A-LevelBiologyEnzymes9700/21 · May/June 2017 · Q28 Marks

Phosphatases are enzymes that catalyse the removal of phosphate groups from organic compounds. Some students investigated the effect of substrate concentration on the rate of the reaction catalysed by an acid phosphatase (enzyme A). The results are shown in Fig. 2.1. [Figure 2.1]

(a)The students used Fig. 2.1 to derive the Michaelis-Menten constant (Km) for enzyme A as 0.3 mmol dm⁻³. Explain how they derived Km.[2]
(b)The students investigated a different phosphatase enzyme (enzyme B) and found the value of Km to be higher than 0.3 mmol dm⁻³. Explain the difference between the values of Km for these two phosphatase enzymes.[2]
(c)The students repeated their investigation on enzyme A with a competitive inhibitor. They used the same concentrations of substrate as before, but added a competitive inhibitor to each reaction mixture. They used the same concentration of the inhibitor in each reaction mixture. The students found that Vmax was the same as before, but Km was higher than 0.3 mmol dm⁻³. Explain how the addition of the competitive inhibitor results in the same value for Vmax but a higher value for Km.[4]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.

half Vmax/AW, = 7 (μmol dm⁻³ min⁻¹) / take half of Vmax of 14 (μmol dm⁻³ min⁻¹) ;
A description of using the graph to find ½ Vmax without reference to figures
read (substrate concentration) from x-axis / AW ;
alternative plot 1 / [S] = x
allow phosphate group(s)/organic compound for substrate
if affinity not used, accept idea of ability to form ESC
check for ora I ref. to competitive inhibition
1 enzyme B has a lower affinity for its substrate (than enzyme A) or the higher the Km the lower the affinity of the enzyme for its substrate ;
R if substrate has affinity for the enzyme
2 enzyme B needs a higher concentration of substrate to reach, Vmax / ½Vmax/ Km (than enzyme A) ;
3 AVP; e.g. enzyme B forms fewer ESC in the same unit of time enzyme B active site is a less good fit for substrate idea that in normal cell enzyme A is saturated (with substrate) so works at a constant rate variations in substrate concentration will have less effect on the rate of formation of product by enzyme A I ref. to turnover number(s)
marks can be taken from a sketch graph
1 competitive inhibitor, occupies / competes with substrate for / AW, active site (of the enzyme);
2 reduces frequency of collisions (with substrate) / fewer ESCs form ; R no ESCs form
3 reduces reaction rate at low substrate concentrations ;
4 idea that curve with inhibitor is to the right of the curve without inhibitor;
5 at high substrate concentration / with increasing substrate concentration, the inhibitor has, no / less, effect; A idea that substrate outcompetes inhibitor at high substrate concentration
6 therefore Vmax is the same as it is determined by the enzyme concentration / AW ; A explanation in terms of active sites, saturated / fully occupied
7 idea of intercept to curve gives a higher value for Km;