Unit 2

Question 1

This is a reactive oxygen species that is damaging to biological materials.

Answer 1

superoxide

Question 2

This is the organic portion of the heme group in hemoglobin.

Answer 2

protoporphyrin

Question 3

This is a genetic disease due to the decreased production of one of the subunits of hemoglobin.

Answer 3

Thallassemia

Question 4

This is the chemical form in which most of the carbon dioxide is transported in the blood.

Answer 4

bicarbonate

Question 5

This substance is produced when carbon dioxide reacts with water.

Answer 5

carbonic acid

Question 6

This type of hemoglobin is composed of two α chains and two γ chains.

Answer 6

fetal

Question 7

This is the molecule whose function is to store oxygen is muscle cells.

Answer 7

myoglobin

Question 8

This oxidized hemeprotein does not reversibly bind oxygen.

Answer 8

metmyoglobin

Question 9

This type of binding is indicated by a sigmoidal-shaped binding curve

Answer 9

sigmoidal

Question 10

This condition is a result of a single point mutation in the β chain of hemoglobin.

Answer 10

sickle cell anemia

Question 11

Under normal conditions, the heme iron in myoglobin and hemoglobin is in the ____________ oxidation state.

Answer 11

ferrous, or Fe⁺²

Question 12

The ability of myoglobin to bind oxygen depends on the presence of a bound prosthetic group called _____________.

Answer 12

heme

Question 13

In hemoglobin, the iron of the heme is bonded to the four nitrogens of porphyrin and to the proximal ______________ residue of the globin chain.

Answer 13

histidine

Question 14

The binding of 2-3-bisphosphogycerate to hemoglobin ____________ (increases, decreases) its affinity of oxygen binding.

Answer 14

decreases

Question 15

The effect of pH on oxygen-binding of hemoglobin is referred to as the _____________.

Answer 15

Bohr Effect

Question 16

Carbon dioxide reacts with the amino terminal groups of hemoglobin to form carbamate groups, which carry a ______________ charge.

Answer 16

negative

Question 17

The T-state of hemoglobin is stabilized by a salt bridge between β₁ Asp 94 and the C-terminal ___________________ of the β₁ chain.

Answer 17

histidine

Question 18

In normal adult hemoglobin, HbA, the β6 position is a glutamate residue, whereas in sickle-cell hemoglobin, HbS, it is a ____________ residue.

Answer 18

valine

Question 19

As the pressure of carbon increases, the affinity of oxygen binding to hemoglobin ______________.

Answer 19

decreases

Question 20

2,3-Bisphosphoglycerate binds only to the __________________ form of hemoglobin.

Answer 20

T- Deoxy

Question 21

What factor(s) influence(s) the binding of oxygen to myoglobin?

Answer 21

The partial pressure of oxygen, pO₂

Question 22

Which of the following is correct concerning the differences between hemoglobin and myoglobin?

1. Both hemoglobin and myoglobin are tetrameric proteins.
2. Hemoglobin exhibits a hyperbolic O₂ saturation curve while myoglobin exhibits a sigmoid shaped curve.
3. Hemoglobin exhibits cooperative binding of O₂ while myoglobin does not.
4. Hemoglobin exhibits a higher degree of O₂ saturation at all physiologically relevant partial pressures of O₂ than does myoglobin.

Answer 22

3- Hemoglobin exhibits cooperative binding of O₂ while myoglobin does not.

Question 23

Which of the following is not correct concerning myoglobin?

The globin chain contains an extensive α-helix structure.The heme group is bound to the globin chain by two disulfide bonds to cysteine residuesThe iron of the heme group is in the Fe⁺² oxidation state.The diameter of the iron ion decreases upon binding to oxygen.The function of myoglobin is oxygen storage in muscle.

Answer 23

The heme group is bound to the globin chain by two disulfide bonds to cysteine residues

Question 24

The structure of normal adult hemoglobin can be described as

A)a tetramer composed of four myoglobin molecules.B)a tetramer composed of two αβ dimmers.C)a tetramer composed of two α₂ and two β₂ dimers.D)a tetramer composed of two α₂ and two γ₂ dimers.E)None of these accurately describe hemoglobin.

Answer 24

a tetramer composed of two αβ dimmers.

Question 25

Which of the following is correct concerning fetal hemoglobin?

A)Fetal hemoglobin is composed of two α and two γ subunits.B)Fetal hemoglobin binds 2,3-BPG more tightly than normal adult hemoglobin.C)Fetal hemoglobin binds oxygen less than HbA at all pO₂.D)Fetal hemoglobin does not exist in the T-form.E)None of the above.

Answer 25

Fetal hemoglobin is composed of two α and two γ subunits.

Question 26

Hemoglobin-binding of oxygen is best described as a

A)concerted model. B)Michaelis-Menten model. C)sequential model. D)combination of sequential and concerted models. E)None of the above.

Answer 26

combination of sequential and concerted models.

Question 27

2-3 Bisphosphoglycerate

A)binds in the central cavity in the T-form of hemoglobin. B)preferentially binds to deoxyhemoglobin and stabilizes it. C)is present in the red blood cells. D)All of the above. E)None of the above.

Answer 27

All of the above.

Question 28

What is the Bohr effect?

A)the ability of hemoglobin to retain oxygen when in competition with myoglobin B)the regulation of hemoglobin-binding by hydrogen ions and carbon dioxide C)the alteration of hemoglobin conformation during low oxygen stress D)All of the above. E)None of the above.

Answer 28

the regulation of hemoglobin-binding by hydrogen ions and carbon dioxide

Question 29

Which of the following statements is correct for hemoglobin and oxygen transport?

A)The oxygen binds to the proximal histidine residue of the globin chain.B)Bonding of carbon dioxide to hemoglobin molecules increases the binding of oxygen.C)Hemoglobin binds more oxygen as the pH is lowered.D)Hemoglobin binds more oxygen at higher [BPG] concentrations.E)The binding of each O₂ molecule to hemoglobin increases its affinity for the next O₂ .

Answer 29

The binding of each O₂ molecule to hemoglobin increases its affinity for the next O₂ .

Question 30

Which of the following describes the Bohr Effect?

Lowering the pH results in the release of O₂ from oxyhemoglobin.Increasing the pressure of CO₂ results in the release of O₂ from oxyhemoglobin.Increasing the pH increases the T-form of hemoglobin. All of the above.a and b

Answer 30

a and b

Question 31

Which of the following is correct concerning the following equilibria?

CO₂ + H₂O >> H₂CO₃

A)An increase in the pressure of CO₂ will result in a decrease of pH.B)This reaction is catalyzed by carbonic anhydrase.C)The H₂CO₃ dissociates to H⁺ and bicarbonate ion, HCO₃^-.D)The majority of CO₂ is transported to the lungs in the form of HCO₃^-.E)All of the above.

Answer 31

All of the above.

Question 32

Carbon dioxide forms carbamate groups in proteins by reaction with

Answer 32

N-terminal amino groups.

Question 33

Sickle-cell anemia is caused by

Answer 33

a substitution of a Val residue for a Glu residue at the β6 position.

Question 34

Which of the following is correct concerning the oxygenation plot of proteins X and Y shown in the attached graph?

A)Protein Y exhibits tighter oxygen-binding than protein X.B)Protein Y corresponds to fetal hemoglobin, and protein X corresponds to normal adult hemoglobin.C)Protein X corresponds to fetal hemoglobin, and protein Y corresponds to normal adult hemoglobin.D)Protein X corresponds to myoglobin, and protein Y corresponds to hemoglobin.E)None of the above.

Answer 34

Protein X corresponds to fetal hemoglobin, and protein Y corresponds to normal adult hemoglobin.

Question 35

Which of the following is not correct concerning the oxygenation plot of proteins X and Y shown below?

A)Protein X exhibits tighter oxygen binding than protein Y.B)Protein Y would function as a better transport protein than protein X.C)Protein X exhibits cooperative binding, whereas Y does not.D)Protein X corresponds to myoglobin, and protein Y corresponds to hemoglobin.E)Protein Y contains multiple binding sites.

Answer 35

Protein X exhibits cooperative binding, whereas Y does not.

Question 36

Which is not correct concerning the models that are accepted to describe cooperative binding?

A)In the sequential model, the binding of a ligand changes the conformation of the subunit to which it binds, which in turn induces a change in neighboring subunits.B)All known allosteric proteins exhibit either the concerted or sequential model exclusive of the other.C)Both models incorporate a low affinity T-state and a higher affinity R-state.D)Both models explain the sigmoid-shaped binding curve.E)In the concerted model, all molecules exist either in the T-state or the R-state.

Answer 36

All known allosteric proteins exhibit either the concerted or sequential model exclusive of the other.

Question 37

Consider the oxygen-binding profile at three different pH values of 7.6, 7.4, and 7.2. Which statement is most correct?

A)Curve X most likely corresponds to pH 7.2.B)Curve Z most likely corresponds to pH 7.6.C)Hb has a higher affinity for oxygen at the pH of curve Z.D)Curve Y most likely corresponds to pH 7.4.E)pH has no effect on the oxygenation of hemoglobin.

Answer 37

Curve Y most likely corresponds to pH 7.4.

Question 38

What would be the expected result of a Lys residue being substituted with a Ser residue in the BPG binding site of hemoglobin?

Answer 38

BPG would bind less tightly because of the loss of a positive charge

Question 39

Why is it advantageous for hemoglobin to have allosteric properties?

Answer 39

Hemoglobin binds oxygen in a positive cooperative manner. This allows it to become saturated in the lungs, where oxygen pressure is high. When the hemoglobin moves to tissues, the lower oxygen pressure induces it to release oxygen and thus deliver oxygen where it is needed.

Question 40

What is fetal hemoglobin? How does it differ from adult hemoglobin?

Answer 40

Fetal hemoglobin contains two a and two g chains, in contrast to adult hemoglobin with two a and two b chains. The fetal hemoglobin g chain is probably a result of gene duplication and divergence. The difference in the chains results in a lower binding affinity of 2-3 BPG to fetal hemoglobin. Thus, the fetal hemoglobin has a higher affinity for oxygen, and the oxygen is effectively transferred from the mother’s hemoglobin to fetal hemoglobin.

Question 41

What is metmyoglobin?

Answer 41

Metmyoglobin is formed when the heme iron ion, which is normally in the +2 oxidation state, is oxidized to the +3 oxidation state. This oxidized form of myoglobin does not bind dioxygen and is not functional.

Question 42

Describe the octahedral coordination sphere of the iron ion in hemoglobin and myoglobin.

Answer 42

The Fe⁺² ion is coordinated to the four nitrogens in the center of the protoporphyrin of the heme. The fifth coordination site is occupied by the “proximal histidine” of the globin chain. The oxygen is bound to the sixth coordination site of the iron.

Question 43

What functional role does the “distal histidine” play in the function of myoglobin and hemoglobin?

Answer 43

The bonding between the iron and oxygen can be described as a combination of resonance structures, one with Fe²⁺ and dioxygen and another with Fe³⁺ and superoxide. The “distal histidine” donates a hydrogen bond to this complex stabilizing the complex and inhibits the oxidation of the iron to the ferric state.

Question 44

Draw the oxygen-binding curve of myoglobin and that of hemoglobin. Indicate the partial pressure of oxygen in the lungs and the range of pressure in tissue.

Answer 44

See graph

Question 45

Describe the structure of normal adult hemoglobin.

Answer 45

Normal adult hemoglobin, HbA, is a tetramer. It is composed of two α subunits and two β subunits. Each subunit has a structure very similar to myoglobin. It can be best described as a pair of identical αβ dimers. Each subunit contains a heme group. So, each molecule of hemoglobin can bind up to four molecules of oxygen.

Question 46

Briefly describe cooperative binding.

Answer 46

Cooperative binding occurs in multi-subunit proteins that possess multiple binding sites. The binding of a ligand to one site causes a conformational change that influences the binding of the ligand to the next site. The binding sites are not independent, but each binding event affects the affinity of the next binding event.

Question 47

Describe the concerted model to explain allosteric cooperative binding.

Answer 47

The protein exists in two conformations, a T-state (for tense) that has a lower affinity for the ligand and an R-state (for relaxed) that has a higher affinity for the ligand. In the concerted model, all of the molecules exist either in the T-state or in the R-state. At each ligand concentration, there is an equilibrium between the two states. An increase in the ligand concentration shifts the equilibrium from the T- to the R-state.

Question 48

Describe the role of 2,3-bisphosphoglycerate in the function of hemoglobin.

Answer 48

2,3-bisphosphoglycerate, 2,3-BPG, is a relatively small, highly anionic molecule found in the RBC. 2,3-BPG only binds to the center cavity of deoxyhemoglobin (T-state). The size of the center cavity decreases upon the change to the R-form so that it cannot bind to the R-state. Thus, the presence of 2,3-BPG shifts the equilibrium toward the T-state. T-state is unstable, and without BPG, the equilibrium shifts so far toward the R-state that little oxygen would be released under physiological conditions.

Question 49

Describe the chemical basis of the Bohr effect.

Answer 49

The effect observed by Christian Bohr is that hemoglobin becomes deoxygenated as the pH decreases. In deoxyhemoglobin, three amino acid residues form two salt bridges that stabilize the T-state. One of these is formed between the C-terminal His β146 and an Asp residue (β94). As the pH increases, this stabilizing salt bridge is broken because His becomes deprotonated and loses its positive charge. At lower pH values, this His is positively charged. The formation of the salt bridge shifts the equilibrium from the R-state to the T-state, thus releasing oxygen.

Question 50

Describe how carbon dioxide affects the oxygenation of hemoglobin.

Answer 50

Increased levels of carbon dioxide cause hemoglobin to release oxygen. The more active the tissue, the more fuel is burned and the more CO₂ is produced. These active tissue cells have the greatest need for oxygen to produce more energy. The CO₂ combines with the N-terminal amino groups to form negatively charge carbamate groups. The negatively charge carbamate groups form salt bridges that stabilize the T-state. Thus, the increase of carbon dioxide causes the conversion of the R-state to the T-state, releasing the bound oxygen to the tissues producing the most CO₂.

Question 51

Briefly describe the cause of sickle-cell anemia.

Answer 51

Sickle-cell anemia is a genetic disorder that is the result of a single substitution of β6 Glu with a Val. This changes a negatively charged side chain to a nonpolar, hydrophobic side chain. This Val binds into a hydrophobic pocket on the β chain of an adjacent molecule whose β6 Val binds to another molecule, thus hemoglobin aggregates. These aggregates form long fibers that strain the RBC and force into a sickled shape. The distorted red blood cells clog capillaries and impair blood flow, resulting in the sickle-cell crisis. The sickled cells are then destroyed, resulting in the anemia.

Question 52

What is thalassemia?

Answer 52

Thalassemia is caused by the substantial decreased production of one of the subunits of hemoglobin. In α-thalassemia, the decreased production of the α chain results in the formation of tetramers of only the β chain. This β₄ binds oxygen more tightly than HbA and does not exhibit cooperative binding. In β-thalassemia, the α chains form insoluble aggregates in the immature red blood cells.

Question 53

What is the role of α-hemoglobin stabilizing protein?

Answer 53

Four genes express the α chains, and only two genes express the β chain. Thus, there is an excess of α chains, which if allowed, would aggregate and become insoluble. Red blood cells produce α-hemoglobin stabilizing protein (AHSP), which binds to the α chain monomers to from a soluble complex. This prevents the aggregation of the α subunits.

Question 54

In many enzyme assays, the natural substrate and product are not used. Why?

Answer 54

Many products are difficult to measure accurately. Some are simply difficult to measure, while others are difficult to discern against the background of other molecules present in the reaction. Instead, substrates are chosen that the enzyme can still process but that result in products that can be easily measured. For example, substrates are chosen that result in products that are colored and can be detected spectrophotometrically.

Question 55

Give an example of a reaction catalyzed by a hydrolase.

Answer 55

Hydrolases catalyze a reaction where a water molecule is added across some linkage that was originally formed by the removal of water. Some examples are the hydrolysis of an ester, the hydrolysis of a peptide bond, the hydrolysis of a glycoside bond, etc. A diagram similar to textbook Section 8.1 are examples.

Question 56

There are six basic categories of enzymes. List the categories and define the type of reaction.

Answer 56

The categories include oxidoreductases (oxidation-reduction reactions), transferases (group transfers), hydrolases (hydrolysis reactions), lyases (addition or removal across a double bond), isomerases (intramolecular group transfer), and ligases (ATP-dependent ligation of substrates).

Question 57

How is the substrate bound to the active site?

Answer 57

The active site is a small part of the total enzyme structure. It is usually a three-dimensional cleft or crevice, which is formed by amino acid residues from different regions of the polypeptide chain. The substrate is bound by multiple noncovalent attractions such as electrostatic interactions, hydrogen bonds, van der Waals forces, and hydrophobic interactions. The specificity is dependent on the precise arrangement of the various functional groups in the binding site.

Question 58

You believe a substrate fits into a cleft like a key into a lock, but your roommate does not. Who is right?

Answer 58

You are both partially correct. Like a lock and key, the substrate fits precisely into the enzyme. However, the site is not a rigid cleft, but is flexible. Thus, it is possible for the substrate to actually modify the shape of the site a bit, a hypothesis known as induced fit.

Question 59

In an enzymatic reaction in a test tube, the reaction will eventually reach equilibrium. Why does this not happen in living organisms?

Answer 59

In a cell, the product may be utilized for a subsequent reaction, thus the reaction may not reach equilibrium.

Question 60

What is the Michaelis-Menten equation? Define all parameters.

Answer 60

V₀ = V_max(S/(S + K_M))

Initial velocity

V₀

Maximum velocity

V_max

Substrate concentration

S

Michaelis constant

K_M

Question 61

How do the intermediate steps in multi-substrate enzyme mechanisms differ?

Answer 61

In a sequential displacement reaction, both substrates bind and a ternary complex of all three is formed. In a double displacement (ping-pong), one or more products are released prior to binding of all substrates. Thus, a substituted enzyme intermediate is formed.

Question 62

Would you expect the order of substrate binding to be critical for enzyme catalysis?

Answer 62

Yes, in some cases. For example, in ping-pong reactions, the proper substrate would have to bind to form the right substituted enzyme intermediate form. In sequential displacement, both conditions are observed. Substrates may need to bind in a particular order (lactate dehydrogenase) or the enzyme may bind substrates and release products in random order (creatine kinase).

Question 63

How are the types of inhibition kinetically distinguishable?

Answer 63

Competitive inhibition can be overcome by the presence of large amounts of substrate. However, the apparent K_M is increased. In noncompetitive inhibition, substrate can bind to the EI complex, however, the V_max is decreased. In mixed inhibition, both values may be altered.

Question 64

How do enzymes work?

Answer 64

They lower the activation energy by stabilizing the transition state of the substrates.

Question 65

Describe transamination

Answer 65

Transamination refers to the transfer of an amine group from one molecule to another. This reaction is catalyzed by a family of enzymes called transaminases.

The reaction results in the exchange of an amine group on one acid with a ketone group on another acid. It is analogous to a double replacement reaction.

Question 66

Derive the Michaelis-Menten Equation

Answer 66

1) The Briggs-Haldane steady-state assumption
is that [ES] stays essentially unchanged
kon [E] [S] = (koff + kcat)[ES]

Question 67

Name 2 reasons why oxygen cannot be carried through the blood without hemoglobin?

Answer 67

1. Oxygen is poorly soluble in aqueous solutions and cannot be carried to tissues in sufficient quantity if it is simply dissolved in blood serum.
2. Diffusion of oxygen through tissues is ineffective over distances greater than a few millimeters.

Question 68

Amino acids can also bind oxygen. Why are they not used to transport oxygen through the body?

Answer 68

However, none of the amino acid side chains in proteins are suited for the reversible binding of oxygen molecules.

Question 69

Name the complex ring structure that heme consists of? What molecule sits in the middle?

Answer 69

Polyporphyrin; Ferrous iron (Fe²⁺)

Question 70

Why is carbon monoxide highly toxic to aerobic organisms?

Answer 70

Carbon monoxide binds to heme just like O₂ but with greater affinity. When it does, O₂ can’t bind.

Question 71

What are the functions of hemoglobin and myoglobin in the body?

Answer 71

The tetrameric hemoglobin is responsible for oxygen transport in the bloodstream. The monomeric neuroglobin is expressed largely in neurons and helps to protect the brain from hypoxia (low oxygen) or ischemia (restricted blood supply).

Question 72

Explain the location of the proximal His residue in myoglobin.

Answer 72

The proximal His — is His⁹³ (the ninety-third residue from the amino-terminal end of the myoglobin polypeptide sequence) and is also called His F8 (the eighth residue in α helix F).

Question 73

How are equilibrium and rate constants denoted?

Answer 73

Equilibrium constants are denoted with a capital K and rate constants are denoted with a lowercase k.

Question 74

Two proteins, A and B, bind to the same ligand, L, with the binding curves shown below. What is the dissociation constant, K_d, for each protein? Which protein (A or B) has a greater affinity for ligand L?

Answer 74

Because Y represents the fraction of binding sites occupied by ligand, the concentration of ligand at which half the binding sites are occupied — that is, the point where the binding curve crosses the line where Y = 0.5 — is the dissociation constant. Because A is half-saturated at a lower [L], it has a higher affinity for the ligand.

Question 75

Myoglobin has a hyperbolic curve for oxygen. What does a hyperbolic curve illustrate? What type of curve does hemoglobin have with oxygen and how is this different from myoglobin?

Answer 75

A hyperbolic curve illustrates non-cooperativity. There usually is only one E-S complex so it follows the E + S –> E-S –> E-P –> E + P scheme. Hemoglobin has a sigmoidal curve. A sigmoidal plot indicates cooperativity. Myoglobin only has one binding site for oxygen and hemoglobin has 4.

Question 76

Why can’t a single-subunit protein with a single ligand-binding site produce a sigmoid binding curve?

Answer 76

A single-subunit protein with a single ligand-binding site cannot produce a sigmoid binding curve — even if binding elicits a conformational change — because each molecule of ligand binds independently and cannot affect ligand binding to another molecule.

Question 77

A protein that bound O₂ with high affinity would bind it efficiently in the lungs but would not release much of it in the tissues. If the protein bound oxygen with a sufficiently low affinity to release it in the tissues, it would not pick up much oxygen in the lungs. How does hemoglobin solve this problem?

Answer 77

Hemoglobin solves the problem by undergoing a transition from a low-affinity state (the T state) to a high-affinity state (the R state) as more O₂ molecules are bound.

Question 78

How is the binding of H and CO2 related to the binding of oxygen?

Answer 78

the binding of H⁺ and CO₂ is inversely related to the binding of oxygen.

At the relatively low pH and high CO₂ concentration of peripheral tissues, the affinity of hemoglobin for oxygen decreases H⁺ and CO₂ are bound, and O₂ is released to the tissues. Conversely, in the capillaries of the lung, as CO₂ is excreted and the blood pH consequently rises, the affinity of hemoglobin for oxygen increases and the protein binds more O₂ for transport to the peripheral tissues.

This effect of pH and concentration on the binding and release of oxygen by hemoglobin is called the Bohr effect

Question 79

The interaction of 2,3-bisphosphoglycerate (BPG) with hemoglobin molecules further refines the function of hemoglobin and provides an example of heterotropic allosteric modulation. What is the effect on the affinity of oxygen to hemoglobin?

Answer 79

2,3-bisphosphoglycerate (BPG) lowers the affinity of oxygen to hemoglobin stabilizing the T state (deoxyhemoglobin) state. This allows more oxygen to be delivered to the tissues when environments with low oxygen concentrations.

Question 80

What is the site of binding of 2,3-BPG?

Answer 80

The site of BPG binding to hemoglobin is the cavity between the β subunits in the T state. This cavity is lined with positively charged amino acid residues that interact with the negatively charged groups of BPG.

Question 81

Does the fetus have a greater or lesser affinity for oxygen than adults? Why?

Answer 81

Greater; Because a fetus must extract oxygen from its mother’s blood, fetal hemoglobin must have greater affinity than the maternal hemoglobin for O₂. The fetus synthesizes γ subunits rather than β subunits. This tetramer has a much lower affinity for BPG than normal adult hemoglobin, and a correspondingly higher affinity for O₂.

Question 82

What causes sickle cell anemia and what is it effect on hydrogen binding?

Answer 82

The altered properties of HbS result from a single amino acid substitution, a Val instead of a Glu residue at position 6 in the two β chains.

When hemoglobin from sickle cells (called hemoglobin S, or HbS) is deoxygenated, it becomes insoluble and forms polymers that aggregate into tubular fibers

Replacement of the Glu residue by Val creates a “sticky” hydrophobic contact point at position 6 of the β chain, which is on the outer surface of the molecule. These sticky spots cause deoxyHbS molecules to associate abnormally with each other, forming the long, fibrous aggregates characteristic of this disorder.

Question 83

Describe each term of the Michaelis - Menten equation.

Answer 83

V₀ - initial velocity of enzyme and substrate reaction

V_max - maximum velocity of the enzyme and substrate reaction

[S] - Concentration of substrate in free form

[E] - concentration of enzyme in free form

[ES] - enzyme-substrate concentration

K_m - Michaelis constant - substrate concentration in which V₀ is half V_max

Question 84

Draw the hyperbolic curve of an enzyme reaction and label each variable in the Michaelis - Menten equation.

Answer 84

See image

Question 85

How is V₀ determined in enzyme reactions?

Answer 85

V₀ is determined by the breakdown of ES to form product

V₀ = k₂ [ES]

Question 86

What is the Michaelis - Menten equation?

Answer 86

It is the rate equation for a one-substrate enzyme-catalyzed reaction. It is a statement of the quantitative relationship between the initial velocity, the maximum velocity, and the initial substrate concentration [S], all related through a constant, Km, called the Michaelis constant.

Question 87

What is the Briggs-Haldane steady state assumption?

Answer 87

The assumption that the formation of the [ES] complex is equal to its breakdown.

k₁ [E][S] = (k-₁ + k₂) [ES]

Question 88

Describe step 1 of the Michaelis-Menten equation derivation.

Answer 88

Step 1: The rates of formation and breakdown of ES are determined by the steps governed by the rate constants k₁ (formation) and (k-₁ + k₂) (breakdown to reactants and products, respectively), according to the expressions

Rate of ES formation = k₁ ([E_t] – [ES])][S] = k₁ [E][S]

Rate of ES breakdown = (k-₁ + k₂) [ES]

Question 89

Describe step 2 of the Michaelis-Menten equation derivation.

Answer 89

Step 2: The initial rate of reaction reflects a steady state in which [ES] is constant — that is, the rate of formation of ES is equal to the rate of its breakdown. This is called the steady-state assumption.

k₁ ([E_t] – [ES])[S] = (k-₁ + k₂) [ES]

Question 90

Describe step 3 of the Michaelis-Menten equation derivation.

Answer 90

Step 3: In a series of algebraic steps, we now solve Step 2 equation for [ES]. First, the left side is multiplied out.

1. k₁ [E_t] [S] – k₁[ES] [S] = (k-₁ + k₂) [ES] (Left side multiplied out)
2. k₁ [E_t] [S] = (k-₁ + k₂) [ES] + k₁[ES] [S] (k₁[ES] [S] added to both sides)
3. k₁ [E_t] [S] = (k-₁ + k₂ + k₁ [S]) [ES] (Right Side simplified)
4. Solve for [ES]
5. Simply further by combining rate constants
6. All rate constants are the Michaelis constant K_m = (k-₁ + k₂) /k₁

See picture

Question 91

Describe step 4 of the Michaelis-Menten equation derivation.

Answer 91

Step 4: We can now express V0 in terms of [ES]. Remember, V0 is determined by the breakdown of ES to form product. V0 = k2 [ES]

See picture

Question 92

Which of the following best describes the change to a Lineweaver Burk plot when a competitive inhibitor is added?

a) The x-intercept shifts further from zero
b) The y-intercept increases
c) The x-intercept shifts closer to zero
d) The y-intercept decreases

Answer 92

C is correct: The x-intercept shifts closer to zero.

Lineweaver Burk plots are a graphical method of analyzing the Michaelis–Menten equation and the enzyme-substrate-inhibitor relationship. The x-axis in the graph is 1/[S], and the y-axis is 1/V. In competitive inhibition, the inhibitor decreases the binding affinity for the enzyme to the substrate by competing with the substrate for the active site. This competition increases the K_M, since it will take more substrate to outcompete the inhibitor and move the reaction forward. Increasing the K_M will cause the x-intercept of the Lineweaver-Burk plot to shift closer to zero (choice C is correct; choice A is incorrect). The y-intercept will not change (choices B and D are incorrect).

Question 93

What variables represents the slope in a Lineweaver Burke plot?

Answer 93

K_m / V_max

Question 94

What variables represents the y and x-intercepts in a Lineweaver Burke plot?

Answer 94

x-intercept = -1/Km

y-intercept = 1/Vmax

Question 95

Which axes are K_m and V_max represented on a Lineweaver Burke plot?

Answer 95

Km is the x-axis and Vmax is y-axis

Question 96

How will K_m and V_max respond on a Lineweaver Burke plot in the presence of a competitive inhibitor?

Answer 96

In terms of competitive inhibitors, recall that they compete directly with the substrate to bind to the active site of the enzyme. In turn, they will decrease the affinity of the enzyme for the substrate, which will increase the K_M. As stated above, the x-intercept is equal to -1 / K_M. Increasing the value of the K_M will then decrease the value of the x-intercept. Decreasing the value of the x-intercept will shift the line closer to zero. Also, the y-intercept will stay the same because competitive inhibitors do not affect the maximum velocity of the reaction. In competitive inhibitors, the slope of the graph increases, the x-intercept shifts to the right, and the y-intercept is unchanged.

Question 97

How will K_m and V_max respond on a Lineweaver Burke plot in the presence of a uncompetitive inhibitor?

Answer 97

For uncompetitive inhibitors, both the K_M and the V_max will be reduced. Recall that since uncompetitive inhibitors increase the affinity of the enzyme for the substrate, K_M decreases. Also, since the x-intercept is equal to -1 / K_M, a smaller K_M would give a larger value for the x-interceptor. In this way, the x-intercept will shift to the left. Also, since uncompetitive inhibitors also decrease the maximum reaction velocity, the value of the y-intercept, or 1 / V_max , will increase. Increasing the y-intercept will shift the line upwards. Therefore, uncompetitive inhibitors will result in a parallel line to the original graph, in which both the x and y-intercepts have changed.

Question 98

How will K_m and V_max respond on a Lineweaver Burke plot in the presence of a noncompetitive inhibitor?

Answer 98

For noncompetitive inhibitors, the K_M is unaffected, and the V_max decreases. Recall that noncompetitive inhibitors are a type of mixed inhibitors. As KM is unchanged, the x-intercept stays the same; however, since the V_max decreases, the y-intercept shifts upward.

Question 99

What is the benefit of the Lineweaver Burke plot in enzyme kinetics?

Answer 99

The Lineweaver Burke plot illustrates enzyme kinetics in a linear fashion.

y=mx + b

See picture

Question 100

What is the Km?

Answer 100

Km is equivalent to the substrate concentration at which V0 is one-half Vmax

Question 101

For a process at equilibrium, ΔG,

Answer 101

ΔG=0, When ΔG is zero, the forward and reverse driving forces are equal, and so the process occurs in both directions at the same rate (the system is at equilibrium).

Question 102

How do living organisms obey the second law of thermodynamics?

Answer 102

In short, living organisms preserve their internal order by taking from their surroundings free energy in the form of nutrients or sunlight, and returning to their surroundings an equal amount of energy as heat and entropy.

Heterotrophic cells acquire free energy from nutrient molecules, and photosynthetic cells acquire it from absorbed solar radiation. Both kinds of cells transform this free energy into ATP and other energy-rich compounds capable of providing energy for biological work at constant temperature.

Question 103

For an organism, when is equilibrium reached?

Answer 103

In the case of an organism, equilibrium is reached only after death and complete decay.

Question 104

If the equilibrium constant for a given chemical reaction is 1.0, what is the standard free energy change for that reaction? What if Keq is greater than or less than 1, what is the standard free energy change?

Answer 104

The standard free-energy change of that reaction is 0.0 and the reaction is at equilibrium (the natural logarithm of 1.0 is zero)

If Keq is >1.0, standard free energy is negative and the reaction proceeds forward, if Keq is <1.0, standard free energy is positive and the action proceeds in the reverse direction.

Question 105

Substrate level phosphorylation refers to

Answer 105

Transfer of a phosphoryl group from a high energy donor to ADP to make ATP

Question 106

What makes a compound high energy? Give examples of high energy phosphate compounds.

Answer 106

Acyl phosphates (1,3 bisphosphoglycerate), enol phosphates (phosphoenolpyruvate), and creatine phosphate.

Question 107

Enzymes, acting as catalysts, accelerate the rates of reactions by

Answer 107

Lowering the free energy of activation of a reaction for the reaction to proceed

Question 108

Km, the Michaelis constant in the Michaelis-Menten model of enzyme kinetics, is

Answer 108

The substrate concentration at which half of the active sites are occupied

Question 109

The ___of an enzyme is the number of substrate molecules converted to product per
enzyme molecule per unit time when the enzyme is saturated with substrate

Answer 109

turnover number

Question 110

In competitive enzyme inhibition, the inhibitor binds to

Answer 110

the active site and prevents the substrate from binding there.

Question 111

In uncompetitive enzyme inhibition, the inhibitor binds to

Answer 111

only to the ES complex

Question 112

Hemoglobin and myoglobin bind oxygen using

Answer 112

Fe+2 that is bound to a heme prosthetic group

Question 113

What are the 8 steps of the chymotrypsin mechanism?

Answer 113

1. His 57 catalyzes removal of H from Ser 195 hydroxyl
2. Ser 195’s nucleophilic O attacks carbonyl C of substrate
3. His 57 donates H to N of scissile peptide bond, tetrahedral intermediate decomposes
4. the portion (C-terminal end) of original substrate with the new amino terminus diffuses away
5. water donates H to His 57
6. resulting OH attacks carbonyl of remaining substrate
7. His 57 donates H to Ser 195 O, leading to collapse of tetrahedral intermediate
8. the portion (N-terminal end) of original substrate with the new carboxylate terminus diffuses away

Question 114

In hemoglobin a ___ histidine donates a ______ to oxygen, in part to prevent the oxidation of the heme iron.

Answer 114

Distal/hydrogen bond

Question 115

The oxygen binding of hemoglobin is most easily approximated using which binding model? Explain the two models.

Answer 115

Neither model adequately describes the oxygen binding of hemoglobin

1. Monod, Wyman, Changeux two state model - The concerted model assumes that the subunits of a cooperatively binding protein are functionally identical, that each subunit can exist in (at least) two conformations, and that all subunits undergo the transition from one conformation to the other simultaneously.
2. The sequential model of Koshland, Nemethy and Filmer - A conformational change in one subunit makes a similar change in an adjacent subunit more likely, and makes the binding of a second ligand molecule more likely as well.

Question 116

The affinity of hemoglobin for oxygen is affected by various ligands. The binding of which of the following increases the affinity of hemoglobin for oxygen?

1. 2,3 bis-phosphoglycerate
2. Protons (ie, lower pH)
3. Carbon dioxide
4. Oxygen
5. Nitric oxide

Answer 116

Oxygen

Nitric oxide and CO bind to the heme group and make O₂ unable to bind.

Question 117

1. Sickle cell anemia is the direct result of a single amino acid change in the beta chain of Hb. What is the cause (more hydrophobic/more hydrophilic) and effect (oxyHb or deoxyHb) becomes less soluble?

Answer 117

More hydrophobic / deoxyHb

Question 118

1. When hemoglobin is 50% saturated with oxygen the predominant forms (oxy vs deoxy) of hemoglobin are

Answer 118

There are approximately equal amounts of oxy and deoxy Hb but the oxy Hb contains approximately 3-4 oxygens and the deoxy Hb contains approximately 0-1 oxygen molecule.

Question 119

1. Explain why hemoglobin has a sigmoidal oxygen binding curve while myoglobin has a simple hyperbolic binding curve.

Answer 119

The reason hemoglobin has a sigmoidal or S-shaped oxygen binding curve whereas myoglobin has a hyperbolic oxygen binding curve is because hemoglobin is a tetramer whereas myoglobin is a monomer.

Question 120

Explain the mechanisms by which an increase in [H⁺] and [CO₂] alter the binding of oxygen to hemoglobin.

Answer 120

Hemoglobin binds to H+, CO2 and oxygen. Oxygen binds to the heme group while CO2 and H+ bind to amino acid residues. When CO2 and H+ bind, they change the affinity of hemoglobin to oxygen, allowing it to release the oxygen in areas of low pH and high CO2 concentration.

CO2 binds to the amino-terminal residue to form a carbaminohemoglobin. The reaction produces H+, contributing to the Bohr effect.

When the oxygen concentration is high, as in the lungs, hemoglobin binds O2 and releases protons. When the oxygen concentration is low, as in the peripheral tissues, H+ is bound and O2 is released.

Question 121

What is the physiological significance of the change in oxygen binding in response to pH and carbon dioxide?

Answer 121

The pH of blood is 7.6 in the lungs and 7.2 in the tissues. Experimental measurements on hemoglobin binding are often performed at pH 7.4.

Carbon dioxide in the tissues is high, some CO2 binds to hemoglobin and the affinity for O2 decreases.