You are working with a 60 kDa protein that you want to visualize on a SDS-PAGE. What % separating gel will you use in your SDS-PAGE?
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Proteins
We generally tend to think of proteins only from a dietary lens, as a component of what we eat. However, they are among the most important and abundant organic macromolecules in the human body, with diverse structures and functions. Every cell contains thousands and thousands of proteins, each with specific functions. Some help in the formation of cellular membrane or walls, some help the cell to move, others act as messages or signals and flow seamlessly from one cell to another, carrying information.
Protein Expression
The method by which living organisms synthesize proteins and further modify and regulate them is called protein expression. Protein expression plays a significant role in several types of research and is highly utilized in molecular biology, biochemistry, and protein research laboratories.
Question based on SDS-PAGE
You are working with a 60 kDa protein that you want to visualize on a SDS-PAGE. What % separating gel will you use in your SDS-PAGE?
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- Question with regards to SDS-PAG You are working with a unique protein that has no basic amino acids and contains a lot of acidic amino acids. You run a SDS-PAGE gel, you stain and destain it. You get nice bands for your ladder but no bands for your sample proteins. After doing an immunoblot specific for your protein (using the SDS-PAGE gel), you do get a band identifying your protein. You know that your concentration of the protein is sufficient to be visualized by staining it. The immunoblot clearly identifies that your protein is in your gel, why can you not see it after staining and destaining the gel?Subject: Biochemistry, chemistry, polyacrylamide gel electrophoresis Differentiate native PAGE and SDS-PAGE in terms of the relative molecular weight of protein bands obtained. Explain your answer.Question: Which base (A, C, T or G) corresponds to X in the unknown? I did the experiment, got the data below, and calculated the binding constants. But I am TOTALLY lost as to how to figure this out! I don't even know what steps I would take. Base pairs Data – all had Temp = 250C PH = 7 Binding Constant A & X [A] = 0.00373221M [X] = 0.00373221M [AX] = 0.0462678M 3.322 C & X [C] = 0.0469007M [X] = 0.0469007M [CX] = 0.00309935M 1.409 T & X [T] = 0.0452279M [X] = 0.0452279M [TX] = 0.00477212M 2.333 G & X [G] = 0.0469554M [X] = 0.0469554M [GX] = 0.00304456M 2.633
- Question:- What is the average radius of a piece of double-stranded DNA in water that has a link length of 5.9 nm and and is 3019228 links long? If the two strands of the double-stranded DNA are separated, each chain now has a radius of 871 nm. What is the link length for single-stranded DNA?RECALL Which of the following statements is (are) true? (a) Bacterial ribosomes consist of 40S and 60S subunits. (b) Prokaryotic DNA is normally complexed with histones. (c) Prokaryotic DNA normally exists as a closed circle. (d) Circular DNA is supercoiledQuestion- Determine the exact molecular weight of the following oligonucleotide and calculate the average molecular weight of an equivalently long oligonucleotide. What is the percent error between the two? Sequence (5’-3’) : GCT TTA GCG GTC AAG TGC TGT GCC ATT T M.W. of a nucleotide in ssDNA = 303.7 + 79.0 M.W. of base pair in dsDNA = 607.4 + 158
- 2.7 Problem in the Determination of Molecular Welght by Gel Electrophoresis A puri- fied recombinant protein is analyzed for molecular weight by SDS-PAGE at pH 8.5. From the protein sequence deduced from the gene that was expressed in bacteria, the protein is expected to have a molecular weight of 44,000. However, the molecular weight of the pro- tein is found by SDS-PAGE to be 52.000. Explain the reason or reasons for this difference in molecular weight. What calculation could you make to help explain this discrepancy?SDS-PAGE gels separate proteins by charge. Therefore proteins that are 10,000 daltons (g/mol) will move the same distance as proteins that are 80,000 daltons (g/mol). This statement is: True or False (And why)Calculations for the Purification of a Recombi- nant Protein The purification of a recombinant protein is carried out starting with 100 liters of a clarified cell lysate (i.e., the cells have been lysed, and the cell debris has been removed to give a clar- ified solution), which has a total protein concentra- tion of 0.36 mg/ml and a recombinant protein con- centration of 2.2 U/ml, where U denotes units of biological activity of the recombinant protein. It is known that the completely pure recombinant pro- tein has a specific activity of 40.0 U/mg. Purifica- tion is continued until a chromatography step that yields 2.0 liters of a fraction containing the protein, with a total protein concentration of 1.11 mg/ml and a recombinant protein concentration of 43.2 U/ml.. For the recombinant protein, calculate the. starting and ending purity, the starting and ending specific activity, and the percentage yield and fold purification through the chromatography step. :
- Affinity chromatography You have created a fusion protein tagged with Glutathione-S-Transferase (GST). Your lab mate tells you that the affinity columns for this type of tagged protein are very similar to that of Histadine tagged proteins. Using the following elements set up a purification column and construct a protocol for an affinity purification using this tag. A large amount of glutathione is usually used to elute the tagged protein off the column. How might this work?Part II: Information Transfer Background Information - Key Points The background information provided for this lab has given you a general overview of some of 24 the key terms and definitions necessary to understand the transfer of information from gene to protein. The information included below will help you work through the specific problems included in your Tutorial 4 Assignment. When working on the problems remember the base mu to pairing rules (Table 3). Table 3: Rules for nucleotide base pairing. cytosine (C) - guanine (G) adenine (A)- thymine (T) DNA RNA For Transcription: ● ● ● ● cytosine (C) - guanine (G) adenine (A)- uracil (U) Initiation is determined by the recognition of the promoter sequence in the DNA by the RNA polymerase. Stef The transcription start site is downstream of the promoter and is designated as the +1 site. Aspartic acid Alanina Valine Arginine Serine Lysine Asparagine Glutamic TEOPO|0C|AGUCAG|UC|AG/DCAG/3G/ CAGUC UGU A C A Threonine G Methionine Isoleucine…FLUORESCENCE BINDING a) In the DAPI DNA lab, concentration of DAPI and DNA was determined to carry out analysis. Here is the scenario: 3.50 mL of DAPI at 25 micromolar was placed in the cuvette. It was titrated with a 150 microgram/mL DNA stock. It took 230 microliters of DNA stock to reach saturation. What is the DAPI molarity at saturation in the cuvette? What is the DNA molarity (in terms of base, 1 base = 325 g/mol) at saturation in (i) (ii) cuvette? im DNA = 150 × 10 g/mL 3.5-CAP 25 μM the 230 μL