Proteins - Mark Scheme Answers (study Mind)

Question 1

(a) Describe the induced-fit model of enzyme action and how an enzyme acts as a catalyst.

Answer 1

1. Substrate binds to the active site/enzyme
2. Active site changes shape (slightly) so it is complementary to
   substrate
3. Reduces activation energy;

Question 2

(c) A competitive inhibitor decreases the rate of an enzyme-controlled
reaction.
Explain how.

Answer 2

1. Boil
2. Denatures the enzyme/ATP synthase;
3. Put in ice/fridge/freezer;
4. Lower kinetic energy so no enzyme-substrate complexes form;
5. Add high concentration of inhibitor;
6. Enzyme-substrate complexes do not form;

Question 3

(d) When bread becomes stale, the structure of some of the starch is
changed. This changed starch is called retrograded starch.
Scientists have suggested retrograded starch is a competitive inhibitor of
amylase in the small intestine.
Assuming the scientists are correct, suggest how eating stale bread could
help to reduce weight gain.

Answer 3

1. (With) increasing Pi concentration, more enzyme-
   substrate complexes are formed;
2. At or above 40 (mmol dm-3) all active sites occupied

Question 4

Describe how the structure of a protein depends on the amino acids it
contains.

Answer 4

1. Structure is determined by (relative) position of amino acid/R group/interactions;
2. Primary structure is sequence/order of amino acids;
3. Secondary structure formed by hydrogen bonding (between amino acids);
4. Tertiary structure formed by interactions (between R groups);
5. Creates active site in enzymes
6. Quaternary structure contains >1 polypeptide chain - Quaternary structure formed by interactions/bonds between polypeptides;

Question 5

Figure 2 shows the SGLT1 polypeptide with NH2 at one end and COOH at
the other end.
Describe how amino acids join to form a polypeptide so there is always
NH2 at one end and COOH at the other end.

You may use a diagram in your answer.

Answer 5

1. One amine/NH2 group joins to a carboxyl/COOH group to form a peptide bond;
2. (So in chain) there is a free amine/NH2 group at one end
   and a free carboxyl/COOH group at the other

Question 6

Explain how the active site of an enzyme causes a high rate of reaction.

Answer 6

1. Lowers activation energy;
2. Induced fit causes active site (of enzyme) to change shape;
3. (So) enzyme-substrate complex causes bonds to form/break;

Question 7

Describe a biochemical test to confirm the presence of protein in a
solution.

Answer 7

1. Add biuret (reagent);
2. (Positive result) purple/lilac/violet /mauve;

Question 8

A dipeptide consists of two amino acids joined by a peptide bond.
Dipeptides may differ in the type of amino acids they contain.
Describe two other ways in which all dipeptides are similar and one way in
which they might differ.

Answer 8

Similarities - 1. Amine/NH2 (group at end), 2. Carboxyl/COOH (group at end), 3. Two R groups, 4. All contain C and H and N and O

Difference - 5. Variable/different R group(s);

Question 9

A solution contained a mixture of three different amino acids. A scientist passed
an electric current through the solution to separate the amino acids.
placed drop of the mixture at one end of a piece of filter paper, attached an electrode to each end of the paper and switched on the current. She switched
off the current after 20 minutes and stained the paper to show spots of the amino acids at new positions.

Answer 9

1. Moved to negative (electrode) because positive(ly charged);
2. (Spots move) different distances/rates because (amino acids)
   different charge/mass;
3. Two spots (not three) because (amino acids) same
   charge/mass

Question 10

Describe how a non-competitive inhibitor can reduce the rate of an
enzyme-controlled reaction.

Answer 10

1. Attaches to the enzyme at a site other than the active site;
2. Changes (shape of) the active site
3. (So active site and substrate) no longer complementary so less/no substrate can fit/bind;

Question 11

Describe how a peptide bond is formed between two amino acids to form a dipeptide.

Answer 11

1. Condensation (reaction) / loss of water;
2. Between amine / NH2 and carboxyl / COOH;

Question 12

The secondary structure of a polypeptide is produced by bonds between amino acids.
Describe how

Answer 12

1. Hydrogen bonds;
2. Between NH (group of one amino acid) and C=O (group);

Question 13

Two proteins have the same number and type of amino acids but different
tertiary structures.
Explain why.

Answer 13

1. Different sequence of amino acids
2. Forms ionic / hydrogen / disulfide bonds in different places;

Question 14

Formation of an enzyme-substrate complex increases the rate of reaction.
Explain how.

Answer 14

1. Reduces activation energy;
2. Due to bending bonds

Question 15

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.

Answer 15

1. Named protein / enzyme (in blood) sensitive to / affected by change in pH;
2. (Resultant) change of charge / shape / tertiary structure;
3. Described effect on named protein or enzyme.