A level june specimen paper 1

Question 1

Give two other factors the technician would have controlled.

[1 mark]

Answer 1

• Concentration of enzyme
• Volume of substrate
• pH

Question 2

Draw a tangent on each curve to find the initial rates of reaction.
Use these values to calculate the ratio of the initial rates of reaction at 60 °C : 37 °C.
Show your working.
[2 marks]

Answer 2

At 37 degrees - 8.75/10 = 0.875
At 60 degrees - 9/2= 4.5
4.5/0.875 = 5.14

Question 3

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

Answer 3

At 60 C there is more kinetic energy within the particles so there are more frequent collisions between enzyme and substrate and so more enzyme - substrate complexes are form.

Question 4

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

Answer 4

At 60C, this high temperature has caused denaturation of all enzymes causing a permanent change to the active site causing the reaction to stop. However there is still some substrate available, when all the enzymes have denatured but aren’t converted into product so concentration remains constant.

Question 5

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

Answer 5

Air is taken in through the mouth and nose. Air moves through the trachea and then the bronchi, and then the bronchioles down the pressure gradient. Air then moves down the diffusion gradient from an area of high concentration to low concentration across the diffusion gradient. Air moves across the alveolar epithelium via diffusion as well as across the capillary endothelium and into the alveoli.

Question 6

Calculate the percentage drop in FEV for group C compared with the healthy
people.
[1 mark]

Answer 6

Percentage decrease = X2-X1/X1 X1=4.2
0.8-4.2/4.2 X100 = -80.952
= 81%

Question 7

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]

Answer 7

Group B are the ones containing people with fibrosis of their lungs as they have a similar FEV to group A. Therefore the bronchioles aren’t affected. However the total volume breathed out has been reduced, and so this provides evidence to suggest its group B.

Question 8

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

Answer 8

Species richness measures only the number of different species and does not measure the number of individuals.

Question 9

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 9

The trees varied in height

Question 10

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 10

The zaretis itys difference in distribution is most likely due to chance as the probability is greater than 5%. All the other species have a very unlikely chance of being distributed differently due to the low P value that holds high signficance.

Question 11

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

Answer 11

• Starch is formed from a- glucose monomers, but cellulose is formed from b-glucose monomers
• Each adjacent glucose molecule is inverted 180 due to a change in the position of the hydrogen and hydroxyl groups

Question 12

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 12

Starch molecules are helical which enables compactness, in order to save space.
Insoluble - doesn’t affect teh water potential gradient
Large molecule - cannot leave the cell

Question 13

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

Answer 13

The adjacent inverted glucose monomers within the cellulose molecule provide for long and straight chains to be produced. The inversion of glucose molecules allow for hydrogen bonding to form being the hydroxyl groups of adjacent parallel chains. This allows for fibrils to form, that increases the structural stability of the cellulose molecule, allowing it to be used to strengthen the cell wall in plant cells

Question 14

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

Answer 14

Facilitated diffusion is a passive process and doesn’t involve the use of ATP, whereas active transport is an active process and involves ATP. Facilitated diffusion takes place down a concentration gradient from high to low, whereas active transport works against the concentration gradient from low to high. Facilitated diffusion involves both carrier and channel proteins, but active transport only involves carrier proteins.

Question 15

Calculate the ratio of the mean rate of uptake of chloride ions in the first hour to the
rate of uptake of chloride ions in the second hour for group B plants.
[2 marks]

Answer 15

60 min - 360/60 = 6
120 min - 470-360/60=1.83
ratio is 3.3:1

Question 16

Explain the results shown in Figure 4.

[4 marks]

Answer 16

Group A’s initial uptake of chloride ions was much slower than group B’s, as only diffusion occurred. Group B’s uptake was much higher as active transport as well as diffusion occurred. Group A’s uptake of chloride ions eventually levelled off as an equilibrium had been reached. Group B failed to level off as no equilibrium had been reached. Group B failed to level off as no equilibrium had been reached as active transport is an active transport.
- Group B’s rate slowed down as the respiratory substrate was eventually used up.

Question 17

What is meant by genetic diversity ?

Answer 17

The number of different alleles of each gene.

number of different base sequences found in each gene

Question 18

The 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 18

The bluethroat showed greater genetic diversity as the percentage of genes showing diversity was much higher. In fact, the bluethroat had 34.9 % genes of genetic diversity compared to the willow flycatcher at 27.8 %

Question 19

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 19

1. To diagnose AIDS, scientists need to check for AIDS- related symptoms
2. To diagnose AIDs, Scientists need to check the number of helper T cells

Question 20

The solution will remain yellow if a person i snot infected with HIV. explain why. (2)

Answer 20

The HIV antibody isnt present and so the second antibody and its attached enzyme wont be present, and so the person is not infected with HIV.

Question 21

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 21

The maternal antibodies are passed on the child, which the mother has conceived. As the mother is infected with HIV, the 2nd antibody which the enzyme attached binds and therefore causes a blue postive test.

Question 22

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 22

1. To make sure that the washings have been effective and that unbound antibodies have been washed away
2. Only the enzyme and nothing else has caused a colour change in the test.

Question 23

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]

Answer 23

The enzyme, used in DNA replication is DNA polymerase. The enzyme active site is extremely specific in terms of what it can bind to. In this case DNA polymerase is complementary to the 5’end of the DNA strand, and so as a result, enzyme movement can only occur from the 5’ to 3’ direction. As the shape of the 5’end and the 3’end differ, the enzyme must move from a 5’ to 3’ direction in order to remain complementary to its. substrate i.e the DNA strand

Question 24

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

Answer 24

At the source of the plant, photosynthesis is continuously ocurring, and glucose is produced. Glucose is converted into sucrose and at a particular point, the companion cells actively transports sugars and sucrose into the phloem. This in fact lowers the water to move into the phloem from the xylem by osmosis. This increases the hydrostatic pressure at the top of the phloem and causes mass movement of substances to the sink of the plant. At the sink of the plant, the substances are used for respiration and storage

Question 25

Name the process that produced the 14CO2 released from the trunk.
[1 mark]

Answer 25

Respiration

Question 26

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 26

26- 30 hours we found the time between which the 14c label moved from the peak at the top and bottom of the trunk.

Question 27

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

Answer 27

We need to know the length of the trunk from both the top and bottom

Question 28

Place stages A to E in the correct order. Start with stage D.
[1 mark]

Answer 28

D, C, B, E, A

Question 29

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 29

1. Obtain an area where cell division is occurring.

2. To allow for light to penetrate through

Question 30

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

Answer 30

The DNA increased due to DNA and chromosome replication. First decrease is due to the homogenous chromosomes separating and forming 2 daughter cells. THe second decrease is due to the sister chromatids separating.

Question 31

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

Answer 31

The DNA would double and go to 2 arbitrary units

Question 32

Messenger RNA (mRNA) is used during translation to form polypeptides.
Describe how mRNA is produced in the nucleus of a cell.
[6 marks]
*Talks about transcription

Answer 32

Helicase enzyme breaks the hydrogen bonds between the DNA strands. This unzips the DNA, exposing the nitrogenous bases one of the strands acts a template strand, to which free RNA nucleotides are attracted to the exposed bases. The free RNA nucleotides and exposed bases are attracted to each other by the complementary base pairing rule that adenine - thymine and cytosine - Guanine are attracted to each other. After the bases have aligned, RNA polymerase joins RNA nucleotides together. Pre- mRNA is spliced to remove introns.

Question 33

Describe the structure of proteins.

[5 marks]

Answer 33

The primary structure is an order of amino acids that have undergone a condensation reaction to form a peptide bond with an adjacent amino acid. The secondary structure is formed from the folding of the polypeptide chain due to hydrogen bonding. The secondary structure takes the form of either an alpha helix or B- pleated sheet. The tertiary structure introduces 3D folding due to hydrogen bonding as well as ionic and disulfide bonds. The quaternary structure has 2 or more polypeptide chains that are bonded together with a possible introduction of a prothetic group i.e iron.
- polymer of amino acids.

Question 34

Describe how proteins are digested in the human gut.

[4 marks]

Answer 34

Endopeptidases break polypeptides into smaller peptide chains by hydrolysing peptide bonds between each amino acid. Exopeptidases remove the end of chain amino acids at the same time as endopeptidases, which increases the efficiency of protein digestion. After endo / exopeptidases have done their work, dipeptidases hydrolyse dipeptides into amino acids which can be assimilated by the body.