Exercise Answers 3 Asal Biology WB
Exercise Answers 3 Asal Biology WB
Exercise Answers 3 Asal Biology WB
Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Workbook answers
Chapter 3
Exercise 3.1 Answering questions Exercise 3.2 Effects of temperature
about graphs and pH on enzyme activity
1 a As time increases, the total volume of 1 a At temperatures from 20 to 44 °C, the rate
oxygen collected also increases. The rate of activity remains constant at about 15
of increase is fastest at the start of the arbitrary units. As temperature increases
reaction, gradually, decreasing, until from from 44 °C to just over 60 °C, the rate of
about 330 s onwards there is no further activity increases steeply, reaching a peak
increase and the graph levels off at a value of 140 a.u. at about 61 °C. At temperatures
of 11.6 cm3. higher than this optimum value, activity
decreases very steeply, reaching 0 at 75 °C.
b The concentration of the substrate,
hydrogen peroxide, is greatest at the start. b At low temperatures, the kinetic energy
This is when collisions between enzyme of enzyme and substrate molecules is
and substrate will be most frequent, and relatively low, and collisions are infrequent,
therefore when the rate of reaction is resulting in a relatively low rate of reaction.
greatest. This explains the high rate of However, we would normally expect the
production of oxygen during this stage. rate of activity to increase as temperature
As time increases, the concentration increases, but this does not happen until
of substrate decreases, because it is the temperature reaches 44 °C. Perhaps
being changed into product. The rate of the activation energy for the reaction is not
reaction therefore gradually slows, until it reached at temperatures below this so, even
becomes zero at 330 s onwards, because when enzyme and substrate collide, the
all of the substrate has been converted to reaction only rarely takes place.
product.
As temperature increases above 44 °C,
2 he curve for catalase from apple is similar
T increased frequency of collision between
to that for carrot, in that it shows a rate of enzyme and substrate, which happens
reaction that decreases with time. However, with more energy, result in an increased
the rate (that is, the slope of the curve) is rate of reaction.
always less than that for carrot. The maximum
volume of oxygen given off in 360 s is only 6.8 At temperatures above 61 °C, the high
cm3, which is 4.8 cm3 less than the maximum kinetic energy of the molecules causes
volume for carrot. Unlike the curve for carrot, vibrations in the enzyme molecules,
the apple curve has still not levelled off at breaking hydrogen bonds and causing the
this time, showing that the reaction is not enzymes to lose their three-dimensional
yet complete and there is still more substrate shape. The active site no longer fits the
present. substrate molecule and, at a temperature
of 75 °C, denaturation of the enzyme
3 he value that you obtain will depend on
T molecules is complete and no reaction
exactly how you draw the tangent, which is occurs.
tricky to do precisely. Your answer should be
somewhere close to 0.13 cm3 s–1. 2 a We would normally expect activity to
be greatest at an optimum pH and to
decrease at pHs lower or higher than that.
1 Cambridge International AS & A Level Biology – Jones & Parkin © Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL BIOLOGY: WORKBOOK
2 Cambridge International AS & A Level Biology – Jones & Parkin © Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL BIOLOGY: WORKBOOK
to each of the test tubes containing the enzyme, Exercise 3.5 Calculating actual
in the water bath.
and percentage error
easure a known volume of urea solution – say
M
5 cm3 – into several test tubes. Place these test 1 ±5 cm3
tubes in the water bath. 2 he uncertainty or error for each
T
eave all the tubes for at least ten minutes,
L measurement is ± 1 cm3, so the total error
to allow their contents to come to the when calculating the difference between two
temperature of the water bath. measurements is ± 2 cm3.
One possible design could be: he percentage error is therefore (0.01 ÷ 0.07)
T
× 100 = 14.3%.
Concentration pH after 5 minutes
of inhibitor / % Trial 1 Trial 2 Trial 3 Mean otice that we ignore the minus sign when
N
calculating actual error or percentage error –
it makes no difference whether the mass has
gone up or down.
3 Cambridge International AS & A Level Biology – Jones & Parkin © Cambridge University Press 2020