{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}


KeyMidterm1 - NAME Last First BIOLOGICAL SCIENCES 102...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
NAME: ____________________________ . Last, First BIOLOGICAL SCIENCES 102 MIDTERM EXAMINATION I January 26, 2009 Winter Quarter Dr. Marilynn Etzler and Dr. Ken Hilt Check the Appropriate Categories Below: Undergraduate ________ Graduate __________ Concurrent __________ INSTRUCTIONS: WRITE YOUR NAME AT THE TOP OF EACH PAGE There are six pages to this examination. Please count them before you start to make sure they are all present. Place your answer in the space provided beneath the question. Do not use any other paper; if you need more room, use the back of these sheets, but clearly state that you are continuing your answer on the back. PAGE VALUE SCORE 2 20 3 20 4 27 5 14 6 19 TOTAL 100
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Name: _Key_ _________ 1. Imidazole is often used as a laboratory buffer. It is a weak acid with a pK a of 7.0, and its structure is shown below. HN NH + H + + HN N Please answer the following questions: a) (3 points) Name the amino acid that has an “imidazole” group as part of its side chain. __Histidine __ b) An enzyme catalyzed reaction was carried out in 0.05 M imidazole buffer. The pH of the buffer at the beginning of the reaction was 7.0. At the end of the reaction the pH had decreased to 6.6. i) (3 points) Were H + ions consumed or produced during the reaction? __Produced __ ii) (10 points) How many moles of H + were consumed or produced per liter by the enzyme catalyzed reaction? (You must clearly show your work below if you would like partial credit for this question.) At beginning: 7.0 = 7.0 + log [Base]/[Acid] 0 = log [Base]/[Acid] 1 = [Base]/[Acid] 2x = .05 M x = .025 M = [Base] = [Acid] At end: 6.6 = 7.0 + log [Base]/[Acid] -0.4 = log [Base]/[Acid] .398 = .4 = log [Base]/[Acid] x + .4 x = .05 M 1.4 x = .05 M x = .036 M = [Acid] .036 M - .025 M = .011Moles acid produced per liter during reaction.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 6

KeyMidterm1 - NAME Last First BIOLOGICAL SCIENCES 102...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online