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Unformatted text preview: – Chemical and Biochemical Processes – Biochemical Engineering Section Paula Mendes Tutorial Problems A. Enzyme Kinetics 1- The enzyme-catalyzed 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 10-3 M s-1 when the substrate concentration is 0.110 M. What is the maximum velocity of this reaction? Solution: 1- 3--3 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 1x10-3 M was assayed using an initial substrate concentration of 3x10-5 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 =2.64x10-5 M min-1 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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