BCH 361 Lecture Notes - Lecture 1: Catalytic Triad, Activation Energy, Hydrolysis

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29 Nov 2020
1. The types of catalysis I observed include two catalysis mechanisms. One mechanism
was covalent catalysis, in which there would be nucleophilic attacks, for instance, by
Ser200 on carbonyl carbon of the acetylcholine to initiate the hydrolysis of the
neurotransmitter. The other one was acid/base catalysis, in which the use of amino acid
residues (Glu, Ser, His aka catalytic triad), specifically His440 and Glu327 would act as
acid/base when adding or removing the acetyl group during the hydrolysis of
a. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2919153/
b. .https://www.researchgate.net/publication/233701777_Acetylcholinesterase_Mec
2. The shape and position of the curve tells me that acetylcholinerase performs optimally at
a specific pH range. Furthermore, the downward curvature shown suggests that this
enzyme is fairly pH dependent. For instance, the enzyme rate is 100% at pH 7.5-7.8,
whereas at pH ranges 6.9-7.2 & 8.1-8.3 acetylcholinerase functions at 80%, which is a
20% decrease in enzymatic function when the pH varied by ยฑ0.3 units.
3. Enzymes enhance the rate of chemical reactions by lowering the activation energy in an
effective way that does not change the equilibrium position. The physical ways enzymes
cause reaction rates to get faster include substrate positioning, exclusion of water,
substrate recognition, orientation of catalytic groups, both substrate and enzyme change
4. The structural basis for enzyme specificity can be described as the tendency of certain
enzymes to recognize and choose the proper substrates among other molecules to
catalyze specific chemical reactions. The enzyme specificity is controlled by structure,
the unique fit of substrate molecules with enzyme controls the selectivity for substrate
and the product yield.
5. If an enzyme were to have a greater binding energy for the substrate than for the
transition state, then there would be no catalytic activity. Since if the enzyme-substrate
complex is more stable than the enzyme transition state complex, then the transition
state would not form and thus, catalysis would not take place.
6. The energy required to reach the transition state aka activation energy is returned when
the transition state proceeds to product. ๐›ฅ๐บ
7. Consider the following:
a. Keq = [P]/[S] = 0.19/0.01=19 M
i. ๐›ฅ๐บ๐‘œ = โˆ’๐‘…๐‘‡๐‘™๐‘›(๐พ๐‘’๐‘ž) = (โˆ’8.3145 ร— 298)๐‘™๐‘›(19) = โˆ’7295.5 ๐ฝ =
โˆ’7.30 ๐‘˜๐ฝ
b. ๐›ฅ๐บ= -RT ln Keq = (-8.3145 ร—298 K) (ln[G1P]/[G6P]) = -2477.7 x ln [G1P]/[G6P] =
2.48 x ln [G6P]/[G1P] Keq-1 = 19 or Keq= 0.0475
i. Since [G6P]/[G1P]= 19, there is 1 molecule of G1P for every 19
molecules of G6P. Because we started with 0.1 M, thus, the [G1P] is
1/19(0.1 M)= 0.005 M. Consequently, the reaction does not proceed as
written to a significant extent.
8. Keq = 0.0475 at 25ยฐC (298 K) & pH 7
a. ฮ”Gยฐโ€ฒ=-RTln(Keq) =-8.315 ร— 10^-3(298)(ln(0.0475)) =7.55 kJ/mol
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