tor at day. at 12 da 1.0 x 104 cells Determine 3.42 in gene therapy research, it
ID: 491885 • Letter: T
Question
tor at day. at 12 da 1.0 x 104 cells Determine 3.42 in gene therapy research, it is common to use reporter genes to the of to protein. reporter typically encode either protein fluoresces an convert substrate into a colored, fluorescent, or lumi that substrate ONPG (o nitrophenyl-B D galactopyranoside) into the yellow the absorbance of the cell lysate at 120 can quantitate the amount of this product The consumption of the substrate, ONPG, can be modeled using the Michaelis Menton equation. dt Km [S] (a) Develop an equation for time in terms of ls, and Km. The units and definitions of each variable are shown in lable 315 (b) Given that [E] 3.0 ug/mL and [Slo 2 mM, find the time it takes for the substrate concentration to drop to half its initial value, given that K 0.161 mM and k 0.006 wmol/(jug enzyme min) (c) If [Elo is decreased by one order of magnitude (i.e., [Elo 0.3 g/mL, find [S] after 30 min. (Hint: You cannot explicitly solve for [S] in terms of other variables.) where t is kin d an of 800 pg h? patc Test B. To d so signe assum test,Explanation / Answer
-dS*(Km+S)/S = dt*([E]O K2
when integrated noting that at t=0 , S= So and t= t S= S
KM* ln (S/SO)+ (S0-S) = [E]o*K2*t
Hence t= {KM*ln(S/SO) +(SO-S)}/ (K2[E]0
Given [E]o= 3ug/ml = 3*10-6 g/10-3 L = 3*10-3g/L, SO= 2*10-3M, Km=0.161*10-3M
K2= 0.006 mol/g.enzyme.min, S=2*10-3/2= 1*10-3M
Hence t={0.161*10-3*ln(2/1) + (2-1)*10-3}/(0.006*3*10-3 ) =61 min
3. [Eo]=0.3*10-3g/L ,0.161*10-3* ln (S/2*10-3)+(2*10-3-S) = 0.3*10-3*0.006*30
When solved using excel ,x= 1.95/1000
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.