Question

The enzyme enolase catalyzes the following reaction in the glycolysis pathway (Chapter 14):

2-phosphoglycerate ® phosphoenolpyruvate + H₂O

Dehydration of 2-phosphoglycerate (2PG) results in phosphoenolpyruvate (PEP), which is a high-energy phosphate donor that can be used to phosphorylate ADP to make ATP. Thus, this is an important reaction to set up the final step of the pathway.

Balantidiasis is a disease caused by the ciliated protozoan, Balantidium coli. As it is a eukaryote, antibiotics against bacteria will not affect it. Your research group has recently found that members of the Balantidium genus possess an unusual form of enolase, which may allow you to develop an inhibitor that would affect it (and not the enolase in human cells). You have cloned the gene for the enolase from Balantidium coli, and expressed it in E. coli (which is a bacterium) and then purified the enzyme. You call this BcEno, to distinguish it from the other types of enolase. The polypeptide is predicted to be 32 kDa in size, and protein behaves as a homodimer with 2 active sites in each dimer. It requires Mg²⁺ for activity and is inhibited by EDTA, a chelator of divalent cations.

1. You perform a kinetic analysis on the enzyme by measuring the rate of catalysis in the presence of increasing amounts of 2PG. You initiate each reaction by adding 5 µL of a solution of 0.1 µg/mL enzyme to each assay tube, which already contains buffer and various amounts of 2PG in a total volume of 95 µL. After each reaction has proceeded for 30 seconds, during which time it should still be in the initial phase, you terminate it by adding 10 µL of 1 M EDTA. You then quantify the amount of PEP.

[2PG] in tube before reaction starts (µM)	amount of PEP after reaction (pmol)
10	44.9
33	133
100	320
330	638
1000	880
3300	1020

From these data, calculate k_cat and K_m of BcENO for 2PG:

k_cat =

K_m =

What is its specificity constant for 2PG?

Comment on the efficiency of this enzyme – is it “catalytically perfect”

2. The most promising molecule produced by your synthesis team is code-named BCEI-17. You repeat the assays exactly as in part (a), except that the tubes also contain BCEI-17. Here is the relevant data:

[2PG] in tube before reaction starts (µM)	amount of PEP (pmol) after reaction at [BCEI-17] (in nM)
	1	3	9	27	81	243
10	42.1	39.0	31.3	19.1	9.1	3.5
33	128	117	95.9	61.3	29.0	11.4
100	310	292	246	167	84.3	33.9
330	624	602	539	400	234	105
1000	876	844	820	706	503	265
3300	1010	992	967	920	793	558

What sort of inhibitor is BCEI-17: competitive, uncompetitive, or mixed?

What is the K_I and/or K_I' of BCEI-17?  

Answer 1

Given ;

2PG+H2O->PEP

Here there is increase in subtract concentration to check enzyme activity.

Total =95microL,adding fixed enzyme concentration of 0.1microg/m of 5microL to all test tubes

[S]+[E]<->[S]->[E]+[P]

2PG	Enolase	Time taken	PEP
10microM	0.1microg/ml	30sec	44.9(pmol)
33	0.1microg/ml	30sec	133
100	0.1microg/ml	30sec	320
330	0.1microg/ml	30sec	638
1000	0.1microg/ml	30sec	880
3300	0.1microg/ml	30sec	1020

.Kcat of enzyme =Amount of S->P/Time

.Km=Viacom/2,where Vmax=Kcat[E]

[S]	Kcat	Km
10microM	0.33microM/sec	0.005
33	1.1	0.055
100	3.3	0.165
330	11	0.55
1000	33.3	1.665
3300	110	5.5

.As the concentration of subtrate increasing at fixed enzyme concentration Kcat is increasing ans km is increasing

.Indicating increase in subtrate concentration,decrease the affinity of enzyme.

.Specificity constant of 2PG =Kcat/Km

10microL	66/sec
33	20/sec
100	20/sec
330	20/sec
1000	20/sec
3300	20/sec

.Specificity constant for 2PG is 20/sec

.Increase in concentration of subtrate increases kcat or turnover number and decreases the affinity of enzyme,so catalytic activity perfect at low subtrate concentration ,at high subtrate concentration it is not perfect.

.Now we add BCEI -17 to all the test tubes ,given

2PG	PEP+I	Vmax	Km
10microM	42.1	0.03	0.055
33	128	0.11	0.55
100	310	0.33	0.165
330	624	1.1	0.55
1000	876	3.33	1.665
3300	1010	11.0	5.5

Here increase in concentration of subtrate is favouring the inhibition,suggesting of competitive and uncompetitive enzyme inhibition.

.Here Vmax is unchanged,km is increasing suggestive of competitive enzyme inhibition.

.Ki represents affinity of inhibitor to inhibit the reaction .ki=kcat[I][E]