Corrected exercises: biomolecules (202-SNU-RE)

First convention: the carbon chain is VERTICAL and the most oxidized carbon, here the carbonyl, is placed at the TOP. Second convention: each asymmetric carbon sits at the intersection of a cross, its two side substituents being carried by the horizontal lines. The spatial rule, which is the heart of the convention and is often forgotten, is that the HORIZONTAL bonds point TOWARD the viewer, in front of the plane of the page, while the VERTICAL bonds point away from the viewer, toward the back. A Fischer projection is therefore not a flat drawing but the flattened representation of a molecule in which each carbon is seen in an imposed orientation. A strict handling rule follows: a projection can be rotated by $180$ degrees in the plane of the page without changing anything, but a rotation by $90$ degrees or simply flipping the page over INVERTS the configuration of every center, which is the classic error.

The only asymmetric carbon is the middle one, and it is also the one farthest from the carbonyl: it therefore decides the series. In structure I, its $-OH$ group is on the RIGHT: this is D-glyceraldehyde. In structure II it is on the LEFT: this is L-glyceraldehyde. The two structures are mirror images of each other and not superimposable, and since the molecule has only one stereogenic center, they are ENANTIOMERS. Note that the letters D and L say nothing about the direction in which polarized light is rotated, contrary to what their historical origin suggests: they are descriptors of CONFIGURATION, not to be confused with the signs $+$ and $-$, which describe a measured property.

An aldohexose has six carbons. The first carries the aldehyde and is not asymmetric; the sixth is a terminal $CH_{2}OH$ and is no more so. That leaves carbons $2$, $3$, $4$ and $5$, each bearing four different substituents: FOUR asymmetric carbons. The number of stereoisomers is therefore $2^{4}=16$, split into $8$ of the D series and $8$ of the L series, since the series is fixed by carbon $5$ alone and the other three remain free, that is $2^{3}=8$ possibilities for each series. For a ketohexose, the carbonyl occupies carbon $2$, which is therefore not asymmetric; only carbons $3$, $4$ and $5$ remain, that is THREE centers, hence $2^{3}=8$ stereoisomers, of which $4$ are of the D series. A ketose therefore systematically has half as many stereoisomers as the aldose of the same molecular formula, which is a good consistency check.

Two EPIMERS are diastereomers that differ only in the configuration of ONE SINGLE asymmetric carbon, all the others being identical. The term is therefore more precise than diastereomer, which designates any pair of stereoisomers that are not mirror images of each other, whatever the number of centers that differ. And it contrasts with enantiomer, which designates two molecules in which ALL the centers are inverted simultaneously. To sum up: all centers inverted gives enantiomers, a single center inverted gives epimers, and an intermediate number gives diastereomers that are not epimers. Glucose and mannose, which differ only at $C_{2}$, are therefore epimers at $C_{2}$, and consequently diastereomers. Glucose and galactose, which differ only at $C_{4}$, are epimers at $C_{4}$, and diastereomers as well. One practical consequence deserves emphasis: unlike enantiomers, epimers have DIFFERENT physical properties, with distinct melting points, solubilities and optical rotations, and they can be separated by ordinary methods.

The question touches an essential distinction between laboratory chemistry and the chemistry of living things. It is true that D-glucose and L-glucose have strictly the same internal energy, the same melting point and the same solubility in water, and that no reaction run with achiral reagents can distinguish them: this is the rule established back in the chapter on aldehydes and ketones, according to which an achiral medium can neither produce nor distinguish an enantiomeric excess. But enzymes are not achiral reagents. They are proteins built from L-series amino acids, hence themselves chiral, and their active site is an asymmetric three-dimensional cavity. A substrate enters only if it matches its shape, exactly as a right hand fits only a right-hand glove. The enzymes of glucose metabolism therefore recognize only the D enantiomer, and L-glucose, if ingested, would pass through the body without being metabolized: it does taste sweet, since taste receptors are less selective, but it provides no calories. As for the origin of this choice, it goes back to the emergence of life and remains debated; what is certain is that once a first set of chiral enzymes was in place, the whole metabolism became locked onto the same series, because a chiral enzyme can only make chiral products of a single handedness.

The rule is simple and worth memorizing as it stands: what is on the RIGHT in the Fischer projection ends up BELOW the plane of the ring in the Haworth representation, and what is on the left ends up above. It follows from the vertical Fischer chain wrapping around on itself to close the ring, an operation that tips the right-hand substituents downward. As for the terminal $CH_{2}OH$, it is the sixth carbon, which is not part of the ring: it becomes a substituent of carbon $5$, and for a D-series sugar it is systematically placed ABOVE the plane. This position also serves as the reference point for reading any Haworth formula: you identify it first, then locate the others relative to it.

The anomeric carbon is carbon $1$, the one that carried the aldehyde and that becomes, after cyclization, a hemiacetal carbon bearing both an $-OH$ and the ring oxygen. The $\alpha$ anomer is the one whose anomeric hydroxyl is on the side OPPOSITE the $CH_{2}OH$, hence below the plane for a D-series sugar; the $\beta$ anomer is the one whose anomeric hydroxyl is on the SAME side as the $CH_{2}OH$, hence above. These two forms interconvert in solution because the cyclic hemiacetal is in EQUILIBRIUM with the open-chain form: the ring opens, carbon $1$ becomes a planar carbonyl again, and it recloses on either side indifferently. This is the phenomenon of mutarotation studied in the chapter on aldehydes and ketones, and the reasoning is the one already met three times in the course, namely that a PLANAR intermediate erases the stereochemical information of the center concerned.

Cyclization is an intramolecular reaction, and like all those met in the course, it favors five- and six-membered rings, the only ones free of significant strain and sufficiently probable geometrically. Both are therefore accessible, and a finer trade-off decides. For GLUCOSE, an aldose whose carbonyl is at $C_{1}$, the pyranose form resulting from attack by the $-OH$ of carbon $5$ allows, as the next question will show, every substituent to be placed equatorial: it is therefore remarkably stable and accounts for more than $99$ percent of the molecules. For FRUCTOSE, a ketose whose carbonyl is at $C_{2}$, attack by the $-OH$ of carbon $5$ closes a FIVE-membered ring, and the arrangement of the substituents there is more favorable than in the corresponding six-membered ring. The furanose form is therefore well represented, notably when fructose is engaged in sucrose, where it is locked in the furanose form. The general lesson is that ring size is not decided by the five-and-six rule alone, but by the detail of the steric interactions of the sugar in question.

$\beta$-D-glucopyranose carries four bulky substituents on the ring: the hydroxyls of carbons $1$, $2$, $3$ and $4$, plus the $CH_{2}OH$ of carbon $5$. In the Haworth representation, the hydroxyl of carbon $1$ is above since the anomer is $\beta$, that of carbon $2$ below, that of carbon $3$ above, that of carbon $4$ below, and the $CH_{2}OH$ above. This regular alternation is precisely the condition for all of them to end up EQUATORIAL when we move to the chair conformation: in a chair, equatorial positions alternate from one side of the mean plane to the other from one carbon to the next. All the bulky groups then point outward from the ring, in the plane, as far from one another as possible, and no repulsive axial interaction appears. This is exceptional: $\beta$-D-glucopyranose is the ONLY aldohexose to achieve this perfect arrangement, all its epimers having at least one substituent forced into an axial position. Mannose thus has its hydroxyl at $C_{2}$ axial, and galactose the one at $C_{4}$.

Since $\beta$-D-glucopyranose is the most stable form a hexose can adopt, it is also the molecule that pays the least in strain and crowding energy, and therefore the most abundant at equilibrium. A more stable compound is also more resistant to uncontrolled degradation, which makes it a good reserve fuel and a good building material: it is found both as a cellular fuel and as the building block of starch and cellulose. The proportion of $64$ percent of $\beta$ anomer at equilibrium, calculated from the optical rotations in the chapter on aldehydes and ketones, directly confirms the reasoning: the $\beta$ anomer wins because it places its anomeric hydroxyl in the equatorial position, whereas the $\alpha$ anomer must put it axial. The gap is not huge, about $64$ against $36$ percent, which corresponds to a modest free energy difference, of the order of $1.5$ kJ/mol; but it goes in the predicted direction, and it is a fine example of a conformational prediction confirmed by a polarimetry measurement.

The anomeric carbon of a cyclic sugar is a HEMIacetal carbon: it carries an $-OH$ and an $-OR$, the latter being the ring oxygen. When a hydroxyl of a second sugar replaces this $-OH$, the carbon ends up bearing TWO $-OR$ groups and no $-OH$: it has become a full ACETAL, in the exact sense of the chapter on aldehydes and ketones. The consequence is immediate and underlies all of question c): an acetal can no longer reopen in neutral or basic medium and therefore no longer releases an aldehyde form. The equation for the formation of maltose is $2\,C_{6}H_{12}O_{6}\rightarrow C_{12}H_{22}O_{11}+H_{2}O$: two glucoses of $180.16$ g/mol give a disaccharide of $342.30$ g/mol with loss of one molecule of water, which is checked by $2\times 180.16-18.02=342.30$.

The Greek letter $\alpha$ or $\beta$ designates the CONFIGURATION of the anomeric carbon of the first sugar, the one that contributes its carbon $1$ to the bond: it indicates on which side of the plane the oxygen involved lies. The first number designates the number of the ANOMERIC carbon of the first sugar, almost always carbon $1$ for an aldose. The second number designates the number of the carbon of the SECOND sugar whose hydroxyl reacted. The arrow indicates the DIRECTION of the bond, from the anomeric carbon to the hydroxyl: it is therefore not decorative, it says which of the two sugars used its anomeric carbon and which merely supplied an ordinary hydroxyl. A notation $\alpha(1\rightarrow 4)$ thus means: carbon $1$ of the first glucose, in the $\alpha$ configuration, is linked to the oxygen carried by carbon $4$ of the second glucose.

A sugar is reducing if it can release an aldehyde group, that is, if it keeps at least one anomeric carbon in HEMIacetal form, able to open. In maltose, cellobiose and lactose, the arrow is one-way: carbon $1$ of the first sugar is engaged, but the second sugar supplied only the hydroxyl of its carbon $4$, and its OWN anomeric carbon remains free. These three disaccharides therefore still have a hemiacetal, open partially in solution, show mutarotation and reduce the Tollens and Fehling reagents. Sucrose is built quite differently, and this is precisely what the double arrow of the notation $\alpha(1\leftrightarrow 2)\beta$ signals: the bond joins the anomeric carbon of glucose, its carbon $1$, to the anomeric carbon of fructose, its carbon $2$. BOTH anomeric carbons are therefore engaged simultaneously, no hemiacetal remains, no open form is accessible: sucrose is not reducing and shows no mutarotation. The double arrow is thus the notation to look for; it means head to head and by itself announces the absence of reducing power.

The difference is not in the skeleton but in the CONFIGURATION of the anomeric carbon involved: $\alpha$ in maltose, $\beta$ in cellobiose. This inversion of a single center completely changes the geometry of the molecule: an $\alpha$ bond imposes an angle that gives the chain a bent shape, tending to coil it, whereas a $\beta$ bond gives a stretched, regular chain. Now digestive enzymes are, as seen in exercise $1$, chiral cavities of very precise geometry. Human maltase recognizes the $\alpha$ geometry and hydrolyzes maltose without difficulty; our enzymatic equipment contains no $\beta$-glucosidase capable of attacking the cellobiose bond. It is therefore not a question of energy, since the two bonds have practically the same thermodynamic stability and are hydrolyzed equally well in hot acid, but a question of enzymatic RECOGNITION. Ruminants and termites do digest cellobiose and cellulose, not because they produce the enzyme but because they harbor symbiotic microorganisms that produce it.

Hydrolysis releases one mole of glucose and one mole of fructose per mole of sucrose. The optical rotation of the equimolar mixture is the average of the two contributions, that is $\frac{52.7+(-92.0)}{2}=\frac{-39.3}{2}=-19.65$ degrees. We start from $+66.5$ and arrive at about $-19.7$: the sign has been INVERTED in the course of the reaction, hence the name invert sugar and that of the enzyme that catalyzes it, invertase. The inversion is entirely explained by fructose, whose strongly negative rotation outweighs that of glucose; this is also why fructose is sometimes called levulose, from the Latin for left. This invert sugar is the main constituent of honey, where the invertase supplied by the bee has hydrolyzed the sucrose of the nectar, and it is used in confectionery because it crystallizes less readily than sucrose and gives softer textures. Measuring the optical rotation here provides a convenient, non-destructive kinetic follow-up of the progress of the hydrolysis, which makes it a great classic of laboratory work.

AMYLOSE is a strictly linear chain of glucoses linked by $\alpha(1\rightarrow 4)$ bonds, typically comprising from several hundred to a few thousand units. AMYLOPECTIN has the same $\alpha(1\rightarrow 4)$ backbone, but it is regularly interrupted by branch points. These branches are $\alpha(1\rightarrow 6)$ bonds: instead of engaging the hydroxyl of carbon $4$ of the next glucose, the anomeric carbon bonds to the hydroxyl carried by carbon $6$, that is, by the terminal $CH_{2}OH$, which is free and available on each unit. A new chain then starts perpendicular to the first. Natural starch contains about $20$ to $30$ percent amylose and $70$ to $80$ percent amylopectin, proportions that vary with the plant and determine the cooking properties of flour or rice.

Everything comes down to the geometry imposed by the configuration of the anomeric carbon. An $\alpha$ bond places the bridging oxygen in an AXIAL position on the first glucose, which imposes a marked angle between two consecutive units: the chain cannot extend in a straight line, it turns slightly at each link, and the accumulation of these small rotations coils it into a HELIX, with about six glucoses per turn. A $\beta$ bond, on the contrary, places this oxygen in an EQUATORIAL position, in line with the ring: two consecutive units line up, provided that each is turned over by $180$ degrees relative to the previous one, which nature does indeed do. The resulting chain is then perfectly STRAIGHT and forms a flat ribbon. We see here that the chair conformation of exercise $2$ is not a theoretical refinement: it decides the architecture of two of the most abundant materials on the planet.

The straight chain of cellulose regularly exposes its hydroxyls on either side of the ribbon. Neighboring chains can therefore line up and form between them a very dense network of HYDROGEN BONDS, to which are added hydrogen bonds within each chain that stiffen it. The result is crystalline microfibrils, where the chains are so tightly packed and so well bonded that water cannot get in between them: cellulose is insoluble, hardly swells, and has a tensile strength comparable to that of steel at equal mass. It is the structural material of plants. Amylose, coiled into a helix, cannot pair up in this way: its hydroxyls point in all directions and the helix occupies a large volume. Hot water therefore penetrates between the chains, hydrates and disperses them, which gives the familiar viscous paste, the basis of sauces and culinary thickeners. Note that the inside of the amylose helix is nonpolar and can house a guest molecule: this is where diiodine slips in during the blue-violet test used to recognize starch.

Each chain end is an attack point for enzymes, which detach glucoses one by one from the ends. The more branched a molecule is, the more ends it has, and the more QUICKLY glucose can be mobilized or stored, since many enzymes work in parallel on the same molecule. An animal has a need for speed that a plant does not: it must be able to release glucose within seconds during exertion or hypoglycemia, whereas a plant mobilizes its reserves over days or seasons. Branching every $8$ to $12$ units, against $24$ to $30$, therefore multiplies the number of work fronts. A second advantage comes with it: branching makes the molecule more compact and more soluble, which allows a lot of it to be stored in the cytoplasm of liver and muscle cells without disturbing the osmotic pressure, whereas the same amount of free glucose would be osmotically unsustainable.

The hydrolysis is written $\left(C_{6}H_{10}O_{5}\right)_{n}+n\,H_{2}O\rightarrow n\,C_{6}H_{12}O_{6}$: each glycosidic bond broken consumes one molecule of water, exactly the reverse of the condensation that formed it. The amount of units contained in $100$ g of starch is $\frac{100}{162.14}=0.6168$ mol, and each unit gives one glucose, that is $0.6168\times 180.16=111.1$ g of glucose. The mass therefore increases by $11.1$ percent. This result often surprises, and it must not be taken as a violation of conservation of mass: the extra $11.1$ g is exactly the mass of WATER consumed by the reaction, which was not included in the starting $100$ g. This can be checked: $0.6168\times 18.02=11.1$ g, and $100+11.1=111.1$ g. It is a classic exam error to calculate a mass of glucose while forgetting the water of hydrolysis and to find $100$ g. The remark also applies to nutrition labeling: $100$ g of starch provide the energy of $111$ g of glucose, which explains why complex carbohydrates are counted at about $4.2$ kilocalories per gram against $3.75$ for free glucose.

The reaction is a triple ESTERIFICATION: propane-$1,2,3$-triol, or glycerol, has three hydroxyls, and each reacts with the carboxyl group of a fatty acid, eliminating one molecule of water. The overall equation reads glycerol plus three fatty acids gives a triglyceride plus three molecules of water, and the functional group created is three times an ESTER group. Nothing here differs from the Fischer esterification studied in the chapter on carboxylic acids, and the reverse reaction is the familiar one: basic hydrolysis of a triglyceride is SAPONIFICATION, which gives back glycerol and three carboxylates, that is, soap. In the body, the same hydrolysis is carried out by pancreatic lipases.

A is the SATURATED chain, entirely straight, made of a regular succession of tetrahedral carbons; all its bonds are single and free rotation around each allows the chain to adopt the stretched zigzag conformation, which is the most stable. B is the UNSATURATED chain of CIS configuration, recognizable by the double line in the middle and above all by the marked BEND that follows. The origin of the bend is geometric: the two carbons of a double bond are $sp^{2}$, hence planar, with angles of $120$ degrees, and rotation is blocked there. In a cis configuration, the two chain fragments leave from the SAME side of the double bond, which imposes on the whole an angle of about $30$ degrees that the molecule cannot get rid of. The chain is thus broken into two misaligned segments, which the figure shows.

Melting means disorganizing a packing of molecules held together by London forces, which grow with the contact SURFACE between neighboring chains. The straight chains of stearic acid stack in parallel, like pencils in a box, with maximal contact along their whole length: a lot of energy is therefore needed to separate them, hence $69$ °C. The bend of oleic acid prevents this tight stacking: the chains get in each other's way, the crystal is loose, contacts are reduced, and the compound melts as low as $13$ °C, that is $56$ °C lower. Elaidic acid sheds light on the mechanism, because it makes it possible to separate the two possible causes. It has exactly the same molecular formula and the same double bond as oleic acid, hence the same number of electrons and the same polarizability, but in the TRANS configuration: the two fragments leave on either side of the double bond and the chain remains globally STRAIGHT, barely offset. It therefore stacks almost as well as a saturated chain and melts at $45$ °C. The comparison proves that the lowering of the melting point is not due to the presence of the double bond in itself, but solely to the cis GEOMETRY it imposes. This is a control-experiment line of reasoning that you must know how to carry out.

The iodine value is the mass of diiodine, expressed in grams, taken up by $100$ g of fat by addition to its double bonds. Each carbon-carbon double bond adds exactly one molecule of $I_{2}$, so this value directly measures the degree of UNSATURATION. Calculation for triolein: it has three oleic chains, hence THREE double bonds, and takes up three moles of diiodine per mole. For $100$ g, the amount of triolein is $\frac{100}{885.46}=0.1129$ mol, which takes up $3\times 0.1129=0.3388$ mol of diiodine, that is $0.3388\times 253.81=86.0$ g. The iodine value of triolein is therefore $86$, consistent with the $80$ to $88$ measured on olive oil, of which oleic acid is the major constituent. A HIGH iodine value signals a strongly unsaturated oil, hence liquid at room temperature, but also more sensitive to oxidation and rancidity, since the positions next to the double bonds are precisely those attacked by atmospheric oxygen. The drying oils used in paint, such as linseed oil, exceed $170$; solid animal fats fall below $50$.

The process consists of reacting the oil with dihydrogen under pressure, in the presence of a metal catalyst, most often finely divided nickel. Hydrogenation ADDS $H_{2}$ across the double bonds and turns them into single bonds, which straightens the chains and solidifies the oil. It is deliberately stopped before completion, hence the name partial hydrogenation. Now the addition of hydrogen is reversible at the surface of the catalyst: a double bond can attach there, partially break, rotate, then re-form, and it re-forms statistically in the TRANS configuration, more stable than cis. The process therefore produces, as an unavoidable by-product, trans fatty acids that did not exist in the starting oil. Industry used it for three reasons: to obtain semi-solid, spreadable fats such as margarine from cheap vegetable oils, to greatly improve shelf life since fewer double bonds remain and oxidation is therefore slower, and to obtain a texture suited to industrial pastry. The regulation came from epidemiology: trans fatty acids raise low-density cholesterol while lowering high-density cholesterol, a particularly unfavorable combination that neither saturated fats nor cis unsaturated fats produce. Their use has been banned or strictly capped in many countries, including Canada since $2018$, which led industry to turn back to other processes, notably interesterification.

The form present is $^{+}H_{3}N-CH_{2}-COO^{-}$, called a ZWITTERION or dipolar ion. The reasoning is a simple internal proton transfer: the $COOH$ group is an acid with $pK_{a}$ $2.34$, so at neutral pH it is very largely deprotonated to $COO^{-}$; the $NH_{2}$ group has a conjugate acid with $pK_{a}$ $9.60$, so at neutral pH it is very largely protonated to $NH_{3}^{+}$. Both conditions are satisfied simultaneously, and the molecule therefore carries TWO opposite charges while remaining globally neutral. This is not a notational curiosity: it explains why amino acids are crystalline solids with very high melting points, often above $200$ °C with decomposition, and very soluble in water but insoluble in ether, the behavior of a SALT and not of an ordinary organic molecule.

First equilibrium, starting from the fully protonated form in very acidic medium: $^{+}H_{3}N-CH_{2}-COOH\rightleftharpoons {}^{+}H_{3}N-CH_{2}-COO^{-}+H^{+}$, with $pK_{a1}=2.34$, which concerns the CARBOXYL group. Second equilibrium: $^{+}H_{3}N-CH_{2}-COO^{-}\rightleftharpoons H_{2}N-CH_{2}-COO^{-}+H^{+}$, with $pK_{a2}=9.60$, which concerns the AMMONIUM group. The order is logical, the carboxyl always being the more acidic of the two. As for the gap with ethanoic acid, it is explained by the INDUCTIVE effect of the neighboring ammonium group: it carries a full positive charge and sits on the carbon immediately adjacent to the carboxyl, so it strongly withdraws electrons and stabilizes the negative charge of the carboxylate formed. The conjugate base being better stabilized, the acid is stronger, by $4.76-2.34=2.42$ units, a factor of $260$. This is exactly the reasoning applied to the chloroethanoic acids in the chapter on carboxylic acids, except that here the electron-withdrawing group is a full charge and not a mere halogen.

The isoelectric point is the pH at which the molecule carries a NET charge of zero on average, and therefore does not migrate in an electric field. For a neutral amino acid, it lies midway between the two $pK_{a}$ values, because that is where the positively and negatively charged forms exactly balance: $pI=\frac{2.34+9.60}{2}=5.97$. On the figure, it corresponds to the middle vertical line, and what characterizes it is that it falls on the PLATEAU of the curve, at exactly ONE equivalent of base added. This is consistent: one equivalent has served to neutralize the carboxyl, the second has not yet begun to deprotonate the ammonium, and the molecule is therefore entirely in zwitterion form. The two $pK_{a}$ values can also be read directly on the figure at the inflection points, at half an equivalent and at one and a half equivalents, where the curve is steepest.

For ASPARTIC ACID, the side chain carries a second carboxyl, with $pK_{a}$ $3.65$. The molecule is therefore acidic, and the two $pK_{a}$ values that bracket the form of zero net charge must be averaged: these are the two carboxyls, $1.88$ and $3.65$, because once the first is deprotonated the molecule is already neutral, and deprotonating the second would make it negative. We obtain $pI=\frac{1.88+3.65}{2}=2.77$, a distinctly acidic value. For LYSINE, the side chain carries a second amine, with $pK_{a}$ $10.53$. The molecule is basic, and the two $pK_{a}$ values that bracket the neutral form are the two ammoniums, $8.95$ and $10.53$, since as long as both are protonated the charge is positive. We obtain $pI=\frac{8.95+10.53}{2}=9.74$, a distinctly basic value. The general rule to remember is therefore that you NEVER mechanically average the first and the last $pK_{a}$: you first identify the form of zero charge, then average the two $pK_{a}$ values that bracket it.

The rule is that an amino acid carries a net positive charge if the pH is BELOW its $pI$, and a negative one if it is above. At $pH=6.0$: glycine, with $pI=5.97$, is practically at its isoelectric point and therefore migrates ALMOST NOT AT ALL, staying at the point of application; aspartic acid, with $pI=2.77$, is at a pH well above its $pI$, so it is NEGATIVELY charged and migrates toward the ANODE, the positive pole; lysine, with $pI=9.74$, is at a pH well below its $pI$, so it is POSITIVELY charged and migrates toward the CATHODE. The three thus separate into three distinct spots, which can be revealed with ninhydrin. The value of the method is twofold. First, it separates molecules that are very similar in size and solubility on the sole criterion of CHARGE, where ordinary chromatography would fail. Second, by choosing the pH of the buffer, you choose what you want to separate: this is the basis of protein electrophoresis in clinical biology, and of isoelectric focusing, where a pH gradient immobilizes each protein exactly at its own $pI$.

The reaction joins the carboxyl group of one to the amino group of the other with elimination of one molecule of water, and the functional group created is an AMIDE, called here a peptide bond. Two dipeptides are possible because the reaction is not symmetric: either glycine supplies its carboxyl and alanine its amine, which gives glycylalanine written Gly-Ala; or the reverse, which gives alanylglycine, Ala-Gly. These two compounds have the same molecular formula but are not identical: they are constitutional isomers, with different physical and biological properties. This is why a peptide is always written in a CONVENTIONAL order, from the $N$-terminal end, the one whose amine is free, to the $C$-terminal end, the one whose carboxyl is free. In synthesis, obtaining one of the two selectively requires protecting the unused functional groups, exactly following the protection logic seen in the chapter on amines.

With three distinct amino acids each used once, the number of tripeptides is the number of permutations of three elements, that is $3!=6$. With the twenty natural amino acids and six positions, each position can receive any of the twenty, independently of the others: the number of peptides is $20^{6}=6.4\times 10^{7}$, that is sixty-four million. The order of magnitude deserves a comment, because it explains a fundamental property of living things: with only six residues we already exceed tens of millions of possibilities, and an average protein has three hundred. The number of imaginable sequences then far exceeds the number of atoms in the observable universe. It is this combinatorics, and not some exotic chemistry, that gives proteins their unlimited functional diversity.

The peptide bond is an amide, and the nitrogen lone pair delocalizes toward the neighboring carbonyl, as the chapter on carboxylic acids established: the $C-N$ bond acquires partial double-bond character, shortens to $133$ pm, the nitrogen becomes nearly $sp^{2}$, and rotation around this bond costs about $80$ kJ/mol, which blocks it at room temperature. The six atoms of the unit, the two alpha carbons, the carbonyl carbon, the oxygen, the nitrogen and its hydrogen, are therefore COPLANAR. The consequence for folding is decisive: of the three backbone bonds in each residue, ONE is locked. Only two degrees of freedom per amino acid remain, the rotations around the two bonds flanking the alpha carbon. The folding of a protein is thus described by only two angles per residue, which enormously reduces the space of possible conformations and makes reproducible regular motifs conceivable. Note also that the conformation adopted is almost always TRANS, at more than $99$ percent, proline being the exception.

The PRIMARY structure is the sequence of amino acids, from the $N$-terminal end to the $C$-terminal end; it is maintained by COVALENT bonds, the peptide bonds themselves. The SECONDARY structure is the local, regular folding of the backbone, alpha helix or beta sheet; it is maintained by HYDROGEN BONDS between the $C=O$ of one residue and the $N-H$ of another, located four positions further along in the helix, or on a neighboring strand in the sheet. The TERTIARY structure is the overall folding of a single chain in space; it is maintained by a set of interactions between SIDE CHAINS: the hydrophobic effect that buries nonpolar residues in the core of the protein, salt bridges between oppositely charged chains, hydrogen bonds, and covalent DISULFIDE bridges between two cysteines. The QUATERNARY structure is the assembly of several distinct chains into a functional complex, like the four subunits of hemoglobin; it is maintained by the same noncovalent interactions as the tertiary structure, but between different chains. Only the primary level and, through disulfide bridges, the tertiary level involve covalent bonds: everything else is held by weak interactions, which is the key to the next question.

Heating an egg white supplies thermal energy that agitates the molecules and breaks the WEAK interactions: hydrogen bonds of the secondary structure, hydrophobic interactions and salt bridges of the tertiary structure. The proteins, essentially ovalbumin, lose their compact folding and DENATURE, exposing to the outside the hydrophobic residues that were until then buried in the core of the molecule. These hydrophobic surfaces, now in contact with water, then associate between neighboring chains to escape the solvent: the proteins aggregate into a three-dimensional network that traps water, hence the white, opaque solid. No peptide bond is broken, and the PRIMARY structure is fully preserved; it is indeed a matter of folding and not of sequence. The irreversibility lies in the aggregation itself: the tangled chains are held by a multitude of weak but very numerous interactions, so that a return to the native state would require undoing them all simultaneously, which is entropically inaccessible. On cooling, we merely freeze the aggregate. Note that a gentle denaturation, in dilute solution and without aggregation, can on the contrary be reversible, which Anfinsen demonstrated on ribonuclease and which earned him the Nobel Prize: the primary structure does contain, by itself, the information needed for folding.

The six residues are all distinct and each used once: the number of possible sequences is the number of permutations of six elements, that is $6!=720$. It is a modest number, but one that no method could explore one by one; and note that this is only the favorable case where the COMPOSITION is already known. If we did not know which residues are present and were searching for them among the twenty natural ones, there would be $20^{6}$, that is sixty-four million possibilities.

EDMAN degradation attacks specifically the $N$-terminal residue, the one whose amine group is free: phenylisothiocyanate attaches to it, then the residue is cleaved off in mild acid as an identifiable derivative, without the rest of the chain being touched. It therefore identifies the FIRST residue, and it can be repeated cycle after cycle to read the sequence from the beginning. Here it releases glycine: the sequence begins with Gly. CARBOXYPEPTIDASE is an enzyme that hydrolyzes the peptide bond closest to the $C$-terminal end, the one whose carboxyl is free: it therefore identifies the LAST residue. Here it releases valine: the sequence ends with Val. We therefore already know Gly-?-?-?-?-Val, which reduces the number of possibilities to $4!=24$.

TRYPSIN cleaves after lysine and arginine; since there is no arginine, it cleaves only after lysine. It gives two fragments of THREE residues each, which means the cleavage takes place exactly in the middle of the chain: lysine therefore occupies position $3$. CHYMOTRYPSIN cleaves after aromatic residues, and the only one present is phenylalanine. It gives a fragment of five residues and one isolated residue, so it cleaves between positions $5$ and $6$: phenylalanine occupies position $5$, and the isolated residue released is necessarily valine, which confirms the carboxypeptidase result. We now have Gly-?-Lys-?-Phe-Val, and only two positions remain to be filled with alanine and serine, that is two possibilities.

The dipeptides are joined by their common ends. Gly-Ala requires alanine to follow glycine immediately, so alanine is in position $2$. Ala-Lys confirms that lysine follows alanine, so it is in position $3$, in agreement with trypsin. Lys-Ser requires serine to follow lysine, so it is in position $4$. Ser-Phe requires phenylalanine to follow serine, so it is in position $5$, in agreement with chymotrypsin. The chain of overlaps reads as follows: Gly-Ala, then Ala-Lys connects through Ala, which gives Gly-Ala-Lys; then Lys-Ser connects through Lys, which gives Gly-Ala-Lys-Ser; then Ser-Phe connects through Ser, which gives Gly-Ala-Lys-Ser-Phe; and valine, identified as $C$-terminal, closes the sequence. The complete sequence is therefore GLY-ALA-LYS-SER-PHE-VAL.

Point-by-point check. Edman indeed releases the first residue, Gly. Carboxypeptidase indeed releases the last, Val. Trypsin cleaves after the lysine in position $3$ and gives Gly-Ala-Lys and Ser-Phe-Val, that is indeed two fragments of three residues. Chymotrypsin cleaves after the phenylalanine in position $5$ and gives Gly-Ala-Lys-Ser-Phe and Val, that is indeed one fragment of five and one isolated residue. The four dipeptides Gly-Ala, Ala-Lys, Lys-Ser and Ser-Phe all correspond to bonds really present. All the results agree. As for partial hydrolysis alone, it would not have been enough for a precise reason: the four dipeptides supplied cover the bonds $1$-$2$, $2$-$3$, $3$-$4$ and $4$-$5$, but NONE contains valine, so nothing indicates where it attaches. We would know that the fragment Gly-Ala-Lys-Ser-Phe exists, without knowing whether valine comes before or after it. It is precisely carboxypeptidase, or failing that chymotrypsin, that removes the ambiguity. The general lesson of a sequencing problem lies there: always check that the fragments COVER all the bonds, and spot which one remains undetermined.

R is insoluble in water and soluble in a nonpolar solvent, which is the behavior of a LIPID. Basic hydrolysis giving glycerol and three long-chain acids is the very definition of the saponification of a TRIGLYCERIDE, a triester of glycerol. R is therefore a triglyceride. An iodine value of $5$ is very low: it signals that the fat takes up almost no diiodine, hence that it has practically no carbon-carbon double bonds. R is therefore a SATURATED triglyceride, with straight chains that stack well, which implies a high melting point: it is a fat that is solid at room temperature, of the tallow or cocoa butter type, and not an oil. S gives a purple color with ninhydrin, a reagent specific to the primary amines of amino acids, and does not migrate in electrophoresis at $pH=5.97$, which means that this pH is exactly its ISOELECTRIC point. Now exercise $6$ calculated $pI=5.97$ for glycine. S is therefore GLYCINE.

First clue, Fehling's solution is negative: Q has no free anomeric carbon, hence no hemiacetal group able to open into an aldehyde. It is a NON-reducing sugar, which requires both its anomeric carbons to be engaged in the glycosidic bond. Second clue, hydrolysis gives two different sugars, one of them a KETOSE: this points to the pair glucose and fructose. Third clue, the optical rotation changes sign: this is exactly the inversion described in exercise $3$, where we go from $+66.5$ degrees for the disaccharide to about $-19.7$ for the equimolar mixture, strongly levorotatory fructose outweighing dextrorotatory glucose. Q is therefore SUCROSE, and the phenomenon responsible is the INVERSION of sucrose, which produces invert sugar and gives its name to the enzyme invertase.

A reducing disaccharide from the chapter necessarily has a single anomeric carbon engaged, which leaves three candidates: maltose, cellobiose and lactose. The decisive clue is that acid hydrolysis gives ONLY ONE sugar: lactose is therefore ruled out, since it would release two different sugars, galactose and glucose. That leaves MALTOSE and CELLOBIOSE, both made of two glucoses and therefore releasing only glucose on hydrolysis. Note that the mutarotation observed on the fresh solution independently confirms the reducing character: it can occur only if an anomeric carbon remains free and allows the ring to open.

The two candidates are MALTOSE, whose bond is $\alpha(1\rightarrow 4)$, and CELLOBIOSE, whose bond is $\beta(1\rightarrow 4)$. They differ only in the configuration of the anomeric carbon involved, and no ordinary chemical test distinguishes them since they have the same functional groups and release the same sugar. The experiment that tells them apart is ENZYMATIC, and that is the whole point of the question. The solution is split into two tubes: into the first we add MALTASE, an $\alpha$-glucosidase that cleaves only $\alpha$ bonds; into the second a $\beta$-glucosidase, for example emulsin extracted from almonds. We then look for the glucose released with the glucose oxidase test, specific to free glucose. If P is maltose, only the first tube contains any; if it is cellobiose, only the second. This double trial has the advantage of giving a POSITIVE result in both cases, instead of concluding from a mere absence of reaction. Be careful not to replace maltase with saliva: salivary amylase cleaves the INTERNAL $\alpha(1\rightarrow 4)$ bonds of starch and releases precisely maltose, which it hardly hydrolyzes at all; it is the maltase of the intestinal brush border that finishes the job, as will be seen in exercise $10$. We are exploiting here the stereochemical selectivity established in exercise $3$: what chemistry does not distinguish, the enzyme does, because its active site is chiral. An instrumental alternative would be to measure the optical rotation, the two disaccharides having very different values, or an NMR measurement that would separate the anomeric hydrogens by their coupling constant.

Each of the two amino acids has an amine group and an acid group, so each can play either the role of the acid partner or that of the amine partner. We therefore obtain FOUR dipeptides: the two mixed dipeptides, in the two possible orders, and the two homogeneous dipeptides formed by an amino acid with itself. Tripeptides and longer chains would moreover be added, since each dipeptide formed keeps a free amine and a free acid. This is exactly the difficulty met with the crossed aldol reaction and with the crossed Claisen condensation: when both partners can play both roles, all the combinations are obtained. Directed synthesis avoids it by PROTECTION, a method already met for acetals and for aniline: the amine of the amino acid that must supply the carboxyl is blocked, typically with a carbamate group, and the carboxyl of the one that must supply the amine is blocked, typically as an ester. Each partner then has only one possible role, the coupling can give only one product, and deprotection follows. This is the principle of Merrifield peptide synthesis, where the chain is moreover anchored to a solid support so that each excess of reagent can simply be rinsed away.

Digestion begins in the MOUTH with salivary amylase, called ptyalin, which hydrolyzes the INTERNAL $\alpha(1\rightarrow 4)$ bonds of starch and cuts it into shorter fragments, dextrins and maltose. Its action is brief because stomach acidity inactivates it. It resumes in the SMALL INTESTINE with pancreatic amylase, of the same specificity, which continues the cutting. The short fragments are finally finished off at the surface of the intestinal cells, the brush border, by membrane-bound enzymes: MALTASE hydrolyzes maltose into two glucoses, SUCRASE sucrose, LACTASE lactose. The final product absorbed is free glucose, which crosses the intestinal wall through a specific transporter and reaches the blood. Note that only MONOSACCHARIDES can be absorbed: no disaccharide and no longer fragment crosses the intestinal wall, which makes complete hydrolysis compulsory.

Amylases recognize only the geometry of $\alpha(1\rightarrow 4)$ bonds and stop a few residues from a branch point. What remains after their action is therefore a set of short, highly branched fragments called LIMIT DEXTRINS, in which all the $\alpha(1\rightarrow 6)$ bonds remain intact, as well as the few $\alpha(1\rightarrow 4)$ bonds surrounding them. The enzyme that finishes the job is ISOMALTASE, also called $\alpha$-dextrinase or debranching enzyme, also located at the brush border: it specifically hydrolyzes the $\alpha(1\rightarrow 6)$ bonds and releases the branches, which amylase can then take up again. The logic of the system shows here: a single enzyme is not enough, one is needed per TYPE of bond, which is the direct consequence of the geometric selectivity of active sites.

Exercise $4$ established that $100$ g of starch give $111.1$ g of glucose, the surplus being the mass of water consumed by the hydrolysis. The amount of glucose is $\frac{111.1}{180.16}=0.6167$ mol, whose combustion releases $0.6167\times 2803=1729$ kJ. Per gram of starting starch, this makes $\frac{1729}{100}=17.29$ kJ/g, that is $\frac{17.29}{4.184}=4.13$ kilocalories per gram. The usual nutritional factor is $4$ kilocalories per gram: the agreement is excellent, to within three percent. The small gap is explained without mystery, the Atwater factors used in nutrition being rounded values that take into account incomplete absorption and metabolic losses, whereas our calculation assumes total combustion. Note that this calculation also explains why complex carbohydrates are counted at about $4.2$ kilocalories per gram and free glucose at only $3.75$: the polymer gains mass on hydrolysis, so each gram of starch provides more energy than a gram of glucose.

There is no contradiction, provided we distinguish the energy CONTAINED from the energy ACCESSIBLE. Cellulose does contain the same energy of combustion as starch, give or take little, and a log of cellulose burns very well in a fireplace. But the human body can release this energy only by first hydrolyzing the polymer into glucose, and our enzymatic equipment contains NO $\beta$-glucosidase capable of attacking the $\beta(1\rightarrow 4)$ bonds. Amylases, tailored to the $\alpha$ geometry, do not recognize the $\beta$ geometry: cellulose therefore passes through the digestive tract intact. The nutritional value of $0$ kilocalories per gram is not a property of the molecule, it is a property of the ENCOUNTER between this molecule and our toolkit of enzymes. This does not make it a useless compound: fibers increase the bulk of the food mass, regulate transit and slow the absorption of sugars, and a fraction is fermented by the colon microbiota into short-chain fatty acids which do provide a little energy.

The molecule is indeed the same, but it is not equally ACCESSIBLE. In a whole wheat grain, the starch is enclosed in compact, partially crystalline granules, themselves trapped in a protein matrix and surrounded by a cellulosic fiber envelope. Amylases must therefore physically reach the chains before they can cut them, and this diffusion is slow: hydrolysis is spread over hours and glucose is released gradually. In mashed potatoes, prolonged cooking in the presence of water has made the granules SWELL and burst, a phenomenon called gelatinization that unwinds the chains and exposes them completely, while the mashing has destroyed any residual structure. The surface accessible to the enzymes is then immense, hydrolysis is very fast, and glucose reaches the blood in waves. This is exactly the notion of GLYCEMIC INDEX, which measures the speed and not the quantity: mashed potatoes exceed $80$ on a scale where glucose is $100$, whereas al dente pasta stays below $50$ for the same carbohydrate intake. The limiting factor is therefore not the chemistry but the physical access to the substrate, a remark that also applies to any heterogeneous catalysis.

The ruminant produces no more cellulase than we do: it HOUSES, in its rumen, an ecosystem of bacteria, protozoa and fungi that produce it. These microorganisms hydrolyze cellulose into glucose, which they immediately ferment into short-chain fatty acids, acetate, propionate and butyrate, which the animal absorbs through the wall of the rumen and uses as its main energy source. It also digests the microorganisms themselves further along its digestive tract, which supplies its proteins. This solution is impractical for humans for three cumulative reasons. First the VOLUME: the rumen holds several hundred liters in a cow, a capacity no human abdomen could house. Next the TIME: fermentation of cellulose takes tens of hours, whereas human transit is much faster. Finally the anatomical POSITION: the rumen comes before the true stomach, which allows the animal to digest the microorganisms afterwards, whereas our own fermentative flora is confined to the colon, downstream of any significant absorption of nutrients. A purely cellulosic diet would therefore leave a human starving in the midst of an abundance of chemical energy, which is the best possible illustration of the thesis of this chapter: in living things, it is stereochemistry, and not energy, that decides.

Only carbon 1 changes: the aldehyde becomes a carboxylic acid. We obtain D-gluconic acid, $C_{6}H_{12}O_{7}$, $M=6\times 12.01+12\times 1.008+7\times 16.00=196.16$ g/mol.

Two hydrogens add across the carbonyl: $C_{6}H_{14}O_{6}$. Sorbitol no longer has a carbonyl or a hemiacetal: it can no longer open to an aldehyde and does not reduce Fehling's solution.

In the diacid, the two ends have become identical, two $COOH$ groups. For D-galactose, the $OH$ groups are on the right, left, left, right: the projection is symmetric about a horizontal plane passing between carbons 3 and 4. The molecule is superimposable on its mirror image: it is a meso compound, inactive. For D-glucose, right, left, right, right does not have this symmetry: the diacid is chiral and active. This is one of the arguments Emil Fischer used to assign the configuration of sugars.

The $OH$ of carbon 2 must be on the same side as that of carbon 5, on the right for the D series, and those of carbons 3 and 4 must be on the same side as each other. Two choices remain for these two carbons, both on the right, or both on the left: D-allose and D-galactose. So 2 aldohexoses.

No. Galactitol also has two identical ends, two $CH_{2}OH$ groups, and the same symmetry as galactaric acid: it is a meso compound, optically inactive.

Each guanine is paired with a cytosine: 22 % cytosine. That leaves $100-44=56$ % for the A-T pairs, shared equally: 28 % adenine and 28 % thymine. These are Chargaff's rules.

GC pairs: $0.44\times 1000=440$, with 3 bonds each, that is 1,320. AT pairs: 560, with 2 bonds each, that is 1,120. Total: 2,440 hydrogen bonds.

$T_{m}=69.3+0.41\times 44=87.3$ °C; for 60 % GC, $69.3+0.41\times 60=93.9$ °C. A GC pair is held by three hydrogen bonds instead of two, and the stacking of GC pairs is also more stabilizing: a GC-rich DNA requires more energy to separate.

Adenine does not equal thymine, nor guanine cytosine: Chargaff's rules are not obeyed. The bases are therefore not paired: the DNA of this virus is SINGLE-stranded. This is the case of bacteriophage $\varphi$X174.

In basic medium, the $OH$ on carbon 2, immediately next to the phosphate bonded to carbon 3, is deprotonated; the alkoxide attacks the phosphorus and forms a cyclic phosphate, cutting the chain. This is a fast intramolecular reaction, which DNA, lacking this $OH$, cannot undergo. DNA is therefore much more stable: it is the molecule for storing information, RNA that of short-lived messages.

$M=27\times 12.01+46\times 1.008+16.00=386.64$ g/mol. $DU=\frac{2(27)+2-46}{2}=5$: four rings and one double bond, exactly the structure described.

$2^{8}=256$ stereoisomers would be possible, with no internal symmetry to subtract any. Nature makes only one: each step of its biosynthesis is catalyzed by a chiral enzyme that imposes the configuration of each center.

It is a very unbalanced amphiphilic molecule: a small polar $OH$ group and a large rigid hydrocarbon skeleton. In the lipid bilayer, the $OH$ sits near the polar heads of the phospholipids, on the water side, and the flat skeleton sinks in among the fatty chains, which it stiffens.

Esterification: $C_{27}H_{46}O+C_{18}H_{34}O_{2}\rightarrow C_{45}H_{78}O_{2}+H_{2}O$. $M=386.64+282.46-18.02=651.08$ g/mol. The ester has lost the only $OH$ group of cholesterol: it no longer has any hydrogen-bond donor site and is even more hydrophobic. It is in this form that it is carried in the core of lipoproteins.

$5.2\times 10^{-3}\times 386.64=2.0$ g/L.

A primary amine, a carboxylic acid, an amide, the peptide bond, and a methyl ester. The two alpha carbons, that of aspartic acid and that of phenylalanine, are asymmetric: 2 centers, $2^{2}=4$ stereoisomers. Only the L,L form is sweet; forms containing a D residue are bitter or tasteless, because the sweet taste receptor is a chiral protein.

Hydrolysis cleaves the amide and the ester: aspartic acid, phenylalanine and methanol. $n=\frac{0.200}{294.30}=6.80\times 10^{-4}$ mol, the same amount of methanol, that is $6.80\times 10^{-4}\times 32.04=0.0218$ g, 21.8 mg. This is a quantity of the same order as that naturally provided by a glass of fruit juice.

People with phenylketonuria cannot break down phenylalanine, which accumulates and is toxic to the brain. Hydrolysis of aspartame releases it: they must avoid it.

The amino group of aspartic acid attacks the carbonyl of the phenylalanine ester, and methanol is expelled: intramolecular nucleophilic acyl substitution. The ring contains the two nitrogens, the two carbonyl carbons and the two alpha carbons: 6 atoms, a diketopiperazine. This is why aspartame is not suitable for baking.

$\frac{40}{200}=0.20$ g of aspartame. Energy: $40\times 4=160$ kcal for sugar, $0.20\times 4=0.8$ kcal for aspartame, two hundred times less, at equal sweetening power.

Carbon 1, the anomeric carbon: the hemiacetal is oxidized to a cyclic ester. Gluconolactone is a LACTONE, which is then hydrolyzed to gluconic acid.

The enzyme recognizes only the $\beta$ anomer. A fresh solution of $\alpha$-glucose contains almost none at the start, and spontaneous mutarotation takes hours to reach 64 %: the standard would appear less concentrated than it is. Mutarotase catalyzes the $\alpha\rightleftharpoons\beta$ equilibrium: the $\alpha$ anomer is converted as it goes, and all the glucose is measured.

Absorbance is proportional to concentration: $c=5.00\times\frac{0.588}{0.420}=7.00$ mmol/L.

$7.00\times 10^{-3}\times 180.16=1.26$ g/L, that is 126 mg/dL. The fasting blood glucose reaches the threshold of 7.0 mmol/L: the result suggests diabetes, to be confirmed by a second measurement.

Fehling's solution reacts with any reducing sugar and with other reducing substances in blood, such as ascorbic acid or uric acid: it would overestimate blood glucose. The enzyme, chiral and specific to a single stereoisomer, sees only glucose.