BIO PAPER 1 SPECIMEN

Question 1

A technician investigated the effect of temperature on the rate of an enzyme-controlled
reaction. At each temperature, he started the reaction using the same concentration
of substrate.
Figure 1 shows his results.
Give two other factors the technician would have controlled.
[1 mark]

Answer 1

Any two of the following;
Concentration of enzyme
Volume of substrate solution
pH

Question 2

1.3 Explain the difference in the initial rate of reaction at 60 °C and 37 °C.
[2 marks]

Answer 2

At 60 °C:

1. More kinetic energy;
2. More E–S complexes formed

Question 3

Explain the difference in the rates of reaction at 60 °C and 37 °C between 20 and
40 minutes.
[4 marks]

Answer 3

Different times:
1. Higher temperature / 60 °C causes
denaturation of all of enzyme;
2. Reaction stops (sooner) because shape of
active site changed;
Different concentrations of product (at 60 °C)
3. Substrate still available (when enzyme
denatured);
4. But not converted to product;
4
Accept converse for
37 °C
2. Reject if active
site on substrate

Question 4

2.1 -Describe how oxygen in the air reaches capillaries surrounding alveoli in the lungs.
Details of breathing are not required.
[4 marks

Answer 4

1. Trachea and bronchi and bronchioles;
2. Down pressure gradient;
3. Down diffusion gradient;
4. Across alveolar epithelium;
5. Across capillary endothelium/epithelium;

Question 5

2.3- Asthma affects bronchioles and reduces flow of air in and out of the lungs.
Fibrosis does not affect bronchioles; it reduces the volume of the lungs.
Which group, B or C, was the one containing people with fibrosis of their lungs?
Use the information provided and evidence from Figure 2 to explain your answer.
[3 marks]
0 2 . 2
0

Answer 5

1. (Group B because) breathe out as quickly as
   healthy / have similar FEV to group A;
2. So bronchioles not affected;
3. FVC reduced / total volume breathed out
   reduced;

Question 6

Species richness and an index of diversity can be used to measure biodiversity within
a community

What is the difference between these two measures of biodiversity?
[1 mark]

Answer 6

Species richness measures only number of
(different) species / does not measure number of
individuals;

Question 7

Scientists investigated the biodiversity of butterflies in a rainforest. Their investigation
lasted several months.
The scientists set one canopy trap and one understorey trap at five sites.
• The canopy traps were set among the leaves of the trees 16–27 m above ground
level.
• The understorey traps were set under trees at 1.0–1.5 m above ground level.
The scientists recorded the number of each species of butterfly caught in the traps.
Table 1 summarises their results.
Table 1

The traps in the canopy were set at 16–27 m above ground level. Suggest why there
was such great variation in the height of the traps.
[1 mark]

Answer 7

Trees vary in height;

Question 8

3.3- By how many times is the species diversity in the canopy greater than in the
understorey? Show your working.
Use the following formula to calculate species diversity.
𝑑 = 𝑁(𝑁 − 1)
∑ 𝑛 (𝑛 − 1)
where 𝑁 is the total number of organisms of all species and 𝑛 is the total number of
organisms of each species.
[3 marks]

Answer 8

1. Index for canopy is 3.73;
2. Index for understorey is 3.30;
3. Index in canopy is 1.13 times bigger;

Question 9

3.4-The scientists carried out a statistical test to see if the difference in the distribution
of each species between the canopy and understorey was due to chance.
The P values obtained are shown in Table 1.
Explain what the results of these statistical tests show.
[3 marks]

Answer 9

1. For Zaretis itys, difference in distribution is
   probably due to chance / probability of being
   due to chance is more than 5%;
2. For all species other than Zaretis itys,
   difference in distribution is (highly) unlikely to
   be due to chance;
3. Because P < 0.001 which is highly significant/is
   much lower than 5%

Question 10

Starch and cellulose are two important plant polysaccharides.
Figure 3 shows part of a starch molecule and part of a cellulose molecule

4.1-Explain the difference in the structure of the starch molecule and the cellulose
molecule shown in Figure 3.
[2 marks]

Answer 10

1. Starch formed from α-glucose but cellulose
   formed from β-glucose;
2. Position of hydrogen and hydroxyl groups on
   carbon atom 1 inverted;

Question 11

4.2- Starch molecules and cellulose molecules have different functions in plant cells. Each
molecule is adapted for its function.

Explain one way in which starch molecules are adapted for their function in plant cells.
[2 marks]

Answer 11

1. Insoluble;
2. Don’t affect water potential;
   OR
3. Helical;
4. Compact;
   OR
5. Large molecule;
6. Cannot leave cell;

Question 12

4.3 - Explain how cellulose molecules are adapted for their function in plant cells.
[3 marks]

Answer 12

1. Long and straight chains;
2. Become linked together by many hydrogen
   bonds to form fibrils;
3. Provide strength (to cell wall);

Question 13

Contrast the processes of facilitated diffusion and active transport.
[3 marks]

Answer 13

1. Facilitated diffusion involves channel or carrier
   proteins whereas active transport only involves
   carrier proteins;
2. Facilitated diffusion does not use ATP / is
   passive whereas active transport uses ATP;
3. Facilitated diffusion takes place down a
   concentration gradient whereas active transport
   can occur against a concentration gradient;

Question 14

Students investigated the uptake of chloride ions in barley plants. They divided the
plants into two groups and placed their roots in solutions containing radioactive
chloride ions.
• Group A plants had a substance that inhibited respiration added to the solution.
• Group B plants did not have the substance added to the solution.
The students calculated the total amount of chloride ions absorbed by the plants every
15 minutes. Their results are shown in Figure 4.

Explain the results shown in Figure 4.
[4 marks]

Answer 14

1. Group A – initial uptake slower because by
   diffusion (only);
2. Group A – levels off because same
   concentrations inside cells and outside cells /
   reached equilibrium;
3. Group B – uptake faster because by diffusion
   plus active transport;
4. Group B fails to level off because uptake
   against gradient/no equilibrium to be reached;
5. Group B – rate slows because few/fewer
   chloride ions in external solution/respiratory
   substrate used up;

Question 15

n this investigation, what is meant by genetic diversity?

[1 mark]

Answer 15

Number of different alleles of each gene;

Question 16

he scientists concluded that the bluethroat showed greater genetic diversity than the
willow flycatcher. Explain why they reached this conclusion. Use calculations to
support your answer.
[2 marks]

Answer 16

1. Has greater proportion of genes / percentage of
   genes showing diversity;
2. Percentage is 35% compared with 28% /
   proportion is 0.35 compared with 0.28;

Question 17

Figure 5 shows a test that has been developed to find out if a person has antibodies
to the human immunodeficiency virus (HIV) antigen.
Figure 5
HIV antigens are attached to
a test well in a dish.
A sample of blood plasma is
added to the well.
If HIV antibodies are present,
they bind to the HIV antigen.
The well is washed.
A second antibody with an
enzyme attached is then
added.
This binds specifically to the
HIV antibody.
The well is washed again.
A yellow solution is added,
which changes to blue if the
enzyme is present. A blue
colour shows that the person
has HIV antibodies.

7.1- Figure 5 shows a test that has been developed to find out if a person has antibodies
to the human immunodeficiency virus (HIV) antigen.
Figure 5
This test only detects the presence of HIV antibodies. Give two reasons why it cannot
be used to find out if a person has AIDS.
[2 marks]

Answer 17

(To diagnose AIDS, need to look for/at)

1. (AIDS-related) symptoms;
2. Number of helper T cells;

Question 18

The solution will remain yellow if a person is not infected with HIV. Explain why.
[2 marks]

Answer 18

1. HIV antibody is not present;
2. (So) second antibody/enzyme will not bind/is
   not present;

Question 19

7.2-The solution will remain yellow if a person is not infected with HIV. Explain why.
[2 marks]

Answer 19

1. HIV antibody is not present;
2. (So) second antibody/enzyme will not bind/is
   not present;

Question 20

7.3-A mother who was infected with HIV gave birth to a baby. The baby tested positive
using this test. This does not prove the baby is infected with HIV.
Explain why.
[2 marks]

Answer 20

1. Children receive (HIV) antibodies from their
   mothers/maternal antibodies;
2. (So) solution will always turn blue/will always
   test positive (before 18 months);

Question 21

7.4- A control well is set up every time this test is used. This is treated in exactly the same
way as the test wells, except that blood plasma is replaced by a salt solution.
Use information from Figure 5 to suggest two purposes of the control well.
[2 marks]

Answer 21

1. Only the enzyme/nothing else is causing a
   colour change;
2. Washing is effective/all unbound antibody is
   washed away;

Question 22

8.3- During replication, the two DNA strands separate and each acts as a template for the
production of a new strand. As new DNA strands are produced, nucleotides can only
be added in the 5’ to 3’ direction.
Use Figure 6 and your knowledge of enzyme action and DNA replication to explain
why new nucleotides can only be added in a 5’ to 3’ direction.
[4 marks]
Turn over for the next question
0

Answer 22

1. Reference to DNA polymerase;
2. (Which is) specific;
3. Only complementary with/binds to 5’ end (of
   strand);
4. Only complementary with/binds with phosphate
   end (of the developing strand);

Question 23

Describe the mass flow hypothesis for the mechanism of translocation in plants.
[4 marks]

Answer 23

1. In source/leaf sugars actively transported into
   phloem;
2. By companion cells;
3. Lowers water potential of sieve cell/tube and
   water enters by osmosis;
4. Increase in pressure causes mass movement
   (towards sink/root);
5. Sugars used/converted in root for respiration for
   storage

Question 24

Scientists measured translocation in the phloem of trees. They used carbon dioxide
labelled with radioactive 14C.
They put a large, clear plastic bag over the leaves and branches of each tree and
added 14CO2. The main trunk of the tree was not in the plastic bag.
At regular intervals after adding the 14CO2 to the bag, the scientists measured the
amount of 14CO2 released from the top and bottom of the main trunk of the tree. On
the surface of the trunk of these trees, there are pores for gas exchange.
Figure 7 shows the scientists’ results.

Name the process that produced the 14CO2 released from the trunk. (1)

Answer 24

Respiration

Question 25

9.3-How long did it take the 14C label to get from the top of the trunk to the bottom of the
trunk? Explain how you reached your answer.
[2 marks]

Answer 25

1. (About) 30 hours;
2. Time between peak 14C at top of trunk and
   bottom;

Question 26

9.4-What other information is required in order to calculate the mean rate of movement of
the 14C down the trunk?
[1 mark]

Answer 26

Length of trunk (between top and bottom);

Question 27

Place stages A to E in the correct order. Start with stage D.
(1)

Answer 27

(D)CBEA;

Question 28

Complete Table 5 to give one reason why each of these steps was necessary.
[2 marks]

Taking cells from the root tip
Firmly squashing the root tip

Answer 28

(Taking cells
from the root
tip)
Region where
mitosis/cell division
occurs;
(Firmly
squashing
the root tip)
To allow light through /
make tissue layer thin;

Question 29

Figure 9 shows how the amount of DNA per cell changed during interphase and
meiosis in an anima
l

Explain how the behaviour of chromosomes causes these changes in the amount of
DNA per cell between F and G.
[3 marks]

Answer 29

(Increase)
1. Chromosomes/DNA replicates;
(First decrease)
2. Homologous chromosomes separate;
(Second decrease)
3. Sister chromatids separate;

Question 30

What would happen to the amount of DNA per cell at fertilisation of cell G?
[1 mark]

Answer 30

1. (DNA would) double/go to 2 (arbitrary units);

Question 31

11.1- Messenger RNA (mRNA) is used during translation to form polypeptides.
Describe how mRNA is produced in the nucleus of a cell.
[6 marks]
1

Answer 31

1. Helicase;
2. Breaks hydrogen bonds;
3. Only one DNA strand acts as template;
4. RNA nucleotides attracted to exposed bases;
5. (Attraction) according to base pairing rule;
6. RNA polymerase joins (RNA) nucleotides
   together;
7. Pre-mRNA spliced to remove introns;

Question 32

Describe the structure of proteins (5)

Answer 32

1. Polymer of amino acids;
2. Joined by peptide bonds;
3. Formed by condensation;
4. Primary structure is order of amino acids;
5. Secondary structure is folding of polypeptide
   chain due to hydrogen bonding;
6. Tertiary structure is 3-D folding due to hydrogen
   bonding and ionic/disulfide bonds;
7. Quaternary structure is two or more polypeptide
   chains;

Question 33

Describe how proteins are digested in the human gut.
[4 marks]
END OF QUESTIONS

Answer 33

1. Hydrolysis of peptide bonds;
2. Endopeptidases break polypeptides into
   smaller peptide chains;
3. Exopeptidases remove terminal amino acids;
4. Dipeptidases hydrolyse/break down dipeptides
   into amino acids