Answers to questions on Proteins & enzymes (HL)

1. (a) Enzymes provide an alternative pathway for the biological reaction with lower activation energy. They possess an active site where the substrate is able to bind to the enzyme. This site either fits or can be induced to fit the substrate thus bringing the reacting species into close contact with each other.

(b) A competitive inhibitor has a similar shape to the substrate so competes with it for the active site. This lowers the rate of the reaction. A non-competitive inhibitor bonds elsewhere on the enzyme which causes the enzyme to change shape so the substrate can no longer bind which also lowers the rate of the reaction.

(c) K_m = ½ V_max = 7.0 μ mol dm^-3

(d) (i) For a competitive inhibitor K_m is higher as there is competition for the active sites but V_max is unaltered as the concentration of the substrate becomes much larger than the competitor concentration.

(ii) For a non-competitive inhibitor K_m is the same as the uninhibited enzyme molecules are not affected in the way that they react with the substrate but V_max is lower as the enzymes shape is still altered no matter how large the substrate concentration.

2. Amino acids exist in solution as a mixture of their zwitterion and either their cationic or anionic form. In an amino acid buffer the conjugate base of the buffer will tend to neutralise the effect of adding a strong acid whereas the conjugate acid will tend to neutralise the effect of adding a strong base.

H₂NCH(R)COO⁻ + H⁺ ⇌ H₃N⁺CH(R)COO⁻

H₃N⁺CH(R)COOH + OH⁻ ⇌ H₃N⁺CH(R)COO⁻ + H₂O

The zwitter ion can react with either H⁺ or OH⁻ to help maintain the pH

3. (a) The cation (conjugate acid) can lose a proton to form the conjugate base in the form of the zwitter ion. This gives pK_a1
H₃N⁺CH(R)COOH ⇌ H₃N⁺CH(R)COO⁻ + H⁺

or the zwitter ion can act as the conjugate acid to lose a proton to give the anion, which is the conjugate base. This gives pK_a2
H₃N⁺CH(R)COO⁻ ⇌ H₂NCH(R)COO⁻ + H⁺

(b) At pH 9.7 the conjugate acid is H₃N⁺CH(CH₃)COO⁻ and the conjugate base is H₂NCH(CH₃)COO⁻.

K_a = [H⁺] x [[conjugate base] ÷ [conjugate acid])

pH = pK_a + log₁₀([conjugate base] ÷ [conjugate acid])

Since pH = pK_a, log₁₀ ([conjugate base] ÷ [conjugate acid]) = 0 so

Ratio of [H₃N⁺CH(CH₃)COO⁻] to [H₂NCH(CH₃)COO⁻] = 1

(c) From the Henderson-Hasselbalch equation the pH of the solution before the NaOH is added
= 2.3 + log₁₀ (0.500 ÷ 0.200) = 2.30 + 0.40 = 2.70
Amount of NaOH added to 1 dm³ = 0.500 g / 40.00 g mol^-1 = 0.0125 mol
In 1 dm³ this will react with 0.0125 mol of the cationic form to produce 0.0125 mol of the zwitter ion.
The new concentration of H₃N⁺CH(CH₃)COOH will be 0.200 – 0.0125 = 0.1875 mol dm^-3
The new concentration of H₃N⁺CH(CH₃)COO⁻ will be 0.500 + 0.0125 = 0.5125 mol dm^-3
The pH after the NaOH has been added = 2.3 + log₁₀ (0.5125 ÷ 0.1875) = 2.3 + 0.44 = 2.74
The change in pH is 0.04.

4. (a) λ_max is the wavelength where the greatest absorbance of the solution occurs in the UV-Vis spectrum.

(b) By interpolation the concentration of the assayed protein = 75 mg dm^-3

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