The kinetics of chymotrypsin w
The kinetics of chymotrypsin were studied and yielded a KM of 5mM and a Vmax of 0.2 mM/min. The enzyme concentration was 0.20μg/ml. Calculate the turnover number, assuming one binding site perenzyme.
Answer:
Ans. Given,
Vmax = 0.2 mM / min ; [1 min = 60 s]
= 0.00333 mM/ s ; [1 mM = 1000 uM]
= 3.33 uM/ s
[E] = 0.20 ug/ mL
# Step 1: Convert [E] in terms of uM
Molar mass of Chymotrypsin = 25.6 kDa = 25600 g/ mol
Now,
[E] = 0.20 ug/ mL
= 2.0 x 10-7 g / mL
= (2.0 x 10-5 g / 25600.0 g mol-1) / mL
= 7.8125 x 10-12 mol /mL ;[1 mL = 0.001 L]
= 7.8125 x 10-12 mol / 0.001 L
= 7.8125 x 10-9 M
= 7.8125 x 10-3 uM
Therefore, 0.2 ug/ mL chymotrypsin is equivalent to7.8125 x 10-3 uM chymotrypsin.
# Now,
Turnover number, Kcat = Vmax /[E]
= 3.33 uM s-1 / (7.8125 x 10-3 uM)
= 426.24s-1
Hence, turnover number = 426.24 s-1
# Note: The molar mass of chymotrypsin = 25.6kDa (Wilcox 1970)
Change in molar mass of the enzyme would cause change in Kcatvalue, too.
"Our Prices Start at $11.99. As Our First Client, Use Coupon Code GET15 to claim 15% Discount This Month!!"
