I was wondering if the methods of enzyme immobilisation reduce enzymatic activity. I am aware that everything dependes on enzyme kinetics and the type of immobilisation, ¿but do the enzymes immobilised via covalent or ionic bonding experience a reduction in their catalytic activity?
Yes. Simple adsorption does not cause as much loss of activity as during covalent bonding. There are several reasons. The enzyme can undergo conformational change, experience hydrophobic interactions with support and partial denaturation. Immobilization also adds an additional factor of diffusion resistance, the substrate has to pass a barrier caused by support to bind. In this case the loss of enzyme activity is only an "apparent" loss. But sometimes the enzyme can exhibit "superactivity", mostly due to conformational change as reported for enzymes in reverse micelles.
- The equilibrium constants that describe binding of it substrate to the enzyme and dissociation from (k1 and k2 and hence Km) it will change, simply because the substrate can no longer approach the enzyme from all angles. One way to look at this is that the effective substrate concentration seen by the enzyme will be lower than the molar concentration.
- The electrical charges of the protein and the counterions (some of them on the matrix) will create an electrical double layer which charged substrates have difficulty crossing.
- Some enzyme reactions produce or consume protons, creating a pH gradient between the surface and the bulk medium.
- Binding of the enzyme to the matrix may change its conformation and/or its flexibility, which would affect kcat and hence Vmax.
- The velocity of the enzyme inside a matrix will increase with [S] only until diffusion of substrate and product through the pores of the matrix become rate limiting, this lowers apparent Vmax. Increasing [E] beyond an optimal value will increase enzyme cost w/o the benefit of increased reaction rates. This effect depends on the Sherwood number Sh of the system.
For all these reasons (and may be some more) the measured enzymatic activity will be lower by a stationary efficiency factor ηs = turnover of immobilised enzyme / turnover by free enzyme ≤ 1, or by the operational efficiency ηo = τf / τi ≤ 1 (where τ is the relaxation constant, the time for [S] to drop to 1/e ≈ 37% of the starting value). Note that the proper way to determine τf is after destruction of the matrix by ultrasound.
They do. But I'm not sure if one or both. Maybe ionic bonds. Or maybe more ionic than covalent.
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