Chapter 4 Flashcards

1
Q

Which parts of amino acids are involved in a peptide bond?

  • carboxyl group of one amino acid and side chain of the other
  • amino group of one amino acid and carboxyl group of the other
  • carboxyl groups of both amino acids
  • side chains of both amino acids
  • amino group of one amino acid and side chain of the other
  • amino groups of both amino acids
A

Amino group of one amino acid and carboxyl group of the other

(The peptide bond always comprises both a nitrogen atom and a carbon atom, where the nitrogen atom from the amino group and the carbon atom from the carboxyl group undergo a condensation reaction and eliminate a water molecule in the process. The amino acid side chains do not participate in the peptide bond, meaning that all types of amino acids can form peptide bonds with all other types.)

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2
Q

Which part of an amino acid gives it its unique properties?

  • peptide bond
  • carboxyl group
  • side chain
  • amino group
  • α-carbon
A

Side chain.

(The side chain of an amino acid is what gives the amino acid its unique chemical properties; the side chain is sometimes also called the R-group. All 20 naturally occurring amino acids are identical except in the collections of atoms composing these side chains.

Each amino acid contains an amino group (consisting of nitrogen and hydrogen atoms) and a carboxyl group (consisting of carbon, oxygen, and hydrogen atoms) that are covalently bonded to an α-carbon. The side chain is also covalently bonded to the α-carbon. Individual amino acids of all types are covalently linked together into a linear polypeptide by the peptide bond, which is formed between the amino and carboxyl groups of neighboring amino acids.)

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3
Q

What is the best type of model for visualizing the surface of a protein?

  • backbone
  • space-filling
  • ribbon
  • wire
A

Space-filling

(The space-filling model is the best type of model for visualizing the surface of a protein. This model provides a contour map of a protein’s surface, which reveals which amino acids are exposed on the surface and shows how the protein might look compared to a small molecule such as water or to another macromolecule in the cell. The backbone model shows the overall organization of the polypeptide chain and provides a straightforward way to compare the structures of related proteins. The ribbon model shows the polypeptide backbone in a way that emphasizes its most conspicuous folding patterns like α helices and β sheets. Finally, the wire model includes the positions of all the amino acid side chains; this view is especially useful for predicting which amino acids might be involved in the protein’s activity.)

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4
Q

What are the two types of β sheets?

  • helical and pleated
  • soluble and insoluble
  • parallel and antiparallel
  • primary and secondary
A

parallel and antiparallel

(The two types of β sheets are parallel and antiparallel. In a β sheet, several segments (strands) of an individual polypeptide chain are held together by hydrogen bonding between peptide bonds in adjacent strands. The amino acid side chains in each strand project alternately above and below the plane of the sheet. The adjacent chains run in opposite directions, forming an antiparallel β sheet. Parallel β sheets have more elongated loops that double back in the structure to maintain the parallel nature of the sheet.)

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5
Q

What does the primary structure of a protein refer to?

  • the locations of the peptide bonds that form the protein’s backbone
  • the structure that forms first as the protein folds into its most stable form
  • the locations of the protein’s α helices and β sheets
  • the linear amino acid sequence of the protein
  • the overall 3 dimensional shape of the protein
A

The linear amino acid sequence of the protein.

(Because a protein’s structure begins with its amino acid sequence, this is considered its primary structure. That is, the primary structure of a protein refers to the linear amino acid sequence of the protein. The chain of linear polymers of amino acids that compose proteins is termed a polypeptide. The locations of the peptide bonds that form the protein’s backbone are between each of the amino acids of the protein. The peptide bonds are involved in maintaining primary structure, but the location of the peptide bonds does not specify the primary structure. The primary structure does determine the secondary and tertiary structures.)

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6
Q

A protein domain is another phrase describing what type of structure of a protein?

  • primary
  • secondary
  • tertiary
  • quaternary
  • none of these
A

None of these

(The protein domain is an organizational unit that is distinct from the primary, secondary, tertiary, and quaternary levels of organization. Studies of the conformation, function, and evolution of proteins have also revealed the importance of a level of organization distinct from these four levels of protein structure. Usually, a single domain is responsible for a single function of the protein and some proteins can be composed of multiple domains. Figure 4–20, shown below, highlights protein domain structure using the example of catabolite activator protein (CAP), a bacterial transcriptional activator protein with two distinct domains, each with a unique function.)

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7
Q

What determines the specificity an antibody has for its antigen?

  • Its Y-shaped, bivalent structure
  • polypeptide loops in its constant domain
  • polypeptide loops of its heavy chains
  • polypeptide loops in its variable domains
  • polypeptide loops of its light chains
A

Polypeptide loops in its variable domains

(The polypeptide loops in its variable domains determine the specificity an antibody has for its antigen. A detailed examination of antibody structure reveals that the antigen-binding sites are formed from several loops of polypeptide chains that protrude from the ends of a pair of closely juxtaposed protein domains. The amino acid sequences in these loops can vary greatly without altering the basic structure of the antibody. An enormous diversity of antigen-binding sites can therefore be generated by changing only the length and amino acid sequence of these “hypervariable loops,” which is how the wide variety of different antibodies is formed. With their unique combination of specificity and diversity, antibodies are not only indispensable for fighting off infections, they are also invaluable in the laboratory, where they can be used to identify, purify, and study other molecules.)

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8
Q

Consider the thermodynamic properties of chemical reactions. Even though enzymes do not affect the overall energy of the reactants or the products (i.e., the thermodynamics), they alter the speed of the reaction. Enzymes accomplish this by doing which of the following?

  • supplying the activation energy for a reaction
  • reducing the activation energy of a reaction
  • not altering the activation energy of a reaction
  • eliminating the activation energy of a reaction
  • Increasing the activation energy of a reaction
A

Reducing the activation energy of a reaction.

(Enzymes reduce the activation energy of a reaction. The activation energy is an energy barrier to reactions. For a colliding water molecule to break a bond linking two sugars, the polysaccharide molecule has to be distorted into a particular shape—the transition state—in which the atoms around the bond have an altered geometry and electron distribution. Conditions are thereby created in the microenvironment of the enzyme active site that greatly reduce the activation energy necessary for the hydrolysis to take place. Other enzymes use similar mechanisms to lower the activation energies and speed up the reactions they catalyze. In reactions involving two or more substrates, the active site acts like a template or mold that brings the reactants together in the proper orientation for the reaction to occur.)

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9
Q

For a given protein, hydrogen bonds can form between which of the following?

  • atoms in the polypeptide backbone
  • atoms of two peptide bonds
  • atoms in two side chains
  • a side chain and water
  • all of the above
  • none of the above
A

All of the above

(For a given protein, hydrogen bonds can form between atoms in the polypeptide backbone, between atoms of two peptide bonds, between atoms in two side chains, and also between a side chain and water. The ability of a protein to bind selectively and with high affinity to a ligand is due to the formation of a set of weak, noncovalent interactions—hydrogen bonds. An α helix is generated when a single polypeptide chain turns around itself to form a structurally rigid cylinder. A hydrogen bond is made between every fourth amino acid, linking the C=O of one peptide bond to the N–H of another. β sheets are maintained by hydrogen bonds as well.)

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10
Q

Which statement concerning feedback inhibition is false?

  • Feedback inhibition is difficult to reverse.
  • Feedback inhibition can work almost instantaneously.
  • Feedback inhibition regulates the flow through biosynthetic pathways.
  • Feedback inhibition is a feedback system for controlling enzyme activity.
  • In feedback inhibition, an enzyme acting early in a reaction pathway is inhibited by a later product of that pathway.
A

Feedback inhibition is difficult to reverse.

(Feedback inhibition is not difficult to reverse. Rather, it is very easy to do so. In feedback inhibition, for example, an enzyme acting early in a reaction pathway is inhibited by a molecule produced later in that pathway. Thus, whenever large quantities of the final product begin to accumulate, the product binds to an earlier enzyme and slows down its catalytic action, limiting further entry of substrates into that reaction pathway. Where pathways branch or intersect, there are usually multiple points of control by different final products, each of which regulates its own synthesis. Feedback inhibition can work almost instantaneously and is rapidly reversed when product levels fall.)

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11
Q

How does an allosteric inhibitor work?

  • It binds to a site other than the active site, causing a conformational change in the enzyme that makes the active site less accommodating to the substrate.
  • It binds to a site other than the active site, causing a conformational change in the enzyme that forces the product to leave the active site.
  • It outcompetes the substrate molecule and binds to the active site, preventing substrate molecules from binding there.
  • It interacts covalently with the substrate, preventing it from fitting into the enzyme’s active site.
A

It binds to a site other than the active site, causing a conformational change in the enzyme that makes the active site less accommodating to the substrate.

(To regulate enzyme activity, an allosteric inhibitor binds to a second site, causing a conformational change in the enzyme that makes the active site less accommodating to the substrate. Unlike competitive inhibition, allosteric inhibition cannot be overcome by experimentally elevating the concentration of the substrate. With allosteric inhibition, there is no direct competition between inhibitor and substrate as both molecules are binding to the enzyme at different locations. There is also no direct interaction between the product of an enzyme and allosteric inhibition of that enzyme. Instead, products from reactions later in the pathway are more likely to act as inhibitors.)

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12
Q

How does phosphorylation control protein activity?

  • The phosphate group, with its positive charges, temporarily relieves feedback inhibition.
  • The phosphate group alters the primary structure of the protein.
  • The phosphate group induces a change in the protein’s conformation.
  • The phosphate group serves as an added source of energy for a protein.
  • The phosphate group, with its negative charges, prevents other negatively charged molecules from interacting with the protein.
A

The phosphate group induces a change in the protein’s
conformation.

(Proteins are commonly controlled by phosphorylation and dephosphorylation. When added to the protein, the phosphate group induces a change in the protein’s conformation. Regulation of protein activity in this manner involves attaching a phosphate group covalently to one or more of the protein’s amino acid side chains. Because each phosphate group carries two negative charges, the enzyme-catalyzed addition of a phosphate group can cause a conformational change by, for example, attracting a cluster of positively charged amino acid side chains from somewhere else in the same protein. This structural shift can, in turn, affect the binding of ligands elsewhere on the protein surface, thereby altering the protein’s activity. Removal of the phosphate group by a second enzyme will return the protein to its original conformation and restore its initial activity.)

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13
Q

What kind of enzyme adds a phosphate group to another protein?

  • GTPase
  • phosphorylase
  • ATPase
  • phosphatase
  • kinase
A

Kinase

(Protein phosphorylation involves the enzyme-catalyzed transfer of the terminal phosphate group of ATP to the hydroxyl group on a serine, threonine, or tyrosine side chain of the protein. This reaction is catalyzed by a protein kinase. The reverse reaction—removal of the phosphate group, or dephosphorylation—is catalyzed by a protein phosphatase. GTPases and ATPases are a class of enzymes that catalyze the decomposition of GTP into GDP and a free phosphate ion and ATP into ADP and a free phosphate ion, respectively. The removal is of a small molecule, not a protein.)

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14
Q

What kind of enzyme removes a phosphate group from a protein?

  • GTPase
  • phosphorylase
  • ATPase
  • phosphatase
  • kinase
A

Phosphatase

(The addition or removal of a phosphate group is a common mechanism by which the function of proteins are regulated. Protein phosphorylation involves the enzyme-catalyzed transfer of the terminal phosphate group of ATP to the hydroxyl group on a serine, threonine, or tyrosine side chain of the protein. This reaction is catalyzed by a protein kinase. The reverse reaction—removal of the phosphate group, or dephosphorylation—is catalyzed by a protein phosphatase. GTPases and ATPases are a class of enzymes that catalyze the decomposition of GTP into GDP and a free phosphate ion and ATP into ADP and a free phosphate ion, respectively. The removal is of a small molecule, not a protein.)

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15
Q

Enzymes can have both active and regulatory sites. What is the purpose of these sites?

  • The binding of CTP at a regulatory site on the protein causes decreased production of carbamoyl aspartate.
  • The binding of CTP at the active site on the protein causes increased production of carbamoyl aspartate.
  • The binding of CTP at the active site on the protein causes decreased production of carbamoyl aspartate.
  • The binding of CTP at a regulatory site on the protein causes increased production of carbamoyl aspartate.
A

The binding of CTP at a regulatory site on the protein causes decreased production of carbamoyl aspartate.

(Aspartate transcarbamoylase catalyzes the first step in the pyrimidine biosynthetic pathway, the conversion of L-aspartate and carbamoyl phosphate to form carbamoyl aspartate. One of the end products in this pathway, CTP, binds at a regulatory site (a location distant from the active site) on the enzyme, resulting in decreased production of carbamoyl aspartate. CTP is able to achieve this regulatory effect because its binding to aspartate transcarbamoylase causes a conformational change that renders the active site inaccessible to substrate, as shown in the figure. Note that as an allosteric regulator, CTP is a noncompetitive inhibitor and does not compete with substrate for binding to the enzyme’s active site.)

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16
Q

Electrophoresis separates proteins on the basis of what factor(s)? Choose all that apply.

  • the protein’s size
  • the protein’s cell of origin
  • the protein’s net charge
  • the protein’s affinity for a ligand molecule
A

The protein’s size and the protein’s net charge.

(Electrophoresis separates proteins on the basis of the size and the net charge of proteins. In this technique, a mixture of proteins is loaded onto a polymer gel and subjected to an electric field; the polypeptides then migrate through the gel at different speeds depending on their size and net charge. Denatured protein will travel faster than folded protein due to easier migration of linear sequence as opposed to a folded globular protein.)

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17
Q

Which of the following are methods for isolating a protein of interest?

  • chromatography and electrophoresis
  • chromatography and mass spectrometry
  • nuclear magnetic resonance and crystallography
  • electrophoresis and nuclear magnetic resonance
  • electrophoresis and crystallography
A

Chromatography and electrophoresis.

(Both chromatography and electrophoresis are methods for isolating a protein of interest. The most efficient forms of protein chromatography separate polypeptides on the basis of their ability to bind to a particular molecule—a process called affinity chromatography. Proteins can also be separated by electrophoresis. In this technique, a mixture of proteins is loaded onto a polymer gel and subjected to an electric field; the polypeptides then migrate through the gel at different speeds depending on their size and net charge. Mass spectrometry, crystallography, and nuclear magnetic resonance are each important in experimentally determining structural properties of the protein, but they are not used for isolating a protein.)

18
Q

Which of the following statements is true regarding protein structure determination?

  • A protein’s three-dimensional structure can be reliably predicted from its amino acid sequence by computer algorithms that access the vast quantities of structural information archived in public databases.
  • Nuclear magnetic resonance can be used to determine the structure of proteins that are too large to crystallize.
  • Determination of a protein’s structure by X-ray crystallography does not require prior knowledge of its amino acid sequence.
  • Mass spectrometry is the fastest way to determine the three-dimensional structure of a polypeptide or protein, regardless of its size.
  • Mass spectrometry can be used to determine the amino acid sequences of a complex mixture of different proteins.
A

Mass spectrometry can be used to determine the amino acid sequences of a complex mixture of different proteins.

(To achieve the resolution needed to distinguish between peptides from different proteins, such mixtures are often subjected to tandem mass spectrometry. In this case, peptides that pass through the first spectrometer are digested into even smaller fragments and then analyzed by a second mass spectrometer. Mass spectrometry is used to determine the amino acid sequence of a protein, but it provides no information regarding three-dimensional structure. Nuclear magnetic resonance is used to determine the structure of smaller proteins. Currently, there are no computer programs that can reliably predict the three-dimensional structure of a protein from its amino acid sequence alone.)

19
Q

Predict what would happen to the secondary structure of a protein if an alcohol that disrupts hydrogen-bonding were added.

  • The α helices would unfold, disrupting protein structure.
  • Individual amino acids would be hydrolyzed from the protein, disrupting protein structure.
  • Nothing would happen to the protein; hydrogen-bonding is not important for secondary structure.
  • The β sheets would unfold, disrupting protein structure.
A

The α helices would unfold, disrupting protein structure.
The β sheets would unfold, disrupting protein structure.

(Both α helices and β sheets result from hydrogen bonding of backbone atoms within the protein. Disrupting hydrogen bonding will cause both structures to unfold.)

20
Q

True or False:

Cylindrical α helices and planar β sheets can be formed by hydrogen-bonding of many different sequences.

A

True

21
Q

Describe the characteristics of secondary structures in alpha helix only.

A
  • Cylindrical structure

- One full turn every 3.6 amino acids

22
Q

Describe the characteristics of secondary structures in beta sheet only.

A
  • Consists of antiparallel or parallel strands

- Side chains alternating above and below the structure.

23
Q

Describe the characteristics of secondary structures in both alpha helix and beta sheet.

A
  • Can be formed by many sequences

- Formed by hydrogen-bonding between backbone atoms

24
Q

Mutations in the nucleic acid sequence of a gene can sometimes direct the substitution of one amino acid for another in the encoded protein. Which amino acid substitution would be most likely to severely disrupt the normal structure of a protein?

  • asparagine to threonine
  • leucine to isoleucine
  • alanine to glycine
  • methionine to arginine
  • tryptophan to phenylalanine
A

Methionine to arginine

(Arginine and methionine have different chemical properties. Methionine has a nonpolar side chain that would likely be buried in the protein’s interior; arginine, on the other hand, is a positively charged amino acid that would likely be facing the protein’s exterior. Replacing methionine with arginine would likely disrupt a protein’s structure.)

25
Q

You want to test pentapeptides (short peptides with only five amino acids) for their ability to bind to and inhibit a particular receptor. To do this, you set out to synthesize all possible pentapeptides and test each individually. Assuming you’ll use just the 20 common amino acids, how many different pentapeptides will you have to test for receptor binding?

A

3,200,000

That’s over 3 million different pentapeptides!

26
Q

To identify genes coding for essential proteins, researchers can create temperature-sensitive mutations. These mutations allow proper protein folding and cell proliferation at the permissive temperature of 22ºC, but they cause protein misfolding and reduced cell proliferation at a higher restrictive temperature, such as 37ºC. Which of the following mutations might increase protein flexibility and lead to a temperature-sensitive phenotype?

  • mutation of a lysine (that was involved in an ionic bond with a glutamic acid) to a glycine
  • a premature stop codon that truncates a protein 10 amino acids from the amino terminus
  • mutation of an alanine to a cysteine, leading to the formation of a new disulfide bond
  • mutation of a bulky isoleucine that was buried in the protein interior to a glycine (side chain = H)
A

Mutation of a lysine (that was involved in an ionic bond with a glutamic acid) to a glycine.
Mutation of a bulky isoleucine that was buried in the protein interior to a glycine (side chain = H)

(Protein folding is maintained by both noncovalent interactions (electrostatic attractions, hydrogen bonds, and van der Waals attractions) and covalent disulfide bonds. Loss of any of these will contribute to proteins unfolding at a temperature lower than they otherwise would. Generation of temperature-sensitive mutations in the yeast S. cerevisiae helped uncover genes involved in the cell division cycle and in protein secretion through the endomembrane system.)

27
Q

Which of the following best describes the stable protein–ligand interaction that is represented in the image? Be sure to use the image as a guide but apply your knowledge regarding how protein and ligand commonly interact with each other. A note of caution though: the red lines in the figure are merely representing interactions and are not meant to be quantified.

  • The formation of a set of many weak, noncovalent interactions maintains the interaction between protein and ligand.
  • The formation of a set of very few weak, covalent interactions maintains the interaction between protein and ligand.
  • The formation of a set of many weak, covalent interactions maintains the interaction between protein and ligand.
  • The formation of a set of very few weak, noncovalent interactions maintains the interaction between protein and ligand.
A

The formation of a set of many weak, noncovalent interactions maintains the interaction between protein and ligand.

(The formation of a set of many weak, noncovalent interactions maintains the interaction between protein and ligand.)

28
Q

Which of the following chemical group interactions may be represented by the red lines in the image?

  • an –OH of the ligand interacting with an –SH of the protein
  • an –OH of the ligand interacting with a –CH2CH3 of the protein
  • a COO– of the ligand interacting with a –CH2CH3 of the protein
  • a –CH2CH3 of the ligand interacting with a –NH3+ of the protein
A

An –OH of the ligand interacting with an –SH of the protein

(One of the red lines in the image could represent an –OH of the ligand interacting with an –SH of the protein. These two chemical groups are both polar, and thus would favorably interact when protein binds with ligand.)

29
Q

Organisms that thrive in extremely cold climates often produce proteins that act as “antifreeze.” Given that all proteins bind to other molecules, how might such antifreeze proteins work?

  • They bind to metal ions and each other, which attracts water molecules to form a thick gel that expands.
  • They bind by way of their β sheets to tiny ice crystals, preventing their growth.
  • They bind to proteins in the cell nucleus, promoting the expression of antifreeze genes.
  • They bind to channels in the cell membrane, preventing water from seeping into or leaking out of cells.
A

They bind by way of their β sheets to tiny ice crystals, preventing their growth.

(Organisms that thrive in extremely cold climates often produce proteins that act as “antifreeze.” Given that all proteins bind to other molecules, such antifreeze proteins bind by way of their β sheets to tiny ice crystals, preventing their growth. Prevention of the growth and expansion of ice crystals can prevent the cell from rupturing.)

30
Q

What are the steps of polysaccharide chain cleavage by lysozyme?

A
  1. Lysozyme and substrate form an enzyme-substrate complex, forcing one sugar molecule into a strained conformation.
  2. Glutamic acid donates a proton to one sugar as aspartic acid attacks the C1 carbon of a second sugar.
  3. A covalent bond forms between the aspartic acid and the sugar, and the sugar-sugar bond is hydrolyzed.
  4. Glutamic acid polarizes a water molecule, drawing a proton away from the water.
  5. The water oxygen attacks the C1 carbon, breaking the sugar-aspartate bond.
  6. Lysozyme and products dissociate.
30
Q

What are the steps of polysaccharide chain cleavage by lysozyme?

A
  1. Lysozyme and substrate form an enzyme-substrate complex, forcing one sugar molecule into a strained conformation.
  2. Glutamic acid donates a proton to one sugar as aspartic acid attacks the C1 carbon of a second sugar.
  3. A covalent bond forms between the aspartic acid and the sugar, and the sugar-sugar bond is hydrolyzed.
  4. Glutamic acid polarizes a water molecule, drawing a proton away from the water.
  5. The water oxygen attacks the C1 carbon, breaking the sugar-aspartate bond.
  6. Lysozyme and products dissociate.

(Enzymes first bind to the substrate, forming an enzyme–substrate complex. The enzyme then catalyzes the reaction. Finally, the enzyme is restored to its original starting state and the products are released.)

31
Q

What happens to glutamic acid 35 and aspartic acid 52 at the end of the reaction?

  • Lysozyme is destroyed and recycled as the product is released.
  • Aspartic acid has formed a covalent bond with the first sugar.
  • Glutamic acid is deprotonated and negatively charged.
  • Both amino acids are restored to their original forms.
A

Both amino acids are restored to their original forms.

Enzymes are restored to their starting point so that they can continue to catalyze many rounds of the reaction.

32
Q

Which statement concerning feedback inhibition is false?

  • Feedback inhibition is difficult to reverse.
  • Feedback inhibition can work almost instantaneously.
  • Feedback inhibition regulates the flow through biosynthetic pathways.
  • Feedback inhibition is a feedback system for controlling enzyme activity.
  • In feedback inhibition, an enzyme acting early in a reaction pathway is inhibited by a later product of that pathway.
A

Feedback inhibition is difficult to reverse.

(Feedback inhibition is not difficult to reverse. Rather, it is very easy to do so. In feedback inhibition, for example, an enzyme acting early in a reaction pathway is inhibited by a molecule produced later in that pathway. Thus, whenever large quantities of the final product begin to accumulate, the product binds to an earlier enzyme and slows down its catalytic action, limiting further entry of substrates into that reaction pathway. Feedback inhibition can work almost instantaneously and is rapidly reversed when product levels fall.)

33
Q

Ras is a GTP-binding protein involved in cell proliferation (division). In its active form, with GTP bound, Ras activates cell signaling pathways that promote cell division. Mutations in the gene that encodes Ras can lead to cancer. How might mutations in the gene encoding Ras lead to the uncontrolled proliferation characteristic of cancer cells?

  • They increase the protein’s affinity for GDP.
  • They decrease the protein’s affinity for GTP.
  • They prevent Ras from being made.
  • They increase the rate at which Ras hydrolyzes GTP.
  • They decrease the rate at which Ras hydrolyzes GTP.
A

They decrease the rate at which Ras hydrolyzes GTP.

(A mutation that decreases the rate at which Ras hydrolyzes GTP would be a logical mutation in the gene encoding Ras that could lead to the uncontrolled proliferation characteristic of cancer cells. Ras is a G protein that is activated upon GTP binding. If a mutation happened that caused a decrease in the ability of Ras to hydrolyze that GTP, then cell-proliferative Ras signaling would lead to cancer.)

34
Q

How do most motor proteins ensure their movements are unidirectional?

  • They couple a conformational change to the formation of an ATP molecule from ADP and Pi.
  • Their asymmetrical structures support movement in only one direction.
  • They couple a conformational change to the hydrolysis of an ATP molecule.
  • They couple a conformational change to a thermodynamically unfavorable reaction.
  • They hydrolyze their bound GTP, effectively preventing movement in the reverse direction.
A

They couple a conformational change to the hydrolysis of an ATP molecule.

(To achieve such directionality, one of the steps must be made irreversible. For proteins that are able to move in a single direction for long distances, this irreversibility is achieved by coupling one of the conformational changes to the hydrolysis of an ATP molecule that is tightly bound to the protein—which is why motor proteins are also ATPases. A great deal of free energy is released when ATP is hydrolyzed, making it very unlikely that the protein will undergo a reverse shape change, as required for moving backward.)

35
Q

Investigators wish to purify an enzyme—a serine protease—using affinity chromatography. They attach to the matrix of an affinity column an antibody molecule that binds specifically to a short sequence of amino acids located in the enzyme’s active site. When they apply a mixture of proteins to the column, the protease adheres to the column and the other proteins pass through. To extract their purified enzyme from the column, the investigators add a large excess of the peptide that the antibody recognizes. What should they expect to occur after this treatment?

  • The peptide will bind to the antibodies on the column, displacing the purified enzyme; they should obtain pure enzyme in the subsequent fractions.
  • The peptide will displace the enzyme and the subsequent fractions should contain a protein complex that includes the enzyme bound to the peptide.
  • The peptide will displace the enzyme, which will then be free to cleave the peptide; subsequent fractions will contain the enzyme and the peptide fragments, the latter of which will emerge first due to their smaller size.
  • The enzyme will remain bound to the column and none of the fractions will contain the enzyme of interest; only flow-through will leave the column.
  • The peptide will disrupt the binding of the enzyme to the antibody because antibodies are generally much larger than enzymes; the enzyme will emerge in the next fractions, followed by the antibody.
A

The enzyme will remain bound to the column and none of the fractions will contain the enzyme of interest; only flow-through will leave the column.

(The enzyme will remain bound to the column and none of the fractions will contain the enzyme of interest. Only flow-through will leave the column. Antibodies bind to their target molecules extremely tightly. A concentrated salt solution or a buffer with a very different pH is usually used to release proteins that are bound to an affinity column. These conditions weaken the binding of the antibody and protein so that the protein can be eluted.)

36
Q

Investigators are studying a protein that must be phosphorylated to be activated. Which method could be used to separate the phosphorylated form of the protein from the form that lacks an activating phosphate group?

  • Ion-exchange chromatography
  • Gel-filtration chromatography
  • differential centrifugation
  • equilibrium sedimentation
  • velocity sedimentation
A

Ion-exchange chromatography

(Ion-exchange chromatography could be used to separate the phosphorylated form of the protein from the form that lacks an activating phosphate group. Ion-exchange columns are packed with small beads carrying either positive or negative charges that retard proteins of the opposite charge. The association between a protein and the matrix depends on the pH and ionic strength of the solution passing down the column. These can be varied in a controlled way to achieve an effective separation. The addition of the phosphate group changes the charge of the protein and therefore can be used to separate the two forms of the protein.)

37
Q

Investigators decide to analyze the purity of a preparation of antibody molecules using SDS polyacrylamide-gel electrophoresis (SDS-PAGE). On Lane 1 of the gel, they load a sample of the antibody. On Lane 2, they load an antibody sample that has been treated with a reducing agent called mercaptoethanol, which breaks disulfide linkages. Following electrophoresis, they see distinct bands representing polypeptides with molecular weights of 50 kD and 25 kD in Lane 2 and only one band weighing 150 kD in Lane 1. What can the investigators conclude about their antibody based on the results of this experiment?

  • Their antibody preparation was not pure because it contains polypeptides of different molecular weights in Lane 2.
  • Their antibody consists of a single polypeptide chain of 150 kD; treatment with mercaptoethanol cleaves the polypeptide backbone at specific sites.
  • Their antibody preparation was pure; the 50-kD and 25-kD polypeptides are subunits of the mature antibody that were held together by noncovalent bonds.
  • Their antibody consists of a single polypeptide chain of 150 kD; treatment with mercaptoethanol cleaves the polypeptide backbone at random sites.
  • Their antibody is composed of subunits (50 kD and 25 kD in molecular weight) that each must include at least one cysteine residue.
A

Their antibody is composed of subunits (50 kD and 25 kD in molecular weight) that each must include at least one cysteine residue.

(Their antibody is composed of subunits (50 kD and 25 kD in molecular weight) that each must include at least one cysteine residue. The untreated antibody in Lane 1 of the gel migrates as a single, distinct band with a molecular weight of 150 kD. Thus, the preparation is pure. Treatment with mercaptoethanol, Lane 2, reduces the disulfide linkages that hold together the subunits (those of 50 kD and 25 kD) that are part of the mature antibody molecule. Disulfide linkages only form between cysteine residues.)

38
Q

What techniques can help determine protein tertiary structure and quaternary structure?

A
  • Nuclear magnetic resonance (NMR) spectroscopy
  • X-ray crystallography
  • Cyroelectron microscopy

(NMR spectroscopy, X-ray crystallography, and cryo-EM are all used to determine the three-dimensional structure of proteins. NMR spectroscopy can be used with small proteins (those less than about 50,000 daltons). X-ray crystallography requires crystals of the protein of interest. Cryo-EM, the newest technique of the three, allows atomic resolution of biological macromolecules in a near-native state.)

39
Q

What techniques cannot help determine protein tertiary structure and quaternary structure?

A
  • Mass spectrometry

- Two-dimensional polyacrylamide-gel electrophoresis