Last day for SUBMISSION OF REGRADES is Friday April 6, 2001. 
No regrade requests will be considered after this date.
Submit regrade in writing to Dr. Lukas Buehler, York Hall 3080C (mail box).
Starting April 9, exams will be left outside Bonner Hall 2402. Exams without signature can be picked up by appointment only after April 6 (e-mail to L.Buehler)

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Class Average  121 points (±33 points)
(Final Exam average: 77±21 points)

Note: Grades have been determined on a scale from 0 to 200 points. Bonus points are added after this calculation. No plus or minus assigned. Grade is written on front page of exam. The following lower thresholds (cutoffs) are needed to receive a particular grad:

155 points or more for A
120 to 154 for B
85 to 119 for C
50 to 84 for D

Please check your exam carefully and compare it to key provided below before requesting a regrade.

How many cystein pairs do you find in this fragment? 3p
6 SS bonds (also accepted 2 source Branden&Tooze) 

What non-protein structure is there and how many copies of this structure do you find? 2p
carbohydrates, N-linked glycosylation; 2 copies

Give the full name, three letter, and single letter code for this amino acid. 6p
Proline, Pro, P

a.)  Which experimental technique used for high resolution determination would you choose to characterize Protein X?  The final end product of this technique would be a(n) ________________________ protein structure.

X-ray crystallography (3pts)
Frozen, static, crystalline structure (3pts)

b.)  Upon further analyses of the amino acid sequence of Protein X, 4 regions were identified to be completely homologous to each other.  These regions contain about 24 amino acids, mostly composed of A, L, and M residues and are each flanked by a combination of Y and K residues.  Predict the location of these 4 regions in a mammalian cell.

4 transmembrane regions.  I would also accept a topology diagram. (2pts)

c.)  Determine the effects of multiple amino acid substitutions of aspartate residues in regions containing A, L, and M residues.  What would be the effects of this same point mutation (an aspartate residue) if it occurred in the flanking regions?

A to D substitution would inhibit the regions from forming transmembrane domains (1pt)
Y to D would have no effect- both polar and could flank the hydrophobic domains (1pt)

Sequence 1: gctagcctagcctttatacg 

Sequence 2: cgaattagaactgtatataa

B. Because nothing in life is that easy, please explain how you arrived at this conclusion by describing a major property of B-DNA that affects Tm (5 points). Be sure to include how you calculated which one has the lower Tm (4 points).

Answer:
A major factor that affects Tm is the amount of guanines and cytosines base pairs in the DNA molecule. The more GC base pairs there are, the harder it is to pull the two strands of DNA apart, because a GC pair forms three H-bonds (compared an AT base pair which form 2 H-bonds.) Therefore, a DNA molecule with more AT base pairs will be much easier to denature/melt than a DNA molecule which contains a high percentage of GC base pairs. (5 points) 
By counting the number of G's and C's in each sequence and comparing each, you can see Sequence 2 has the lower GC content and a lower Tm. (4 points)

Sequence 1: % GC= 50% (#of g and c/total # of bases) (10gc /20 total bases) 
Sequence 2: %GC =25%  (#of g and c/total # of bases)(5gc /20 total bases) 
Sequence 2 has a lower Tm. 
 
 

B) What secondary structural motif found in structure A? (2 points)
Antiparallel b-barrel (2 points)

C) Describe structure B in terms of the number of subunits, and their orientation relative to each other and to the membrane. (4 points)
H, L, M, and cytochrome. (1 point) 
L & M are in the membrane. (1 point)
H has one transmembrane helix, and associates with one face of  L & M. (1 point)
Cytochrome is bound to the opposite face of L & M.  It has no transmembrane segments. (1 point)
 
 

structure A = beta sheet; 
upper left quadrant in ramachandran plot

structure B= right handed alpha helix
lower left quadrant in ramachandran plot

B)  For which of these structures would hydropathy plotting have given an accurate estimate of transmembrane segments?  Why is this – i.e. what structural elements are recognizable in a hydropathy plot? (5 points)
Photosynthetic reactions center. (3 points)
Hydropathy plots identify hydrophobic a-helices (2 points)
 

Ligand binding site is 25 angstrom apart from channel gate (open to closed transition); channel is pentameric complex spanning the membrane; opening and closing of channel is controlled by ligand binding to receprtor; the ligand binding site is different from the channel gate and the control mechanism has to be transmitted across the protein structure from binding site to membrane located channel opening.
 

Cooperativity = sigmoidal curve
Non-cooperativity = linear curve

Short helix is part of pore loop structure folded into the center of a tetrameric channel comples and shielded from lipids by transmembrane alpha helical segments; contributes K-binding through helical dipole moment;
 

Sialic acid removal from glycosylation triggers binding to asialoglycoprotein receptor and endocytosis (part of protein turnover); 
N-linked glycosylation

A. Why is it important to have both Hemoglobin and Myoglobin in the body?  Your answer should include a description of how oxygen is unloaded at tissues, as well as a binding curve for Mb and Hb. (6 points)
At high arterial pressure, O2 has a high affinity for  Hb and Mb.  As pressure decreases (venous blood), O2 loses affinity for Hb, but still saturates Mb.  (1.0 pts)(This is illustrated by binding curve).  The result:  at muscle, O2 is unloaded from Hb, and trapped in muscle by Mb for use by the actively respiring tissues. (0.5 pts)"

0.5 pts for Hb and Mb are O2 carriers in the body (OR
that O2 is insoluble in blood)
0.5 pts for Hb in blood
0.5 pts for Mb in muscle

3.0 pts for binding curve (1 pt for axes, 1 pt for Hb
curve, 1 pt for Mb curve)

1.5 pts for full explanation of unloading.

 

B. In active tissues, the environment becomes more acidic.  How does this affect your above binding curve?  Draw a NEW binding curve showing the normal pH (7.4) AND showing the change, if any, to Hb and Mb at a lower pH.  (3 points)

1.5 pts for Hb curve shift right (= decrease O2
affinity)
1.5 pts for Mb curve no change

C. Name one disease that can be caused by  mutated Hemoglobin.  (1 point)
1 pt for sickle cell or other
acceptable thalassemia
 

An anemic individual, whose blood has only half the normal hemoglobin content, may appear in good health.  Yet, a normal individual is incapacitated by exposure to sufficient carbon monoxide (CO) to occupy half his heme sites (Note: carbon monoxide behavior is in all aspects identical to that of oxygen except that it binds to Hb with 200 times greater affinity than does O2). 

Explain this:

For anemia, the Hb present functions normally and is presumably present in sufficient quantity to carry enough O2 (at least under low exertion).  For CO poisoning, half the Hb is “irreversibly” bound to CO (affinity 200x greater for CO than O2).  Because CO occupies  binding sites in Hb, most Hb is converted to the R state, which has increased affinity for O2 over normal Hb (remember your graph above?)  Therefore, in tissues, little of the O2 that is carried can be released.  Therefore, the victim asphyxiates.