BL/CH401 EXAM 2 October 28^th 1996

1. 50 Pts. Enzyme Kinetics Problem

Draw 2 plots: (1) Vo vs [S]; and (2) 1/Vo vs 1/[S], using the data given below - 25 Pts
Determine Km and Vmax for uninhibited reaction (include units) - 10 Pts
Determine what type of Inhibitor A and B are using the Km' and Vmax' for the inhibitors (show your reasoning by comparing Vmax and Vmax' and Km and Km') - 10 Pts
Calculate the Ki for binding of Inhibitor A and B to the enzyme (include units) - 5 Pts

Use these equations for calculating the Ki

Competitive Inhibitor Km' = Km (1 + [I]/Ki)

Noncompetitive Inhibitor Vmax' = Vmax / (1 + [I]/Ki)

[S]	Vo (No I)	Vo (I = 1 mM A)	Vo (I = 2 mM B)
mM	µmol/min	µmol/min	µmol/min
0.0	0.0	0.0	0.0
1.0	30.0	15.0	15.0
2.0	45.0	22.5	25.7
5.0	64.3	32.1	45.0
10.0	75.0	37.5	60.0
50.0	86.5	43.3	81.8

2. 40 Pts. Protein Structure Problem (Use Words and Diagrams)

Describe the four levels of protein structure and the bonds that hold these structures together. Be sure to illustrate each level of the structure of a protein with a drawing of the bonds or interactions involving the polypeptide backbone and/or the amino acid side chains.

3. 10 Pts. Thought Question

Design an alpha-helix (list an Amino Acid Sequence with at least 12 amino acid residues) which has one side facing the aqueous environment surrounding a protein and one side facing the interior of the protein. Remember that alpha helixes have ~4 residues per turn, so that the 1^st, 4^th, 5^th, 8^th, 11^th, and 12^th amino acid side chains are on one side of the helix and 2^nd, 3^rd, 6^th, 7^th, 9^th, and 10^th on the other side.

ANSWERS:

1. 50 Pts. Enzyme Kinetics Problem

Vo vs [S] Plot:

1/Vo vs 1/[S] Plot

Table of Results:

Reaction	Km or Km'	Vmax or Vmax'	Type of Inhibitor	Ki
(Units)	mM	µmol/min		mM
No Inhibitor	Km = 2.0	Vmax = 90
I = A (1 mM)	Km' = 2.0	Vmax' = 45	NonCompetitive	Ki for A = 1
I = B (2 mM)	Km' = 5.0	Vmax' = 90	Competitive	Ki for B = 1.3

2. 40 Pts. Protein Structure

See Lecture Notes for Lectures 8 and 9

3. 10 Pts. Thought Question

Basically the aqueous environment surrounding the protein will require the amino acid side chains facing it on the alpha helix to be hydrophilic amino acids (polar ones) while the side of the helix faciing the interior of the protein will need to display hydrophobic side chains (non-polar ones).

The design of the alpha helix with one face hydrophilic and one face hydrophobic could be accomplished by placing hydrophilic amino acids on one side (ie. at positions 1, 4, 5, 8, 11 and 12 of the amino acid sequence) and hydrophobic amino acids on the other side (ie. at positions 2, 3, 6, 7, 9 and 10 of the amino acid sequence) or you could start with the hydrophobic side chains and follow with the hydrophilic on the other face of the helix. Either way is OK but my example below is done with hydrophilics first, then the hydrophobics:

1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9 || 10 || 11 || 12

Asp Val Ala Ser Gln Leu Ile Glu Trp Phe Thr Asn

Another way to illustrate this is with the Helix Wheel, which simulates the distribution of Amino Acid side chains in an alpha helix:

Basically a view looking down on helix from the central axis with side chains array around the helix.

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