The resulting rate (kcat [e]tot) is only. You use this if the question is asking about which enzyme has the greatest efficiency. Web you will get vmax and km under your assay conditions’ used concentration of enzyme (e.g. Web km is a substrate concentration and is the amount of substrate it takes for an enzyme to reach v max /2. Web kcat, or k2 or turnover number (they all mean the same thing) is a measure of how many substrates one (1) enzyme can convert into a product per second. To solve your question, (1)calculate kcat, i.e kcat=vmax/ [et] where [et]= total enzyme. That substrate with the highest value is the best substrate for the enzyme, accounting for the name specificity constant. Web the reciprocal form of the expression of kcat / km shows that kcat / km is a harmonic sum of kinetic terms that correspond to the heights of the transition states relative to. Web as should be clear, the $k_ {cat}$ is the rate constant for the reaction that occurs after substrate is bound to the enzyme. The rate of any reaction is limited by the rate at which reactant.
Web the kcat /k m value, or specificity constant, of the various substrates can be compared. Let suppose our vmax = 4mm/min calculation of kcat (the turnover. As km is constant, the affinity of the enzyme for the substrate should not change. Web the kcat /km ratio, where kcat is the catalytic constant for the conversion of substrate into product, and km is the michaelis constant, has been widely used as a measure of. Web kcat/km = measure of catalytic efficiency. Web kcat/km is the catalytic efficiency of the enzyme. This is a good interpretation. Web k m is a dependent variable, and its value can change due to many reasons, including the ph level of the system, temperature, or any other condition. Kcat/km is a useful measure of the efficiency. You use this if the question is asking about which enzyme has the greatest efficiency. Also may be given vmax and enzyme.
