exam4key

BCH 5425 Molecular Biology and Biotechnology

Spring 1998

Exam IV (100 Points)

1. Regarding protein assays:

A. (10 points) What are the criteria which an ideal protein assay should meet?

1. assay should be specific for protein of interest

2. assay should be rapid

3. assay should be sensitive

4. assay should be quantitative

B. (10 points) Few assays are able to meet all desired criteria. For each the criteria listed in A above, describe the consequences if the assay does not meet the criteria

1. Can get false positives

2. Purification will take a long time

3. A significant portion of your desired protein will be used up in the assay and yield will be reduced due to this loss

4. You will be unable to determine yields for each purification step and will be unable to evaluate losses at each step

2. (20 points) The following data is from an ammonium sulfate precipitation experiment of a crude extract of E. coli cells which are expressing a protein of interest. The starting sample contains 15.5 mgs of total protein. An assay for the protein of interest indicates the starting sample has 153 units of activity. Using this information, fill in the blanks in the table below.

% (NH₄)₂ SO₄ in soln	Total mgs Protein in soln	units of Activity in soln	Total mgs Protein in pellet	units of Activity in pellet	Specific activity (soln)	Specific activity (pellet)	% yield of activity (pellet)	Purif.
0	15.5	153	0	0	9.9	-	0	-
10	15.2	130	0.3	23	8.6	76.7	15	8
20	14.7	64	0.8	89	4.4	111.3	58	11
30	13.7	29	1.8	124	2.1	68.9	81	7
40	11.4	11	4.1	142	1.0	34.6	93	4
50	7.8	5	7.7	148	0.6	19.2	97	2
60	4.2	1	11.3	152	0.2	13.5	99	1
70	2.3	0	13.2	153	0.0	11.6	100	1
80	1.7	0	13.8	153	0.0	11.1	100	1
90	1.5	0	14	153	0.0	10.9	100	1

3. (10 points) Using the information from the above table, indicate how you could incorporate an ammonium sulfate precipitation step in a purification scheme for this particular protein of interest under the following restrictions:

A. The E. coli expression system does not make a lot of the protein of interest. Therefore you need to achieve a better than 90% yield for this step.

You would add ammonium sulfate to 40% and take the pellet. This will have 93% of the protein activity (yield 93%) with an expected specific activity of 34.6, for a purification of 4-fold

B. The E. coli system efficiently expresses the protein of interest. Therefore, yield is not a concern and you need to maximize purity.

The highest specific activity is in the 20% ammonium sulfate pellet. Therefore, you would add ammonium sulfate to 20% and take the pellet. We expect a yield of 58% (89 units) with a specific activity of 111.3, for a purification of 11-fold

4. (15 points) Provided with the following characteristics of a given protein and chromotographic resins, indicate the appropriate resin to try to use to purify the protein. (Assume the buffer for such experiments is pH 7.0) DRAW A LINE FROM THE PROTEIN TO THE RESIN.

Protein Resin

A. pI = 5.6 CM-52 (Carboxymethyl cellulose)

B. pI = 3.8 DE-52 (Diethylaminoethyl cellulose)

C. pI = 8.7 SP-sephadex (Sulphopropyl sephadex)

D. pI = 10.5 QAE-sephadex (Quaternary amine sephadex)

pI 5.6 -> QAE-sephadex

pI 3.8 -> DE-52

pI 8.7 -> SP-sephadex

pI 10.5 -> CM-52

5. (10 points) A protein sample elutes from an ion exchange column at a NaCl concentration of 1.2 M. The pooled protein has a volume of 50 mls (and a NaCl concentration of 1.2 M). We would like to proceed with the next step of purification, but first we must reduce the NaCl concentration to 5 mM. You will dialyze the sample versus water.

A. If you wanted to perform this dialysis in a single step, how large a volume of dialysis buffer (i.e. water) would you need?

(50ml)*(1.2M) + (Xml dialyis buffer)*(0M) = (X+50)*(.005M)

X=11,950 ml (or 11.95 liters)

B. If you only had a 1 liter sized container in which to perform the dialysis, how many dialysis steps would be required to get the final salt concentration less than or equal to 5mM (assume you can fit 2 liters of buffer and the sample in the container)?

Step 1:

(50ml)*(1.2M) + (1000ml dialyis buffer)*(0M) = (1000+50)*(XM)

X=0.057M (or 57mM)

Step 2:

(50ml)*(0.057M) + (1000ml dialyis buffer)*(0M) = (1000+50)*(XM)

X=0.0027M (or 2.7mM)

Therefore, two 1L dialyses will reduce salt to within acceptable limit

6. (10 points) The bacteriophage M13 can infect only "male" E. coli cells, i.e. cells with a cellular structure called a pilus to which the M13 attaches in order to infect. The pilus in such cells forms in response to the presence of ‘F’ factor which is coded for by an extrachromosomal element known as an F’ episome. In order to work with M13, certain E. coli strains (e.g. JM101) have been developed in conjunction with F’ episomes in such a way that selective pressure exists to maintain these F’ episomes. Explain how this selective pressure works.

The host cells chromosome has suffered a deletion in the region of the proline biosynthetic genes (proAB). Thus, the host is unable to synthesize proline. The proAB gene region is subsequently inserted into the F’ episome. Thus, when grown in proline deficient media the E.coli host requires the F’ episome (with proAB genes) to survive (i.e. to produce proline).

7. (15 points) An investigator is performing a SELEX type experiment. The would like to know what RNA sequences can bind to bacteriophage M2 RNA polymerase. They have a putative promoter region for M2 RNA polymerase:

5’ GACTGAAGACGGCTCCAGGT 3’

and have identified a critical region of six contiguous nucleotides (GACGGC, underlined above) which they will randomize. Since all possible sequences at this site would constitute 4⁶ , or 4096 uniques sequences, they will initial construct a library containing all these sequences. The figure below shows sequence information for the starting library and after three rounds of enrichment (selecting RNA’s which bind the polymerase). What is your conclusion from this data?

1. They have selected for one unique sequence which binds to the polymerase. This sequence in the six nucleotide region is: 5’ A A T C C C 3’

2. This sequence is distinctly different from the original putative promoter region.

3. They did not regenerate the putative promoter sequence.

4. However, their starting library indicates that the original gene synthesis was in error: No G base was present at position #1 or #5 and no T base was present at position #2 and #4. Thus, there is no way they could regenerate the putative sequence (GACGGC).

5. Thus, while they have identified one unique sequence, there may well be others (with G at positions 1 and/or 5, and T at 2 and/or 4) which bind with equivalent or better affinity.