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BIOL 2083 > Exam 2 > Flashcards

Exam 2 Flashcards

1
Q

the ratio of enzyme activity relative to total protein

A

specific activity

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2
Q

a type of purification that is based on the attraction of the protein for a particular chemical gorup

A

affinity chromatography

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3
Q

proteins can be separated from small molecules and ions through a semipermeable membrane by

A

dialysis

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4
Q

in gel filtration chromatography, in what order to proteins come off of the column

A

large -> medium -> small

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5
Q

allows the detection of small amounts of target proteins as well as the ability to determine the size of target proteins

A

western blotting

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6
Q

what is the gel matrix for protein electrophoresis

A

polyacrylamide

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7
Q

the pH of a protein at which its net charge is zero

A

pI or isoelectric point

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8
Q

an immunological technique for sensitive quantification of a target protein

A

ELISA

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9
Q

What is the advantage of adding SDS to gel electrophoresis

A

SDS allows proteins to be separated on the basis of approximate mass

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10
Q

Two-dimensional electrophoresis is a combination of which two techniques

A

isoelectric focusing and SDS-PAGE

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11
Q

Which conditions could cause changes in the proteome of a cell?

A

the developmental stage, the environmental condition, enzymatic modification

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12
Q

Why is an assay necessary for protein-purification studies

A

an assay allows researchers to accurately measure the amount of the desired protein. This is important in determining if particular purification steps are effective in isolating the protein from the other cellular material.

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13
Q

Compare and contrast gel filtration and ion-exchange chromatography

A

Although both are used in purification, the properties of the column material determine how the separation is accomplished. Gel filtration is based on porous beads, and molecules are separated by size. In ion-exchange chromatography, the column material is changed with either positively or negatively charged molecules. Separation is base don the proteins’s charge and affinity for the column media.

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14
Q

What is the purpose of determining the specific activity, yield, and purification level of a protein-purification protocol

A

The measurements allow one to determine whether the individual steps were effective at selectively isolating the protein while maintaining its presence and activity. In order to successfully purify protein, both the yield and the purification level must remain high.

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15
Q

List four possible steps in a protein purification strategy. mention the reason for each.

A

Differential centrifugation: Separating the cellular components by size
Salting Out: Separating proteins on the basis of their solubility
Gel Filtration Chromatography: Separating proteins based on size
Ion-Exchange Chromatography: separating proteins based on overall charge
Affinity chromatography: separating proteins on the basis of an interaction with a specific chemical.

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16
Q

A protein that on an SDS-PAGE runs as a single 25kilodalton band runs as a 75kilodalton band on a native gel (or gel filtration). Why?

A

The protein is really a trimer (3 subunits) of 25kDa which gives a native molecular weight of 75 kDa. On an SDS-page, all quaternary structure interactions will be disrupted and it will run with an apparent molecular weight of 25 kDa. However, when the trimer is intact for a native gel or gel filtration column, it runs at 75 kDa

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17
Q

You perform the separation of a protein mixture (ABC) on an ion-exchange column with a buffer at pH 7.4 and get the following elution order: ProteinC, ProteinA, ProteinB. You perform a separation of the same mixture using the same column using a buffer with pH 5.6. The order of elution changes to ProteinA, ProteinC, ProteinB. Why?

A

The pH of the buffer solution passed over the ion-exchange chromatography has changed between the two experiments. The PH of the solution can modulate the overall charge of the proteins in the mixture and alter the elution profile of the proteins from the ion-exchange column.

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18
Q

What are the advantages of the HPLC over the other chromatographies discussed in chapter 5?

A

The HPLC method offer higher resolution and faster run times because the resin is made of finer beads and the high pressure allows for faster elution.

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19
Q

Explain three reasons why it is important to know the primary structure of a protein.

A
  1. Primary structures from different proteins can be compared to infer knowledge about structure and function. .
  2. Primary structure comparison of similar proteins from different species provides information about evolution.
  3. Primary structure can be searched for internal repeats that may yield information on the history of the individual protein.
  4. Primary structure can reveal the presence of amino acid sequences that regulate protein function and location.
  5. Primary structure can provide insight into the molecular basis of disease.
  6. Primary structure can be used as a guide to explore nucleic acid information.
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20
Q

What are two methods for identifying the amino acid sequence of a protein?

A

Edman degradation and mass spectrometry techniques

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21
Q

the site on the enzyme where the reaction occurs

A

active site

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22
Q

the substance that the enzyme binds and converts to a product

A

substrate

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23
Q

enzyme that do not have the required cofactor bound

A

apenzymes

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24
Q

enzymes will decrease the energy of activation but do not change the _______ of a chemical reaction

A

delta G or Equilibrium

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25
Q

enzymes that cleave molecules by the addition of water

A

hydrolases

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26
Q

how do enzymes accelerate the rate of a chemical reaction

A

lowering the free energy of activation of the reaction

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27
Q

a reaction can occur spontaneously only if delta G is

A

negative

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28
Q

when delta G for a system is zero, the system is

A

at equilibrium

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29
Q

the differences in values for delta G and delta G not prime

A

concentrations of reactants and products

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30
Q

What is the common strategy by which catalysis occurs

A

stabilization of transition state

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31
Q

An enzyme will specifically bind its substrate because of

A

a large number of weak interactions at the active site

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32
Q

the difference between the substrate and the transition state

A

Gibbs free energy of activation

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33
Q

the molecular structure that is short-lived and neither substrate nor product is known as the

A

transition state

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34
Q

How is the substrate bound to the active site

A

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 acids residues from different regions of the polypeptide chain. The substrate is bound by multiple non covalent 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.

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35
Q

While some enzymes have very specific substrates, others are more promiscuous. What would you suspect is the reason for this?

A

Specificity of binding is separate from catalysis. The specificity of the enzyme for its substrate is due to many weak interactions between the substrate and the amino acids of the protein. Thus for the less specific binding protein, there must be fewer required interactions for binding

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36
Q

List the six major classes of enzymes

A
  1. Oxidoreductases
  2. Transferases
  3. Hydrolases
  4. Lyases
  5. Isomerases
  6. Ligases
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37
Q

when there is no net change in the concentration of substrate or product

A

equilibrium

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38
Q

turnover number

A

kcat

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39
Q

enzymes that do not obey michaelis-menten kinetics

A

allosteric

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40
Q

the symmetry rule for the concerted model states that

A

all subunits in an allosteric enzyme must be either in the T or the R state; there can not be hybrids

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41
Q

the type of inhibition by a product of one enzyme on another enzyme in an earlier protein in a metabolic pathway is considered a

A

feedback inhibitor

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42
Q

a low Km value means that the enzyme has

A

a high affinity for the substrate

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43
Q

stabilizes the T state of the enzyme

A

allosteric inhibitor

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44
Q

the Vmax and Km are _______ on one another

A

independent

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45
Q

the y intercept of a lin weaver-burk plot is equal to

A

1/Vmax

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46
Q

A critical feature of the Michaelis-Menten model of enzyme catalysis

A

the formation of an ES complex

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47
Q

Contain distinct regulatory sites and have multiple functional sites and display cooperativity

A

allosteric proteins

48
Q

The Km is equal to the substrate concentration when

A

the reaction rate is half its maximal value

49
Q

When a substrate concentration is much greater than Km, the rate of catalysis is almost equal to

A

Vmax

50
Q

Multiple substrate enzyme reactions are divided into two classes

A

sequential displacement and double displacement

51
Q

the equation for catalytic efficiency

A

kcat/Km

52
Q

An enzyme can use either substrate A or B. The catalytic efficiency for substrate A is 0.002 and the catalytic efficiency for substrate B is 2. Which is the preferred substrate?

A

B

53
Q

Describe the difference between the concerted and the sequential model of allosteric regulation

A

The concerted model describes a situation where a multisubunit enzyme can only assume an R or T conformation, whereas the sequential model assumes that the subunits can assume a conformation different from the neighboring subunits (R/T hybrids). In the former, substrate influences the equilibria between each subunit’s R or T conformation and in the latter, substrate can have an intermediate impact on affinity and conformation.

54
Q

Explain how allosteric activators and inhibitors affect the T/R equilibrium

A

Allosteric activators bind to and stabilize the R form of the enzyme. Once the R form is stabilized it is taken out of equilibrium with the T state. This means it takes less substrate to yield a greater enzymatic rate. Allosteric inhibitors bind to and stabilize the T form, which takes it out of equilibrium with the R state. This means it takes more substrate to yield a greater enzymatic rate.

55
Q

List the four common catalytic strategies

A

Covalent catalysis, general acid-base catalysis, metal ion catalysis, catalysis by approximation and orientation

56
Q

the inhibitor which binds only to the ES complex and lowers the Vmax and Km

A

uncompetitive

57
Q

the enzyme inhibition that can be overcome by increasing the concentration of substrate

A

competitive

58
Q

an enzyme catalyst mechanism that uses a metal cation to stabilize a negative charge in the active site

A

metal ion catalysis

59
Q

penicillin is an example of what type of inhibitor

A

irreversible

60
Q

what types of inhibition can be reversed

A

competitive, noncompetitive, uncompetitive

61
Q

where does a competitive inhibitor bind?

A

the active site

62
Q

In noncompetitive inhibition, the uninhibited and inhibited double-reciprocal plots will have _______ x-intercept

A

the same

63
Q

which amino acids in chymotrypsin are found in the active site and are participants in substrate cleavage

A

his, ser, asp

64
Q

Explain the effect of temperature on enzymatic activity. Use an example

A

An increase in temperature will increase the rate of an enzymatic reaction because the thermal energy increases Brownian motion and increases the likelihood of enzyme and substrate interaction. However, this relationship occurs only up to a certain point. Beyond the temperature optimum an increase in temperature will result in the disruption of the weak interactions holding the enzyme together. An example is the enzyme tyrosinase, which produces the dark-colored fur in Siamese cats. It has an optimum temperature lower than core body temperature and thus only produces pigment at the extremities of these animals (ears, feet, nose, tail).

65
Q

What is a substrate analog/affinity label?

A

This is a substrate analog that is structurally similar to the substrate, binds to the active site, and chemically reacts with a residue in the active site. It is used to study enzyme structure and mechanism.

66
Q

The initial reaction kinetics of some enzymes results in a quick burst of product in a short period of time, followed by a slower but sustained increase in product formation over time. What does this type of kinetic response tell an enzymologist about the mechanism of the catalysis?

A

The catalytic mechanism occurs in two steps.

67
Q

Describe the mechanism for the proteolysis catalyzed by chymotrypsin

A
  1. The peptide substrate binds to chymotrypsin at the active site.
  2. Serine 195 performs a nucleophilic attack of the peptide carbonyl
  3. There is a collapse of the tetrahedral intermediate to form acyl-enzyme intermediate
  4. The amine component is released
  5. Water binds at the active site
  6. Water performs a nucleophilic attack on the acyl-enzyme intermediate
  7. There is a collapse of the tetrahedral intermediate
  8. The carboxylic acid component is released.
68
Q

the shape of the myoglobin binding curve that shows that it is not regulated allosterically

A

rectangular hyperbola

69
Q

the type of hemoglobin that is composed of two alpha chains and two gamma chains

A

fetal

70
Q

the molecule whose function is to facilitate diffusion of oxygen in muscle cells

A

myoglobin

71
Q

this condition is a result of a single point mutation in the beta chain of hemoglobin

A

sickle cell anemia

72
Q

the ability of myoglobin and hemoglobin to bind oxygen depends not the presence of a bound prosthetic group called

A

heme

73
Q

the form of hemoglobin found in the R state is called

A

oxygenated

74
Q

the binding of 2,3-biphosphoglycerate to hemoglobin _____ its affinity of oxygen binding

A

decreases

75
Q

the effect of pH on oxygen binding of hemoglobin is referred to as

A

Bohr

76
Q

in normal adult hemoglobin, HbA, the Beta6 position is a glutamate residue, whereas in sickle-cell hemoglobin, HbS, it is a ____ residue

A

valine

77
Q

As the partial pressure of carbon dioxide increases, the affinity of oxygen binding to hemoglobin

A

decreases

78
Q

What factor influences the binding of oxygen to myoglobin

A

the partial pressure of oxygen

79
Q

Which of the following is correct concerning the differences between hemoglobin and myoglobin in regards to binding of O2

A

Hemoglobin exhibits cooperative binding of O2, while myoglobin does not

80
Q

The structure of normal adult hemoglobin can be described as

A

a tetrameter composed of two alphabeta dimers

81
Q

binds the central cavity in the T form of hemoglobin, preferentially binds to deoxyhemoglobin and stabilizes it, is present in the red blood cells

A

2,3-biphosphoglycerate

82
Q

Why is it advantageous for hemoglobin to have allosteric properties?

A

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. The sigmoidal shape of the binding curve ensures a wide range of oxygen release across physiologically relevant oxygen pressures (between the lungs and the tissues).

83
Q

Compare and contrast fetal hemoglobin and adult hemoglobin.

A

Fetal hemoglobin contains two  and two  chains, in contrast to adult hemoglobin, which has two  and two  chains. The difference in the chains results in a lower binding affinity of 2,3-BPG to fetal hemoglobin. Thus, fetal hemoglobin has a higher affinity for oxygen, and the oxygen is effectively transferred from the mother’s hemoglobin to fetal hemoglobin.

84
Q

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

A

2,3-bisphosphoglycerate (2,3-BPG), is a relatively small, highly anionic molecule found in the red blood cells. 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. The 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. (Compare graph of pure hemoglobin and hemoglobin from red blood cells)

85
Q

Describe the effect of pH and carbon dioxide on oxygen binding to hemoglobin.

A

A decrease in the pH will stimulate the release of oxygen from hemoglobin. Structural changes that occur upon protonation of hemoglobin at lower pH shifts the equilibrium from the R state to the T state, thus releasing oxygen. 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. 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.

86
Q

What are the 6 coordination sites for iron in oxygenated hemoglobin?

A

The iron atom is coordinated by 4 nitrogen atoms of the heme group, the proximal histidine residue and the last site is bound to molecular oxygen.

87
Q

Cells maintain a ______ concentration of intracellular potassium as compared to the extracellular concentration

A

higher

88
Q

the type of amino acid found in the transmembrane helix of an integral protein

A

hydrophobic

89
Q

the non covalent force which favors close packing of the tails of the lipids in a membrane

A

van der Waals

90
Q

membranes are said to be ______ because their two faces always differ from each other

A

asymmetrical

91
Q

inserts into lipid bilayers, disrupting interactions between fatty acids, thereby helping to maintain membrane fluidity

A

cholesterol

92
Q

the temperature at which a phospholipid membrane transitions from a rigid to a fluid state

A

melting temperature

93
Q

an increase in the ratio of saturated to unsaturated fatty acid chains in a membrane ______ the fluidity of the membrane

A

decreases

94
Q

the energy for _____ transport comes from the gradient itself

A

passive

95
Q

how many molecules this are membranes

A

2

96
Q

which of the following membranes would be the most fluid

A

a bilayer made of lipids with polyunsaturated 16-carbon fatty acids

97
Q

an anti porter and a supporter are examples of

A

a secondary transporter

98
Q

the most common motif found in membrane-spanning proteins is

A

alpha helices of non polar amino acids that pass through the membrane

99
Q

the fluoresensce recover after photobleaching (FRAP) has been used to study

A

the lateral diffusion in membranes

100
Q

What determines the relative permeability of molecules across a lipid bilayer?

A

Their hydrophobicity; the more hydrophobic, the easier they will be able to pass through the membrane. The more polar or charged the molecule, the more impermeable it will be to the lipid bilayer. Ex: indole vs. tryptophan

101
Q

What is the function of prostaglandin H2 synthase-1? How does its position in the membrane facilitate its activity?

A

Prostaglandin H2 synthase-1 converts arachidonic acid into prostaglandin H2. The protein is embedded in the membrane, with a hydrophobic channel submerged about halfway through the bilayer. The arachidonic acid is a product of membrane lipid hydrolysis, and enters the protein channel from within the membrane, successfully avoiding any interaction with aqueous environments.

102
Q

How do secondary transporters drive the transport of a substance up its concentration gradient?

A

the thermodynamically uphill flow of one molecule is coupled to the downhill flow of another

103
Q

Describe the molecular basis for the selectivity of the potassium channel

A

The protein structure of the potassium channel has a small pore that is the exact size of a potassium ion. As the potassium ions move through the channel, they become desolvated. As they lose their water molecules, carbonyl groups on the channel replace the water molecules. The structure is such that only a potassium ion will fit and not a sodium ion.

104
Q

Describe two techniques that could be used to purify a target protein.

A

One technique is differential centrifugation. The cells are disrupted by being spun in a centrifuge at low speeds to yield a pellet consisting of nuclei and a supernatant. This is repeated a few times to yield a series of pelleted enriched in the cell materials and supernatant and form a homogenate. Another technique is dialysis. Separation in dialysis is based on the size of the molecules for buffer exchange. The solution is placed in a cellophane bag with pore that are too small to allow protein to diffuse but also big enough to let salt equilibrate.

105
Q

Describe three techniques or measurement used to evaluate protein purification protocol for effective isolation of a pure enzyme.

A

One technique is measurement of specific activity. Total activity is divided by total protein, which enables us to measure the degree of purification by comparing specific activities after each purification step. Another technique is measurement of yield. Yield is a measure of the total activity retained after each purification step as a percentage of the activity in the crude extract. Another technique is measurement of purification level. It measures the increase in purity and is obtained by dividing the specific activity after each purification step by the specific activity if the initial extract

106
Q

Explain what the term delta G tells you about a chemical reaction

A

Delta G is the change in free energy when a reaction occurs. It provides information on the spontaneity of the reaction, but not the rate of the reaction. When delta G is negative, the reaction is exergonic or spontaneous. When delta G is positive, the reaction is endergonic or nonspontaneous. When delta G is neutral, the reaction is at equilibrium and there is no net change in the amount of reactants and products in the reaction.

107
Q

Discuss the formation of the enzyme-substrate complex and how this facilitates catalysis of biochemical reactions.

A

Enzymes bring together substrates in enzyme-substrate complexes. The substrates bind to specific regions on the enzyme called the active sites. When the enzyme and substrate interact, it promotes the formation of the transition state. This facilitates catalysis of biochemical reactions because it lowers the delta G of the reaction, providing the rate-enhancement characteristic of enzyme action.

108
Q

What is the parameter Km in Michaelis-Menten kinetics? What is its significance and how can it be used to compare different enzymatic reactions?

A

Km is the substrate concentration at half of Vmax, which is the fastest possible rate for an enzyme at fixed concentration and for a specific substrate. Km is significant because it is a measure of the affinity an enzyme possesses for a specific substrate. It can be used to compare different enzymatic reactions by determining which substrate an enzyme has a higher affinity for from looking at the Km value. If it is high, there is a greater amount of substrate required to reach half of Vmax, and vice versa.

109
Q

Compare and contrast the kinetics (as a function of substrate concentration) of a Michaelis-Menten enzyme and an allosteric enzyme. Include the physical basis of each.

A

Allosteric enzymes have an initial lag in reaction velocity in response to the substrate, whereas Michaelis-Menten enzymes do not have that initial lag. Velocity increases instantly in response to the substrate. Allosteric enzymes display a sigmoidal curve on a graph comparing velocity to substrate concentration, and Michaelis-Menten enzymes increase rapidly and level off sooner. Allosteric enzymes are also more complex than Michaelis-Menten because they consists of a committed step and remaining steps, which are facilitated by Michaelis-Menten enzymes.

110
Q

Describe how allosteric effectors influence enzyme function.

A

Positive effectors, also called activators, bind to the enzyme at the regulatory site and stabilize it by transforming it into the R form. This lowers the substrate threshold for activity. An example of this would be ATP. Negative effectors, also called inhibitors, bind to the enzyme at the regulatory site and stabilize it by transforming it into the T form. This increases the substrate threshold for activity. An example of this would be CTP.

111
Q

What is the molecular basis of gout?

A

Gout is the result of consuming foods rich in purine nucleotides, for example, adenine and guanine. They are degraded into urate, which forms crystals in the fluid and lining of your joints. It can also be caused by loss of allosteric regulation by an important enzyme in purine synthesis, PRS. It is normally feedback inhibited by purine nucleotides, however, in certain people the regulatory site has undergone a mutation that renders PRS insensitive to feedback inhibition. This leads to a glut of purine nucleotides, which are then converted into urate. Excess urate will accumulate and cause gout.

112
Q

Describe the effects of the three different types of reversible inhibitors on enzymes

A

Competitive inhibitors are structurally similar to substrates and bind to the active site, which prevents the actual substrate from binding. This causes the Km value to increase but does not have an effect on the Vmax value. Uncompetitive inhibitors will bind only to the enzyme-substrate complex somewhere other than the active site. It causes both the Km value and Vmax value to decrease. Noncompetitive inhibitors will bind to either the enzyme or the enzyme-substrate complex somewhere other than the active site. This causes the Vmax value to decrease but has no effect on the Km value.

113
Q

Describe the chymotrypsin active site and how these residues contribute to its catalytic mechanism.

A

Chymotrypsin contains the highly reactive serine 195, histidine 75, and aspartic acid 102. Serine 195 produces a two-step hydrolysis in the enzyme. The first step is the burst phase because the covalent attachment is very rapid, and the second step is the steady phase because the hydrolysis release is much slower. Histidine 75 plays a role in the modification by TPCK. The reactive group in TPCK covalently modifies one of the ring nitrogens of histidine 75. It also serves to position the serine side chain and to polarize its hydroxyl group so that it is poised for deprotonation. Aspartic acid 102 helps orient the histidine residue and make it a better proton acceptor through hydrogen bonding and electrostatic effects

114
Q

Describe the general structure of adult hemoglobin, Highlight the structural differences between deoxy- and oxy-hemoglobin.

A

Hemoglobin has a protoporphyrin ring with a central iron atom coordinated by four nitrogen atoms of heme. It also contains four heme groups. Deoxyhemoglobin is hemoglobin before oxygen is found, and the iron atom is slightly out of the plane of the heme ring. Oxyhemoglobin is hemoglobin after oxygen is found, and the iron atom is pulled into the plane the heme ring is in.

115
Q

What are the three allosteric effectors for hemoglobin? Describe their effects on the allosteric state of hemoglobin and illustrate how they change the oxygen-binding curve.

A

2,3-BPG is an allosteric effectors that stabilizes the T state and facilitates the release of oxygen. CO2 and hydrogen are also allosteric effectors and are produced by actively respiring tissues to enhance oxygen release by hemoglobin. The stimulation of oxygen release by CO2 and H+ is called the Bohr effect.

116
Q

Describe the molecular basis as to how membrane channels are selective for certain ions use the potassium channel as an example.

A

Membrane channels are selective for certain ions because the channel is just the right size for K+ to fit. Na+ cannot fit because it is too small, and other larger molecule physically cannot fit. The opening to the outside and the cavity are filled with water, and K+ can fit without losing its shell of bound water molecules. When K+ moves farther into the cavity, its gets more narrow and the ions must remove their water molecules and interact with groups from the protein. After the bound water molecules are removed, another K+ is moved into the holding site and the K+ is pushed to the next site. This process of pushing K+ ions further into the channel as others enter eventually kicks the K+ ions out and into the opposite side of the membrane.

117
Q

Compare and contrast pumps and secondary transporters.

A

Pumps facilitate the movement of molecules against their concentration gradient and require energy to do so. This is called active transport. Secondary transporters use the power of one concentration gradient to drive the transportation of a different molecule against its concentration gradient. This does not require energy but does require two different molecules.

BIOL 2083 (10 decks)

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