Describe the basic steps involved in protein digestion, protein denaturation, and electrophoresis

Describe the basic steps involved in protein digestion, protein denaturation, and electrophoresis.

How “Chicken” Are You?
Using peptide mapping to determine amino acid sequence similarities1
After this lab exercise you should be able to:

  1. Describe the basic steps involved in protein digestion, protein denaturation, and
  2. Explain how SDS gel electrophoresis works to separate polypeptide pieces based on
  3. Describe the importance of SDS and the reducing agent to this method.
  4. Explain the role of the tracking dye, the stain, and molecular markers in
  5. Describe what proteases are and list some examples of their practical uses.
  6. Explain how this method may be used to determine structural relatedness of a protein
    in different species, and perhaps evolutionary relationships among species.
  7. Explain the limitations of this type of study.
    Before the lab review the following topics in your textbook
    • The structure of proteins and denaturation and sequence homology (section 3.4)
    • Gel electrophoresis
    As you go through the lab, answer the questions in the handout. You will submit the
    questions that are assigned point values as your assignment for this lab. The assignment
    is due before the beginning of lab 6.
    Proteins carry out many different functions in living cells. They can be structural
    components, enzymes that carry out specific reactions, receptors that receive external
    signals, etc. The function of a protein depends on its three-dimensional shape, which in
    turn depends on the protein’s primary structure = its amino acid sequence. When the
    protein loses its three-dimensional shape = it denatures, it will also lose its function.
    After a polypeptide is synthesized in the cell, it has to be folded into its proper shape.
    This involves the formation of bonds that lead to the proper folding of the polypeptide
    into secondary and tertiary structures. The folding of the polypeptide into the secondary
    structure of the protein backbone such as an -helix or a -pleated sheet involves the
    formation of weak hydrogen bonds. These bonds can be easily broken by heat or changes
    in pH, for example. The overall three-dimensional structure of the polypeptide is the
    tertiary structure. Tertiary structure is formed by interactions among amino acid side
    chains and involves weak interactions including hydrophobic / hydrophilic interactions
    and hydrogen bonds, ionic bonds, and covalent bonds such as disulfide bridges formed
    between two cysteine residues. If the protein contains more than one polypeptide, the
    structure is referred to as quaternary structure.
    1 This exercise uses a kit from Modern Biology Inc.
    Proteins that carry out essential functions for the cell tend to change very little in a
    population through mutations over time. That means their amino acid sequences are
    highly conserved. Therefore, we can look at the amino acid sequences for the same
    protein from different species, and we can use the level of similarities and differences
    among different species to determine how closely related they are to each other; that is,
    we can assess their evolutionary relationships based on these sequences. Please review
    the case of cytochrome c described in your textbook (section 3.4) as one such example.
    Proteases are enzymes that cut (cleave) polypeptides at specific amino acid sequences.
    The digestive enzyme chymotrypsin, for example, cuts peptide bonds formed from the
    carbonyl groups of tryptophan, phenylalanine, and tyrosine. Another protease, papain
    (derived from Papaya) cuts peptide bonds formed from leucine and glycine. Thus,
    depending on which enzyme is used, and where the target amino acids are located, a
    polypeptide could be cut to generate different size fragments. As an example, consider
    the polypeptides in Figure 1 below. They all have the same starting length. They are
    almost identical in molecular weight. The potential cut sites for the enzyme
    chymotrypsin have been marked.
    Protein 1: val glu glu gln arg trp val leu ala his thr gln arg glu gln met val ala
    Protein 2: val glu glu gln arg trp val leu ala his thr gln phe glu gln met val ala
    Protein 3: val thr gln phe glu leu arg glu trp val ala met gln phe gln glu val ala
    Figure 1: The amino acid sequences for three small proteins. Arrows mark the potential
    cut sites for the enzyme chymotrypsin.
    How many fragments would be produced by cutting these small proteins with
    chymotrypsin? Will any of the fragments be identical in size among the three proteins?
    If polypeptides from closely related species have the same amino acid sequence with the
    same amino acid residues at the same location, then treatment with the same enzyme
    should lead to synthesis of similar length fragments. Now consider a mutation in one
    species alters one of the amino acids that is the target of the enzyme so that a cut site is
    lost due to the mutation, or a new cut site is formed. In this case, it is likely that when the
    polypeptide extracted from this species is treated with the same enzyme, different size
    fragments will be formed compared to other related species without such a mutation. As
    an example, compare proteins 1 and 2 in Figure 1 above. The pattern of fragments
    generated can be used to determine how closely related different species are based on
    amino acid sequence similarities for a given polypeptide. For this approach to work,
    these peptide fragments have to be separated according to their sizes.
    One common method to separate peptide fragments based on size is SDS gel
    electrophoresis. You can learn about gel electrophoresis and its use in separating DNA
    fragments by reading chapters 14 and 17 in your textbook. The method used to separate
    peptide fragments is similar in principle. An electric field is applied to the fragments,
    which will then move away from the negative electrode towards the positive electrode.
    DNA naturally has negative charges (why?). The electric charge of polypeptides depends
    on the ionization of the amino acids, which itself is dependent on the pH of the solution.
    Peptide fragments thus have to be coated with uniform charges before gel electrophoresis
    can work. An agarose or polyacrylamide gel is typically used as the medium through
    which the fragments move. The gel is made with depressions (wells) in which the
    fragments are loaded before the electric field is applied. When the electric field is
    applied, the fragments then have to move through the gel from the negative to the
    positive poles. The gel serves as a “sieve” through which the fragments have to pass.
    The larger the fragment is, the more slowly it will move through the “sieve”, and thus gel
    electrophoresis allows us to separate the fragments based on size. Molecular markers
    (peptide fragments) of known size, also called a ladder, are included in the procedure for
    comparison with the fragments of interest. The gel is stained with some sort of stain to
    visualize the fragments. The locations of the fragments on the gel can be compared with
    those of the fragments in the ladder that have also passed through the same gel with the
    same electric field in order to determine the size of the fragments of interest.
    For this method to work with polypeptides, they have to be denatured and linearized.
    Any disulfide bridges have to be cut to help linearize the polypeptides. This is carried
    out by boiling the samples in the presence of reducing agents, such as -mercaptoethanol,
    as well as the anionic detergent SDS (sodium dodecyl sulfate). The detergent SDS also
    covers the native charge of the amino acids in polypeptides with its own negative
    charges. Thus, any peptide fragment will be linear and uniformly negatively charged.
    Therefore, neither the original three dimensional shape nor the charges of ionizable
    amino acids will interfere with the movement of the fragments through the gel. The only
    thing that will matter will be the size of the fragment. Since the pH of the solution
    matters as far as ionization of amino acids is concerned, the pH of the solution (buffer)
    used during electrophoresis is important to make sure the denatured polypeptides are
    uniformly negatively charged.
    5 points: As an example, consider albumin, a protein made of a single polypeptide
    weighing 66,000 daltons (66 kDa). On the other hand, gamma globulin has quaternary
    structure: is made of four polypeptides, two of which weigh 23,000 daltons (23 kDa)
    each, and two of which weigh 53,000 daltons (53 kDa) each. When treated with reducing
    agent and SDS, the subunits separate and they all linearize. If albumin and gamma
    globulin were run through gel electrophoresis, which polypeptides would move the
    fastest? Which would move the slowest?
    10 points: Consider the proteins in Figure 1. Assume they are treated with chymotrypsin
    to cut them into fragments, and then the fragments are separated by gel electrophoresis.
    What would the fragment patterns look like in the gel for the three different proteins,
    assuming we can separate polypeptides that differ in size by very small amounts?
    Complete Figure 2 below to show the location of the uncut polypeptides (- enzyme lanes)
    and the cut fragments (+ enzyme lanes).
    Figure 2: Gel electrophoresis of proteins from Figure 1 (- enzyme lanes) as well as cut
    fragments from those proteins generated through treatment by chymotrypsin (+ enzyme
    lanes). The location of the band for the uncut protein 1 is shown.
    In this experiment we will be comparing the structural similarity of serum albumin from
    several different vertebrate species and thus examining how closely related the proteins
    from these species are to each other. These proteins have similar molecular weights. We
    will first cut the proteins with the enzyme chymotrypsin (enzyme digest). We will then
    separate the fragments based on their relative sizes through gel electrophoresis so that we
    can compare the fragments generated by the enzyme digest for the different species.
    4 points: Your instructor will let you know which species will be studied in your lab.
    Based on that, write the hypothesis and prediction for the experiment.
    • Serum albumin from several different vertebrate species
    • Chymotrypsin
    • Sample buffer containing glycerol, SDS, -mercaptoethanol, and bromphenol blue
    (tracking dye)
    • Coomassie blue stain (40% methanol, 10% glacial acetic acid, 0.25% Coomassie
    brilliant blue R-250)
    • De-staining solution (20% methanol, 5% glacial acetic acid)
    • Electrophoresis buffer
    • Standard protein markers (BioLabs P7706S/L)
    • Incubator at 37C
    • Ice bath
    • Boiling water bath
    • Bench-top shaker
    • Gel electrophoresis chambers
    • 5% agarose gels
    • Micropipettors and tips
    • Gloves and goggles
    Basic Procedure
    Part I: Enzyme digest
  8. Wear goggles and gloves.
  9. Each group will be given the serum albumin from one species. For your group, label
    3 microcentrifuge tubes 1, 2, and 3.
  10. To each microcentrifuge tube add 10 μL serum albumin for the species assigned to
    your group.
  11. Add 10 μL distilled water to tube 1 and 10 μL chymotrypsin to each of tubes 2 and 3.
  12. Add 20 μL sample buffer to tube 1, mix and place in an ice bath.
  13. Place tubes 2 and 3 in the 37C incubator. Set a timer for 5 minutes for tube 2. Set
    another timer for 30 minutes for tube 3.
  14. Remove tube 2 from the incubator after 5 minutes. Add 20 μL sample buffer to tube
    2, mix and place in an ice bath.
  15. Remove tube 3 from the incubator after 30 minutes. Add 20 μL sample buffer to tube
    3 and mix.
  16. Mix the contents of each tube well before boiling the samples.
  17. Either use cap locks for the microcentrifuge tubes, or use a straight pin to pierce the
    lid of the tubes to provide a vent for steam to escape the tubes during boiling. Care
    should be taken to make sure the tubes are not inverted in the water bath so that the
    contents are not lost or diluted!
  18. Once the water is boiling vigorously, place the tubes in a rack and place the rack in
    the boiling water bath for 5 minutes.
    Part II: Analyzing peptide fragments by gel electrophoresis
  19. Your instructor will prepare the gel and place the gel into the electrophoresis chamber
    and will add enough buffer to cover the surface of the gel.
  20. Practice loading mock samples (dye solution) into the wells in a practice gel. You
    need to be comfortable with inserting the micropipettor tip into a well without
    pushing too far to puncture the bottom of the well. You also need to be comfortable
    injecting the sample into the well and removing the micropipettor tip without sucking
    the sample back into the pipette or releasing the sample into the buffer solution
    instead of the well.
  21. One gel will be used to analyze the peptide fragments from the tubes for 2 groups.
    Depending on the number of groups, the serum albumin from one or two species may
    be analyzed by more than one group.
  22. Follow Table 1 below to load 10 μL of the samples/markers into each well of a gel.
    Table 1: The order of samples loaded into the wells of the gel for electrophoresis.
    Sample well # Sample
    1 Standard proteins (ladder)
    2 Protein sample from Group 1 Tube 1
    3 Protein sample from Group 1 Tube 2
    4 Protein sample from Group 1 Tube 3
    5 Standard proteins (ladder)
    6 Protein sample from Group 2 Tube 1
    7 Protein sample from Group 2 Tube 2
    8 Protein sample from Group 2 Tube 3
  23. 5 points: After all of the samples are loaded, connect the electrodes to the gel box.
    Where must the negative electrode be in relation to the wells in the gel? Why?
  24. Turn on the power supply and run the gel at 100 V until the blue dye has migrated to
    within 1 cm of the positive electrode end of the gel (approximately 3-4 hours).
  25. Turn off the power supply. Disconnect the electrical leads.
  26. Remove the gel from the unit and transfer to a staining box.
  27. The staining solution contains Coomassie blue stain as well as acetic acid. The acid
    will precipitate and immobilize the proteins in the structure of the gel matrix so that
    the protein bands do not become blurry due to diffusion. Cover the gel with staining
    solution and allow it to stand for 1-2 hours.
  28. After the staining step, the unbound stain must be removed from the gel. This is done
    through the de-staining step. The de-staining solution is a dilute mixture of acetic
    acid and methanol. Pour out the extra stain (can be saved for future use) and add destaining
    solution to cover the gel. De-stain for 4 hours with gentle shaking.
  29. Examine the gel to observe the bands. Record the relative migration of the standards
    and the bands observed in the lanes for tubes 1-3 for each species studied.
  30. Interpret each lane of the gel:
  31. 6 points: What do you expect in the lane for #1 tubes? Explain the reason for your
    expectation. Do the results match the expectations?
  32. 5 points: Do you observe differences in the lanes for tubes 2 and 3? What does that
    tell you?
  33. 10 points: Are there differences among the species? What does that tell you about
    how closely related the proteins are from these species? Do the results match the
    prediction and support the hypothesis?
  34. 5 points: How would your results have been different if an enzyme with a different
    amino acid sequence preference had been used?

Describe the basic steps involved in protein digestion, protein denaturation, and electrophoresis

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