SLE212 Study Guide - Final Guide: Enzyme, Stereospecificity, V12 Engine
13 Jun 2018
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I) Enzymes
1) Active holoenzymes are formed from ________ in the presence of ________.
a. apoenzymes; cofactors.
b. apoenzymes; proteins.
c. proteins; cofactors.
d. apoenzymes; inactive holoenzyme.
Answer: a
2) Unlike typical catalyzed reactions in organic chemistry enzyme reactions are
a. reaction specific.
b. usually stereospecific.
c. modulated to change activity levels.
d. essentially 100% efficient.
e. All of the above.
Answer: e
3) Which of these statements about enzyme-catalyzed reactions is false?
a. At saturating levels of substrate, the rate of an enzyme-catalyzed reaction is
proportional to the enzyme concentration.
b. If enough substrate is added, the normal Vmax of a reaction can be attained even in the
presence of a competitive inhibitor.
c. The rate of a reaction decreases steadily with time as substrate is depleted.
d. The activation energy for the catalyzed reaction is the same as for the uncatalyzed
reaction, but the equilibrium constant is more favorable in the enzyme-catalyzed
reaction.
e. The Michaelis-Menten constant Km equals the [S] at which V = 1/2 Vmax.
Answer: d)
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4) Line-Weaver Burke plot practice question
A biochemistry student wished to estimate the kinetic parameters of acid phosphatase. To do
this they measured the hydrolysis of p-nitrophenol phosphate at different substrate
concentrations by the enzyme. The activity of the enzyme was calculated in µmol of p-
nitrophenol produced per minute and the concentration of the substrate, p-nitrophenol, was in
mmol L-1. This data were inverted and plotted on a Line-Weaver burke plot (1/Vo against
1/[S]) (depicted below). The linear equation for this plot was:
y= 0.167x + 0.667
Line-Weaver Burke plot
of acid phosphatase
1/ [p-nitrophenol phosphate]
(L mmol-1)
-10 -5 0 5 10 15 20 25 30 35 40 45
1/ acid phosphatase activity
(min mol-1)
0
1
2
3
4
5
6
7
8
y=0.167x + 0.667
6) From the equation above, calculate the Vmax of this enzyme.
Answer: Vmax = 1.5 µmol min-1
7) From the equation above, calculate the Michaelis-Menton constant (Km) of this enzyme.
Answer: Km = 0.25 mmol L-1
8) If an non-competitive inhibitor C was added to this reaction then how would the values for
Km and Vmax change?
Answer: Vmax would decrease while Km would not change.
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II) Buffers
1) You are given a buffer solution, which contains a compound with a pKa 3.8. This solution
has a total volume of 1L. The concentration of conjugate acid is 85 mmol/L and the
concentration of conjugate base is 15 mmol/L. To this solution is added 80 mL of 0.5 mol/L
NaOH. Which of the following statements is incorrect about this solution?
a. Before the addition of sodium hydroxide, the solution was better able to buffer any
potential drop in pH because the concentration of conjugate base was much higher
than that of the acid. Thus there was plenty of base to accept excess protons.
b. Before the addition of sodium hydroxide, the solution was better able to buffer any
potential rises in pH because the concentration of conjugate acid was much higher
than that of the base. Thus there was plenty of acid to donate protons.
c. Before the addition of the sodium hydroxide, the solution would not buffer any
potential drop in pH well because there was a low concentration of conjugate base in
which to react with the excess protons.
d. After the addition of sodium hydroxide, the solution was a much better buffer than
before because the concentration of the conjugate base was 1.2 times that of acid, and
hence similar.
Answer: a, This is the only incorrect answer. You are told that you have 85 mmol of
acid and 15 mmol of base. However in option (a) the following clause is contradictory to
this “the concentration of conjugate base was much higher than that of the acid”. The other
options are indeed true.
Questions 64 and 65 relate to the following text:
It is your first day in the lab at your new work. You are given Tris acid and base powders and
told to make up 1 L of a buffer with a total concentration of 0.2 M and a pH=7.8.
The pKa of Tris is 8.1. The molecular mass of Tris base is 121.1 g.mol-1. The molecular
mass of Tris acid is 157.6 g.mol-1.
2) What mass (gram) of Tris base would you need to make this buffer?
a. 31.5
b. 24.2
c. 21
d. 8.1
Answer: d
3) What mass (gram) of Tris acid would you need to make up this buffer?
a. 31.5
b. 24.2
c. 21
d. 8.1
Answer: c
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Document Summary
Enzymes, active holoenzymes are formed from ________ in the presence of ________, apoenzymes; cofactors, apoenzymes; proteins, proteins; cofactors, apoenzymes; inactive holoenzyme. Answer: a: unlike typical catalyzed reactions in organic chemistry enzyme reactions are, reaction specific, usually stereospecific, modulated to change activity levels, essentially 100% efficient, all of the above. Answer: d: line-weaver burke plot practice question. A biochemistry student wished to estimate the kinetic parameters of acid phosphatase. To do this they measured the hydrolysis of p-nitrophenol phosphate at different substrate concentrations by the enzyme. The activity of the enzyme was calculated in mol of p- nitrophenol produced per minute and the concentration of the substrate, p-nitrophenol, was in mmol l-1. This data were inverted and plotted on a line-weaver burke plot (1/vo against. The linear equation for this plot was: y t i v i t c a e s a t a h p s o h p d c a.
