First things first, we have to write out what is given to us.
Given
#[E] = 4 nM#
#K_m = 2.8 muM#
#V_m = (60 muM)/min#
Next, we need to use two formulas to figure this problem out
#color(white)(aaaaaaaaaaa)##color(red)((k_(cat))/K_m color(white)(aaaaaaaa) k_(cat) = V_m/[[E]]#
We are going to figure out our #k_(cat)# or the turnover number of the enzyme, but first, our answer is supposed to be represented in #muM# so we will first convert #4nM -> muM#.
#(4 cancel(nM))/(1) * (1*10^-9 M)/(1 cancel(nM)) = (4 * 10^-9 cancel(M))/(1) * (1 muM)/(1 * 10^-6 cancel(M)) = darr#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)(4 * 10^-3 muM)#
Plugin
#color(red)(k_(cat) = V_m/[[E]]) -> [(60 cancel(muM))/min]/[(4*10^-3cancel(muM))/1] = (15000)/min#
Now that we have our #k_(cat)#, we will use it to find out the #"catalytic efficiency"#, which basically tells us how good the enzyme is at catalyzing the reaction at low substrate concentration.
#color(red)((k_(cat))/K_m) = [(15000)/min]/[(2.8muM)/1] = (5357.14)/(min*muM)#
Take this answer and convert #min -> s#
#(5357.14)/(cancelmin*muM) * (1 cancelmin)/(60 s) = color(blue)((89)/(s*muM)#
#Answer: 89/(s*muM)#
