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Homework 5 Key

In class we have discussed how enzyme assays can be used to determine the enzyme kinetic constants Vmax and Km but in this assignment you will do the reverse. You will use known Vmax and Km values to calculate expected rates for a pair of enzyme catalyzed reactions.

Suppose that you are provided data about phenylalanine hydroxylase and phenylalanine aminotransferase (the enzyme that converts phenylalanine to phenylpyruvate in patients with PKU). You are told that the Vmax and Km values for phenylalanine hydroxylase in a standard enzyme assay are 100 µmol min-1 and 10 µM respectively, and those for phenylalanine aminotransferase are 10 µmol min-1 and 1000 µM respectively.

Prepare a table that reports expected values of enzyme velocity (v) at the following substrate concentrations ([s]): 1, 2, 6, 10, 20, 40, 60, 100, 500, 1000, 2400, 5000 (all values are in µM).

Use the Michaelis-Menten equation (V=(Vmax[S] )/(km + [S]), where Vmax = 100 µmol min-1 and Km = 10 µM) for each [S] given for phe hydroxylase and Vmax = 10 µmol min-1 and Km = 1000 µM for aminotransferase.

	phenylalanine hydroxylase	phenylalanine aminotransferase
[S] (µM)	V(µmol min-1)	V(µmol min-1)
1	9.09	0.00999001
2	16.67	0.01996008
6	37.50	0.059642147
10	50.00	0.099009901
20	66.67	0.196078431
40	80.00	0.384615385
60	85.71	0.566037736
100	90.91	0.909090909
500	98.04	3.333333333
1000	99.01	5
2400	99.59	7.058823529
5000	99.80	8.333333333

Prepare Michaelis-Menten plots that illustrate these kinetic data. Don’t forget to carefully label the axes.

Now consider this pseudo-clinical case of two patients with PKU. Assume that the production of phenylpyruvate by phenylalanine aminotransferase at a rate of less than 1 µmol min-1 as measured by this assay can be handled by the body (the phenylpyruvate is converted to something non-toxic or is excreted) and does NOT cause PKU.

A patient with mild PKU comes to the clinic with a blood phenylalanine concentration of 500 µM. When his phenylalanine hydroxylase is assayed as described above, Vmax and Km values are found to be 100 µmol min-1 and 5000 µM respectively. Prepare a Michaelis Menten plot for his enzyme as described above using the same substrate concentration range. To what extent has the change in Km affected his enzyme’s ability to convert phenylalanine to tyrosine, assuming that the substrate concentration it “sees” is the same as the patient’s blood phenylalanine concentration? At this blood concentration of phenylalanine, at what rate is his phenylalanine aminotransferase producing phenylpyruvate? Is this rate consistent with a diagnosis of PKU (that is, is it above 1 µmol min-1)?

Phenylalanine hydroxylase from the patient shows a 10 fold reduction in the velocity of the reaction it catalyzes; 98 µmol min-1 in the normal form vs 10 µmol min-1 in the patient with mild PKU.

Given that the patient’s phenylalanine aminotransferase has a Vmax of 10 µmol min-1 and a Km of 1000 µM and a blood [phenylalanine] of 500 µM, this patient’s phe aminotransferase will make phenylpyruvate at

V=(Vmax[S] )/(km + [S])

= (10 µmol min-1 * 500 µM)/(1000µM + 500 µM)

= 5000/1500 µmol min-1

= 3.33 µmol min-1

A patient with strong PKU comes to the clinic with a blood phenylalanine concentration of 2400 µM. When her phenylalanine hydroxylase is assayed as described above, Vmax and Km values are found to be 2 µmol min-1 and 10 µM respectively. Prepare a Michaelis Menten plot for her enzyme as described above using the same substrate concentration range. To what extent has the change in Vmax affected her enzyme’s ability to convert phenylalanine to tyrosine, assuming that the substrate concentration it “sees” is the same as the patient’s blood phenylalanine concentration? At this blood concentration of phenylalanine, at what rate is her phenylalanine aminotransferase producing phenylpyruvate? Is this rate consistent with a diagnosis of PKU that is more severe than the first patient?

Phenylalanine hydroxylase from the patient shows a 50 fold reduction in the velocity of the reaction it catalyzes; 99.59 µmol min-1 in the normal form vs 1.99 µmol min-1 in the patient with severe PKU at 2400 µM of phe.

Given that the patient’s phenylalanine aminotransferase has a Vmax of 10 µmol min-1 and a Km of 1000 µM and a blood [phenylalanine] of 2400 µM, this patient’s phe aminotransferase will make phenylpyruvate at

V=(Vmax[S] )/(km + [S])

= (10 µmol min-1 * 2400 µM)/(1000µM + 2400 µM)

= 24000/3400 µmol min-1

= 7.05 µmol min-1

Hydroxylase Michaelis-Menten plot

V(umol/min) 1.0 2.0 6.0 10.0 20.0 40.0 60.0 100.0 500.0 1000.0 2400.0 5000.0 9.09090909090909 16.66666666666667 37.5 50.0 66.66666666666667 80.0 85.7142857142857 90.9090909090909 98.0392156862745 99.00990099009901 99.58506224066385 99.80039920159675

[S] (uM)

V(umol/min)

M-M of aminotransferase

V of aminotransferase 1.0 2.0 6.0 10.0 20.0 40.0 60.0 100.0 500.0 1000.0 2400.0 5000.0 0.00999000999000999 0.0199600798403194 0.0596421471172962 0.099009900990099 0.196078431372549 0.384615384615385 0.566037735849057 0.909090909090909 3.333333333333333 5.0 7.058823529411764 8.333333333333332

[S] (uM)

V (umol/min)