2) (5 pts) Explain why typical long-chain fatty acids (shown below) form soluble

Question

2) (5 pts) Explain why typical long-chain fatty acids (shown below) form soluble micelles in solutions with pH> 7 but are insoluble in solutions with pH 3. Use the Henderson-Hasselbalch equation to aid your explanation pk 5 Long-chain fatty acid (R long alkyl group) 3) (5 pts) Calculate the percentage of histidine (shown below in fully protonated form) that has its imidazole ring protonated at pH -7.3 imidazole HN ring NH pko 1.8 4) (5 pts) At 30 °C and pH-2.5, the change in enthalpy and change in entropy for the denaturation of the enzyme ribonuclease are 240 kJ/mol and 0.780 kJ/mol K, respectively. What is the temperature at which the native (folded) and denatured forms of ribonuclease are in equilibrium? Assume enthalpy and entropy do not change with temperature

Explanation / Answer

Fatty acids have both polar and nonpolar regions which form aggregates in aqueous solution called miscelles.In missiles the polar head forms the outer surface and nonpolar faces toward the interior .To form miscelle it must have a charged polar head group like Acetate ion this will only happen in if pH is greater than pk and when pH is less than 5 no acid is ionized to give charge head hence no miscelle is formed. This can be understood by Henderson Equation

pH = pKa + log [Ionized form]/[unionized form]

If ph> 7 let say 8

8 = 5 + log [IF]/[UF]

3 = log [IF]/[UF]

Log (1000) = log [IF]/[UF]

1000 = [IF]/[UF]

% IF = IF/IF+UF

IF=(IF/UF)/(IF/UF+1)=1000/1001×100 =99%

We can use the Henderson-Hasselbach equation to determine the percentage of histidine that has its imidazole side chain protonated at pH 7.3. pKa value for the imidazole group is 6.

pH = pKa + log ([A]/[HA+])

7.3 = 6 + log ([A]/[HA+])

log ([A]/[HA+]) = 1.3

([A]/[HA+]) = 101.3 = 19.95

Percentage of histidine = 1/(1+19.95) = 1/20.95 = 4.77%

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