# 3 - Lecture 3 Announcements 1 Quiz next Wednesday 9/8 Dont...

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Lecture 3 August 30, 2010 Announcements: 1. Quiz next Wednesday 9/8. Don’t be late—we start just after 10:10. Be sure to talk to Prof. F. if you have any special problems. 2. Biology Open House: Wednesday, 9/1 from 4 - 5:30PM, 2 nd floor of Stimson Hall. Talk to faculty, student advisors, reps from pre-med, pre-vet, research clubs. Discuss courses, research, Life; and pizza…. 3. PyMOL: Office hours + reviews for PyMOL will normally happen in two places: Carpenter red computer lab: Sundays 3 - 5PM; “review session” in Malott Hall room 251 Tuesdays 4:30 - 6PM 202 Biotech PyMOL individual office hours, Tuesdays 8:45 - 10:45AM The first PyMOL assignment is the tutorial, pp. 302-311 of the LG. There you will find ALL instructions you need. You also need the CD. Go through it systematically. 4. Health careers orientation for juniors and seniors applying in 2010 (all talks at 4:35PM): HumEc: Sept 16 in 280 MVR CALS: Sept 20 in 145 Warren Hall Arts: Sept 21 in the HEC Auditorium of Goldwin Smith Hall Eng: Sept 22 in the Upson Hall Lounge Friday’s lecture: Properties of ionizable groups How to use the Henderson-Hasselbalch Equation to calculate the ratio [unprotonated]/[protonated] Notice that higher pKa means higher affinity for H + . Easy to convert from the ratio [unprotonated]/ [protonated] to the fraction, [unprotonated]/{ [protonated] + [unprotonated]}; e.g. from H-H Equation, find a ratio, which in Monday’s lecture was [COO - ]/[COOH] = 3. Re-write as [COO - ]/[COOH] = 3/1, which means that [COO - ] = 3 and [COOH] = 1, so the total [COO - ] + [COOH] = 4. Then the fraction of carboxyls that are protonated is [COOH]/{ [COO - ] + [COOH]} = 1/4. Prof showed that ionizable groups in proteins show a range of pKa values. Why should an ionization process have different tendencies?! Consider one example, the ionization of a carboxyl group. Look at three cases (3 different environments in a protein): COOH COO - + H+ pKa = 4 for this rxn in water (“isolated carboxyl”). Remember that the Ka is an equilibrium constant: the larger the equilibrium constant, the more the rxn tends to go to the right as written. Since the pKa is the negative log of the equilibrium constant, the smaller the pKa the more the ionization rxn goes to the right.

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Now consider the same rxn, but inside a protein that places NH 3 + next to the COO - . The positive charge of the amino makes forming the negatively charged COO - easier, so the pKa in this situation is 3 (think about it…). Now consider the same rxn, but inside a protein that places another COO - next to the COO - of our original ionization rxn. In this situation, forming a negative charge by ionization of the carboxyl group is more difficult than for the isolated carboxyl, because it would place a negative charge next to another negative charge. Now the pKa is 5. As another example of ionization behavior, H
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3 - Lecture 3 Announcements 1 Quiz next Wednesday 9/8 Dont...

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