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!

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.

The Km is just a measure of the dissociation of the enzyme.

1. To determine whether the enzyme will have a higher Vmax for S1 or S2, we need to understand the concept of Vmax. Vmax is the maximum velocity or rate at which an enzyme can catalyze a reaction when all of its active sites are saturated with substrate. When the enzyme is functioning at Vmax, it is said to be working at maximum capacity.

In this case, the Km values give us an indication of how readily the enzyme binds to its respective substrates. Km is the substrate concentration at which the enzyme works at half of its maximum velocity. A lower Km value indicates that the enzyme has a higher affinity for the substrate.

Since the Km for S2 is lower (10 mM) compared to the Km for S1 (100 mM), it means that the enzyme has a higher affinity for S2. This suggests that S2 will reach Vmax at a lower substrate concentration compared to S1. Therefore, the enzyme will have a higher Vmax for S2.

2. To determine which substrate is a better substrate for the enzyme, we need to consider the affinity of the enzyme for each substrate. As explained earlier, the Km value is a measure of the substrate affinity for the enzyme. A lower Km indicates a higher affinity.

In this case, since the Km for S2 is lower (10 mM) compared to the Km for S1 (100 mM), it means that the enzyme has a higher affinity for S2. Therefore, S2 is a better substrate for the enzyme.

3. The kcat is a measure of the catalytic efficiency of an enzyme, representing the number of substrate molecules converted to product per unit time by a single enzyme molecule when it is fully saturated with substrate.

The kcat value is calculated as Vmax divided by the total enzyme concentration ([E]). Since the enzyme concentration is not given, we cannot directly compare the kcat values for S1 and S2. However, we can infer that the enzyme will have a higher kcat for the substrate it has a higher Vmax, which in this case is S2.

In summary:

1. The enzyme will have a higher Vmax for S2 because it has a lower Km value for S2, indicating a higher affinity for that substrate.
2. S2 is a better substrate for the enzyme because it has a lower Km value, indicating a higher affinity for S2.
3. It is likely that the enzyme will have a higher kcat for S2, as the enzyme will have a higher Vmax for this substrate.