-citrate
1. 10uM Mean 9.08 SD = 0.024
2. 5uM Mean 7.68 SD = 0.016
3. 3.33uM Mean 6.67 SD = 0.026
4. 2.5uM Mean 5.88 SD = 0.02
5. 1.67uM Mean 4.34 SD = 0.04
6. 1.25uM Mean 3.77 SD = 0.026
+ citrate
1. 10uM Mean 11.75 SD = 0.079
2. 5uM Mean 9.52 SD = 0.041
3. 3.33uM Mean 8.33 SD = 0.012
4. 2.5uM Mean 7.41 SD = 0.028
5. 1.67uM Mean 6.67 SD = 0.036
6. 1.25uM Mean 6.25 SD = 0.049
S | V -citrate | V +citrate | 1/S | 1/V -Citrate | 1/V +Citrate |
10 | 9.08 | 11.75 | 0.1 | 0.1101322 | 0.08510638 |
5 | 7.68 | 9.52 | 0.2 | 0.1302083 | 0.10504202 |
3.33 | 6.67 | 8.33 | 0.3003 | 0.149925 | 0.12004802 |
2.5 | 5.88 | 7.41 | 0.4 | 0.170068 | 0.13495277 |
1.67 | 4.34 | 6.67 | 0.598802 | 0.2304147 | 0.14992504 |
1.25 | 3.77 | 6.25 | 0.8 | 0.265252 | 0.16 |
The PFK1 enzyme is regulated by a number of factors in the cell, including the level of ATP. PFK1 is a homo- tetramer (four identical subunits) and ATP binds to multiple sites. The binding of ATP can...
BACKGROUND The first part of this assignment focuses on the thermodynamics of enzyme-catalysed reactions. To answer the questions posed you will need apply the following two equations. Note that you may need to re- arrange these equations to determine the value you need. You also need to take care with the units of the values you use and determine. The first equation defines the relationship between the change in free energy (AG ") and the equilibrium between the products and...
Question 1 - Linked Reactions (10 marks) The reaction catalysed by pyruvate kinase is: Phosphoenolpyruvate+ ADPpyruvateATP Keg 3.63 x 105 a) Calculate the AG" for this reaction. Show your working. 3 marks b) The hydrolysis of ATP has following equation: AG 30.5 kJ/mol Calculate the AG" for the following reaction: Phosphoenolpyruvate pyruvateP Show your working. 2 marks c) At 37 °C, the steady-state concentrations of phosphoenolpyruvate, ATP, and ADP have been measured to be 23 HM, 1.85 mM and 140...