What is the impact of the lower value Vmax on the affinity for enzyme for substrate?And what is impact of the lower V max on the amount of product formed ? If the lower value of black resulting in the new plot (red curve) is due to presence of enzyme inhibitor is the inhibitor reversible or irreversible ? And why? Michaelis Menten Plot.ds (Compatibility Mode] - Microsoft ExcelHomeInsertPage LayoutFormulasDataReviewViewAdd-InsAcrobatCalibri11Insert -GeneralPasteDelete -Conditional FormatCellClipboardFormatting- as Table Styles- Format-FontAlignmentNumberStylesCellsF24ABDGHM1 [Ethanol)VoCalc. Vo delta^20.0070.060.0327 0.0007450.3530.0150.110.057444 0.0027624.0.0310.160.0872640.005290.30.0680.210.118696 0.0083360.10.230.131384 0.0097250.250.20.280.148218 0.0173660.20.30.290.154831 0.018271• Vo0.40.280.158364 0.0147950.15Calc. Vo100.07729211Km0.0293910.112Vmax0.170.051314160.10.2030.40.51718Fig. 11. Michaelis-Menten Plot using a value of 0.17 for Vmax.Notice: 1) that a new set of calculated values in column C has been generated and2) that by using these new data a new best fit Michaelis-Menten plot (red curve)has been generated.Fig. 11 shows that the rates for the new enzyme-catalyzed reaction (red curve) aresignificantly lower than those originally obtained (blue symbols).

The relationship between Km and substrate concentration is that Km corresponds to the substrate concentration where the reaction rate of the enzyme-catalysed reaction is half of the maximum reaction rate Vmax. As the substrate concentration increases, the reaction rate will approach Vmax, but Km remains unchanged.Uncompetitive and non competitive inhibitors decreases the vmax.