1.4- Proteins and enzymes

Question 1

Maltose is hydrolysed by the enzyme maltase.

Explain why maltase catalyses only this reaction. (3)

Answer 1

1. Active site (of enzyme) has (specific) shape / tertiary structure / active
   site complementary to substrate / maltose;
   Reject active site on substrate.
   Must have idea of shape
   Assume “it” = maltase
   Accept (specific) 3D active site
   Reject has same shape
2. (Only) maltose can bind / fit;
   Accept “substrate” for “maltose”
3. To form enzyme substrate complex.
   Accept E−S complex

Question 2

Cow’s milk contains the sugar lactose. Many cats are unable to digest cow’s milk because
they are lactose intolerant.
Cow’s milk can be made suitable for these cats by treating it with the enzyme lactase to
hydrolyse lactose. This makes the cow’s milk lactose-free. Beads are coated with lactase
and placed in a tube, as shown in the diagram below. Cow’s milk flows over the beads
and the lactose is hydrolysed.

Attaching lactase to the beads is a more efficient use of lactase than adding the
lactase directly to cow’s milk.
Suggest three reasons why it is more efficient to attach lactase to the beads.
(3)

Answer 2

1. (Lactase / beads) can be reused / not washed away;
2. Accept lactase / beads not wasted
3. Less lactase used is insufficient
4. No need to remove from milk;
5. Accept lactase not present in milk.
6. Allows continuous process;
7. The enzyme is more stable;
8. Avoid end-product inhibition.

Question 3

Monosaccharides and disaccharides taste sweet.
The lactose-free milk made after hydrolysis with lactase tastes sweeter than the
cow’s milk containing lactose.
Suggest why. (2)

Answer 3

1. (Lactose hydrolysed to) galactose and glucose;
2. (So) more sugar molecules;
3. Idea of more sugars essential
4. (So) more / different receptors stimulated / sugars produced are sweeter
   (than lactose).

Question 4

A stomach ulcer is caused by damage to the cells of the stomach lining. People with
stomach ulcers often have the bacterium Helicobacter pylori in their stomachs.
A group of scientists was interested in trying to determine how infection by H. pylori
results in the formation of stomach ulcers.
The scientists grew different strains of H. pylori in liquid culture.
The table below shows the substances released by each of these strains.

Describe and explain how centrifuging the culture allowed the scientists to obtain a
cell-free liquid. (3)

Answer 4

1. Large / dense / heavy cells;
2. Form pellet / move to bottom of tube (when centrifuged);
3. Liquid / supernatant can be removed.

Question 5

The scientists measured cell damage by measuring the activity of lysosomes.
Give one function of lysosomes. (1)

Answer 5

Break down cells / cell parts / toxins.

Question 6

H. pylori cells produce an enzyme that neutralises acid.

Suggest one advantage to the H. pylori of producing this enzyme. (2)

Answer 6

1. To stop / reduce them being damaged / destroyed / killed;
   Reject (to stop) bacteria being denatured.
2. By stomach acid.

Question 7

What do these data suggest about the damage caused to human cells by the toxin
and by the enzyme that neutralises acid?
Explain your answer. (3)

Answer 7

1. More cell damage when both present / A;
2. Some cell damage when either there on their own / some cell damage in
   B and C;
   MP1 and MP2 − figures given from the graph are insufficient.
3. Standard deviation does not overlap for A with B and C so
   difference is real;
   MP3 and MP4 both aspects needed to gain mark.
4. Standard deviations do overlap between B and C so no real
   difference.

Question 8

The scientists carried out a further investigation. They treated the liquid from strain
A with a protein-digesting enzyme before adding it to a culture of human cells. No
cell damage was recorded.
Suggest why there was no damage to the cells. (3)

Answer 8

1. Enzyme (a protein) is broken down (so no enzyme activity);
   Accept hydrolyse / digested for ‘broken down’.
2. No toxin (as a result of protein-digesting enzyme activity);
   Must be in the correct context.
3. (So) toxin is protein.
   This must be stated, not inferred from use of
   ‘protein−digesting enzyme’.

Question 9

4.A principle of homeostasis is the maintenance of a constant internal environment. An
increase in the concentration of carbon dioxide would change the internal environment
and blood pH.
Explain the importance of maintaining a constant blood pH (3)

Answer 9

1. Named protein / enzyme (in blood) sensitive to / affected by change in
   pH;
   Accept converse for MP2 and MP3.
   Named example should be a protein that might be affected
   (by change in pH) eg haemoglobin, carrier protein in plasma
   membrane.
   Accept ‘change in H+
   concentration’ for ‘change in pH’.
2. (Resultant) change of charge / shape / tertiary structure;
   The change in charge idea relates to the enzyme / protein
   and not the blood (plasma) or red blood cells.
   ‘Denaturation’ alone is insufficient.
3. Described effect on named protein or enzyme.
   e.g. less oxygen binds with haemoglobin / less transport across membranes /
   fewer substrates can fit active site / fewer enzyme-substrate complexes.
   Idea of ‘less’ or ‘fewer’ required. Ignore suggestion of ‘no’ or
   ‘none

Question 10

Cyanide is poisonous. Cyanide binds to cytochrome oxidase, which is an enzyme in the
electron transport chain in mitochondria. This stops the movement of electrons to oxygen.
As a result, ATP cannot be made via aerobic respiration. If a person or animal is exposed
to cyanide, a substance that acts as an antidote can reduce or prevent poisoning. This
substance binds to cyanide.
Scientists investigated the effect of cyanide on the rate of respiration of cells in different
animal organs and in organs from different animals. They extracted organs from animals
that had just been killed. For each animal organ they set up 3 dishes. Each dish
contained:
• phosphate solution
• saline (sodium chloride) solution
• cyanide solution of known concentration.
They measured the mean amount of oxygen used by the slices of organs in one hour.
Their results are shown in Table 1.
T

Suggest how binding of cyanide to cytochrome oxidase affects the enzyme. (3)

Answer 10

1. Inhibition;
   Accept either competitive or non-competitive inhibition or a
   description of either.
2. Changes tertiary structure (of enzyme);
3. Changes shape of / blocks active site (of enzyme);
   The active site must be in the context of the enzyme /
   cytochrome oxidase.
4. Enzyme cannot bind to its substrate / no enzyme-substrate
   complex formed.
   Accept ‘ES’. Accept ‘substrate cannot attach to enzyme’.

Question 11

Suggest how the antidote can reduce poisoning by cyanide. (1)

Answer 11

(Antidote reacts with / binds to cyanide) so cyanide cannot bind to enzyme /
cytochrome oxidase
OR
(Antidote reacts with / binds to cyanide) so causing cyanide to be released
from the enzyme / cytochrome oxidase.
Key idea is how the antidote affects the cyanide.

Question 12

Table 1 shows the scientists’ results for different trials. The trials could be put into
groups to allow comparisons to be made within each group.
(i) As an example of how trials could be grouped, Group 1 has been completed
in Table 2 below. Complete Table 2 to show three other possible ways that
the scientists’ trials could be grouped. (2)

Answer 12

1. A + C + E / all liver (trials)
2. B + D + F / all kidney (trials)
3. D + E / all rat (trials);;
   Accept a description of any trial letter.
   All 3 groups correct = 2 marks.
   Any 2 groups correct = 1 mark.
   1 group / no groups correct = 0 mark.

Question 13

What is the effect of cyanide on Group 1 trials in Table 2? Use evidence from
Table 1 to support your answer. (3)

Answer 13

1. Cyanide reduces oxygen use / rate of respiration in A and B
   / in both
   OR
   as concentration of cyanide increases, the use of oxygen
   decreases in both;
   Accept use of letters or description of the animal and organ
   Reference to ‘both’, in some way, is required.
2. Greater effect of cyanide (on oxygen use) on sheep kidney / B
   than on sheep liver / A;
   Comparison required in the statement. The statement should
   not be inferred from MP3.
3. Appropriate calculations of mean oxygen use from the data
   E.g. 1 liver falls by 74% whereas kidney falls by 87%
   OR
   liver falls to 0.26 / to 26% whereas kidney falls to 0.13 / to 13%
   E.g. 2 liver falls by 2.0(au) whereas kidney falls by 12.2(au);

Question 14

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 volume of substrate
solution and the same volume of enzyme solution.
The figure below shows his results.

Give one other factor the technician would have controlled. (1)

Answer 14

Concentration of substrate solution / of enzyme solution / pH.

Question 15

Describe and explain the differences between the two curves.(5)

Answer 15

1. Initial rate of reaction faster at 37 °C;
2. Because more kinetic energy;
3. So more E–S collisions / more E–S complexes formed;
4. Graph reaches plateau at 37 °C;
5. Because all substrate used up.
   Allow converse for correct descriptions and explanations for
   curve at 25 °C

Question 16

7.Figure 1 shows one base pair of a DNA molecule.

Name part F of each nucleotide. (1)

Answer 16

Deoxyribose.

Question 17

During replication, the two strands of a DNA molecule separate and each acts as a
template for the production of a new strand.
Figure 2 represents DNA replication.

Name the enzyme shown in Figure 2. (1)

Answer 17

DNA polymerase

Question 18

The arrows in Figure 2 show the directions in which each new DNA strand is being
produced.
(d) Use Figure 1, Figure 2 and your knowledge of enzyme action to explain why the
arrows point in opposite directions.

Answer 18

1. (Figure 1 shows) DNA has antiparallel strands / described;
2. (Figure 1 shows) shape of the nucleotides is different / nucleotides
   aligned differently;
3. Enzymes have active sites with specific shape;
4. Only substrates with complementary shape / only the 3’ end can bind
   with active site of enzyme / active site of DNA polymerase.

Question 19

Mucus also contains glycoproteins. One of these glycoproteins is a polypeptide with
the sugar, lactose, attached.

Describe how lactose is formed and where in the cell it would be attached to a
polypeptide to form a glycoprotein. (4)

Answer 19

1. Glucose and galactose;
2. Joined by condensation (reaction);
3. Joined by glycosidic bond;
4. Added to polypeptide in Golgi (apparatus);

Question 20

A change from Glu to Lys at amino acid 300 had no effect on the rate of reaction
catalysed by the enzyme. The same change at amino acid 279 significantly reduced
the rate of reaction catalysed by the enzyme.

Use all the information and your knowledge of protein structure to suggest reasons for
the differences between the effects of these two changes.
(3)

Answer 20

1. (Both) negatively charged to positively charged
   change in amino acid;
2. Change at amino acid 300 does not change the
   shape of the active site
   OR
   Change at amino acid 300 does not change the
   tertiary structure
   OR
   Change at amino acid 300 results in a similar
   tertiary structure;
3. Amino acid 279 may have been involved in a
   (ionic, disulfide or hydrogen) bond and so the
   shape of the active site changes
   OR
   Amino acid 279 may have been involved in a
   (ionic, disulfide or hydrogen) bond and so the
   tertiary structure changed;
   OR
   Amino acid 279 may be in the active site and
   be required for binding the substrate;

Question 21

Draw the general structure of an amino acid.

[1 mark]

Answer 21

H           
             I           
H2N----C--C-COOH
             I          
            R

Question 22

Washing powders often contain enzymes from bacteria. These enzymes include
proteases that hydrolyse proteins in clothing stains.
Figure 1 shows the effect of temperature on a protease that could be used in
washing powder.

Explain the shape of the curves at 50 °C and 60 °C.
(4)

Answer 22

1. Both denatured (by high temperature);
2. Denaturation faster at 60 °C due to more
   (kinetic) energy;
3. Breaks hydrogen/ionic bonds (between amino
   acids/R groups);
4. Change in shape of the active site/active site no
   longer complementary so fewer enzymesubstrate complexes formed/substrate does not
   fit;

Question 23

(a) Induced fit and lock and key are two models used to explain the action of enzymes.
(i) Describe the induced fit model of enzyme action. (2)

Answer 23

(i) Active site / enzyme not complementary;

Active site changes (shape) / is flexible;

(Change in enzyme allows) substrate to fit / E-S complex to form;
Active site becomes complementary / wraps around substrate =
2 marks
For mark point 2. allow ‘binding site’ but not ‘enzyme’
For mark point 2. can only have enzyme changes (shape) if
active site has been mentioned earlier
Final mark point must have context
Reject: active site on substrate for second marking point only
Accept: diagrams only if suitably labelled or annotated

Question 24

Describe one way that the lock and key model is different from the induced fit
model. (1)

Answer 24

Active site does not change (shape) / is fixed (shape) / is rigid / does not

wrap around substrate / (already) fits the substrate / is complementary
(before binding);
Assume that ‘it’ refers to lock and key