lecture 7.105 Structures, MB_HB HO

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Biological Sciences 105 Lecture 7, October 23, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 1 On to higher order structures How do we know proteins fold into unique structures? Early evidence came from the ability to crystallize a protein. In order to make a crystal, need to have many identical molecules. Therefore, even though have a large molecule (protein), it must have a single, unique structure. Why do proteins fold into unique structures? Take all possible conformations and calculate the change in free energy ( G) compared to the completely unfolded random coil. A plot of the folding pathway may look something like this: Final conformation is likely to be the global minimum energy configuration. The G for the final folded structure is generally low, -20 to - 65 kJ/mol, which I call low compared to the -200 to - 400 kJ/mol for the formation of a single covalent bond. Most of the stabilizing force comes from the many small, non-covalent interactions discussed earlier. What’s the destabi lizing force? Answer is … Then the G of stabilization is the balance between the large increase in Gibbs free energy due to entropy and the large decrease in Gibbs Free Energy due to the many weak interactions. Two general rules apply to protein structure: 1. Put the hydrophobic residues in the interior away from water, 2. the number of hydrogen bonds should be maximized. The four levels of protein structure: Primary structure is the amino acid sequence. Secondary structure is the residue-by-residue conformation of the backbone. Tertiary structure is the three dimensional conformation of a single polypeptide. Quaternary structure is the three dimensional structure of a multisubunit protein.
Biological Sciences 105 Lecture 7, October 23, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 2 Secondary Structure - = local structure. Five distinct types Recall that the C-N bond has partial double bond character due to sharing of the e - in the O double bond. This constrains the rotation of the C with respect to the N. Planar. Can get rotation around the carbon. A couple of residues look like this: Linus Pauling (who is this?) realized that short term interactions between nearby residues could result in hydrogen bonds. Number of different types of defined 2 structure. Not all rotations about the carbon result in stabilizing H-bonds, nor are they all allowed. The angle refers to the angle between the carbon and the carbon of the carboxyl (on the right), and the angle refers to the angle between the carbon and the nitrogen on the amine (on the left). One representation of the shape of a protein backbone in 3D space is to give the position in 3D space of each of the atoms. But another, equally valid representation would be to give the position of the first atom, and then all the ψ an d angles thereafter. This is sometimes used. 1. Right-handed -helix . Many ways to draw.
Biological Sciences 105 Lecture 7, October 23, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 3 Characteristics of an -helix: 1. Rises 0.54 nm per complete turn 2. One turn goes 3.6 residues. 5.4 A/3.6 = 1.5 A/residue. 3. The R groups stick out of the helix center. They don't participate in the hydrogen bonds, but they do influence them. 4. Hydrogen bonds are formed between fourth residues. Go to the computer and show an α -helix in MbO. Rainbow color A, find polar, just main chan Select resn pro, sticks 2. Anti-parallel -sheet.
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Biological Sciences 105 Lecture 7, October 23, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 4 Can create a continuous sheet of antiparallel -sheet structure by having the protein loop back upon itself many times, like so; Go to the computer and show a -sheet, 3. The turn region of the antiparallel -sheet has a particular conformation, called a -turn or -bend . Go to the computer and show a -sheet 4. Parallel -sheet . Cannot be formed by adjacent sections of the protein, must be connected through distant regions. Like so;
Biological Sciences 105 Lecture 7, October 23, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 5 5. Random coil Most other structures are commonly referred to as random coil . They are not necessarily random, but simply not one of usually defined 2 structural types. The different R groups influence whether a particular 2 structure is stable. If this were not the case, all proteins would be made of only -helices sheets. For example: Proline does not fit well into an -helix because it has constraints on the rotation of the bonds coming from the -carbon. Proline is a helix breaker. It also doesn't fit well into -sheets, but does fit well in - turns. Glycine's R group (H) is too small to exclude water from the backbone region, so it H-bonds to water instead of the residue 4 positions away. It is allowed in -helices, but it does not stabilize them. The particular 2 structure finally adopted depends on a balance between stabilizing and destabilizing forces. Know many of the important forces, so can predict with some success different 2 structural elements from the 1 structure. There are numerous programs on the internet that do this, none of which are perfect.
Biological Sciences 105 Lecture 7, October 23, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 6 Tertiary structure Most proteins are globular, with hydrophobic residues buried inside, away from water. The position of every atom in 3D space gives you the tertiary structure of the protein. Two general rules apply to tertiary protein structure: 1. Put the hydrophobic residues in the interior away from water, 2. the number of hydrogen bonds must be maximized. Show folding surface on computer. For a long time it was thought that the particular conformation a protein adopts is determined solely by its 1 structure, and that the 3 structure was achieved spontaneously. Self-folding and self -assembly. This is certainly true for some (many) proteins. However, Levinthal’s Paradox: Can protein randomly search for their global energy minimum? Take a protein with 100 aa’s. Assume there are only 2 conformations possible for each aa. Then there are 2 100 = 1.27 x 10 30 conformational possibilities. Allowing the conformations to be sampled at the speed of molecular vibrations, one for every 10 -13 s, then the time to test all the conformations is 10 -13 s x 1.27 x 10 30 = 1.27 x 10 17 sec = 4 x 10 9 years, which is about the age of the Earth. What? It is now known that many other proteins require additional factors in the cell to fold in the correct manner. Molecular chaperones. Folding pathways.
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Biological Sciences 105 Lecture 7, October 23, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 7 Correct vs. incorrect folding pathways. You'll probably hear about this in Bis104. 3 structure results from a combination of weak interactions. 1. Hydrophobic interactions between R-groups and water, most important. 2. H-bonds between side chain and other side chains or the backbone. 3. Salt bridges between side chains. 4. Disulfide bridges stabilize, that is, lock in certain interactions. When a protein is heated, or plunged into strong acid or base, or exposed to non- polar solvents like acetone, etc., the interactions between the different groups that stabilized the 3 structure are destroyed. Why acid or base? What happens? Why acetone? The protein whose structure has been destroyed in this manner is referred to as denatured . Some proteins can renature spontaneously. Lots can't, and that's where chaperones come in. But the fact that some can proves that the 1 structure leads to the 3 structure. How is 3D structure determined? Originally, by x-ray diffraction. Recall what diffraction is from physics. Can tell from the pattern what the shape, size, etc. of the original slit is. Can rotate the slit with respect to the light and get a new set of diffraction patterns from which one can determine 3D information, i.e., the thickness of the material in which the slit is made. Can do the same thing with crystals and x-rays. In crystals there are spaces
Biological Sciences 105 Lecture 7, October 23, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 8 between the molecules. Spaces too small to be examined with visible light (400 nm wavelength), but can be examined with x-rays (5 nm wavelength). Rotate the sample, get new pattern, repeat. Then calculate what the position of all the atoms must be to give the observed diffraction patterns. Looks hard? It is. The real problem is getting a reference point in the molecule. This is done by substituting heavy metals into the structure, then getting new diffraction patterns. You know the position of the metal substitute, and you can see its influence on the other positions on the diffraction picture. I heard this good analogy. It's like trying to figure out the position of every branch and leaf on a tree by examining its shadows. The first person to undertake this was Max Perutz, using hemoglobin. Mid 1930's. He wanted to do this for his Ph.D. project. It took 25 years. They did give him his Ph.D. before the solution was finished. Awarded Nobel Prize when finally finished. The first protein solved was actually myoglobin, not hemoglobin. Solved by a student of Perutz, Kendrew. Kendrew shared the Nobel with Perutz. The reason myoglobin was solved first is because it is a single polypeptide, whereas hemoglobin has four subunits. The two rules for protein structure given earlier, hydrophobics on the inside and maximize hydrogen bonds, were first seen with the structure of myoglobin.
Biological Sciences 105 Lecture 7, October 23, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 9 It has held up with the solution of the structure to hundreds of other proteins. Although the other proteins look quite different from myoglobin and each other, the same principles apply. Hemoglobin harder to do than myoglobin because it has four subunits, two kinds. Two and two subunits, arranged The subunits are held to one another by the same forces that stabilize a tertiary structure. Each different subunit is rather like myoglobin. Quaternary Structure Many proteins have subunits. Quaternary structure refers to the 3D placement of its subunits as well. The four levels of protein structure: Primary structure is the amino acid sequence. Secondary structure is the residue-by-residue conformation of the backbone. Tertiary structure is the three dimensional conformation of a single polypeptide. Quaternary structure is the three dimensional structure of a multisubunit protein. Search for RCBS protein database. Look at PyMol, 1MBO, 2HHB, 1GFL
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Biological Sciences 105 Lecture 7, October 23, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 10 Myoglobin and Hemoglobin We will use the binding properties of myoglobin and hemoglobin to illustrate a number of important points about the way proteins (and enzymes) interact with other molecules. Myoglobin and hemoglobin are O 2 binding proteins. Myoglobin stores O 2 in muscles, making it available for quick release during muscle activity. Why needed? Carbohydrate + O 2 energy + CO 2 + H 2 O; what is this? Hemoglobin circulates in the blood and supplies tissues with O 2 . Both myoglobin and hemoglobin use a prosthetic group to bind O 2 . A prosthetic group is a non-polypeptide addition to a protein that adds to the protein's ability to function. Many different kinds. The prosthetic group used by myoglobin and hemoglobin is called a heme . This ring structure is called a tetrapyrrole or a porphyrin ring . It can be substituted at all the X positions, making molecules with different properties. Very hydrophobic. When the porphyrin ring has an Fe atom in the center, it is a heme. Other substitutions give rise to other types of molecules, i.e., chlorophyll has a Mg in the porphyrin.
Biological Sciences 105 Lecture 7, October 23, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 11 The Fe in heme is Fe +2 , this referring to the oxidation state of the atom. Fe +2 has six binding orbitals. That is, it can bind to six different atoms, or it has a coordination number of 6. In heme, 4 of the 6 are contributed by the porphyrin nitrogens. A fifth ligand is a nitrogen in a nearby His residue, called a proximal His ligand. This is out of the plane of the page, i.e., So heme is held into place in the protein by hydrophobic interactions and by the covalent bond to Histidine. The last coordination site is used to bind O 2 . Why use a protein at all. Why not free heme? O 2 oxidizes Fe +2 to Fe +3 , which can no longer bind O 2 . The protein keeps the Fe in the +2 oxidation state, and makes the binding of O 2 reversible . This is important. Both myoglobin and hemoglobin must be able to both bind and release O 2 , or they are no good for storage. Describe the binding equilibria for O 2 and myoglobin: K a is the association constant = K eq for association. The higher the affinity of the binding site for the ligand, the higher is K a . This K a is not the acid dissociation constant, which was unfortunately also called K a . K d is the dissociation constant = K eq for the dissociation reaction. The higher the affinity of the binding site for the ligand, the lower is K d .
Biological Sciences 105 Lecture 7, October 23, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 12 What follows is the derivation of the hyperbolic relationship between the fraction of total binding sites that are occupied as a function of the concentration of the ligand . P = protein, A = ligand P + A PA; K eq = K a = association constant (not acid dissociation constant!) ] ][ [ ] [ A P PA K a = ] [ ] [ ] [ P PA A K a = and ] ][ [ ] [ P A K PA a = Dissociation constant , Kd, is K eq for PA P + A So K d = 1/K a Define Y = fraction of binding sites occupied = ] [ ] [ ] [ P PA PA + Substituting [PA] from above, d a a a a a a a K A A Y K A A K K A K A K P P P P A K P A K Y + = + = + = + = ] [ ] [ 1 ] [ ] [ ] [ 1 ] [ ] [ ] [ ] [ ] [ ] ][ [ ] ][ [
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Biological Sciences 105 Lecture 7, October 23, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 13 So what is Y really? Let Y = ½ Then [A] 1/2 is the value of [A] that gives Y= ½. Y 1/2 = 0.5 = [A] 1/2 / ([A] 1/2 + K d ) 0.5[A] 1/2 +0.5 K d = [A] 1/2 K d = ([A] 1/2 - 0.5[A] 1/2 ) / 0.5 = 0.5[A] 1/2 / 0.5 K d = [A] 1/2 So K d = ligand concentration at which ½ of all binding sites are occupied. For O 2 -binding proteins like Mb, [A] = [O 2 ] = the partial pressure of O 2 above the solution. This is proportional to [O 2 ]. Partial pressure is easier to measure than [O 2 ], and we use it instead. Now, 50 2 2 P pO pO Y + = where P 50 is the pO 2 giving half saturation of Mb. This is the equation of a hyperbolic line, approaches 1.0 asymptotically
Biological Sciences 105 Lecture 7, October 23, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 14 In capillaries, pO 2 = 20-30 mm Hg. So myoglobin is saturated with O 2 when muscles at rest. When muscles work hard and use up O 2 , pO 2 drops and O 2 is released from myoglobin. A simple storage system that delivers extra O 2 when needed. Affinity Look at what K d means. Kd is the equilibrium constant for the dissociation reaction. Kd is the concentration of ligand that gives ½ saturation of the protein. This means that Kd measures the affinity of the protein for the ligand. Low Kd means a low concentration of ligand to push the equilibrium far towards the PA complex. So a low Kd means high affinity. Hemoglobin is different. Hemoglobin Not a storage protein, but a carrier protein. Hemoglobin has a molecular mass of 65 kDa, which is approximately 4 x 17 kDa for myoglobin. Consists of 4 subunits, 2 2 . These and have nothing to do with -helix and ß-sheet 2 structure. Y
Biological Sciences 105 Lecture 7, October 23, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 15 Hemoglobin binds 4 O 2 per molecule, or 1 O 2 per subunit. The quaternary structure of hemoglobin is such that the hydrophobic regions of the individual subunits are buried from water. Salt bridges between chains. and are about 50% the same. They were obviously derived from the same ancestral gene. Subunits about 30% similar to myoglobin. High sequence homology indicates an evolutionary reason to keep certain parts of the protein 1° structure unchanged. Genes have evolutionary trees, like family trees, just like organisms. The 4 structure of hemoglobin gives it different binding properties than myoglobin. Look at O 2 binding curve. Properties of curve: inflection point at 26 mm Hg. pO 2 in capillaries is 20 - 30 mm Hg. pO 2 in lungs is 100 mm Hg.
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Biological Sciences 105 Lecture 7, October 23, 2018 Copyright Steven M. Theg, 2017. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 16 The curve indicates that hemoglobin is completely saturated with O 2 in the lungs, and releases O 2 to the tissues (or myoglobin) in capillaries. Just what you want for an O 2 carrier protein.

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