Enzymes - Chapter 7, 8, 9

Question 1

How does 2, 3 BPG affect oxygen transport from lungs to tissue?

Answer 1

It is able to create salt-bridges which stabilise the T (tense) state of hemoglobin and thus makes the release of oxygen more favourable. Furthermore, its stabilisation of the T state also facilitates new oxygen transport. Usually, 2,3 BPG is present at a similar concentration to hemoglobin.

Question 2

What is “cooperative” binding?

Answer 2

Essentially, cooperativity refers to the phenomenon of one oxygen molecule binding to a heme group in hemoglobin making a second binding more likely (favourable). This is caused by an induced 15 degree rotation of the two dimers at the alpha-beta dimer interface, caused by the heme iron first moving into the protoporphyrin plane, which results in the proximal histidine moving which in turn is connected via an alpha helix to the interface. The rotation produces a more relaxed state in freer rotation, and the increased possibility of moving raises the opportunity of a second oxygen binding.

Question 3

How does the structure of the heme group facilitate oxygen binding?

Answer 3

At the centre of a functional heme group lies an iron (II) ion which is bonded four times in the same plane to protoporphyrin. Above and below the plane lie the 5th and 6th coordination sites. In myoglobin, the 5th site is taken up by the imidazole of the proximal histidine. These free sites are where oxygen binds as it is attracted to the positive charge of the iron.

Question 4

What happens in the heme group upon oxygen binding?

Answer 4

When oxygen binds to the iron in the heme group, it takes on some of its electron density, resulting in it changing oxidation state (becoming an Fe3+), while the oxygen is converted to superoxide (very reactive). The loss of electron density reduces the size of the iron and allows it to move 0.4 Å to fit into the protoporphyrin plane (which ultimately facilitates cooperative binding).

Question 5

What is the superoxide ion and what are its effects?

Answer 5

Superoxide is an oxygen molecule with additional electron density. It is extremely unstable and thus both reactive and damaging to organic tissues. Thus, it is important to prevent its release. This is achieved by a distal histidine coordinating to the superoxide, stabilising it. This is an example of the protein component of hemoglobin affecting the heme group’s activity.

Question 6

Describe the structure of hemoglobin:

Answer 6

Hemoglobin consists of four myoglobin chains (2 alpha and 2 beta, which are related by divergent evolution). Each of these chains contains only alpha helices, and has a tertiary structure referred to as a global fold. An alpha and a beta chain coordinate together to form a dimer, and one hemoglobin is made up of two such dimers connected via an interface. Iron-iron distances between heme groups lie between 24 to 40 Å apart.

Question 7

How do changes in partial oxygen pressure affect the uptake and release of oxygen by hemoglobin in tissues?

Answer 7

The partial pressure of oxygen in the lungs is 100 torr, while in tissues it is only 20 torr. When looking at the sigmoidal oxygen binding curve of hemoglobin, it can be seen that the difference in fractional saturation of hemoglobin between these concentrations is 66%. For myoglobin, which does not follow a sigmoidal curve but a steep inverse exponential curve, this difference is only 7%. This makes hemoglobin better adapted to transport oxygen from lungs to tissues, and means that 66% of oxygen taken up in the lungs gets released into tissues.

Question 8

What are the differences between hemoglobin and myoglobin?

Answer 8

1. myoglobin is comprised of a single globin fold, while hemoglobin consists of two alpha-beta dimers connected at an interface. Each alpha and beta chain is essentially a myoglobin chain.
2. hemoglobin transports oxygen from lungs to tissues while myoglobin accepts oxygen in the tissues.
3. Myoglobin: P50 = 2 torr ; hemoglobin: P50 = 26 torr
   -> myoglobin attaches to oxygen more efficiently, while the sigmoidal shape of the hemoglobin oxygen binding curve allows for maximisation of oxygen release into tissues (66% vs 7% if myoglobin were the transport-protein)

Question 9

How is the oxygen transport adapted to exercise?

Answer 9

During exercise, the partial oxygen pressure in resting muscle is 40 torr while it is 20 torr in exercising muscle. Thus, when looking at hemoglobin’s oxygen binding curve, it can be seen that 45% of oxygen is transported to exercising muscle while 21% is transported to resting muscle.

Question 10

Explain the structural differences between deoxyhemoglobin and oxyhemoglobin:

Answer 10

In deoxyhemoglobin, the entire enzyme is in the tense (T) state; this means that there is little free rotation around the dimer interface and the enzyme is constrained by subunit-subunit interactions. In oxyhemoglobin (all heme groups oxygenated) however, there is a lot of free rotation and less constraint. The free rotation furthermore allows for more opportunity to bind more oxygen in the case that the hemoglobin is only partially oxygenated, meaning that with every oxygen binding, another becomes more likely.
Upon oxygen binding, the iron ion is able to move 0.4 Å into the protoporphyrin plane as it loses some electron density to the oxygen, making the proximal histidine move, causing the connected alpha helix to move as well which ultimately results in changes at the dimer interface.

Question 11

Why is the oxygen binding curve sigmoidal?

Answer 11

The sigmoidal shape results from the T and R state binding curves contributing to form the overall hemoglobin binding curve. While the R state contributor resembles the myoglobin curve, the T state contributor is a straight line beginning at the origin.

Question 12

Why is carbon monoxide dangerous to the human organism?

Answer 12

Carbon monoxide is able to bind to hemoglobin at the oxygen binding site irreversibly and around 200 times tighter than oxygen. Thus, if carbon dioxide is taken in, it can permanently reduce our ability to take in oxygen and transport sufficient oxygen for tissues to keep alive. CO poisoning is treated in high oxygen pressure chambers.;

Question 13

How does pH affect hemoglobin’s oxygen binding ability?

Answer 13

Additional protons are able to coordinate with different amino chemical groups in the protein component which have a pKa of around 7. They are able to facilitate the formation of additional salt bridges which stabilise the T state, making oxygen release more favourable. Thus, at a lower pH, hemoglobin’s oxygen binding curve will have a less steep rise. This ultimately means that oxygen transport rates respond to changes in pH.

Question 14

How does carbon dioxide presence affect hemoglobin’s oxygen binding ability?

Answer 14

CO2 can react to form carbonic acid inside the red blood cells. This leads to a drop in pH which stabilises the T state (explained on another card). Furthermore, the carbonic acid can react with terminal amino groups at the interface to form carbamate, which are able to form additional salt bridges stabilising the T state.
Intermezzo: most CO2 transport results from HCO3- being taken back to the lungs by red blood cells.

Question 15

What is sickle cell anaemia and why does it occur?

Answer 15

In sickle cell anaemia, a single base-mutation in the genetic code for the beta chain of hemoglobin results in an amino acid substitution from glutamate to valine which changes the quaternary structure of the hemoglobin and results in the red blood cells taking on a crooked sickle shape. These differently shaped cells tend to aggregate and restrict blood flow, which causes the anaemia.

Question 16

What features allow proteases to hydrolyse peptide bonds?

Answer 16

Proteases tend to have an amino acid residue able to activate a nucleophile to attack the carbonyl carbon, as well as a component which polarises the peptide carbonyl group, making it more susceptible to attack.
Furthermore, they contain a feature which stabilises the tetrahedral intermediate which is formed as a result of nucleophilic attack on the carbonyl, i.e. and oxyanion hole.

Question 17

Apart from serine proteases, what other types of proteases are there and how do they function?

Answer 17

In cysteine proteases, a cysteine residue activated by a histidine acting as a general base catalyst gets activated to enable it to nucleophilically attack a carbonyl.
Aspartyl proteases tend to be symmetric enzymes with two aspartate residues in its active site, one of which activates a water molecule to make it a good nucleophile while the other polarises the peptide carbonyl.
Lastly, in metalloproteases, a metal ion (usually zinc (II)) activates a water molecule to enable it to attack the carbonyl.

Question 18

Explain how proteases in eukaryotes and bacteria are an example of convergent evolution.

Answer 18

Chymotrypsin, a protease present in eukaryotes, and subtilisin, found in bacteria, both contain a catalytic triad of serine, histidine, and aspartate, all connected by hydrogen bonds, the purpose of which is to deprotonate the serine group upon substrate binding to make it a better nucleophile and ultimately hydrolyse the peptide bond. These two proteases do not have a common ancestor, and rather are an example of convergent evolution arriving at the same solution for a problem in two different domains of organisms. The only difference between chymotrypsin and subtilisin relevant to peptide hydrolysis is a small incongruence in their oxyanion holes.

Question 19

Explain why product formation from chymotrypsin catalysing the hydrolysis of a peptide bond occurs in two phases.

Answer 19

Chymotrypsin-catalysed peptide bond hydrolysis occurs in two phases, a burst phase and a steady-state phase. This is because chymotrypsin catalyses reactions via covalent catalysis, meaning that it binds covalently to the acyl component temporarily while immediately releasing the amine component. This was discovered by measuring the absorbance and thus concentration of the yellow p-nitrophenolate. In the burst phase, p-nitrophenolate is released as one of the first fast steps of the process. Then the acyl-enzyme intermediate is hydrolysed to release the carboxylic acid component and a free enzyme. At this point, the reaction is subject to kinetic laws, and it slowly moves towards a steady-state with a Km = 20 uM and kcat = 77 s-1.

Question 20

What is the structure and function of the catalytic triad?

Answer 20

The catalytic triad consists of three amino acid residues interacting with each other. Ser 195’s side chain is H-bonded to the imidazole ring of his 57, which in turn is H-bonded to the Asp 102 carboxylate. The histidine positions serine’s side chain and polarises its hydroxyl for deprotonation, later on accepting a proton and acting as a general base catalyst. Deprotonation makes serine more nucleophilic and allows for a better attack on the carbonyl in the peptide. Aspartate facilitates histidine to accept said proton through electrostatic effects and H-bonding.

Question 21

Describe the steps of peptide bond hydrolysis by chymotrypsin:

Answer 21

To start with, the substrate binds to the active site. Then, the alkoxide serine binds to the carbonyl carbon via nucleophilic attack, producing a tetrahedral intermediate which is quite unstable. To prevent it from reacting along other pathways, it is stabilised by an oxyanion hole. The intermediate then collapses to an acyl-enzyme, while the amine component is formed from the proton on the histidine and subsequently departs. Following this, histidine causes a water molecule to dissociate, and the formed hydroxy binds to the carbonyl, again creating an unstable tetrahedral intermediate stabilised by the oxyanion hole. The carboxylic acid side-product is then released, and the catalytic triad returns to its original state.

Question 22

Which physical components of a reaction (equilibrium, kinetics, and enthalpy) are affected by enzymes? Explain:

Answer 22

Enzymes are not able to influence the equilibrium or enthalpy of a reaction. They act as catalysts, making them incapable of shifting the equilibrium or changing the energy change occurring. Enzymes only affect a reaction’s kinetics, by providing an alternative pathway and utilising binding energy to lower the reaction’s energy barrier as it is split into a multi-step process, as well as thus increasing a reaction’s rate. This results in the original equilibrium being achieved more rapidly.

Question 23

Define the Michaelis constant and the turnover frequency:

Answer 23

The Michaelis constant (Km) is defined as the substrate concentration at which the reaction proceeds at half the maximum rate. In real life, a lot of enzymes have developed a Km equal to the substrate concentration of their medium to maximise activity while remaining sensitive to environmental changes. The turnover frequency (kcat) is the maximum rate at which the reaction can proceed. If r = kcat, and increase in substrate concentration will not increase the rate, since at kcat all enzymatic active sites are always occupied.

Question 24

What types of cofactors are they and how do they influence enzymatic reactions?

Answer 24

Cofactors are compounds which bind to the enzyme along with the substrate to facilitate the reaction. Oftentimes they are either metals or coenzymes derived from vitamins.
Cofactors can either be prosthetic groups if they are tightly bound to the enzyme and remain with it permanently (i.e. the heme group in hemoglobin) or cosubstrates which only bind reversibly.

Question 25

Explain what the concept of a maximum rate / turnover frequency suggests about the enzymatic reaction mechanism.

Answer 25

Since at a certain substrate concentration the initial rate of the reaction achieves a maximum and remains at this rate even if the substrate concentration is increased suggests that the enzyme gets used up. It follows logically that while the enzyme is involved in the reaction, its catalytic abilities are focused on one substrate, insinuating that a reversible enzyme-substrate complex is formed, which can then react and dissociate into enzyme and product.

Question 26

What is an active site?

Answer 26

Active sites are 3D clefts into which substrate can bind via non-covalent interactions (note that they might bind covalently later in the case of covalent catalysis). While active sites and substrate fit together in a lock and key principle in theory, most of the time a process of dynamic recognition will allow them to bind in an induced fit.

Question 27

How do enzymes lower the energy barrier of a reaction?

Answer 27

Enzymes may lower a reaction’s energy barrier by hastening collision in correct orientation by binding substrates in proximity to one another. Furthermore, the binding energy released when a substrate attaches to an active site is used to bridge the activation energy.