Tryptophan is one
of few residues that participates in biological
electron transfer reactions. Upon substitution of the native Cu2+ center with Zn2+ in the blue-copper protein azurin,
a long-lived tryptophan neutral radical can be photogenerated. We
report the following quantum yield values for Zn-substituted azurin
in the presence of the electron acceptor Cu(II)-azurin: formation
of the tryptophan neutral radical (Φrad), electron
transfer (ΦET), fluorescence (Φfluo), and phosphorescence (Φphos), as well as the efficiency
of proton transfer of the cation radical (ΦPT). Increasing
the concentration of the electron acceptor increased Φrad and ΦET values and decreased Φphos without affecting Φfluo. At all concentrations
of the acceptor, the value of ΦPT was nearly unity.
These observations indicate that the phosphorescent triplet state
is the parent state of electron transfer and that nearly all electron
transfer events lead to proton loss. Similar results regarding the
parent state were obtained with a different electron acceptor, [Co(NH3)5Cl]2+; however, Stern–Volmer
graphs revealed that [Co(NH3)5Cl]2+ was a more effective phosphorescence quencher (K
SV = 230 000 M–1) compared to
Cu(II)-azurin (K
SV = 88 000 M–1). Competition experiments in the presence of both
[Co(NH3)5Cl]2+ and Cu(II)-azurin
suggested that [Co(NH3)5Cl]2+ is
the preferred electron acceptor. Implications of these results in
terms of quenching mechanisms are discussed.