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Spring 2013 Lecture 15
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Spring 2013 Lecture 15
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![SOLVED: Using your Vmax =2.225 umol/min from the uninhibited data, calculate kcat (kcat = Vmax/[ET], where ET is the total enzyme concentration. The concentration of stock solution of enzyme used was 0.35mg/mL](https://cdn.numerade.com/ask_previews/37951912-45bb-4020-9a65-3c82938b63de_large.jpg "SOLVED: Using your Vmax =2.225 umol/min from the uninhibited data, calculate kcat (kcat = Vmax/[ET], where ET is the total enzyme concentration. The concentration of stock solution of enzyme used was 0.35mg/mL")
SOLVED: Using your Vmax =2.225 umol/min from the uninhibited data, calculate kcat (kcat = Vmax/[ET], where ET is the total enzyme concentration. The concentration of stock solution of enzyme used was 0.35mg/mL
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SOLVED: Using the Lineweaver-Burk plot calculate Vmax, Km and Kcat, 0.102 4= vtak 4210 Kcct 4. 79 " fhio Vrer 3.1ry Jrfan -9.1656 umfhv) Km - 0.486 Y 9.7656 Using the Michaelis-Menten
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