Please see the attached image. The Michaelis-Menten equation models the hyperbolic relationship between [S] and the initial reaction rate Vo for anenzyme-catalyzed, single-substrate reaction E + S = ES → E + P. The model can be more readily understood whencomparing three conditions: [S] << Km, [S] = Km, and [S] >> Km-Match each statement with the condition that it describes.Note that "rate" refers to initial velocity Vo where steady state conditions are assumed. [Etotal] refers to the total enzymeconcentration and [Efree] refers to the concentration of free enzyme.[S] << Km[S]:= Km[S] >> KmNot true for any of theseconditions[ES] is much lower than [Efree].[Efree] is about equal to [Eotal].Almost all active sites willIncreasing [Etotal] will increasebe filled.Adding more S will not increaseThe rate is half of theKm.the rate.maximum rate.Answer Bank

Those proteins or biological catalysts which help to speed up the chemical reaction are termed enzymes. Enzymes aid in building some substrates and breaking down some substrates. The factors that likely affect the working of an enzyme include pH, the enzyme’s concentration, temperature, and substrate concentration.The rate of reaction when the enzyme is completely saturated by the substrate is referred to as Vmax or maximum reaction rate. The value which represents substrate concentration when the rate of reaction is half of the Vmax is referred to as Km (Michaelis-Menten constant). According to Michaelis Menten equation: V = Vmax [S]/ Km+ [S] Where, V = reaction velocity Vmax = maximum velocity Km = Michaelis-Menten constant [S] = substrate concentration For a single substrate reaction: E +S ⇌ ES → E + P If the substrate concentration will be much lower than the Km the rate of reaction will be very small. Almost all the active sites of enzymes will remain unoccupied and a meager amount of enzyme-substrate complex will be formed. The total free enzyme will be almost equal to the total amount of enzyme. The free enzyme will possess the highest concentration. If the substrate concentration will become equal to Km. The rate of reaction will become half. V0 = Vmax/2 If the substrate concentration is much higher than the Km almost all the free enzymes will be utilized and their active sites will be occupied. Once the reaction reaches Vmaxthen addition of further substrate will not affect it. In any of these conditions, the enzyme concentration will have no effect on Km. So, increasing or decreasing it will not alter the value of Km. Hence the correct matches of statements with given conditions are: [S]＜＜Km [Efree] is about equal to [Etotal]. [ES] is much lower than [Efree]. [S] = Km The rate is half of the maximum rate. [S]＞＞Km Adding more S will not increase the rate. Almost all active sites will be filled. Not true for any of these conditions. Increasing [Etotal] will increase Km.