Cambridge Pre-U Specimen Papers and Mark Schemes Cambridge International Lev Level el 3 Pre-U Certificate in BIOLOGY
For use from 2011 onwards
Cambridge Pre-U Specimen Papers and Mark Schemes Schemes
Specimen Mater Materials ials Biology (9790) Cambridge International Level 3 Pre-U Certificate in Biology (Principal) For use from 2011 onwards
QAN 500/380/72
www.cie.org.uk/cambridgepreu
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Cambridge Pre-U Specimen Papers and Mark Schemes
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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Level 3 Pre-U Certificate Principal Subject
BIOLOGY
9790/01
Paper 1 Structured Questions
For Examination from 2013
SPECIMEN PAPER 2 hours 30 minutes
Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST
Write your name, Centre number and candidate number on all the work you hand in. Write in dark blue or black pen. You may use a pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. Section A Twenty questions for which you must choose what you consider to be the right answer and record your choice in the appropriate space provided. Marks will not be deducted for any wrong answers. Write your answers in the spaces provided on the Question Paper.
For Examiner's Use Section A
Section B Write your answers in the spaces provided on the question paper.
21 22 23
The number of marks is given in brackets [ ] at the end of each question. 24 25 26 Total
This document consists of 31 printed pages and 1 blank page.
© UCIE 2011
[Turn over
2 Section A
For Examiner's
1
Use
The resolving power of a microscope depends on the wavelength used by the system. Table 1.1 shows the wavelengths and resolving powers of three types of microscope. Table 1.1 type of microscope
wavelength / µm
resolving power / µm
light microscope
0.8
0.4
ultra-violet microscope
0.2
0.1
0.005
0.0025
electron microscope
Table 1.2 gives details of four biological structures which are investigated using microscopes. The ticks ( ) and crosses () indicate whether or not each structure can be clearly seen with each microscope. Table 1.2
light microscope
chloroplast, length 5 µm
Escherichia coli bacterium, length 2 µm
ribosome, diameter 25 nm
thickness of plasma membrane 10 nm
ultra-violet microscope electron microscope
Which row correctly completes Table 1.2 to show which structures can be clearly seen with an ultra-violet microscope? chloroplast, length 5 µm
Escherichia coli bacterium, length 2 µm
ribosome, diameter 25 nm
thickness of plasma membrane 10 nm
A
B
C
D
answer
[1]
3 2
Use your knowledge and the information provided to work out which description of telomeres and telomerase reverse transcriptase (TERT) is correct.
For Examiner's Use
telomeres
telomerase reverse transcriptase (TERT)
A
present in eukaryotes
uses RNA as a template to make single stranded DNA
B
present in eukaryotes
inhibits the loss of telomeres from DNA during semiconservative replication
C
present in prokaryotes
inhibits the loss of telomeres from DNA during semiconservative replication
D
present in prokaryotes
uses RNA as a template to make single stranded DNA
answer
3
[1]
The enzyme phosphofructokinase is involved in phosphorylation of hexose phosphate sugars during glycolysis. It is involved in control of the rate of glycolysis and thus respiration, by end-product inhibition. Deduce which of the following is a description of this enzyme. shape of binding site(s)
substrate
products
hexose
hexose phosphate
A
no allosteric site, active site complementary to ATP and hexose
B
allosteric site complementary to glucose, active site complementary to hexose phosphate
hexose phosphate
hexose phosphate
C
allosteric site complementary to ATP, active site complementary to ATP and hexose phosphate
hexose phosphate
hexose bisphosphate
D
no allosteric site, active site complementary to hexose bisphosphate
hexose bisphosphate
two triose phosphate
answer
© UCIE 2011
9790/01/SP/13
[1]
[Turn over
4 4
An action potential arrives at the synaptic knob increasing the permeability of the membranes to ……1……, which diffuse in and cause vesicles to move to the pre-synaptic membrane and fuse with it. ……2…… occurs and the ……3…… moves across the synaptic cleft by ……4…… and attaches to receptors on the post-synaptic membrane, causing …… 5…… channels to open and a post-synaptic potential to be generated. Which words correctly complete the numbered gaps? 1
2
3
4
5
A
acetylcholine
endocytosis
transmitter substance
active transport
calcium ion
B
calcium ions
exocytosis
acetylcholine
diffusion
sodium ion
C
calcium ions
exocytosis
transmitter substance
active transport
sodium ion
D
sodium ions
endocytosis
acetylcholine
diffusion
calcium ion
answer
5
[1]
The diagram shows the major components of the lac operon. 3
2
DNA 1 Which statement is correct? A
1 is a ribosome, 2 is a t-RNA molecule and 3 is a phosphorylated amino acid, the activator, so lactose-digesting enzyme can be made.
B
1 is mRNA polymerase, 2 is β-galactosidase, the inducer and 3 is the repressor, so lactose-digesting enzyme cannot be made.
C
1 is mRNA polymerase, 2 is the repressor and 3 is lactose, the inducer, so lactosedigesting enzyme can be made.
D
1 is the repressor, 2 is a β-galactosidase molecule and 3 is lactose, the promoter, so lactose-digesting enzyme can be made. answer
[1]
For Examiner's Use
5 6
Small samples from crime scenes can be genetically profiled (DNA finger printed).
For Examiner's Use
Which are necessary parts of a successful genetic profiling process? crime scene sample
PCR
ethidium bromide and X-rays
A
red blood cells
B
saliva
C
semen
D
skin cells
Key
= used = not used
answer
7
[1]
A symbiont may be defined as a species in which individuals live in a long-term, intimate and beneficial relationship with hosts of a different species. As the name suggests, endosymbionts live within their hosts. Which statement endosymbionts?
provides
evidence
that
mitochondria
and
chloroplasts
A
Proteins encoded by the nucleus are exported to these organelles.
B
Their inner membrane has different structure from other intracellular membranes.
C
They are surrounded by double membrane.
D
They contain their own ribosomes.
answer
© UCIE 2011
9790/01/SP/13
are
[1]
[Turn over
6 8
Approximately half of the total protein in a pea seed consists of the storage protein vicilin.
For Examiner's Use
Each molecule of vicilin is made up of three identical polypeptides.
•
Each polypeptide is made up of two β-pleated sheet regions with linking α-helix regions, folded into the shape shown to the right.
•
β-pleated sheet regions
This allows the three polypeptides to pack together into a compact, flat storage molecule, as shown below.
•
α-helix regions
Which row correctly describes the structure of vicilin?
primary structure
secondary structure
tertiary structure
quaternary structure
A
amino acid sequence of one polypeptide
α-helix and β-pleated sheet regions of each polypeptide
association of three polypeptides
folding of each polypeptide
B
amino acid sequence of one polypeptide
α-helix and β-pleated sheet regions of each polypeptide
folding of each polypeptide
association of three polypeptides
association of three polypeptides
amino acid sequence of one polypeptide
α-helix and β-pleated sheet regions of each polypeptide
folding of each polypeptide
association of three polypeptides
amino acid sequence of one polypeptide
folding of each polypeptide
α-helix and β-pleated sheet regions of each polypeptide
C
D
answer
[1]
7 9
The graphs represent the frequency of alleles in species X, Y and Z during and after selection.
For Examiner's Use
key characteristics selected for characteristics selected against X
Y
Z during selection
allele frequency
range of characteristic(s) regulated by alleles
after selection
allele frequency
range of characteristic(s) regulated by alleles In which species does evolution take place? A
X only
B Y only C Y and Z D
none of X, Y nor Z
answer
© UCIE 2011
9790/01/SP/13
[1]
[Turn over
8 10 Curve X shows the oxygen dissociation curve for human haemoglobin. Under certain conditions this curve becomes displaced to the right. This is termed the Bohr effect and is shown by curve Y. X
100 80 percentage 60 saturation of haemoglobin 40
Y
20 0 0
2
4
6
8
10 12 14
partial pressure of oxygen / kPa Which change is responsible for the Bohr effect? A
a decrease in the partial pressure of oxygen
B
a decrease in the temperature of the blood
C
an increase in pH of the blood
D
an increase in the partial pressure of carbon dioxide
answer
[1]
For Examiner's Use
9 11 Which statements correctly describe the structure and function of prokaryote ribosomes?
For Examiner's Use
1
Prokaryote ribosomes are smaller than eukaryote ribosomes and sediment at 70 S.
2
A prokaryote ribosome consists of two subunits, one of 50 S and one of 30 S.
3
In prokaryotes, ribosomes translate mRNA in the same cellular compartment in which it is transcribed.
4
In prokaryotes, ribosomes can begin translating mRNA before its synthesis has been completed.
5
A prokaryote ribosome can accommodate only one amino acyl-tRNA at a time.
A
1, 2, 3, 4 and 5
B
1, 2, 3 and 4 only
C
1, 3 and 5 only
D
2, 4 and 5 only
answer
© UCIE 2011
9790/01/SP/13
[1]
[Turn over
10 12 A snake venom causes death by leading to paralysis of muscles. It exerts its effect at synapses.
For Examiner's Use
The statements below were put forward by scientists as possible explanations for the effects of this venom. 1
It interferes with the binding of neurotransmitter vesicles to the membranes.
2
It binds with neurotransmitter receptor sites.
3
It blocks calcium and sodium channels.
4
It destroys the myelin sheath of the neurone.
5
It binds with neurotransmitter.
Which statements should be investigated further? A
1, 2, 3 and 5 only
B
2, 4 and 5 only
C
4 only
D
1, 2, 3, 4 and 5
answer
[1]
13 What happens during the light-dependent reactions of photosynthesis? 1
ADP is hydrolysed.
2
ADP is phosphorylated.
3
ATP is hydrolysed.
4
ATP is phosphorylated.
5
NADP is oxidised.
6
NADP is reduced.
A
1 and 5 only
B
2 and 6 only
C
1, 4 and 5 only
D
2, 3 and 6 only answer
[1]
11 14 One of the many recessive mutations of the CFTR gene changes one amino acid in the region of the CFTR protein that binds ATP. The graph shows the effect of different concentrations of ATP on normal and mutant CFTR proteins.
For Examiner's Use
normal CFTR
100 80 percentage of functioning ion channels
mutant CFTR
60 40 20 0 concentration of ATP
Which correctly describes individuals who are homozygous for this mutation? 1
Their CFTR protein cannot bind ATP and cannot act as an ion channel.
2
Their CFTR protein binds ATP less readily than normal CFTR protein.
3
They produce CFTR protein that must bind ATP to function as an ion channel.
4
They produce a mixture of normal and mutant CFTR protein, both of which can act as an ion channel.
A
1 only
B
2 only
C
2 and 3 only
D
2 and 4 only
answer
© UCIE 2011
9790/01/SP/13
[1]
[Turn over
12 15 Many plants are not fertilised by pollen from their own flowers. This is known as self-incompatibility. In any individual species a single gene, the S gene, is responsible and it may have many different alleles. If a pollen grain has an S allele which matches an allele in the genotype of the stigma then the pollen grain fails to germinate or the pollen tube fails to grow through the style. The genotype of the stigma of a flower is S3S4. Which pollen grains would germinate? 1
2
3
4
S1
S1S3
S3
S2S4
pollen grain A
1 only
B
2 and 4 only
C
3 only
D
3 and 4 only
answer
[1]
16 Molecules can be transported in several ways. 1
cohesion/tension
2
diffusion
3
mass flow
4
osmosis
Which row shows the correct method of transport? into capillaries
in phloem
out of stomata
in xylem
A
1
2
3
4
B
2
1
4
3
C
3
4
1
2
D
4
3
2
1
answer
[1]
For Examiner's Use
