M. Henderson Pozzi, P.F. Fitzpatrick / Archives of Biochemistry and Biophysics 498 (2010) 83–88
87
R
N
to neutralize the negative charge developing on oxygen as it ac-
cepts electrons from the flavin [22,40,45–47]. The role of Lys315
in the PAO reaction is clearly different, since it must be unproto-
nated for optimal oxygen reactivity. The effect of the K315M muta-
N
O
O
NH
N
H
m
tion on the kcat/K value of oxygen is much less than has been
observed upon mutating the positively charged residue in other
flavin oxidases, suggesting that another residue in PAO must play
that role.
H O
H
While the present results provide insight into the role of
Lys315 in the reaction of a mammalian PAO and the effects of
mutating this residue, they do not provide an obvious rationale
for the results with maize PAO and LSD1. The characterization
of the K661A mutant of LSD1 was quite qualitative [33], so the
explanation may be that a decrease of ꢂ25-fold in the reaction
would have resulted in no detectable activity in the assay. How-
ever, the characterization of the K300M mutation in the maize
enzyme was done by single-turnover kinetic analyses that mea-
sured the rate constant for flavin reduction directly [32]. It may
be that this lysine has an additional role in the reductive half-
reaction of that enzyme. The sequence identity between maize
and mouse PAO is only 21%, and they oxidize different sides of
the secondary nitrogen in the substrate, so that some differences
in the roles of active site residues would not be remarkable. An
alternative explanation is that the mutation has a structural effect
in the maize enzyme, such that the resting enzyme is in an inac-
tive conformation and the rate constant for reduction reflects the
rate constant for a conformational change to an active enzyme.
Such a model was recently proposed to explain the effects of
mutating a conserved active site histidine in the lactate dehydro-
LYS NH2
Scheme 2.
sistent with a model in which the role of the lysine is to properly ori-
ent the water molecule hydrogen-bonded to the flavin N(5), so that a
free lone pair of electrons is available to facilitate the transfer of the
proton from the flavin N(5) to oxygen. The linear proton inventory
would then result from the movement of the N(5) proton. An alter-
native explanation for the solvent isotope effect and the role of
Lys315 is that the proton is transferred from the flavin to the bridg-
ingwateras asecondprotonistransferredfromthewaterto theneu-
tral lysine. This would require another amino acid residue to act as
the source of the proton for oxygen. Either model yields the hydro-
gen bond interactions shown in Scheme 2. The latter model could re-
sult in a curved proton inventory if both protons were in flight in the
transition state for flavin oxidation. The linear proton inventory seen
withthewild-typeenzymewouldbeduetoasynchronicityintheex-
tent of transfer of the two protons. While the data are fit better by a
model for a single proton, the precision of the data cannot rule out
the involvement of two protons with different isotope effects if the
isotope effect arising from one is 1.1 or less.
genase flavocytochrome b
2
[48].
In contrast to the result with the wild-type enzyme, the solvent
inventory for the mutant protein is clearly bowed, indicating that
multiple protons are in flight in the transition state for flavin oxi-
dation in this case. Our data do not allow us to determine the num-
ber of protons involved with any accuracy beyond stating that the
best fits occur with three or more protons. This is a general prob-
lem with the proton inventory method [38]. The elimination of
the amino moiety of Lys315 by the mutation is likely to alter the
interaction of the conserved water molecule with the flavin N(5)
and result in an additional water molecule in the active to fill the
cavity left by the amino group. If the additional water serves as
the proton donor to oxygen as the N(5) proton is lost, a curved pro-
ton inventory would result. Moreover, transfer of an additional
proton between the two water molecules would be required to
avoid formation of adjacent hydroxide and hydronium ions. This
model thus predicts that there would be at least three protons in
flight in the key transition state for the mutant protein, consistent
with the proton inventory. The alternative model in which the ly-
sine accepts a proton from the bridging water as a second proton is
transferred from the flavin is also consistent with the curved pro-
ton inventory in the mutant protein. The role of proton acceptor
from the bridging water would be taken by the additional water
molecule in the active site. The need to avoid formation of hydrox-
ide suggests that a chain of water molecules would be needed, pro-
References
[
[
1] P.F. Fitzpatrick, Arch. Biochem. Biophys. 493 (2010) 13–25.
2] A. Mattevi, M.A. Vanoni, F. Todone, M. Rizzi, A. Teplyakov, A. Coda, M.
Bolognesi, B. Curti, Proc. Natl. Acad. Sci. USA 93 (1996) 7496–7501.
[3] P. Trickey, M.A. Wagner, M.S. Jorns, F.S. Mathews, Structure 7 (1999) 331–345.
[
4] E.C. Settembre, P.C. Dorrestein, J. Park, A.M. Augustine, T.P. Begley, S.E. Ealick,
Biochemistry 42 (2003) 2971–2981.
[
5] D. Leys, J. Basran, N.S. Scrutton, EMBO J. 22 (2003) 4038–4048.
[6] J. Ma, M. Yoshimura, E. Yamashita, A. Nakagawa, A. Ito, T. Tsukihara, J. Mol.
Biol. 338 (2004) 103–114.
[
7] C. Binda, P. Newton-Vinson, F. Hubalek, D.E. Edmondson, A. Mattevi, Nat.
Struct. Biol. 9 (2002) 22–26.
[8] C. Binda, A. Coda, R. Angelini, R. Federico, P. Ascenzi, A. Mattevi, Structure 7
(1999) 265–276.
[
9] P. Stavropoulos, G. Blobel, A. Hoelz, Nat. Struct. Mol. Biol. 13 (2006) 626–632.
[
10] P.D. Pawelek, J. Cheah, R. Coulombe, P. Macheroux, S. Ghisla, A. Vrielink, EMBO
J. 19 (2000) 4204–4215.
[11] G. Palmer, V. Massey, in: T.P. Singer (Ed.), Biological Oxidation, John Wiley and
Sons, New York, 1968, pp. 263–300.
[
[
12] N.S. Scrutton, Nat. Prod. Rep. 21 (2004) 722–730.
13] D.E. Edmondson, C. Binda, A. Mattevi, Arch. Biochem. Biophys. 464 (2007)
269–276.
[
[
14] P.F. Fitzpatrick, Acc. Chem. Res. 34 (2001) 299–307.
15] A. Faust, K. Niefind, W. Hummel, D. Schomburg, J. Mol. Biol. 367 (2007) 234–
248.
[16] E.C. Ralph, J.S. Hirschi, M.A. Anderson, W.W. Cleland, D.A. Singleton, P.F.
Fitzpatrick, Biochemistry 46 (2007) 7655–7664.
[
17] E.C. Ralph, M.A. Anderson, W.W. Cleland, P.F. Fitzpatrick, Biochemistry 45
2006) 15844–15852.
[18] K.A. Kurtz, M.A. Rishavy, W.W. Cleland, P.F. Fitzpatrick, J. Am. Chem. Soc. 122
2000) 12896–12897.
(
4
(
viding an explanation for the curvature of the proton inventory.
For several flavoprotein oxidases, rapid flavin oxidation requires
the presence of a positively charged active site residue, presumably
[
19] H. Gaweska, M. Henderson Pozzi, D.M.Z. Schmidt, D.G. McCafferty, P.F.
Fitzpatrick, Biochemistry 48 (2009) 5440–5445.
[20] M. Henderson Pozzi, V. Gawandi, P.F. Fitzpatrick, Biochemistry 48 (2009)
2305–12313.
1
[
[
[
[
21] V. Massey, J. Biol. Chem. 269 (1994) 22459–22462.
22] A. Mattevi, Trends Biochem. Sci. 31 (2006) 276–283.
23] T. Wu, V. Yankovskaya, W.S. McIntire, J. Biol. Chem. 278 (2003) 20514–20525.
24] M. Cervelli, F. Polticelli, R. Federico, P. Mariottini, J. Biol. Chem. 278 (2003)
4
The only structure of a reduced member of the monoamine oxidase family is that
of maize PAO. This shows the water bridging the flavin and Lys300, as shown in Fig. 1.
However, no product is bound in this structure. If the reaction with oxygen involves
the reduced enzyme-product complex, as suggested by the low value of kcat compared
to kred, the possibility must be considered that the product displaces the bound water.
In that case the nitrogen in the newly oxidized carbon-nitrogen bond could fulfill the
proposed role of the oxygen in the water molecule.
5271–5276.
[
25] Y. Wang, T. Murray-Stewart, W. Devereux, A. Hacker, B. Frydman, P.M. Woster,
R.A. Casero Jr., Biochem. Biophys. Res. Commun. 304 (2003) 605–611.
[26] M. Sebela, A. Radova, R. Angelini, P. Tavladoraki, I. Frebort, P. Pec, Plant Sci. 160
(2001) 197–207.