Enzymes

9 methods · 15 questions

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In an enzyme reaction the rate rises while the varied factor limits it, then plateaus once something else becomes limiting. Read the plateau, not just the slope: reaching a flat maximum tells you the varied factor is no longer the limiting one, and for a product-against-time curve the height of the plateau is set by the total substrate supplied.

Trigger: a curve that levels off, or a question naming a limiting factor.

Instances:
(i) rate against substrate concentration flattens when every active site is occupied (saturation): the rate is at its maximum, so more substrate changes nothing and doubling the substrate does not always double the rate;
(ii) rate against enzyme concentration flattens when the enzyme is no longer limiting: the substrate now limits, so adding substrate would raise the rate;
(iii) a product-against-time curve levelling off means the substrate has been used up (or the product now inhibits the enzyme), never that the enzyme was used up;
(iv) two product-against-time curves reaching the same final height were given the same total substrate, whatever their speeds.

Linked questions (3)

NSAA 2016 Section 1, Part D (Biology questions) · Q71

The graph shows the effect of increasing the substrate concentration on an enzyme-controlled reaction when all the other variables were kept constant.

Which of the following labels, if any, could be correct for the

y

-axis?

1. rate of substrate loss /

mg min^{- 1}

2. rate of enzyme-substrate complex formation / number of complexes

s^{- 1}

3. rate of product formed per enzyme molecule /

mg min^{- 1}

A graph with the x-axis labelled concentration of substrate / arbitrary units, marked 10 at the left end and 20 at the right end, and an unlabelled y-axis with 0 at the origin. The plotted line is horizontal at a constant high value across the whole range.

Anone of them
B1 only
C2 only
D3 only
E1 and 2 only
F1 and 3 only
G2 and 3 only
H1, 2 and 3

Worked solution

Answer:H

The line is horizontal: the quantity on the

y

-axis does not change as the substrate concentration rises from 10 to 20 arbitrary units.

With the enzyme concentration and every other variable fixed, this is the saturated region of the enzyme: every active site is occupied all the time, so the reaction already runs at its maximum rate and adding more substrate makes no difference.

Each of the three quantities is a rate of this same reaction. The rate at which substrate is used up, the rate at which enzyme-substrate complexes are formed and the rate at which product is made per enzyme molecule are all constant in the saturated region, so any of the three labels could be correct.

The answer is H.

NSAA 2019 Section 1, Part D (Biology questions) · Q71

An investigation was carried out on the effect of substrate concentration on an enzyme-controlled reaction.

Four different concentrations of substrate were tested. In each case, the concentration of product was measured at regular intervals following the introduction of the substrate.

All the other variables were kept constant.

The results obtained are shown on the graph.

Which of the following statements is/are correct?

1. As substrate concentration increases, the time taken to produce 0.2 arbitrary units of product increases.
2. Doubling the substrate concentration always doubles the rate of the reaction.
3. The average rate of reaction for a substrate concentration of

1.0 mmol dm^{- 3}

is 0.005 arbitrary units per second.

A key (solid line 5.0 mmol per dm cubed, long dashes 1.0, short dashes 0.50, dotted 0.25) and a graph on a grid of product concentration in arbitrary units (y-axis, 0 to 0.8) against time in seconds (x-axis, 0 to 120). Four straight lines start at 15 s and rise steadily: the 5.0 line from 0.1 at 15 s to 0.8 at 120 s, the 1.0 line from about 0.075 to 0.6, the 0.50 line from 0.05 to 0.4 and the 0.25 line from 0.025 to 0.2, with points marked every 15 s.

Anone of them
B1 only
C2 only
D3 only
E1 and 2 only
F1 and 3 only
G2 and 3 only
H1, 2 and 3

Worked solution

Answer:D

Reading the graph, 0.2 units of product are reached after about 30 s at

5.0 mmol dm^{- 3}

, about 40 s at

1.0

, 60 s at

0.50

and 120 s at

0.25

: the time decreases as the substrate concentration increases, so statement 1 is wrong.

Doubling the concentration from

0.25

0.50 mmol dm^{- 3}

doubles the product formed in 120 s (0.2 to 0.4 units), but doubling from

0.50

1.0

raises it only from 0.4 to 0.6 units, so the rate does not always double: statement 2 is wrong.

For

1.0 mmol dm^{- 3}

the product reaches 0.6 units after 120 s, an average rate of

0.6 \div 120 = 0.005

units per second: statement 3 is correct.

The answer is D.

NSAA 2020 Section 1, Part D (Biology questions) · Q71

The graph shows the mass of product produced over time for an enzyme-controlled reaction in two different conditions, X and Y. All other variables were kept constant.

Which of the following could be correct for this graph?

1. Condition X is a higher pH than condition Y.
2. Condition X has less substrate supplied than condition Y.
3. Condition X is a lower pH than condition Y.
4. Condition X has more substrate supplied than condition Y.

A graph of mass of product produced (y-axis) against time (x-axis), with no scale values. Curve X (solid) rises steeply from the origin and levels off at a plateau; curve Y (dashed) rises more slowly from the origin and levels off later at the same plateau.

Anone of them
B1 and 2 only
C1 and 3 only
D1 and 4 only
E2 and 3 only
F2 and 4 only
G3 and 4 only
H1, 2, 3 and 4

Worked solution

Answer:C

Both curves level off at the same final mass of product, so the same amount of substrate was supplied in each condition: statements 2 and 4, which need different amounts of substrate, cannot be correct.

The reaction in X is faster than in Y, so X must be closer to the enzyme's optimum pH. That is possible whether X is at a higher pH than Y (1) or at a lower one (3), depending on which side of the optimum Y lies, so either statement could be correct.

The answer is C.