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# tutorial3_ans - Bioc 300 Tutorial 3 Bioc 300 Tutorial 3...

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Bioc 300 - Tutorial 3 Bioc 300 - Tutorial 3 – July 26 th 2009 Topics: - Michaelis-Menten kinetics, lineweaver burk plots - Glycolysis - The Citric Acid Cycle

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1 (#2 pg 238) Penicillin is hydrolyzed by penicillinase is some resistant bacteria. The mass of this enzyme is is 29.6 kd. The amount of penicillin hydrolyzed in 1 minute in a 10 ml solution containing 10 -9 g of purified penicillinase was measured as a function of the concentration of penicillin (measurements on next page) a) Plot V o vs [s] and 1/V o vs 1/[S] for these data.Does penicillinase obey Michaelis- Menten kinetics? If so, what is K m ? b) What is the value of V max ?
#1 continued. [Penicillin] uM Amount hydrolyzed (nanomoles) 1 0.11 3 0.25 5 0.34 10 0.45 30 0.58 50 0.61

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#1 continued. [Penicilli n] uM 1/uM Amount hydrolyzed (nanomole s) V (mol/minute) 1/V 1 1/1 = 1 0.11 0.11x10 -9 = 1.8 × 10 -12 1/(1.8…) 3 1/3 = 0.33 0.25 0.25x10 -9 = 4.2 x 10 -12 1/(4.2…) 5 1/5 = 0.2 0.34 0.34x10 -9 = 5.6 x 10 -12 1/(5.6…) 10 1/10 = 0.1 0.45 0.45x10 -9 = 7.5 x 10 -12 1/(7.5…) 30 1/30 = 0.033 0.58 0.58x10 -9 = 9.6 x 10 -12 1/(9.6…) 50 1/50 = 0.02 0.61 0.61x10 -9 = 10.1 x 10 -12 1/(10.1…)
Michaelis-Menten Graph

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Lineweaver-Burk Y intercept = 1/Vmax = 1.46 =, therefore Vmax= 1/(1.46*10 9 ) = 6.84*10 10 X intercept = -1/Km = 1.46/7.6225, therefore Km= 5.2*10 -6
2 (#4 pg 238) The kinetics of an enzyme are measured as a function of substrate concentration in the presence and in the absence of 2mM inhibitor (I). (measurements on next page) a) What are the values of Vmax and Km in the absence and presence of the inhibitor? b) What type of inhibition is this?

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#2 continued. [S] uM V - No inhibitor V – with inhibitor 3 10.4 4.1 5 14.5 6.4 10 22.5 11.3 30 33.8 22.6 90 40.5 33.8 [S] uM 1/[S] V - No inhibitor 1/V V – with inhibitor 1/V 3 1/3 10.4 1/10.4 4.1 1/4.1 5 1/5 14.5 1/14.5 6.4 1/6.4 10 1/10 22.5 1/22.5 11.3 1/11.3 30 1/30 33.8 1/33.8 22.6 1/22.6 90 1/90 40.5 1/40.5 33.8 1/33.8
Lineweaver-Burk

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#2 continued Y intercept = 1/Vmax = 0.0215, therefore Vmax = 46.5 umol/min Vmax is basically the same with and without inhibitor, therefore its competitive inhibition Km for without = 1.1 x 10 -5 Km for with = 3.1 x 10 -5
3. The first reaction in glycolysis that results in the formation of an energy-

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## This note was uploaded on 12/13/2011 for the course BIOLOGY 300 taught by Professor Asdf during the Spring '08 term at UBC.

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tutorial3_ans - Bioc 300 Tutorial 3 Bioc 300 Tutorial 3...

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