appendix i practiceproblems s2016

8/18/2019 Appendix I PracticeProblems S2016

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Practice Problems: Experiment 1

Consult as needed: Table 1.1 pK a values for buffers in dilute solution

1.1. What volume of glacial acetic acid (17.6 N) and weight of sodium acetate!3 H2O (F.W. = 136) is

required to make 100 ml of 0.20 M buffer, pH 4.06? (Note: F.W. stands for formula weight in g/mol).

1.2. a) What volume of glacial acetic acid (17.6 N) and weight of sodium acetate!3 H2O (F.W.= 136) isrequired to make 100 ml of 0.20 M buffer, pH 5.06?

b) How is 100 ml of 0.050 M buffer, pH 5.06, prepared from a stock of 0.20 M buffer, pH 5.06?

1.3 What quantities of reagents are needed to make 250 ml of a 0.050 M acetate buffer at pH 5.5?

Available reagents: 1.0 N HCl, 1.0 N KOH, CH3COOK (FW 98.15, pKa 4.76)

1.4. a) What weights of sodium carbonate (FW = 106) and sodium bicarbonate (FW = 84) are requiredto make 100 ml of 0.20 M buffer, pH. 10.38?

b) How is 100 ml of a 0.080 M buffer, pH 10.38, prepared from a stock of 0.20 M buffer, pH 10.38?

1.5. What volume of hydrochloric acid (11.7 N) and weight of Tris (MW = 121) are required to make

100 ml of 0.20 M buffer, pH 8.71?

1.6. What volume of concentrated HCl (11.7 N) and weight of Tris base (MW = 121 g/mole) is neededto make 100 ml of 0.20 M buffer, pH 8.45?

1.7. a) Glycine has both weak acid (the primary amine) and weak base (the carboxyl group) properties,

making it a useful as a buffer. What are the pKa’s of glycine’s amino and carboxyl groups? What isthe structure of pure glycine in H2O?

b) What is the pH of 0.2M glycine in H2O?

c) If 50 mls of 0.20 M glycine is mixed with 50 mls of 0.1 M KOH, what is the resulting pH?

d) How will the pH change if the temperature is increased above 200C?

1.8. a) If 50 mls of 0.20 M glycine is mixed with 50 mls of 0.1M HCl, what is the resulting pH?

b) How will the pH change if the temperature is increased above 200C?

1.9. a) Phenol Red is a weakly acidic dye which is yellow in the HA form and red in the A form. What

concentration of Phenol Red is required to obtain an absorbance of 0.50 at 550 nm at pH 7.8?

Note: "550 of A = 2.5 x 104 M-1 cm-1; "550 of HA = 0; pKa = 7.8..

b) What would the absorbance of the solution in part (a) be if the pH was 8.8?


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1.10. Determine the pKa of a dye. The dye was diluted into several different buffers and the A450 was

measured (Table I). Using the data in Table I along with the "450 (anionic form) = 50,000 M-1

cm-1

find the pKa of the dye.

Table I

1.11. Calculate the pKa of m-ultra red when equimolar amounts of the pH indicator m-ultra red are

mixed in separate test tubes with large molar excesses of buffers at seven different pH values. All

buffer plus dye solutions were diluted to the same final volume and the absorbances at 550 nm wereread.

equimolar concentrations of

m-ultra red at various pH values

Challenge Problem

1.12. When an enzyme catalyzes the hydrolysis of X to Y, one mole of protons is released per mole ofsubstrate utilized. When this reaction was carried out at 24°C in a 3 ml reaction mixture buffered only

by Tris buffer at an initial pH of 8.5, 20 µmoles of X were convertedto Y. What is the minimum concentration of Tris buffer required to prevent the pH from decreasing

below 8.3?

Buffer Buffer volume,ml

Dye stock volume,ml

A450

0.01 M acetate, pH 4.5 4.5 0.5 0.00

0.01 M HEPES, pH 7.0 4.5 0.5 0.32

0.01 M Tris, pH 8.0 4.5 0.5 0.81

0.01 M carbonate, pH 10.0 4.5 0.5 1.00

0.01 M NaOH 4.5 0.5 1.00

pH A550

0 0.00

1 0.00

2 0.00

6 0.54

10 1.40

11 1.40

12 1.40


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2.1. What are the units of absorbance, A?

2.2. A 6.00 mg/l solution of adenine in water has an absorbance of 0.590 at 260 nm in a 1-cm cuvette.

Adenine has a molecular weight of 135.1. What is its molar extinction coefficient, !, at 260 nm?

2.3. a). A 1.00 mg/ml solution of bovine serum albumin (BSA) in neutral phosphate buffer has an A260

of 0.600. If the molecular weight of BSA is 66,000 Daltons, what is its molar extinction coefficient?

b). A 1.00 mg/ml solution of bovine serum albumin (BSA) in neutral phosphate buffer has an A280of 0.667. If the molecular weight of BSA is 66,000 Daltons, what is its molar extinction coefficient?

c). At what wavelength is the absorbance of the 1.00 mg/ml solution of BSA the most sensitive?

2.4. Calculate the molar absorption coefficient of NADH at 340 nm from the spectral data collected on a

16.5 mg/l solution of NADH (see figure below). Molecular weight of NADH = 665 g/mol.

2.5. For guanosine, !275 = 8400 M –1 cm –1 and !245 = 9150 M

–1 cm – 1 in 0.1 M HCl. In the same

solvent, tyrosine has an !275 = 1340 and !245 = 170. A mixture of these compounds in 0.1 M HCl hasan A275 = 0.69 and an A245 = 0.49. What are the molar concentrations of the two solutes?

2.6. One ml of a guanosine solution was mixed with 4 ml of 0.1 M HCl. When 0.07 ml of this solution

was added to 1.5 ml of 0.1 M HCl in a cuvette, the absorbance at 275 nm was 0.50. What was themillimolar concentration of guanosine in the original solution? The appropriate molar extinction

coefficient is given in problem 2.5.


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2.7. A pure zinc-containing protein was found to have an absorption coefficient of 1.5 (mg/ml) –1 cm –1

at 280 nm. A concentrated solution of the protein in water was found to contain 7.2 µg/ml zinc. A 0.1ml aliquot of the same solution was diluted with 0.9 ml of water (dilution A) and 0.2 ml of dilution Awas added to 1.0 ml of water (dilution B). Dilution B had a measured A280 = 0.375. What are the

minimum molecular weight and minimum molar absorption coefficient of the protein? The atomicweight of zinc is 65.4 g/mol. ( Hint: minimum molecular weight assumes only one mole of Zn binds

to one mole of protein; so, finding the moles of Zn in the protein solution will give the moles of protein).

2.8. A solution of unknown protein concentration was assayed by the Bradford protein assay, in 0.5 ml

assay volumes, in a 0.5 ml cuvette with a 1 cm path-length. Absorbances were read in a Shimadzuspectrophotometer at 595 nm. Below are the data obtained from the assay. Fill in the table, plot the

standard curve, and determine the protein concentration from the data.

Bradford Protein Assay

tubesBSA standard

or unknown sampleH2Oµl

reagentµl

averageA595

averageA595 –

blank

µg / µl protein

(sample vol)

1,2 0 µl 0.2 µg/µl BSA 250 250 0.151

3,4 10 µl 0.2 µg/µl BSA 240 “ 0.203

5,6 20 µl 0.2 µg/µl BSA 230 “ 0.258

7,8 30 µl 0.2 µg/µl BSA 220 “ 0.311

9,10 40 µl 0.2 µg/µl BSA 210 “ 0.351

11,12 50 µl 0.2 µg/µl BSA 200 “ 0.401

13,14 250 µl unk, 1/20 dil 0 “ 0.155

15,16 250 µl unk, 1/6 dil 0 “ 0.271

17,18 250 µl unk, 1/2 dil 0 “ 0.434


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2.9. A solution of unknown protein concentration (UNK) was assayed by the Bradford protein assay

using the microplate assay protocol. Below are the data obtained from the assay. Fill in the table, plotthe standard curve, and determine the protein concentration from the data.

2.10. A new dye-binding protein assay is compared with the biuret protein assay. The new procedurecalls for mixing 1.5 ml of protein solution with 1.5 ml of reagent and reading the A630 after 3minutes. To 0.5 ml of a BSA solution was added 1.5 ml of water. The assay was performed on 1.5 ml

of this diluted BSA solution, giving an A630 of 0.470. One-half ml of the original BSA solution wasmixed with 0.5 ml of water and 4.0 ml of biuret reagent. After 30 minutes the A540 of the solution

was 0.305. Assuming the same optical paths for both readings, proper use of blanks, both assays intheir linear ranges, etc., what is the relative sensitivity of the dye-binding and biuret assays?

Problems 2.11 – 2.14 are challenge problems.

2.11. It is possible to study tight binding of a small cofactor to an enzyme by mixing the two in solution

and then centrifuging to pellet the enzyme-cofactor complex. The experiment is normally performedwith excess cofactor. Unbound cofactor remains in solution after centrifugation and washed away. It

is assumed that any cofactor found with the enzyme pellet is bound.

This study was performed using 20.0 mg of an FMN binding enzyme with M.W. 200,000 and anexcess of FMN. The entire pellet of enzyme-FMN was dissolved in water to give a total volume of 15

ml. The absorbance of this solution at both 260 nm and 280 nm were A260 = 0.95; A280 = 1.33.

How many moles of FMN are bound per mole of enzyme in the pellet?

sample volume

Bradford

reagent

µg / µl BSA

(sample

volume)

A595

average

A595 average

minus blankduplicates

BSA (0.024 µg/µl)

or UNK sample

H2O

0.0 µl BSA 125 µl 125 µl 0.398

10.4 µl BSA 114.6 µl “ 0.464

20.8 µl BSA 104.2 µl “ 0.518

41.7 µl BSA 83.3 µl “ 0.612

62.5 µl BSA 62.5 µl “ 0.689

83.3 µl BSA 41.7 µl “ 0.753

105 µl BSA 20 µl “ 0.818

125 µl BSA 0.0 µl “ 0.848

125 µl (1/3) dilution of UNK “ 1.211









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If the enzyme bound one FMN per subunit, then what is the estimated protein subunit molecular

weight?

2.12. In (A) Spectrum of Protein or Nucleic Acid , are the separate spectra of aqueous solutions of known

concentrations of protein or nucleic acid; in (B) Spectrum of a Mixture, the spectrum is of a mixtureof protein and nucleic acid. Calculate the protein and nucleic acid concentrations in the mixture (B).

(A) Spectrum of Protein or Nucleic Acid


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(B) Spectrum of a Mixture (of protein and nucleic acid)

2.13. A new procedure (an assay) for the determination of inorganic phosphate is available. A bottle ofthe assay reagent, Phosphate Reagent (a liquid), includes a brief method written on it’s side:

"Mix one part of reagent with two parts of water. Add one part of diluted reagent to one part of a

phosphate containing solution, heat at 100oC for ten minutes, cool to room temperature, and read at660 nm. The reagent+phosphate chromophore is stable at room temperature; and has an absorption

coefficient, a = 1.24 x 105 (mg P/ ml) –1 cm –1 (P is elemental phosphorus --- atomic weight = 31)".

Write-up a protocol to assess the useful range of phosphate for the assay. Consider the following points:

a) What range of molar potassium phosphate concentrations (in the cuvette) is used for the standardcurve? Explain and show calculations.

b) If one working stock solution of phosphate is made for the preparation of the standard curve, whatconcentration would be chosen? Why?

c) Using the answer to b above, set up a protocol table for a standard curve.d) How might the method be modified to increase its sensitivity so that the phosphate content of very

dilute solutions cab be determined accurately?

2.14. In Dr. Fairclough’s lab"

-Bungarotoxin ("

-Btx, MW = 8000 protein) has been labeled with a dye:carboxytetramethylrhodamine hydroxysuccinimide ester (Rh). The unreacted dye is removed with aP-2 gel filtration column. Assume that this procedure removes all non-covalently associated dye

from the protein; only covalently attached dye remains and the protein is highly fluorescent.

An initial characterization of the labeled Btx derivative, Rh-"-Btx, involved recording a visible

absorption spectrum, Rh-"-Btx Spectrum, of a 1/50 dilution of the Rh-"-Btx solution. Also, using a

1.0 mg/ml unlabeled "-Btx stock solution, a standard curve was prepared for a novel protein assay,


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read at 700 nm. The protocol for this assay is shown in Table I (next page) as well as the tabulation

of the A700 of the standards and several dilutions of the Rh-"-Btx unknown.

Rh- -Btx Spectrum

Wavelength (nm)

400 450 500 550 600 650 700

A b s o r b a n c e

0.0

0.0

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Table I

l Btx stock l H2O ml BCA g -Btx A700

0.0 20.0 1.0 0.0 0.100

4.0 16.0 1.0 4.0 0.130

8.0 12.0 1.0 8.0 0.160

12.0 8.0 1.0 12.0 0.190

16.0 4.0 1.0 16.0 0.220

20.0 0.0 1.0 20.0 0.25020.0 (1/2 dilution Rh-"-Btx unknown) 0.0 1.0 0.310

20.0 (1/5 dilution Rh-"-Btx unknown) 0.0 1.0 0.176

20.0 (1/20 dilution Rh-"-Btx unknown) 0.0 1.0 0.109

From the data above, work-up the following points:

a) First, prepare a standard curve for the protein assay and get an equation for A700 vs. µg "-Btx.

b) Second, determine the protein concentration of the Rh-"-Btx derivative in mg/ml (show calculations)

c) Third, determine the molar "-Btx concentration (in units such as M, mM, µM, nM, pM, etc).

d) Fourth, determine the molar dye concentration in the Rh-"

-Btx derivative assuming a molarabsorption coefficient, !550 = 80,000 M –1 cm

–1 (show calculations).

e) Fifth, in this preparation, calculate the average number of moles of dyes / moles of "-Btx.

f) Sixth, how many µl of the Rh-"-Btx are needed to add to 1.0 ml of water to have a solution with anA550 = 1.0?

g) Lastly, what is the [protein] in this A550 = 1.0 solution (in µM)?


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3.1 From the continuous assays performed with different enzyme concentrations shown below:

a) What are the initial velocities of the reactions for Curves A#E?

b) Relative to Curve B, how much enzyme was used to obtain Curves D and E?

c) Draw the curves that would be obtained with 0.50X and 16X the amount of enzyme.

d) Fixed-time assays are those performed by stopping the reaction at a given time instead of

continuously monitoring the production of product. The reactions are stopped by denaturing the enzyme(heat, acid, organic solvents, etc.), or by the addition of an inhibitor. The amount of product formed atthe time is then determined and a velocity calculated. If fixed-time assays are run with the amount of

enzyme shown in Curves A through E, what time interval are chosen for each enzyme concentration?Why?

e) For routine fixed-time assays, which concentration of enzyme would be chosen? Why?

f) If Curve C was obtained by addition of 65 µg of protein, what is the specific activity of the enzyme

preparation? What assumption is made in order to say that this is the specific activity of the enzyme?

g) Assume that Curve E was obtained with 12 picomoles of pure enzyme that has a molecular weight of80,000 and two identical subunits, each with one catalytic site. Calculate the catalytic center activity and

molecular activity of the enzyme. Calculate the specific activity of the enzyme (IU/mg protein).


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h) Assume that Curve B was obtained as follows. A stock solution of enzyme was diluted 1/10 to

produce Dilution #1. Dilution #1 was then diluted 1/25 to produce Dilution #2. Dilution #2 was againdiluted 1/5 to produce Dilution #3. A 0.10 ml aliquot of Dilution 3 was used to produce Curve B. A 0.50

ml sample of Dilution 2 was found to contain 32 µg of protein by the Coomassie Blue assay. Calculatethe specific activity of the enzyme using the same assumption as in part g above.

i) Calculate the total number of units of enzyme (total IU) contained in 7.0 ml of the stock enzymesolution in part h above.

j) Suppose that Curve B was obtained with an enzyme with a Km = 40 µM for the substrate and that thesubstrate concentration was 160 µM. Draw the initial rate portions of the curves that would be obtained

with 40 µM and 10 µM substrate.

3.2. Glyceraldehyde-3-phosphate dehydrogenase (G-3-P) catalyzes the following reaction:

G-3-P + Pi + NAD+ $ 1,3-diphosphoglycerate NADH + H

+

To purify this enzyme, chicken liver is homogenized with isolation buffer, and the cellular debris isremoved by centrifugation. The protein concentration of the supernatant is determined to be 2.5 mg/ml.

G-3-P assays are performed on the supernatant fluid (0.1 ml of enzyme solution in a 3.0 ml final volumereaction mixture containing appropriate concentrations of substrates and buffer). The results are:

a) Which dilution is the best for estimating the amount of enzyme present? Why?

b) What is the specific activity of the original extract? (" at 340 nm, NADH = 6220 M-1 cm-1)


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3.3. A new lactate dehydrogenase (LDH) is isolated from the muscle of a Torpedo fish. Fixed-time

assays are conducted for LDH in a 1 ml final reaction volume, with appropriate substrate concentrationsand buffer. For the kinetic assays, 2 ml of a 1/10 dilution of the enzyme solution was prepared for a

working stock solution. The reactions are started by adding 30 %l of enzyme working stock solution to970 %l reaction mix cocktail, and measured at 340 nm, for 20 seconds. The original enzyme solution had

a protein concentration of 0.063 mg/ml. The results are tabulated below and are used to determine the

turnover number for this Torpedo LDH (MW 38,400 Daltons) and to compare these results to beefmuscle LDH. (" at 340 nm, NADH = 6220 M-1 cm-1).

a) Calculate mg of enzyme in each sample and the number of IUs of enzyme assayed in each sample(complete the table below, showing a sample calculation for each column).

b) Prepare a graph of IU vs mg of enzyme. Using this graph, calculate the specific activity of the

Torpedo LDH. (Hint: before proceeding with the IU calculations, make sure the rates are linear).

c) Now use the specific activity to calculate the turnover number for Torpedo LDH.

d) Is the Torpedo LDH faster or slower than bovine muscle LDH? Compare it to the bovine LDH:140,000 Daltons with a turnover number of 7000 min –1

.

3.4. A mutant acetylcholine esterase is isolated from the Torpedo fish. The enzyme is assayed bymeasuring the rate of production of choline in the following reaction:

The rate of choline production also is assayed in the presence of a competitive inhibitor of the

esterase: 15 %M compound I:

Sampleworking stock

solution of enzyme,ml (dilution)

mgenzyme

reaction mixcocktail, ml A340/20sec %mole P/min

1 0.03 (undil.) 0.970 0.0270

2 0.03 (1/2) 0.970 0.01353 0.03 (1/3) 0.970 0.0090

4 0.03 (1/6) 0.970 0.0045

5 0.03 (1/9) 0.970 0.0030

6 0.03 (1/12) 0.970 0.0022

H2O +

Choline

Acetylcholine

esterase

H H

HHHO–C–C–N–(CH

3)3

++

Acetylcholine

CH3 –C–O–C–C–N–(CH

3)3

O

+H

H H

H

Acetate

CH3 –C–O

–

O

Compound I

CH3 –C–N–C–C–N–(CH

3)3

O

+H

H

H

H

H


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The results are in the table below:

a) Prepare data for a Lineweaver-Burk plot (ie complete the table). Show units.

b) Draw the Lineweaver-Burk plot of the data, label the axes, and indicate the units.

c) Estimate Vmax and K m for the enzyme from the Lineweaver-Burk plot.

d) Estimate K mapp from the Lineweaver-Burk plot using the definition of competitive inhibition.

e) Estimate the Ki for the inhibitor from the Lineweaver-Burk results.

Sample acetylcholine

!M

1/[sub]

!M-1

Inhibitor

!M

Vo

nM min-1

1/Vo

nM-1

min

1 8.33 ----- 9.52

2 12.50 ----- 13.33

3 25.00 ----- 22.224 100.00 ----- 44.44

5 25.00 15.00 13.33


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4.1. Formulate the equilibrium expression for the reaction below, in the direction written. What is meant

by the apparent equilibrium constant? Why is it pH dependent?

lactate + NAD+ # pyruvate + NADH + H+

4.2. LDH activity can be assayed in either direction. What are the advantages and disadvantages of

assaying LDH in one direction or the other (ie +d(NADH)/dt or –d(NADH)/dt )? What practicaldifferences between the assays need be considered?

4.3. To measure the concentration of NADH in a cell homogenate, what wavelength would be chosen

for maximum sensitivity (explain), or for optimizing both sensitivity and specificity (explain)?

4.4. What is the isoelectric point of a protein? Why is it important to know the isoelectric point of a protein that is being purified by crystallization?

4.5. A hypothetical enzyme, unisilase, molecular weight 35,000 daltons, hydrolyzes its substrate, unisil producing the blue-colored compound nisil. The molar absorption coefficient of nisil at 595 nm is

1.24 x 104. The usual assay procedure is to follow the production of nisil at 595 nm. A typical

reaction mixture contains 50 mM phosphate buffer, pH 7.2; 2 mM unisil; and 0.1 ml enzyme, all ina total volume of 2.0 ml; using cuvettes with a 1 cm path-length. The data in the table below were

collected in the course of purifying unisilase. Complete the table:

4.6. The enzyme catalase is to be purified from beef liver. Ammonium sulfate fractionation of the crudehomogenate is used as a first step in the purification scheme. The fractionation of catalase by

ammonium sulfate is not known, so a systematic examination of increasing ammonium sulfate

Crude

Extract

Pure Unisilase

Volume of sample 85 ml 6 ml

Dilution for protein determination 1/200 none

Protein concentration in diluted sample 67 µg/0.2 ml 205 µg/0.1 ml

Total protein (mg)

Dilution for enzyme assay 1/40 1/300

A595 15 sec. 0.090 0.023

30 sec. 0.170 0.050

45 sec. 0.245 0.075

60 sec. 0.325 0.101

75 sec. 0.375 0.127

90 sec. 0.415 0.153

Total activity (I.U.)

Yield of enzyme activity 100%

Specific activity (I.U./mg)

Fold purification 1

Turnover number


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concentrations was performed. After adding ammonium sulfate, each fraction was centrifuged and the

supernatants were assayed for catalase. The results are shown in the table below.

a) How is the purest enzyme using ammonium sulfate fractionation obtained? Provide an explanation

Note: several answers to this question are possible. b) What is the specific activity resulting from the selected purification procedure?

c) What is the fold purification that is expected?

4.7 Two different ammonium sulfate fractionation schemes for use as a first step in the purification ofan enzyme are shown below. Complete the table and answer: Which of the two fractionation procedures

would be the best to incorporate as the first step in a multi-step purification procedure for the enzyme?

Crude

Scheme I

0.5-0.8 saturated pellet

dissolved in buffer

Scheme II

0.6-0.7 saturated pellet

dissolved in buffer

Protein (mg/ml) 20 5.0 5.0

I.U./ml 100 300 1000

Volume (ml) 50 12 2.5

Specific activityFold purification

Total I.U.

% yield of activity

4.8. The electron transport catalyst, plastocyanin, from the unicellular cyanobacterium, Microcystis

aeruginosa, has an isoelectric point of 4.8, a pH optimum of 8.0, and is most stable to long term storage

at pH 7.0 (4oC). In the purification of this plastocyanin, anion exchange chromatography is examinedTwo columns are prepared, one at pH 8.0 and one at pH 5.0; the solution containing the plastocyaninis

loaded onto the columns and each uses a linear KCl gradient as eluant. The results appear in thefollowing two plots on the next page.

Which of these two procedures should be adopted, or should another column be examined at pH 7.0?

Tube % Saturation

(NH4)2SO4

[Protein],

mg/ml

Activity,

I.U. ml –1

Spec. Activity,

I.U. mg –1

% Recovery

of Activity

crude extract ---------- 19.3 1230

1 20 19.4 12002 30 16.3 1191

3 40 14.0 1000

4 50 11.7 224

5 60 5.0 79

6 70 3.2 41

7 80 2.1 6.0


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4.9. Four groups (A, B, C, & D) of MCB 120L students purified LDH from steak. A continuous

assay for LDH in a 1 ml final reaction volume was used, in cuvettes with a 1 cm path-length,reactions started by adding 30 %l enzyme solution. Below are the data for the “peak” LDHfraction from the AMP-affinity column for each group. Also tabulated are the results of the

Bradford protien assay.

a) Given !340 = 6220 M –1cm –1 for NADH complete the chart.

b) Which preparation of LDH has the highest specific activity?c) Which preparation has the most IUs of LDH?

B

BB

B

B BBB

B B B BBB B B B BB

Ñ

Ñ

Ñ

Ñ

Ñ

ÑÑ

Ñ

Ñ

ÑÑ

Ñ

Ñ

Ñ

ÑÑ

Ñ

Ñ

Ñ0.0

0.5

1.0

1.5

2.0

2.5

0.0

0.2

0.4

0.6

0.8

1.0

0 5 10 15 20 25 30

Fraction Number

H 5 Data

A 2 8 0

Ñ

Ñ

_ _

_ _

_ _

E n z y m e A c t i v i t y

B

B

B BBB B BBB B B B BBB B

B

B

B

BÑ

Ñ

ÑÑ

Ñ Ñ

Ñ

Ñ

ÑÑ

Ñ

Ñ

Ñ

Ñ

Ñ Ñ

Ñ

ÑÑ

0.0

0.5

1.0

1.5

2.0

2.5

0.0

0.2

0.4

0.6

0.8

1.0

0 5 10 15 20 25 30Fraction Number

H 8 Data

A 2 8 0

Ñ

Ñ

_ _

_ _

_ _

E n z y m e

A c t i v i t y

B

B


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d) Which preparation has the most protein in it?

4.10. A small neuropeptide is newly isolated from the brain. Molecular biological study revealed that agene family of three genes codes for this peptide. The expressed peptides vary by one amino acidat position 5:

Peptide I: M·A·V·H·E·I·C·GPeptide II: M·A·V·H·K·I·C·G

Peptide III: M·A·V·H·L·I·C·G

Design an ion exchange chromatography protocol that would separate these three peptides given buffers of your choice, DEAE-Sepharose, CM-Sepharose, and salt.

Table of pK a values

a) First calculate the net charge of each peptide at pH values pH 6.5, 7.5 and 8.5. Helpful hint: if

the pKa of an ionizable species is equal to or greater than ± 2 pH units from the pH, then thespecies is assumed fully protonated or unprotonated.

b) Predict the order of elution using DEAE (anion-exchange) or CM (cation-exchange) at the

various pH values.

c) Which column and pH achieves the best (theoretical) separation?

Group A B C D

Volume of “pure” enzyme 5 3 2.5 4

Enzyme dilution for assay 1/500 1/800 1/400 1/1000

&A340/min 0.040 0.074 0.059 0.049

I.U./assay

I.U./ml of “pure enzyme”

Total I.U.

Dilution for protein assay 1/50 1/100 1/100 1/100

Volume assayed for protein, ml 0.5 0.5 0.5 0.5

%g protein in assay 5.86 7.85 4.68 8.30

mg/ml undiluted “pure” enzyme

Total protein, mg

Specific activity, I.U./mg

amino acid -NH3+ -COOH side chain

E 9.67 2.19 4.25

H 9.17 1.82 6.00

C 10.78 1.71 8.18

K 8.95 2.18 10.53

M 9.21 2.28 ---

A 9.69 2.34 ---

L 9.60 2.36 ---

G 9.60 2.34 ---

I 9.68 2.36 ---


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MCB120L A

17

4.11 Malate dehydrogenase (MDH) coverts malate to oxaloacetate as shown in the reaction:

malate + NAD+ $ oxaloacetate + NADH

Keq' = 6.2 x 10-6 = [oxaloacetate] [NADH] / [malate] [NAD+]

a) Enterprising MCB 120L students wish to study MDH from yeast and purify it using AMP-affinitychromatography as was done for lactate dehydrogenase (LDH). Why is AMP-affinity column a

reasonable first hypothetical purification method?

b)

There is no known transition state analog to elute MDH from the AMP-affinity column as wasavailable for LDH (ie the NAD+-pyruvate adduct). From the tables of molecules and their Km values

or Ki values for competitive inhibitors of MDH below, choose a hypothetical elution condition.

molecule Km (mM) molecule Ki (mM)

malate 0.014 ATP 1.36

NAD+ 0.31 ADP 1.34

oxaloacetate 0.023 AMP 0.95

NADH 0.030 cAMP 0.56

citrate 0.021

thyroxine 0.40

c) The AMP-affinity column did not bind MDH as hypothesized; a parallel control experiment resulted

in binding LDH. Propose an explanation for the lack of MDH binding to the AMP-affinity column.

d) From the tables above, choose a ligand and an eluant couple for affinity chromatography that potentially can be used to purify MDH.

citrate thyroxine AMP NAD+


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MCB120L A

18

Practice Problems: Experiment 5/6

5.1. A PCR reaction is used to amplify a 750 bp (bp = “base pairs”) DNA fragment, starting with 2000

copies of double stranded template DNA. The mass of one mole of base pairs is 660 g.

a. After 35 PCR cycles, how many copies are theoretically expected?

b. How many %g of product are theoretically expected?c. But, only 14 %g of PCR product was isolated. Assuming no losses in isolation, how efficient was the

PCR reaction?

d. To clone the 750 bp PCR product into a 2500 bp expression plasmid, 100 ng of restricted plasmid isligated with with a 5-fold molar excess of restricted PCR product. How many ng of PCR product is

needed?

5.2. A mass ruler (ie nucleic acid molecular weight standards) has the following characteristics:

Band Bandlength, bp

Bandmass, ng

A 2000 500B 1000 250

C 400 100

D 100 25

a. A restriction digest shows two bands, X and Y. X’s length is the same as mass ruler band A and its

intensity is the same as band C. Y’s length is the same as band B and its intensity is the same as bandC. Fill in the table below. It may help to draw a sketch of the gel.

Band Length Mass Mass/Length ratio

X

Y

What is the molar ratio of X/Y?

b. On another gel, band W is the same size as band D and as intense as band B. Band Z is the same sizeas band B and as intense as band A.

Band Length Mass Mass/Length ratio

W

Z

What is the molar ratio of W/Z?

5.3 The cloning of genes that code for proteins from bacteria is simpler than from eukaryotes since prokaryotic genes do not have introns and do not involve (extensive) RNA processing to express a

protein. The primary transcript of a prokaryotic gene is typically the mRNA that is translated directly (iewithout modification of the primary transcript). So, the cloning of bacterial genes by PCR is

accomplished by creating primers that recognize gene of interest, and using genomic DNA as the source


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MCB120L A

19

Below are the protein and gene sequences, respectively, for Malate Dehydrogenase (MDH) from the

cyanobacterium, Synechocystis PCC 6803.

MNILEYAPIA CQSWQVTVVG AGNVGRTLAQ RLVQQNVANV VLLDIVPGLP QGIALDLMAA 60

QSVEEYDSKI IGTNEYEATA GSDVVVITAG LPRRPGMSRD DLLGKNANIV AQGAREALRY 120

SPNAILIVVT NPLDVMTYLA WKVTGLPSQR VMGMAGVLDS ARLKAFIAMK LGACPSDINT 180

LVLGGHGDLM LPLPRYCTVS GVPITELIPP QTIEELVERT RNGGAEIAAL LQTGTAYYAP 240

ASSAAVMVES ILRNQSRILP AATYLDGAYG LKDIFLGVPC RLGCRGVEDI LEVQLTPEEK 300

AALHLSAEAV RLNIDVALAM VSDG 333

atgaatattt tggagtatgc tccgatcgcc tgtcagtcct ggcaggttac cgtggtcggc 60

gctggcaatg tggggcggac ccttgcccag aggttagtgc agcaaaatgt cgccaacgta 120

gttttgttgg acattgtgcc aggcttaccc cagggcattg ccttggattt gatggccgcc 180

cagagcgtgg aggaatacga cagcaaaatc attggcacca atgaatacga ggccaccgcc 240

ggctccgatg tggtggtaat taccgctggt ctaccccgca ggcccggcat gagtcgggat 300

gatttgttgg gcaaaaacgc caacattgtg gcccaggggg cccgggaagc attgcgttat 360

tcccccaacg ccattttgat tgtggtcacc aatcccctgg atgtaatgac ctatttggcc 420

tggaaagtaa ctggtttacc ttcccaacgg gttatgggca tggcgggggt gttggactcg 480

gctcggctca aggccttcat tgcgatgaaa ttaggggcct gtccttctga tatcaacacc 540

ttagtgctgg gcgggcacgg agatttgatg ctgcccttgc cacgatactg caccgtcagc 600

ggggttccca ttaccgaatt aatacccccc caaaccattg aagagttggt ggagcgtacc 660cgtaacggtg gggctgaaat tgccgcctta ctacaaacgg gcacagccta ttatgcgccg 720

gcctcttccg ctgcggtgat ggtggagtcc attttacgca atcagtctag aattctcccc 780

gccgccacct accttgatgg tgcctatgga ttgaaggaca ttttccttgg agtgccctgc 840

cgtttggggt gtcgaggagt ggaagatatt ctcgaagtgc aattaacccc tgaagaaaaa 900

gctgccctcc atctttctgc agaagcagtt cgccttaata ttgatgtggc gttggccatg 960

gttagcgacg ggaagatatt cgcgatggtt agtgacgggt aa 1002

Design both the forward (aka sense) primer and the reverse (aka anti-sense) primer to amplify the entireMDH coding region by PCR. Make the primers at least 18 nucleotides long, and include the stop codon.

5.4 Most eukaryotic genes have introns, so the cloning of protein coding genes from eukaroyotes

typically involves creating a DNA copy of the mRNA since RNA processing removes the introns fromthe primary RNA transcript. Poly(A+) RNA is isolated from the tissue of choice, then copied by the

enzyme, Reverse Transcriptase to produce DNA copies of (all) the mRNA present, which is calledcDNA. The gene of interest is copied from this population of cDNA by the usual PCR method. These

two steps are used so often that they are coupled together in a process called: RT-PCR (ReverseTranscriptase – PCR); the researcher supplies the specific primers and target mRNA.

Also, many other proteins include transit or signal peptide sequences at the N-terminus, and the proteins

are referred to as pre-proteins (eg pre-plastocyanin), which are later removed to produce the maturefunctioning, cellular protein. The transit and signal sequences are used in the cellular localization of the protein. The transit sequences direct proteins to either the mitochondria or to chloroplasts after

translation is complete; signal sequences direct proteins during translation to the endoplasmic reticulum(the beginning of the secretory pathway). The important point for the cloning: the transit and signal

sequences remain on the mRNA after RNA processing since these protein sequences are needed. If theexpression of the mature protein is the goal, then the primers are designed to begin copying the mRNA

at the N-terminus of the mature protein, and often, a methionine codon is added to provide a start codonfor the cloned protein.

Design both the forward (aka sense) primer and the reverse (aka anti-sense) primer to amplify the

plastocyanin coding for the mature protein region by PCR. Make each primer 18 nucleotides long.


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MCB120L A

20

Below is the sequence information for the protein, the gene, and the mRNA of plastocyanin from

spinach (from the Expasy.org website: Swiss-Prot/trEMBL query).

Protein sequence:Molecule processing

Transit peptide 1 – 69

Chain 70 – 168

10 20 30 40 50 60

MATVASSAAV AVPSFTGLKA SGSIKPTTAK IIPTTTAVPR LSVKASLKNV GAAVVATAAA

70 80 90 100 110 120

GLLAGNAMAV EVLLGGGDGS LAFLPGDFSV ASGEEIVFKN NAGFPHNVVF DEDEIPSGVD

130 140 150 160

AAKISMSEED LLNAPGETYK VTLTEKGTYK FYCSPHQGAG MVGKVTVN

Gene sequence:gagcacaacc aatgaaaaac tagataatac cctttattgg cccacacatc acaaatcctt 60

taatccaatg ccactaaaaa tcccacaaat gaaaaccaca caaaaccatg taaacagaca 120

tcacctcatc tatcctccat tttatcacct cttattaaat cccatcctcg ctctcagcac 180

tctttggcaa ttgtcatttc ttaattgcat tccacttaaa cccctccaaa atcctcctcc 240

ttccattaca atggccaccg tcgcttcctc cgctgccgta gccgtcccct cctttaccgg 300

ccttaaggcc tcaggatcca tcaagcctac caccgcaaaa atcatcccaa ccaccaccgc 360

cgtcccaaga ttgtctgtca aggcttcctt gaagaatgtc ggagccgccg tggtagcaac 420

cgcagccgcc ggacttctag ccggaaacgc catggccgtt gaggtgttgc tcggaggggg 480

tgacggatca ttggcattcc ttccaggaga cttcagcgta gcctcgggcg aggagatcgt 540

attcaagaac aatgccggat tcccccacaa cgtagtgttt gacgaggacg agattccctc 600

cggtgtcgac gccgcgaaga tttcgatgtc cgaggaggat ttgctgaatg caccagggga 660

aacttacaaa gttaccctta ctgagaaagg aacttacaag ttctactgct caccccacca 720

gggtgctggt atggtgggaa aagtaactgt caactaatat tttaatgaaa aattgaattt 780

ttatttatat gactttgttg tagagctatt tatataaacg atctcatcaa ttggtacgtt 840

tagcacattt gtgtagagct atttattcga aatttggttt agatttgatc tcataaatca 900taatccaaaa ttggattata ccggtttaga tttgaattga actacagtac tacactccta 960

agataaaatg ccaccattca atctttatct gaaagaattc 1000

cDNA sequence (derived from mRNA):atggccaccg tcgcttcctc cgctgccgta gccgtcccct cctttaccgg ccttaaggcc 60

tcaggatcca tcaagcctac caccgcaaaa atcatcccaa ccaccaccgc cgtcccaaga 120

ttgtctgtca aggcttcctt gaagaatgtc ggagccgccg tggtagcaac cgcagccgcc 180

ggacttctag ccggaaacgc catggccgtt gaggtgttgc tcggaggggg tgacggatca 240

ttggcattcc ttccaggaga cttcagcgta gcctcgggcg aggagatcgt attcaagaac 300

aatgccggat tcccccacaa cgtagtgttt gacgaggacg agattccctc cggtgtcgac 360

gccgcgaaga tttcgatgtc cgaggaggat ttgctgaatg caccagggga aacttacaaa 420

gttaccctta ctgagaaagg aacttacaag ttctactgct caccccacca gggtgctggt 480

atggtgggaa aagtaactgt caactaa 507


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MCB120L A

21


7.1. The enzyme catalase contains approximately 116 residues of arginine, 204 of aspartic acid, 144 of

glutamic acid, 68 of histidine, 120 of lysine, and 84 of tyrosine per molecule. Make a rough estimateof the net ionic charge of catalase at pH 7.5 and pH 9.5.

7.2. The lower reservoir buffer is about 250 ml of a Tris•HCl buffer (0.05 M) pH 8.3. Gels are often runfor about 3 hours at 40 ma. A century ago, Michael Faraday discovered the relationship between theamount of current passing from electrode to solution and the amount of chemical converted at that

electrode. He found that a transfer of 96,500 coulombs (now called a Faraday in his honor) of chargecorresponds to a release of 1 equivalent of reactant. An ampere is a current of 1 coulomb/second

Remembering the reactions that take place at the anode, decide how much the pH will change in thelower reservoir if a gel is run for 3 hours at 40 ma. The pK a of the amino group of Tris is 8.1. Do

you think you could reuse the buffer for a second run?

7.3. A solution contains the following three pure proteins:

Protein Native MW (daltons) pI

A 100,000 (dimer, identical units) 5.8

B 80,000 10

C 50,000 7.1

a) If this mixture were run on a Native-PAGE system identical to the conditions used in the lab (nostacking gel), how many bands would be visible after Coomassie Blue staining? Why?

b) If a 10% SDS-PAGE system were run identical to the conditions used in lab, how many bands would

be visible after Coomassie Blue staining? Why?

c) If a 10% SDS-PAGE system were run similar to the conditions used in lab, with the exception that 2-mercaptoethanol was omitted from the 2X SDS-sample buffer, then what might be the results after

staining with Coomassie Blue? Why?

7.4 Below is the reaction for the enzyme: theobromine demethylase (TD), which was isolated from themuscle tissue of the Lesser Zimbu Monkey. Scientists at the UCD Genome Center have sequenced

the LZM genome and two isoforms of TD were found in the databanks.TD

Theobromine + NADH + H+ + O2 NAD+ + H2O + CH2O + 3-Methylxanthine

Compute pI/MW (from the Expasy web site: protparam)

TD_A_ZIMBU (Q9BE24) TD_B _ ZIMBU (Q4R5B6)

A chain (EC 1.1.1.27) B chain (EC 1.1.1.28)

(TD-A: Theobromine Demethylase). (TD-B: Theobromine Demethylase).

(Lesser Zimbu Monkey). (Lesser Zimbu Monkey)

FT CHAIN 2 - 318 TD-A chain. FT CHAIN 2 - 320 TD-B chain.

Molecular weight (average): 35600.48 Molecular weight (average): 36507.30

Theoretical pI: 8.47 Theoretical pI: 5.76

Samples from the TD purification steps were analyzed by 6% Native-PAGE and histochemically

stained. Below is a copy of the histochemically stained native gel.


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MCB120L A

22

a). Briefly explain the most likely composition of the three TD isozymes observed in the histochemical

stained bands (show any necessary work used to arrive at your conclusions).

b). In addition to the reagents PMS and NBT for the histochemical staining, what else is needed tospecifically identify TD?

7.5 The sea slug, Elysia chlorotica, eats algae, and captures and stores the algal chloroplasts in its owncells. The captured chloroplasts continue to photosynthesize and fix carbon into glucose, which isused by the sea slug. Former MCB 120L students chose to investigate two soluble cytoplasmic

dehydrogenases from the sea slug involved in glucose utilization paths: Glyceraldehyde-3-PDehydrogenase (GPDH) and Lactate Dehydrogenase (LDH).

GPDH triose-3-P + NAD+$ 1,3-bis-P-glycerate + NADH + H

+

LDH lactate + NAD+ $ pyruvate + NADH + H

+

a) These enzymes have never been purified from the sea slug. Prior to running a Native-PAGEanalysis, briefly explain what could be done to make an educated estimation of the possible molecular

weights and isoelectric points of LDH and GPDH from sea slugs.

b) Briefly explain how to distinguish between LDH and GPDH by Native-PAGE when using phenazinemethosulfate and nitroblue tetrazollum in the histochemical staining technique.

c) The Native-PAGE revealed which band is GPDH and LDH (only one isozyme for each), each band

is cut out of the gel, prepared for, and analyzed by, 10% SDS-PAGE. Only one band is observed foreach enzyme with Coomassie blue staining. Determine the subunit molecular weight for each enzyme.

lanes

1 2 3 4 5

top of gelLanes assignments

1. crude extract

2. 20% supernatant

3. 40% pellet

4. affinity column flow-through

5. affinity column combined

eluted fractionsdye-front


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MCB120L A

23

PROTEIN STANDARDS MW (DALTONS) R M SAMPLES R M

bovine serum albumin 66,000 0.25 GPDH 0.41

ovalbumin 45,000 0.37 LDH 0.45

carbonic anhydrase 29,000 0.51

(-lactoglobulin 18,400 0.67

lysozyme 14,300 0.75

7.6 An analysis of protein composition by SDS-PAGE using a mini-gel apparatus as is Experiment 5(approximately 1 mm gel thickness) usually requires from 0.5 %g to 1.0 %g of a single purified

protein for sufficient visualization after staining with Coomassie Blue, whereas 10 %g to 20 %g of protein in a crude extract is required in order to see the compliment of stained bands.

a) For a crude extract of 16 mg/ml, describe the preparation and loading of 10 %g protein in a finalloading volume of 20 %l using 2X SDS-sample buffer.

b) For a purified protein of 0.5 mg/ml, describe the preparation and loading of 1 %g protein in a fina

loading volume of 10 %l using 2X SDS-sample buffer.


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Practice Problems AI-24

24


8.1. The optimum time for harvesting E. coli cultures is when the cells have grown to “late log phase”,which is usually determined by measuring the A600 of a culture in a spectrophotometer. An A600 of 0.6

is approximately 5 x 108 cells/ml, which corresponds to late log phase.

a. A measure at A600 of an E. coli culture is 0.15. If the doubling time of E. coli is 1/2 hour, how muchlonger will it take for the culture grow to get to late log phase? (Doubling time is the amount of time

it takes for one bacterium to divide into two.)

b. After 4 ml of an E. coli culture is transformed with a 4000 bp plasmid (2600 bp vector + 1400 bpinsert) in late log phase, the cells are harvested and 18 %g of plasmid is isolated from the cells. What

is the average number of plasmids per E. coli? (~660 g/mol base pairs.)

c. The 18 %g of plasmid from (b) above is suspended in 50 ul of TE to create a stock of plasmid. To

recover the insert from the plasmid, 20 %l of this stock is digested with 30 %l of restriction enzymemix (final volume is 50 %l). After 6 hours, 10 %l of the digest is removed and 2 %l of loading dye is

added; the 12 %l mixture is run on an agarose gel. Assuming digestion is complete, how many bandsand what band lengths should be seen? How many ng of DNA will be in each band?

8.2. The MCB 120L students are asked to test a new antisense (ie reverse) primer for BVR expressionThe vector is pASK75. The PCR products were cloned into the Xba I and Sal I sites.

The sequences from the Xba I site through the strep-tag of the clones made from the PCR products given

by the old primer, the new primer, and non-recombinant pASK75 are shown below.

pASK75:

Strep Tag $Ser Ala Trp Arg His Pro Gln Phe Gly Gly stopstop

TCT AGA ATG [ 96 bp] TCG AGG TCG ACC TGC AGG C AGC GCT TGG CGT CAC CCG CAG TTC GGT GGT TAA TAA GCT T

Xba I Sal I Pst I Hind III

old primer:


TCT AGA ATG [900 bp] GAA AAT GGT CGA CCT GCA GGC AGC GCT TGG CGT CAC CCG CAG TTC GGT GGT TAA TAA GCT T

Xba I Sal I Pst I Hind III

new primer:


TCT AGA ATG [900 bp] GAA GCT TGT CGA CCT GCA GC AGC GCT TGG CGT CAC CCG CAG TTC GGT GGT TAA TAA GCT T

Xba I Hind III Sal I Pst I Hind III


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25

To make sure the clones have the correct inserts, minipreps of the plasmid DNA are prepared from each

and digested with Pst I, or Xba I and Hind III (ie double digest, two enzymes). The digests are run on anethidium bromide stained agarose gel, shown in Figure 8.1.

Figure 8.1. Agarose gel

Figure 8.2. SDS polyacrylamide gel

a. The mass ruler is an equimolar mixture of DNA fragments. There are 85 ng in the 750 bp band.

How many ng are in the 3000 bp band? ____________ ng


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26

b. The pASK75 PstI digest (band labeled “A”) is the same intensity as the 750 bp band in the mass ruler.There is 100 %l of the PstI digest; 4.0 ul of 6X loading dye is added to 20.0 %l of the PstI digest, and

then 14.0 %l of this mixture is loaded onto the gel.

How much pASK75 is in the original 100 %l of digest? _______ ng

c. In the spaces below the agarose gel, to the left of “Clone ID”, identify which of the Xba/Hind digestsare from: “pASK75”, “old”, or “new”.

Crude extracts of induced E. coli cultures transformed with each of the three plasmids were loaded onto

an affinity column for BVR made by linking biliverdin to Sephadex beads. The eluted fractions were pooled and loaded on the SDS gel shown in Figure 8.2. The gel has been stained with Coomassie

Blue.

d. In the spaces below the SDS gel, to the left of “clone ID”, identify each lane as “pASK75”, “old” or“new”.

e. Briefly explain which clone(s) will be detected by a Western blot that recognizes the strep tag?

8.3. A gene of 1000 bp was isolated by PCR. The 5’-end of both primers had EcoRI sites attached. Thus,

the PCR fragment is inserted into the EcoRI site of the MCS randomly in either direction (Figure 8.3:orientation A or B). There is a PstI restriction enzyme site within the PCR fragment; the relative location

of the PstI site is shown below (Figure 8.3). Devise a restriction mapping experiment that definitivelydistinguishes between orientation A and B. Assume the distances between the restriction enzyme sites of

the MCS region are essentially negligible when analyzed by agarose gel electrophoresis.

Figure 8.3. Insertion of a PCR fragment into a plasmid MCS

MCS

plasmid

3000bp

EcoRIPstI

SalI

XbaI

EcoRI

EcoRI digest and ligation

PCR fra ment, 1000b

EcoRI

EcoRI

PstI

SalI

XbaI

plasmid

4000bp

PstI EcoR I

PstI

PstI

SalI

plasmid

4000bp

PstI

EcoRI

PstI

SalI

XbaI

plasmid4000bp PstI

SalI

plasmid

4000bp

PstI

EcoRI

orientation

A

orientation

B

0 500 1000

bp


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27

8.4. MCB 120L students, TA’s, and instructors had their DNA analyzed for neurotransmitter geneexpression. One gene associated with intelligence, and named the Smart Gene, is expressed at high

levels in all MCB 120L students, TA’s, and instructors (or so we thought). The Smart Gene DNAsequence is known, and the ORF sequence does not contain any introns (only the sense strand is shown).

An antibody to the Smart Gene protein can be purchased commercially.

Smart Gene sequence: 66 nucleotides

ATGATCTGTCTAGATTTGCTACGCTCGAGCGGGTCGGAATTCTCGGTGGTTATGGATCCCAATTGA (Met) (XbaI) (XhoI) (EcoRI) (BamHI) (stop)

a) MCB 120L students want to isolate the Smart Gene by PCR and clone it into the pASK75 vector(shown below). Fill in the restriction enzyme sequence chosen for the primers (more than one

combination is possible; and fill in the proper nucleotides of the Smart Gene sequence (only 10 areneeded); the CC are spacers.

Restriction Enzyme Site 10 nucleotides of the Smart Genesense primer: 5’— (forward primer)

Restriction Enzyme Site 10 nucleotides of the Smart Geneantisense primer: 5’—

(reverse primer)

Restriction enzymerecognition sequences

b) Was the stop codon from the Smart Gene included in the antisense primer, why or why not?

C C

C C

name Recognition Sequence

BamHI 5’-GGATCC-3’EcoRI GAATTC

KpnI GGTACC

HindIII AAGCTT

SalI GTCGAC

XbaI TCTAGA

XhoI CTCGAG


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c) Below is a restriction map of the smart gene isolated by PCR from one MCB 120L student and one ofthe instructors (Instructor X, not to be named). What do the data reveal?

d) Offer a conclusion and an explanation for the results (there is more than one plausible explanation).

pASK75 vector

XbaI 118)

EcoRI (210) KpnI (222)

BamHI (231)

XhoI (237)

SalI 243)

HindIII (291)

PvuI (1379)

NsiI (2219)

StyI (2370)

SpeI (2492)pASK75(3266)

tetA promoter

strep-tag

tet R

Amp R

Ori

(3266 bp)

amp r

bp

100

80

60

40

20

Student (1 – 3) Instructor X (4 – 6)

Lanes: 1 2 3 4 5 6 7

Lane assignments:

1. PCR fragment (uncut)

2. XhoI digest

3. EcoRI digest

4. PCR fragment (uncut)

5. XhoI digest

6. EcoRI digest

7. MW markers

appendix i practiceproblems s2016

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