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Title: Edexcel a level biology B question paper 2 june 2024 salters Nuffield + mark scheme
Description: Edexcel a level biology B question paper 2 june 2024 salters Nuffield + mark scheme
Description: Edexcel a level biology B question paper 2 june 2024 salters Nuffield + mark scheme
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Edexcel as level biology
B question paper 1 june
2024 salters Nuffield +
mark scheme
Please check the examination details below before entering your candidate information
Candidate surname
Centre Number
Other names
Candidate Number
Pearson Edexcel Level 3 GCE
Monday 13 May 2024
Morning (Time: 1 hour 30 minutes)
Paper
reference
Biology B
8BI0/01
Advanced Subsidiary
PAPER 1: Core Cellular Biology and Microbiology
You must have:
Scientific calculator, HB pencil, ruler
Total Marks
Instructions
Use black ink or ball-point pen
...
Answer all questions
...
Information
The total mark for this paper is 80
...
question(s) marked with an asterisk (*), marks will be awarded for your ability to
• Instructure
your answer logically, showing how the points that you make are related or
follow on from each other where appropriate
...
•
Try to answer every question
...
Turn over
P74474A
©2024 Pearson Education Ltd
...
These inorganic
ions are dissolved in the water in the soil
...
Which row in the table shows the molecules that incorporate these ions?
amino acids
chlorophyll
nucleotides
A magnesium ions
nitrate ions
magnesium ions
B
magnesium ions
nitrate ions
nitrate ions
C
nitrate ions
magnesium ions
nitrate ions
D
nitrate ions
magnesium ions
magnesium ions
(ii) Explain why these ions dissolve in water
...
...
...
...
If you change your mind about an
answer, put a line through the box and then mark your new answer with a cross
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(b) The graph shows the uptake of nitrate ions by a plant in the presence and
absence of oxygen
...
(2)
1
...
...
...
(Total for Question 1 = 5 marks)
*P74474A0328*
3
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(a) The diagram shows part of a karyotype from a female with Down’s syndrome
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(2)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
(b) The diagram shows two chromosomes, C1 and C2, before translocation
...
4
*P74474A0428*
(1)
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2 Chromosome mutations include non-disjunction and translocation
...
Control group
n = 11
With Turner’s syndrome
n = 37
Mean mass / kg ± SD
45
...
3
36
...
8
Mean height / cm ± SD
146
...
3
134
...
0
Feature
Body mass index (BMI)
20
...
(4)
...
...
...
...
...
...
(Total for Question 2 = 9 marks)
*P74474A0528*
5
Turn over
(a) (i) Which pair of structures is found in these bacteria?
(1)
A
nucleoid and 70S ribosomes
B
nucleoid and 80S ribosomes
C
nucleus and 70S ribosomes
D
nucleus and 80S ribosomes
(ii) Name three other structures that are surrounded by the cell wall of bacteria
...
2
...
(b) A scientist determined the thickness of the cell wall of these bacteria
...
subtilis is 55
...
aeruginosa is 2
...
(i) Calculate the ratio of the thickness of the cell wall of B
...
aeruginosa
...
(1)
Answer
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(ii) Some antibiotics affect only Gram positive bacteria
...
subtilis or P
...
(2)
...
...
...
(c)
Bacillus subtilis was grown in a culture
...
7 × 105 cells
...
Give your answer in cells per minute
...
cells per minute
(Total for Question 3 = 8 marks)
*P74474A0728*
7
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(a) The photograph shows meiosis in a bluebell plant
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(ii) Devise a practical procedure to determine the total percentage of cells in both
metaphase I and metaphase II in a bluebell
...
...
...
...
...
...
...
...
*P74474A0928*
9
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(b) Explain how meiosis results in genetic variation in a flowering plant
...
...
...
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BLANK PAGE
Turn over
*P74474A01128*
11
A
C
(Source: © CNRI/SCIENCE PHOTO LIBRARY)
(a) Name the parts labelled A, B and C
...
B
...
(b) Draw a diagram of this mitochondrion
...
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(c) The actual length of this mitochondrion is 0
...
(i) What is the magnification of this photograph?
(1)
A
× 1200
B
× 8200
C
× 12 000
D
× 82 000
(ii) The magnification of this photograph can be shown as a scale bar
...
(1)
1 cm scale bar
Answer
...
(2)
...
...
...
(Total for Question 5 = 8 marks)
*P74474A01328*
13
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*P74474A01428*
14
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6 The diagram shows the structure of the influenza virus
...
Which virus also has a helical capsid and an envelope?
(1)
A
Ebola
B
human immunodeficiency virus
C
λ (lambda) phage
D
tobacco mosaic virus
(b) Neuraminidase is an enzyme
...
(3)
...
...
...
...
...
Neuraminidase and haemagglutinin are important in the lytic cycle of this virus
...
Sialic acid is found on glycoproteins present in the cell membrane, mucus and the
envelope of the virus
...
The lytic cycle of the influenza virus includes the following steps:
1 the virus attaches to a cell
2 the virus moves until it attaches to the correct cell surface receptor
3 new virus particles are made inside the host cell
4 new virus particles bud from the host cell
5 new virus particles are attached to the outside of the host cell membrane and
to each other
6 new virus particles infect new host cells
...
These cells are
coated in mucus
...
(6)
...
...
...
...
...
...
...
...
...
(Total for Question 6 = 10 marks)
*P74474A01728*
17
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*P74474A01828*
18
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7 The synthesis of DNA takes place during the cell cycle
...
(3)
DNA content
per cell
Time
mitosis
*P74474A01928*
19
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The diagram shows the stages involved in one experiment
...
The heavier DNA molecules form bands lower down the gradient than lighter
DNA molecules
...
20
*P74474A02028*
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*(b) The experiments of Meselson and Stahl contributed to our current understanding
of DNA replication
...
DNA taken after
stage 1
DNA taken after
stage 2
DNA taken after
stage 3
DNA taken after
stage 4
Analyse the data to explain the results of this experiment and how it contributed
to our current understanding of DNA replication
...
...
...
...
...
...
...
...
(Total for Question 7 = 9 marks)
*P74474A02128*
21
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(a) The table gives some information about polysaccharides
...
(3)
Type of glycosidic bond
Polysaccharide
both
1–4 and 1–6
1–4 only
1–6 only
neither
1–4 nor 1–6
cellulose
glycogen
starch
(b) Monosaccharides and disaccharides are either reducing sugars or
non‑reducing sugars
...
The method can be used to determine the concentration of the
reducing sugar
...
Step 1
Add Benedict’s solution to the sugar solution in a test tube
Step 2
Place the test tube in a hot water bath until the colour changes
Step 3
Determine the concentration of reducing sugar from a colour chart
22
*P74474A02228*
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8 Carbohydrates can be classified as monosaccharides, disaccharides,
or polysaccharides
...
Colour after heating
Concentration of reducing
sugar / mg dm–3
blue
green
yellow
orange
red
0
<1000
1000 to 1500
1500 to 2000
>2000
(i) Describe two limitations of using this method for determining the
concentration of reducing sugars
...
...
2
...
...
A student was given a solution containing both glucose and sucrose
...
(4)
...
...
...
...
...
...
(Total for Question 8 = 9 marks)
24
*P74474A02428*
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Non-reducing sugars, such as sucrose, must be heated in acid before step 1
...
(a) Phospholipids are one type of lipid found in cell membranes
...
(3)
...
...
...
...
...
O
O P OH
OH
O
O
O
O
*P74474A02528*
25
Turn over
Give your answer in standard form
...
(b) The graphs show the percentage of four types of phospholipid, A, B, C and D,
in the cell membrane and in the membranes of endoplasmic reticulum,
mitochondria and Golgi apparatus
...
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The cross-sectional area of this cylinder is 0
...
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(i) Describe how the structure of one type of phospholipid can be different from
the structure of another type of phospholipid
...
...
...
...
(3)
...
...
...
...
...
(2)
...
...
...
(Total for Question 9 = 13 marks)
TOTAL FOR PAPER = 80 MARKS
28
*P74474A02828*
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(iii) The scientists who produced the data did not expect the phospholipid
content in the membranes of the endoplasmic reticulum and the Golgi
apparatus to be different
...
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Summer 2024
Question Paper Log Number P74474A
Publications Code 8BI0_01_2406_MS
All the material in this publication is copyright ©
Pearson Education Ltd 2024
General Marking Guidance
•
All candidates must receive the same treatment
...
•
Mark schemes should be applied positively
...
•
Examiners should mark according to the mark scheme not
according to their perception of where the grade boundaries
may lie
...
All marks on the mark
scheme should be used appropriately
...
Examiners should always award full marks if deserved, i
...
if
the answer matches the mark scheme
...
•
Where some judgement is required, mark schemes will provide
the principles by which marks will be awarded and
exemplification may be limited
...
•
Crossed out work should be marked UNLESS the candidate has
replaced it with an alternative response
...
463 × 1
...
140369
Accept rounding – 2
...
1
• BMI calculated (1)
21
...
1178539)
Accept (if rounded to 2
...
1214953
Accept (if rounded to 2
...
5238095/21
...
0 : 0
...
043
• 23 : 1
...
subtilis, because it {is Gram positive / has a thick
peptidoglycan cell wall} (1)
• antibiotics interfere with the synthesis of peptidoglycan
(crosslinks) / inhibit formation of peptidoglycan (1)
Question
Number
3(c)
Answer
(2)
Additional Guidance
Mark
An answer that makes reference to the following:
• increase in cell number calculated (1)
(1
...
7 / 1366
...
66) gains 1 mark
Correct answer with no working gains 2 marks
(2)
Question
Number
4(a)(i)
Answer
Mark
The only correct answer is C
A
is incorrect because P is in prophase which comes before R which
is in late anaphase
B
is incorrect because Q is in early anaphase which comes before R
which is in late anaphase
D
is incorrect because S is in metaphase which comes before R
which is in late anaphase
Question
Number
4(a)(ii)
Additional Guidance
Answer
An answer that makes reference to the following:
•
(1)
Additional Guidance
Mark
Sequence should be logical
select the { anther / filament / carpel / ovary / embryo sac } (1)
and any four from:
•
place in {acid / HCl} and {heat / place in hot / warm water bath} (1)
Accept stated temperature
•
add acetic orcein (1)
Accept other appropriate named stains e
...
Feulgen’s, Toluene blue
•
tease the cells apart (with a mounted needle) / macerate (1)
•
place a coverslip on top and gently squash the preparation (1)
•
count total number of cells in each stage (of meiosis) / count (all)
cells and count number of cells at metaphase I and metaphase II (1)
•
divide the number of cells in metaphase (multiplied by 100) by the
total number of cells (1)
(5)
Question
Number
4(b)
Answer
Mark
An explanation that makes reference to the following:
•
independent assortment (1)
•
crossing over between homologous chromosomes (1)
•
different combinations of alleles (in the gametes) / different
combinations of maternal and paternal chromosomes (1)
Question
Number
5(a)
Additional Guidance
Answer
(3)
Additional Guidance
Mark
An answer that makes reference to the following:
A : matrix (space)
B : outer membrane / envelope
Accept intermembrane space / double
membrane
C : crista / cristae / (folded) inner membrane
All three correct = 2 marks
1 or 2 correct = 1 mark
(2)
Question
Number
5(b)
Answer
Additional Guidance
Mark
An answer that makes reference to the following:
•
diagram that (roughly) shows the shape and position of the cristae
(minimum of 7 cristae at least half width of drawing) (1)
•
diagram that has no overlapping lines, breaks in lines, shading (1)
Question
Number
Answer
(2)
Additional Guidance
Mark
5(c)(i)
The only correct answer is D
A
is incorrect because 74 000 ÷ 0
...
9 = 82 222
(1)
C
Question
Number
5(c)(ii)
is incorrect because 74 000 ÷ 0
...
12 (195
...
12 (162…) µm (1)
Additional Guidance
Mark
Accept ECF from Q5ci)
A 8
...
21(951…) or 1
...
83(3333) µm
(1)
Question
Number
5(d)
Answer
Additional Guidance
Mark
An explanation that makes reference to the following:
•
because (many) specimens / cells, will be colourless / not be
visible (without a stain) (1)
Accept stains provide a contrast
•
because electron microscope uses electrons to show (electron)
dense areas (1)
Accept electrons are not absorbed/scattered by
coloured stains / electron microscopes need to use
heavy metals
Question
Number
Answer
6(a)
Additional Guidance
(2)
Mark
The only correct answer is A
B
is incorrect because Ebola has a helical capsid and an envelope
whereas HIV has a polyhedral capsid
C
is incorrect because Ebola has a helical capsid and an envelope
whereas λ (lambda) phage does not have an envelope and has a
complex capsid
D
is incorrect because Ebola has a helical capsid and an envelope
whereas TMV does not have an envelope
(1)
Question
Number
6(b)
Answer
Additional Guidance
Mark
A description that makes reference to two of the following:
•
globular / 3D / tertiary protein (1)
•
held together by {hydrogen bonds / disulfide bridges / ionic
bonds/hydrophobic (interactions)}
(1)
•
between R groups (1)
•
has an active site (1)
Accept hydrophobic R groups on inside /hydrophilic
R groups on outside = mp 2 & 3
(3)
EXP
Question Indicative content
Number
*6(c)
Indicative content:
Level 1:
Lytic Cycle (L)
• virus attaches/binds to (host) cells
• (viral) genetic material / provirus, inserted (into host cell)
• viral proteins/capsid made
• genetic material /(viral) RNA, replicates
• new viral particles assembled
• cell ruptures/viral particles, released
Haemagglutinin (H)
•
•
•
binds to sialic acid groups on the {glycoproteins / cell membrane / host cell / mucus}
so that it can infect host cell
needed by new virus particles to attach to more (host) cells
1 mark = 1 point from L, H or N
2 marks = 2 points from L, H or N
Level 2 :
3 marks = 3 points from at least
two of L, H or N
4 marks = 4 points from at least
two of L, H or N
Neuraminidase (N)
•
•
•
•
•
•
•
•
needed to separate virus particles from the {(non-host) cells / membrane} that they attach to
breaks down mucus
cleave the haemagglutinin from the sialic acid group on cell
needed to separate the new virus particles from each other
needed to separate the new virus particles from the host cell membrane (that they have budded
out of)
needed to release the new virus particles from mucus
so that they are free to attach to new host cells
so that the infection is spread
Level 3 :
5 marks = 5 points from L, H and
N
6 marks = 6 points from L, H and
N
Question
Number
7(a)
Answer
Additional Guidance
Mark
An answer that makes reference to the following:
•
a {vertical / upward sloping} line that shows the DNA content increasing
(before mitosis) and a {vertical / downward sloping} line that shows the
DNA content decreasing (after/towards end of mitosis) (1)
•
a horizontal line extending across all of mitosis (1)
•
DNA doubles after replication and returns to same level indicated at start
(halves) (1)
(3)
Question Indicative content
Number
*7(b)
Indicative content:
DNA replication (D)
• reference to semi-conservative replication
• helicase {separates the DNA strands / break hydrogen bonds between DNA strands}
• each DNA strand acts as a template
• bases complementary base pair with the exposed bases (on template strands)
• DNA polymerase joins together (adjacent) DNA nucleotides
• each new molecule (of DNA) has one old strand and one new strand
• number of molecules doubles each time so total width of bands doubles each time
Stage 1 (S1)
• bases contain {heavy nitrogen / 15 N }
• only one band near bottom of tube, as all the DNA has {heavy nitrogen / 15N }
Stage 2 (S2)
• DNA replication occurs using the { light nitrogen / 14N }
• new strands will contain light nitrogen / 14N }
• only one band as all new molecules / DNA have one heavy and one light strand
• band higher up in the tube as it is lighter than the {original / stage 1} DNA
Stage 3 (S3)
• new strands will contain { light nitrogen / 14N }
• (two bands because) {original / stage 1 / heavy nitrogen / 15N strands} will bind with lighter strands to form
{medium weight DNA molecules / band in same position as stage 2 }
• (two bands because) light strands from stage 2 molecules will bind with new light strands to form lightest
molecules, so will be higher up the tube
• bands of equal widths as equal number of light and heavy strands in stage 2 DNA
Level 1:
1 mark = 2 details from
D, S1, S2, S3 or S4
2 marks = 3 details from D,
S1, S2, S3 or S4
Level 2:
3 marks = 4 details from
two of D, S1, S2 S3 or S4
4 marks = 4 details from
three of D, S1, S2 S3 or S4
Level 3:
5 marks = 5 details from
four of D, S1, S2 S3 or S4
6 marks = 5 details from
all of D, S1, S2 S3 and S4
Stage 4 (S4)
• all new strands contain { light nitrogen / 14N }
• (two bands because) {original / stage 1 / heavy nitrogen / 15N strands} will bind with lighter strands to form
{medium weight DNA molecules / band in same position as stage 2 / band in same position as the last stage}
• (two bands because) light strands from stage 2 molecules will bind with new light strands to form lightest
molecules, so will be { higher up the tube / in same position as the last stage }
• more light strands in stage 3 DNA that heavy strands so the width of band for lightest DNA will be wider
• in a ratio of 3 : 1
Question
Number
Answer
Additional Guidance
Mark
8(a)
Type of glycosidic bond
Polysaccharide
both
1-4 and 1-6
cellulose
1-4 only
1-6 only
neither
1-4 nor 1-6
X
glycogen
X
starch
X
(3)
Question
Number
8(b)(i)
Answer
Additional Guidance
Mark
A description that makes reference to the following:
•
deciding on colours is subjective / people may judge colours
differently / hard to distinguish between some colours (1)
•
Accept idea that 1500 is both yellow and orange
the values for the concentration of sugars are ranges /
semiquantitative / an { accurate / exact } sugar concentration can Accept idea that the ranges of concentration are
large
not be determined (1)
Question
Answer
Number
8(b)(ii)
An answer that makes reference to four of the following:
Additional Guidance
•
carry out test for reducing sugar (1)
Accept description of test (Benedict’s, plus heat –)
for reducing and non-reducing
•
use chart to determine concentration of glucose / sucrose (1)
Accept alternative method e
...
3 = 3 333 333
...
3 / 3 333 333
...
3333 × 2 = 6 666 666
...
67
2 marks for 6 666 667 /6 666 666
...
3 x106
6
...
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Title: Edexcel a level biology B question paper 2 june 2024 salters Nuffield + mark scheme
Description: Edexcel a level biology B question paper 2 june 2024 salters Nuffield + mark scheme
Description: Edexcel a level biology B question paper 2 june 2024 salters Nuffield + mark scheme