a' K. 'm 긍 Double Reciprocal Plots and uncompetitive Inhibition uTu[wt Vo Km 1 a' + Vm [S] Vmax max a'=1.5 a' = 2 Vm max 1 [S] mM a'=1 [0]
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I have a question, what are these ''a'' means in this plot? Normally its 1/vmax or -1/km.
Please be specific. Thank you.
![-
a'
Double Reciprocal Plots and uncompetitive Inhibition
V. - (V) 11 + V
=
Vo
max
Voμm/min
[S] /max
a' = 1.5
a' = 2,
a'
Vmax
[S]
MM
a' = 1
[0]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F276c085d-4976-4cc2-ba9f-ed8f400f3309%2Fce927c46-ca9a-4130-82b3-6a8635b7430a%2Fohe65b_processed.png&w=3840&q=75)
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- PTP1B Substrate kcat Km. kcat/Km UM 10-7 x (s-1 M) DADEPYLIPQQG DADAPYLIPQQG DAAEP YLIPQQG AAAAPYLIPQQG 44.6 + 1.8 39.8 + 0.32 3.9 + 0.9 13.7 + 0.46 1.1 + 0.25 0.29 + 0.01 35.3 + 0.22 6.6 + 0.22 0.53 + 0.02 34.7 + 0.25 52.7 + 0.7 0.066 + 0.001 ) The units for kcat/KM in the above are given according to standard scientific notation. On this (d) ( basis what is the value of this kinetic parameter for the DADEPYLIPQQG substrate?A = Control = 100 B = Treatment = 10 FoldChange (FC) = 10/100 = 0.1 log2FC = log2(10)-log2(100) = 3.32 - 6.64 = -3.32 Please describe why log2FC is more informative?For a gradient system with a gradient of 5-90% in 50 min, flow 2 mL/min, column 100 x 4.6 mm i.d., the first peak elutes at 20 min, and the last peak elutes at 50 min. Calculate k* for this system. 2. Can isocratic elution be used for this sample? If so, what is the range of k values to be expected. 3.Propose a way to shorten the gradient run time by eliminating the wait for the first peak to elute at 20 min. 4. For the original conditions of this question, change the conditions necessary to use a 150 x 4.6 mm column while maintaining the same k* value.
- -Inhibitor +Inhibitor [S] (mM) Vη&νβσπ: (μmol/sec). ν0&νβσπ;&νβσπ:(μmol/sec) 0.0001 33 17 0.0005 71 50 0.001 83 67 0.005 96 91 0.01 98 95 What is the TYPE of iinhibitor?Calculate, or make a best estimate of, the unknown factors in the situations listed in Table 1. Table 1. Data for problem 1. To find Data (i) Ek [K+]out = 4 mM; [K+]in = 130 mM (ii) [Cl-]in [Cl-]out = 570 mM; ECl = -65 mV (iii) ECl [Cl-]out = 150 mM; [Cl-]in = 8 mM; in a mammal (iv) [Na+]in Overshoot of action potential = +35 mV; saline [Na+] = 112 mM (v) [K+]in Blood [K+] = 3.2 mM; undershoot of action potential = -87 mV (vi) ECa [Ca2+]out = 5.6 mM; free [Ca2+]in = 0.8 mM (vii) [K+]out [K+]in = 350 mM; Ek = -82 mVhi can you find the maximum binding capacity(b max) using the table
- The order in the MAR is for Atropine 15 mg IV q6h hours PRN. How many mL will you give if the dosage strength is 75mg/7.5mL?The Quikchange cycling parameters are listed below: 1.95 degrees, 2 min 2. 95 degrees, 20 sec 3. 60 degrees, 10 sec 4. 68 degrees, 2 min 5. Repeat steps 2-4 18 times 6. 4 degrees, indefinitely Match each step above to the correct function(s) listed below. Write your selection(s) for each step. You may have more than one answer for each. 1.initial denaturation 2. dNTP pairing 3. renaturation 4. Primer annealing 5.new DNA extension 6. polymerase unwinding 7. refrigeration 8.secondary denaturation 9. degradation of excess DNAPTP1B Substrate kcat Km kcat/Km UM 10-7 x (s-1 M) DADEPYLIPQQG DADAPYLIPQQG DAAEPYLIPQQG AAAAPYLIPQQG 44.6 + 1.8 39.8 + 0.32 3.9 + 0.9 13.7 + 0.46 1.1 + 0.25 0.29 + 0.01 35.3 + 0.22 6.6 ± 0.22 0.53 + 0.02 34.7 + 0.25 52.7 ± 0.7 0.066 + 0.001 (d) (. ) The units for kcat/KM in the above are given according to standard scientific notation. On this basis what is the value of this kinetic parameter for the DADEPYLIPQQG substrate?
- 2. The two diagrams to the right il- lustrate plots of steady-state ki- netic studies to characterize the in- teraction of heart muscle phos- phofructokinase-1 with a non-phy- siological, synthetic substrate fruc- tose-6-sulfate. Because the kcat is smaller than that for the natural 5 0.8- NH = 3.5 0.6- 0.4- 0.2- 10 μΜ 20 μΜ 48 µM substrate, higher enzyme concen- trations could be used. The results show the influence of increasing 2 0.2- 0.4- concentrations of ATP on the initial -0.6- > velocity of the enzyme catalyzed reaction in the presence of no -0.8F 4 12 20 28 36 44 52 60 68 76 84 92 .2 2.0 2.4 ΑMP () , 10 μΜ AMP (+ ) 20 μΜ AMP (-), and 48 µM AMP (-). [AΤP] (μΜ) log[ATP] (µM) (a) . Write the reaction in words catalyzed by the enzyme for the alternative substrate, describe how ATP interacts with the enzyme in the case of no AMP (•). (b) ! with respect to the binding of AMP and ATP to the allosteric effector sites on the enzyme. Explain the physical significance of the displacement…2.5kg convert to my- Why Michalis plot is not used, instead double reciprocal plot is used to determine KM and Vmax?