What percentage of max is obtained when the substrate is present at 80% of the Km? Use two digits in your answer.
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- You are attempting to determine KM by measuring the reaction velocity at different substrate concentrations, but you do not realize that the substrate tends to precipitate under the experimental conditions you have chosen. How would this affect your measurement of KM?calculate the reaction velocity at saturating substrate concentrations. Your numerical answer is assumed to be in units of M sec-1. [S] = 100 mM k1 = 10 sec-1 k2 = 3000 sec-1 k-1 = 20 sec-1 [E]T = 1 \muμMWhat effect does a high substrate concentration have on the rate order of the Michaelis-Menten equation?
- If Vmax for a reaction is 10 μM ⋅ s-1 and the KM is 0.5 μM, what is the reaction velocity when the substrate concentration is 2 μM?What is the parameter in the enzyme measurements?At steady state,[Es} stays the same although substrate concentraion is steadily decreasing and product concentraion is increasing .how do you explain this?
- If the enzyme lactase has a Vo of 0.40 mM per minute when [S] = 1.25 mM, and a Vo of 1.0 mM per minute when [S] = 5.0 mM, what is its Vmax, Km and the slope on a linewver-burk plot?The equation of the double reciprocal plot is y = 0.5294 x + 1.4960. What is the value of vmax (in M/s)? The substrate concentration is given in units of molarity (M) and reaction velocity has units of molarity per second (M/s). (Report to three significant figures)when the substrate concentration of certain enzyme is quarter than Michaelis constant, the velocity of enzyme will be... * 3/4 V max 1/5 V max Non of these 2/5 V max 1/4 V max 3/5 V max
- Calculate the Km of the enzyme with these parameters. kcat = 130s^-1 Vo = 3.0 μMs-1 S = 10 μM Et = 0.09 µMExplain why the maximum initial reaction rate cannot be reached at low substrate concentrations.For some Enzyme, the Vmax is 18 micromols/min, Km is 400 microM. If the substrate concentration is 100 microM, what is the velocity of the reaction?