lecture 8.105 Enzymes HO

Biological Sciences 105 Lecture 8, October 25, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 1 When we stopped last time we had just finished looking at the binding equilibria for myoglobin. 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. 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. The 4  structure of hemoglobin gives it different binding properties than myoglobin. Look at O 2 binding curve. hemoglobin is much smaller than myoglobin

Biological Sciences 105 Lecture 8, October 25, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 2 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. 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. It is sigmoidal because hemoglobin binds oxygen cooperatively . How does cooperativity make a hyperbolic binding curve into a sigmoidal one? Look at most extreme case of cooperativity: Must bind n molecules of ligand A to a protein, or none. P + n  A  PA n n D n n A n n a n a n a n a n n n n a A K A Y A K A A K A K A P K P A P K PA P PA Y A P PA K ] [ ] [ ] [ 1 ] [ ] [ 1 ] [ ] ][ [ ] [ ] ][ [ ] [ ] [ ] [ ] ][ [ ] [ + = + = + = + = + = = Plot this for n = 4, the extreme for hemoglobin, and get a sigmoidal curve. High affinity state = T state vs. low affinity state = R state.

Biological Sciences 105 Lecture 8, October 25, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 4 A Hill plot, with slope = Hill coefficient. Look at the y-axis intercepts How is cooperativity achieved? Conformational change of hemoglobin upon binding of O 2 . The heme group wants to be planar, but is pulled into pucker by the proximal His ligand. The binding of O 2 relieves this pucker by causing the rearrangement of positions of the amino acid residues near the O 2 binding site, making the heme planar. This causes the breaking of specific H-bonds and salt bridges in the deoxy form, and the formation of new H-bonds and salt bridges in the oxy form. This ultimately leads to rearrangement of the 4  structure. The  -  pairs move as log {y/(1-y)}

Biological Sciences 105 Lecture 8, October 25, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 5 a unit, but rotate 15  with respect to each other. You can think of it as the two  subunits sliding closer together. Play Lan- Xin’s PowerPoint slide 23 and the next few. The 4th O 2 is bound 300 x more easily than is the 1st. Just the reverse occurs when releasing O 2 . Salt bridges reform, heme puckers more and more until it is easy to release the last O 2 . An allosteric modulator is a small molecule that affects the activity of a protein in some way. Can be positive or negative. O 2 is an allosteric modulator of the affinity of Hb for O 2 . There are more allosteric modulators of the affinity of hemoglobin for O 2 .

Biological Sciences 105 Lecture 8, October 25, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 7 2,3-bisphosphoglycerate is also an allosteric effector of hemoglobin. Usually present in equimolar amounts with hemoglobin. Binds in the center of deoxyhemoglobin (can't fit into oxy-Hb) and stabilizes deoxy structure. Stabilizes T structure, not S. So what effect on the affinity of Hb for O 2 ? BPG decreases affinity of hemoglobin for O 2 .

Biological Sciences 105 Lecture 8, October 25, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 8 As we’ll see when we look at structure of Hb, there is a highly positively charged cleft at the top of the protein at a  -  interface where the negatively charged BPG binds. When a sea-level human goes to high altitudes, BPG in blood increases. This decreases affinity of hemoglobin for O 2 , facilitating the unloading of O 2 in the tissue. Fetal hemoglobin is slightly different from adult hemoglobin. His 143 in the  subunit is replaced by Ser. This gives fetal Hb a lower affinity for BPG, because two of the + charges in the BPG cleft are lost. That gives a higher affinity of Hb for O 2 at a given BPG concentration. This helps the fetal Hb extract O 2 from the maternal blood.

Biological Sciences 105 Lecture 8, October 25, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 10 Recall what the enzyme has to do here is speed the reaction and then be regenerated at the end. That is, act as a true catalyst. An energy diagram which defines  G and E a Recall that the effect of the catalyst is to lower the activation energy barrier for the reaction, not to change the position of equilibrium. That is, just because the reaction has a negative value for  G does not mean that the product will be formed fast. In fact, it says nothing about the rate of the reaction, only something about the position of the equilibrium according to the equation:  G = - RT ln K eq . In a similar vein, the energy of activation says nothing about the position of the equilibrium for the reaction, only how quickly the equilibrium positions will be reached. The important point here is that enzyme or catalyst speeds both the forward and reverse reactions, because it is the equilibrium position that is being sought by the reaction. The catalyst simply allows that equilibrium to be reached sooner. To be more quantitative, there is a relation that can be derived in physical chemistry between the rate of a reaction and the E a . We will see in more detail next time that the speed of any reaction can be described by rate equations using rate constants . For a unimolecular reaction: B A k  , the rate at which that occurs is: V = k  A, where k is a proportionality constant, or rate constant. It is the rate constant which can be related to E a by the following equation:

Biological Sciences 105 Lecture 8, October 25, 2018 Copyright Steven M. Theg, 2018. All federal and state copyrights reserved for all original material presented in this course through any medium, including lecture or print. 11 RT E a e h T k − = k k = Boltzmann constant (1.38E-23 J/K) and h is the Plank's constant (6.63E-34 J s). The point is that k is inversely and exponentially proportional to E a , meaning that the higher E a is, the lower k is. If you look at the x-axis as the reaction coordinate, you will see there is a point at which you can no longer say you have the substrate, nor do you have the product. What you have now is the transition state . This is an elusive and fleeting chemical state, completely unstable. Where does the transition state occur on the diagram? At the top of the E a peak. From the transition state, the reaction proceeds both forward to products or backwards to substrates with equal probability. Again, there is nothing about the E a that dictates the direction of the reaction. Why does more product form (before equilibrium) in the presence of a catalyst for a reaction with a negative  G? Because the distance from product to the transition state is larger than the distance of the substrate to the transition state. However, there is nothing inherently directional about the action of an enzyme on a reaction.