Question

The N enzyme and its two substrates, D and W

a. Use the following datat and the attached graph paper to determine Km and Vmax for the N hydrolysis of its D substrate and of its W substrate.

b. Choose ONE substrate and use the following data to determine Km and Vmax from a linear graph, then calculate the Kcat and Kcat/Km if the enzyme concentration is 1.5ng/microliter.

c. For which of these two substrates does the N enzyme have least affinity?

[D] (mM) | Vo (nmoles/min) | [W] (mM) | Vo (nmoles/min) |

2.1 | 2 | 1.3 | 1 |

4.2 | 3.4 | 2.6 | 1.7 |

8.4 | 5.5 | 5.2 | 2.8 |

16.8 | 8.5 | 10.4 | 4.3 |

25 | 10 | 15 | 5 |

50 | 11 | 30 | 5.5 |

Answer #1

From the given data

Drawn below are two plots doe D and W

a. for D

Km = 80 mM

Vmax = 2.07 nmol/min

for W

Km = 33.33 mM

Vmax = 1.0367 nmol/min

b. For D

Kcat = Vmax/[Et]

= 2.07/1.5

= 1.38

Kcat/Km = 1.38/80 x 10^6

= 1.725 x 10^-8

Plots

Enzyme efficiency is measured by
KM
b. kcat/
KM
c. kcat
d. ½ Vmax
The KM values in mM for a hexokinase with different
substrates is:
glucose = 3.4
fructose
=2.3
galactose =
0.5
gulose = 1.3
The highest activity for the enzyme
takes place with the substrate
glucose
b.
fructose
c. galactose
d. gulose

3. An enzyme can catalyze a reaction with either of two
substrates, S1 or S2. The Km for S1 was found to be 2.0
mM, and the Km, for S2 was found to be 20 mM. A student determined
that the Vmax was the same for the two
substrates. Unfortunately, he lost the page of his notebook and
needed to know the value of Vmax. He carried out
two reactions: one with 0.1 mM S1, the other with 0.1 mM...

The following data were obtained for the variation of initial
velocity and substrate for a reaction catalyzed by chymotrypsin
(5nM) at pH 8, 37 °C.
1. Using a linear plot, determine the KM and Vmax of the
chymotrypsin given the data below: (12)
[S] (μM) V0 (μmol/min)
0 0
0.5 200
1.0 400
1.5 580
2.0 750
2.5 840
3.0 860
4.0 875
6.0 890
The kcat/KM parameter is a measure of...

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