3 - Lecture 3 Announcements 1 Quiz next Wednesday 9/7 Dont...

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Lecture 3 August 29, 2011 Announcements: 1. Quiz next Wednesday 9/7. Don’t be late—we start just after 10:10. Be sure to talk to Prof. F. if you have any special problems. One copy of text is being put on 2-hour reserve in Uris Library (should arrive soon). 2. Biology Open House: Wednesday, 8/31 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. Interested in joining TRP The Research Paper? Organizational meeting Thursday 9/1/11 from 4:30 - 5:15PM in 115 Rockefeller Hall. This magazine describes research of undergraduates at Cornell. Positions are open! 4. 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 The first PyMOL assignment is the tutorial, pp. 303-312 of the LG. There you will find ALL instructions you need. You also need the CD. Go through it systematically. 5. Health careers orientation for juniors and seniors applying in 2012 (all talks at 4:35PM): HumEc: Sept 6 in 153 MVR CALS: Sept 7 in 145 Kennedy Hall Arts: Sept 8 in the Kaufmann Auditorium of Goldwin Smith Hall Eng: Sept 12 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 about 3 (think about it why it should be less than 4).
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3 - Lecture 3 Announcements 1 Quiz next Wednesday 9/7 Dont...

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