Question
Analysis of amino acid and protein concentration is a key area of biological research.
The titration curve of aqueous glycine zwitterions with aqueous sodium hydroxide is shown from pH 6.0 to 13.0. Refer to section 33 of the data booklet.
Deduce the pH range in which glycine is an effective buffer in basic solution.
Enzymes are biological catalysts.
The data shows the effect of substrate concentration, [S], on the rate, v, of an enzyme-catalysed reaction.
Determine the value of the Michaelis constant (Km) from the data. A graph is not required.
Outline the action of a non-competitive inhibitor on the enzyme-catalysed reaction.
The sequence of nitrogenous bases in DNA determines hereditary characteristics.
Calculate the mole percentages of cytosine, guanine and thymine in a double helical DNA structure if it contains 17% adenine by mole.
Answer/Explanation
Markscheme
«pH range» 8.6–10.6
Accept any value between 8.2 and 11.0.
[1 mark]
«Km =» 0.67 «mmol dm–3»
Do not penalize if a graph is drawn to determine the value.
[1 mark]
does not compete for active site
OR
binds to allosteric site/away from «enzyme» active site
OR
alters shape of enzyme
reduces rate/Vmax
[2 marks]
«% cytosine + % guanine = 100% – 17% – 17% = 66%»
Cytosine: 33 «%» AND Guanine: 33 «%»
Thymine: 17 «%»
[2 marks]
Examiners report
[N/A]
[N/A]
[N/A]
[N/A]
Question
The structures of the amino acids cysteine, glutamine and lysine are given in section 33 of the data booklet.
An aqueous buffer solution contains both the zwitterion and the anionic forms of alanine. Draw the zwitterion of alanine.
Calculate the pH of a buffer solution which contains 0.700 mol dm–3 of the zwitterion and 0.500 mol dm–3 of the anionic form of alanine.
Alanine pKa = 9.87.
Answer/Explanation
Markscheme
Penalize incorrect bond linkages or missing hydrogens once only in 8 (a) and 8 (c) (i).
[1 mark]
«pH = 9.87 + log\(\left( {\frac{{0.500}}{{0.700}}} \right)\)»
«= 9.87 – 0.146»
= 9.72
pH can be deduced by an alternative method.
[1 mark]
Examiners report
[N/A]
[N/A]
Question
The graph of the rate of an enzyme-catalyzed reaction is shown below.
Determine the value of the Michaelis constant, Km, including units, from the graph.
Sketch a second graph on the same axes to show how the reaction rate varies when a competitive inhibitor is present.
Outline the significance of the value of Km.
Answer/Explanation
Markscheme
«Km = [substrate] at \(\frac{1}{2}\) Vmax»
4.2 x 10–3
mol dm–3
Accept answers in the range of 3.5 x 10–3 to 5.0 x 10–3 mol dm–3.
M2 can be scored independently.
[2 marks]
graph to right of curve AND finish at same Vmax
Do not penalize if curve does not finish exactly at same Vmax as long as it is close to it (since drawn curve does not
flatten out completely at Vmax = 0.50).
[1 mark]
Km is inverse measure of affinity of enzyme for a substrate / Km is inversely proportional to enzyme activity
OR
high value of Km indicates higher substrate concentration needed for enzyme saturation
OR
low value of Km means reaction is fast at low substrate concentration
Idea of inverse relationship must be conveyed.
Accept “high value of Km indicates low affinity of enzyme for substrate/less stable ES complex/lower enzyme activity”.
Accept “low value of Km indicates high affinity of enzyme for substrate/stable ES complex/greater enzyme activity”.
[1 mark]
Examiners report
[N/A]
[N/A]
[N/A]
Question
An enzyme catalyses the conversion of succinate to fumarate ions in a cell, as part of the process of respiration.
The rate of the reaction was monitored and the following graph was plotted.
Determine the value of the Michaelis constant, Km, by annotating the graph.
The malonate ion acts as an inhibitor for the enzyme.
Suggest, on the molecular level, how the malonate ion is able to inhibit the enzyme.
Draw a curve on the graph above showing the effect of the presence of the malonate ion inhibitor on the rate of reaction.
Answer/Explanation
Markscheme
Km labelled on x-axis as the [succinate ion] at ½Vmax
OR
horizontal line at ½Vmax AND vertical line down to x-axis
«Km ＝» 6.5 x 10–3 «mol dm–3»
Annotation of graph required for M1.
Accept any specific value in the range 6.0 x 10–3 to 7.5 x 10–3 «mol dm–3».
similar shape/size/structure «as succinate ion/substrate»
competes for the active site «with the succinate ion/substrate»
Accept “competitive inhibitor” for M2.
Award [1 max] if non-competitive inhibition is correctly described.