Question

The tetracoordinated copper center in the oxidized and reduced forms of plastocyanin exhibits longest bond with

• A. cysteine-S and methionine-S, respectively
• B. methionine-S and cysteine-S, respectively
• C. cysteine-S and cysteine-S, respectively
• D. methionine-S and methionine-S, respectively

Hint:

- In blue copper proteins like plastocyanin, the axial Cu–S(Met) bond is always longer than the equatorial Cu–S(Cys) and Cu–N(His) bonds.
- Remember: \(\mathrm{Cys{-}S}\) is strong, \(\mathrm{Met{-}S}\) is weak.
- The entatic state enforces a geometry optimal for fast electron transfer, but the longest bond remains Cu–S(Met).

Answer 1

The Correct Option is D

Step 1: Structure of plastocyanin.
Plastocyanin is a blue copper protein with a \(\text{Cu}^{2+/+}\) center coordinated by two histidine imidazoles (N donors), one cysteine thiolate (S), and one methionine thioether (S). This gives a distorted tetrahedral geometry around the Cu center. Step 2: Bonding considerations.
- The cysteine thiolate provides a strong covalent bond with Cu due to strong \(\pi\)-donation.
- The histidine N donors also form strong bonds.
- The methionine S is a weak thioether donor and interacts more weakly with the copper center. Step 3: Oxidized vs reduced state.
In both oxidation states (\(\text{Cu}^{2+}\) and \(\text{Cu}^{+}\)), the \(\mathrm{Cu{-}S_{Met}}\) bond remains the longest bond because of weak donation from methionine compared to cysteine. The geometry is often described as “entatic state” where bond lengths are tuned for efficient electron transfer, but still the Cu–S(Met) bond is the longest in both states. Step 4: Conclusion.
Therefore, in both oxidized and reduced plastocyanin, the longest bond is with methionine-S. \[ \boxed{\text{Answer: (D) methionine-S and methionine-S, respectively}} \]