Question

1.Which step of glycolysis is catalyzed by phosphoglucose isomerase? What is the chemical “reason” for this reaction in glycolysis? (2)

2. Calculate values of Km and Vmax for PGI. (4)

3. How long (in seconds) does a single catalytic event take? (2)

The enzyme phosphoglucose isomerase (PGI) catalyzes the following reaction in glycolysis сH,ОРО,?- Phosphoglucose2-03POH2C сH,он н Н isomerase но Н Н он Н он он но но Н он Н Glucose 6-phosphate Fructose 6-phosphate Kinetics experiments were performed on PGI. Enzyme activity (initial velocity, Vo) was measured at varying concentrations of Glucose-6-phosphate (G6P). The enzyme concentration used in all experiments was 1.5 HM. The data are shown below. Vo (mM/s) [G6PI (mM) 0.01 00 0.360 0.0200 0.600 0.0400 0.900 0.0800 1.200 0.160 1.440

Answer 1

Solution:

1) Second step

Phosphoglucose isomerase catalysed the second step of glycolysis. It converts glucose-6-phosphate to fructose-6-phosphate.

Isomerase enzymes are used to convert one isomer into another isomer. Since, glucose-6-phosphate and fructose-6-phosphate are isomers, hence phosphoglucose isomerase enzyme is used for this conversion.

3) Given, Vmax = 1.440 mM/s

Thus, according to Mechalis-menten equation,

V = Vmax [S] / Km + [S]

Thus, for first trial:

V = 0.360 mM/s

[S] = 0.0100 mM

Hence,

0.360 mM/s = 1.440 mM/s x 0.0100 mM / Km + 0.0100 mM

Km + 0.0100 = 0.0400  

Km = 0.0400 - 0.0100 = 0.0300 mM

3) Kcat = Vmax / Et

Kcat = (1.440 mM/s ) / 1.5 uM

Kcat = (1440 uM/s) / 1.5 uM

Kcat = 960 s-1

Thus,

Time required to a single event take place= 1/kcat

= 1 / 960 s-1 = 0.00104 sec