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1 Exercise Building Blocks
Before attempting this tutorial, you must have read chapter 1 (Building Blocks) and chapter 2 (Nutrients and Energy)
Problem 1:
Leucine has two dissociable protons: one with a pK1 of 2.3, the other with a pK2 of 9.7. Thus leucine will have different dissociation states at pH 1, pH 7 and at pH 11.
Task: Draw leucine with the correct dissociation states at each pH using the hint below.
Hint: Leucine has a total of 6 carbon atoms, the side-chain is branched at carbon-4 (counting from the carboxylgroup).
Once you have drawn the three structures you can look up the solution below.
Structural formulas can be found at the corresponding tabs.
Problem 2:
The concepts of free energy and equilibrium constant still appear very theoretical to you. To get more insight you order “phosphoglucomutase” from a chemical supplier. This enzyme converts glucose-1-phosphate into glucose-6-phosphate (this is an important step in the breakdown of glycogen). You make up a 0.1M solution of glucose-1-phosphate and add a small amount of phosphoglucomutase (a few grains freeze-dried enzyme). After equilibration has occurred, you determine the following concentrations: glucose-1-phosphate 4.5 mM; glucose-6-phosphate 95.5 mM.
Where is position 1 in the glucose molecule?
Rule:
- Number the carbon atoms consecutively from the end nearest the highest priority functional group or principal functional group (the one that will appear as a suffix).
- The functional group priority follows a strict hierarchy (e.g., carboxylic acids > aldehydes > ketones > alcohols > amines). Carbon bearing the highest priority functional group gets the lowest possible number (usually carbon 1).
How much glucose-1-phosphate (Na+ salt) do you have to weigh in to prepare 1L of a 0.1M solution (atomic weights C=12, H=1, O=16, P=31, Na = 23)?
Glucose-1-phosphate has 6 Carbon (12), 11 Hydrogen (1), 9 oxygen, (16), 2 Na (23), 1 phosphate (31); Total molecular weight = 304; 30.4g/L for a 0.1M solution.
What is the equilibrium constant of the reaction?
K’eq = Products/Substrates = 0.0955M/0.0045M = 21.2 (unit cancels out)
What is the free energy of the reaction?
Equation required: ΔG0’ = -RTlnK’eq ; R = 8.314J/mol K; T=298K
-8.314 J/(mol*K) *298K*ln 21.2 = -7567 J/mol
Do you spot the mistake in the figure?
Urate is an acid, the proton is leaving behind an oxygen with a negative charge (-), which is missing.
Do you think uric acid is a strong acid? Give a reason for your answer.
Probably not, strong acids dissociate completely. Urate can occur in different forms (see figure), of which the keto-form cannot dissociate a proton. Thus, urate cannot be completely dissociated.
Compare uric acid to building blocks in the body, from what molecule might it be derived from?
It is most similar to purine bases (see lecture). In fact, it is a breakdown product of purine bases.
Consider its low solubility, what problem might this cause if there is too much uric acid in the body?
Uric acid can precipitate in the urine and may cause kidney stones. It can also precipitate in the fluid surrounding joints causing painful gout.
