10 Study Guide Prot 14-3

8/11/2019 10 Study Guide Prot 14-3

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Study Guide for the Proteins Section of MBioS 303

by Ronald Brosemer

GENERAL COMMENTS:

What is the major theme of the lectures on proteins? -- the relationship between the chemical structures of

amino acids and polypeptides and the function of proteins in biological processes.

What are you responsible for? Everything covered in lecture and narrated PowerPoint slides: no more and noless. Notes to the PP slides contain a potpourri of info. Unless specified otherwise, you are not directly

responsible for the info in the PP notes, but you may find them especially helpful.

If you have a question about proteins, what do you do? (This is essentially a repeat of the same topic in the

course introduction.) In no particular order:

(NOTE that not all options are available for students taking this course online. Actually contacting the

instructor via email is -- I have found by experience -- the best method for asking and answering.)

1. Review your organic chemistry.

2. Consult the text (which is an excellent reference source). Note that the text covers far more material than

you are responsible for. You may have to use the table of contents to find the information you are seeking.

3. Consult the "Supplementary Topics on Proteins" file at the MBioS 303 website.

4. Ask during your recitation session. This allows the answer to be shared with your colleagues (who likely are

too shy to ask the same question).


6. Send an email to the instructor. I have found this to be an especially effective way to ask a question, since

you must be able to phrase the question lucidly and the instructor must be able to respond in a concise and

clear manner.

7. Ask the instructor in person.

8. Although one must be extremely careful, I have found Wikipedia to be surprisingly helpful on biochemical

topics. But remember: Reader beware! You never know who posted the information you are reading!!!


10. I have found "Molecular and Cell Biology for Dummies" to be a good introduction to biochemical topics.

Do you have to learn the structures and names of the twenty amino acids? Yes, but .... The "but" is that you

will have one page attached to the exam with the structures of the twenty amino acids we discuss; however,

no names will accompany these structures. (Refer to the sample exam below to see what I mean.) Thus, you

must be able to recognize (but not write) the structures of all twenty amino acids and to name them. Recall

that the structure of amino acids and polypeptides depends upon the pH, a topic you should understand.Reminder (from below): A copy of this table accompanies the first exam:

pK VALUES FOR FREE AMINO ACIDS

Gly: 2.4, 9.8 Ala: 2.3, 9.9 Phe: 1.8, 9.1

Pro: 2.0, 10.6 Trp: 2.4, 9.4 Met: 2.1, 9.3

Thr: 2.1, 9.1 Asn: 2.1, 8.8 Gln: 2.2, 9.1

Ile: 2.3, 9.8 Leu: 2.3, 9.7

Ser: 2.1, 9.2 Val: 2.4, 9.6 His: 1.8, 6.0 (side chain), 9.2

Asp: 2.0, 3.9 (side chain), 10.0 Tyr: 2.2, 9.1, 10.9 (side chain) Glu: 2.2, 4.3 (side chain), 9.7

Arg: 1.8, 9.0, 12.5 (side chain) Lys: 2.2, 9.2, 10.8 (side chain) Cys: 1.8, 8.3 (side chain), 10.8


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Study Guide for the Proteins Section pg. 2

Emphasis in this course will be placed on the topics outlined below.

The major chemical bonds found in Nature.

Strong and weak acids.

pH and the Henderson-Hasselbalch equation.

The effect of pH on the charge of an acid-base functional group.

The general structure of alpha-amino acids. All alpha-amino acids have a C atom to which is bonded an alpha-amino group (in the case of proline, this is a secondary amine) and one alpha-carboxylic acid group. In addition

a H atom and a side group are attached to the C atom.

All nineteen chiral amino acids found in proteins are of the L-configuration. (Glycine does not have a chiral

carbon atom because two H atoms are attached to the carbon atom.)

The structures of the twenty amino acids that serve as the building blocks for proteins.

The three-letter abbreviations of each of the twenty amino acids. Although the single-letter abbreviations are

used extensively in this course, you are not responsible for memorizing them; you are responsible for the

three-letter abbreviations.

The pK of acid-dissociable groups in the twenty amino acids. You should know the approximate pK values for

all acid-dissociable groups in amino acids (and peptides). However, if an exact value is required, that will be

supplied with an exam or quiz.

The amino acids with acid-dissociable side groups are: C, D, E, H, R, K, Y. The dissociation of the side group of

cyteine is not emphasized in this course.

The effect of pH on the net (or average) ionic charge of an amino acid.

As a reminder: in the H.H. equation there are four variables: pH, pK, (acid form), (base form). If any three are

known, the fourth can be calculated. pK is indeed a constant; it has remained invariant for several billion

years (although the exact value does depend upon many variables such as temperature and ionic strength).

In most of the discussions in this course, it is the [(base form)/(acid form)] ratio that is most relevant, since this

allows calculation of the net or average charge of an acid-dissociable functional group. That means that the

H.H. equation then has only three variables: pH, pK (which is constant for each of the functional groups), and

[(base form)/(acid form)]; i.e., if two of these variables are known, the other variable can be calculated from

the H.H. equation.

SOME QUESTIONS:

What is the difference between covalent , hydrogen and van der Waals bonds? What is the relative strength

of these interactions?


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What is the difference between a strong and a weak acid?

What is the difference between pH and pK? Why is it important to determine the pK of dissociable groups in

biological molecules?

Can you recognize all twenty of the amino acid structures and assign the appropriate three-letter

abbreviation?

The condensation of an alpha-carboxylic acid group of one amino acid

and the alpha-amino group of another amino acid, together with elimination of the elements of water,

forms a peptide bond.

The amino acid in a polypeptide is termed an "amino-acid residue" since it has lost the elements of water

when forming the peptide. Often this is simply shortened to "amino acid", even though, strictly speaking,

this is not correct. "Amino acid" should refer only to a free amino acid, i.e., an amino acid that has not

formed a peptide bond with another amino acid.

The primary structure of a polypeptide or protein is the amino-acid sequence of that molecule.

Every polypeptide (with the rare exception of cyclic polypeptides) has one N-terminal amino-acid residue

(which has a free alpha-amino group) and one C-terminal amino-acid residue (which has a free alpha-

carboxylic acid group).

By convention, the numbering of amino-acid residues in polypeptides begins with the N-terminal amino-acid

residue. The number of the C-teminal amino-acid residue thus equals the number of amino-acid residues in

the polypeptide.

The role of oxidizing two sulfhydryl (-SH) groups to a disulfide (-S-S-) bond in protein structure.

The structure of the polypeptide hormone insulin in different mammals.

A protein = polypeptide with a molecular weight > ~10,000 = ~10,000/~110 = ~90 amino-acid residues

The number of amino-acid residues in a protein can be estimated from the molecular weight (and vice versa)

by dividing the MW by the "average" MW of an amino-acid residue, ~110 D. Some examples of the MW ofamino-acid residues are: G = 57; V = 99; D = 115; W = 186.

Conformation is the three-dimensional structure of a molecule. For a protein, this means how the polypeptide

chain folds in 3-D space.

Secondary structure, which is part of the conformation, = the polypeptide's local conformation for its

backbone. This incudes, but is not limited to, such structural motifs as alpha-helices and beta-sheets.

Six atoms form a planar structure around a peptide bond. These six atoms are the alpha C atoms of both the

amino acid residues, the C=O atoms in the carbonyl group of the peptide bond, and the N-H atoms of the

amino group of the peptide bond. This planar structure is a consequence of the double-bond character of


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the tautomerization of the peptide bond. This planar structure constrains the number of possible

conformations that a polypeptide or protein can form.

One of the major secondary-structural motifs in proteins is the alpha helix. The right-handed helical structure

(i.e., clockwise as viewed from the N-terminal end) is stabilized by hydrogen bonds between the alpha-

amino group of an amino-acid residue and the carbonyl-oxygen of an amino-acid residue that is four

positions away. E.g., a hydrogen bond in an alpha helix could be formed between amino-acid residues at

positions 100 and 104, 123 and 127, etc.

There are about 3.6 amino-acid residues per turn of the alpha helix. This is related to the facts that the pitch

(the lateral distance along one turn) is 5.4 Angstroms and the rise (the lateral distance along one amino-acid

residue) is 1.5 Angstroms.

The percent of amino-acid residues in alpha helices varies with the protein; it can vary from 0 to 100 percent,

depending upon the protein. The values can only be determined experimentally.

Hydrophobic ("fear water") interactions are weak van der Waals forces between groups that have low

solubility in water and thus higher solubility in non-polar solvents. The side chains of amino-acid residues of

peptides can interact with one another to form hydrophobic interactions. These interactions are individually

so weak that they are usually not called "bonds". However, many of these weak hydrophobic interactionscombine to form the major factor that determines the conformation of proteins.

Hydrophobic side chains tend to overlap with one another in order to exclude water as a solvent. These

interactions are a major driving force in the 3-D structure of a globular (= shape something like a drop or

globule) protein with the result that hydropobic amino-acid residues tend to be buried in the interior where

they can avoid water and hydrophilic (= love water) amino acids tend to be on the surface of the protein

where they can interact with water. Note that the locations of amino-acid residues are a "tendency", not a

hard-and-fast rule.

There are several hydrophobicity scales that are based upon the relative solubility of a amino-acid side group

in water and in some organic solvent. Although the scales result in different rankings of amino acids in term

of hydrophobicity, a general concensus emerges: F, M, I, L, V are hydrophobic, H, Q, N, E, K, D, R arehydrophilic, and the rest are amphipathic (i.e., neither here nor there). The hydrophobic side groups are

largely non-polar; the hyrdrophilic side groups largely polar. (The amide side groups of N and Q attain

charge-separation via resonance.)

Usually (but not always) when a polypeptide traverses the lipid layer of a membrane, the polypeptide assumes

an alpha-helical structure for a length of ~ 20 hydrophobic amino-acid residues. This ~ 20-amino-acid-

residue segment is of sufficient lateral length to span the lipid portion of a membrane. This is due to the

facts that: the side groups in an alpha-helical structure stick out away from the helical polypeptide

backbone; the side groups of hydrophobic amino-acid residues tend to "be happy" in a non-polar lipid

solvent; the thickness of the lipid layer of a membrane is about 30 Angstroms; the rise (the lateral distance

along one amino-acid residue in an alpha helix) is 1.5 Angstroms; polypeptides tend to traverse a membranein a perpendicular direction. Note that the hydrophobic side groups do not actively interact all that much

with the lipid solvent; it is more a matter of avoiding interaction with the polar solvent water.

The beta-adrenergic receptor is a nice example of the alpha-helical structure of transmembrane segments of

proteins. This receptor is a membrane-bound protein that binds to epinephrine (also called adrenaline) and

transmits the hormonal signal from the extracellular space to the cytosol. The beta-adrenergic receptor has

a seven-membrane-spanning motif, i.e., seven segments that traverse the membrane, with each having an

alpha-helical conformation.


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The second major secondary-structural motif in proteins is the beta sheet (often called the beta pleated sheet

because of the pleated nature of the polypeptide backbone). Here the carboxyl-oxygen atoms and amino-

nitrogen atoms that form the hydrogen bonds stabilizing this structure are on amino acids that may be quite

distant from one another in the primary sequence. (This is in contrast to the alpha helix, where the hydrogen

bonds are between amino-acid residues that are separated by exactly three amino-acid residues.)

Two forms of beta sheets occur: the anti-parallel beta sheet where the two polypeptide segments run in

opposite directions (i.e., N to C and C to N); and the parallel beta sheet where the two polypeptide segmentsrun in the same direction (i.e., both N to C). The former is a bit more stable than the latter because the H

bonds are close to linear in the former and more bent in the latter. Anti-parallel beta sheets occur in

proteins more commonly than do parallel beta sheets. Often more than two polypeptide segments are

involved in making beta-sheet segments of proteins.

There are other less important secondary-structural components found in proteins, but these are not

considered in this course.

The tertiary structure of a protein describes folding of the secondary-structural elements plus the spatial

arrangements of the amino-acid side chains.

Most proteins ina an aqueous environment assume a globular 3-D shape. The major driving force ishydrophobic interactions. I.e., the best way for hydrophobic side chains to avoid the aqueous polar solvent

is to interact with one another in the middle of the protein; this can occur only if the proteins assumes a ~

spherical (i.e., globular) shape.

Some proteins have a fibrous shape; these proteins tend to have a lot of alpha helices and beta sheets.

Denaturation is the loss of the native conformation without altering the primary structure (i.e., not peptide

bonds are broken). Generally, denatured proteins are less soluble in water because the usually-buried

hydrophobic amino-acid residues are now exposed to the aqueous solvent.

Denaturing agents disrupt the weak bonds that stabilize the secondary and tertiary structure of a protein.

Denaturing agents include:

Heat: The increased thermal motion of atoms breaks the weak non-covalent bonds.Cooking is a great example

of this type of denaturation. It not only sterilizes food but also precipitates many of the once-soluble

proteins.

Ethanol: Since alcohol is less polar than water, hydrophobic groups are more soluble in alcohol. The only way

that buried hydrophobic groups in proteins can interact with the less polar alcohol solvent is by unfolding

the protein to expose the once-buried groups. This results in denaturation. The reason that 70 percent

alcohol is used to sterilize skin instead of 95 percent alcohol is that the latter precipitates proteins so fast

that a solid outer barrier can form in some microorganisms, resulting in protection of the genetic apparatus.

Thus some microorganisms survive and can multiply. Seventy percent alcohol acts more slowly and thus can

penetrate into the core of most microorganisms, resulting in protein denaturation throughout the cell.Nowadays isopropyl alcohol is used more often than ethanol.

How newly-synthesized proteins fold is an area of contemporary research. Folding during and after synthesis

occurs in a time scale of seconds. The primary sequence of the folding proteins plays a prime role. Some (but

probably no all) protein folding is aided by interactions with other proteins termed "chaperones". Details

remain murky. Misfolding of proteins can lead to debilitating conditions, such as cystic fibrosis, Alzheimer's

disease, and mad cow disease.

Proper protein folding (or lack thereof) can make or break a biotechnology firm.


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Many (but not all)


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in muscle (hence its name). Most textbooks state that the function of Mb is to store oxygen in tissues; this is

an important function in some animals, such as diving mammals. However, myoglobin has other functions

involved in transporting oxygen through tissues and cells.

Hb is a tetramer; i.e., it has four polypeptides with, therefore, 4 N-terminals and 4 C-terminals. Since Hb has

more than one polypeptide, it has quaternary structure. The MW of the most common form of hemoglobin

in human adults, HbA, is ~68,000. The four monomeric polypeptide units in HbA are two alpha subunits and

two beta subunits. Hb is found exclusively in erythrocytes (= red blood cells) where it functions to transportoxygen from lungs or gills to tissues.

An allosteric protein = a protein whose function can be increased or decreased by binding of a small molecule

(termed the "allosteric ligand" or "allosteric effector") to a site that differs from the functional site of the

protein. The former site is termed the "allosteric site". Two examples of functional sites are the oxygen-

binding sites (iron atoms) in Hb and the catalytic site of an enzyme.

Hb is an allosteric protein and is an excellent model of how allosteric proteins and allosteric effectors work.

Almost all allosteric proteins have quaternary structure (and even the few exceptions can be explained by

different semi-independent domains in a single polypeptide chain).

Oxygen binding curves are a plot of percent of Mb or Hb iron atoms binding oxygen as a function of the partialpressure of oxygen.The oxygen-binding curve for Mb is hyperbolic, for Hb sigmoidal. The major reason for

the difference is the absence of quaternary structure in Mb and its presence in Hb.

The steepest slope of the oxygen-binding curve for Hb is in the range of oxygen partial pressures normally

found in organisms where Hb is functional (e.g., vertebrates). The Hb oxygen-binding curve fits the function

of Hb to bind oxygen in the lungs (where the partial pressure of oxygen is high) and deliver it to tissues

(where the partial pressure is lower).

The sigmoidal oxygen-binding curve for Hb is typical of an allosteric interaction. Oxygen is unusual in that it is

both a functional molecule (it binds to Hb to be transported) and an allosteric effector (or ligand). As

allosteric effector, the binding of one oxygen molecule to Hb enhances the binding of the next oxygen

molecule; i.e., the second oxygen molecule binds more tightly than the previous one. Hb is also exceptionalin that the functional site (where oxygen binds to be transported) is the same as the allosteric site (where

oxygen binds to control the binding of other oxygen molecules)

The model that is used in this course to explain the allosteric effect of oxygen binding to Hb may not be the

actual mechanism, but it is a relatively simple explanation of basic principles.

When no oxygen is binding to the Fe iron atom of a heme, the Fe atom has only 5 coordinate covalent bonds:

4 to N atoms in the porphyrin ring, one to a N atom of a side group of one of the histidine residues. The

porphyrin ring structure has a bulge in it since the histidine pulls the Fe iron slightly out of the plane of the

porphyrin ring.

When oxygen binds to the Fe atom of one of the four hemes in Hb, the porphyrin ring around the iron is nowplanar. Fe has 6 coordinate covalent bonds: 4 to N atoms in the porphyrin ring, one to a N atom of a side

group of one of the histidine residues in Hb and one to oxygen. Therefore, the binding of the oxygen moves

the histidine residue from its location in deoxyHb. Since that histidine residue is part of the polypeptide

backbone, the whole polypeptide conformation is slightly altered. The conformational alteration of one

subunit is transmitted to the other subunits via non-covalent interactions between the subunits. In the other

deoxy subunits, the heme Fe atom is pushed slightly into a more planar line with the porphyrin ring. This

makes it easier for the next oxygen to bind since the Fe is closer to the planar structure found in oxyHb. With

the binding of each successive oxygen atom (up to the maximum of four), the binding of the next oxygen

atom becomes tighter. This is reflected in the sigmoidal oxygen-binding curves for Hb.


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The deoxyHb has more non-covalent bonds, particularly ionic bonds, between amino-acid residues. Thus

deoxyHb is said to have the T (= tense, taut, tight) structure and oxyHb the R (= relaxed) form.

The distortion of the porphyrin ring plane in the deoxy form also occurs in Mb, but since Mb does not have

quaternary structure, there is no transmission of the conformational change from one polypeptide to

another. Hence Mb does not show sigmoidal oxygen-binding curves.

Another important allosteric effector of Hb is 2,3-bisphosphoglycerate (BPG and whose structure you can

recognize). This compound is found only in erythrocytes, which is also the location of hemoglobin.

The negative charges on BPG interact postitive charges on side chains of specific H and K residues as well as

the N-terminals of Hb beta subunits. The binding site is at the junction of the four Hb subunits (and distant

from the oxygen-binding sites, which are the Fe atoms of heme). Since more ionic bonds are formed, the T

form of Hb is favored. I.e., the binding of oxygen is decreased = deoxyHb is more stable = oxyHb is less

stable. The allosteric sigmoidal oxygen-binding curve of Hb is shifted to the right in the presence of

increasing concentrations of BPG (= deoxyHb becomes more stable). The level of BPG in erythrocytes thus is

an important control of oxygen binding and release. E.g., one of the initial adaptations to high altitude is an

increase in synthesis of BPG which shifts the oxygen-binding curve to the right, resulting in increased

delivery of oxygen to tissues at the lower oxygen tensions found at high altitude.

Since the fetus obtains its oxygen only from the mother's blood, fetal forms of Hb must have a tighter binding

of oxygen than does HbA; i.e., fetal forms of Hb must have more of the R form of Hb than does HbA. The

major form of fetal Hb in the later stages of pregnancy is HbF, a tetramer of two alpha chains and two

gamma chains. Gamma chains differ from beta chains at several sites. Of particular interest is position 143

where the gamma chain has a S residue in place of the H residue in beta chains. H143 is one of the sites to

which BPG binds. Since the S143 in the gamma chain cannot form an ionic bond with a negative charge from

BPG (as does H143 in the beta chain), BPG binds less tightly to gamma chains than to beta chains. BPG does

bind to HbF because there are other sites in the gamma chain that form ionic bonds with BPG, but since

there are fewer ionic bonds formed than in beta chains,deoxy HbF is less stable than deoxy HbA. That is, the

R or oxy form of HbF is more stable than the R form of HbA. This allows fetal blood to "steal" oxygen away

from the mother's blood. Soon after birth, the gene making gamma chains is almost turned off, the gene

making beta chains is turned on. Over the first year after birth, HbF is replaced almost completely with HbA.

However, a small percentage of HbF remains throughout life to remind ourselves of our mothers.

Why is position F8 occupied by H in all globin polypeptides? It is this side group that covalently interacts with

the Fe of heme and binds it to the proteins. Any other of the 19 amino acids would fail to bind the heme,

resulting in an inactive globin molecule and thus elimination of that organism from the gene pool.

Why is position B6 occupied by G in all globin polypeptides? G has the smallest possible side group. The B and

E alpha helices of globins cross very close to one another at site B6. Any other of the 19 amino acids has a

side group that would be too bulky, disrupting the close encounter of the third kind and thus the entire

conformation of the globin molecule. This would negatively effect the physiological function of that globinmolecule, resulting in elimination of that replacement experiment from the gene pool.

These invariant sites illustrate the principle that some amino-acid residues in a protein are absolutely required

for the physiological function of that protein, while other amino-acid residues can be changed via evolution

with impunity.

SOME QUESTIONS:

What is the structural basis for the different oxygen-binding curves for myoglobin and hemoglobin? Why are

the differences in oxygen binding important in the whole organism?


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What is 2,3-bisphosphoglycerate? Where is it found in humans? Why is it important? How does it interact

with hemoglobin? Does it alter the properties of myoglobin? Why?

What is the structural basis for the effect of pH on the oxygen-binding capacity of hemoglobin? Why is this

important biologically?

What are the structural differences between HbA and HbF? How do these differences affect the function of

these two molecules? Why is this important?


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binding it at the substrate-binding site. However, the competitive inhibitor cannot undergo reaction to

products. As long as the competitive inhibitor is binding to the enzyme, substrate cannot bind and react.

Since the substrate and inhibitor are competing for the same site, increasing concentration of the substrate

can outcompete the inhibitor (such as flooding banks with money to avoid a recession). Competitive

inhibition is characterized by the fact that excessive concentrations of substrate can overcome the

inhibition; this is not the case for most other types of inhibition (which are not considered in this course.)

Examples of applying competitive enzyme kinetics to everyday issues are the use of ethanol to overcomepoisoning by ethylene glycol and of oxygen to overcome poisoning by carbon monoxide.

Enzymes display optimal temperatures. In general, as temperature increases, the rate of all reactions (whether

enzyme-catalyzed or not) increases. As the temperature continues to increase, proteins tend to denature,

resulting in loss of catalytic activity for enzymes. These two factors combine to result in an optimal

temperature of each enzyme. Thermophiles (e.g., microorganisms in hot springs) tend to have enzymes with

elevated optimal temperatures, humans with optimal temperatures above 37 degrees C, psychrophiles (e.g.,

in Antarctica) with low optimal temperaures. The structural bases for these differences remain obscure.

SOME QUESTIONS:

What is catalysis? Why is it important for biological systems?What is the difference between Vmaxand Km? Why are both important in defining the catalytic activity of an

enzyme?

What is the active site of an enzyme? Why is it important? Do all enzymes have the same active site? Why?

What is a turnover number? Why is it important? Do all enzymes have the same turnover number? Why?

What is a competitive inhibitor and how does it work? How can competitive inhibition be reversed?

What is meant by the optimal temperature and the optimal pH for an enzyme? What biological significance

do these parameters have?


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So why is chymotrypsin specific in the sense that it catalyzes the cleavage of a peptide bond only if that bond

is to the C-terminal side of F, Y, or W (and is not a terminal peptide bond)? What is common about F, Y and

W? They all have bulky aromatic hydrophobic side groups. (Yes, Y has an acidic side group, but in the typical

biological pH range, most of the phenolic side group is in the uncharged acidic form.)

Answer: Adjacent to the catalytic triad is a binding pocket in chymotrypsin. This binding pocket is lined with

hydrophobic amino-acid residues. Of the twenty amino acids, only the side groups of F, Y and W fit into this

pocket and interact (via hydrophobic interactions) with the pocket. Since an enzyme can catalyze a reactiononly after the substrate(s) have bound to the substrate-binding site, chymotrypsin has its specific specificity.

Two other examples of serine proteases with other specificies are given. Trypsin has the same catalytic

mechanism as chymotrypsin, but it catalyzes the cleavage of a peptide bond only if that bond is to the C-

terminal side of K or R (and is not a terminal peptide bond)? What is common about K and R? They both

have side chains that are long and carry a positive charge. The binding pocket for trypsin is similar to that for

chymotrypsin, except that at the bottom of the pocket is a D residue with its - charge. This - charge interacts

with the + charge on the side groups of K and R, resulting in the different substrate specificity for trypsin.

Both trypsin and chymotrypsin are synthesized in the pancreas, released as an enzymically inactive precursor

(or else pancreatic proteins would be degraded) into the small intestine, where they are activated. They are

involved, together with many other enzymes with their individual substrate specificities, in the complete

enzymic degradation of dietary proteins to free amino acids, which are then absorbed by the intestine and

carried by the blood to various tissues.

Elastase is another member of the serine proteases. Elastase has the same catalytic mechanism as

chymotrypsin, but it catalyzes the cleavage of a peptide bond only if that bond is to the C-terminal side of G

or (sometimes) A (and is not a terminal peptide bond)? What is specific about G (and to some extent A)? It

has the smallest side group of any of the 20 amino acids (and A has the second smallest). The binding pocket

has the side groups of two amino-acid residues sticking out to form a physical block, resulting in a shallow

binding pocket. Only G (and to some extent A) have side groups that are small enough to fit into the shallow

pocket.

Organofluorophosphates are a class of inhibitors of enzymes that have the catalytic triad. The inhibitor binds

at the substrate binding site, but the reactive F group then removes the hydroxyl H atom of the catalytic S

group, resulting in covalent modification of the S and thus irreversible inhibition of the enzyme. Of particular

importance are nerve poisons, such as sarin. They inhibit acetylcholine esterase (also termed choline

esterase), an important enzyme in neural function because it inactivates acetylcholine after release of this

neurotransmitter from synaptic terminals.

SOME QUESTIONS:

What is a serine protease?

Why do pK values play such a crucial role in the mechanism of the serine proteases?

What are the crucial structural differences between trypsin and chymotrypsin? Why do these differences

lead to different biological functions of these two enzymes? How are the two enzymes similar and how does

this impact their physiological function?

Give some specific examples of how the action of drugs can be explained by the mechanism of an enzyme-

catalyzed reaction.


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One example of the use of an enzyme inhibitor in chemotherapy is 5-

fluorouracil. It results in irreversible inhibition of the enzyme thymidylate synthase; the inhibitor is not 5-

fluorouracil itself but rather a metabolic product of 5-fluorouracil, namely 5-FdUMP Thymidylate synthase

catalyzes a key step in the synthesis of DNA, since its product is thymidylate (dTMP, thymine + deoxyribose +

phosphate at the 5' postition), a necessary precursor of DNA. The vitamin derivative tetrahydrofolate is a

cofactor for this reaction; during the thymidylate-synthase reaction, a derivative of tetrahydrofolate is

oxidized to dihydrofolate. The relatively complicated mechanism (which you should understand) includes, as

the last step, removal of the 5-H atom from the substrate dUMP (uracil + deoxyribose + phosphate at the 5'

position) to form a proton and a nucleophilic C atom.5-Fluorouracil enters the metabolic pathway for conversion of uracil to dUMP. Thus 5-fluorouracil, via several

enzyme catalyzed reactions, is converted to 5-FdUMP. This compound begins reaction at the active site of

thymidylate synthase. However, since fluorine is such a strong nucleophilic atom (F is very stable, F+is

essentially hardly ever formed), the C-F bond cannot be broken during the enzyme mechanism (because F+

would be the unstable product). This results in a dead-end covalent complex that irreversibly inhibits the

catalytic activity of thymidylate synthase.

5-FdUMP is an example of a suicide inhibitor = mechanism-based inhibitor = Trojan Horse inhibitor = binding

of a substrate analogue to the binding site of an enzyme with formation of an irreversible complex via a

covalent bond during the "normal" catalytic reaction

Since the synthesis of an important precursor of DNA is inhibited, DNA synthesis itself is inhibited due to

deficiency of the precursor dTMP. Cancer cells (as well as some other cells, such as hair follicles) require

rapid DNA synthesis to support the rapid cell proliferation. Thus 5-fluorouracil tends to inhibit spread of

cancer but at the risk of also inhibiting DNA synthesis in normal cells. This is an example of the usual

balancing act in chemotherapy: sufficient drug to inhibit cancer-cell division but not enough to result in side

effects due to inhibition of DNA synthesis by normal cells.

As a side remark, 5-fluorouracil is used in treating colon and rectal cancer, breast cancer, gastrointestinal

cancers, head and neck cancer, ovarian cancer, and basal cell carcinoma of the skin and actinic keratoses.

SOME QUESTIONS:

Give a specific example of how the action of a drug can be explained by the mechanism of an enzyme-

catalyzed reaction.

What are the chemical properties of the fluorine atom that make it so important in the suicide inhibition of

thymidylate synthase?

Why is inhibition of thymidylate synthase so important in our everyday lives?


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THE HOUR EXAM ON PROTEINS:

The exam will consist of 25 multiple-choice questions worth 4 points each. The sample exam below is an

accurate reflection of the exam format. It is critical to understand why an answer is correct; it is not

particularly helpful to simply come up with the correct answer and then move on.

The exam will emphasize understanding, not memorization. Yes, it is necessary to know a lot of nuts and bolts

about proteins; however, it is also important to understand the principles governing the relationship

between chemical structure and biological function.

Keep asking yourself: Why did Nature choose this structure for this function? If you can answer this question

more often than not, you will ace this section of the course.

Not all topics in the sample exam may have been covered in the current lectures (including any assigned

PowerPoint files with narration); therefore you are not responsible for those novel questions or answers.

However, an important facet of learning is knowing what you should know and what you do not have to

know. It is your responsibility to realize which topics in this sample exam you are responsible for and which

topics you do not have to address.

NOTE: As shown on the sample exam, there will be a page with the structures of the 20 amino acids

accompanying the exam. The structure-page will be similar to the one shown at the end of the sample examquestions. However, the order of amino acids will likely differ from that shown below.

In addition, as shown on the sample exam, a table entitled "pK VALUES FOR FREE AMINO ACIDS" will

accompany the exam.

REMINDER: You are responsible for the 3-letter abbreviations of amino acids. In addition, you are responsible

for what is covered in lecture (including any assigned PowerPoint files with narration): no more and no less.

----------------------------------

Instructions on the first page of any exam administered by Dr. B will be similar to the following:

ENTER YOUR NAME AND ID NUMBER ON EACH ODD-NUMBERED PAGE OF THIS EXAM.You have until the scheduled end of class to complete this examination. Enter your answers with a #2 pencil

on the answer sheet.

Print your eight-digit I.D. number in the boxes A through H under the section IDENTIFICATION NUMBER in the

lower left portion of the answer sheet. Fill in the bubbles corresponding to your I.D. number. Also fill in the

boxes and bubbles for your name (last name first) in case there is a problem with your ID number.

There are 25 (or 24 or 26 -- depending upon the exam) questions worth 4 points each. For each question

choose the single best answer on the answer sheet. Only the answers entered on the answer sheet are

considered to be your intended answer.

It is the students responsibility to read the entire exam. Sometimes page breaks occur in positions wheresome of the answer choices for questions are at the top of the next page; be sure to check! Each question has

either four or five possible choices. In case a typographical error or interpretation of the wording of a question

leads to clarification announcements during the examination, you are responsible for any changes announced

up to 20 minutes before the scheduled end of the exam. If such an announcement is made after that time, you

are not responsible for the change and suitable adjustments will be made to the acceptable correct answers

for all students.


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SAMPLE EXAM ON PROTEINS:(ANSWERS ARE SHOWN AT THE END OF THE STUDY EXAM.)

Choose the single best answer for each of the following questions.

pK VALUES FOR FREE AMINO ACIDS

Gly: 2.4, 9.8 Ala: 2.3, 9.9 Phe: 1.8, 9.1

Pro: 2.0, 10.6 Trp: 2.4, 9.4 Met: 2.1, 9.3

Thr: 2.1, 9.1 Asn: 2.1, 8.8 Gln: 2.2, 9.1Ile: 2.3, 9.8 Leu: 2.3, 9.7

Ser: 2.1, 9.2 Val: 2.4, 9.6 His: 1.8, 6.0 (side chain), 9.2

Asp: 2.0, 3.9 (side chain), 10.0 Tyr: 2.2, 9.1, 10.9 (side chain) Glu: 2.2, 4.3 (side chain), 9.7

Arg: 1.8, 9.0, 12.5 (side chain) Lys: 2.2, 9.2, 10.8 (side chain) Cys: 1.8, 8.3 (side chain), 10.8

1 An example of quaternary structure in proteins is the:

A binding of a substrate to the active site of an enzyme.

B presence of two subunits in an enzyme.

C presence of disulfide bonds.D presence of both alpha helices and beta sheets in the same polypeptide.

2 Titin, the largest known protein, is responsible for a muscle's ability to spring back into shape after beingstretched. Titin is a single polypeptide with 26,929 amino-acid residues. What is the approximate

molecular weight of titin?

A 27,000B 280,000

C 3,000,000

D 2,500

E 15,000

3 Alzheimer's disease is associated with aggregation of amyloid proteins. A 42-amino-acid peptide may be

involved in this aggregation. The last few residues of this peptide are immersed in the cell membrane.

Based on your knowledge about membrane proteins, which of the following sequences most probablyidentifies the last four amino acids in this 42-amino-acid peptide?

A Gly-Asp-Tyr-LysB Val-Val-Ile-Ala

C Met-Lys-Tyr-His

D Asp-His-Ser-Arg

4 In myoglobin and hemoglobin, oxygen binds directly to:

A Fein the heme.

B the hole where the two beta chains come close to one another.C the proximal histidine.

D the hydroxyl group of serine in the catalytic triad.

5 Fluorouracil is an effective chemotherapeutic agent because:

A fluorine is so electropositive that it does not form a negatively charged F ion.

B fluorine is so electronegative that it does not form a positively charged F ion.C H can displace F to convert fluorouracil to uracil.

D it can bind to the catalytic triad and interfere with the splitting of a bond.

6 Protein secondary structures such as alpha helices and beta sheets are stabilized mainly by:


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A an attraction between positive and negative charges.

B disulfide bonds.C hydrogen bond formation.

D hydrophobic interactions.

7 The role of the aspartate residue in the reactive triad in chymotrypsin is to:

A bind the Phe, Tyr, or Trp side group on the substrate.

B donate a proton to the peptide bond that is broken.C stabilize the acid form of an adjacent His residue.

D accept a proton from an adjacent Ser residue.

E bind to a + charged Lys group on the substrate.

8 A common physiological adaptation to high altitude in humans is an increase in the blood of:

A B C

D E

9 What is the following peptide?

A Glu-Lys-Ala-Leu

B Ala-Asp-Ile-Met

C Asn-Glu-Leu-Cys

D Asp-Gln-Val-MetE Thr-Arg-Ile-Trp

10 Which of the following increases the binding of oxygen to hemoglobin? A(n):

A increase in the concentrtion of BPG

B decrease in the partial pressure of oxygen

C decrease of pHD binding of oxygen to the first heme site

11 All of the following statements about myoglobin are correct EXCEPT:

CH3

C

H

O

C

O

O-

C

CH2

H

OPO3

=

=O

3PO

CH3 CH2 OH

CH3

C C

O

O-

NH3

+

H

C

O

O-

C

CH2

OH

OH H


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A Myoglobin has a high content of alpha-helix.

B Most of the charged residues are located on the outside of the molecule.C Many of the amino acids are in beta sheets.

D Myoglobin has only one C-terminal amino acid.

12 Consider the enzyme-catalyzed reaction AB. You determine experimentally that the maximal initial

velocity, Vmax, for the amount of enzyme you have added is 50 mM of B appearing per minute. You also

determine that the Michaelis constant, Km, for this enzyme is 20 mM of A. What concentration of A willgive an initial velocity of 25 mM of B appearing per minute? (HINT: No calculations are required.)

A 1,000 mM

B 40 mM

C 10 mMD 20 mM

13 A major advantage of using genetically-engineered human insulin rather than porcine (pig) insulin to treatdiabetes in humans is that:

A human insulin has two polypeptide chains and porcine insulin one polypeptide chain.

B the patient may recognize pig insulin to be a foreign protein.

C human insulin has one polypeptide chain and porcine insulin two polypeptide chains.D the bacteria in which the genetically-engineered human insulin is produced, also contain other

compounds that help control diabetes.

14 A major driving force for protein folding is that the hydrophobic groups tend to interact with one another

in order to:

A maximize the amount of interaction of hydrophobic side groups with water molecules.B minimize the amount of interaction of hydrophobic side groups with water molecules.

C increase the solubility of the hydrophobic groups in water.

D allow water to form many hydrogen bonds with the hydrophobic groups.

15 Consider a protease that specifically catalyzes the hydrolysis of a peptide bond between an aspartic acidresidue and the adjacent amino acid. Based solely upon this information, which of the following statementis most likely? (NOTE: the answer may not be correct in the actual case, but it is logical based upon

principles we have considered. NOTE also that the question asks what is in the enzyme, not what is in the

substrate!)

The binding site on the enzyme contains a(n) _____ residue:A Glu

B Arg

C GlyD Leu or Ile

16 In the diagram below, the plane (outlined by the dotted lines) illustrates the:


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A presence of hydrophobic interactions beween the O atom and the N atom.

B location of rotation around the bond between the central C and N atoms because the bond is acovalent bond.

C presence of hydrogen bonding between the O atom and the N atom in the plane.

D absence of rotation around the bond between the central C and N atoms because of its partialdouble-bond character.

17 Some drugs used in cancer chemotherapy are suicide inhibitors (= Trojan Horse inhibitors = mechanism-based inhibitors). Why are these drugs so termed? These drugs:

A are often used in rescuing people who attempted suicide.

B do not become an enzyme inhibitor until it undergoes at least part of the catalyzed reaction at

the active site of the enzyme.C were used by the Greeks to fool the Trojans.

D are usually inhibitors that bind at a site separate from the active site but still result in inhibiting

the mechanism of the enzyme-catalyzed reaction.E are taken up only by cancer cells, thus resulting in killing only these cells.

18 The initial treatment of carbon monoxide poisoning is an example of:

A binding of an allosteric ligand.B the effect of pH on the activity of a protein.

C competitive inhibition.

D denaturing a protein.

19 Many nerve poisons, such as sarin, contain a fluorine atom that reacts with the:

A iron atom of hemoglobin or myoglobin.B histidine nearest the iron atom of hemoglobin or myoglobin.

C serine side group in the catalytic triad of an enzyme.

D histidine side group in the catalytic triad of trypsin or chymotrypsin.

20 The reason that a number of people develop a red face after consuming only one or two alcoholic drinks isthat, in these individuals, the rate of oxidation of:A alcohol to acetaldehyde is abnormally fast.

B alcohol to acetaldehyde is abnormally slow.

C acetaldehyde to acetic acid is abnormally fast.

D acetaldehyde to acetic acid is abnormally slow.

21 In a cell, the concentration of S is about 70 mM. The Vmaxfor the enzyme that catalyzes the reaction SP is 250 units. (For this question, ignore what the units are for initial velocity.) The value for Kmis 50 mMWhat is the approximate initial velocity for this enzyme-catalyzed reaction in the cell?

HINT: You need do no calculations; you can apply reasoning to arrive at the correct answer. Note that

units are not shown in the answers; they will be addressed in the next question.)A 70

B 250

C 125

D 145E 35

F 100

G 108

22 What are appropriate units for the correct answer in the previous question?


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A (liters per mmoles)/minute

B (grams per liter)/secondsC seconds/mM

D minutes/liter

23 An allosteric interaction between a ligand and a protein is one in which:

A the binding of a molecule to a binding site affects the binding of an additional molecule to the

same site.B multiple molecules of the same ligand can bind to the same binding site.

C the binding of a molecule to a binding site affects the binding properties of another site on the

protein.

D two different ligands can bind to the same binding site.

24 One of the main reasons that alcohol can be used to sterilize surgical instruments is that alcohol:

A stabilizes hydrophobic interactions in the interior of proteins.B is a good solvent for hydrophobic groups in proteins.

C is metabolized to a substance that is toxic to bacteria.

D decreases the binding of oxygen to hemoglobin.

THE PROTEINS EXAM WILL HAVE TWO QUESTIONS SIMILAR TO THE FOLLOWING TWO --

GUARANTEED!!!

25 The following is a titration curve for which amino acid?

A glutamic acid

B lysineC histidine

D alanine

HINT: Estimate the 2 or 3 pK values from the graph and check your estimates with the values in the table "pKVALUES FOR FREE AMINO ACIDS" at the beginning of the exam.

26 Refer to the previous question. What is the charge on the amino acid at pH 9.2?


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A -1 B -0.5 C 0 D +0.5 E +1

HINT: You have to apply your vast knowledge about the ionic forms of this amino acid.

STRUCTURES OF AMINO ACIDS

Answers to study questions for the section on proteins:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

B C B A B C C B D D C D B B B D B C C D D B C B B D

CH2

CH2

S CH3

C

H

H3

N+ CO O-

C

OH

H H

C COO-H3

N+

H

+ CO O-C

HN

H2

H2

C

H2

C CH2 CCH 3 CH 3

H

CH2

C

H

H3

N+ CO O- C COOH3

N+

H

CH2

CH2

CH2

NHC

NH2

NH 2+

C COO-

H3 N+

H

CH2

C

O

NH2

CCH3

CH3

H

C

H

H3 N+

CO O-

CH3

C

H

H 3 N + CO O -

CH CH3

CH2

CH3

C

H

H3 N+

CO O-

C

H

H3 N+

CO O

CH2

OH

CH2

C

H

H3

N+ CO O- C COO-H3

N+

H

CH2

CH2

CH2

CH2

NH3

+

H

C

H

H3

N+ CO O- C

H

H3

N+ CO O-

CH2

CH2

CO O-

C COOH3

N+

H

N

N

H

H+

CH2

CH2

CO O-

C

H

H3

N+ CO O-

CH2

SH

C

H

H3

N + CO O -

CH2

N

H

C COO-H3

N+

H

C COO-H3

N+

H

CH2

CH2

C

O

NH2

C

OH

CH3

H

C COOH3

N+

H

10 Study Guide Prot 14-3

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