please help me, really need to get this right. Table 2: Initial rates of reactio
ID: 904265 • Letter: P
Question
please help me, really need to get this right.
Table 2: Initial rates of reaction between Beta-galactosidase and lactose at different lactose concentrations a. Determine the KM and L values of lactose from Table 2, assuming that the enzyme concentration used was 0.5 mg/mL. b. Explain the differences in reaction kinetics that you are observing between your experiment substrate (ONPG) and lactose (what differences do you observe between Km and ka and what can you conclude based on your findings?). c. Describe a practical application of -galactosidase. Which substrate is used or most relevant for that application? Discuss how the Km and k cat would impact that application.Explanation / Answer
[E]o = 0.5 mg/mL
We have,
v = Vmax[S] / KM + [S]
when [S] = 0.31 mM , v = 0.0009 mM/s
and [S] = 0.63 mM , v = 0.0018 mM/s
0.0009 = Vmax * 0.31 / KM + 0.31 .........(1)
0.0018 = Vmax * 0.63 / KM + 0.63 .........(2)
From eq (1) and (2)
0.0009 / 0.0018 = 0.31 * (KM + 0.63) / 0.63 * (KM + 0.31)
1/2 = 0.31KM + 0.1953 / 0.63KM + 0.1953
0.63KM + 0.1953 = 0.62KM + 0.3906
0.01KM = 0.1953
KM = 19.53 M
From equation (1)
0.0009 = Vmax * 0.31 / 19.53 + 0.31
0.0009 = Vmax * 0.31 / 19.84
0.0009 = Vmax * 0.015625
Vmax = 0.0576 mM/s
Vmax = kcat [E]o
0.0576 = kcat * 0.5
kcat = 0.1152 1/s
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