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THis problem is part of my final review guide for PchemFive solutions of substrate (S) were prepared with the concentrations give in the first column below, and each one was divided into five equal volumes (total of 25 solutions). The same concentration of enzyme was present in each one. An inhibitor (I) was then added in different concentrations to the samples, and the initial rate of formation of product was determined with the results given below. Does the inhibitor act competitively, uncompetitively, mixed, or non-competitively? Calculate the binding constants for the substrate (KM) and the inhibitor (Kc and Ku) with correct units. Initial rate (uM/s) 0 0.20 0.40 0.60 0.80 0.050 0.020 0.015 0.012 0.010 0.008 0.10 0.035 0.026 0.021 0.017 0.015 [S] (mM) 0.20 0.056 0.042 0.033 0.028 0.024 0.40 0.080 0.059 0.047 0.039 0.034 0.60 0.0930.069 0.055 0.046 0.039

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No inhu bite KM0.012 83.33333333 0.02147.61904762 0.03330.3030303 0.04721.27659574 0.055 0.01 0.017 0.028 0.039 100 58.82352941 35.71428571Whe flatted文vs TSİ data frouw.e. A eb áve. Cve 2.14 ア mmA Yanベー ヌ:53 mer -competžva 28 98 + 9763 b) = 2-848 K M 2 0 2 0 608 n

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