Key HW 3 - BIS 102 HW 3 Answers for Dr. Hilt PDF

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BIS 102 HW 3 Answers for Dr. Hilt...

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Biological Sciences 102 Winter, 2014 K. Hilt

Answer Key Homework #3

Assigned text problems: Biochemical Calculations (Segel)

4-4 on pp. 223-224 and 4-9 on pp. 237-238 pp. 319 #2, 3, 6, 10, and 12. You need to be able to plot and interpret Lineweaver-Burk graphs. 1. Enzyme Z catalyzes the reaction: A + NAD+ ↔ B + NADH + H+. Besides following the [H+] using a pH meter, the reaction can be monitored spectroscopically at 340 nm. This is the wavelength at which NADH absorbs light. Some of the first enzymes that were worked on in biochemistry were dehydrogenases, like our enzyme Z, because of this advantage. Radioisotopes were not yet available. a) Diagram what the kinetic assay results would be if the reaction was measured from left to right. What would it be if the assay was measured from right to left?

[NADH]

left to right (red line) right to left (green line) Time

b) Describe how you would accurately determine enzyme Z's Km for NAD+. Describe the experimental set-up and the analysis of data. Assay a series of tubes where the concentrations of buffer, compound A, and enzyme Z were constant. Vary the concentration of NAD+. Make a Lineweaver-Burk plot and determine the Km from the x-intercept:

1/vo

1/[S]

c) Imagine that twenty I.U.’s of enzyme Z were catalyzing the above reaction for one minute, under Vmax conditions, in a 3.00 ml assay volume. The assay is buffered with 20 mM phosphate buffer, pH 7.60. What will the pH be at the end of that one minute? pH 7.0 Use two Henderson-Hasselbalch equations. This question is very similar to questions #40 and #41 in Biochemical Calculations (Segel), on page 93, which were assigned in homework #1.

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2. You are going to measure the enzyme activity of enzyme “T”. Enzyme “T” catalyzes the reaction: S ↔ P. Enzyme “T” has a Km of 1.0 x 10-6 M for S. You mix the following solutions together: 2.60 ml 10 mM phosphate buffer, pH 7.2 0.30 ml 30 mM S 0.10 ml enzyme “T”. The results of this enzyme assay are plotted below:

nmol P produced in assay

30

25

20

15

10

5

0 0

10

20

30

40

50

Time (seconds)

a) What is vo for the above reaction? Take any two points and calculate the slope. b) How many International Units (I.U.’s) of enzyme “T” were present in the assay? 0.030 I.U. c) Carefully sketch in, on the above graph, the expected initial velocity if the substrate concentration had initially been 1.5 x 10-6 M. Draw a line on the graph with a slope of 18 nmol/min. Note: this line can begin anywhere, but needs to have this slope. d) Let’s assume that 1.0 x 10-12 mol of enzyme “T” was present in both assaysi.e. in part “a” and in part “c”. Calculate the turnover number for enzyme “T”. Turnover number = k2 = 30,000 min-1 Note: the turnover number always refers to the optimal velocity, i.e. Vmax; optimal substrate concentration, optimal pH, etc. The turnover number for the reverse reaction would be k -1.

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3.

The following table lists properties of four different proteins:

Protein chymotrypsinogen hemoglobin lysozyme ovalbumin

pI 9.4 7.0 11.1 4.9

MW 23,200 64,500 14,100 45,000

a) What is the order of elution of the above proteins from a sizing column at pH 7? Hemoglobin, ovalbumin, chymotrypsinogen, lysozyme b) What is the order of elution from an anion exchange column at pH 7? Lysozyme, chymotrypsinogen, hemoglobin, ovalbumin c) What is the order of migration of the above proteins in a SDS-PAGE gel? Lysozyme (fastest), hemoglobin, chymotrypsinogen, ovalbumin Note: on a SDS-PAGE gel, hemoglobin will dissociate into its four subunits. We must assume that the other proteins do not have subunits, which is the case.

d) How will these proteins band in an agarose IEF gel? (+ electrode) Ovalbumin, hemoglobin, chymotrypsinogen, lysozyme (- electrode) Note: the positive electrode always has HCl present as a H+ reservoir. Hence, the pH gradient begins at a low pH at the + electrode. 4. Part of the H+-portion of the Bohr effect occurs at histidine 146 at the C−termini of the β-subunits in hemoglobin. In hemoglobin Hiroshima, that histidine has been changed to an aspartic acid residue. a) How will hemoglobin Hiroshima migrate (band), compared to normal hemoglobin, in a native PAGE gel? Why? Hb Hiroshima will run faster. In our gel, negatively charged proteins are migrating towards the positive pole at the other end of the gel. The substitution of D for H will increase the q/m ratio of Hb Hiroshima, but will have no noticeable effect on the mass of the protein. b) How will hemoglobin Hiroshima migrate (band), compared to normal hemoglobin, in an SDS- PAGE gel? Why? They will migrate the same distance. The substitution of D for H will not noticeably affect how many SDS molecules bind to the denatured protein. Remember that thousands of SDS molecules bind to each denatured subunit of Hb. Each SDS molecule that binds brings with it a

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full -1 charge. The charges from the bound SDS overwhelm whatever the initial native charge was of the protein. c) How will hemoglobin Hiroshima migrate (band), compared to normal hemoglobin, in an isoelectric focusing gel? Explain. Hb Hiroshima will band closer to the (+) electrode. Remember, in IEF gels, proteins retain their native 4o structure.

5.

Calculate the missing values (X) in the following enzyme purification table:

Fraction 1. crude 2. resuspended 40% (NH4)2SO4 pellet 3. affinity column

Volume (ml) 1000 200

2

[Protein] (mg/ml) 10 20

1

I.U./ml 2 12

Total I.U. 2000 2400

Activity Yield (%) 100 120

Specific activity 0.2 0.6

Fold Purification 1.00 3

200

400

20

200

1000

How do you explain the value you calculate for X2 (i.e. why is it greater than 2000 I.U.)? The ammonium sulfate step may have removed an inhibitor of the enzyme. Alternatively, the ammonium sulfate step may have removed a competing enzyme, one that converted the product that we were trying to measure to something else.

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6. A plant is growing in the jungles of South America. Would the fatty acid composition of its membrane lipids contain more 18:3 (9,12,15) or more 18:0? Explain. 18:0. The jungles are hot. Heat can make the membranes too fluid. To reduce fluidity, the plant needs fatty acid components with fewer cis double bonds and hence more van der Waals interactions between the lipid tails.

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What is this all about?

Image from Voet and Voet, Biochemistry, 3rd ed., c2004

A beautiful model of a membrane bilayer. Water molecules are shown in blue. Notice how some of the water molecules have momentarily slipped into the phospholipid bilayer. This reminds us of the dynamic nature of the bilayer. Small gaps are always occurring between neighboring phospholipids and proteins in the membrane....