Summary -
Proteins have four levels of structure. Primary structure describes a protein's amino acid sequence; secondary structure describes how segments of the protein chain orient into regular patterns-either $\boldsymbol{\alpha}$ helix or $\boldsymbol{\beta}$-pleated sheet; tertiary structure describes how the entire protein molecule coils into an overall three-dimensional shape; and quaternary structure describes how individual protein molecules aggregate into larger structures.

Proteins are classified as either globular or fibrous. Fibrous proteins such as $\alpha$-keratin are tough, rigid, and water-insoluble; globular proteins such as myoglobin are water-soluble and roughly spherical in shape. Many globular proteins are enzymes-substances that act as catalysts for biological reactions. Enzymes are grouped into six classes according to the kind of reaction they catalyze. In addition to their protein part, many enzymes contain cofactors, which can be either metal ions or small organic molecules called coenzymes.


Summary of Reactions -
1. Amino acid synthesis (Section 26.3)
a. From $\alpha$-bromo acids

b. Diethyl acetamidomalonate synthesis

c. Reductive amination of an $\alpha$-keto acid

d. Enantioselective synthesis

$A(Z)$ enamido acid
$\xrightarrow[\text { 2. } \mathrm{NaOH}, \mathrm{H}_{2} \mathrm{O}]{\text { 1. } \mathrm{H}_{2},[\mathrm{Rh}(\text { DiPAMP })(\mathrm{COD})]^{+} \mathrm{BF}_{4}^{-}}$

An (S)-amino acid
2. Peptide sequencing by Edman degradation (Section 26.6)

3. Peptide synthesis (Section 26.7)
a. Amine protection

Boc-protected amino acid
b. Carboxyl protection


Additional Problems -
Visualizing Chemistry
PROBLEM Identify the following amino acids:


26-19 (a) -
(b)

(c)

PROBLEM Give the sequence of the following tetrapeptide (yellow = $S$ ):
26-20

PROBLEM Isoleucine and threonine are the only two amino acids with two chirality centers. Assign $R$ or $S$
26-21 configuration to the methyl-bearing carbon atom of isoleucine.

PROBLEM Is the following structure a D amino acid or an L amino acid? Identify it.
26-22

PROBLEM Give the sequence of the following tetrapeptide:
26-23


Mechanism Problems -
PROBLEM The reaction of ninhydrin with an $\alpha$-amino acid occurs in several steps.
26-24 (a) The first step is formation of an imine by reaction of the amino acid with ninhydrin. Show its structure and the mechanism of its formation.
(b) The second step is a decarboxylation. Show the structure of the product and the mechanism of the decarboxylation reaction.
(c) The third step is hydrolysis of an imine to yield an amine and an aldehyde. Show the structures of both products and the mechanism of the hydrolysis reaction.
(d) The final step is formation of the purple anion. Show the mechanism of the reaction.

Ninhydrin
PROBLEM The chloromethylated polystyrene resin originally used for Merrifield solid-phase peptide
26-25 synthesis was prepared by treatment of polystyrene with chloromethyl methyl ether and a Lewis acid catalyst. Propose a mechanism for the reaction.

PROBLEM An Fmoc protecting group can be removed from an amino acid by treatment with the amine base 26-26 piperidine. Propose a mechanism.

Fmoc-protected amino acid

PROBLEM Proteins can be cleaved specifically at the amide bond on the carboxyl side of methionine residues 26-27 by reaction with cyanogen bromide, $\mathrm{BrC} \equiv \mathrm{N}$.


Mechanism Problems -
The reaction occurs in several steps:
(a) The first step is a nucleophilic substitution reaction of the sulfur on the methionine side chain with BrCN to give a cyanosulfonium ion, $\left[\mathrm{R}_{2} \mathrm{SCN}\right]^{+}$. Show the structure of the product, and propose a mechanism for the reaction.
(b) The second step is an internal $\mathrm{S}_{\mathrm{N}} 2$ reaction, with the carbonyl oxygen of the methionine residue displacing the positively charged sulfur leaving group and forming a five-membered ring product. Show the structure of the product and the mechanism of its formation.
(c) The third step is a hydrolysis reaction to split the peptide chain. The carboxyl group of the former methionine residue is now part of a lactone (cyclic ester) ring. Show the structure of the lactone product and the mechanism of its formation.
(d) The final step is a hydrolysis of the lactone to give the product shown. Show the mechanism of the reaction.

PROBLEM A clever recent method of peptide synthesis involves formation of an amide bond by reaction of an
26-28 $\alpha$-keto acid with an $N$-alkylhydroxylamine:

An $\boldsymbol{\alpha}$-keto acid $\quad$ A hydroxylamine $\quad$ An amide
The reaction is thought to occur by nucleophilic addition of the $N$-alkylhydroxylamine to the keto acid as if forming an oxime (Section 19.8), followed by decarboxylation and elimination of water. Show the mechanism.


Amino Acid Structures and Chirality -
PROBLEM Except for cysteine, only $S$ amino acids occur in proteins. Several $R$ amino acids are also found in 26-29 nature, however. ( $R$ )-Serine is found in earthworms, and ( $R$ )-alanine is found in insect larvae. Draw Fischer projections of $(R)$-serine and $(R)$-alanine. Are these D or L amino acids?

PROBLEM Cysteine is the only amino acid that has L stereochemistry but an $R$ configuration. Make up a 26-30 structure for another $L$ amino acid of your own creation that also has an $R$ configuration.

PROBLEM Draw a Fischer projection of ( $S$ ) -proline. 26-31

PROBLEM Show the structures of the following amino acids in their zwitterionic forms:
26-32 (a) $\operatorname{Trp}$
(b) Ile
(c) Cys
(d) His

PROBLEM Proline has $\mathrm{p} K_{\mathrm{a} 1}=1.99$ and $\mathrm{p} K_{\mathrm{a} 2}=10.60$. Use the Henderson-Hasselbalch equation to calculate the
26-33 ratio of protonated and neutral forms at $\mathrm{pH}=2.50$. Calculate the ratio of neutral and deprotonated forms at $\mathrm{pH}=9.70$.

PROBLEM Using both three- and one-letter codes for amino acids, write the structures of all possible peptides
26-34 that contain the following amino acids:
(a) Val, Ser, Leu
(b) $\mathrm{Ser}, \mathrm{Leu}_{2}$, Pro

PROBLEM Look at the side chains of the 20 amino acids in Table 26.1, and then think about what is not 26-35 present. None of the 20 contain either an aldehyde or a ketone carbonyl group, for instance. Is this just one of nature's oversights, or is there a likely chemical reason? What complications might an aldehyde or ketone carbonyl group cause?


Amino Acid Synthesis and Reactions -
PROBLEM Show how you could use the acetamidomalonate method to prepare the following amino acids:
26-36 (a) Leucine (b) Tryptophan
PROBLEM Show how you could prepare the following amino acids using a reductive amination:
26-37 (a) Methionine (b) Isoleucine
PROBLEM Show how you could prepare the following amino acids enantioselectively:
26-38 (a) Pro (b) Val
PROBLEM Serine can be synthesized by a simple variation of the amidomalonate method using formaldehyde 26-39 rather than an alkyl halide. How might this be done?

PROBLEM Predict the product of the reaction of valine with the following reagents:
26-40 (a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$, acid (b) Di-tert-butyl dicarbonate (c) $\mathrm{KOH}, \mathrm{H}_{2} \mathrm{O}$
(d) $\mathrm{CH}_{3} \mathrm{COCl}$, pyridine; then $\mathrm{H}_{2} \mathrm{O}$

PROBLEM Draw resonance forms for the purple anion obtained by reaction of ninhydrin with an $\alpha$-amino acid 26-41 (see Problem 26-24).


Peptides and Enzymes -
PROBLEM Write full structures for the following peptides:
26-42 (a) C-H-E-M (b) P-E-P-T-I-D-E

PROBLEM Propose two structures for a tripeptide that gives Leu, Ala, and Phe on hydrolysis but does not react 26-43 with phenyl isothiocyanate.

PROBLEM Show the steps involved in a synthesis of Phe-Ala-Val using the Merrifield procedure. 26-44

PROBLEM Draw the structure of the PTH derivative product you would obtain by Edman degradation of the 26-45 following peptides:
(a) I-L-P-F (b) D-T-S-G-A

PROBLEM Which amide bonds in the following polypeptide are cleaved by trypsin? By chymotrypsin?
26-46
Phe-Leu-Met-Lys-Tyr-Asp-Gly-Gly-Arg-Val-Ile-Pro-Tyr
PROBLEM What kinds of reactions do the following classes of enzymes catalyze?
26-47 (a) Hydrolases (b) Lyases (c) Transferases

PROBLEM Which of the following amino acids are more likely to be found on the exterior of a globular protein, 26-48 and which on the interior? Explain.
(a) Valine
(b) Aspartic acid
(c) Phenylalanine
(d) Lysine

PROBLEM Leuprolide is a synthetic nonapeptide used to treat both endometriosis in women and prostate 26-49 cancer in men.

(a) Both C-terminal and N -terminal amino acids in leuprolide have been structurally modified. Identify the modifications.
(b) One of the nine amino acids in leuprolide has D stereochemistry rather than the usual L. Which one?
(c) Write the structure of leuprolide using both one- and three-letter abbreviations.
(d) What charge would you expect leuprolide to have at neutral pH ?


General Problems -
PROBLEM The $\alpha$-helical parts of myoglobin and other proteins stop whenever a proline residue is encountered
26-50 in the chain. Why is proline never present in a protein $\alpha$ helix?
PROBLEM Arginine, the most basic of the 20 common amino acids, contains a guanidino functional group in
26-51 its side chain. Explain, using resonance structures to show how the protonated guanidino group is stabilized.

Arginine

PROBLEM Cytochrome $c$ is an enzyme found in the cells of all aerobic organisms. Elemental analysis of 26-52 cytochrome $c$ shows that it contains $0.43 \%$ iron. What is the minimum molecular weight of this enzyme?

PROBLEM Evidence for restricted rotation around amide $\mathrm{O}=\mathrm{C}-\mathrm{N}$ bonds comes from NMR studies. At room 26-53 temperature, the ${ }^{1} \mathrm{H}$ NMR spectrum of $\mathrm{N}, \mathrm{N}$-dimethylformamide shows three peaks: $2.9 \mathrm{\delta}$ (singlet, 3 H ), $3.0 \mathrm{\delta}$ (singlet, 3 H ), and $8.0 \delta$ (singlet, 1 H ). As the temperature is raised, however, the two singlets at $2.9 \delta$ and $3.0 \delta$ slowly merge. At $180^{\circ} \mathrm{C}$, the ${ }^{1} \mathrm{H}$ NMR spectrum shows only two peaks: $2.95 \delta$ (singlet, 6 H ) and $8.0 \delta$ (singlet, 1 H ). Explain this temperature-dependent behavior.


General Problems -
PROBLEM Propose a structure for an octapeptide that shows the composition Asp, Gly $y_{2}$, Leu, Phe, $\mathrm{Pro}_{2}$, Val 26-54 on amino acid analysis. Edman analysis shows a glycine N-terminal group, and leucine is the C-
terminal group. Acidic hydrolysis gives the following fragments:
Val-Pro-Leu, Gly, Gly-Asp-Phe-Pro, Phe-Pro-Val
PROBLEM Look at the structure of human insulin, and indicate where in each chain the molecule is cleaved by 26-55 trypsin and chymotrypsin.


Insulin -
PROBLEM What is the structure of a nonapeptide that gives the following fragments when cleaved?
26-56
Trypsin cleavage: Val-Val-Pro-Tyr-Leu-Arg, Ser-Ile-Arg
Chymotrypsin cleavage: Leu-Arg, Ser-Ile-Arg-Val-Val-Pro-Tyr

PROBLEM Oxytocin, a nonapeptide hormone secreted by the pituitary gland, functions by stimulating uterine
26-57 contraction and lactation during childbirth. Its sequence was determined from the following evidence:

1. Oxytocin is a cyclic compound containing a disulfide bridge between two cysteine residues.
2. When the disulfide bridge is reduced, oxytocin has the constitution Asn, Cys ${ }_{2}$, Gln, Gly, Ile, Leu, Pro, Tyr.
3. Partial hydrolysis of reduced oxytocin yields seven fragments: Asp-Cys, Ile-Glu, Cys-Tyr, Leu-Gly, Tyr-Ile-Glu, Glu-Asp-Cys, and Cys-Pro-Leu.
4. Gly is the C-terminal group.
5. Both Glu and Asp are present as their side-chain amides (Gln and Asn) rather than as free sidechain acids.

What is the amino acid sequence of reduced oxytocin? What is the structure of oxytocin itself?

PROBLEM Aspartame, a nonnutritive sweetener marketed under such trade names as Equal, NutraSweet, and 26-58 Canderel, is the methyl ester of a simple dipeptide, Asp-Phe-OCH3.
(a) Draw the structure of aspartame.
(b) The isoelectric point of aspartame is 5.9. Draw the principal structure present in aqueous solution at this pH .
(c) Draw the principal form of aspartame present at physiological $\mathrm{pH}=7.3$.


Insulin -
PROBLEM Refer to Figure 26.5 and propose a mechanism for the final step in Edman degradation-the acid-
26-59 catalyzed rearrangement of the ATZ derivative to the PTH derivative.
PROBLEM Amino acids are metabolized by a transamination reaction in which the $-\mathrm{NH}_{2}$ group of the amino
26-60 acid changes places with the keto group of an $\alpha$-keto acid. The products are a new amino acid and a new $\alpha$-keto acid. Show the product from transamination of isoleucine.

PROBLEM The first step in the biological degradation of histidine is formation of a 26-61 4-methylidene-5-imidazolone (MIO) by cyclization of a segment of the peptide chain in the histidine ammonia lyase enzyme. Propose a mechanism.

4-Methylidene-5-imidazolone (MIO)
PROBLEM The first step in the biological degradation of lysine is reductive amination with $\alpha$-ketoglutarate to
26-62 give saccharopine. Nicotinamide adenine dinucleotide phosphate (NADPH), a relative of NADH, is the reducing agent. Show the mechanism.

Lysine
$+$

$\alpha$-Ketoglutarate


CHAPTER 27 -
Biomolecules: Lipids

FIGURE 27.1 Soap bubbles, so common yet so beautiful, are made from animal fat, a lipid. (credit: "Reflections in soap bubbles" by Image Picture Photography/Flickr, CC BY 2.0)


CHAPTER CONTENTS - 27.1 Waxes, Fats, and Oils
27.2 Soap
27.3 Phospholipids
27.4 Prostaglandins and Other Eicosanoids
27.5 Terpenoids
27.6 Steroids
27.7 Biosynthesis of Steroids

WHY THIS CHAPTER? We've now covered two of the four major classes of biomolecules-proteins and carbohydrates-and have two remaining. In this chapter, we'll cover lipids, the largest and most diverse class of biomolecules, looking both at their structure and function and at their metabolism.

Lipids are naturally occurring organic molecules that have limited solubility in water and can be isolated from organisms by extraction with nonpolar organic solvents. Fats, oils, waxes, many vitamins and hormones, and most nonprotein cell-membrane components are some examples. Note that this definition differs from the sort used for carbohydrates and proteins in that lipids are defined by a physical property (solubility) rather than by structure. Of the many kinds of lipids, we'll be concerned in this chapter with only a few: triacylglycerols, eicosanoids, terpenoids, and steroids.

Lipids are classified into two broad types: those like fats and waxes, which contain ester linkages and can be hydrolyzed, and those like cholesterol and other steroids, which don't have ester linkages and can't be hydrolyzed.

Animal fat-a triester ( $\mathrm{R}, \mathrm{R}^{\prime}, \mathrm{R}^{\prime \prime}=\mathrm{C}_{11}-\mathrm{C}_{19}$ chains)

Cholesterol


CHAPTER CONTENTS - 27.1 Waxes, Fats, and Oils
Waxes are mixtures of esters of long-chain carboxylic acids with long-chain alcohols. The carboxylic acid usually has an even number of carbons from 16 to 36 , while the alcohol has an even number of carbons from 24 to 36 . One of the major components of beeswax, for instance, is triacontyl hexadecanoate, the ester of the $\mathrm{C}_{30}$ alcohol (1-triacontanol) and the $\mathrm{C}_{16}$ acid (hexadecanoic acid). The waxy protective coatings on most fruits, berries, leaves, and animal furs have similar structures.


Triacontyl hexadecanoate (from beeswax) -
Animal fats and vegetable oils are the most widely occurring lipids. Although they appear different-animal fats like butter and lard are solids, whereas vegetable oils like corn and peanut oil are liquid-their structures are closely related. Chemically, fats and oils are triglycerides, or triacylglycerols-triesters of glycerol with three long-chain carboxylic acids called fatty acids. Animals use fats for long-term energy storage because they are far less highly oxidized than carbohydrates and provide about six times as much energy as an equal weight of stored, hydrated glycogen.


A triacylglycerol -
Hydrolysis of a fat or oil with aqueous NaOH yields glycerol and three fatty acids. The fatty acids are generally unbranched and contain an even number of carbon atoms between 12 and 20 . If double bonds are present, they have mostly, although not entirely, $Z$, or cis, geometry. The three fatty acids of a specific triacylglycerol molecule need not be the same, and the fat or oil from a given source is likely to be a complex mixture of many different triacylglycerols. TABLE 27.1 lists some of the commonly occurring fatty acids, and TABLE 27.2 lists the approximate composition of fats and oils from different sources.

TABLE 27.1 Structures of Some Common Fatty Acids

| Name | No. of carbons | Melting point $\left({ }^{\circ} \mathrm{C}\right)$ | Structure |
| :--- | :---: | :--- | :--- |
| Saturated |  |  |  |
| Lauric | 12 | 43.2 | $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{10} \mathrm{CO}_{2} \mathrm{H}$ |
| Myristic | 14 | 53.9 | $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{12} \mathrm{CO}_{2} \mathrm{H}$ |
| Palmitic | 16 | 63.1 | $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{14} \mathrm{CO}_{2} \mathrm{H}$ |
| Stearic | 18 | 68.8 | $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{16} \mathrm{CO}_{2} \mathrm{H}$ |
| Arachidic | 20 | 76.5 | $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{18} \mathrm{CO}_{2} \mathrm{H}$ |

Unsaturated


A triacylglycerol -
Unsaturated

| Palmitoleic | 16 | -0.1 | $(Z)-\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{CH}=\mathrm{CH}\left(\mathrm{CH}_{2}\right)_{7} \mathrm{CO}_{2} \mathrm{H}$ |
| :--- | :---: | :---: | :--- |
| Oleic | 18 | 13.4 | $(Z)-\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{7} \mathrm{CH}=\mathrm{CH}\left(\mathrm{CH}_{2}\right)_{7} \mathrm{CO}_{2} \mathrm{H}$ |
| Linoleic | 18 | -12 | $(Z, Z)-\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{4}\left(\mathrm{CH}=\mathrm{CHCH}_{2}\right)_{2}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{CO}_{2} \mathrm{H}$ |
| Linolenic | 18 | -11 | $($ all $Z)-\mathrm{CH}_{3} \mathrm{CH}_{2}\left(\mathrm{CH}=\mathrm{CHCH}_{2}\right)_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{CO}_{2} \mathrm{H}$ |
| Arachidonic | 20 | -49.5 | $($ all $Z)-\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{4}\left(\mathrm{CH}=\mathrm{CHCH}_{2}\right)_{4} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$ |

TABLE 27.2 Composition of Some Fats and Oils


A triacylglycerol -
TABLE 27.2 Composition of Some Fats and Oils

|  | Saturated fatty acids (\%) |  |  | Unsaturated fatty acids (\%) |  |  |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| Source | $\mathrm{C}_{12}$ lauric | $\mathrm{C}_{14}$ myristic | $\mathrm{C}_{16}$ palmitic | $\mathrm{C}_{18}$ stearic | $\mathrm{C}_{18}$ oleic | $\mathrm{C}_{18}$ linoleic |

Animal fat

| Lard | - | 1 | 25 | 15 | 50 | 6 |
| :--- | ---: | ---: | ---: | ---: | ---: | ---: |
| Butter | 2 | 10 | 25 | 10 | 25 | 5 |
| Human fat | 1 | 3 | 25 | 8 | 46 | 10 |
| Whale blubber | - | 8 | 12 | 3 | 35 | 10 |

| Vegetable oil |  |  |  |  |  |  |
| :--- | ---: | ---: | ---: | ---: | ---: | ---: |
| Coconut | 50 | 18 | 8 | 2 | 6 | 1 |
| Corn | - | 1 | 10 | 4 | 35 | 45 |
| Olive | - | 1 | 5 | 5 | 80 | 7 |
| Peanut | - | - | 7 | 5 | 60 | 20 |


A triacylglycerol -
More than 100 different fatty acids are known and about 40 occur widely. Palmitic acid ( $\mathrm{C}_{16}$ ) and stearic acid $\left(\mathrm{C}_{18}\right)$ are the most abundant saturated fatty acids; oleic and linoleic acids (both $\mathrm{C}_{18}$ ) are the most abundant unsaturated ones. Oleic acid is monounsaturated because it has only one double bond, whereas linoleic, linolenic, and arachidonic acids are polyunsaturated fatty acids because they have more than one double bond. Linoleic and linolenic acids occur in cream and are essential in the human diet; infants grow poorly and develop skin lesions if fed a diet of nonfat milk for prolonged periods. Linolenic acid, in particular, is an example of an omega- $\mathbf{3}$ fatty acid, which has been found to lower blood triglyceride levels and reduce the risk of heart attack. The name omega-3 means that there is a double bond three carbons in from the noncarboxyl end of the chain.

Stearic acid

Linolenic acid, an omega-3 polyunsaturated fatty acid
The data in TABLE 27.1 show that unsaturated fatty acids generally have lower melting points than their saturated counterparts, a trend that is also true for triacylglycerols. Since vegetable oils generally have a higher proportion of unsaturated to saturated fatty acids than animal fats (TABLE 27.2), they have lower melting points. The difference is a consequence of structure. Saturated fats have a uniform shape that allows them to pack together efficiently in a crystal lattice. In unsaturated vegetable oils, however, the $\mathrm{C}=\mathrm{C}$ bonds introduce bends and kinks into the hydrocarbon chains, making crystal formation more difficult. The more double bonds there are, the harder it is for the molecules to crystallize and the lower the melting point of the oil.


A triacylglycerol -
The $\mathrm{C}=\mathrm{C}$ bonds in vegetable oils can be reduced by catalytic hydrogenation, typically carried out at high temperature using a nickel catalyst, to produce saturated solid or semisolid fats. Prior to 2015 , industrially produced margarine and shortening were produced by hydrogenating soybean, peanut, or cottonseed oil until the proper consistency is obtained. As of 2015, trans fats are no longer considered safe by the FDA and have since been phased out of industrially produced foods. Unfortunately, the hydrogenation reaction is accompanied by some cis-trans isomerization of the remaining double bonds, producing fats with about 10\% to $15 \%$ trans unsaturated fatty acids. Dietary intake of trans fatty acids increases cholesterol levels in the blood, thereby increasing the risk of heart problems. The conversion of linoleic acid into elaidic acid is an example.

PROBLEM Carnauba wax, used in floor and furniture polishes, contains an ester of a $\mathrm{C}_{32}$ straight-chain alcohol
27-1 with a $\mathrm{C}_{20}$ straight-chain carboxylic acid. Draw its structure.
PROBLEM Draw structures of glyceryl tripalmitate and glyceryl trioleate. Which would you expect to have a
27-2 higher melting point?


A triacylglycerol - 27.2 Soap
Soap has been in use for nearly 5000 years. As early as 2800 BC, the Babylonians boiled fats with ashes to create a soap-like substance. Ancient Egyptian medical papyri dating from 1550 BC reveals that Egyptians bathed regularly with soap made from a mixture of animal fats, vegetable oils, and alkaline salts. Chemically, soap is a mixture of the sodium or potassium salts of the long-chain fatty acids produced by hydrolysis (saponification) of animal fat with alkali. Wood ash was used as a source of alkali until the early 1800s, when the LeBlanc process for making $\mathrm{Na}_{2} \mathrm{CO}_{3}$ by heating sodium sulfate with limestone became available.


A fat ( $\mathrm{R}=\mathrm{C}_{11}-\mathrm{C}_{19}$ Aliphatic chains) -
Glycerol

Crude soap curds contain glycerol and excess alkali as well as soap but can be purified by boiling with water and adding NaCl or KCl to precipitate the pure carboxylate salts. The smooth soap that precipitates is dried, perfumed, and pressed into bars for household use. Dyes are added to make colored soaps, antiseptics are added for medicated soaps, pumice is added for scouring soaps, and air is blown in for soaps that float. Regardless of these extra treatments and regardless of price, though, all soaps are basically the same.

Soaps act as cleansers because the two ends of a soap molecule are so different. The carboxylate end of the longchain molecule is ionic and therefore hydrophilic (Section 2.12), or attracted to water. The long hydrocarbon portion of the molecule, however, is nonpolar and hydrophobic, avoiding water and therefore more soluble in oils. The net effect of these two opposing tendencies is that soaps are attracted to both oils and water and are therefore useful as cleansers.

When soaps are dispersed in water, the long hydrocarbon tails cluster together on the inside of a tangled, hydrophobic ball, while the ionic heads on the surface of the cluster protrude into the water layer. These spherical clusters, called micelles, are shown schematically in FIGURE 27.2. Grease and oil droplets are solubilized in water when they are coated by the nonpolar, hydrophobic tails of soap molecules in the center of micelles. Once solubilized, the grease and dirt can be rinsed away.

FIGURE 27.2 A soap micelle solubilizing a grease particle in water. An electrostatic potential map of a fatty acid carboxylate shows how the negative charge is located in the head group.


A fat ( $\mathrm{R}=\mathrm{C}_{11}-\mathrm{C}_{19}$ Aliphatic chains) -
FIGURE 27.2 A soap micelle solubilizing a grease particle in water. An electrostatic potential map of a fatty acid carboxylate shows how the negative charge is located in the head group.

As useful as they are, soaps also have drawbacks. In hard water, which contains metal ions, soluble sodium carboxylates are converted into insoluble magnesium and calcium salts, leaving the familiar ring of scum around bathtubs and a gray tinge on white clothes. Chemists have circumvented this problem by synthesizing a class of synthetic detergents based on salts of long-chain alkylbenzenesulfonic acids. The mechanism of synthetic detergents is the same as that of soaps: the alkylbenzene end of the molecule is attracted to grease, while the anionic sulfonate end is attracted to water. Unlike soaps, though, sulfonate detergents don't form insoluble metal salts in hard water and don't leave an unpleasant scum.

A synthetic detergent
( $R=$ various $C_{12}$ chains)
PROBLEM Draw the structure of magnesium oleate, a component of bathtub scum.
27-3
PROBLEM Write the saponification reaction of glyceryl dioleate monopalmitate with aqueous NaOH .
27-4


A fat ( $\mathrm{R}=\mathrm{C}_{11}-\mathrm{C}_{19}$ Aliphatic chains) - 27.3 Phospholipids
Just as waxes, fats, and oils are esters of carboxylic acids, phospholipids are esters of phosphoric acid, $\mathrm{H}_{3} \mathrm{PO}_{4}$.

A phosphoric acid monoester

A phosphoric acid diester

A phosphoric acid triester

Phospholipids are of two general types: glycerophospholipids and sphingomyelins. Glycerophospholipids are based on phosphatidic acid, which contains a glycerol backbone linked by ester bonds to two fatty acids and one phosphoric acid. Although the fatty-acid residues can be any of the $\mathrm{C}_{12}-\mathrm{C}_{20}$ units typically present in fats, the acyl group at C 1 is usually saturated and the one at C 2 is usually unsaturated. The phosphate group at C 3 is also
bonded to an amino alcohol such as choline $\left[\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{~N}\left(\mathrm{CH}_{3}\right)_{3}\right]^{+}$, ethanolamine $\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)$, or serine $\left[\mathrm{HOCH}_{2} \mathrm{CH}\left(\mathrm{NH}_{2}\right) \mathrm{CO}_{2} \mathrm{H}\right]$. The compounds are chiral and have an L , or $R$, configuration at C 2 .

Sphingomyelins are the second major group of phospholipids. These compounds have sphingosine or a related dihydroxyamine as their backbone and are particularly abundant in brain and nerve tissue, where they are a major constituent of the coating around nerve fibers.

Sphingosine

A sphingomyelin


A fat ( $\mathrm{R}=\mathrm{C}_{11}-\mathrm{C}_{19}$ Aliphatic chains) - 27.3 Phospholipids
Sphingosine

A sphingomyelin

Phospholipids are found widely in both plant and animal tissues and make up approximately $50 \%$ to $60 \%$ of cell membranes. Because they are like soaps in having a long, nonpolar hydrocarbon tail bound to a polar ionic head, phospholipids in the cell membrane organize into a lipid bilayer about $5.0 \mathrm{~nm}(50 \AA$ ) thick. As shown in FIGURE 27.3, the nonpolar tails aggregate in the center of the bilayer in much the same way that soap tails aggregate in the center of a micelle. This bilayer serves as an effective barrier to the passage of water, ions, and other components into and out of cells.

FIGURE 27.3 Aggregation of glycerophospholipids into the lipid bilayer that composes cell membranes.


A fat ( $\mathrm{R}=\mathrm{C}_{11}-\mathrm{C}_{19}$ Aliphatic chains) - 27.4 Prostaglandins and Other Eicosanoids
Prostaglandins are a group of $\mathrm{C}_{20}$ lipids that contain a five-membered ring with two long side chains. First isolated in the 1930s by Ulf von Euler at the Karolinska Institute in Sweden, much of the structural and chemical
work on prostaglandins was carried out by Sune Bergström and Bengt Samuelsson. All three shared Nobel Prizes for their work. The name prostaglandin derives from the fact that the compounds were first isolated from sheep prostate glands, but they have subsequently been found to be present in small amounts in all body tissues and fluids.

The several dozen known prostaglandins have an extraordinarily wide range of biological effects. Among their many properties, they can lower blood pressure, affect blood platelet aggregation during clotting, lower gastric secretions, control inflammation, affect kidney function, affect reproductive systems, and stimulate uterine contractions during childbirth.

Prostaglandins, together with related compounds called thromboxanes and leukotrienes, make up a class of compounds called eicosanoids because they are derived biologically from 5,8,11,14-eicosatetraenoic acid, or arachidonic acid (FIGURE 27.4). Prostaglandins (PG) have a cyclopentane ring with two long side chains; thromboxanes (TX) have a six-membered, oxygen-containing ring; and leukotrienes (LT) are acyclic.

Arachidonic acid

Prostaglandin $\mathrm{E}_{\mathbf{1}}\left(\mathrm{PGE}_{1}\right)$

Thromboxane $\mathbf{B}_{\mathbf{2}}\left(\right.$ TXB $\left._{2}\right)$

Prostaglandin $\mathrm{I}_{2}\left(\mathrm{PGI}_{2}\right)$ (prostacyclin)

Leukotriene $\mathrm{E}_{\mathbf{4}}\left(\mathrm{LTE}_{4}\right)$


A fat ( $\mathrm{R}=\mathrm{C}_{11}-\mathrm{C}_{19}$ Aliphatic chains) - 27.4 Prostaglandins and Other Eicosanoids
Leukotriene $\mathrm{E}_{\mathbf{4}}\left(\mathrm{LTE}_{4}\right)$

FIGURE 27.4 Structures of some representative eicosanoids. All are derived biologically from arachidonic acid.
Eicosanoids are named based on their ring system (PG, TX, or LT), substitution pattern, and number of double bonds. The various substitution patterns on the ring are indicated by letter as in FIGURE 27.5, and the number of double bonds is indicated by a subscript. Thus, $\mathrm{PGE}_{1}$ is a prostaglandin with the "E" substitution pattern and one double bond. The numbering of the atoms in the various eicosanoids is the same as in arachidonic acid, starting with the $-\mathrm{CO}_{2} \mathrm{H}$ carbon as C 1 , continuing around the ring, and ending with the $-\mathrm{CH}_{3}$ carbon at the other end of the chain as C20.

A prostaglandin (PG)

A thromboxane (TX)

A leukotriene (LT)

FIGURE 27.5 The naming system for eicosanoids.

Eicosanoid biosynthesis begins with the conversion of arachidonic acid to $\mathrm{PGH}_{2}$, catalyzed by the multifunctional PGH synthase (PGHS), also called cyclooxygenase (COX). There are two distinct enzymes, PGHS-1 and PGHS-2 (or COX-1 and COX-2), both of which accomplish the same reaction but appear to function independently. COX-1 carries out the normal physiological production of prostaglandins, and COX-2 produces additional prostaglandin in response to arthritis or other inflammatory conditions. Vioxx, Bextra, and several other drugs selectively inhibit the COX-2 enzyme but cause potentially serious heart and gastrointestinal problems in weakened patients. (See the Chapter 15 Chemistry Matters.)


A fat ( $\mathrm{R}=\mathrm{C}_{11}-\mathrm{C}_{19}$ Aliphatic chains) - 27.4 Prostaglandins and Other Eicosanoids
PGHS accomplishes two transformations, an initial reaction of arachidonic acid with $\mathrm{O}_{2}$ to yield $\mathrm{PGG}_{2}$ and a subsequent reduction of the hydroperoxide group $(-\mathrm{OOH})$ to the alcohol $\mathrm{PGH}_{2}$. The sequence of steps involved in these transformations was shown in FIGURE 8.12.

Further processing of $\mathrm{PGH}_{2}$ leads to other eicosanoids. $\mathrm{PGE}_{2}$, for instance, arises by an isomerization of $\mathrm{PGH}_{2}$ catalyzed by PGE synthase (PGES) (FIGURE 27.6).

FIGURE 27.6 Mechanism of the conversion of PGH $_{2}$ into PGE $_{2}$.
PROBLEM Assign $R$ or $S$ configuration to each chirality center in prostaglandin $\mathrm{E}_{2}$ (Figure 27.6), the most 27-5 abundant and biologically potent of mammalian prostaglandins.


A fat ( $\mathrm{R}=\mathrm{C}_{11}-\mathrm{C}_{19}$ Aliphatic chains) - 27.5 Terpenoids
We saw in the Chapter 8 Chemistry Matters that terpenoids are a vast and diverse group of lipids found in all living organisms. Despite their apparent structural differences, all terpenoids contain a multiple of five carbons and are derived biosynthetically from the five-carbon precursor isopentenyl diphosphate (FIGURE 27.7). Although formally a terpenoid contains oxygen, while a hydrocarbon is called a terpene, we'll use the term terpenoid to refer to both for simplicity.

Isopentenyl diphosphate

Patchouli alcohol (a sesquiterpenoid- $\mathrm{C}_{15}$ )

Lanosterol
(a triterpenoid- $\mathrm{C}_{30}$ )

Camphor (a monoterpenoid- $\mathrm{C}_{10}$ )

$\beta$-Carotene
(a tetraterpenoid- $\mathrm{C}_{40}$ )
FIGURE 27.7 Structures of some representative terpenoids.
You might recall from the chapter on Alkenes: Reactions and Synthesis that terpenoids are classified according to the number of five-carbon multiples they contain. Monoterpenoids contain 10 carbons and are derived from two isopentenyl diphosphates, sesquiterpenoids contain 15 carbons and are derived from three isopentenyl diphosphates, diterpenoids contain 20 carbons and are derived from four isopentenyl diphosphates, and so on, up to triterpenoids $\left(\mathrm{C}_{30}\right)$ and tetraterpenoids $\left(\mathrm{C}_{40}\right)$. Lanosterol, for instance, is a triterpenoid from which steroid hormones are made, and $\beta$-carotene is a tetraterpenoid that serves as a dietary source of vitamin A (FIGURE 27.7).


A fat ( $\mathrm{R}=\mathrm{C}_{11}-\mathrm{C}_{19}$ Aliphatic chains) - 27.5 Terpenoids
The terpenoid precursor isopentenyl diphosphate, formerly called isopentenyl pyrophosphate and thus abbreviated IPP, is biosynthesized by two different pathways, depending on the organism and the structure of the final product. In animals and higher plants, sesquiterpenoids and triterpenoids arise primarily from the mevalonate pathway, whereas monoterpenoids, diterpenoids, and tetraterpenoids are biosynthesized by the 1-deoxyxylulose 5-phosphate ( $D X P$ ) pathway, also called the methylerithritol phosphate, or MEP, pathway. In bacteria, both pathways are used. We'll look only at the mevalonate pathway, which is more common and better understood at present.

(R)-Mevalonate

1-Deoxy-D-xylulose 5-phosphate
The Mevalonate Pathway to Isopentenyl Diphosphate
As shown in FIGURE 27.8, the mevalonate pathway begins with the conversion of acetate to acetyl CoA, followed by Claisen condensation to yield acetoacetyl CoA. A second carbonyl condensation reaction with a third molecule of acetyl CoA, this one an aldol-like process, then yields the six-carbon compound 3 -hydroxy-3-methylglutaryl CoA, which is reduced to give mevalonate. Phosphorylation, followed by loss of $\mathrm{CO}_{2}$ and phosphate ion, completes the process.


Step 1 of FIGURE 27.8: Claisen Condensation -
The first step in mevalonate biosynthesis is a Claisen condensation to yield acetoacetyl CoA, a reaction catalyzed by acetoacetyl-CoA acetyltransferase. An acetyl group is first bound to the enzyme by a nucleophilic acyl substitution reaction with a cysteine -SH group. Formation of an enolate ion from a second molecule of acetyl CoA, followed by Claisen condensation, then yields the product.


Step 2 of FIGURE 27.8: Aldol Condensation -
Acetoacetyl CoA next undergoes an aldol-like addition of an acetyl CoA enolate ion in a reaction catalyzed by 3-hydroxy-3-methylglutaryl-CoA synthase. The reaction occurs by initial binding of the substrate to a cysteine -SH group in the enzyme, followed by enolate-ion addition and subsequent hydrolysis to give (3S)-3-hydroxy-3-methylglutaryl CoA (HMG-CoA).


FIGURE 27.8 MECHANISM -
The mevalonate pathway for the biosynthesis of isopentenyl diphosphate from three molecules of acetyl CoA. Individual steps are explained in the text.
(1) Claisen condensation of two molecules of acetyl CoA gives acetoacetyl CoA.

2 Aldol-like condensation of acetoacetyl CoA with a third molecule of acetyl CoA, followed by hydrolysis, gives
(3S)-3-hydroxy-3-methylglutaryl CoA.


Acetyl CoA -
Acetoacetyl CoA

(3S)-3-Hydroxy-3-methylglutaryl CoA
3 Reduction of the thioester group by 2 equivalents of NADPH gives (R)-mevalonate, a dihydroxy acid.

Phosphorylation of the tertiary hydroxyl and diphosphorylation of the primary hydroxyl, followed by decarboxylation and simultaneous expulsion of phosphate, gives isopentenyl diphosphate, the precursor of terpenoids.

Isopentenyl diphosphate

Step 3 of FIGURE 27.8: Reduction
Reduction of HMG-CoA to give ( $R$ )-mevalonate is catalyzed by 3-hydroxy-3-methylglutaryl-CoA reductase and requires 2 equivalents of reduced nicotinamide adenine dinucleotide phosphate (NADPH), a close relative of NADH (Section 19.12). The reaction occurs in two steps and proceeds through an aldehyde intermediate. The first step is a nucleophilic acyl substitution reaction involving hydride transfer from NADPH to the thioester carbonyl group of HMG-CoA. Following expulsion of HSCoA as leaving group, the aldehyde intermediate undergoes a second hydride addition to give mevalonate.

Step 4 of FIGURE 27.8: Phosphorylation and Decarboxylation
Three additional reactions are needed to convert mevalonate to isopentenyl diphosphate. The first two are straightforward phosphorylations by ATP that occur through nucleophilic substitution reactions on the terminal phosphorus. Mevalonate is first converted to mevalonate 5-phosphate (phosphomevalonate) by reaction with ATP; mevalonate 5-phosphate then reacts with a second ATP to give mevalonate 5-diphosphate (diphosphomevalonate). The third reaction results in phosphorylation of the tertiary hydroxyl group, followed by decarboxylation and loss of phosphate ion.


Acetyl CoA -
Mevalonate 5-diphosphate
Isopentenyl diphosphate
The final decarboxylation of mevalonate 5-diphosphate seems unusual because decarboxylations of acids do not typically occur except in $\beta$-keto acids and malonic acids, in which the carboxylate group is two atoms away from an additional carbonyl group. As discussed in Section 22.7, the function of this second carbonyl group is to act as an electron acceptor and stabilize the charge resulting from loss of $\mathrm{CO}_{2}$. In fact, though, the decarboxylation of a $\beta$-keto acid and the decarboxylation of mevalonate 5 -diphosphate are closely related.

Catalyzed by mevalonate-5-diphosphate decarboxylase, the substrate is first phosphorylated on the free - OH group by reaction with ATP to give a tertiary phosphate, which undergoes spontaneous $\mathrm{S}_{\mathrm{N}} 1$-like dissociation to give a tertiary carbocation. The positive charge then acts as an electron acceptor to facilitate decarboxylation in the same way a $\beta$ carbonyl group does, giving isopentenyl diphosphate. (In the following structures, the diphosphate group is abbreviated OPP.)

Mevalonate 5-diphosphate

PROBLEM The conversion of mevalonate 5-phosphate to isopentenyl diphosphate occurs with the following
27-6 result. Which hydrogen, pro- $R$ or pro- $S$, ends up cis to the methyl group, and which ends up trans?

Mevalonate 5-diphosphate
Isopentenyl diphosphate


Conversion of Isopentenyl Diphosphate to Terpenoids -
The conversion of isopentenyl diphosphate (IPP) to terpenoids begins with its isomerization to dimethylallyl diphosphate, abbreviated DMAPP and formerly called dimethylallyl pyrophosphate. These two $\mathrm{C}_{5}$ building blocks then combine to give the $\mathrm{C}_{10}$ unit geranyl diphosphate (GPP). The corresponding alcohol, geraniol, is itself a fragrant terpenoid that occurs in rose oil.

Further combination of GPP with another IPP gives the $\mathrm{C}_{15}$ unit farnesyl diphosphate (FPP), and so on, up to $\mathrm{C}_{25}$.

Terpenoids with more than 25 carbons-that is, triterpenoids $\left(\mathrm{C}_{30}\right)$ and tetraterpenoids ( $\mathrm{C}_{40}$ ) - are synthesized by dimerization of $\mathrm{C}_{15}$ and $\mathrm{C}_{20}$ units, respectively (FIGURE 27.9). Triterpenoids and steroids, in particular, arise from dimerization of farnesyl diphosphate to give squalene.

Geranyl diphosphate (GPP)

Farnesyl diphosphate (FPP)
$\downarrow$ Dimerization

Squalene
FIGURE 27.9 An overview of terpenoid biosynthesis from isopentenyl diphosphate.
The isomerization of isopentenyl diphosphate to dimethylallyl diphosphate is catalyzed by IPP isomerase and occurs through a carbocation pathway. Protonation of the IPP double bond by a hydrogen-bonded cysteine residue in the enzyme gives a tertiary carbocation intermediate, which is deprotonated by a glutamate residue as base to yield DMAPP. X-ray structural studies on the enzyme show that it holds the substrate in an unusually deep, well-protected pocket to shield the highly reactive carbocation from reaction with solvent or other external substances.


Conversion of Isopentenyl Diphosphate to Terpenoids -
Both the initial coupling of DMAPP with IPP to give geranyl diphosphate and the subsequent coupling of GPP with a second molecule of IPP to give farnesyl diphosphate are catalyzed by farnesyl diphosphate synthase. The process requires $\mathrm{Mg}^{2+}$ ion, and the key step is a nucleophilic substitution reaction in which the double bond of IPP behaves as a nucleophile in displacing diphosphate ion leaving group ( $\mathrm{PP}_{\mathrm{i}}$ ) or DMAPP. Evidence suggests that the DMAPP develops a considerable cationic character and that spontaneous dissociation of the
allylic diphosphate ion in an $\mathrm{S}_{\mathrm{N}} 1$-like pathway probably occurs (FIGURE 27.10).

Farnesyl diphosphate (FPP)
FIGURE 27.10 Mechanism of the coupling reaction of dimethylallyl diphosphate (DMAPP) and isopentenyl diphosphate (IPP), to give geranyl diphosphate (GPP).

Further conversion of geranyl diphosphate into monoterpenoids typically involves carbocation intermediates and multistep reaction pathways that are catalyzed by terpene cyclases. Monoterpene cyclases function by first isomerizing geranyl diphosphate to its allylic isomer linalyl diphosphate (LPP), a process that occurs by spontaneous $\mathrm{S}_{\mathrm{N}} 1$-like dissociation to an allylic carbocation, followed by recombination. The effect of this isomerization is to convert the C2-C3 double bond of GPP into a single bond, thereby making cyclization possible and allowing $E / Z$ isomerization of the double bond.

Further dissociation and cyclization by electrophilic addition of the cationic carbon to the terminal double bond then gives a cyclic cation, which might either rearrange, undergo a hydride shift, be captured by a nucleophile, or be deprotonated to give any of the several hundred known monoterpenoids. As just one example, limonene, a monoterpenoid found in many citrus oils, arises by the biosynthetic pathway shown in FIGURE 27.11.


Proposing a Terpenoid Biosynthesis Pathway -
Propose a mechanistic pathway for the biosynthesis of $\alpha$-terpineol from geranyl diphosphate.

Strategy
$\alpha$-Terpineol, a monoterpenoid, must be derived biologically from geranyl diphosphate through its isomer linalyl diphosphate. Draw the precursor in a conformation that approximates the structure of the target molecule, and then carry out a cationic cyclization, using the appropriate double bond to displace the diphosphate leaving group. Since the target is an alcohol, the carbocation resulting from cyclization evidently reacts with water.


Solution -
PROBLEM Propose mechanistic pathways for the biosynthetic formation of the following terpenoids:
27-7

$\alpha$-Pinene
(b)

$\gamma$-Bisabolene


Solution - 27.6 Steroids
In addition to fats, phospholipids, eicosanoids, and terpenoids, the lipid extracts of plants and animals also contain steroids, molecules that are derived from the triterpenoid lanosterol (FIGURE 27.7) and whose structures are based on a tetracyclic ring system. The four rings are designated $A, B, C$, and $D$, beginning at the lower left. The three 6 -membered rings ( $\mathrm{A}, \mathrm{B}$, and C ) adopt chair conformations but are prevented by their rigid geometry from undergoing the usual cyclohexane ring-flips (Section 4.6).

A steroid ( $\mathrm{R}=$ various side chains)

Two cyclohexane rings can be joined in either a cis or a trans manner. With cis fusion to give cis-decalin, both groups at the ring-junction positions (the angular groups) are on the same side of the two rings. With trans fusion to give trans-decalin, the groups at the ring junctions are on opposite sides.

As shown in FIGURE 27.12, steroids can have either a cis or a trans fusion of the $A$ and $B$ rings, but the other ring fusions (B-C and C-D) are usually trans.

An A-B trans steroid

An A-B cis steroid

FIGURE 27.12 Steroid conformations. The three 6-membered rings have chair conformations but are unable to undergo ring-flips. The $A$ and $B$ rings can be either cis-fused or trans-fused.

Substituent groups on the steroid ring system can be either axial or equatorial. As with simple cyclohexanes, equatorial substitution is generally more favorable than axial substitution for steric reasons. The hydroxyl group of cholesterol, for example, has the more stable equatorial orientation.

Cholesterol
PROBLEM Draw the following molecules in chair conformations, and tell whether the ring substituents are 27-8 axial or equatorial:
(a)

(b)


Solution - 27.6 Steroids
Cholesterol
PROBLEM Draw the following molecules in chair conformations, and tell whether the ring substituents are 27-8 axial or equatorial:
(a)

(b)

PROBLEM Lithocholic acid is an A-B cis steroid found in human bile. Draw lithocholic acid showing chair 27-9 conformations, as in Figure 27.12, and tell whether the hydroxyl group at C3 is axial or equatorial.
Lithocholic acid


Steroid Hormones -
In humans, most steroids function as hormones, chemical messengers that are secreted by endocrine glands
and carried through the bloodstream to target tissues. There are two main classes of steroid hormones: the sex hormones, which control maturation, tissue growth, and reproduction, and the adrenocortical hormones, which regulate a variety of metabolic processes.


Sex Hormones -
Testosterone and androsterone are the two most important male sex hormones, or androgens. Androgens are responsible for the development of male secondary sex characteristics during puberty and for promoting tissue and muscle growth. Both are synthesized in the testes from cholesterol. Androstenedione is another minor hormone that has received particular attention because of its use by prominent athletes.

Testosterone

Androsterone

Androstenedione
(Androgens)
Estrone and estradiol are the two most important female sex hormones, or estrogens. Synthesized in the ovaries from testosterone, estrogenic hormones are responsible for the development of female secondary sex characteristics and for regulation of the menstrual cycle. Note that both have a benzene-like aromatic A ring. In addition, another kind of sex hormone called a progestin is essential in preparing the uterus for implantation of a fertilized ovum during pregnancy. Progesterone is the most important progestin.

Estrone

Estradiol

Progesterone
(a progestin)
(Estrogens)


Adrenocortical Hormones -
Adrenocortical steroids are secreted by the adrenal glands, small organs located near the upper end of each kidney. There are two types of adrenocortical steroids, called mineralocorticoids and glucocorticoids. Mineralocorticoids, such as aldosterone, control tissue swelling by regulating cellular salt balance between $\mathrm{Na}^{+}$ and $\mathrm{K}^{+}$. Glucocorticoids, such as hydrocortisone, are involved in the regulation of glucose metabolism and in the control of inflammation. Glucocorticoid ointments are widely used to bring down the swelling from exposure to poison oak or poison ivy.

Aldosterone (a mineralocorticoid)

Hydrocortisone (a glucocorticoid)


Synthetic Steroids -
In addition to the hundreds of steroids isolated from plants and animals, many more have been synthesized in pharmaceutical laboratories in the search for new drugs. Among the best-known synthetic steroids are oral contraceptives and anabolic agents. Most birth control pills are a mixture of two compounds, a synthetic estrogen, such as ethynylestradiol, and a synthetic progestin, such as norethindrone. Anabolic steroids, such as methandrostenolone (Dianabol), are synthetic androgens that mimic the tissue-building effects of natural testosterone.

Ethynylestradiol (a synthetic estrogen)

Norethindrone
(a synthetic progestin)

Methandrostenolone (Dianabol)


Synthetic Steroids - 27.7 Biosynthesis of Steroids
Steroids are heavily modified triterpenoids that are biosynthesized in living organisms from farnesyl diphosphate $\left(\mathrm{C}_{15}\right)$. A reductive dimerization first converts farnesyl diphosphate to the acyclic hydrocarbon squalene ( $\mathrm{C}_{30}$ ), which is converted into lanosterol (FIGURE 27.13). Further rearrangements and degradations then take place to yield various steroids. The conversion of squalene to lanosterol is among the most intensively studied of biosynthetic transformations. Starting from an achiral, open-chain polyene, the entire process requires only two enzymes and results in the formation of six carbon-carbon bonds, four rings, and seven chirality centers.

2 Farnesyl diphosphate
Dimerization

Squalene

Lanosterol
FIGURE 27.13 An overview of steroid biosynthesis from farnesyl diphosphate.
Lanosterol biosynthesis begins with the selective epoxidation of squalene to give ( $3 S$ )-2,3-oxidosqualene, catalyzed by squalene epoxidase. Molecular $\mathrm{O}_{2}$ provides the epoxide oxygen atom, and NADPH is required,
along with a flavin coenzyme. The proposed mechanism involves reaction of $\mathrm{FADH}_{2}$ with $\mathrm{O}_{2}$ to produce a flavin hydroperoxide intermediate ( ROOH ), which transfers an oxygen to squalene in a pathway initiated by nucleophilic attack of the squalene double bond on the terminal hydroperoxide oxygen (FIGURE 27.14). The flavin alcohol formed as a by-product loses $\mathrm{H}_{2} \mathrm{O}$ to give FAD, which is reduced back to FADH 2 by NADPH. As noted in Section 8.7, this biological epoxidation mechanism is closely analogous to the mechanism by which peroxyacids $\left(\mathrm{RCO}_{3} \mathrm{H}\right)$ react with alkenes to give epoxides in the laboratory.


Synthetic Steroids - 27.7 Biosynthesis of Steroids
The second part of lanosterol biosynthesis is catalyzed by oxidosqualene-lanosterol cyclase and occurs as shown in FIGURE 27.15. Squalene is folded by the enzyme into a conformation that aligns the various double bonds for a cascade of successive intramolecular electrophilic additions, followed by a series of hydride and methyl migrations. Except for the initial epoxide protonation/cyclization, the process is probably stepwise and appears to involve discrete carbocation intermediates that are stabilized by electrostatic interactions with electron-rich aromatic amino acids in the enzyme.

Squalene

(3S)-2,3-0xidosqualene

FIGURE 27.14 Proposed mechanism of the oxidation of squalene by flavin hydroperoxide.

Steps 1, 2 of FIGURE 27.15: Epoxide Opening and Initial Cyclizations
Cyclization begins in step 1 with protonation of the epoxide ring by an aspartic acid residue in the enzyme. Nucleophilic opening of the protonated epoxide by the nearby 5,10 double bond (steroid numbering; Section 27.6) then yields a tertiary carbocation at C10. Further addition of C10 to the 8,9 double bond in step 2 next gives a bicyclic tertiary cation at C8.

(3S)-2,3-Oxidosqualene
Step 3 of FIGURE 27.15: Third Cyclization
The third cationic cyclization is somewhat unusual because it occurs with non-Markovnikov regiochemistry and gives a secondary cation at C13 rather than the alternative tertiary cation at C14. There is growing evidence, however, that the tertiary carbocation may in fact be formed initially and that the secondary cation arises by subsequent rearrangement. The secondary cation is probably stabilized in the enzyme pocket by the proximity of an electron-rich aromatic ring.

Mechanism of the conversion of 2,3-oxidosqualene to lanosterol. Four cationic cyclizations are followed by four rearrangements and a final loss of $\mathrm{H}^{+}$from C9. The steroid numbering system is used for referring to specific positions in the intermediates (Section 27.6). Individual steps are explained in the text.


Synthetic Steroids - 27.7 Biosynthesis of Steroids
Protonation on oxygen opens the epoxide ring and gives a tertiary carbocation at C4. Intramolecular electrophilic addition of C4 to the 5,10 double bond then yields a tertiary monocyclic carbocation at C10.

The C10 carbocation adds to the 8,9 double bond, giving a C8 tertiary bicyclic carbocation.

Further intramolecular addition of the C8 carbocation to the 13,14 double bond occurs with non-Markovnikov regiochemistry and gives a tricyclic secondary carbocation at C13.

4 The fourth and final cyclization occurs by addition of the C13 cation to the 17,20 double bond, giving the protosteryl cation with $17 \beta$ stereochemistry.

Hydride migration from C17 to C20 occurs, establishing $R$ stereochemistry at C20.

(3S)-2,3-Oxidosqualene

2 $\downarrow$

(3)

(4)

Protosteryl cation
(5)

FIGURE 27.16


Protosteryl cation -
©

©
from C13 to C17, establishing the final $17 \beta$ stereochemistry of the side chain.

Methyl migration from C14 to C13 occurs.
(0)

8 A second methyl migration occurs, from C8 to C14.
(8)

Loss of a proton from C9 forms an 8,9 double bond and gives lanosterol.
©

Step 4 of FIGURE 27.15: Final Cyclization
The fourth and last cyclization occurs in step 4 by addition of the cationic center at C13 to the 17,20 double bond, giving what is known as the protosteryl cation. The side-chain alkyl group at C17 has $\beta$ (up) stereochemistry, although this stereochemistry is lost in step 5 and then reset in step 6.

Steps 5-9 of FIGURE 27.15: Carbocation Rearrangements
Once the tetracyclic carbon skeleton of lanosterol has been formed, a series of carbocation rearrangements occur (Section 7.11). The first rearrangement, hydride migration from C 17 to C 20 , occurs in step 5 and results in establishment of $R$ stereochemistry at C20 in the side chain. In step 6, a second hydride migration occurs from C13 to C17 on the $\alpha$ (bottom) face of the ring and reestablishes the $17 \beta$ orientation of the side chain. Finally, two methyl migrations, the first from C 14 to C 13 on the top $(\beta)$ face and the second from C 8 to C 14 on the bottom $(\alpha)$ face, place the positive charge at C8. A basic histidine residue in the enzyme then removes the neighboring $\beta$ proton from C9 to give lanosterol.

From lanosterol, the pathway for steroid biosynthesis continues on to yield cholesterol. Cholesterol then becomes a branch point, serving as the common precursor from which all other steroids are derived.

PROBLEM Compare the structures of lanosterol and cholesterol, and catalog the changes needed for the 27-10 transformation shown.


Saturated Fats, Cholesterol, and Heart Disease -
We hear a lot these days about the relationship between saturated fats, cholesterol, and heart disease. What are the facts? It's well established that a diet rich in saturated animal fats often leads to an increase in blood serum cholesterol, particularly in sedentary, overweight individuals. Conversely, a diet lower in saturated fats and higher in polyunsaturated fats leads to a lower serum cholesterol level. Studies have shown that a serum cholesterol level greater than $240 \mathrm{mg} / \mathrm{dL}$ (a desirable value is $<200 \mathrm{mg} / \mathrm{dL}$ ) is correlated with an increased incidence of coronary artery disease, in which cholesterol deposits build up on the inner walls of coronary arteries, blocking the flow of blood to the heart muscles.

A better indication of a person's risk of heart disease comes from a measurement of blood lipoprotein levels. Lipoproteins are complex molecules with both lipid and protein components that transport lipids through the body. They can be divided into three types according to density, as shown in TABLE 27.3. Very-low-density lipoproteins (VLDLs) act primarily as carriers of triglycerides from the intestines to peripheral tissues, whereas low-density lipoproteins (LDLs) and high-density lipoproteins (HDLs) act as carriers of cholesterol to and from the liver.

FIGURE 27.17 It's hard to resist, but a high intake of saturated animal fat doesn't do much for your cholesterol level. (credit: modification of work "food-chicken-meat-outdoors" by www.pixel.la/Flickr, Public Domain)

Evidence suggests that LDLs transport cholesterol as its fatty-acid ester to peripheral tissues, whereas HDLs remove cholesterol as its stearate ester from dying cells. If LDLs deliver more cholesterol than is needed, and if insufficient HDLs are present to remove it, the excess is deposited in arteries. Thus, a low level of low-density lipoproteins is good because it means that less cholesterol is being transported, and a high level of high-density lipoproteins is good because it means that more cholesterol is being removed. In addition, HDL contains an enzyme that has antioxidant properties, offering further protection against heart disease.


Saturated Fats, Cholesterol, and Heart Disease -
As a rule of thumb, a person's risk drops about $25 \%$ for each increase of $5 \mathrm{mg} / \mathrm{dL}$ in HDL concentration. Normal values are $>40 \mathrm{mg} / \mathrm{dL}$ for men and $<50 \mathrm{mg} / \mathrm{dL}$ for women, perhaps explaining why premenopausal women appear to be somewhat less susceptible than men to heart disease.

TABLE 27.3 Serum Lipoproteins

| Name | Density (g/mL) | \% Lipid | \% Protein | Optimal (mg/dL) | Poor (mg/dL) |
| :---: | :---: | :---: | :---: | :---: | :---: |
| VLDL | $0.930-1.006$ | 90 | 10 | - | - |
| LDL | $1.019-1.063$ | 75 | 25 | $<100$ | $>130$ |
| HDL | $1.063-1.210$ | 67 | 33 | $>60$ | $<40$ |

Not surprisingly, the most important factor in gaining high HDL levels is a generally healthful lifestyle. Obesity, smoking, and lack of exercise lead to low HDL levels, whereas regular exercise and a sensible diet lead to high HDL levels. Distance runners and other endurance athletes have HDL levels nearly $50 \%$ higher than the general population. Failing that-some of us, but not everyone, wants to run 30 miles or bike 80 miles per week-diet is also important. Diets high in cold-water fish, like salmon and whitefish, raise HDL and lower blood cholesterol because these fish contain almost entirely polyunsaturated fat, including a large percentage of omega-3 fatty acids. Animal fat from red meat and cooking fats should be minimized because saturated fats and monounsaturated trans fats raise blood cholesterol.


Key Terms -
- anabolic steroid
- androgen
- eicosanoid
- estrogen
- fatty acid
- hormone
- lipid
- lipid bilayer
- lipoprotein
- micelle
- mineralocorticoid
- omega-3 fatty acid
- phospholipid
- polyunsaturated fatty acid
- progestin
- prostaglandin
- saponification
- steroid
- terpenoid
- triacylglycerol
- wax


Summary -
Lipids are the naturally occurring materials isolated from plants and animals by extraction with a nonpolar organic solvent. Animal fats and vegetable oils are the most widely occurring lipids. Both are triacylglycerols-triesters of glycerol with long-chain fatty acids. Animal fats are usually saturated, whereas vegetable oils usually have unsaturated fatty acid residues.

Phospholipids are important constituents of cell membranes and are of two kinds. Glycerophospholipids, such as phosphatidylcholine and phosphatidylethanolamine, are closely related to fats in that they have a glycerol backbone esterified to two fatty acids (one saturated and one unsaturated) and to one phosphate ester. Sphingomyelins have the amino alcohol sphingosine for their backbone.

Eicosanoids and terpenoids are still other classes of lipids. Eicosanoids, of which prostaglandins are the most abundant kind, are derived biosynthetically from arachidonic acid, are found in all body tissues, and have a wide range of physiological activity. Terpenoids are often isolated from the essential oils of plants, have an immense diversity of structure, and are produced biosynthetically from the five-carbon precursor isopentenyl diphosphate (IPP). Isopentenyl diphosphate is itself biosynthesized from 3 equivalents of acetate in the mevalonate pathway.

Steroids are plant and animal lipids with a characteristic tetracyclic carbon skeleton. Like the eicosanoids, steroids occur widely in body tissues and have a large variety of physiological activities. Steroids are closely related to terpenoids and arise biosynthetically from the triterpenoid lanosterol. Lanosterol, in turn, arises from cationic cyclization of the acyclic hydrocarbon squalene.


Visualizing Chemistry -
PROBLEM The following model is that of cholic acid, a constituent of human bile. Locate the three hydroxyl
27-11 groups, and identify each as axial or equatorial. Is cholic acid an A-B trans steroid or an A-B cis steroid?

PROBLEM Propose a biosynthetic pathway for the sesquiterpenoid helminthogermacrene from farnesyl 27-12 diphosphate.

PROBLEM Identify the following fatty acid, and tell whether it is more likely to be found in peanut oil or in red
27-13 meat:


Mechanism Problems -
PROBLEM Propose a mechanistic pathway for the biosynthesis of caryophyllene, a substance found in clove 27-14 oil.


Caryophyllene -
PROBLEM Suggest a mechanism by which $\psi$-ionone is transformed into $\beta$-ionone on treatment with acid. 27-15

PROBLEM Isoborneol is converted into camphene on treatment with dilute sulfuric acid. Propose a 27-16 mechanism for the reaction, which involves a carbocation rearrangement.

Isoborneol

Camphene

Fats, Oils, and Related Lipids
PROBLEM Fatty fish like salmon and albacore are rich in omega-3 fatty acids, which have a double bond three
27-17 carbons in from the noncarboxyl end of the chain and have been shown to lower blood cholesterol levels. Draw the structure of $5,8,11,14,17$-eicosapentaenoic acid, a common example. $($ Eicosane $=$ $\mathrm{C}_{20} \mathrm{H}_{42}$.)

PROBLEM Fats can be either optically active or optically inactive, depending on their structure. Draw the
27-18 structure of an optically active fat that yields 2 equivalents of stearic acid and 1 equivalent of oleic acid on hydrolysis. Draw the structure of an optically inactive fat that yields the same products.

PROBLEM Spermaceti, a fragrant substance from sperm whales, was widely used in cosmetics until it was
27-19 banned in 1976 to protect whales from extinction. Chemically, spermaceti is cetyl palmitate, the ester of cetyl alcohol ( $n-\mathrm{C}_{16} \mathrm{H}_{33} \mathrm{OH}$ ) with palmitic acid. Draw its structure.


Caryophyllene -
PROBLEM Show the products you would expect to obtain from reaction of glyceryl trioleate with the following
27-20 reagents:
(a) Excess $\mathrm{Br}_{2}$ in $\mathrm{CH}_{2} \mathrm{Cl}_{2}$
(b) $\mathrm{H}_{2} / \mathrm{Pd}$
(c) $\mathrm{NaOH} / \mathrm{H}_{2} \mathrm{O}$
(d) $\mathrm{O}_{3}$, then $\mathrm{Zn} / \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$
(e) $\mathrm{LiAlH}_{4}$, then $\mathrm{H}_{3} \mathrm{O}^{+}$
$\mathrm{CH}_{3} \mathrm{MgBr}$, then $\mathrm{H}_{3} \mathrm{O}^{+}$

PROBLEM How would you convert oleic acid into the following substances?
27-21 (a) Methyl oleate
(b) Methyl stearate (c) Nonanal
(d) Nonanedioic acid
(e) 9-Octadecynoic acid (stearolic acid) (f) 2-Bromostearic acid
(g) 18-Pentatriacontanone, $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{16} \mathrm{CO}\left(\mathrm{CH}_{2}\right)_{16} \mathrm{CH}_{3}$

PROBLEM Plasmalogens are a group of lipids found in nerve and muscle cells. How do plasmalogens differ 27-22 from fats?


A plasmalogen -
PROBLEM What products would you obtain from hydrolysis of a plasmalogen (Problem 27-22) with aqueous 27-23 NaOH ? With $\mathrm{H}_{3} \mathrm{O}^{+}$?

PROBLEM Cardiolipins are a group of lipids found in heart muscles. What products would be formed if all ester 27-24
bonds, including phosphates, were saponified by treatment with aqueous NaOH ?

A cardiolipin

PROBLEM Stearolic acid, $\mathrm{C}_{18} \mathrm{H}_{32} \mathrm{O}_{2}$, yields stearic acid on catalytic hydrogenation and undergoes oxidative
27-25 cleavage with ozone to yield nonanoic acid and nonanedioic acid. What is the structure of stearolic acid?

PROBLEM How would you synthesize stearolic acid (Problem 27-25) from 1-decyne and
27-26 1-chloro-7-iodoheptane?
Terpenoids and Steroids
PROBLEM Without proposing an entire biosynthetic pathway, draw the appropriate precursor, either geranyl
27-27 diphosphate or farnesyl diphosphate, in a conformation that shows a likeness to each of the following terpenoids:

Guaiol
(b)

Sabinene

PROBLEM Indicate by asterisks the chirality centers present in each of the terpenoids shown in Problem
27-28 27-27. What is the maximum possible number of stereoisomers for each?
PROBLEM Assume that the three terpenoids in Problem 27-27 are derived biosynthetically from isopentenyl
27-29 diphosphate and dimethylallyl diphosphate, each of which was isotopically labeled at the diphosphate-bearing carbon atom (C1). At what positions would the terpenoids be isotopically labeled?

PROBLEM Assume that acetyl CoA containing a ${ }^{14} \mathrm{C}$ isotopic label in the carboxyl carbon atom is used as
27-30 starting material for the biosynthesis of mevalonate, as shown in Figure 27.8. At what positions in mevalonate would the isotopic label appear?


A plasmalogen -
PROBLEM Assume that acetyl CoA containing a ${ }^{14} \mathrm{C}$ isotopic label in the carboxyl carbon atom is used as
27-31 starting material and that the mevalonate pathway is followed. Identify the positions in $\alpha$-cadinol where the label would appear.

PROBLEM Assume that acetyl CoA containing a ${ }^{14} \mathrm{C}$ isotopic label in the carboxyl carbon atom is used as
27-32 starting material and that the mevalonate pathway is followed. Identify the positions in squalene
where the label would appear.

Squalene
PROBLEM Assume that acetyl CoA containing a ${ }^{14} \mathrm{C}$ isotopic label in the carboxyl carbon atom is used as 27-33 starting material and that the mevalonate pathway is followed. Identify the positions in lanosterol where the label would appear.

Lanosterol


General Problems -
PROBLEM Flexibilene, a compound isolated from marine coral, is the first known terpenoid to contain a
27-34 15-membered ring. What is the structure of the acyclic biosynthetic precursor of flexibilene? Show the mechanistic pathway for the biosynthesis.

Flexibilene

PROBLEM Draw the most stable chair conformation of dihydrocarvone.
27-35

PROBLEM Draw the most stable chair conformation of menthol, and label each substituent as axial or
27-36 equatorial.

Menthol (from peppermint oil)

PROBLEM As a general rule, equatorial alcohols are esterified more readily than axial alcohols. What product
27-37 would you expect to obtain from reaction of the following two compounds with 1 equivalent of acetic anhydride?
(a)

(b)

PROBLEM Propose a mechanistic pathway for the biosynthesis of isoborneol. A carbocation rearrangement is
27-38 needed at one point in the scheme.


Isoborneol -
PROBLEM Digitoxigenin is a heart stimulant obtained from the purple foxglove Digitalis purpurea and used
27-39 in the treatment of heart disease. Draw the three-dimensional conformation of digitoxigenin, and identify the two -OH groups as axial or equatorial.

PROBLEM What product would you obtain by reduction of digitoxigenin (Problem 27-39) with $\mathrm{LiAlH}_{4}$ ? By
27-40 oxidation with the Dess-Martin periodinane?
PROBLEM Vaccenic acid, $\mathrm{C}_{18} \mathrm{H}_{34} \mathrm{O}_{2}$, is a rare fatty acid that gives heptanal and 11-oxoundecanoic acid
27-41 $\left[\mathrm{OHC}\left(\mathrm{CH}_{2}\right)_{9} \mathrm{CO}_{2} \mathrm{H}\right]$ on ozonolysis followed by zinc treatment. When allowed to react with $\mathrm{CH}_{2} \mathrm{I}_{2} / \mathrm{Zn}(\mathrm{Cu})$, vaccenic acid is converted into lactobacillic acid. What are the structures of vaccenic and lactobacillic acids?

PROBLEM Eleostearic acid, $\mathrm{C}_{18} \mathrm{H}_{30} \mathrm{O}_{2}$, is a rare fatty acid found in the tung oil used for finishing furniture. On
27-42 ozonolysis followed by treatment with zinc, eleostearic acid furnishes one part pentanal, two parts glyoxal ( $\mathrm{OHC}-\mathrm{CHO}$ ), and one part 9 -oxononanoic acid $\left[\mathrm{OHC}\left(\mathrm{CH}_{2}\right)_{7} \mathrm{CO}_{2} \mathrm{H}\right]$. What is the structure of eleostearic acid?

PROBLEM Diterpenoids are derived biosynthetically from geranylgeranyl diphosphate (GGPP), which is itself
27-43 biosynthesized by reaction of farnesyl diphosphate with isopentenyl diphosphate. Show the structure of GGPP, and propose a mechanism for its biosynthesis from FPP and IPP.


Isoborneol -
PROBLEM Diethylstilbestrol (DES) has estrogenic activity even though it is structurally unrelated to steroids.
27-44 Once used as an additive in animal feed, DES has been implicated as a causative agent in several types of cancer. Show how DES can be drawn so that it is sterically similar to estradiol.

PROBLEM Cembrene, $\mathrm{C}_{20} \mathrm{H}_{32}$, a diterpenoid hydrocarbon isolated from pine resin, has a UV absorption at 245 27-45 nm, but dihydrocembrene $\left(\mathrm{C}_{20} \mathrm{H}_{34}\right)$, the product of hydrogenation with 1 equivalent of $\mathrm{H}_{2}$, has no UV absorption. On exhaustive hydrogenation, 4 equivalents of $\mathrm{H}_{2}$ react, and octahydrocembrene, $\mathrm{C}_{20} \mathrm{H}_{40}$, is produced. On ozonolysis of cembrene, followed by treatment of the ozonide with zinc, four carbonyl-containing products are obtained:

Propose a structure for cembrene that is consistent with its formation from geranylgeranyl diphosphate.

PROBLEM $\alpha$-Fenchone is a pleasant-smelling terpenoid isolated from oil of lavender. Propose a pathway for
27-46 the formation of $\alpha$-fenchone from geranyl diphosphate. A carbocation rearrangement is required.

PROBLEM Fatty acids are synthesized by a multistep route that starts with acetate. The first step is a reaction
27-47 between protein-bound acetyl and malonyl units to give a protein-bound 3 -ketobutyryl unit. Show the mechanism, and tell what kind of reaction is occurring.

3-Ketobutyryl-protein
PROBLEM Propose a mechanism for the biosynthesis of the sesquiterpenoid trichodiene from farnesyl 27-48 diphosphate. The process involves cyclization to give an intermediate secondary carbocation, followed by several carbocation rearrangements.

Farnesyl diphosphate (FPP)

Trichodiene


CHAPTER 28 -
Biomolecules: Nucleic Acids

FIGURE 28.1 If these Afghan hounds look similar, that's because they're identical-all cloned from somatic cells of the same donor. (credit: modification of work "Figure 1" in "Birth of clones of the world's first cloned dog" by Kim, M.J., Oh, H.J., Kim, G.A. et al./Sci Rep 7, 15235 (2017), CC BY 4.0)


CHAPTER CONTENTS -
28.1 Nucleotides and Nucleic Acids
28.2 Base Pairing in DNA
28.3 Replication of DNA
28.4 Transcription of DNA
28.5 Translation of RNA: Protein Biosynthesis
28.6 DNA Sequencing
28.7 DNA Synthesis
28.8 The Polymerase Chain Reaction

WHY THIS CHAPTER? Nucleic acids are the last of the four major classes of biomolecules we'll consider. So much has been written and spoken about DNA in the media that the basics of DNA replication and transcription are probably known to you. Thus, we'll move fairly quickly through the fundamentals and then look more closely at the chemical details of DNA sequencing, synthesis, and metabolism. This field is moving rapidly, and there's a lot you may not be familiar with.

The nucleic acids, deoxyribonucleic acid (DNA) and ribonucleic acid (RNA), are the chemical carriers of a cell's genetic information. Coded in a cell's DNA is the information that determines the nature of the cell, controls its growth and division, and directs biosynthesis of the enzymes and other proteins required for cellular functions.

In addition to nucleic acids themselves, nucleic acid derivatives such as ATP are involved as phosphorylating agents in many biochemical pathways, and several important coenzymes, including NAD ${ }^{+}$, FAD, and coenzyme A, have nucleic acid components. See TABLE 26.3 for their structures.


CHAPTER CONTENTS - 28.1 Nucleotides and Nucleic Acids
Just as proteins are biopolymers made of amino acids, the nucleic acids are biopolymers made of nucleotides, joined together to form a long chain. Each nucleotide is composed of a nucleoside bonded to a phosphate group, and each nucleoside is composed of an aldopentose sugar linked through its anomeric carbon to the nitrogen atom of a heterocyclic purine or pyrimidine base.

The sugar component in RNA is ribose, and the sugar in DNA is $2^{\prime}$-deoxyribose. (In naming and numbering nucleotides, numbers with a prime superscript refer to positions on the sugar and numbers without a prime superscript refer to positions on the heterocyclic base. Thus, the prefix $2^{\prime}$-deoxy indicates that oxygen is missing from C2' of ribose.) DNA contains four different amine bases: two substituted purines (A, adenine, and G, guanine) and two substituted pyrimidines ( $\mathbf{C}$, cytosine, and $\mathbf{T}$, thymine). Adenine, guanine, and cytosine also occur in RNA, but thymine is replaced in RNA by a closely related pyrimidine base called $\mathbf{U}$, uracil.

Ribose

2-Deoxyribose

Purine

Thymine (T) DNA

Uracil (U) RNA

The structures of the four deoxyribonucleotides and the four ribonucleotides are shown in FIGURE 28.2. Although similar chemically, DNA and RNA differ dramatically in size. Molecules of DNA are enormous, containing as many as 245 million nucleotides and having molecular weights as high as 75 billion. Molecules of RNA, by contrast, are much smaller, containing as few as 21 nucleotides and having molecular weights as low as 7000 .

2'-Deoxyadenosine 5'-phosphate
Deoxyribonucleotides

C

2'-Deoxycytidine 5'-phosphate

2'-Deoxyguanosine 5'-phosphate

T

Thymidine 5'-phosphate

Adenosine 5'-phosphate
Ribonucleotides

Cytidine 5'-phosphate


CHAPTER CONTENTS - 28.1 Nucleotides and Nucleic Acids
2'-Deoxyguanosine 5'-phosphate

T

Thymidine 5'-phosphate

Adenosine 5'-phosphate
Ribonucleotides

Cytidine 5'-phosphate

Guanosine 5'-phosphate

U

Uridine $5^{\prime}$-phosphate

FIGURE 28.2 Structures of the four deoxyribonucleotides and the four ribonucleotides.
Nucleotides are linked together in DNA and RNA by phosphodiester bonds [ $\mathrm{RO}-\left(\mathrm{PO}_{2}{ }^{-}\right)-\mathrm{OR}$ '] between phosphate, the $5^{\prime}$-hydroxyl group on one nucleoside, and the $3^{\prime}$-hydroxyl group on another nucleoside. One end of the nucleic acid polymer has a free hydroxyl at C3' (the $\mathbf{3}^{\prime}$ end), and the other end has a phosphate at $\mathbf{C 5}^{\prime}$ (the $\mathbf{5}^{\prime}$ end). The sequence of nucleotides in a chain is described by starting at the $5^{\prime}$ end and identifying the bases in order of occurrence, using the abbreviations G, C, A, T (or U in RNA). Thus, a typical DNA sequence might be written as TAGGCT.

PROBLEM Draw the full structure of the DNA dinucleotide AG. 28-1

PROBLEM Draw the full structure of the RNA dinucleotide UA.
28-2


CHAPTER CONTENTS - 28.2 Base Pairing in DNA
Samples of DNA isolated from different tissues of the same species have the same proportions of heterocyclic bases, but samples from different species often have greatly differing proportions of bases. Human DNA, for example, contains about $30 \%$ each of adenine and thymine and about $20 \%$ each of guanine and cytosine. The bacterium Clostridium perfringens, however, contains about $37 \%$ each of adenine and thymine and only $13 \%$ each of guanine and cytosine. Note that in both examples the bases occur in pairs. Adenine and thymine are present in equal amounts, as are cytosine and guanine. Why?

In 1953, Rosalind Franklin, Maurice Wilkins, James Watson, and Francis Crick published scientific reports describing the secondary structure of DNA. According to their model, DNA under physiological conditions consists of two polynucleotide strands, running in opposite directions and coiled around each other in a double helix like the handrails on a spiral staircase. The two strands are complementary rather than identical and are held together by hydrogen bonds between specific pairs of bases, A with $T$ and $C$ with $G$. That is, whenever an A base occurs in one strand, a T base occurs opposite it in the other strand; when a C base occurs in one, a G occurs in the other (FIGURE 28.3). This complementary base-pairing thus explains why A and $T$ are always found in equal amounts, as are $G$ and $C$.

A
T

G

FIGURE 28.3 Hydrogen-bonding between base pairs in the DNA double helix. Electrostatic potential maps show that the faces of the bases are relatively neutral, while the edges have positive and negative regions. Pairing $G$ with $C$ and $A$ with $T$ brings together oppositely charged regions.


CHAPTER CONTENTS - 28.2 Base Pairing in DNA
A full turn of a DNA double helix is shown in FIGURE 28.4. The helix is $20 \AA$ wide, there are 10 base pairs per turn, and each turn is 34 Å in length. Notice in FIGURE 28.4 that the two strands of the double helix coil in such a way that two kinds of "grooves" result, a major groove $12 \AA$ wide and a minor groove $6 \AA$ wide. The major groove is slightly deeper than the minor groove, and both are lined by flat heterocyclic bases. As a result, a variety of other polycyclic aromatic molecules are able to slip sideways, or intercalate, between the stacked bases. Many cancer-causing and cancer-preventing agents function by interacting with DNA in this way.

FIGURE 28.4 A turn of the DNA double helix in both space-filling and wire-frame formats. The sugar-phosphate backbone runs along the outside of the helix, and the amine bases hydrogen bond to one another on the inside. Both major and minor grooves are visible.

An organism's genetic information is stored as a sequence of deoxyribonucleotides strung together in the DNA chain. For the information to be preserved and passed on to future generations, a mechanism must exist for
copying DNA. For the information to be used, a mechanism must exist for decoding the DNA message and implementing the instructions it contains.

What Crick called the "central dogma of molecular genetics" says that the function of DNA is to store information and pass it on to RNA. The function of RNA, in turn, is to read, decode, and use the information received from DNA to make proteins. This view is greatly oversimplified but is nevertheless a good place to start. Three fundamental processes take place.

- Replication is the process by which identical copies of DNA are made so that information can be preserved and handed down to offspring.
- Transcription is the process by which genetic messages are read and carried out of the cell nucleus to ribosomes, where protein synthesis occurs.
- Translation is the process by which the genetic messages are decoded and used to synthesize proteins.


Predicting the Complementary Base Sequence in Double-Stranded DNA -
What sequence of bases on one strand of DNA is complementary to the sequence TATGCAT on another strand?


Strategy -
Remember that A and G form complementary pairs with T and C , respectively, and then go through the sequence replacing A by T, G by C, T by A, and C by G. Remember also that the $5^{\prime}$ end is on the left and the $3^{\prime}$ end is on the right in the original strand.


Solution -
Original: (5') TATGCAT (3')
Complement: (3') ATACGTA (5') or (5') ATGCATA (3')

PROBLEM What sequence of bases on one strand of DNA is complementary to the following sequence on 28-3 another strand?
(5') GGCTAATCCGT (3')


Solution - 28.3 Replication of DNA
DNA replication is an enzyme-catalyzed process that begins with a partial unwinding of the double helix at various points along the chain, brought about by enzymes called helicases. Hydrogen bonds are broken, the two strands separate to form a "bubble," and bases are exposed. New nucleotides then line up on each strand in a complementary manner, A to T and G to C , and two new strands begin to grow from the ends of the bubble, called the replication forks. Each new strand is complementary to its old template strand, so two identical DNA double helices are produced (FIGURE 28.5). Because each of the new DNA molecules contains one old strand and one new strand, the process is described as semiconservative replication.

FIGURE 28.5 A representation of semiconservative DNA replication. The original double-stranded DNA partially unwinds, bases are exposed, nucleotides line up on each strand in a complementary manner, and two new strands begin to grow. Both strands are synthesized in the same $5^{\prime} \rightarrow 3^{\prime}$ direction, one continuously and one in fragments.

Addition of nucleotides to the growing chain takes place in the $5^{\prime} \rightarrow 3^{\prime}$ direction and is catalyzed by the DNA polymerase enzyme. The key step is the addition of a nucleoside $5^{\prime}$-triphosphate to the free $3^{\prime}$-hydroxyl group of the growing chain with the loss of a diphosphate leaving group.


Solution - 28.3 Replication of DNA
Because both new DNA strands are synthesized in the $5^{\prime} \rightarrow 3^{\prime}$ direction, they can't be made in exactly the same way. One new strand must have its $3^{\prime}$ end nearer a replication fork, while the other new strand has its $5^{\prime}$ end nearer the replication fork. What happens is that the complement of the original $5^{\prime} \rightarrow 3^{\prime}$ strand is synthesized continuously in a single piece to give a newly synthesized copy called the leading strand, while the complement of the original $3^{\prime} \rightarrow 5^{\prime}$ strand is synthesized discontinuously in small pieces called Okazaki fragments (named for Tsuneko Okazaki, in recognition of her discovery of them) that are subsequently linked by DNA ligases to form the lagging strand.

The magnitude of the replication process is staggering. The nucleus of every human cell contains 2 copies of 22 chromosomes plus an additional 2 sex chromosomes, for a total of 46 . Each chromosome consists of one very large DNA molecule, and the sum of the DNA in each of the two sets of chromosomes is estimated to be 3.0 billion base pairs, or 6.0 billion nucleotides. Despite the size of these enormous molecules, their base sequence
is faithfully copied during replication. The entire copying process takes only a few hours and, after proofreading and repair, an error gets through only about once per 10 to 100 billion bases. In fact, only about 60 of these random mutations are passed on from parent to child per human generation.


Solution - 28.4 Transcription of DNA
As noted previously, RNA is structurally similar to DNA but contains ribose rather than deoxyribose and uracil rather than thymine. RNA has three major types, each of which serves a specific purpose. In addition, there are a number of small RNAs that appear to control a wide variety of important cellular functions. All RNA molecules are much smaller than DNA, and all remain single-stranded rather than double-stranded.

- Messenger RNA (mRNA) carries genetic messages from DNA to ribosomes, small granular particles in the cytoplasm of a cell where protein synthesis takes place.
- Ribosomal RNA (rRNA) complexed with protein provides the physical makeup of the ribosomes.
- Transfer RNA (tRNA) transports amino acids to the ribosomes, where they are joined together to make proteins.
- Small RNAs, also called functional RNAs, have a variety of functions within the cell, including silencing transcription and catalyzing chemical modifications of other RNA molecules.

The genetic information in DNA is contained in segments called genes, each of which consists of a specific nucleotide sequence that encodes a specific protein. The conversion of that information from DNA into proteins begins in the nucleus of cells with the synthesis of mRNA by transcription of DNA. In bacteria, the process begins when RNA polymerase recognizes and binds to a promoter sequence on DNA, typically consisting of around 40 base pairs located upstream ( 5 ') of the transcription start site. Within the promoter are two hexameric consensus sequences, one located 10 base pairs upstream of the start and the second located 35 base pairs upstream.

Following formation of the polymerase-promoter complex, several turns of the DNA double helix untwist, forming a bubble and exposing 14 or so base pairs of the two strands. Appropriate ribonucleotides then line up by hydrogen-bonding to their complementary bases on DNA, bond formation occurs in the $5^{\prime} \rightarrow 3^{\prime}$ direction, the RNA polymerase moves along the DNA chain, and the growing RNA molecule unwinds from DNA (FIGURE 28.6). At any one time, about 12 base pairs of the growing RNA remain hydrogen-bonded to the DNA template.


Solution - 28.4 Transcription of DNA
FIGURE 28.6 Biosynthesis of RNA using a DNA segment as a template.
Unlike what happens in DNA replication, where both strands are copied, only one of the two DNA strands is transcribed into mRNA. The DNA strand that contains the gene is often called the sense strand, or coding strand, and the DNA strand that gets transcribed to give RNA is called the antisense strand, or noncoding strand. Because the sense strand and the antisense strand in DNA are complementary, and because the DNA antisense strand and the newly formed RNA strand are also complementary, the RNA molecule produced during transcription is a copy of the DNA sense strand. That is, the complement of the complement is the same as the original. The only difference is that the RNA molecule has a U everywhere that the DNA sense strand has a T.

Another part of the picture in vertebrates and flowering plants is that genes are often not continuous segments of the DNA chain. Instead, a gene will begin in one small section of DNA called an exon, then be interrupted
by a noncoding section called an intron, and then take up again farther down the chain in another exon. The final mRNA molecule results only after the noncoded sections are cut out of the transcribed mRNA and the remaining pieces are joined together by spliceosome enzymes. The gene for triose phosphate isomerase in maize, for instance, contains eight noncoding introns accounting for approximately $70 \%$ of the DNA base pairs and nine coding exons accounting for only $30 \%$ of the base pairs.

PROBLEM Show how uracil can form strong hydrogen bonds to adenine.
28-4
PROBLEM What RNA base sequence is complementary to the following DNA base sequence?
28-5
(5') GATTACCGTA (3')

PROBLEM From what DNA base sequence was the following RNA sequence transcribed?
28-6
(5') UUCGCAGAGU (3')


Solution - 28.5 Translation of RNA: Protein Biosynthesis
The primary cellular function of mRNA is to direct the biosynthesis of the thousands of diverse peptides and proteins required by an organism-as many as 150,000 in a human. The mechanics of protein biosynthesis take place on ribosomes, small granular particles in the cytoplasm of a cell that consist of about $60 \%$ ribosomal RNA and $40 \%$ protein.

The specific ribonucleotide sequence in mRNA forms a message that determines the order in which amino acid residues are to be joined. Each "word," or codon, along the mRNA chain consists of a sequence of three ribonucleotides that is specific for a given amino acid. For example, the series UUC on mRNA is a codon directing incorporation of the amino acid phenylalanine into the growing protein. Of the $4^{3}=64$ possible triplets of the four bases in RNA, 61 code for specific amino acids and 3 code for chain termination. TABLE 28.1 shows the meaning of each codon.

|  |  | Third base ( 3 ' end) |  |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: |
| First base (5' end) | Second base | U | C | A | G |
| U | U | Phe | Phe | Leu | Leu |
|  | C | Ser | Ser | Ser | Ser |
|  | A | Tyr | Tyr | Stop | Stop |
|  | G | Cys | Cys | Stop | Trp |
| C | U | Leu | Leu | Leu | Leu |
|  | C | Pro | Pro | Pro | Pro |
|  | A | His | His | Gln | Gln |
|  | G | Arg | Arg | Arg | Arg |
| A | U | Ile | Ile | Ile | Met |
|  | C | Thr | Thr | Thr | Thr |


Solution - 28.5 Translation of RNA: Protein Biosynthesis
| TABLE 28.1 Codon Assignments of Base Triplets |  |  |  |  |  |  |  |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Third base (3' end) |  |  |  |  |  |  |  |
| First base (5' end) | Second base | U | C | A | G |  |  |
|  | A | Asn | Asn | Lys | Lys |  |  |
|  | G | Ser | Ser | Arg | Arg |  |  |
| G | U | Val | Val | Val | Val |  |  |
|  | C | Ala | Ala | Ala | Ala |  |  |
|  | A | Asp | Asp | Glu | Glu |  |  |


Solution - 28.5 Translation of RNA: Protein Biosynthesis
The message embedded in mRNA is read by transfer RNA (tRNA) in a process called translation. There are 61 different tRNAs, one for each of the 61 codons that specify an amino acid. A typical tRNA is single-stranded and has roughly the shape of a cloverleaf, as shown in FIGURE 28.7. It consists of about 70 to 100 ribonucleotides and is bonded to a specific amino acid by an ester linkage through the $3^{\prime}$ hydroxyl on ribose at the $3^{\prime}$ end of the tRNA. Each tRNA also contains on its middle leaf a segment called an anticodon, a sequence of three ribonucleotides complementary to the codon sequence. For example, the codon sequence UUC present on mRNA is read by a phenylalanine-bearing tRNA having the complementary anticodon base sequence GAA. [Remember that nucleotide sequences are written in the $5^{\prime} \rightarrow 3^{\prime}$ direction, so the sequence in an anticodon must be reversed. That is, the complement to $\left(5^{\prime}\right)$-UUC- $\left(3^{\prime}\right)$ is ( $\left.3^{\prime}\right)$-AAG- $\left(5^{\prime}\right)$, which is written as $\left(5^{\prime}\right)$-GAA- $\left(3^{\prime}\right)$.]

FIGURE 28.7 Structure of a tRNA molecule. The tRNA is a roughly cloverleaf-shaped molecule containing an anticodon triplet on one "leaf" and an amino acid unit attached covalently at its $\mathbf{3 '}^{\prime}$ end. The example shown is a yeast tRNA that codes for phenylalanine. The
nucleotides not specifically identified are chemically modified analogs of the four common nucleotides.
As each successive codon on mRNA is read, different tRNAs bring the correct amino acids into position for enzyme-mediated transfer to the growing peptide. When synthesis of the proper protein is completed, a "stop" codon signals the end, and the protein is released from the ribosome. This process is illustrated in FIGURE 28.8.


Solution - 28.5 Translation of RNA: Protein Biosynthesis
FIGURE 28.8 A representation of protein biosynthesis. The codon base sequences on mRNA are read by tRNAs containing complementary anticodon base sequences. Transfer RNAs assemble the proper amino acids into position for incorporation into the growing peptide.


Predicting the Amino Acid Sequence Transcribed from DNA -
What amino acid sequence is coded by the following segment of a DNA coding strand (sense strand)?
(5') CTA-ACT-AGC-GGG-TCG-CCG (3')


Strategy -
The mRNA produced during translation is a copy of the DNA coding strand, with each T replaced by U . Thus, the mRNA has the sequence
(5') CUA-ACU-AGC-GGG-UCG-CCG (3')
Each set of three bases forms a codon, whose meaning can be found in TABLE 28.1.


Solution -
Leu-Thr-Ser-Gly-Ser-Pro.

PROBLEM List codon sequences for the following amino acids:
28-7 (a) Ala
(b) Phe
(c) Leu
(d) Tyr

PROBLEM List anticodon sequences on the tRNAs carrying the following amino acids.
28-8
(a) Ala
(b) Phe
(c) Leu
(d) Tyr

PROBLEM What amino acid sequence is coded by the following mRNA base sequence?


CUU-AUG-GCU-UGG-CCC-UAA -
PROBLEM What is the base sequence in the original DNA strand on which the mRNA sequence in Problem 28-10 28-9 was made?


CUU-AUG-GCU-UGG-CCC-UAA - 28.6 DNA Sequencing
A scientific revolution is now under way in molecular biology, as scientists learn how to manipulate and harness the genetic machinery of organisms. None of the extraordinary advances of the past several decades would have been possible, however, were it not for the discovery in 1977 of methods for sequencing immense DNA chains.

The first step in DNA sequencing is to cleave the enormous chain at known points to produce smaller, more manageable pieces, a task accomplished by the use of restriction endonucleases, or restriction enzymes. Each different restriction enzyme, of which more than 4000 are known and approximately 600 are commercially available, cleaves a DNA molecule at a point in the chain where a specific base sequence occurs. For example, the restriction enzyme AluI cleaves between $G$ and $C$ in the four-base sequence AG-CT. Note that the sequence is a palindrome, meaning that the sequence ( $5^{\prime}$ )-AGCT-( $3^{\prime}$ ) is the same as its complement ( $3^{\prime}$ )-TCGA-( $5^{\prime}$ ) when both are read in the same $5^{\prime} \rightarrow 3^{\prime}$ direction. The same is true for other restriction endonucleases.

If the original DNA molecule is cut with another restriction enzyme having a different specificity for cleavage, still other segments are produced whose sequences partially overlap those produced by the first enzyme. Sequencing all the segments, followed by identification of the overlapping regions, allows for complete DNA sequencing.

A dozen or so different methods of DNA sequencing are now available, and many others are under development. The Sanger dideoxy method is among the most frequently used and was the method responsible for first sequencing the entire human genome of 3.0 billion base pairs. In commercial sequencing instruments, the dideoxy method begins with a mixture of the following:


CUU-AUG-GCU-UGG-CCC-UAA - 28.6 DNA Sequencing
- The restriction fragment to be sequenced
- A small piece of DNA called a primer, whose sequence is complementary to that on the $3^{\prime}$ end of the restriction fragment
- The four 2'-deoxyribonucleoside triphosphates (dNTPs)
- Very small amounts of the four $2^{\prime}, 3^{\prime}$-dideoxyribonucleoside triphosphates (ddNTPs), each of which is labeled with a fluorescent dye of a different color (A $2^{\prime}, 3^{\prime}$-dideoxyribonucleoside triphosphate is one in which both $2^{\prime}$ and $3^{\prime}-\mathrm{OH}$ groups are missing from ribose.)

A 2'-deoxyribonucleoside triphosphate (dNTP)

A 2', 3'-dideoxyribonucleoside triphosphate (ddNTP)

DNA polymerase is added to the mixture, and a strand of DNA complementary to the restriction fragment begins to grow from the end of the primer. Most of the time, only normal deoxyribonucleotides are incorporated into the growing chain because of their much higher concentration in the mixture, but every so often, a dideoxyribonucleotide is incorporated. When that happens, DNA synthesis stops because the chain end no longer has a $3^{\prime}$-hydroxyl group for adding further nucleotides.

When reaction is complete, the product consists of a mixture of DNA fragments of all possible lengths, each terminated by one of the four dye-labeled dideoxyribonucleotides. This product mixture is then separated according to the size of the pieces by gel electrophoresis (Section 26.2), and the identity of the terminal dideoxyribonucleotide in each piece-and thus the sequence of the restriction fragment-is determined by
noting the color with which the attached dye fluoresces. FIGURE 28.9 shows a typical result.

FIGURE 28.9 The sequence of a restriction fragment determined by the Sanger dideoxy method can be read simply by noting the colors of the dye attached to each of the various terminal nucleotides.


CUU-AUG-GCU-UGG-CCC-UAA - 28.6 DNA Sequencing
FIGURE 28.9 The sequence of a restriction fragment determined by the Sanger dideoxy method can be read simply by noting the colors of the dye attached to each of the various terminal nucleotides.

So efficient is the automated dideoxy method that sequences up to 1100 nucleotides in length, with a throughput of up to 19,000 bases per hour, can be sequenced with $98 \%$ accuracy. After a decade of work and a cost of about $\$ 500$ million, preliminary sequence information for the entire human genome of 3.0 billion base pairs was announced early in 2001 and complete information was released in 2003 . More recently, the genome sequencing of individuals, including that of James Watson, one of the discoverers of the double helix, has been accomplished. The sequencing price per genome is dropping rapidly and is currently less than $\$ 1,000$, meaning that the routine sequencing of individuals is within reach.

Remarkably, our genome appears to contain only about 21,000 genes, less than one-fourth the previously predicted number and only about twice the number found in the common roundworm. It's also interesting to note that the number of genes in a human $(21,000)$ is much smaller than the number of kinds of proteins (perhaps 500,000). This discrepancy arises because most proteins are modified in various ways after translation (posttranslational modifications), so a single gene can ultimately give many different proteins.


CUU-AUG-GCU-UGG-CCC-UAA - 28.7 DNA Synthesis
The ongoing revolution in molecular biology has brought with it an increased demand for the efficient chemical synthesis of short DNA segments, called oligonucleotides, or simply oligos. The problems of DNA synthesis are similar to those of peptide synthesis (Section 26.7) but are more difficult because of the complexity of the nucleotide monomers. Each nucleotide has multiple reactive sites that must be selectively protected and deprotected at specific times, and coupling of the four nucleotides must be carried out in the proper sequence. Automated DNA synthesizers are available, however, that allow the fast and reliable synthesis of DNA segments up to 200 nucleotides in length.

DNA synthesizers operate on a principle similar to that of the Merrifield solid-phase peptide synthesizer (Section 26.8). In essence, a protected nucleotide is covalently bonded to a solid support, and one nucleotide at a time is added to the growing chain by the use of a coupling reagent. After the final nucleotide has been added, all the protecting groups are removed and the synthetic DNA is cleaved from the solid support. Five steps are needed: