Problems based on biochemical calculations

Question 1.:

Aldolase catalyzes the glycolytic reaction

Fructose 1,6-bisphosphate       →            glyceraldehydes 3-phosphate + dihydroxyacetone phosphate

The standard free-energy change for this reaction in the direction written is +23.8 kJ/mol. The concentrations of the three intermediates in the hepatocytes of a mammal are: fructose 1,6-bisphosphate, 1.4x10^-5M; glyceraldehydes 3-phosphate, 3x10^-6M; and dihydroxyacetone phosphate, 1.6x10^-5M. At body temperature (37°C), what is the actual free-energy change for the reaction?

Solution:

Fructose 1,6-bisphosphate             →             glyceraldehydes 3-phosphate + dihydroxyacetone phosphate

[A]                                                                   [B]                                           [C]

1.4x10^-5M                                                  3x10^-6M                                  1.6x10^-5M

ΔG’°= +23.8 kJ/mol

T= 37+273= 310 K

ΔG= ΔG’°+ RT ln [B][C]/[A]

Put all given values in above equation

Product of R and T will divide by 1000 to get value in kJ/mol

After putting all values

ΔG = 23.8+2.58 ln 3.43×10^˗6

Value of ln 3.43×10^-6 = ˗12.6

ΔG = 23.8+2.58 × (˗12.6)

ΔG = 23.8 + (˗32.46)

ΔG = ˗8.6 kJ/mol

After calculation we are able to find actual free-energy change i.e. ˗8.6 kJ/mol for given reaction.

Question 2.: protein has one tryptophan and one, tyrosine in its sequence. Assume molar extinction coefficients at 280 nm of tryptophan and tyrosine as 3000 and 1500 M^-1cm^-1respectively. What would be the molar concentration of that protein if its absorption at 280 nm is 0.90?

Solution:

A= εcl

A= absorbance; ε= molar extinction coefficients; c= concentration; l= pathlength

A= 0.90

ε in this question is

3000+1500= 4500 

l= 1 cm

Therefore,

c= A/ε

c= 0.90/4500

                                    c=.0002M= 0.2mM (Answer)