. Recall your study of equilibria and kinetics from general chemistry. You used equations with upper caseKand lower case k during the study of equilibria and kinetics respectively. What do the upper and lowercase letters refer to? affinity of enzymes for substrate. Explain why this approximation makes sense. Does a large KM in4,5Graph 13.53.KmV, V LST V2.5Equatilon 11.50.5Equation 2 E+S ES E+P0.10[S). mM15Equation 3 K th,to[S), and number of substrates per reaction. Generate a finalized list of assumptione3. Define the term Var in your own wordsmax4. Solve the Michaelis-Menten equation for KM when o = Vmax /2. What does this tell you about therelationship between (substrate] and enzymes with high KM values? With low KM values? Makeageneralization.%3D5. a. What is the Ky value as you can estimate it from Graph 1?b. On Graph 1, draw a curve that has the same V, but a smaller KM value.3. Examine Equation 2. When k, is very small compared to k the KMmaxhigh or low affinity of enzyme for substrate?becomes an approximation oindicateVelocity, mM product/ min

Hello. Since your question has multiple parts, we will solve the first question for you. If you want the remaining questions to be solved, then please resubmit the whole question and specify those questions you want us to solve. 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). For a single substrate reaction: E +S ⇌ ES → E + P where , E = Enzyme S = Substrate ES = Enzyme-substrate complex P = Product Every enzyme catalyzes a specific reaction or reactions that are very closely related. At any point in time enzyme does not disturb the equilibrium of the reaction. It speeds up the forward and backward reaction by an equal factor. K represented by Keq is the equilibrium constant and it is equal to the ratio of rate constants of both forward and backward reactions. Keq = Kf/Kb Keq = e −∆Go / RT for any enzyme-catalyzed reaction. k is used to represent the rate constants of both forward and backward reactions. k1 and k2 represent the rate constant forward reactions and k-1 represents the rate constant backward reaction. Km = k2 + k-1/k1 kcat = Vmax/ Etotal [turnover number]