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EK In-class Exam 7
Divya_4978
#1 Posted : Tuesday, June 16, 2020 1:15:43 AM
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Hi,
I do not seem to understand passage 4 question 152 (page 309) in the in-class EK topic acids and bases. I do not understand how to go about finding the pKa of arginine's side chain according to the graph.

I also do not get question 155. I am finding it hard to determine the answer between A and B.

Finally, question 156. In the solutions I don't get how we know that lysine has a lower pKa than arginine and how it leads to concluding the answer.

Thanks in advance.
INSTR_Molly_59
#2 Posted : Tuesday, June 16, 2020 5:16:52 PM
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Hello Divya,

152: We see a drastic decrease in fluorescence of enzyme after pH of 12. This tells me that something has changed with the enzyme. Likely, an effect on the charge as a result of pH. pH effects comes into play if pH is less than pKa or pH is more than pKa, so I know that between these two values, the pH was at one point lower than the pKa, and at one point higher than the pKa (so between 12 and 13). This suggests that the pKa of the enzyme is 12.5 Answer is C.

155: The mutant GFP14R strain is more resistant to the pH change at pH of 13 (bigger difference between the two bars), suggesting that the pKa is likely closer to 13 than the original strain, and higher than the wildtype. Answer is A.

156: This is a follow-up of the last question. We know that arginine has a higher pKa. We don't talk about stability of conjugate acid; we only look at stability of conjugate base to determine acidity. Since we know that arginine is higher pKa than lysine, arginine is more basic, lysine is more acidic. That means, lysine would have a more stable conjugate base. Answer is A.

#156 was tricky! If you managed to follow the logic all the way through, then you definitely got pKas/pHs down pat! Message again if you need more assistance :)
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