Recommended problems from chapter 15: 5,7,8,10,13,15,17,20,21

Recognition of substrates by

See chemical mechanisms notes on web page, covered in chapter 14 in your textbook. Also see chapter 13 for discussion of the “lock and key” vs. the “induced fit” substrate binding modes.

2 hypotheses on how substrates fit within the of enzymes:

1. “Lock and key” •Exact fit between substrate and

2. “Induced fit” •Enzyme active site is “flexible” or “dynamic” and can assume distinct (but probably related) conformations. A corollary is that it can probably accommodate distinct (but probably somewhat similar) substrates. • “Good” substrates can fit into the active site in such a way that the transition state is approximated.

1 General considerations in the regulation of enzymes

1. Genetic control. At this level of regulation the amount of the enzyme is changed in response to environmental conditions. For example, in response to an abundance of glucose, the production of metabolic enzymes that cleave large polysaccharides stops. In the case of lac repressor, this regulation is on the transcription level.

2. Control of enzymatic activity. At this level the amount of an enzyme is not varied. Rather, the activity of the enzyme is regulated (up or down). A. For S ↔ P, as the amount of product builds up the reverse reaction becomes more pronounced, until equilibrium is reached, where there is no net change in the amounts of product and substrate.

B. Enzymatic rates can be modulated by the amount of available substrates and their Km values.

C. Covalent modification can alter the affinity of enzyme to substrate (alter the Km). Phosphorylation and de-phosphorylation are such control systems. Nobel prize in 1992 was awarded to Edmond Fischer and Edwin Krebs for this discovery.

D. Allosteric effectors are molecules that alter the affinity of an enzyme to a substrate. These are non-covalent modifications.

Regulation of enzyme activity by phosphorylation

Phosphorylation can occur on serine, threonine, or tyrosine residues.

What are the predicted physical consequences of this modification on an enzyme active site?

2 Aspartate transcarbamoylase as a model system of allosteric control The reaction catalyzed by ATCase is diagrammed below. This is one of the initial reactions in pyrimidine synthesis.

O O- C

NH2 CH2 + C C 2- O OPO3 + - H3N COO H Carbamoyl phosphate Aspartic acid

aspartate transcarbamoylase

O O- C

NH CH - 2 2 + H2PO4

C C

O N COO- H H N-Carbamoyl aspartate

Feedback inhibition of ATCase by CTP

3 Allosteric models

•Allosteric interactions affect the activity of an enzyme by binding at a site distinct from the active site (allos = different; stereo = place or solid).

•Allosteric effectors can be the substrate(s); this is a homotropic effect.

•The effector(s) can be distinct from the substrate(s) molecules; this is a heterotropic effect.

•An allosteric effector can be positive (increases enzymatic activity) or negative (decreases enzymatic activity). Often both effects are found for a given enzyme.

For example, in the system: O2 is a homotropic positive effector (increases affinity for at the oxygen ). Bisphosphoglycerate (BPG) and CO2 are negative heterotropic effectors (decrease affinity for oxygen).

In general, allosteric effects result from interactions among subunits of oligomeric . It is possible, however, that single-subunit proteins can exhibit allosteric effects if there are multiple binding sites for substrates within the single polypeptide.

The Monod, Wyman, and Changeux allosteric model:

Consider a that may assume two states. The “taut” state binds substrate very poorly, whereas the “relaxed” state has a high affinity for substrate. The subscript 0 denotes an unbound protein, and these two states exist in equilibrium, such that:

R0 ↔ T0

Further assume that the equilibrium of R0 and T0 lies substantially to the right, such that in the absence of substrate there is a lot more taut form than relaxed. If L is the association

equilibrium constant, L = [T0] / [R0] and L is large.

If KR is the equilibrium dissociation constant of substrate from the relaxed form, and KT is the equilibrium dissociation constant of substrate from the taut form, then by the conditions above

KR << KT.

The extreme case is KR / KT = 0, and in this case there is no affinity of T0 to substrate.

4 The Monod, Wyman, and Changeux allosteric model: cooperative binding

R0 ↔ T0

If substrate is added to this system, the equilibrium is perturbed. R0 binds the substrate and no longer exchanged with T0. Now the reaction above goes to the left, more R0 is made, and, in turn, more substrate is bound by R0, and so on. This is a positive homotropic effect.

The Monod, Wyman, and Changeux allosteric model: cooperative binding

c = KR / KT ; L = [T0] / [R0]; n = number of substrate binding sites; Y = [filled substrate binding sites] / [total number of substrate binding sites].

This substrate binding behavior is cooperative. Filling of a binding site alters the apparent overall affinity for substrate. All binding sites, however, are equivalent.

5 The Monod, Wyman, and Changeux allosteric model; positive heterotropic effects

Each subunit has an allostery site and a substrate binding site. If an allosteric activator (A) is present, then the R0 ↔ T0 reaction goes to the left, and there are more substrate binding sites available. This equilibrium shift results in an increase of the apparent (measured) affinity for substrate at a fixed activator concentration. Overall, the of substrate binding decreases, because there are more substrate binding sites available. The allosteric activator has the same effect as a lower equilibrium constant L.

The Monod, Wyman, and Changeux allosteric model; negative heterotropic effects

In the presence of inhibitor, the equilibrium R0 ↔ T0 goes to the right. This reduces the number of substrate binding sites available, and lowers the apparent affinity for substrate. In the presence of a fixed concentration of inhibitor, a titration of substrate would show an increase in the cooperativity of binding, but a decrease in the apparent affinity.

6 The Monod, Wyman, and Changeux allosteric model; heterotropic effects

Plot of v0 versus [Aspartate] for the ATCase reaction

Is ATP an inhibitor or activator? How does it affect cooperativity of aspartate binding? Is UTP an inhibitor or activator? How does it affect cooperativity of aspartate binding? Does this make sense if you consider the role of ATCase in pyrimidine synthesis?

7 Structure of E. coli ATCase

•A view along the 3-fold axis of the 300 kDa protein complex. Composition is c6r6. •The 6 catalytic subunits are in red and blue, and consist of 2 trimers stacked atop each other. •The 6 regulatory subunits are in yellow, there are 3 dimers that bind the two trimers. •Each trimer has catalytic activity that is higher than the catalytic activity of the complex. The trimer is neither affected by ATP nor CTP. What does this argue about the role of the regulatory subunits? Where do you expect the regulatory allostery effectors to bind?

Structure of E. coli ATCase

The T-state conformation The R-state conformation

A view perpendicular to the previous slide, along the molecular two-fold symmetry axis. Given the information about ATCase in the previous slides, what reagent would you add to the protein in order to obtain these two distinct structure?

8 Structure of E. coli ATCase The T-state conformation The R-state conformation

O O PALA is a compound that C C PO 2- Carbamoyl binds the enzyme but 3 H N C O PO 2- H2 2 3 phosphate does not react. NH + NH3 Predict to which -OOC C C COO- - - H2 H OOC C C COO Asp artate conformation PALA H2 H would bind. N-(Phosphonaetyl)-L-aspartate (PAL A)

The Koshland, Nemethy, and Filmer model, AKA the sequential model

•The symmetry model assumes a constant, fixed substrate binding site (Emil Fischer’s “lock and key” model).

•In contrast, the sequential model allows for changed in the conformation of the other binding sites as a result of substrate binding at a particular site on the oligomeric protein. The binding sites are NOT always equivalent.

•This means that substrate binding sites are “coupled,” such that filling one binding site changes the other binding sites on the oligomeric protein.

•The degree of “coupling” among the subunits can vary. At the extreme, when the subunits are completely coupled, the sequential model actually resembles the symmetry model.

9 The symmetry model of allostery

•A tetramer protein is represented.

•All the subunits of a given protein can be either in the T or the R form.

•No combinations of R and T subunits are allowed within the protein.

•The T and R states are at equilibrium regardless of the number of substrates bound.

The sequential model of allostery

•Binding of S to one subunit of the enzyme can induce a in the other subunits of the enzyme.

•The greatest change occurs at the subunits that have bound .

•The symmetry of the protein subunits is not preserved as was the case with the symmetry model.

•At the heart of this model are the distinct conformations that the subunits assume as a consequence of substrate binding.

10 Hemoglobin as a model system for

Hemoglobin and serve similar functions. They bind O2 and deliver it as needed. Hemoglobin is found in blood, whereas myoglobin is found in muscle. Each polypeptide binds a single group, a porphyrin ring structure that binds Fe2+. The porphyrin ring system also binds O2.

Oxygen binding by myoglobin

If the heme group contains Fe2+, oxygen binding is efficient. In contrast, if the heme group contains Fe3+, oxygen binding is poor. The function of the myoglobin polypeptide, in part, can be viewed as: (1) to provide structure that carries the heme group. (2) to protect the heme group from oxidation. (3) to provide a binding site for oxygen.

11 Oxygen binding by myoglobin

In oxymyoglobin, O2 is the sixth binding ligand of the iron ion. In deoxymyoglobin, the 6th binding position is empty. In metmyoglobin (containing Fe3+), water fills the 6th binding position.

Oxygen binding by myoglobin

His E7 “crowds” the binding site of CO, such that it binds to myoglobin with 100 fold lower affinity in comparison to the free heme group with imidazole (shown in a).

The Myoglobin-CO affinity is 250 fold higher than the myoglobin-O2 affinity.

12 Oxygen binding by myoglobin Oxygen binding to myoglobin can be written simply as a reversible equilibrium

Mb + O2↔ MbO2

[Mb][O2 ] The equilibrium dissociation constant of oxygen from myoglobin is Kd = [MbO 2 ] The fractional saturation of myoglobin, Yoxygen, is the fraction of filled O2 binding sites divided by the total number of O 2 € binding sites. [MbO ] [O ] Y 2 2 O2 = = [MbO2] +[Mb] Kd + [O2]

This works out to be a hyperbolic curve, just like the binding of substrate to € enzyme under steady-state conditions (or binding of to receptor).

pO2 in blood is 100 torr arterial and 30 torr venous. What are the venous and arterial fractional saturation values of myoglobin?

Deoxyhemoglobin structure

Hemoglobin is a tetramer (contains 4 polypeptides), 2 α and 2 β, and can be written α2β2. The X-ray crystal structure of hemoglobin was solved in 1968 by , after having worked on this problem for approximately 30 years (what an optimist!).

•Notice the two-fold symmetry between the α1β1 and α2β2 subunits.

•Mb, Hb-α, and Hb-β are not very similar in terms of amino acid sequence. In terms of tertiary structure, however, the three polypeptides assume a remarkably similar structure.

13 Structure of oxyhemoglobin

Note how the subunits are closer than they are in deoxyhemoglobin. The grey arrows in the center area indicate the shifting of the subunits.

Oxygen binding to Hb

The p50 of hemoglobin is 26 torr.

The dashed line indicates a hyperbolic, hypothetical binding curve with a half- maximal saturation of 26 torr. The dashed line indicates how simple binding between hemoglobin and oxygen would look.

Clearly, there is binding cooperativity going on here!

14 The Hill equation and the Hill coefficient

In 1910 performed an analysis of the sigmoidal Hb oxygen binding curve. Consider a protein E with n subunits, each containing a single binding site for a ligand S. Assume that the protein binds with infinite cooperativity, such that

E + nS ↔ ESn Here the protein has all the sites filled or none of the sites filled.

The dissociation constant of this reaction is

[E][S]n K = [ESn ]

As before, the fractional saturation YS is given by €

[ESn ] YS = [E]+[ESn ]

€

The Hill equation and the Hill coefficient [E][S]n Substituting [ES ] = into the fractional saturation equation gives the Hill equation: n K [E][S]n [E][S]n [ES ] [S]n Y = n = K = K = S [E]+[ES ] [E][S]n [E]K +[E][S]n K + [S]n € n [E]+ K K

Having an infinite cooperativity is physically impossible. The value of n, however, can be used to measure the degree of cooperativity of the system. Here n is not the actual number of binding sites,€ rather it can be taken to mean the degree of cooperativity between binding sites within the system. The variable n is the Hill coefficient.

For a particular system, if n = 1 there is no observed cooperativity. If n < 1 there is negative cooperativity (binding ligand to a site reduces the ligand affinity of subsequent sites on the same protein. If n > 1, there is positive cooperativity in the system. Also, if you know that the number of binding sites is 3, and your hill coefficient calculations give a number greater than 3, then you know that somewhere along the line you’ve made a large error!

15 Hill Plot of hemoglobin and myoglobin [S]n Y K +[S]n [S]n The Hill equation can be arranged to give S = = 1− Y [S]n K S 1- K +[S]n

Taking the log of both sides yields € Y log( S ) = nlog[S]− logK 1− YS

€ which is a linear plot with a slope of n and an X-intercept of logK/n.

Movement of Hb helix F as a consequence of O2 binding

Unbound structure is in blue, bound structure is in red. In the unbound structure, the iron ion is 0.6Å above the ring structure. In the bound conformation, the iron ion is within the plane of the porphyrin ring assembly. As the ring moves, the histidine that is bound to it moves, and, in turn, the entire helix F moves. This allows higher affinity in the other 3 oxygen binding sites of Hb.

16 In 1904 Christian Bohr (father of Niels) Modulation of hemoglobin O2 binding: reported that O2 the and BPG affinity for hemoglobin increases with increasing pH (reduced [H+]). This is called the Bohr effect.

For every O2 molecule Hb binds, ~0.6 protons are released due to the T→ R transition.

BPG (2,3 bisphosphoglycerate) is a negative allosteric effector of Hb. It lowers Hb affinity for O2 and increases the degree of cooperativity.

CO2 and pH effects on Hb O2 binding

Respiring tissues produce CO2, which is converted to bicarbonate in the erythrocyte by the activity of : + - CO2 + H2O ↔ H + HCO3

+ The presence of CO2 in respiring tissues results in an increase in [H ], lower pH, and a lower affinity of O2 to Hb. This results in the unloading of oxygen in respiring tissues and the - formation of the T state. As a result, additional HCO3 is made.

In the lungs, where pO2 is high, Hb binds O2, becomes converted into the R state, and unloads + H . This results in the production of CO2, which is exhaled.

17 Abnormal Hb is found in human diseases 1. Most Hb mutations occur on the surface of the protein and do not have clinical manifestations, e.g., HbE. HbS is a glaring exception.

2. Internal residue mutations destabilize Hb, result in release of the heme group (Heinz bodies), and lysis of the red blood cells. This is termed hemolytic anemia.

3. Other internal residue mutations destabilize Hb but don’t result in release of the heme group. Rather, the function of the heme group is disrupted. In Hb Bristol, valine→aspartate at position 11 of Hbβ results in positioning of a polar group in contact with the heme, thus facilitating the access of water to the subunit’s hydrophobic interior, and destabilization of oxygen binding to the heme. Perturbation of heme binding results in lower O2 affinity.

4. Changes in the α-β contact surfaces sometime result in an increase in oxygen binding. This results in a lower O2 saturation in the tissues. Individuals compensate for this condition by having increased amount of RBCs. This is termed polycythemia.

5. Sickle-cell anemia (HbS) is caused by a protein surface mutation of glutamate to valine. The polar→non-polar residue substitution on the surface of the protein results in aggregates of deoxyHbS into rigid fibers that extend through the surface of the erythrocyte and give it an inflexible, “sickled” appearance.

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