1. I believe that there is no way to know this for sure just looking at the Km values. Although Km=k2+k3/k1, their is know way of knowing unless more information is provided. k2 is the rate constant for the dissociation of the ES complex to E+S. k3 is the rate constant for the dissociation of the ES complex to form E + P. The Km in noway can provide this information unless more information is provided.
2. Lets take a closer look at Michealis-Mendels equation to answer this question. Vo=Vmax(S/S+Km). If the substrate concentration is equal to Km then Vo will be equal to 1/2(Vmax). Since the Km of S2 is lower than S1, less substrate is required to reach 1/2 of Vmax for S2 compared to S1. So, in general, it is plausible that S2 will have a higher affinity for the enzyme because it requires less substrate to reach 1/2(Vmax) or Vmax, but the only way to know for sure is to look at the kcat/Km ratio. But without knowing the kcat, their is noway to tell for sure.
3. In general, a higher Km corresponds to a higher kcat, but this is not always the case. Therefore, choosing between S1 and S2 based on Km values can not be done based upon my knowledge, or without additional information. It is very plausible that both kcals for both enzymes will be the same, but their is noway to know for sure.
An enzyme works on two substrates, S1 and S2. Its Km for S1 is 100mM and its Km for S2 is 10mM. Based on this information, answer the following questions:
1. For a given [E] will it have a higher Vmax for S1 or S2? Explain your answer
2. Is S1 or S2 a better substrate for the enzyme. Explain your answer
3. Will the enzyme have a higher kcat for S1 or S2? Explain your answer
Thanks!
2 answers
The Km is just a measure of the dissociation of the enzyme.