Question

2. We want to measure kcat, which is equal to VmaxIE). Explain why in our experiment the concentration of the substrate and the cofactor must be much greater than then concentration of the enzyme (saturating concentrations). Think about the shape of the Michaelis Menten curve.

Answer 1

Ets ES , = enzyme K-1 S = substrate Es K2P+E ES = enzyme-substroute Сотчраах = K₂ [es ] us Rate of enzyme catalysis. di Asconding to stedy state approximation, DIES) – KLE) [s] - K,[ES] -K2 CES) = 0 K CE(S) k on) tk 2 greater " After time t. change in concentration of [E)= ().-[ES). [5] = 750.-(45) neentration of substrate much the [ES] & CE)) -- [Es) - k{te).-tes)}ts). Kitka (ES) (k_,+K2) = K (F).)– TE9[5.] .:C85] {[%,+Ka+ K [8} = K CE), (3), [ES) = KTEO] [5.] K,+K2 +K, 15). RateK2 ES] .: Rate = K2RCE). L50 3,+Ką +6,05). Rate = K2 [ E). [SO] _ K,+K2 + to Ri .: Rale - K2 [€] s] Hore K2 [E.) = Vmax Km+tsa)

greator, First orboder Overall onder . The substroate and the co-faeton much 1 km <<« [s) wAt high substroate so, Rates ko (F). VA - Concentration t so Rate = K2 CEO] = constant =KCE] Vmax = K₂ [eo] | Kz = V max on Jung Ang S o Rate t ny 'high conem braction tso KV Low Concentrati ts. ? Rm+ to Rate = K2 FOTs.] 52 , 09). «s km - Kan [sa] [F I Rater const 1so] isolto poboustijo