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Unformatted text preview: – Chemical and Biochemical Processes – Biochemical Engineering Section Paula Mendes Tutorial Problems A. Enzyme Kinetics 1 The enzymecatalyzed conversion of a substrate at 25 o C has a K m of 0.035 M. The rate of the reaction is 1.15 x 103 M s1 when the substrate concentration is 0.110 M. What is the maximum velocity of this reaction? Solution: 1 33 S S m s max S m S max S s M 1.52x10 110 . ) 110 . 035 . ( 1.15x10 C ) C (K r V C K C V r = + = + = ⇔ + = 2 An enzyme with a K m of 1x103 M was assayed using an initial substrate concentration of 3x105 M. After 2 min, 5 percent of the substrate was converted. What is the maximum velocity of this reaction? Solution: For ti=0 and tf=2 min, C Sf = 0.95C Si ti) (tf ) C C ( C C ln K ti) (tf C C ln K C C ln K V ti) (tf V C C ln K C C ln K ti) (tf V ] ) C C ln K ( dt V dC ) 1 C K ( dt V dC C C K C K C V dt dC r Sf Si Sf Si m Si Si m Sf Sf m max max Si Si m Sf Sf m max C C S S m tf ti max S C C S m tf ti max S C C S S m S m S max S S Sf Si Sf Si Sf Si + = + + = = + + = ⇔ = ⇔ = + ⇔ + = = ∫ ∫ ∫ ∫ 2 ) 5x3x10 9 . 3x10 ( 0.95x3x10 3x10 ln 1x10 ti) (tf ) C C ( C C ln K V 5 5 5 5 3 Sf Si Sf Si m max + = + = V max =2.64x105 M min1 B. Cell Growth Kinetics 1 A microorganism is growing at its maximum specific growth rate. Conditions are the following: a) Calculate μ max . b) What will the population size after 25 days? Solution: a) Since the microorganism is growing at its maximum specific growth rate μ = μ max 1 max Ci Cf max max Cf Ci max Ci Cf max max Ci Cf tf ti max C C C C max C C G d .4 0) (1 50 75 ln μ ti) (tf C C ln μ : μ calculate can we ly respective , 75 C and 50 C for 1 tf and ti employing by example, For μ calculate to used be can table the from data of...
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This note was uploaded on 06/06/2011 for the course CHEM 3040 taught by Professor Reddy during the Spring '10 term at Taylor's.
 Spring '10
 reddy
 Enzyme, Kinetics

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