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Biochem > Unit 3 > Flashcards

Unit 3 Flashcards

1
Q

Ligand

A

Small molecule that binds to a larger one. Ligands are bound reversibly to a protein

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2
Q

Binding site

A

A ligand binds a protein at a binding site that is complementary to the ligand in size, shape, charge, and hydrophobic or hydrophilic character

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3
Q

Induced Fit

A

The binding of a protein and ligand is also paired with a conformational change in the protein that makes the binding more complementary to the ligand, permitting tighter binding

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4
Q

Why do multicellular organisms transport oxygen on FE2+ incorporated into a heme group?

A
  • Oxygen is poorly soluble in aqueous solution and cannot be carried to tissues in sufficient quantity if put in the bloodstream because oxygen diffusion is ineffective over greater distances.
  • Multicellular organisms cannot use protein alone to transport oxygen because no amino acid side chains are suited for the reversible binding of oxygen molecules
  • Metals like Fe2+ can bind to oxygen but having free FE2+ promotes the formation of highly reactive oxygen species. (like hydroxy radical)
  • Fe2+ must be sequestered to be less reactive. It is incorporated into a protein bound prosthetic group called a heme to do this
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5
Q

Pyrrole Ring (draw structure)

A

There are four pyrrole rings

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6
Q

What are the 6 coordination points to iron? (draw)

A

Iron has 6 places where it can bond. 4 of them are in the plane. One is going up and the other is going down.

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7
Q

What is the oxidation state of Fe2+ in myoglobin and hemoglobin?

A

The oxidation state is Fe2+.
*Fe3+ doesn’t bind oxygen and it is formed when there is a O2 molecule on each side of the Fe.

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8
Q

Protoporphyrin

A

Refers to the the porphyrins lacking metal ions in the structure

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9
Q

Porphyrins

A

Pyrrole rings with metal inside the structure. A heme group is a specific type of poryphyrin with Fe2+

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10
Q

What is a globin?

A

Globins are a family of proteins that have heme groups

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11
Q

Describe the secondary and tertiary structure of myoglobin

A
  • Myoglobin secondary structure is composed entirely of alpha helices and non helical residues. There are around 8 alpha helices in myoglobin
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12
Q

Where is the location of the heme group in myoglobin?

A

The heme group is in the core of the protein

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13
Q

How does myoglobin folding occur? Describe the polar or nonpolar character of the inside and outside of the native protein. The oxygen binding site in myoglobin involves the heme group and two histidine residues. Is it surprising to find histidine residues in the interior of a protein?

A

Myoglobin folding occurs through the hydrophobic effect. It makes it favorable to fold the protein into a tertiary structure. The inside of a protein is typically non=polar and the outside is polar. It is surprising to find histidine residues in the interior of a protein because it is hydrophobic/polar which means that it should be on the outside of the protein.

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14
Q

What is the physiological role of myoglobin?

A

Facilitates oxygen diffusion in muscle tissue. Oxygen storage protein

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15
Q

What is the meaning of P50

A

The pressure at which half of the possible binding spots are bound to the myoglobin. The myoglobin is 50% saturated.

To put it simply, if we have a population of myoglobin molecules, at the P50 value, half of these myoglobin molecules will have oxygen bound to their binding sites, while the other half will not have oxygen bound.

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16
Q

What is the P50 of oxygen binding to myoglobin

A

0.26 kPa

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17
Q

What does the hyperbolic shape of the binding curve tell you about the sensitivity of myoglobin to changes in ligand concentration?

A
  • It is relatively insensitive to small changes in concentration of dissolved oxygen, so it functions well as an oxygen-storage protein
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18
Q

Discuss the role of the distal histidine in the myoglobin ligand binding site

A

The role of the distal histidine is to create a bond with the oxygen and to prevent the bonding of carbon monoxide.
- Due to the linear conformation of carbon monoxide, there would be too much steric rubbing with the histidine. This is effective at preventing the bonding of carbon monoxide. The oxygen is bent which prevents steric hindrance.
- There is an H on Histidine N at certain pH’s, permitting it ot create hydrogen bonds with oxygen which stabilize the structure

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19
Q

Discuss the role of the globin protein in preventing heme iron oxidation

A

Free heme molecules leave the Fe2+ with two “open” coordination points. having simultaneous reaction of O2 molecules on both coordinate points can irreversibly convert Fe2+ to Fe3+. This reaction is sequestered if there is a protein taking up one coordinated bond

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20
Q

What is “molecular breathing” and why is it important in terms of myoglobin function

A
  • If the protein were rigid, O2 could not readily enter or leave the heme pocket. however, rapid molecular flexing of the amino acid side chains produces transient cavities in the protein structure of O2 so it can make its way in and out
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21
Q

Is there a myoglobin T and R state?

A

Not in the way it is for hemoglobin. Myoglobin does undergo conformational changes upon oxygen binding, but these changes are limited to the tertiary structure of the protein, rather than the quaternary structure. When oxygen binds to the heme group within myoglobin, it induces a slight rotation in the heme group and associated helices, leading to a more compact and stable conformation. This conformational change enhances myoglobin’s affinity for oxygen and promotes oxygen binding and release within muscle tissues.
*salt bridges aren’t involved in this

ASK ABOUT THIS ITS CONFUSING

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22
Q

Hemoglobin’s role

A

In red blood cells, transports oxygen from teh lungs to the tissues

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23
Q

Discuss the quaternary structure of hemoglobin

A
  • Contains 4 heme prosthetic groups, one associated with each polypeptide chain
  • Has 2 alpha and Beta chains. Their interactions are mostly salt bridges between each other.
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24
Q

Compare the tertiary structure of the single subunit of hemoglobin to myoglobin

A

They are very similar to each other reflecting their evolution. essentially 4 myoglobins make up a hemoglobin

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25
Q

How similar is the primary sequence of the hemoglobin to that of the myoblobin

A

They are not as similar. There are a few times in which all three have the same code

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26
Q

Where are the heme groups located in each subunit of hemoglobin?

A

Located in the center of each subunit

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27
Q

In a hemoglobin molecule, are the oxygen binding sites far apart of close together?

A

The oxygen binding sites are far apart, since they are each located in the center of the subunit.
- The binding of oxygen to one of he heme groups causes a conformational change in salt bridges which makes it easier for another oxygen to enter a nearby heme group

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28
Q

What forces hold the subunits together?

A

The forces that hold the subunits together are the salt bridges, THe Asp, His, and Lys is the salt bridge that stabilizes the tense conformation.
(Asp isn’t actually involved in the salt bridge but it is changing the environment)

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29
Q

How does the binding of O2 in hemoglobin lead to conformational changes in the ligand binding site?

A

In the T state, the heme group is puckered (draw). Once O2 binds to the Fe2+, this causes a conformational change that makes the heme group more planar. This change transmits to the subunit interface by decreasing the number of salt bridges on the interface.

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30
Q

Compare the conformation of the hemoglobin T and R states

A

The T-state has many more of the salt bridges than the R state does which is more relaxed. Another dramatic result of the T–> R transition is a narrowing of the pocket between the Beta subunits

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31
Q

Hemoglobin binds oxygen ______

A

Cooperatively

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32
Q

Name the shape the hemoglobin oxygen binding

A

Sigmoidal

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33
Q

What is meant by the statement that the binding of O2 by hemoglobin is cooperative?

A

Hemoglobin being cooperative means that the subunits work together to make the environment more or less favorable for binding O2. Having one subunit with oxygen will conformationally change the proteins which will decrease the interaction of salt bridges. The binding of one ligand affects the affinity of any remaining unfilled binding sites.

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34
Q

Do isolated hemoglobin exhibit cooperatively in O2 binding?

A

No because if the subunits aren’t interacting with each other, there is no cooperative binding

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35
Q

Estimate the P50 for hemoglobin. Compare it to the P50 for myoglobin. Which protein has a higher affinity for oxygen?

A

3.8 kpa. The P50 for myoglobin is lower,meaning it takes less oxygen to be 50% saturated. This means that myoglobin has a higher affinity for oxygen.

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36
Q

Is myoglobin or hemoglobin more sensitive to small changes in oxygen concentration?

A

Hemoglobin

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37
Q

Hemoglobin is an example of an allosteric protein. Explain

A

Hemoglobin is an example because when oxygen binds to a heme group, it causes a conformational change which affects other subunits in the protein. An allosteric protein is one in which the binding of a ligand on one site affects the binding properties on another.

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38
Q

Modulator

A

Modulators are a type of ligand that induce a change in conformation of a protein

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39
Q

Heterotropic modulator

A

When the modulator is a molecule other than the normal ligand

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40
Q

Homotropic modulator

A

When the normal ligand and the modulator are identical

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41
Q

Oxygen is a __ modulator

A

Homotropic because it is the molecule taht causes the conformational shift that will allow more oxygen to bind

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42
Q

How does H+ produced in rapidly metabolizing tissues, promote delivery of oxygen from the lungs to the tissues?

A

CO2 with H20 increases the H+ concentration. H+ increasing decreases the pH. High levels of H+ binding to the hemoglobin favor the T state and that allows for the release of oxygen.

Therefore, when the red blood cells from the lungs which are packed in O2 due to the high pO2 enter the tissues with a high H+, it will transition to the T state which releases all those oxygen molecules

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43
Q

There are certain proton binding sites in hemoglobin that are of higher affinity n the deoxy from than in the oxy form (protein binding is encouraged in some sites more than others due to the changing pKa). The increase in affinity for H+ must reflect that some groups are higher in deoxyhemoglobin and in oxyhemoglobin. What residues participate to give this Bohr effect?

A
  • H+ binds to several amino acid residues in the protein (the pKa can change to encourage ot discourage proton binding)
  • His 146 forming a pair with Asp 94 when protonated (stabilize the T state)
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44
Q

Closely positioned anion would increase the pK of histidine because…

A

Look at the chart*
If you have His (positively charged) next to an anion that is negatively charged, you want to make sure that your histidine keeps its positive charge because it likes those interactions. You wouldn’t want it to have a neutral charge because it is more favorable to have a negative charge interact with a positive one.

Therefore, the Histidine would like to remain protonated which means that it wants its H. To have an H on a molecule, the pH<pKa. Therefore, the pKa would increase.

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45
Q

Two ways CO2 allosterically lowers oxygen affinity

A

1.
CO2 + H20 –> H+ + HCO3-
When there is more CO2 in the environment, it can be hydrated, leading to the formation of bicarbonate and H+. The increase of H+ in the environment can interact with the side chains on teh hemoglobin. The protonation of H+ can lead to the increase in salt bridges, changing the hemoglobin in the T-state which has a lower affinity for oxygen
2.
Co2 binds as a carbamate group to hemoglobin at the alpha-amino group at the amino-terminal end of each globin chain, forming carbaminohemoglobin. This reaction produces H+, contributing to the Bohr effect. The bound carbamates also form additional salt bridges to stabilize the T state and promote the release of oxygen. (draw this)

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46
Q

When is myoglobin used?

A

During exercise: When muscle tissues are actively contracting during exercise, they require more oxygen to support increased energy production. As a result, oxygen levels in muscle tissues decrease, triggering myoglobin to release stored oxygen. This released oxygen can then be utilized by mitochondria within muscle cells for ATP synthesis, providing energy for muscle contraction.

In hypoxic conditions: When tissues experience low oxygen levels (hypoxia) due to factors such as high altitude, reduced blood flow, or impaired oxygen delivery, myoglobin releases stored oxygen to help meet the tissue’s oxygen needs. Myoglobin’s high affinity for oxygen allows it to efficiently release oxygen even in conditions of low oxygen tension.

During recovery: After periods of increased metabolic activity, such as intense exercise, oxygen levels in muscle tissues may remain elevated to support tissue repair and recovery. Myoglobin can continue to release oxygen during this recovery phase to support ongoing metabolic processes and tissue repair.

47
Q

Can you clarify for me how increasing the H+ concentration encourages the T state?

A

Let’s say you’re a red blood cell that has just traveled from the lungs to the tissues. Having H+ in the environment will bind to amino groups and cause attractions between different amino acid groups which increase more salt bridges. The microenvironment tries to act like a buffer, increasing or decreasing pKa to give the amino acids their preferred forms. But eventually, if there’s too much H+ everything will be protonated causing more H+ bonds.

48
Q

How does BPG play a role in influencing the oxygen affinity of hemoglobin?

A

BGP greatly reduces the affinity of hemoglobin for oxygen - they have an inverse relationship.

49
Q

Describe the physiological significance of BPG

A

For a healthy human at sea level, O2 delivers to the tissues at 40%. If you were then placed at a high altitude where the pO2 was way lower, the BPG concentration will rise so the oxygen can be released in your tissues since the hemoglobin has a lower affinity for oxygen when BPG increases

50
Q

Where in the deoxyhemoglobin structure does the binding of BPG occur?

A

In the cavity between the two B-subunits in the T state

51
Q

What types of interactions are involved in getting the BPG to bind to the deoxyhemoglobin?

A

Salt Bridges/Ionic interactions. The cavity is lined with positively charged amino acid residues that interact with the negatively charged groups of BPG

52
Q

How does the interaction of BPG affect hemoglobin’s oxygen binding affinity?

A

BPG lowers hemoglobin affinity for oxygen by stabilizing the T state. In the absence of BPG, hemoglobin is primarily present in the R state, where it binds O2 efficiently in the lungs but fails to release it in the tissues

53
Q

Explain what happens to the BPG binding site in oxygenated hemoglobin

A

The binding pocket for BPG disappears on oxygenation, following transition to the R state (remember how in the R state the B-subunit pocket gets narrower? this prevents teh BPG from binding)

54
Q

T or F: As a response to low oxygen levels (since oxygen is consumed in the tissues), RBCs produce BPG

A

True

55
Q

Distinguish between a coenzyme and a cofactor?

A
  • Cofactor: chemical component that is either one or more inorganic ion such as Fe2+, Mg2+, Mn2+ or a coenzyme. (chemical component is not covalently attached)
  • Coenzyme: complex organic or metallorganic molecule

Cofactors and coenzymes describe different types of prosthetic groups that are attached to enzymes to help them function

56
Q

Distinguish between an apoprotein/apoenzyme and holoprotein/holoenzyme

A

Holoenzyme: complete, catalytically active enzyme together with its bound coenzyme and/or metal ions

Apoenzyme: The protein part of such an enzyme is called the apoenzyme or apoprotein

57
Q

Describe the key features of the active site of an enzyme

A
  • The surface of the active site is lined with amino acid residues with substituent groups that bind the substrate and catalyze its chemical transformation
  • The active site encloses a substrate, sequestering it completely from solution
58
Q

Distinguish between delta Gº and delta G’º

A
  • delta Gº: the standard free energy change. Describes the free energy for reactions with standards like 298 K, pp=1 atm, and [S]=1M
  • delta G’º: the biochemical standard free energy change. It means the pH=7
59
Q

What determines the rate of a reaction?

A

The energy barrier between the Substrate and Product

60
Q

Is the standard free energy of the reaction positive or negative?

A
  • The standard free energy of a reaction is the difference between where the S and P are in the ground state.
  • If the P is lower than the S, then the standard free energy is negative.
  • If the S is lower than the standard free energy is positive
    *draw this
61
Q

Draw figure 6-2 (p.180)

A
62
Q

Is the equilibrium constant greater or less than 1

A
  • If the P has a lower free energy than the S, then that means there is more product that substrate. If there is more P than S, the equilibrium constant is greater than 1.
  • If the P has a higher free energy than the S, then that means there is more substrate than product. This means the equilibrium constant is less than 1.

K= [P]/[S]

63
Q

Does the equilibrium favor reactants or products?

A
  • If the P has a lower free energy than S, the products are favored.
  • If the P has a higher free energy than the S, the substrate is favored
64
Q

Does a favorable equilibrium mean that the rate of the forward reaction (reactants going to products) is fast?

A

No. Favorable equilibrium means that equilibrium lies towards the formation of products rather than reactants.

65
Q

Does the presence of a catalyst have any effect on the position of the equilibrium?

A

The function of a catalyst is to increase the rate of reaction (i.e. lower the transition state), they do not affect reaction equilibria

66
Q

Does a higher standard free energy of activation correspond to a slower or faster reaction rate?

A

A slower reaction rate

67
Q

T or F: Reaction rates can be increased by raising the temperature which increases the number of molecules with sufficient free energy to overcome this free energy barrier

A

True

68
Q

Is temperature control a realistic variable in a biological system?

A

No. We cannot realistically alter the temperature of our bodies and increase it for a reaction because it could denature other proteins and disrupt homeostasis. Therefore, we must seek alternative methods.

69
Q

What is the effect of an enzyme on the standard free energy of activation? Explain how this is accomplished

A

Enzymes provide an alternative path to lower the standard free energy of activation.
Two ways they can achieve this:
- Noncovalent interactions: formation of a weak interaction in the Enzyme Substrate (ES) complex is accompanied by release of a small amount of free energy that stabilizes the interaction so that the amount of energy that is required to complete the reaction, is canceled out by the energy from non-covalent interactions
- Covalent bond: covalent interactions between enzyme and substrate lower activation energy by providing an alternative, lower-energy reaction path that is stabilized

(this is true)

70
Q

Why does an enzyme form many, weak non-covalent interactions with the transition states than it forms with the substrate?

A
  • An enzyme forms many interactions with the transition state so it can promote the creation of the product without losing too much energy
  • Without an enzyme, going from the substrate to the product would be very difficult to do and it would require a lot of energy.
  • WIth an enzyme complementary to the substrate, you are going to have interactions but this isn’t encouraging the formation of the product (the bending of the stick). Additionally, the covalent interactions will lower the energy and stabilize it. Making it even harder to enter the transition state.
  • With an enzyme complementary to the transition state, initially, the substrate (metal) will only have a few interactions with the enzyme. It needs to gain energy to enter the transition state so it is closer to breaking. This energy it requires to enter the transition state is compensated by the weak interactions it would form. By getting into the transition state, you are increasing the likelihood of having products. This is why it’s more favorable to have the enzyme complementary to the transition state - you have more products and can compensate for the energy loss

*make sure you can draw these diagram

71
Q

Reaction equilibria are linked to…

A

delta Gº

72
Q

Reaction rates are linked to…

A

delta G(t) (activation energy)

73
Q

What does the rate of the forward reaction equal?

A

kf[A]= rate of the forward reaction

74
Q

What does the rate of the reverse reaction equal?

A

kr[B] = rate of the reverse reaction

75
Q

What is true of the forward and reverse reactions at equilibrium?

A

kf[A]=kr[B]

76
Q

What is the expression for Keq?

A

Keq= [B]/[A]

77
Q

Write an expression for Keq in terms of rate constants

A

Keq = kforward/kreverse

78
Q

What is the effect of an enzyme on kf and kr? on kf/kr?

A

An enzyme can increase the forward reaction but it also will increase the reverse reaction by the same amount, forcing the equilibrium to remain the same. Since Kf/Kr is the equilibrium, this isn’t affected.

Enzymes accelerate both the forward and reverse reactions equally, which means they do not change the position of the equilibrium. Instead, they speed up the attainment of equilibrium by lowering the activation energy barriers for both the forward and reverse reactions. As a result, enzymes increase the rates of both reactions without altering the overall equilibrium constant.

79
Q

Explain that enzymes accelerate reactions by stabilizing the transition state?

A

To lower the activation energy for a reaction to accelerate it, the system must acquire an amount of energy equivalent to the amount by which delta G activation is lowered. Much of this energy come from binding energy contributed by formation of weak noncovalent interactions between substrate and enzyme in transition state

80
Q

V= d[P]/dt =

A

d[S]/dt

81
Q

Rate law for the elementary reaction

A

V=k[S]

82
Q

First order rate law units

A

sec-1

83
Q

Second order rate law units

A

M-1 sec-1 (this is if there’s one concentration raised to the second power or if you have two concentrations multiplied)

84
Q

Third order rate law units

A

M-2 sec-1

85
Q

T or F: Does substrate concentration affect the rate of enzyme-catalyzed reaction

A

True

86
Q

What is meant by Vo? Why is measuring Vo considered to be a simplification?

A

Vo is the initial rate. It is a simplification because [S] changes over time as it is converted to product. Another reason is because the reaction doesn’t only go one way. To disregard the rate of the reverse reaction, we use Vo, before there is any product made

87
Q

Describe how Vo can be determined at a different number of different [S]. Describe how this data is used to generate the Michaelis-Menten plot

A
  • At different substrate levels, the Vo varies. The higher the substrate, the higher/more steep the Vo is. The lower the substrate the less sleep and lower Vo is
  • This generates the Michaelis-Menten plot because it just plots the points out. At a specific concentration, it takes the Vo and plots it on the Y axis which is now initial velocity.
88
Q

What are the x and y axis for the Michaelis-Menten plot

A

x: substrate concentration
y: initial velocity

89
Q

What are the x and y axis for the plot to generate Vo

A

x: time
y: product concentration (line is drawn tangent just at the beginning when no product is formed)

90
Q

Km

A

The substrate concentration at which Vo is 1/2 Vmax

91
Q

Vmax

A

The plateau-like Vo region is the Vmax

92
Q

What are the two fates of the Enzyme-Substrate (ES)

A

The ES. can either become the E+P or it can go back and dissociate into the E+S

93
Q

k1

A

Rate constant of the forward rxn from E+S to ES

94
Q

k-1

A

Rate constant of the reverse rxn from ES to E+S

95
Q

k2

A

Rate constant of the forward rxn from ES to E+P

96
Q

What is the Michaelis-Menten equation?

A

Vo = Vmax[S] / Km+[S]

97
Q

In what part of the Michaelis-Menten plot is the velocity directly proportional to the substrate concentration? What does the equation look like in this form?

A

When the Km»[S] because the [S] in the denominator essentially becomes insignificant.

Vo = Vmax [S]
————
Km

98
Q

In what part of the Michaelis-Menten plot is the velocity independent of the substrate concentration?

A

When the [S]» Km, the Km becomes insignificant and the top and bottom [S] of the equation cancel out.

Vo = Vmax

99
Q

Use the Michaelis-Menten equation to show that the velocity will be half Vmax when [S] = Km

A

Vo = (Vmax[S])/(Km +[S)
Vo = (Vmax[S])/([S] +[S])
Vo = (Vmax[S])/(2[S]). *cancel out the S
Vo= (Vmax/2)

100
Q

Nelson and Cox problem 12 insight

A
  • We know that the highest Vo value we can get will be the Vmax so we can estimate that the highest value we have is Vmax
  • We know that the substrate concentration at half of Vmax is the Km. All we have to do is convert it into micromoles so we would multiply it by 10^6
101
Q

Describe how the Lineweaver Burke plot can be generated from a Michaelis-Menten plot

A

It converts the hyperbolic curve into a linear form by taking 1/ all of the values

102
Q

What shortcomings of the Michaelis-Menten plot is overcome in the Lineweaver-Burke plot

A

Since the Lineweaver-Burke plot is linear, you can obtain the values of Vmax and Km more accurately

103
Q

How can the axis intercepts on a Lineweaver-Burke plot be used to determine Km and Vmax values?

A

The y-intercept is the 1/Vmax value, so you can derive the Vmax from this.

The x-intercept is the -1/Km value so yu can derive the Km from this

104
Q

Question: generate a double-reciprocal plot of the data from Nelson & Cox probem

A
  • Start off by taking the values and converting them into 1 over the values so you can get the correct concentrations.
  • You also want to make sure that the concentrations are the same so convert the [S] into micromoles
  • Then take two random points and find the slope of those converted values - this gives you your ratio of Km/Vm
  • Then you can determine the Vmax since you know it’s the highest value and the Km since you know its half the Vmax’s concentration

BUT ALSO

Plug it into the formula 1/Vo = Km/Vmax (1/[S]) + 1/Vmax. And make 1/Vo, or y, 0 so you can solve for (1/[S]), or the x intercept

105
Q

Define the Michaelis constant, Km, in terms fo the three reactant constants

A

Km = k2 + k-1
————-
k1

106
Q

Define the dissociation constant Kd for the following reaction ( E+S –> ES)

A

Kd = K-1/K1
*this is true because dissociation is with the ES –> E+S so there is no product involved

107
Q

What are the units of k1 and k-1

A

k-1 is sec-1 (this is because its only a matter of time before the ES dissociates)
k1 is M-1s-1 (this is because to go forward in a reaction, it depends on both the time and the moles of the substrate you have)

108
Q

Under what conditions does Km = Kd `

A

They are equal when k2 is not taken into account. The conditions this would occur in would be if your Enzyme had a really strong affinity for your substrate so it would not want to let it go and the k2 value would essentially be 0 or if there wasn’t enough substrate for the k2 value to be significant enough

109
Q

Does a high value of Kd or Km indicate a high or low affinity of E and S for each other?

A

high Kd means that there is a low affinity of E and S because its dissociating

high Km means that there is a low k1 value meaning that there is also a low affinity of E and S. This is a relationship found in the equataion.

Km = k2+ k-1
———–
k1

So if Km increases, k1 decreases, symbolizing that to go from the E+S to ES state is unlikely which means the enzyme and substrate do not have a high affinity for one another.

110
Q

Define the turnover number

A

Kcat is also known as the turnover number. It describes the rate limiting of any enzyme catalyzed reaction at saturation. If the reaction has several steps, one of which is clearly rate-limiting, Kcat is equivalent to the rate constant for that limiting step

Mazimym rate that an enzyme and substrate can be turned into a product

111
Q

Nelson and Cox problem 16 - what is the turnover number?

A

Vmax = Kcat[Et]
OR
Vo = kcat[Et][S]
————–
Km + Es

  • Find the concentration of [Et] in moles using the Mr, molecular mass in g/mol to convert it.
  • Vmax is also concentration/time of a reaction so also convert that
  • Plug into the equation
112
Q

What is the upper limit on kcat/Km

A

It is imposed by the rate at which E and S can diffuse together. This diffusion-controlled limit is 10^8 or 10^9 M-1s-1.

113
Q

What does it mean if an enzyme has achieved catalytic perfection?

A

Enzymes that have a kcat/Km near the range of 10^8 or 10^9 M-1s-1 are considered catalytically perfect.

114
Q

How to correctly do the Lineweaver-Burke Graph:

A
  1. Take the inverse of all the values if you’re given [S[ and Vo
  2. 1/[S] is the x-axis and 1/Vo is the y-axis (THE UNITS WILL ALSO CHANGE TO BE 1 OVER WHATEVER THE UNITS ARE)
  3. Plot all the points and draw a line through it
  4. Take the y and x intercept
  5. Solve for Vmax and Km. (since you already have the value from the intercept just plug it back in teh Vm equation).

Biochem (15 decks)

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