Date | May 2019 | Marks available | 2 | Reference code | 19M.3.hl.TZ1.12 |
Level | HL | Paper | 3 | Time zone | TZ1 |
Command term | Suggest | Question number | 12 | Adapted from | N/A |
Question
Alcohol dehydrogenase (ADH) catalyses the oxidation of methanol. The products of oxidation, methanal and methanoic acid, are toxic.
A Michaelis–Menten plot for an enzyme-catalysed reaction is shown.
Sketch a curve to show the effect of a competitive inhibitor.
Suggest, based on the Michaelis–Menten plot, how a competitive inhibitor such as ethanol reduces the toxicity of methanol.
Enzymatic activity is studied in buffered aqueous solutions.
Calculate the ratio in which 0.1 mol dm−3 NaH2PO4 (aq) and 0.1 mol dm−3 Na2HPO4 (aq) should be mixed to obtain a buffer with pH = 6.10. Use section 1 of the data booklet.
pKa (NaH2PO4) = 7.20
Markscheme
[✔]
Note: Line must start at origin, to the right of original line and bending toward the same Vmax.
Km is higher /same Vmax reached at higher [substrate] [✔]
slower reaction rate
OR
gives time to excrete/eliminate methanol [✔]
«pH = pKa + log / 6.10 = 7.20 + log »
log = «6.10 – 7.20 =» –1.10
OR
= «10–1.10 =» 0.079 [✔]
NaH2PO4 : Na2HPO4 = 12.6 : 1 [✔]
Note: Award [2] for correct final answer.
Examiners report
Most candidates correctly sketched a curve to show the effect of a competitive inhibitor on a Michaelis-Menten plot.
Many scored 1 out of 2 for stating that the inhibitor causes a slower reaction rate.
The calculation of the ratio of a conjugate acid-base pair to create a buffer with a specific pH was poorly done. Some candidates wrote a ratio without indicating which compound each value referred to and thus could not score. Many candidates used the concentrations of 0.1 mol dm−3 for both compounds and could not proceed.