| Date | May 2016 | Marks available | 2 | Reference code | 16M.3.hl.TZ0.12 |
| Level | HL | Paper | 3 | Time zone | TZ0 |
| Command term | Determine | Question number | 12 | Adapted from | N/A |
Question
Enzymes play an important role in the functioning of our bodies.
The graph below shows a Michaelis–Menten plot for an enzyme. Sketch and label two curves on the graph below to show the effect of adding a competitive and non-competitive inhibitor.
Enzyme solutions are prepared in buffers. Determine the pH of a buffer solution containing 2.60×10−3moldm−3 ethanoic acid and 3.70×10−3moldm−3 sodium ethanoate. Refer to sections 1 and 21 of the data booklet.
Markscheme
graph showing competitive inhibitor eventually reaching Vmax
graph showing non competitive inhibitor not reaching Vmax
Curves must be labelled and should not cross given curve.
Penalize one mark if one or both sketched curve(s) cross the given curve.
Award [1 max] if curves are not labelled competitive or non-competitive OR are labelled the wrong way round.
\(\log \frac{{\left( {3.70 \times {{10}^{ - 3}}} \right)}}{{\left( {2.60 \times {{10}^{ - 3}}} \right)}} = 0.153\)
«pH=4.76+0.153=»4.91
Award [2] for correct final answer.
Accept other method of calculation.
