The D-isomer ofglycine is thepredominant isomer in- 123docz.net

Compound I. Because alkoxy groups are donating by resonance, Compound HI

D. The D-isomer ofglycine is thepredominant isomer in humanbeings

Solution

Thecarboxyl terminal has the lowest pKa value for all twenty of the aminoacids forwhich wecode. This can be verified by looking at the amino acids in Figures 7-17 through 7-21. This eliminates choice A. Because the carboxyl terminal has a lower pKa value than the amino terminal, the carboxyl terminal is more acidic than the amino terminal. By being more acidic, the carboxyl terminal loses a

proton more readily than the amino terminal. This means that choice B is valid and thus is eliminated. Choice C requires that you look at the amino acids in Figure 7-21. The pKa of the amino terminal of lysine is 9.2. The pKa of the side chain of lysine is 10.8. The lower pKa is associated with the amino terminal, which implies that the amino terminal is more acidic than the side chain. This makeschoice C valid, and thus it is eliminated. The only choice left is answer D.

There is no D-isomer of glycine, because the side group for glycine is a hydrogen, therefore glycine does not contain a chiral center. Without a chiral center, there can be no D or L configuration associated with the isomer. This confirms that the statement is false, making choice D the best answer. For the other amino acids, the naturally occurring form is the L-isomer. This can be recalled by thinking of the phrase "L is for life." It canalsobe remembered by thinkingamino acids are natural. Either way you chooseto recall the meaning of the L label, amino acids are naturally L. If you are bothered by the word "always" in choice A, sometimes an always is okay.

OrgaillC ChCmiStry Nitrogen Compounds Biological

Example 7.23

What is the isoelectric point for tyrosine?

A. 2.2 B. 5.6 C 6.2 D. 9.1 Solution

Tyrosine is not a basic amino acid, so its isoelectric point is found by averaging pKai and pKa2. The isoelectric point is solved according to the following

mathematics:

j _pKai +pKa2 _ 2.2+ 9.1 -11.3 = 565

2 2 2 *

This makes choice B the best answer. As a general rule, acidic amino acids have pi values less than 5.5 and basic amino acids have pi values greater than 7.5. All other amino acids have pi values within the 5.5 to 7.5range.

Example 7.24

What is the isoelectric point for lysine?

A. 5.7 B. 6.5 C. 9.2 D. 10.0 Solution

Lysine is a basic amino acid, so its isoelectric point is found by averaging pKa2 and pKa3. The isoelectric point is solved according to the following mathematics:

j =pKa2 +pKa3 =9.2 +10.8 =20 = 10

2 2 2

This makes choice D the best answer. As a general rule, basic amino acids have pi values greater than 7.5, so choices A and B are eliminated before any math

needs to be done.

Example 7.25

What is the isoelectric point for the dipeptide alanine-arginine?

A. 7.5 B. 9.0 C 11.1 D. 11.5 Solution

Alanine is a hydrophobic amino acid and arginine is a basic amino acid, so the isoelectric point for the dipeptide is found by averaging pKa2 and pKa3. As written, the peptide linkage is made from the C-terminal of alanine and the N-

terminal of arginine, so there is no pKa(coOH) for alanine andno pKa(NH3+) f°r arginine. For the dipeptide, pKa2 is the N-terminal of alanine and pKa3 is the side chain of arginine. The isoelectric point is solved as follows:

j _pKa2+pKa3 _ 9.9 +13.2 _23.1 = n 55

2 2 2 '

Organic Chemistry Nitrogen Compounds Biological

This makes choice D the best answer. The difficult task in these questions is not determining which pKa terms belong in the equation, but rather assigning a numerical value to each pKa term. If you've grown weary of continually turning back to the amino acids in Figures 7-17 through 7-21, then start making some approximations. Knowing that the N-terminal is pK^ and its value is around 9 should tell you that the pi is greater than 9. This eliminates choices A and B, which helps to some extent. From here, you have to know the values. The MCAT test writers will provide pKa values if the question is this specific.

Example 7.26

What is the isoelectric point for the dipeptide histidine-leucine?

A. 5.5 B. 6.1 C. 7.6 D. 9.2 Solution

Histidine is a basic amino acid and leucine is a hydrophobic amino acid, so the isoelectric point for the dipeptide is found by averaging pKa2 and pKa3. As written, the C-terminal of histidine and the N-terminal of leucine make up the peptide linkage, so pKa3is the N-terminal of histidine and pK^ is the side chain of histidine. The presence of the leucine does not alter the pi from what it would have been with histidine alone. The isoelectric point is solved according to the following mathematics:

pi _PKa2+pKa3 _6.1 +9.2 -15.3 = 765

2 2 2 '

This makes choice C the best answer. You should have been able to solve this

particular question without consulting any other data. The side chain of histidine has a pKa close to physiological pH, so you know that the pi is an average of roughly 7 and 9, which means that the pi is about 8. Only choice C is close to that number. Another method for deterrnining the best answer is to consider that histidine is a basic amino acid. We know that the pi is greater than 7.5, but it must be less than the pKa of the amino terminal, given that pKa3 corresponds to the amino terminal. Only choice C fits in the 7.5 to 9.2 range.

Example 7.27

At what pH is the tripeptide serine-cysteine-isoleucine perfectly neutral?

A. 5.3 B. 6.1 C. 8.4 D. 9.4 Solution

The pH at which it is perfectly neutral is the isoelectric point. Of the three amino acids, none are basic, so the isoelectric point for the tripeptide is found by averaging pKai and pKa2. The tripeptide has three active protons, the N- terminal of serine, the side chain of cysteine, and the C-terminal of isoleucine.

The C-terrninal is always pKai and in this case pKa2 is the side chain of cysteine.

The isoelectric point is found using the following mathematics:

! =pKaj_+pKa2_ _ 2.3+ 8.4 -10.7 = 535

2 2 2 '

OrgaillC CliemiStry Nitrogen Compounds Biological

This makes choice A the best answer. Choices C and D should have been

eliminated, because there are no basic amino acids in the tripeptide. Cysteine is actually an acidic amino acid, but because most cysteine side chains are involved in crosslinking,we rarely consider their acid-base propertiesin proteins.

Example 7.28

Whichof the following tripeptides has the LOWEST pi value?

A. Asp-Lys-Asp B. Glu-Lys-Glu C. Lys-Asp-Lys D. Lys-Glu-Lys

Solution

The lowest pi value corresponds to the most acidic tripeptide. Choices A and B have two acidic amino acids and one basic amino acid, while choices C and D have two basic amino acids and one acidic amino acid. This eliminates choices C

and D. ChoicesA and B each have one lysine, so both pis are found by averaging pKa2 and pKa3- In each tripeptide, there are five active protons (the N-terminal, the C-terminal, and three the side chains). Both pK^ and pKa3correspond to the

side chains of the acidic amino acids. In choice A, the acidic amino acid is

aspartic acid while in choice B the acidic amino acid is glutamic acid. Aspartic acid has a lower side chain pKa than glutamic acid, so the tripeptide in choice A has a lower pi than the tripeptide in choice B. Choice A is the best answer.

Example 7.29

At pH = 5.0, histidine exists with a net charge of:

A. +2.

B. +1.

C. 0.

D. -1.

Solution

To determine the charge of an amino acid at a given pH, first determine whether the sites are protonated or deprotonated. When the pH is lower than the pKa, the site is protonated. When the pH is greater than the pKa, the site is deprotonated. At pH = 5.0, histidine has a deprotonated carboxyl terminal (pKa

= 1.8), a protonated side chain (pKa = 6.1), and a protonated amino terminal (pKa

= 9.2). The carboxyl terminal carries a negative charge, while the side chain and amino terminal are both carry a positive charge. The overall charge is therefore a positive one. The best answer is choice B. Histidine at pH = 5.0 is drawn below:

pH <pKa H3N+ J^ pKa =1.8

protonated ^syT O" pH>pKa

deprotonated

overall charge = +1 +1 -1 = +1

pKa =6.1 f NH

PH<pKa \ /

protonated h N1" ~~^

Organic Chemistry Nitrogen Compounds

Example 7.30

At pH = 9.5, tyrosine exists with a net charge of:

A. +1 B. 0 C. -1 D. -2

Biological

Solution

At pH = 9.5, tyrosine has a deprotonated carboxyl terminal (pKa = 2.2), a deprotonated amino terminal (pKa = 9.1), and a protonated side chain (pKa = 10.1). The carboxyl terminal carries a negative charge, while the amino terminal is uncharged. The side chain is uncharged when protonated, so the overall charge is a negative one. The best answer is choice C.

Proteins

Proteins are biologicalpolymers built from amino acids. They are held together by covalent bonds in peptide linkages. A peptide linkage is formed when the amino terminal of one amino acid attacks the carbonyl carbon of the carboxyl terminal of another amino acid, resulting in the loss of water. Because amino acids have specific stereochemistry, proteins are highly chiral (on average, they have one chiral center for every amino acid in the protein). This large degree of chirality explains why enzymes are so selective.

Protein Structural Features and Levels

The structure of a protein can be broken down into levels. The most elementary levelis the primary structure, which is defined as the sequence (connectivity) of amino acids within the protein. To break down the primary structure, you must cleave the peptide bonds. Figure 7-30 shows a tetrapeptide (four- amino acid protein).

Tetrapeptide (4 amino acid protein)

Peptide linkage (amide bond)

Figure 7-30

In this case, the primary structure is the order in which the four amino acids are arranged from the amino terminal to the carboxyl terminal. The primary structure of a protein or enzyme is determined by sequencing the protein one

amino acid at a time. This will be discussed later. The interactions between the

amino acids within the protein are responsible for the secondary structure, which is defined as the folding of the amino acids into their natural configuration within the protein. The chemical reasons for special features in the secondary structure are most often attributed to hydrogen bonding, and sometimes to cross- linking through disulfide bridges and kinks and turns caused by the presence of proline. Some of the structural features of interest in proteins are the oc-helix and

Organic Chemistry Nitrogen Compounds

fi-pleated sheets. In the oc-helix, two residues near one another in the protein are held together by hydrogen bonds. There are 3.6 amino acids per turn and each

turn is 5.4 A in length. They coil like a phone cord. In 6-pleated sheets, the

amino acids strands are not coiled. They are held together by hydrogen bonds from the hydrogen on nitrogen to the carbonyl oxygen. Figure 7-31 shows a fi- pleated sheet with antiparallel strands

R74 H

H R„ H O

Figure 7-31

A disulfide bridge is formed when two cysteine residues lose the H from the thiol group to form a sulfur-sulfur bond. Disulfide bridging is most often considered to be part of the tertiary structure. Because loss of hydrogen results in an increase in the oxidation state of sulfur, this process is oxidative, so a cross- linked protein is considered more oxidized than the protein without cross- linking. To break a cystine cross-linkage, one can add a reducing agent such as fi-mercaptoethanol, HSCH2CH2OH. As a side note, cross-linking is responsible for an increase in rigidity for the polymer (polypeptide in this case) due to the loss of both flexibility and entropy. In terms of tnermodynamics, the process of cross-linking is driven by enthalpy and not entropy (recall that AG = AH - TAS).

Figure 7-32 shows the reductive cleavage of a disulfide linkage using the chemical reagent S-mercaptoethanol.

Biological

H Rft H

OrgaiUC Chemistry Nitrogen Compounds Biological

Vasopressin (antidiuretic hormone)

S S

I I

Figure 7-32

When any of the molecular interactions that hold together the secondary or tertiary structure of a protein are broken, the protein is said to be denatured (no longer in its natural form). The overall folding can be changed by the environment in which the protein exists. In an aqueous environment, a protein will fold in such a way as to expose the hydrophilic side groups to the solvent (water) while nainimizing the exposure of the hydrophobic (alkyl and phenyl) side groups to the solvent (water). In a hydrophobic environment, a protein will fold in such a way as to expose the hydrophobicside groups to the solvent (lipid) while ininimizing the exposure of the hydrophilic (hydrogen bonding) side groups to the solvent (lipid). This overall folding of the protein is referred to as the tertiary structure. The tertiarystructure is globular for many proteins.

Example 7.31

Taking a protein from a Upid environment and placing it into an aqueous environment would most affect which of the following?

A. Primary structure B. Secondary structure C. Tertiary structure D. Quaternary structure

Solution

A protein in a Upid environment has its hydrophobic side chains exposed to the solvent and its hydrophiUc side chains in the inner core. A protein in an aqueous environment has its hydrophilic side chains exposed to the solvent and its hydrophobic side chains in the inner core. The conversion from lipid environment to aqueous environment breaks apart the internal hydrogen bonding, which denatures the protein. This affects the secondary structure by changing the structural features. The tertiary and quaternary structures are also affected, but only as a consequence of changes in the secondary structure. This question calls for the most specificchange. The best answer is choice B.

Organic Chemistry Nitrogen Compounds

Example 7.32

Turns are caused by which of the following amino acids?

A. Cysteine

B. Histidine C. Proline

D. Glycine

Solution

Because proline is cycUc, it cannot freely rotate about its sigmabonds, forcing the protein chain to turn. Choice C is the best answer. The following pictorial representation shows the impact of the cycUc structure of proline on the turn.

proline residue ^^ O

turn—><*

O /0s

,ằ>x

Bent peptide containing proline

protein turn / O^f

Blown up view of protein turn Example 7.33

Which of the foUowing results in a denatured protein?

A. Cleaving disulfide bonds B. SpUtting base pairs C. Breaking alpha bonds D. Forming of beta bonds

Solution

Breaking disulfide bonds requires the reductive cleavage of the S-S bond. This definitely changes the tertiary structure, thus denaturing the protein. Base pairing is involved in DNA and RNA, not proteins, so choice B is eliminated.

Alpha bonds and beta bonds are throw away answers, so choices C and D are

eliminated. Choice A is the best answer.

Unique Biological Process

A common rule in the chemistry of amino acids is that they exist exclusively as L- stereoisomers in biological systems. An exception to this "aU amino acids are L"

rule is found in the bacterial ceU waU. The ceU wall of a bacteria is held together by a net-like structure that involves a cross-link between D-alanine and glycine.

As a point of biological interest, fi-lactams such as penicillin act in a medicinal fashion by breaking this linkage and destroying the ceU walls of bacteria. Again, these are not facts to be processed and stored, but they are interestingbiological anecdotes that have been presented on previous MCATs. Discussing them here is simply to provide a Uttle background, so if they show up again, there wiU be a smaUair of famiUarity to them. Figure 7-33shows the formation of the cross-link

in the bacterial cell waU.

Biological

Organic Chemistry Nitrogen Compounds Biological

O H O H H

II I II I I

R-C— CH2-NH2+ R'-N-C-C-N-C-CO,"

| l I z

Terminal Residue of the ** C**3 *-H3 pentaglycine bridge Terminal D-Ala-D-Ala unit

O H H H H

II I I I + I

R~C-CH2-N-C-C-N-R' + HoN-C-CO,"

II I I 2

O CH3 CH3

Gly-D-Ala crosslink D-Ala

Figure 7-33 Gel Electrophoresis

Let us consider gel electrophoresis, a biochemistry lab technique based on the idea that charged particles migrate through electric fields. Particles migrate according to their charge (which impacts the strength of the electrical force) and theirsize (which impacts their resistance to flow as they travel through a gel). A polyacrylamide gel is chosen to offer resistance so that the particles do not migrate too rapidly and thus not have the time to separate to a distinguishable amount. Figure 7-34 shows the schematic of an electrophoresis apparatus.

Figure 7-34

In gelelectrophoresis, cations migrate to the cathode and anions migrate to the anode. This aUows for separation of cations from anions. To get separation of Uke charges (cations from cations for instance), the particles must migrate at different speeds. The speed depends on the acceleration, which depends on the

electric force (F =qE) and the resistive force due to drag (Fdrag)- Equation 7.3

shows the relationship ofthe particle's mass and acceleration to itscharge, q, the strength of theelectric field, E, and the resistive force due to the drag as it moves through the gel.

ma = qE-Fdrag (7.3)

The charge of an amino acid, polypeptide, or protein fragment depends on the pH of the environment and the isoelectric point, pi, of the species. When pH is greater than pi, the environment is basic, resulting in a negatively charged species. When pH is less than pi, the environment is acidic, resulting in a positively charged species. The bigger the difference between the pH of the environment and pi of the species, the greater the magnitude of its charge. As the magnitude of charge increases, the electric force increases, and therefore its acceleration increases. This means that the species can be separated by their pi values. The other factor that influences the migration rate is the size of the species. Larger species experience greater resistance as they travel through the medium. As a result, there aretwotypes ofelectrophoresis techniques employed

to separate protein fragments.

OrgaillC ChemiStry Nitrogen Compounds Biological

The first type of electrophoresis employs a pH gradient in the gel and the charged species migrate through the gel until they reach a pH equal to their pi.

At that point they have no net charge, so thereis no electric force. This technique is referred to as isoelectric focusing. The second type of electrophoresis involves adding sodium dodecyl sulfate, SDS, to the protein mixture. The protein fragments incorporate SDS into their secondary structure, resulting in every fragment takingon a negative charge. Larger protein fragments can incorporate more SDS than smaller ones, resulting in a greater magnitude of charge.

However, larger species are more massive. As a result, the incorporation of SDS creates an environment where aU of the protein fragments have the same mass- to-charge ratio. Given that they are experiencing the same electric field, any separation is due to differences in size, where small species migrate faster than larger species,because they experienceless resistance due to drag.

Example 7.34

What is true of the protein fragment in a mixture that migrates fastest in SDS PAGE and migrates to the highest pH value in a gel with a pH gradient?

A. It is the heaviest fragment in the mixture and is most likely to contain aspartic acid.

B. It is the heaviest fragment in the mixture and is most likely to contain lysine.

C. It is the Ughtest fragment in the mixture and is most likely to contain aspartic

acid.

D. It is the Ughtest fragment in the mixture and is most likely to contain lysine.

Solution

In SDS PAGE, the speed at which a species migrates depends on the resistance it experiences moving through the gel. Larger fragments experience greater resistance, so a species that migrates rapidly through the gel must be small. This means that the fastest fragment must be the Ughtest fragment, so choices A and B are eliminated. Because the species migrated to a high pH value, it must have a large pi value. The presence of lysine, arginine, and histidine increases the pi, so the species must have at least one of those three amino acids. Aspartic acid lowers the pi value, so choice C is eliminated and choice D is the best answer.

Affinity Chromatography

Another technique that can separate proteins according to their charge is affinity chromatography. We will look specifically at ion-exchange chromatography.

Ion-exchange chromatography separates by the affinity of charged proteins for oppositely charged sites on the polymer in the column (stationary phase). The stationary phase is made of an insoluble polymer, often polystyrene or ceUulose, to which functional groups capable of carrying a charge have been attached.

Typical anionic groups include sulfate on sulfonated polystyrene and carboxylate on carboxymethyl-ceUulose. Typical cationic groups include diethylaminoethoxy on diethylaminoethoxy-cellulose. Positively charged columns bind anionic species tightly and negatively charged columns bind cationic species tightly. A protein carries a positive charge when the solution pH is less than the isoelectric point, pi, of the protein. When a mixture of proteins is added to the column and aUowed to migrate down the polymer, selected proteins can be separated from the mixture by binding the column. Proteins carrying the same charge as the column elute. To release a bound protein from an ion-exchange column, the polymer can be washed with a solution of varying pH or a solution of gradually increasing salt concentration. Both techniques for releasing the protein could denature it irreversibly, so the preferred technique is not always obvious.

The D-isomer ofglycine is thepredominant isomer in humanbeings

Questions 1-6) Alcohols, Acidity, and Nucleophilicity

Questions 8 -14) Fischer and Haworth Projections