Proteins & Enzymes 1.4

Question 1

Scientists have investigated the effects of competitive and non competitive inhibitors of the enzyme maltase. Describe competitive and non-competitive inhibition of an enzyme (5)

Answer 1

Competitive inhibition:

1. Inhibitors reduce binding of enzyme to substrate / prevent formation of Enzyme-Substrate complex (Competitive inhibition)
2. Inhibitor similar shape to substrate (Induced fit model)
3. Substrate binds to active site (of enzyme)
4. (Inhibition) can be overcome by more substrate (temporarily decreases rate of reactions catalysed)

Non-competitive inhibition:

5. Inhibitor binds to allosteric site (site other than active site)
6. Prevents active site from binding to substrate and forming ESC (conformational change = substrate no longer complementary)
7. Cannot be overcome by adding more substrate (permanent conformational change)

Question 2

In humans, the enzyme maltase breaks down maltose to glucose. This takes place at a normal body temperature. Explain why maltase breaking down maltose allows this reaction to take place at normal body temperature (5)

Answer 2

1. Tertiary structure / 3D shape of enzyme (means);
2. Active site complementary to maltose / substrate / maltose fits into active site / active site and substrate fit as lock and key model;
3. Description of induced fit;
4. Enzyme is a catalyst;
5. Lowers activation energy / energy required for reaction;
6. By forming enzyme-substrate complex; Accept idea that binding stresses/distorts the bonds so more easily broken.
   (Do not award point 6 simply for any reference to E-S complex)

Question 3

Describe the structure of proteins (5)

Answer 3

1. Polymer of amino acids;
2. Joined by peptide bonds in a condensation reaction;
3. Primary structure is number AND sequence/order of amino acids;
4. Secondary structure is folding of polypeptide chain into Alpha helix and Beta-pleated sheets due to hydrogen bonding (dipeptide);
5. Tertiary structure is 3D folding due to hydrogen bonding and ionic bonding and disulfide bridges;
6. Quaternary structure is two or more polypeptide chains joined together.

Question 4

Describe how a peptide bond is formed between two amino acids to form a dipeptide (1)

Answer 4

1. Condensation reaction (loss of water);

2. Between amine and carboxyl group / NH2 and COOH.

Question 5

Describe how an enzyme-substrate complex increases the rate of reaction (1)

Answer 5

1. Reduces activation energy;

2. Due to bending bonds / Without enzyme, very few substrate have sufficient energy for reaction

Question 6

Describe how a change in the base sequence of the DNA coding for an enzyme may result in a non-functional protein

Answer 6

1. Change in primary structure changes sequence of amino acids;
2. Hydrogen / Ionic / Disulfide bonding form in different positions;
3. Alters the tertiary structure of enzyme’s active site;
4. So no Enzyme-Substrate-Complexes can be formed.

Question 7

What is the proteome of a cell?

Answer 7

The proteome is the full range of/different number of proteins that the cell / DNA / Genome is able to code for (at a given time)

Question 8

When a pathogen causes an infection, plasma cells secrete antibodies which destroy this pathogen.
Explain why these antibodies are only effective against a specific pathogen (2)

Answer 8

1. Antigens (on pathogen) have a specific tertiary (3D) structure (GLYCOPROTEINS);
2. Antibody is complementary to antigen so antigen-antibody complex forms.

Question 9

Describe & Explain how you could use the biuret test to distinguish a solution of enzyme, lactase, from a solution of lactose (2)

Answer 9

Add biuret reagent to both solutions -
1. Lactase/enzyme will give off a purple colour;
2. Because lactase is a (tertiary) protein.
OR
1. Lactose/reducing sugar will not give off a purple colour/remains blue
2. Because lactose is a disaccharide.

Question 10

Sucrase does not hydrolyse lactose. Use your knowledge of the way in which enzymes work to explain why (3)

Answer 10

1. Lactose has a different shape/structure;
2. Does not bind to active site of sucrase.
   OR
3. Active site of sucrase has a specific shape/structure;
4. So does not bind to lactose so no ESC’s formed.

Question 11

Describe the induced fit model of enzyme action (2)

Answer 11

1. Active site not complementary;
2. Active site undergoes conformational change in shape;
3. to allow ESCs to form.

Question 12

Describe one way that the lock and key model is different from the induced fit model (2)

Answer 12

1. Active site does not change shape/is fixed shape/rigid;

2. Substrate (already) fits as complementary before binding.

Question 13

An enzyme only catalyses one reaction. Explain why

Answer 13

1. Active site is specific shape;

2. Only one substrate fits/binds (to active site)

Question 14

Diabetes mellitus is a disease that can lead to an increase in blood glucose concentration. Some diabetics need insulin injections. Insulin is a protein so it cannot be taken orally. Suggest why insulin cannot be taken orally (2)

Answer 14

1. Broken down by (salivary) enzymes/digested/denatured (by pH)/too large to be absorbed;
2. Insulin no longer functional (!)

Question 15

What is the effect of substrate concentration on the rate of an enzyme controlled reaction? (3)

Answer 15

1. Increases then plateaus/constant/steady/rate does not change;
2. It plateaus as all active sites occupied/saturated;
3. (Constant rate of reaction) / maximum number of ESCs formed per second.

Question 16

Explain how a competitive inhibitor works:

Answer 16

Inhibitor is a similar shape to substrate;
Inhibitor enters active site / competes with substrate;
Less substrate binds/fewer ESCs formed per second.

Question 17

Describe how a non-competitive inhibitor works:

Answer 17

Attaches to enzyme at a site other than the active site (allosteric site);
Changes (shape of) active site / changes tertiary structure (of enzyme;
(So active site and substrate) no longer complementary so less/no substrate can fit/bind (no longer complementary so less ESCs form)