IB Docs (2) Team MC test: Proteins & enzymes (AHL)
Multiple choice test on B.7 Proteins & enzymes (AHL)
Use the following 'quiz' to test your knowledge and understanding of this sub-topic. As this relates to a sub-topic on the options you may need access to the IB data booklet.
If you get an answer wrong, read through the explanation carefully to learn from your mistakes.
What is the relationship between the Michaelis constant, Km and the maximum rate of an enzyme-catalysed reaction Vmax?
Km is the substrate concentration when the rate of the reaction = ½ the maximum rate.
Which are true concerning the Michaelis constant, Km for an enzyme-catalysed reaction?
I. A particular enzyme will always have the same value of Km with the same substrate.
II. The lower the value of Km the more efficient the enzyme is at catalysing the reaction at low substrate concentrations.
III. Km has the units of rate of reaction.
Since Km is the substrate concentration when the rate of the reaction = ½ the maximum rate, it will have the units of concentration, not the units of rate of reaction.
Which is the effect a non-competitive inhibitor has on both the maximum rate, Vmax and the value of the Michaelis constant, Km when compared to the enzyme-catalysed reaction where there is no inhibitor present?
For non-competitive inhibitors Vmax will be lower, but the concentration of substrate when one half of Vmax is reached, i.e. the value of Km, remains the same.
Which row is correct about where a competitive inhibitor bonds to an enzyme and the effect it has on the maximum rate of the enzyme-catalysed reaction?
| Row | Location of bonding | Effect on maximum rate |
| 1 | Bonds to the enzyme at the allosteric site | Vmax is decreased |
| 2 | Bonds to the enzyme at the allosteric site | Vmax remains the same |
| 3 | Bonds to the enzyme at the active site | Vmax is decreased |
| 4 | Bonds to the enzyme at the active site | Vmax remains the same |
Because competitive inhibitors bond at the active site increasing the substrate concentration will eventually minimise the effect of the competitive inhibitor. The value of the maximum rate will be unaffected although a higher concentration of substrate will be required to reach it.
Which describes how a product at the end of a metabolic pathway can control the rate of formation of the product formed in the first step of the pathway.
The end product can cause negative feedback by acting as a non-competitive inhibitor in the first step which slows down the rate of formation of the initial product.
The pKa of propanoic acid is 4.87. What will be the pH of the resulting solution when 10.0 cm3 of 0.100 mol dm−3 sodium hydroxide solution is added to 20.0 cm3 of 0.100 mol dm−3 propanoic acid solution?
When 10.0 cm3 of 0.100 mol dm−3 sodium hydroxide solution is added to 20 cm3 of the acid solution the half-equivalence point will have been reached, so pH = pKa.
Which explains why adding a small amount of a strong acid to a buffer solution made from a weak acid, HA and its sodium salt, NaA has virtually no effect on the pH of the solution?
The concentration of A− ions in the buffer solution is relatively high as the salt is completely dissociated. When H+ ions are added they combine with A− ions to form undissociated HA.
The isoelectric point of glycine, H2NCH2COOH, is equal to 6. Which statement best explains why a solution of glycine in water has two different buffering regions, one centred at about pH 2.3 and the other at about pH 9.6?
All the statements are correct but the only one that explains why it has two buffer regions is that it contains two different acid/conjugate base pairs, H3N+CH2COOH/ H3N+CH2COO− and H3N+CH2COO−/H2NCH2COO−.
The Beer-Lambert law, 
can be used in the assay of dilute solutions of proteins using UV-VIS spectroscopy.
Which row identifies 
, ε and l correctly?
| Row | |
ε | l |
| 1 | molar absorption coefficient | absorption | path length |
| 2 | absorption | molar absorption coefficient | intensity of incident radiation |
| 3 | absorption | molar absorption coefficient | path length |
| 4 | molar absorption coefficient | absorption | intensity of incident radiation |
is the absorbance, ε is the molar absorption coefficient, which is a constant for a particular protein, and l is the path length of the light passing through the sample. Since the path length can be kept constant it means that the absorbance is directly proportional to the concentration of the protein solution.
What is the significance of λmax when assaying proteins by UV-VIS spectroscopy?
λmax is the wavelength where the maximum absorption of light occurs. Calibration curves are made by measuring the absorption at this wavelength for several different known concentrations of the protein solution and plotting the best fit line form the data.
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