phase) checkpoint passed for dividing or not ;
S phase:
any one from:
DNA (semi-conservative) replication ;
doubles, mass / number of strands, of DNA ;
ref. to each chromosome now comprises two chromatids ;
1(c)(ii) any one from: 1
uncontrolled / increased, cell division / mitosis ;
resting cells could enter mitosis ;
more cells move from the G1 to the S phase ;
(a) Cysteine is an amino acid containing sulfur. Fig. 2.1 shows the structure of the molecule formed by joining two cysteine molecules together.
SH
CH2 O H
H O H
C C N C C N
O H
H H CH2
SH
Fig. 2.1
Draw a circle around an R-group in the molecule shown in Fig. 2.1. [1]
(b) Goblet cells in the human gas exchange system produce proteins called mucins.
(i) The ends of mucin molecules contain many cysteine residues. Mucin strands are formed by joining the ends of mucin molecules together through covalent bonds between
R-groups.
State the name of these covalent bonds.
…… [1]
(ii) Mucin strands are transported out of the goblet cells and then absorb water to form mucus.
Suggest and explain how mucin strands are transported out of the goblet cells.
…… [3]
Cystic fibrosis is a genetic disease caused by a mutation in the human CFTR gene. This results in mucus that is thicker than normal.
(c) Suggest how thicker mucus interferes with the maintenance of healthy gas exchange surfaces in the lungs.
…… [2]
(d) Row 1 and row 2 of Table 2.1 show the DNA base sequences of part of the normal CFTR allele and the same part of a mutated CFTR allele. The base sequences shown are for the
DNA strands used in the synthesis of RNA. When Table 2.1 is completed, row 3 will show the base sequence of the RNA synthesised from the same part of the mutated CFTR allele.
Table 2.1
1 DNA base sequence of part of the normal T A G T A G A A A C C A
CFTR allele
2 DNA base sequence of part of the mutated T A G T A A C C A C A A
CFTR allele
3 RNA base sequence synthesised from the mutated CFTR allele
(i) The difference between the DNA base sequence in row 1 and the DNA base sequence in row 2 of Table 2.1 is caused by a single gene mutation.
State the name of this type of gene mutation.
…… [1]
(ii) Row 1 and row 2 in Table 2.1 show the DNA strands used in the synthesis of RNA.
State the term used to describe the DNA strand that is used in the synthesis of RNA.
…… [1]
(iii) Complete Table 2.1 to show the missing bases in row 3. [1]
(iv) The normal CFTR allele is approximately 189 000 base pairs in length. The CFTR polypeptide consists of only 1480 amino acids.
Explain the reasons for this difference between the number of base pairs and the number of amino acids.
…… [3]
[Total: 13]
Mark scheme
2(a) either R group circled ; 1
SH
CH2 O H
H O H
C C N C C N
O H
H H CH2
SH
2(b)(i) disulfide (bonds) ; 1
2(b)(ii) exocytosis ; 3
plus any two from:
either:
(idea of) vesicles will be large enough to contain many mucins / ref. to bulk transport ;
vesicles forming from Golgi, body / apparatus ;
vesicles moved by microtubules / cytoskeleton (to cell surface membrane) ;
vesicles fuse with cell surface membrane ;
active process / requires ATP ;
AVP ; e.g. mucins are, polar / hydrophilic, so cannot cross, phospholipid bilayer / hydrophobic core of membrane
or:
mucin molecules are hydrophilic ;
can exit via protein channels ;
by facilitated diffusion ;
2(c) any two from: 2
(because mucus is too thick) cilia, have difficulty / AW, moving the mucus (upwards) ;
pathogens, build up / not removed / AW ;
more chance of, infection / disease ;
AVP ;
2(d)(i) deletion ; 1
2(d)(ii) transcribed / template ; 1
R transcription strand
I anti-sense / coding, strand
2(d)(iii) AUC AUU GGU GUU ; 1
2(d)(iv) any three from: 3
1 idea of 3 bases coding for 1 amino acid ;
2 introns removed from, primary transcript RNA / AW ;
A gene splicing / RNA splicing
A mRNA does not contain introns
R if not in correct context (primary transcript to mRNA)
3 introns / non-coding DNA, do not code for amino acids ;
4 (only) exons are joined together (to form mRNA) ;
5 DNA triplet / mRNA codon, for STOP does not code for an amino acid ;
6 methionine at start / first amino acid / amino acid coded for by START codon, removed ;
7 AVP ; e.g. ref. to (upstream) enhancer sequences
ref. to (downstream) terminator sequences
ref. to (non-coding) regulatory sequences / promoter
(a) Fig. 3.1 is a diagram of an area of phloem tissue from a transverse section through the stem of a squash plant, Cucurbita pepo.
companion cell
X
Y
Fig. 3.1
(i) Cell X and cell Y in Fig. 3.1 are sieve tube elements.
Explain why cell X and cell Y have very different appearances in this transverse section.
…… [2]
(ii) Sucrose is formed from the glucose synthesised by mesophyll cells in the leaves of C. pepo.
Explain how companion cells are involved in the transfer of sucrose into phloem sieve tubes.
…… [4]
(b) Hydrogen bonding is important in the movement of water in xylem.
(i) Explain how hydrogen bonding occurs between two water molecules.
…… [2]
(ii) Outline how hydrogen bonding is involved in water transport in the xylem of a plant stem.
…… [3]
(iii) Hydrogen bonding between water molecules gives water a relatively high latent heat of vaporisation.
Suggest why it is important to plants that water has a high latent heat of vaporisation.
…… [2]
[Total: 13]
Question 4 starts on page 12.
Mark scheme
3(a)(i) cell Y shows, a sieve plate / sieve pores ; ora 2
idea that the sieve plates are at different heights in the stem / the section misses the sieve plate in cell X / AW ;
3(a)(ii) any four from: 4
1 (protein in) companion cell surface membrane, moves / pumps, protons into, cell wall / apoplast ;
2 using ATP / by active transport ;
3 establishes a proton gradient ;
4 protons move into companion cell, by facilitated diffusion ;
A described e.g. down the, electrochemical / concentration, gradient
5 protons cotransport sucrose / protons and sucrose pass through a cotransporter (protein), into companion cell ;
6 sucrose moves against its concentration gradient ;
7 diffusion of sucrose into phloem sieve tube, through plasmodesmata ;
A all marking points from an annotated diagram
3(b)(i) this could be answered diagrammatically or in text, e.g. 2
any two from:
idea that oxygen has a small negative charge and hydrogen has a small positive charge ;
idea that a hydrogen bond forms between the oxygen of one water molecule and the hydrogen of a neighbouring water
molecule ;
AVP ; e.g. oxygen is more electronegative than hydrogen
3(b)(ii) any three from: 3
1 ref. to cohesion between water molecules ;
2 ref. to adhesion / hydrogen bonding, between water molecules and, cellulose / hydrophilic parts of lignin (in cell wall) ;
A hydrophilic parts of cell wall
3 formation of continuous column / AW ;
4 (column) pulled up by, transpiration / evaporation / transpiration pull ;
3(b)(iii) any two from: 2
ref. to plant leaves heat up, when light absorbed for photosynthesis / at high (ambient) temperatures ;
(high latent heat of vaporisation leads to large cooling effect / removal of (large quantities of) heat energy, by evaporation ;
reduces, protein / enzyme, denaturation ;
reduces rate of water loss by, transpiration / evaporation, at high temperatures ;
AVP ;
Tuberculosis (TB), influenza and polio are examples of infectious diseases.
(a) (i) Explain what is meant by an infectious disease.
…… [2]
(ii) Name a species of organism that causes TB.
…… [1]
(b) Immunity can be described as artificial or natural and passive or active.
Name the type of immunity that a mother gives to her baby through breast milk.
…… [1]
(c) The influenza virus can mutate frequently to produce different strains of the virus. A new vaccine is often necessary to stimulate the production of new antibodies to these new strains.
Explain why different antibodies need to be produced to give immunity to these new strains.
…… [3]
(d) Polio is a serious viral disease affecting young children. In 1996, polio caused paralysis in more than 75 000 children across Africa. A long-term vaccination programme allowed the
World Health Organization (WHO) to declare that Africa was largely free of polio in 2020.
(i) Explain how vaccination programmes can help to control the spread of infectious diseases, such as polio.
…… [3]
(ii) Antibiotics, such as penicillin, do not help to prevent the spread of viral diseases, such as polio.
Explain why penicillin is not effective against viruses.
…… [1]
[Total: 11]
Mark scheme
4(a)(i) caused by pathogens ; 2
is transmissible / AW ;
4(a)(ii) Mycobacterium tuberculosis / Mycobacterium bovis ; 1
4(b) natural and passive ; 1
4(c) any three from: 3
1 (as influenza virus mutates) antigen (proteins) altered / altered capsid proteins / AW ;
2 antibodies are specific to antigens ;
3 ref. to complementary shape (between variable regions of antibody and antigen) ;
4 ref. to variable regions / Fab, at the ends ;
4(d)(i) for children, allow people 3
for polio allow disease
any three from:
1 ref. to vaccination programme aims to vaccinate, whole populations / large numbers ;
2 vaccine allows (most), children / people, to have an immune response (against antigens of the disease organism) ;
3 develop, long-term immunity / artificial active immunity / memory cells against, polio / disease ;
4 idea that vaccinated individuals cannot, transmit virus / transmit polio / infect others ;
5 herd immunity / described ;
6 AVP ; e.g. ref. to vaccination programme aims to break the transmission cycle
4(d)(ii) any one from: 1
penicillin only acts on, bacteria / prokaryotes ;
virus does not have, cell walls / murein / peptidoglycan ;
virus does not have, transpeptidases / the enzyme that is, inhibited / acted on, by penicillin ;
virus does not grow / penicillin (only) acts on growing cells / AW ;
A when cell wall is, growing (larger) for growing cells
virus does not have cellular structure / virus is acellular ;
Pneumonia is a severe lung disease that can interfere with gas exchange. A person with pneumonia can be connected to an ECMO machine. This machine performs the gas exchange functions of the lungs.
A cannula (tube) is inserted into the right atrium and this takes blood to the ECMO machine. In the
ECMO machine, blood is passed firstly to an artificial pump and then to an oxygenator, where gas exchange occurs. The blood is then warmed and returns by another cannula to the vena cava.
(a) Complete Fig. 5.1 to show how the ECMO machine is connected to the right atrium and to the vena cava. Use a single line to represent each cannula.
artificial pump
ECMO machine oxygenator
heater
Fig. 5.1
[1]
(b) In the oxygenator, a partially permeable membrane separates the blood from air that has been enriched with extra oxygen.
(i) State the name of a structure in the gas exchange system that has the same function as the partially permeable membrane of the oxygenator.
…… [1]
(ii) In the oxygenator, blood and oxygen-enriched air flow in opposite directions.
Suggest and explain how the oxygenator carries out the functions of gas exchange that normally occur in the lungs.
…… [3]
(c) Fig. 5.2 is a photomicrograph showing a transverse section of part of the human aorta.
tunica media
tunica externa
×16
Fig. 5.2
Explain how the structure of the tunica media in Fig. 5.2 is different from the structure of the tunica media in a muscular artery and relate the difference to the function of the aorta.
…… [3]
(d) Some biologists investigated the transport of carbon dioxide in the blood of Caiman latirostris, a type of reptile.
The biologists found that when C. latirostris respires:
• most of the carbon dioxide is converted into hydrogencarbonate ions in red blood cells
• the hydrogencarbonate ions combine with haemoglobin inside the red blood cells
• the hydrogencarbonate ions remain combined with haemoglobin until the blood reaches the lungs.
(i) Explain why the physiology of C. latirostris requires carbonic anhydrase.
…… [1]
(ii) Explain why the physiology of C. latirostris does not require the chloride shift.
…… [1]
[Total: 10]
Question 6 starts on page 18.
Mark scheme
5(a) line drawn from right atrium to the arrow into the pump and line drawn from the arrow out of the heater back to the 1
(inferior / superior) vena cava ;
5(b)(i) alveolar wall / alveolar epithelium ; 1
A alveolus / alveoli
A wall of air sacs
5(b)(ii) any three from: 3
oxygen moves (from the oxygen-enriched air) into the blood and carbon dioxide moves from the blood (into the oxygen-
enriched air) / AW ;
by diffusion / down their concentration gradients ;
(opposite flow arrangement / counter current) maintains a steep diffusion gradient / ref. to equilibrium is not reached / AW ;
short diffusion pathway (across the membrane) ;
AVP ; e.g. oxygenator membrane (will have a) large surface area
5(c) (the tunica media of aorta) contains, more elastin / more elastic fibres / more elastic tissue / less smooth muscle, than the 3
muscular artery ; ora
plus any two from:
allows the aorta to stretch / aorta less likely to burst (as a result of high blood pressure) ;
during ventricular contraction / systole ;
(after stretching) the aorta recoils ;
maintains high blood pressure ;
evens out the flow of blood / AW ;
5(d)(i) for speeding up / as enzyme for / for catalysing, conversion of (water and) carbon dioxide to, carbonic 1
acid / hydrogencarbonate ions ;
5(d)(ii) hydrogencarbonate ions do not leave the red blood cells so no charge imbalance / AW ; 1
(a) Collagen is the most common structural protein in vertebrates. Collagen provides the skin with flexibility and strength.
Explain how the structure of a collagen fibre provides the skin with strength.
…… [3]
(b) The enzyme collagenase breaks down collagen. Collagenase has several important medical uses, such as in the treatment of burnt skin.
Scientists investigated the effect of pH on the activity of collagenase at 37 °C.
The results of their investigation are shown in Fig. 6.1.
100
80
percentage 60 of maximum activity 40
20
0
4.0 5.0 6.0 7.0 8.0 9.0 10.0 pH
Fig. 6.1
Explain why the activity of collagenase is lower at pH 8.0 than at the optimum pH.
…… [2]
[Total: 5]
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Mark scheme
6(a) any three from: 3
covalent bonds between collagen molecules (between R groups / side-chains) ;
covalent bonds are strong ;
(molecules) staggered / overlapping molecules, so no weak areas (gives greater strength) ;
(in skin) fibres line up in layers, not parallel / running in different directions ;
provides tensile strength / strength in different directions ;
6(b) any two from: 2
at pH 8.0:
ionic / hydrogen bonds, (between R groups) broken / altered ;
active site shape altered ;
(active site) no longer / less, complementary to substrate / collagen
or
fewer enzyme–substrate complexes formed ;
AVP ; e.g. the amino acids in the active site affected by the changing pH
ref. to partial denaturation
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Biology 9700/22 for February/March 2023 is out of 60 marks, with 1 hour 15 minutes to answer it, across 6 questions.
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