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Pearson Edexcel a level biology B (9BI0/01) question paper 1 june 2024 salters Nuffield Merged with mark scheme
Typology: Exams
1 / 57
Answer ALL questions.
Some questions must be answered with a cross in a box. If you change your mind about an
answer, put a line through the box and then mark your new answer with a cross.
1 Blood is pumped around the body in blood vessels to supply cells with nutrients
and oxygen.
(a) Which box in each row of the table shows where endothelial cells and valves
are found?
Features of
blood vessels
Type of blood vessel
both capillaries
and veins
capillaries
only
veins only
neither capillaries
nor veins
Endothelial cells
Valves
(b) Nutrients and oxygen pass into cells from tissue fluid.
Explain how tissue fluid is formed.
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(Total for Question 1 = 4 marks)
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2 Glucose is a monosaccharide that is found in sucrose, amylose and amylopectin.
(a) Sucrose is made from glucose and one other molecule.
Which row of the table gives the name of this molecule and the type of reaction
that joins the two molecules together?
(b) Compare and contrast the structure of amylose and amylopectin.
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(Total for Question 2 = 4 marks)
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Name of molecules Type of reaction
fructose condensation
fructose hydrolysis
galactose condensation
galactose hydrolysis
3 Viruses can replicate using the lytic cycle.
Viruses can be grown in a laboratory by culturing them with appropriate cells.
(a) Explain why viruses have to be cultured with ‘appropriate cells’.
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(b) The graph shows the changes in the number of viruses after they had been added
to the cells.
Log 10
number
of viruses
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Time after adding viruses to the cells / min
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(i) Explain why the number of viruses did not start to increase until
after 20 minutes.
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(ii) Calculate how many more viruses are present at 70 minutes than
at 50 minutes.
Give your answer to three significant figures.
Answer..............................................................
(Total for Question 3 = 7 marks)
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4 Photosynthesis results in the production of GALP. This molecule is used by plants to
produce glucose and other organic molecules.
(a) Glucose consists of the elements of carbon, hydrogen and oxygen.
Name the inorganic molecules that supply these elements for photosynthesis.
Carbon...............................................................................................................................................................................................................................................................
Hydrogen........................................................................................................................................................................................................................................................
Oxygen..............................................................................................................................................................................................................................................................
(b) Light is essential for photosynthesis. It is needed for photolysis and for the
excitation of electrons in the photosystems.
Which row in the table shows where photolysis and the excitation of
electrons occurs?
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Photolysis Excitation of electrons
stroma stroma
stroma thylakoid membrane
thylakoid membrane stroma
thylakoid membrane thylakoid membrane
(c) Scientists have developed an experimental technique that uses
artificial photosynthesis.
Electricity generated from solar panels was used to convert carbon dioxide and
water into the organic compound acetate.
Plants were then able to grow in the dark using the acetate.
(i) Explain why these plants were able to grow in the dark.
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(ii) The scientists hope that this technique will be able to produce
plant‑based food.
State one advantage of using this method.
Give a reason for your answer.
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(Total for Question 4 = 7 marks)
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5 Septicaemia is a life‑threatening condition that can arise from bacterial infection.
(a) In one year, 245 000 people were diagnosed with septicaemia and 49 735 of these
people died as a result.
Which percentage of people survived septicaemia in this year?
(b) Two types of bacteria that cause septicaemia are Escherichia coli ( E. coli ) and
Staphylococcus aureus ( S. aureus ).
E. coli is Gram negative and S. aureus is Gram positive.
(i) Which of the following statements about the cell walls of these bacteria
are correct?
A 2 only
B 3 only
C 1 and 2 only
D 1 and 3 only
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(ii) Both E. coli and S. aureus produce toxins.
E. coli produces the same type of toxin as Salmonella bacteria.
Which of the following statements about the toxins produced by these
bacteria are correct?
A 1 only
B 2 only
C 1 and 2 only
D 1, 2 and 3
(c) Vaccines against bacteria are not always effective. The antigens they contain
stimulate only a weak immune response.
Which of the following types of immune response is stimulated by a vaccine?
A artificial active
B artificial passive
C natural active
D natural passive
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(d) There are currently no vaccines for either E. coli or S. aureus.
Scientists are developing vaccines against E. coli and S. aureus which can be
implanted under the skin.
The bacterial antigens are held in a mesh that contains chemicals to attract
tissue macrophages.
The effect of implantation of these vaccines on the immune response in mice
was investigated.
(i) The graph shows the number of T helper cells in the lymph nodes that drain
the part of the body where the vaccine was implanted.
Number of
T helper cells
5
cells per cm
3
Time after implantation of vaccine / days
Explain the results shown in the graph.
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(ii) In part of this investigation, mice had the vaccine implanted 35 days before
being infected with E. coli.
A control group that had not received the vaccine was also infected
with E. coli.
The levels of antibody in the blood of both groups of mice were determined.
The graph shows these results.
Concentration
of antibody / a.u.
Time after infected with E. coli
mice with vaccine implanted
control group of mice
Explain the results shown in the graph.
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(Total for Question 5 = 10 marks)
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6 In a cell, DNA is found in both the nucleus and the mitochondria.
(a) Where is DNA found in mitochondria?
A cytoplasm
B intermembrane space
C matrix
D nucleus
(b) Which row of the table describes DNA found in the mitochondria?
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Shape
Ratio of
phosphodiester bonds : pentose sugars
circular higher than nuclear DNA
circular lower than nuclear DNA
linear higher than nuclear DNA
linear lower than nuclear DNA
(c) Gel electrophoresis can be used to analyse the DNA of a person.
The diagram shows some of the bands produced when the DNA of a family was
analysed using gel electrophoresis.
Position 1
Position 2
Position 3
Position 4
Position 5
Mother Father Daughter Son 1 Son 2
Analyse the information to explain the bands found at positions 1 to 5 in this family.
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(d) A mutation in the tRNA
Lys
gene in the mitochondria replaces one uracil with a
cytosine in the resulting tRNA molecule.
The diagram shows the position of the uracil that is replaced with cytosine in a
tRNA
Lys
molecule.
uracil replaced with cytosine
(i) Name this type of mutation.
....................................................................................................................................................................................................................................................................................
(ii) Explain how this mutation could affect the structure of the tRNA
Lys
molecule
shown in the diagram.
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(iii) Explain how this mutation could affect the role of the tRNA
Lys
molecule.
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(Total for Question 6 = 11 marks)
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7 Coal tits, blue tits and great tits are three species of bird found in the UK all
year round.
The photograph shows a blue tit.
(Source: © Andrew_Howe/Getty Images)
These birds all have mitochondria in their red blood cells.
(a) More than 90% of ATP is produced by mitochondrial respiration.
(i) Describe how the remaining ATP is produced.
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(ii) In the mitochondria, ATP is synthesised on the inner
mitochondrial membrane.
Leakage of protons (H
) across the inner mitochondrial membrane reduces
ATP synthesis and generates more heat. This is called leaked respiration.
Explain why leakage of protons reduces ATP synthesis.
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(b) The table shows some information about adult coal tits, blue tits and great tits.
Species of bird Length from head to tail / cm Wing span / cm Mass / g
Coal tit 11 19 11
Blue tit 12 18 11
Great tit 14 24 18
(i) Calculate the magnification of the photograph of the blue tit.
Measure from the top of the head to the tip of the tail.
Answer..............................................................
(ii) Calculate the length from head to tail : mass ratio for the great tit.
Give your answer to one decimal place.
Answer..............................................................
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Bird feeders containing nuts were set up and refilled each day
Birds were caught daily in a two-week period in both autumn and winter. They
were ringed, measured and a blood sample taken before they were released
Samples of intact red blood cells and red blood cell mitochondria were
separated from the blood
Leaked respiration and the volume of mitochondria were measured
(c) The generation of heat by red blood cell mitochondria in these birds in autumn
and winter was investigated.
The mean air temperature in autumn was 10.7 °C and in winter 7.2 °C.
The flowchart shows some of the stages in this investigation.
(i) Explain why well‑stocked bird feeders containing nuts were available
throughout this investigation.
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*(ii) The graphs show the results of this investigation.
Graph 1: volume of mitochondria
Coal tit Blue tit Great tit
Mean volume of
mitochondria in red
blood cells
Season
Graph 2: respiration producing ATP
Coal tit Blue tit Great tit
Mean rate of
respiration / a.u.
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Season
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Mean rate of leaked
respiration / a.u.
Graph 3: leaked respiration
Coal tit Blue tit Great tit
Season
Explain the results of this investigation.
Use all the information in the question to support your answer.
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(Total for Question 7 = 15 marks)
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8 Changes must occur to mammalian sperm cells as they pass through the female
reproductive tract, if successful fertilisation is to take place.
Capacitation results in changes to the sperm cell membrane and increased motility.
The acrosome reaction (AR) must follow capacitation but not too soon nor too late.
(a) Explain why the timing of the AR is important for successful fertilisation.
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(b) A protein is involved in the timings of capacitation and the AR.
This protein exists in two forms: protein G and protein G‑P.
Protein G‑P results in the stimulation of capacitation and protein G results in the
inhibition of the AR.
Protein G‑P is located in the midpiece (neck) and flagellum of the sperm.
Protein G is located in the head region of the sperm.
The levels of these two forms of protein change just before capacitation, during
capacitation and just before the AR.
Complete the table to show which form of protein is present and which is absent.
If the protein is present put a tick ( ✓
) in the box and if the protein is absent put a
cross (
) in the box.
Event
Presence of protein G Presence of protein G‑P
Head region
Midpiece and
flagellum
Head region
Midpiece and
flagellum
Just before capacitation
During capacitation
Just before the AR
(c) In some infertile males, the gene coding for protein G is methylated.
Explain why DNA‑methylation of this gene could result in infertility.
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(d) Scientists have used ‘knockout’ mice to investigate the effect of protein G
on fertility.
(i) State the meaning of the term ‘knockout’ mice, as used in this context.
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(ii) In an investigation, female mice and ‘knockout’ mice were housed together.
Describe how this investigation should be designed to confirm that protein G
affects fertility.
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(Total for Question 8 = 15 marks)
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