In size exclusion chromatography, which component will elute last in the set up?O Small proteins will elute last because they do not pass through the pores of the gel beads.O Large proteins will elute last because they pass through the pores of the gel beads.O Small proteins will elute last because they pass through the pores of the gel beads.Large proteins will elute last because they do not pass through the pores of the gel beads.

Size exclusion chromatography is a process of separation of molecules based on their size. The…

Answered: In size exclusion chromatography, which…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

In size exclusion chromatography, which component will elute last in the set up? Group of answer choices Large proteins will elute last because they pass through the pores of the gel beads. Small proteins will elute last because they do not pass through the pores of the gel beads. Large proteins will elute last because they do not pass through the pores of the gel beads. Small proteins will elute last because they pass through the pores of the gel beads.
Using ultrafiltration, you concentrate a solution of Protein X as follows: Initial volume protein X = 300 mL Initial concentration protein X = 12 mg/mL Final volume protein X = 15 mL What's the final concentration of protein X in mg/mL?
A protein gives a single band on SDS gel electrophoresis, as shown in lanes 1 and 2 below. There is little, if any, effect from adding β-mercaptoethanol(BME) to the sample; if anything, the protein runs a little bit slower. When treated with the proteolytic enzyme thrombin and electrophoresis in the absence of BME, the protein migrates a bit more rapidly (lane 3). But if BME is present, two much more rapidly migrating bands are found (lane 4). Explain these results in terms of a model for the protein.
Immunofluorescence and Fluorescent livecell imaging techniques can both be used to determine protein localisation.List the advantages and disadvantages of using immunofluorescence andfluorescent live cell imaging
A protein you want to isolate is often associated with the cell membrane. After homogenizing your sample, your protein is left in the supernatant. Which of the following will most likely happen next? O The protein will precipitate once the percent saturation of ammonium sulfate has reached 100% The protein will remain in the supernatant even at 100% saturation. O The protein will precipitate at low ammonium sulfate concentration O The protein will precipitate once the percent saturation of ammonium sulfate has reached 64%.
Sketch the appearance after visualization of a protein mixture containing the seven proteins (fibrinogen, y-globulin, collagen, ovalbumin, myoglobin, hemoglobin, insulin) when subjected to two-dimensional (2D) gel electrophoresis.
a 2D gel electrophoresis was done on the proteins found in the chart. create a chart with the correct parameters on both dimensions using 1-6 to indicate the position of each protein on teh 2D gel.
Which of the following technique is FALSE in microscopy? Osmium tetroxide can be used as a fixative as well as a negative stain in transmission electron microscopy. Two proteins; each tagged with their individual primary antibody followed by one protein with rhodamine conjugated secondary antibody and the other with fluorescein-conjugated secondary antibody can be visualized simultaneously using a fluorescence microscope. The emission spectrum of Rhodamine is 580 nm The emission spectrum of Fluorescein is 521 nm O An objective lens with 100X magnification and projection lens with 10X magnification is going to produce a 1000X total magnification. Phase contrast and differential interference contrast (DIC) do not require staining.
Human Immunodeficiency Virus (HIV) Principle: The assay starts with a sample applied to the sample well. A recombinant HIV antigen conjugated to colloidal gold embedded in the sample pad reacts with the HIV antibody present in serum or plasma forming conjugate / HIV antibody complex. As the mixture is allowed to migrate along the test strip, the conjugate / HIV antibody complex is captured by recombinant HIV antigen immobilized on a membrane forming a colored test band in the test region. A negative sample does not produce a test band due to the absence of colloidal gold conjugate / HIV antibody complex. The antigens used in the conjugate test are recombinant proteins that correspond to highly immunoreactive regions of HIV 1 and HIV 2. A colored control band in the control region appears at the end of test procedure regardless of test result. This control band is the result of colloidal gold conjugate binding to the anti-HIV antibody immobilized on the membrane. The…
Which of the following is TRUE about immunofluorescence microscopy where we localize proteins using fluorescently tagged antibodies? Pick the best answer: We can use it to resolve 2 points that are 10 nanometers apart. The emission wavelength of the fluorophore is typically shorter than the excitation wavelength. The resolution can be further improved with increased magnification. All else being equal, a fluorophore that emits red light will give better resolution than a fluorophore that emits green light. Based on what we discussed in class, we can use the technique on live samples therefore enabling us to see the dynamic movement of cytosolic proteins in living cells. None of the above
SDS is required to unfold proteins during SDS-PAGE electrophoresis and it is included in both the loading and running buffer. If SDS is not included in the running buffer.... the proteins in your sample may partially renature thus affecting migration. the gel will not conduct current in the absence of SDS. there is no effect on the samples. the proteins in your sample will be graded.
protein is purified and at a concentration of 600 μg in 1.75 ml of buffer. You do an assay with your purified protein, and the assay requires 25 μg of biochemisfunase. What volume of the purified biochemisfunase would yield 25 μg? Show your calculation.

SEE MORE QUESTIONS

Recommended textbooks for you

Biochemistry

ISBN:

9781319114671

Author:

Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:

W. H. Freeman

Lehninger Principles of Biochemistry

Biochemistry

ISBN:

9781464126116

Author:

David L. Nelson, Michael M. Cox

Publisher:

W. H. Freeman

Fundamentals of Biochemistry: Life at the Molecul…

Biochemistry

ISBN:

9781118918401

Author:

Donald Voet, Judith G. Voet, Charlotte W. Pratt

Publisher:

WILEY

Biochemistry

ISBN:

9781319114671

Author:

Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:

W. H. Freeman

Lehninger Principles of Biochemistry

Biochemistry

ISBN:

9781464126116

Author:

David L. Nelson, Michael M. Cox

Publisher:

W. H. Freeman

Fundamentals of Biochemistry: Life at the Molecul…

Biochemistry

ISBN:

9781118918401

Author:

Donald Voet, Judith G. Voet, Charlotte W. Pratt

Publisher:

WILEY

Biochemistry

ISBN:

9781305961135

Author:

Mary K. Campbell, Shawn O. Farrell, Owen M. McDougal

Publisher:

Cengage Learning

Biochemistry

ISBN:

9781305577206

Author:

Reginald H. Garrett, Charles M. Grisham

Publisher:

Cengage Learning

Fundamentals of General, Organic, and Biological …

Biochemistry

ISBN:

9780134015187

Author:

John E. McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson

Publisher:

PEARSON

SEE MORE TEXTBOOKS