Lecture 2/3-properties Of Biological Molecules I&II Flashcards
(32 cards)
Standard free-E* change equation
Delta G = -(R)(T)ln(Keq)
Gibbs Free E* equation
Delta G=Delta G*‘+ln([Prod]/[React])
Three types of non-covalent bonds
Hydrophobic interactions, van der Waal’s, electrostatic interactions
What produces the “hydrophobic effect?”
Water has a dipole moment, which results in a slight (+) and (-) charge on water. This allows for hydrogen bonding
How can one drive a reaction to the product side? What is this manipulation called?
Add reactant/remove products - Le Chatlier’s Principle (“Law of Mass Action”); also, add heat
The reaction of Glucose to Glucose-6-phosphate occurs by the enzyme hexokinase. What is the Delta-G prime for this reaction if there is 1 mole of glucose, and 1000 moles of G-6-P?
Delta G’ = -RT(ln[prod]/[react]) = -8.31(310)(ln[1000]/[1]]) = -2576.1xln([1000]/[1]]) = -17.8kJ/mol
The reaction of Fructose-6-P to Fructose-1,6-bisphosphate occurs by the enzyme phosphofructokinase. What is the Delta-G prime for this reaction if there is 100 mole of F-6-P, and 1 mole of F-1,6-BP?
Delta G’ = -RT(ln[prod]/[react]) = -8.31(310)ln([100]/[1]) = -11863.4 Jmol^-1/1000 = -11.9 kJ/mol
The reaction of Fructose-6-P to Fructose-1,6-bisphosphate occurs by the enzyme phosphofructokinase. What is the Delta-G prime for this reaction if there is 10 mole of F-6-P, and 1 mole of F-1,6-BP?
Delta G’ = -RT(ln[prod]/[react])
The reaction of Fructose-6-P to Fructose-1,6-bisphosphate occurs by the enzyme phosphofructokinase. What is the Delta-G prime for this reaction if there is 1 mole of F-6-P, and 1000 mole of F-1,6-BP?
Delta G’ = -RT(ln[prod]/[react])
What is the pH of a solution acetoacetic acid? The [A]=100 and [HA]=1 (pKa=3.58)
pH=pKa+log([A]/[HA] pH=3.58 + log[100]/[1] pH=3.58+2 pH=5.58
What is the pH of a solution acetoacetic acid? The [A]=1 and [HA]=1 (pKa=3.58)
pH=pKa+log([A]/[HA] pH=3.58+log([1]/[1]) pH=3.58+0 pH=3.58
What is the pH of a solution acetoacetic acid? The [A]=1 and [HA]=100 (pKa=3.58)
pH=pKa+log([A]/[HA] pH=3.58+log([1]/[100]) pH=3.58-2 pH=1.58
Energy can neither be created nor destroyed, only changed
First law of thermodynamics
The entropy of an isolated system not in equillibrium must increase
Second law of thermodynamics
First law of thermodynamics
Energy can neither be created nor destroyed, only changed
Second law of thermodynamics
The entropy of an isolated system not in equillibrium must increase
Delta G = -(R)(T)ln(Keq)
Standard free-E* change equation
Delta G=Delta G*‘+ln([Prod]/[React])
Gibbs Free E* equation
Hydrophobic interactions, van der Waal’s, electrostatic interactions
Three types of non-covalent bonds
Water has a dipole moment, which results in a slight (+) and (-) charge on water. This allows for hydrogen bonding
What produces the “hydrophobic effect?”
Add reactant/remove products - Le Chatlier’s Principle (“Law of Mass Action”); also, add heat
How can one drive a reaction to the product side? What is this manipulation called?
Delta G’ = -RT(ln[prod]/[react]) = -8.31(310)(ln[1000]/[1]]) = -2576.1xln([1000]/[1]]) = -17.8kJ/mol
The reaction of Glucose to Glucose-6-phosphate occurs by the enzyme hexokinase. What is the Delta-G prime for this reaction if there is 1 mole of glucose, and 1000 moles of G-6-P?
Delta G’ = -RT(ln[prod]/[react]) = -8.31(310)ln([100]/[1]) = -11863.4 Jmol^-1/1000 = -11.9 kJ/mol
The reaction of Fructose-6-P to Fructose-1,6-bisphosphate occurs by the enzyme phosphofructokinase. What is the Delta-G prime for this reaction if there is 100 mole of F-6-P, and 1 mole of F-1,6-BP?
Delta G’ = -RT(ln[prod]/[react])
The reaction of Fructose-6-P to Fructose-1,6-bisphosphate occurs by the enzyme phosphofructokinase. What is the Delta-G prime for this reaction if there is 10 mole of F-6-P, and 1 mole of F-1,6-BP?
