How can we digest lactose even though it has Beta glycosidic linkages?

How can we digest lactose even though it has Beta glycosidic linkages?

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I have Read that we cannot digest cellulose because we do not have enzymes to digest Beta glycosidic linkages in Cellulose

Then how is it that we have an enzyme called Lactase to digest the Beta glycosidic linkages in lactose?

Also If we have evolutionarily developed Lactase ,Why didnt we develop "Cellulase " too ?

One simple reason is physical interaction. lactose is only composed of two sugars, while cellulose is giant network of interlocked sugars. Cellulose keeps the sugars tightly packed in three dimensions. even getting to the bonding site is difficult.

human and many other lactases are a huge molecules that would never be able to fit on to a cellulose chain, while cellulase of often a tiny molecule as enzymes go and the active site is not buried in the center. Compare the two enzymes.


two different cellulase enzymes

Note the big difference in size. Also look at cellulase in action.

Notice it has to peel up the cellulose strip by strip, while lactose is just floating free in solution. This even limits the speed of cellulose breakdown. which means organisms that breakdown cellulose have to spend a log time doing it, while humans have a relatively short and fast digestive timeframe, so even if we could we would not get much from it.

These functional limits also makes a cellulase enzyme harder to evolve, since it can't be too large or in organisms with fast digestive processes. We often forget the physical necessities of enzyme function. there are plenty of other reasons enzymes do not function on multiple substrates, but this is why is basically impossible to for evolution to change lactase into a cellulose cleaving enzyme.

24.7: Disaccharides and Glycosidic Bonds

  1. identify disaccharides as compounds consisting of two monosaccharide units joined by a glycoside link between the C1 of one sugar and one of the hydroxyl groups of a second sugar.
  2. identify the two monosaccharide units in a given disaccharide.
  3. identify the type of glycoside link (e.g., 1,4&prime‑&beta) present in a given disaccharide structure.
  4. draw the structure of a specific disaccharide, given the structure of the monosaccharide units and the type of glycoside link involved.

Note: If &alpha‑ or &beta‑D‑glucose were one of the monosaccharide units, its structure would not be provided.

Make certain that you can define, and use in context, the key terms below.

Notice that most of the disaccharides discussed in this section contain one unit of D-glucose. You are not expected to remember the detailed structures of maltose, lactose and sucrose. Similarly, we do not expect you to remember the systematic names of these substances.

Previously, you learned that monosaccharides can form cyclic structures by the reaction of the carbonyl group with an OH group. These cyclic molecules can in turn react with another alcohol. Disaccharides (C12H22O11) are sugars composed of two monosaccharide units that are joined by a carbon&ndashoxygen-carbon linkage known as a glycosidic linkage . This linkage is formed from the reaction of the anomeric carbon of one cyclic monosaccharide with the OH group of a second monosaccharide.

The disaccharides differ from one another in their monosaccharide constituents and in the specific type of glycosidic linkage connecting them. There are three common disaccharides: maltose, lactose, and sucrose. All three are white crystalline solids at room temperature and are soluble in water. We&rsquoll consider each sugar in more detail.

How can we digest lactose even though it has Beta glycosidic linkages? - Biology

The biological macromolecules are grouped into four classes of molecules that play important roles in cells and in organisms as a whole. All of them are polymers strings of repeated units (monomers).

This chapter discusses the biomolecules from a biological perspective: what they are made of, how they are put together, and what their roles are in the body. These molecules are also discussed in MCAT Organic Chemistry Review (Chapter 7) from an organic chemistry perspective: nomenclature, chirality, etc.


Proteins are biological macromolecules that act as enzymes, hormones, receptors, channels, transporters, antibodies, and support structures inside and outside cells. Proteins are composed of twenty different amino acids linked together in polymers. The composition and sequence of amino acids in the polypeptide chain is what makes each protein unique and able to fulfill its special role in the cell. Here, we will start with amino acids, the building blocks of proteins, and work our way up to three-dimensional protein structure and function.

Amino Acid Structure and Nomenclature

Understanding the structure of amino acids is key to understanding both their chemistry and the chemistry of proteins. The generic formula for all twenty amino acids is shown below.

Figure 1 Generic Amino Acid Structure

All twenty amino acids share the same nitrogen-carbon-carbon backbone. The unique feature of each amino acid is its side chain (variable R-group), which gives it the physical and chemical properties that distinguish it from the other nineteen.

Classification of Amino Acids

Each of the twenty amino acids is unique because of its side chain. Each amino acid has a three-letter abbreviation and a one-letter abbreviation, which you do not need to memorize. Though they are all unique, many of them are similar in their chemical properties. It is not necessary to memorize all 20 side chains, but it is important to understand the chemical properties that characterize them. The important properties of the side chains include their varying shape, charge, ability to hydrogen bond, and ability to act as acids or bases. These side group properties are important in the structure of proteins.

We now consider the 20 amino acids, organizing them into broad categories:

Hydrophobic (Nonpolar) Amino Acids

Hydrophobic amino acids have either aliphatic (straight-chain) or aromatic (ring structure) side chains. Amino acids with aliphatic side chains include glycine, alanine, valine, leucine, and isoleucine. Amino acids with aromatic side chains include phenylalanine, tyrosine, and tryptophan. Hydrophobic residues tend to associate with each other rather than with water, and therefore are found on the interior of folded globular proteins, away from water. The larger the hydrophobic group, the greater the hydrophobic force repelling it from water.

Polar Amino Acids

These amino acids are characterized by an R-group that is polar enough to form hydrogen bonds with water but not polar enough to act as an acid or base. This means they are hydrophilic and will interact with water whenever possible.

The hydroxyl groups of serine, threonine, and tyrosine residues are often modified by the attachment of a phosphate group by a regulatory enzyme called a kinase. The result is a change in structure due to the very hydrophilic phosphate group. This modification is an important means of regulating protein activity.

Acidic Amino Acids

Glutamic acid and aspartic acid are the only amino acids with carboxylic acid functional groups (pKa &asymp 4) in their side chains, thereby making the side chains acidic. Thus, there are three functional groups in these amino acids that may act as acids or bases&mdashthe two backbone groups and the R-group. You may hear the terms glutamate and aspartate&mdashthese simply refer to the anionic (unprotonated) form of the molecule.

Basic Amino Acids

Lysine, arginine, and histidine have basic R-group side chains. The pKas for the side chains in these amino acids are 10 for Lys, 12 for Arg, and 6.5 for His. Histidine is unique in having a side chain with a pKa so close to physiological pH. At pH 7.4 histidine may be either protonated or deprotonated&mdashwe put it in the basic category, but it often acts as an acid, too. This makes it a readily available proton acceptor or donor, explaining its prevalence at protein active sites (discussed below). A mnemonic is &ldquoHis goes both ways.&rdquo This contrasts with amino acids containing &ndashCOOH or &ndashNH2 side chains, which are always anionic (RCOO &ndash ) or cationic (RNH3 + ) at physiological pH. (By the way, histamine is a small molecule that has to do with allergic responses, itching, inflammation, and other processes. (You&rsquove heard of antihistamine drugs.) It is not an amino acid don&rsquot confuse it with histidine.)

Sulfur-Containing Amino Acids

Amino acids with sulfur-containing side chains include cysteine and methionine. Cysteine, which contains a thiol (also called a sulfhydryl&mdashlike an alcohol that has an S atom instead of an O atom), is actually fairly polar, and methionine, which contains a thioether (like an ether that has an S atom instead of an O atom) is fairly nonpolar.

Proline is unique among the amino acids in that its amino group is bound covalently to a part of the side chain, creating a secondary &alpha-amino group and a distinctive ring structure. This unique feature of proline has important consequences for protein folding (see Section 3.2).

Table 1 Summary Table of Amino Acids

&bull Which of the following amino acids is most likely to be found on the exterior of a protein at pH 7.0? 1


There are two common types of covalent bonds between amino acids in proteins: the peptide bonds that link amino acids together into polypeptide chains and disulfide bridges between cysteine R-groups.

The Peptide Bond

Polypeptides are formed by linking amino acids together in peptide bonds. A peptide bond is formed between the carboxyl group of one amino acid and the &alpha-amino group of another amino acid with the loss of water. The Figure below shows the formation of a dipeptide from the amino acids glycine and alanine.

Figure 2 Peptide Bond (Amide Bond) Formation

In a polypeptide chain, the N&ndashC&ndashC&ndashN&ndashC&ndashC pattern formed from the amino acids is known as the backbone of the polypeptide. An individual amino acid is termed a residue when it is part of a polypeptide chain. The amino terminus is the first end made during polypeptide synthesis, and the carboxy terminus is made last. Hence, by convention, the amino-terminal residue is also always written first.

&bull In the oligopeptide Phe-Glu-Gly-Ser-Ala, state the number of acid and base functional groups, which residue has a free &alpha-amino group, and which residue has a free &alpha-carboxyl group. (Refer to the beginning of the chapter for structures.) 2

&bull Thermodynamics states that free energy must decrease for a reaction to proceed spontaneously and that such a reaction will spontaneously move toward equilibrium. The diagram below shows the free energy changes during peptide bond formation. At equilibrium, which is thermodynamically favored: the dipeptide or the individual amino acids? 3

&bull In that case, how are peptide bonds formed and maintained inside cells? 4

Hydrolysis of a protein by another protein is called proteolysis or proteolytic cleavage, and the protein that does the cutting is known as a proteolytic enzyme or protease. Proteolytic cleavage is a specific means of cleaving peptide bonds. Many enzymes only cleave the peptide bond adjacent to a specific amino acid. For example, the protease trypsin cleaves on the carboxyl side of the positively charged (basic) residues arginine and lysine, while chymotrypsin cleaves adjacent to hydrophobic residues such as phenylalanine. (Do not memorize these examples.)

Figure 3 Specificity of Protease Cleavage

&bull Based on the above, if the following peptide is cleaved by trypsin, what amino acid will be on the new N-terminus and how many fragments will result: Ala-Gly-Glu-Lys-Phe-Phe-Lys? 5

The Disulfide Bond

Cysteine is an amino acid with a reactive thiol (sulfhydryl, SH) in its side chain. The thiol of one cysteine can react with the thiol of another cysteine to produce a covalent sulfur-sulfur bond known as a disulfide bond, as illustrated below. The cysteines forming a disulfide bond may be located in the same or different polypeptide chain(s). The disulfide bridge plays an important role in stabilizing tertiary protein structure this will be discussed in the section on protein folding. Once a cysteine residue becomes disulfide-bonded to another cysteine residue, it is called cystine instead of cysteine.

Figure 4 Formation of the Disulfide Bond

&bull Which is more oxidized, the sulfur in cysteine or the sulfur in cystine? 6

&bull The inside of cells is known as a reducing environment because cells possess antioxidants (chemicals that prevent oxidation reactions). Where would disulfide bridges be more likely to be found, in extracellular proteins, under oxidizing conditions, or in the interior of cells, in a reducing environment? 7

Protein Structure in Three Dimensions

Each protein folds into a unique three-dimensional structure that is required for that protein to function properly. Improperly folded, or denatured, proteins are non-functional. There are four levels of protein folding that contribute to their final three-dimensional structure. Each level of structure is dependent upon a particular type of bond, as discussed in the following sections.

Denaturation is an important concept. It refers to the disruption of a protein&rsquos shape without breaking peptide bonds. Proteins are denatured by urea (which disrupts hydrogen bonding interactions), by extremes of pH, by extremes of temperature, and by changes in salt concentration (tonicity).

Primary (1 o ) Structure: The Amino Acid Sequence

The simplest level of protein structure is the order of amino acids bonded to each other in the polypeptide chain. This linear ordering of amino acid residues is known as primary structure. Primary structure is the same as sequence. The bond which determines 1 &omicron structure is the peptide bond, simply because this is the bond that links one amino acid to the next in a polypeptide.

Secondary (2 o ) Structure: Hydrogen Bonds Between Backbone Groups

Secondary structure refers to the initial folding of a polypeptide chain into shapes stabilized by hydrogen bonds between backbone NH and CO groups. Certain motifs of secondary structure are found in most proteins. The two most common are the &alpha-helix and the &beta-pleated sheet.

All &alpha-helices have the same well-defined dimensions that are depicted below with the R-groups omitted for clarity. The &alpha-helices of proteins are always right handed, 5 angstroms in width, with each subsequent amino acid rising 1.5 angstroms. There are 3.6 amino acid residues per turn with the &alpha-carboxyl oxygen of one amino acid residue hydrogen-bonded to the &alpha-amino proton of an amino acid three residues away. (Don&rsquot memorize these numbers, but do try to visualize what they mean.)

Figure 5 An &alpha Helix

The unique structure of proline forces it to kink the polypeptide chain hence proline residues never appear within the &alpha-helix.

Proteins such as hormone receptors and ion channels are often found with &alpha-helical transmembrane regions integrated into the hydrophobic membranes of cells. The &alpha-helix is a favorable structure for a hydrophobic transmembrane region because all polar NH and CO groups in the backbone are hydrogen bonded to each other on the inside of the helix, and thus don&rsquot interact with the hydrophobic membrane interior. &alpha-Helical regions that span membranes also have hydrophobic R-groups, which radiate out from the helix, interacting with the hydrophobic interior of the membrane.

&beta-Pleated sheets are also stabilized by hydrogen bonding between NH and CO groups in the polypeptide backbone. In &beta-sheets, however, hydrogen bonding occurs between residues distant from each other in the chain or even on separate polypeptide chains. Also, the backbone of a &beta-sheet is extended, rather than coiled, with side groups directed above and below the plane of the &beta-sheet. There are two types of &beta-sheets, one with adjacent polypeptide strands running in the same direction (parallel &beta-pleated sheet) and another in which the polypeptide strands run in opposite directions (antiparallel &beta-pleated sheet).

Figure 6 A &beta-Pleated Sheet

&bull If a single polypeptide folds once and forms a &beta-pleated sheet with itself, would this be a parallel or antiparallel &beta-pleated sheet? 8

&bull What effect would a molecule that disrupts hydrogen bonding, e.g., urea, have on protein structure? 9

Tertiary (3 o ) Structure: Hydrophobic/Hydrophilic Interactions

The next level of protein folding, tertiary structure, concerns interactions between amino acid residues located more distantly from each other in the polypeptide chain. The folding of secondary structures such as &alpha-helices into higher order tertiary structures is driven by interactions of R-groups with each other and with the solvent (water). Hydrophobic R-groups tend to fold into the interior of the protein, away from the solvent, and hydrophilic R-groups tend to be exposed to water on the surface of the protein (shown for the generic globular protein).

Figure 7 Folding of a Globular Protein in Aqueous Solution

Under the right conditions, the forces driving hydrophobic avoidance of water and hydrogen bonding will fold a polypeptide spontaneously into the correct conformation, the lowest energy conformation. In a classic experiment by Christian Anfinsen and coworkers, the effect of a denaturing agent (urea) and a reducing agent (&beta-mercaptoethanol) on the folding of a protein called ribonuclease were examined. In the following questions, you will reenact their thought processes. Figure out the answers before reading the footnotes.

&bull Ribonuclease has eight cysteines that form four disulfides bonds. What effect would a reducing agent have on its tertiary structure? 10

&bull If the disulfides serve only to lock into place a tertiary protein structure that forms first on its own, then what effect would the reducing agent have on correct protein folding? 11

&bull Would a protein end up folded normally if you (1) first put it in a reducing environment, (2) then denatured it by adding urea, (3) next removed the reducing agent, allowing disulfide bridges to reform, and (4) finally removed the denaturing agent? 12

&bull What if you did the same experiment but in this order: 1, 2, 4, 3? 13

The disulfide bridge is not a great example of 3° structure because it is a covalent bond, not a hydrophobic interaction. However, because the disulfide is formed after 2° structure and before 4° structure, it is usually considered part of 3° folding.

&bull Which of the following may be considered an example of tertiary protein structure? 14

I. van der Waals interactions between two Phe R-groups located far apart on a polypeptide

II. Hydrogen bonds between backbone amino and carboxyl groups

III. Covalent disulfide bonds between cysteine residues located far apart on a polypeptide

Quaternary (4 o ) Structure: Various Bonds Between Separate Chains

The highest level of protein structure, quaternary structure, describes interactions between polypeptide subunits. A subunit is a single polypeptide chain that is part of a large complex containing many subunits (a multisubunitcomplex). The arrangement of subunits in a multisubunit complex is what we mean by quaternary structure. For example, mammalian RNA polymerase II contains twelve different subunits. The interactions between subunits are instrumental in protein function, as in the cooperative binding of oxygen by each of the four subunits of hemoglobin.

The forces stabilizing quaternary structure are generally the same as those involved in secondary and tertiary structure&mdashnon-covalent interactions (the hydrogen bond, and the van der Waals interaction). However, covalent bonds may also be involved in quaternary structure. For example, antibodies (immune system molecules) are large protein complexes with disulfide bonds holding the subunits together. It is key to understand, however, that there is one covalent bond that may not be involved in quaternary structure&mdashthe peptide bond&mdashbecause this bond defines sequence (1° structure).

&bull What is the difference between a disulfide bridge involved in quaternary structure and one involved in tertiary structure? 15


Carbohydrates can be broken down to CO2 in a process called oxidation, which is also known as burning or combustion. Because this process releases large amounts of energy, carbohydrates generally serve as the principle energy source for cellular metabolism. Glucose in the form of the polymer cellulose is also the building block of wood and cotton. Understanding the nomenclature, structure, and chemistry of carbohydrates is essential to understanding cellular metabolism. This section will also help you understand key facts such as why we can eat potatoes and cotton candy but not wood and cotton T-shirts, and why milk makes some adults flatulent.

Monosaccharides and Disaccharides

A single carbohydrate molecule is called a monosaccharide (meaning &ldquosingle sweet unit&rdquo), also known as a simple sugar. Monosaccharides have the general chemical formula CnH2nOn.

Figure 8 Some Metabolically Important Monosaccharides

Two monosaccharides bonded together form a disaccharide, a few form an oligosaccharide, and many form a polysaccharide. The bond between two sugar molecules is called a glycosidic linkage. This is a covalent bond, formed in a dehydration reaction that requires enzymatic catalysis.

Figure 9 Disaccharides and the &alpha- or &beta-Glycosidic Bond

Glycosidic linkages are named according to which carbon in each sugar comprises the linkage. The configuration (&alpha or &beta) of the linkage is also specified. For example, lactose (milk sugar) is a disaccharide joined in a galactose-&beta-1,4-glucose linkage (above). Sucrose (Table sugar) is also shown above, with a glucose unit and a fructose unit.

&bull Does sucrose contain an &alpha- or &beta-glycosidic linkage? 16

Some common disaccharides you might see on the MCAT are sucrose (Glc-&alpha-1,2-Fru), lactose (Gal-&beta-1,4-Glc), maltose (Glc-&alpha-1,4-Glc), and cellobiose (Glc-&beta-1,4-Glc). However, you should NOT try to memorize these linkages.

Polymers made from these disaccharides form important biological macromolecules. Glycogen serves as an energy storage carbohydrate in animals and is composed of thousands of glucose units joined in &alpha‑1,4 linkages &alpha‑1,6 branches are also present. Starch is the same as glycogen (except that the branches are a little different), and serves the same purpose in plants. Cellulose is a polymer of cellobiose but note that cellobiose does not exist freely in nature. It exists only in its polymerized, cellulose form. The &beta-glycosidic bonds allow the polymer to assume a long, straight, fibrous shape. Wood and cotton are made of cellulose.

Hydrolysis of Glycosidic Linkages

The hydrolysis of polysaccharides into monosaccharides is favored thermodynamically. Hydrolysis is essential in order for these sugars to enter metabolic pathways (e.g., glycolysis) and be used for energy by the cell. However, this hydrolysis does not occur at a significant rate without enzymatic catalysis. Different enzymes catalyze the hydrolysis of different linkages. The enzymes are named for the sugar they hydrolyze. For example, the enzyme that catalyzes the hydrolysis of maltose into two glucose monosaccharides is called maltase. Each enzyme is highly specific for its linkage.

This specificity is a great example of the significance of stereochemistry. Consider cellulose. A cotton T-shirt is pure sugar. The only reason we can&rsquot digest it is that mammalian enzymes generally can&rsquot break the &beta-glycosidic linkages found in cellulose. Cellulose is actually the energy source in grass and hay. Cows are mammals, and all mammals lack the enzymes necessary for cellulose breakdown. To live on grass, cows depend on bacteria that live in an extra stomach called a rumen to digest cellulose for them. If you&rsquore really on the ball, you&rsquore next question is: Humans are mammals, so how can we digest lactose, which has a &beta linkage? The answer is that we have a specific enzyme, lactase, which can digest lactose. This is an exception to the rule that mammalian enzymes cannot hydrolyze &beta-glycosidic linkages. People without lactase are lactose malabsorbers, and any lactose they eat ends up in the colon. There it may cause gas and diarrhea, if certain bacteria are present people with this problem are said to be lactose intolerant. People produce lactase as children so that they can digest mother&rsquos milk, but most adults naturally stop making this enzyme, and thus become lactose malabsorbers and sometimes intolerant.

Figure 10 The Polysaccharide Glycogen

&bull Which requires net energy input: polysaccharide synthesis or hydrolysis? 17

&bull If the activation energy of polysaccharide hydrolysis were so low that no enzyme was required for the reaction to occur, would this make polysaccharides better for energy storage? 18

Lipids are oily or fatty substances that play three physiological roles, summarized here and discussed below.

1) In adipose cells, triglycerides (fats) store energy.

2) In cellular membranes, phospholipids constitute a barrier between intracellular and extracellular environments.

3) Cholesterol is a special lipid that serves as the building block for the hydrophobic steroid hormones.

The cardinal characteristic of the lipid is its hydrophobicity. Hydrophobic means water-fearing. It is important to understand the significance of this. Since water is very polar, polar substances dissolve well in water these are known as water-loving, or hydrophilic substances. Carbon-carbon bonds and carbon-hydrogen bonds are nonpolar. Hence, substances that contain only carbon and hydrogen will not dissolve well in water. Some examples: Table sugar dissolves well in water, but cooking oil floats in a layer above water or forms many tiny oil droplets when mixed with water. Cotton T-shirts become wet when exposed to water because they are made of glucose polymerized into cellulose, but a nylon jacket does not become wet because it is composed of atoms covalently bound together in a nonpolar fashion. A synonym for hydrophobic is lipophilic (which means lipid-loving) a synonym for hydrophilic is lipophobic. We return to these concepts below.

Fatty Acid Structure

Fatty acids are composed of long unsubstituted alkanes that end in a carboxylic acid. The chain is typically 14 to 18 carbons long, and because they are synthesized two carbons at a time from acetate, only even-numbered fatty acids are made in human cells. A fatty acid with no carbon-carbon double bonds is said to be saturated with hydrogen because every carbon atom in the chain is covalently bound to the maximum number of hydrogens. Unsaturated fatty acids have one or more double bonds in the tail. These double bonds are almost always (Z) (or cis).

&bull How does the shape of an unsaturated fatty acid differ from that of a saturated fatty acid? 19

&bull If fatty acids are mixed into water, how are they likely to associate with each other? 20

The drawing on the next page illustrates how free fatty acids interact in an aqueous solution they form a structure called a micelle. The force that drives the tails into the center of the micelle is called the hydrophobic interaction. The hydrophobic interaction is a complex phenomenon. In general, it results from the fact that water molecules must form an orderly solvation shell around each hydrophobic substance. The reason is that H2O has a dipole that &ldquolikes&rdquo to be able to share its charges with other polar molecules. A solvation shell allows for the most water-water interaction and the least water-lipid interaction. The problem is that forming a solvation shell is an increase in order and thus a decrease in entropy (∆S < 0), which is unfavorable according to the second law of thermodynamics. In the case of the fatty acid micelle, water forms a shell around the spherical micelle with the result being that water interacts with polar carboxylic acid head groups while hydrophobic lipid tails hide inside the sphere.

Figure 11 A Fatty Acid Micelle

&bull How does soap help to remove grease from your hands? 21

Triacylglycerols (TG)

The storage form of the fatty acid is fat. The technical name for fat is triacylglycerol or triglyceride (shown below). The triglyceride is composed of three fatty acids esterified to a glycerol molecule. Glycerol is a three-carbon triol with the formula HOCH2&ndashCHOH&ndashCH2OH. As you can see, it has three hydroxyl groups that can be esterified to fatty acids. It is necessary to store fatty acids in the relatively inert form of fat because free fatty acids are reactive chemicals.

Figure 12 A Triglyceride (Fat)

The triacylglycerol undergoes reactions typical of esters, such as base-catalyzed hydrolysis. Soaps are the sodium salts of fatty acids (RCOO-Na + ). They are amphipathic, which means they have both hydrophilic and hydrophobic regions. Soap is economically produced by base-catalyzed hydrolysis of triglycerides from animal fat into fatty acid salts (soaps). This reaction is called saponification and is illustrated below.

Figure 13 Saponification

Lipases are enzymes that hydrolyze fats. Triacylglycerols are stored in fat cells as an energy source. Fats are more efficient energy storage molecules than carbohydrates for two reasons: packing and energy content.

1) Packing: Their hydrophobicity allows fats to pack together much more closely than carbohydrates. Carbohydrates carry a great amount of water-of-solvation (water molecules hydrogen bonded to their hydroxyl groups). In other words, the amount of carbon per unit area or unit weight is much greater in a fat droplet than in dissolved sugar. If we could store sugars in a dry powdery form in our bodies, this problem would be obviated.

2) Energy content: All packing considerations aside, fat molecules store much more energy than carbohydrates. In other words, regardless of what you dissolve it in, a fat has more energy carbon-for-carbon than a carbohydrate. The reason is that fats are much more reduced. Remember that energy metabolism begins with the oxidation of foodstuffs to release energy. Since carbohydrates are more oxidized to start with, oxidizing them releases less energy. Animals use fat to store most of their energy, storing only a small amount as carbohydrates (glycogen). Plants such as potatoes commonly store a large percentage of their energy as carbohydrates (starch).

Introduction to Lipid Bilayer Membranes

Membrane lipids are phospholipids derived from diacylglycerol phosphate or DG-P. For example, phosphatidyl choline is a phospholipid formed by the esterification of a choline molecule [HO(CH2)2N + (CH3)3] to the phosphate group of DG-P. Phospholipids are detergents, substances that efficiently solubilize oils while remaining highly water-soluble. Detergents are like soaps, but stronger.

Figure 14 A Phosphoglyceride (Diacylglycerol Phosphate, or DGP)

We saw above how fatty acids spontaneously form micelles. Phospholipids also minimize their interactions with water by forming an orderly structure&mdashin this case, it is a lipid bilayer (below). Hydrophobic interactions drive the formation of the bilayer, and once formed, it is stabilized by van der Waals forces between the long tails.

Figure 15 A Small Section of a Lipid Bilayer Membrane

&bull Would a saturated or an unsaturated fatty acid residue have more van der Waals interactions with neighboring alkyl chains in a bilayer membrane? 22

A more precise way to give the answer to the question above is to say that double bonds (unsaturation) in phospholipid fatty acids tend to increase membrane fluidity. Unsaturation prevents the membrane from solidifying by disrupting the orderly packing of the hydrophobic lipid tails. The right amount of fluidity is essential for function. Decreasing the length of fatty acid tails also increases fluidity. The steroid cholesterol (discussed a bit later) is a third important modulator of membrane fluidity. At low temperatures, it increases fluidity in the same way as kinks in fatty acid tails hence, it is known as membrane antifreeze. At high temperatures, however, cholesterol attenuates (reduces) membrane fluidity. Don&rsquot ponder this paradox too long just remember that cholesterol keeps fluidity at an optimum level. Remember, the structural determinants of membrane fluidity are: degree of saturation, tail length, and amount of cholesterol.

The lipid bilayer acts like a plastic bag surrounding the cell in the sense that it separates the interior of the cell from the exterior. However, the cell membrane is much more complex than a plastic bag. Since the plasma bilayer membrane surrounding cells is impermeable to charged particles such as Na + , protein gateways such as ion channels are required for ions to enter or exit cells. Proteins that are integrated into membranes also transmit signals from the outside of the cell into the interior. For example, certain hormones (peptides) cannot pass through the cell membrane due to their charged nature instead, protein receptors in the cell membrane bind these hormones and transmit a signal into the cell in a second messenger cascade (see Chapter 7 for more details on the plasa membrane).

A terpene is a member of a broad class of compounds built from isoprene units (C5H8) with a general formula (C5H8)n.

Terpenes may be linear or cyclic, and are classified by the number of isoprene units they contain. For example, monoterpenes consist of two isoprene units, sesquiterpenes consist of three, and diterpenes contain four.

Squalene is a triterpene (made of six isoprene units), and is a particularly important compound as it is biosynthetically utilized in the manufacture of steroids. Squalene is also a component of earwax.

Whereas a terpene is formally a simple hydrocarbon, there are a number of natural and synthetically derived species that are built from an isoprene skeleton and functionalized with other elements (O, N, S, etc.). These functionalized-terpenes are known as terpenoids. Vitamin A (C20H30O) is an example of a terpenoid.

Steroids are included here because of their hydrophobicity, and, hence, similarity to fats. Their structure is otherwise unique. All steroids have the basic tetracyclic ring system (see below), based on the structure of cholesterol, a polycyclic amphipath. (Polycyclic means several rings, and amphipathic means displaying both hydrophilic and hydrophobic characteristics.)

As discussed above, the steroid cholesterol is an important component of the lipid bilayer. It is both obtained from the diet and synthesized in the liver. It is carried in the blood packaged with fats and proteins into lipoproteins. One type of lipoprotein has been implicated as the cause of atherosclerotic vascular disease, which refers to the build-up of cholesterol &ldquoplaques&rdquo on the inside of blood vessels.

Figure 16 Cholesterol-Derived Hormones

Steroid hormones are made from cholesterol. Two examples are testosterone (an androgen or male sex hormone) and estradiol (an estrogen or female sex hormone). There are no receptors for steroid hormones on the surface of cells because steroids are highly hydrophobic, they can diffuse right through the lipid bilayer membrane into the cytoplasm. The receptors for steroid hormones are located within cells rather than on the cell surface. This is an important point! You must be aware of the contrast between peptide hormones, such as insulin, which exert their effects by binding to receptors at the cell-surface, and steroid hormones, such as estrogen, which diffuse into cells to find their receptors.


Phosphoric acid is an inorganic acid (it does not contain carbon) with the potential to donate three protons. The Kas for the three acid dissociation equilibria are 2.1, 7.2, and 12.4. Therefore, at physiological pH, phosphoric acid is significantly dissociated, existing largely in anionic form.

Figure 17 Phosphoric Acid Dissociation

Phosphate is also known as orthophosphate. Two orthophosphates bound together via an anhydride linkage form pyrophosphate. The P&ndashO&ndashP bond in pyrophosphate is an example of a high-energy phosphate bond. This name is derived from the fact that the hydrolysis of pyrophosphate is thermodynamically extremely favorable. The ∆G° for the hydrolysis of pyrophosphate is about &minus7 kcal/mol. This means that it is a very favorable reaction. The actual ∆G° in the cell is about &minus12 kcal/mol, which is even more favorable.

There are three reasons that phosphate anhydride bonds store so much energy:

1) When phosphates are linked together, their negative charges repel each other strongly.

2) Orthophosphate has more resonance forms and thus a lower free energy than linked phosphates.

3) Orthophosphate has a more favorable interaction with the biological solvent (water) than linked phosphates.

The details are not crucial. What is essential is that you fix the image in your mind of linked phosphates acting like compressed springs, just waiting to fly open and provide energy for an enzyme to catalyze a reaction.

Figure 18 The Hydrolysis of Pyrophosphate


Nucleotides are the building blocks of nucleic acids (RNA and DNA). Each nucleotide contains a ribose (or deoxyribose) sugar group a purine or pyrimidine base joined to carbon number one of the ribose ring and one, two, or three phosphate units joined to carbon five of the ribose ring. The nucleotide adenosine triphosphate (ATP) plays a central role in cellular metabolism in addition to being an RNA precursor. Significantly more information about the structure and function of the nucleic acids RNA and DNA will be provided in Chapter 5.

ATP is the universal short-term energy storage molecule. Energy extracted from the oxidation of foodstuffs is immediately stored in the phosphoanhydride bonds of ATP. This energy will later be used to power cellular processes it may also be used to synthesize glucose or fats, which are longer-term energy storage molecules. This applies to all living organisms, from bacteria to humans. Even some viruses carry ATP with them outside the host cell, though viruses cannot make their own ATP.

Figure 19 Adenosine Triphosphate (ATP)

Chapter 3 Summary

&bull Amino acids (AAs) consist of a tetrahedral a-carbon connected to an amino group, a carboxyl group, and a variable R group, which determines the AA&rsquos properties.

&bull Proteins consist of amino acids linked by peptide bonds, which are very stable. The primary structure of a protein consists of its amino acid sequence.

&bull The secondary structure of proteins (&alpha-helices and &beta-sheets) is formed through hydrogen bonding interactions between atoms in the backbone of the molecule.

&bull The most stable tertiary protein structure generally places polar AA&rsquos on the exterior and nonpolar AA&rsquos on the interior of the protein. This minimizes interactions between nonpolar AA&rsquos and water, while optimizing interactions between side chains inside the protein.

&bull Proteins have a variety of functions in the body including (but not limited to) enzymes, structural roles, hormones, receptors, channels, antibodies, transporters, etc.

&bull The monomer for a carbohydrate is a monosaccharide (simple sugar), with the molecular formula CnH2nOn. The common monosaccharides are glucose, fructose, galactose, ribose, and deoxyribose.

&bull Two monosaccharides joined with a glycosidic linkage form a disaccharide. The common disaccharides are maltose, sucrose, and lactose. Mammals can digest a glycosidic linkages, but generally not blinkages.

&bull Polysaccharides consist of many monosaccharides linked together. Glycogen (animals) and starch (plants) are storage units for glucose and can be broken down for energy. Cellulose is also a glucose polymer, but the beta linkage prevents digestion. It forms wood and cotton.

&bull Lipids are found in several forms in the body, including triglycerides, phospholipids, cholesterol and steroids, and terpenes. Triglycerides and phospholipids are linear, while cholesterol and steroids have a ring structure.

&bull Lipids are hydrophobic. Triglycerides are used for energy storage, phospholipids form membranes, and cholesterol is the precursor to the steroid hormones.

&bull The building blocks of nucleic acids (DNA and RNA) are nucleotides, which are comprised of a pentose sugar, a purine or pyrimidine base, and 2-3 phosphate units.


1. Why is ATP known as a &ldquohigh energy&rdquo structure at neutral pH?

A) It exhibits a large decrease in free energy when it undergoes hydrolytic reactions.

B) The phosphate ion released from ATP hydrolysis is very reactive.

C) It causes cellular processes to proceed at faster rates.

D) Adenine is the best energy storage molecule of all the nitrogenous bases.

2. Which of the following best describes the secondary structure of a protein?

A) Various folded polypeptide chains joining together to form a larger unit

B) The amino acid sequence of the chain

C) The polypeptide chain folding upon itself due to hydrophobic/hydrophilic interactions

D) Peptide bonds hydrogen-bonding to one another to create a sheet-like structure

3. Phenylketonuria (PKU) is an autosomal recessive disorder that results from a deficiency of the enzyme phenylalanine hydroxylase. This enzyme normally converts phenylalanine into tyrosine. PKU results in intellectual disability, growth retardation, fair skin, eczema and a distinct musty body odor. Which of the following is most likely true?

A) Treatment should include a decrease in tyrosine in the diet.

B) The musty body odor is likely caused by a disorder in aromatic amino acid metabolism.

C) Patients with PKU should increase the amount of phenylalanine in their diet.

D) PKU can be acquired by consuming too much aspartame (an artificial sweetener that contains high levels of phenylalanine).

4. A genetic regulator is found to contain a lysine residue that is important for its binding to DNA. If a mutation were to occur such that a different amino acid replaces the lysine at that location, which of the following resulting amino acids would likely be the least harmful to its ability to bind DNA?

5. Increasing the amount of cholesterol in a plasma membrane would lead to an increase in:

B) atherosclerotic plaques.

6. A human space explorer crash-lands on a planet where the native inhabitants are entirely unable to digest glycogen, but are able to digest cellulose. Consequently, they make their clothing out of glycogen-based material. The starving space explorer eats one of the native inhabitants&rsquo shirts and the natives are amazed. Based on this information, which of the following is/are true?

I. The explorer can digest &alpha-glycosidic linkages.

II. The native inhabitants can digest &alpha-glycosidic linkages.

III. The native inhabitants can digest starch.


Photosynthesis is the process plants use to derive energy from sunlight and is associated with a cell&rsquos chloroplasts. The energy is used to produce carbohydrates from carbon dioxide and water. Photosynthesis involves light and dark phases. Figure 1 represents two initial steps associated with the light phase.

The light phase supplies the dark phase with NADPH and a high-energy substrate.

A researcher attempted to produce a photosynthetic system outside the living organism according to the following protocols:

&bull Chloroplasts were extracted from green leaves and ruptured, and their membranes were thereby exposed, then a solution of hexachloroplatinate ions carrying a charge of &minus2 was added.

&bull The structure of the composite was analyzed, and the amount of oxygen produced by the system was measured.

The researcher concluded that the ions were bound to the membrane&rsquos Photosystem 1 site by the attraction of opposite charges. The resulting composite is shown in Figure 2. It was found that the hexachloroplatinate-membrane composite was photosynthetically active.

1. In concluding that the hexachloroplatinate ions were bound to Photosystem 1 due to the attraction of opposite charges, the researchers apparently assumed that the structure of the membrane was:

A) determined solely by hydrophobic bonding.

C) covalently bound to the platinate.

2. Figure 1 indicates that:

A) photoactivation of the chloroplast membrane results in the reduction of the anhydride-containing molecule NADP + .

B) electrons are lost from Photosystem 1 through the conversion of NADPH to NADP + , and are replaced by electrons from Photosystem 2.

C) there is a net gain of electrons by the system.

D) electrons are lost from Photosystem 1 through the conversion of NADP + to NADPH, but are not replaced by electrons from Photosystem 2.

3. In addition to NADPH, the photosynthetic light phase must supply the dark phase with another molecule which stores energy for biosynthesis. Among the following, the molecule would most likely be:

4. If NADP + is fully hydrolyzed to its component bases, phosphates, and sugars, what type of monosaccharide would result?

5. If in a given cell the photosynthetic dark phase were artificially arrested while the light phase proceeded, the cell would most likely experience:

A) decreased levels of NADPH.

B) increased levels of NADPH.

C) increased levels of carbohydrate.

D) increased photoactivation of the chloroplast.

6. To determine the primary structure of the protein portion of Photosystem 1, a series of cleavage reactions was undertaken. To break apart the protein, the most logical action to take would be to:

A) decarboxylate free carboxyl groups.

B) hydrolyze peptide bonds.

C) repolymerize peptide bonds.

D) hydrolyze amide branch points.

7. A researcher examined a sample of the principal substance produced by the photosynthetic dark phase and concluded that he was working with a racemic mixture of glucose isomers. Which of the following experimental findings would be inconsistent with such a conclusion?

A) The sample is composed of carbon, hydrogen, and oxygen only.

B) The sample consists of an aldohexose.

C) The sample rotates the plane of polarized light to the left.

D) The sample is optically inactive.


1. A Choice A is the best because it directly addresses the energetics of ATP hydrolysis. Choice B discusses the reactivity of the released phosphate ion and not the structure of ATP itself, so it can be eliminated. Choice C can be eliminated because it describes the rate of cellular processes not the energy of ATP. Choice D can be eliminated because the structure of adenine is not related to why ATP is a good energy storage molecule.

2. D The secondary structure of proteins is the initial folding of the polypeptide chain into a-helices or b-pleated sheets. Choice A describes the formation of a quaternary protein, choice B can be eliminated because it describes the primary protein structure, and choice C can be eliminated because it describes the tertiary protein structure.

3. B A defect in phenylalanine hydroxylase (or the THB cofactor) would result in a build-up of phenylalanine. This would lead to an excess of phenylalanine byproducts such as phenylacetate, phenyllactate and phenylpyruvate, and a decrease in tyrosine. Therefore, patients with PKU should increase the amount of tyrosine in their diet (it becomes an essential amino acid in this condition choice A is wrong), as well as eliminate phenylalanine from their diet (choice C is wrong). PKU is a genetically acquired disorder (autosomal recessive), as mentioned in the passage, and thus it is not acquired by consuming too much phenylalanine (choice D is wrong). It is true that phenylalanine and its derivatives are aromatic amino acids and that the high levels of these compounds lead to the distinct musty body odor (choice B is correct). Process of elimination (POE) is probably the best method to use in answering this question since it is unclear (without prior knowledge of PKU) what the underlying mechanism of the body odor would be.

4. D While knowing the structures of the different amino acids is unlikely to be important for the MCAT, knowing which of the amino acids are basic (histidine, arginine, lysine) and which are acidic (glutamate, aspartate) is likely to be relevant. In this case, since lysine is basic (and therefore best at binding the negatively charged DNA), one can assume that a mutation resulting in another basic amino acid would cause the least change in its ability to bind DNA. Therefore, a mutation from lysine to arginine would cause the least harm (choice D is correct). A mutation from lysine to glutamate or aspartate (both acidic) would likely cause the most harm to its ability to bind DNA (choices B and C are wrong). Glycine is a neutral amino acid (choice A is wrong).

5. C Plasma membranes can be up to 50% composed of sterols. Sterols help stabilize the membrane at both spectrums of the temperature. At low temperatures, they increase fluidity because the ring structure of cholesterol does not allow for tight phospholipid tail packing. This decreases the temperature at which the membrane would freeze (choice D is wrong). At high temperatures, cholesterol decreases membrane fluidity (the OH group of cholesterol prevents phospholipid dispersion) and permeability (by filling in the &ldquoholes&rdquo between the fatty acid tails, choice A is wrong), thus increasing the temperature at which membranes would melt (choice C is correct). The formation of atherosclerotic plaques, while related to cholesterol, is due to high levels of blood cholesterol, not membrane cholesterol (choice B is wrong).

6. A Item I is true: humans can digest &alpha-glycosidic linkages, such as those found in glycogen. If the natives&rsquo shirts are made of glycogen, our space explorer should have no trouble consuming and digesting them (choice C can be eliminated). Item II Is false: cellulose contains &beta-glycosidic linkages. If the natives can digest cellulose, but not glycogen, then they cannot digest &alpha-glycosidic linkages (choice D can be eliminated). Item III is false: starch also contains a &alpha-glycosidic linkages. If the natives cannot digest glycogen, then they likely cannot digest starch either (choice B can be eliminated and choice A is true).


1. B The passage states that the ion is attracted to Photosystem 1 by the attraction of opposite charges (positively-charged photosystem and negatively-charged hexachloroplatinate ion).

2. A The main result of the light phase, as depicted in Figure 1, is the reduction of NADP + to make NADPH (choice A). Choice B is wrong since NADP + is converted into NADPH, not vice versa. Choice C is incorrect since in any system, mass and charge are conserved. Electrons move from one molecule to another, but they are not created or destroyed in a chemical reaction. Choice D is eliminated since Figure 1 depicts electrons moving from Photosystem 2 to Photosystem 1.

3. D The passage states that the light reactions supply the dark reactions with a &ldquohigh energy substrate&rdquo. The most likely candidate among the choices is ATP.

4. C NADPH contains ribose, a pentose.

5. B The light phase makes NADPH, and the dark phase consumes it. In the absence of the dark phase, NADPH will continue to be produced, but none will be consumed, making NADPH levels rise (choice B). Choice C is wrong since the dark phase is responsible for biosynthesis, such as carbohydrate production, so this will decrease, not increase. Choice D can be eliminated since the amount of light and photoactivation should remain the same.

6. B Proteins are composed of amino acid residues which are joined together by peptide bonds during the translation process. To split the protein into smaller pieces, proteases and chemical reagents act to hydrolyze the peptide bond, reversing the biosynthetic process.

7. C A racemic mixture is one which contains equal quantities of two stereoisomers that rotate plane-polarized light in opposite directions. Since there are equal quantities of both, racemic mixtures are optically inactive. Thus, choice C, which states that the sample rotates light, is inconsistent with the conclusion that the sample is racemic and is the correct answer choice. All other choices are consistent with the conclusion that the sample is a racemic mixture of glucose. Carbohydrates, of which glucose is one, are made of only carbon, hydrogen and oxygen (choice A is consistent and can be eliminated), glucose, with six carbons and a carbonyl group on the 6 th carbon, is an aldohexose (choice B is consistent and can be eliminated), and racemic mixtures do not rotate light (choice D is consistent and can be eliminated).

1 Leucine, alanine, and isoleucine are all hydrophobic residues more likely to be found on the interior than the exterior of proteins. Serine (choice C), which has a hydroxyl group that can hydrogen bond with water, is the correct answer.

2 As stated above, the amino end is always written first. Hence, the oligopeptide begins with an exposed Phe amino group and ends with an exposed Ala carboxyl all the other backbone groups are hitched together in peptide bonds. Out of all the R-groups, there is only one acidic or basic functional group, the acidic glutamate R-group. This R-group plus the two terminal backbone groups gives a total of three acid/base functional groups.

3 The dipeptide has a higher free energy, so its existence is less favorable. In other words, existence of the chain is less favorable than existence of the isolated amino acids.

4 During protein synthesis, stored energy is used to force peptide bonds to form. Once the bond is formed, even though its destruction is thermodynamically favorable, it remains stable because the activation energy for the hydrolysis reaction is so high. In other words, hydrolysis is thermodynamically favorable but kinetically slow.

5 Trypsin will cleave on the carboxyl side of the Lys residue, with Phe on the N-terminus of the new Phe-Phe-Lys fragment. There will be two fragments after trypsin cleavage: Phe-Phe-Lys and Ala-Gly-Glu-Lys.

6 The sulfur in cysteine is bonded to a hydrogen and a carbon the sulfur in cystine is bonded to a sulfur and a carbon. Hence, the sulfur in cystine is more oxidized.

7 In a reducing environment, the S-S group is reduced to two SH groups. Disulfide bridges are found only in extracellular polypeptides, where they will not be reduced. Examples of protein complexes held together by disulfide bridges include antibodies and the hormone insulin.

8 It would be antiparallel because one participant in the &beta-pleated sheet would have a C to N direction, while the other would be running N to C.

9 Putting a protein in a urea solution will disrupt H-bonding, thus disrupting secondary structure by unfolding &alpha-helices and &beta-sheets. It would not affect primary structure, which depends on the much more stable peptide bond. Disruption of 2°, 3°, or 4° structure without breaking peptide bonds is denaturation.

10 The disulfide bridges would be broken. Tertiary structure would be less stable.

11 The shape should not be disrupted if breaking disulfides is the only disturbance. It&rsquos just that the shape would be less sturdy&mdashlike a concrete wall without the rebar.

12 No. If you allow disulfide bridges to form while the protein is still denatured, it will become locked into an abnormal shape.

13 You should end up with the correct structure. In step one, you break the reinforcing disulfide bridges. In step two, you denature the protein completely by disrupting H-bonds. In step four, you allow the H-bonds to reform as stated in the text, normally the correct tertiary structure will form spontaneously if you leave the polypeptide alone. In step three, you reform the disulfide bridges, thus locking the structure into its correct form.

14 This is a simple question provided to clarify the classification of the disulfide bridge. Item I is a good example of 3° structure. Item II is describes 2°, not 3°, structure. Item III describes the disulfide, which is considered to be tertiary because of when it is formed, despite the fact that it is a covalent bond.

15 Quaternary disulfides are bonds that form between chains that aren&rsquot linked by peptide bonds. Tertiary disulfides are bonds that form between residues in the same polypeptide.

16 The anomeric carbon of glucose is pointing down, which means the linkage is &alpha-1,2. So, sucrose is Glc-&alpha-1,2-Fru.

17 Because hydrolysis of polysaccharides is thermodynamically favored, energy input is required to drive the reaction toward polysaccharide synthesis.

18 No, because then polysaccharides would hydrolyze spontaneously (they&rsquod be unstable). The high activation energy of polysaccharide hydrolysis allows us to use enzymes as gatekeepers&mdashwhen we need energy from glucose, we open the gate of glycogen hydrolysis.

19 An unsaturated fatty acid is bent, or &ldquokinked,&rdquo at the cis double bond.

20 The long hydrophobic chains will interact with each other to minimize contact with water, exposing the charged carboxyl group to the aqueous environment.

21 Grease is hydrophobic. It does not wash off easily in water because it is not soluble in water. Scrubbing your hands with soap causes micelles to form around the grease particles.

22 The bent shape of the unsaturated fatty acid means that it doesn&rsquot fit in as well and has less contact with neighboring groups to form van der Waals interactions. Phospholipids composed of saturated fatty acids make the membrane more solid.

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Cellulose, a fibrous carbohydrate found in all plants, is the structural component of plant cell walls. Because the earth is covered with vegetation, cellulose is the most abundant of all carbohydrates, accounting for over 50% of all the carbon found in the vegetable kingdom. Cotton fibrils and filter paper are almost entirely cellulose (about 95%), wood is about 50% cellulose, and the dry weight of leaves is about 10%&ndash20% cellulose. The largest use of cellulose is in the manufacture of paper and paper products. Although the use of noncellulose synthetic fibers is increasing, rayon (made from cellulose) and cotton still account for over 70% of textile production.

Like amylose, cellulose is a linear polymer of glucose. It differs, however, in that the glucose units are joined by &beta-1,4-glycosidic linkages, producing a more extended structure than amylose (part (a) of Figure 5.1.3). This extreme linearity allows a great deal of hydrogen bonding between OH groups on adjacent chains, causing them to pack closely into fibers (part (b) of Figure 5.1.3). As a result, cellulose exhibits little interaction with water or any other solvent. Cotton and wood, for example, are completely insoluble in water and have considerable mechanical strength. Because cellulose does not have a helical structure, it does not bind to iodine to form a colored product.

Figure 5.1.3 : Cellulose. (a) There is extensive hydrogen bonding in the structure of cellulose. (b) In this electron micrograph of the cell wall of an alga, the wall consists of successive layers of cellulose fibers in parallel arrangement.

Cellulose yields D-glucose after complete acid hydrolysis, yet humans are unable to metabolize cellulose as a source of glucose. Our digestive juices lack enzymes that can hydrolyze the &beta-glycosidic linkages found in cellulose, so although we can eat potatoes, we cannot eat grass. However, certain microorganisms can digest cellulose because they make the enzyme cellulase, which catalyzes the hydrolysis of cellulose. The presence of these microorganisms in the digestive tracts of herbivorous animals (such as cows, horses, and sheep) allows these animals to degrade the cellulose from plant material into glucose for energy. Termites also contain cellulase-secreting microorganisms and thus can subsist on a wood diet. This example once again demonstrates the extreme stereospecificity of biochemical processes.

Starch digestion

Starch digestion occurs in reactions catalyzed by enzymes called alpha-amylases. They are endoglycosidases, enzymes which hydrolyze at random α-(1→4) glycosidic bonds inside the chains of both amylopectin than amylose, releasing:

  • maltotriose (a trisaccharide consisted of three units of glucose)
  • alpha-dextrins or alpha-limit dextrins.

Alpha-dextrins are branched oligosaccharides formed by many molecules of glucose linked by α-(1→4) glycosidic bonds and one α-(1→4). Those consisted of 5-6 units, at the end of the digestion of amylopectin by alpha-amylase, account for about one third of the final product.
Only maltose and maltotriose will form from the amylose digestion, not being present branching points.

Glycogen is affected minimally by these reactions because of, after animal death, it goes towards a rapid degradation especially to glucose and then lactic acid.

Alpha-amylase is secreted both by salivary glands, in this case is called “salivary” (sometimes ptyalin), than by exocrine pancreas, in this case is called “pancreatic“.


Starch digestion begins in mouth by salivary alpha-amylase, therefore the rate of mastication and the time of permanence in mouth, however relatively short, are the first factors that affect the interaction between starch and the enzyme and that can improve digestion.


Once in the stomach, that essentially acts as a tank, gastric acidity inactivates salivary alpha-amylase, whose optimal pH is about 7, though the presence of starch may partly protects the enzyme from gastric degradation, allowing the passage with meal into the duodenum, where it may support pancreatic alpha-amylase in the digestive process.
If in adulthood this action has a minimal functional role (see below), in newborn infants, particularly in premature ones, it may be of some utility because of in the first months of life in premature infants the production of pancreatic alpha-amylase is reduced. However, due to the low concentration of salivary alpha-amylase into intestinal lumen, pediatricians and nutritionists recommend to avoid starch from the diet until the baby is about six months of age.

Small intestine

When we pass from the stomach into the small intestine, bicarbonate ion secreted by pancreas (under stimulation of secretin hormone) neutralizes gastric acidity leading pH to about 7, an optimal value for the action of pancreatic enzymes, including alpha-amylase, and intestinal enzymes, and for the residual salivary alpha-amylase. So starch digestion, which occurs mostly in the duodenum, begins again by the action of pancreatic alpha-amylase, secreted in amounts greatly exceeding than the digestive needs (in reply to meals the enzyme is secreted in amounts at least 10 times greater than that needed for optimal starch digestion).
Although pancreatic alpha-amylase acts primarily in the polar milieu of intestinal content, where therefore the most part of the starch digestion occurs, a part adheres to the intestinal mucosa on the brush border surface of enterocytes. It has been proposed that this topographic disposition could be favorable because it would cause the release of the cleavage products of the starch (maltose, maltotriose and alpha-limit dextrins) at the lumen-membrane interface, where the final part of the digestion occurs by the action of brush border enzymes (see below).

Ileum, the final part of the small intestine, is able to digest and absorb carbohydrates, but in a less extend than jejunum and obviously duodenum. In the presence of illness affecting jejunum or of a surgical removal of the upper tract of the small intestine, the ileum can adapt to the new condition and assume an important role in carbohydrate digestion and absorption.

In fruit, it indicates that they are nutritionally rich. However, in soft drinks and other “sweets” it’s just empty calories. Dextrose – Sweetener made from .

For example, glucose, a sugar, is a monosaccharide these are carbohydrates in their simplest form (Timberlake, 2006). When only a few of these monosaccharid.

Each nucleotide is made up of one phosphate group (PO4), a five-carbon sugars (pentose),one of five nitrogenous base molecules, and one or more phosphate gro.

Primary alcohols have one carbon and two hydrogen atoms attached will produce aldehyde and then to carboxylic acid in two steps by oxidation. While secondary.

Disaccharides are condensation products of two monosaccharide units examples are maltose and sucrose. A disaccharide is 2 monosaccharides covalently linked .

The purpose of the experiment was to test the effect of the enzyme lactase on the milk solution. Because milk contains lactose (broken down by lactase to pro.

Diastereo isomeers - It the configurational changes considered C2, C3, or C4 in glucose. Example: Mannose, galactose. Annomerism - It is the spatial configur.

A regioselective synthesis of 6-amino-6-deoxycellulose, 6-N-sulfonated, and 6-N-carboxymethylated deoxycellulose were described by C. Liu et. al, by replace.

Alkaline phosphatase catalyzes the hydrolysis and transphosphorylation of phosphomonoesters (1). A kinetic reaction scheme of AP is presented in Figure 2. Fi.

The fact that soap is made up of fatty acids that are either unsaturated or saturated, they both have a different chemical makeup. Unsaturated fatty acids ha.

ELI5: Why can't humans break beta-glycosidic bonds. Being able to get energy from grass and "Fiber" would be a survival advantage during resource scarcity

Humans can break alpha-glycosidic bonds (starch, glycogen) and have even adapted to cleave branched chain alpha-glycosidic bonds. But why, if humans are truly omnivores, couldn't we graze for calories during resource scarcity?

We lack the enzyme that breaks down beta-glycosidic bonds. Our DNA only codes for glycogen phosphorylase which is specific to alpha 1-4 glycosidic bonds. Where this occurred revolutionarily probably isn't answerable, but is likely due to selective pressures of what was available at the time.

Those bonds are largely broken down by bacteria, not the animal themselves when it is consumed by herbivores. Cattle, for example, have a rumen that allows for those materials to be fermented, which assists in the digestion of those materials.

Those are foods with some of the worst calories spent in digestion vs calories gained ratios around. Animals that rely on them typically have extra large digestive tracts and relatively sedentary lives, with a great deal of time spent eating, all to make the low returns worth it. We can persue higher return foods more easily by not having larger, more generalized digestive systems than we have already, because it's less weight, and fewer cells to support. The calorie economics most likely did not work out in favor of such an adaptation in our case

Yeah, that would be great. So would being able to fly and have ESP.

Evolution doesn't work that way, though. It doesn't magically make stuff work just because it would be cool. Humans can't do those things because, over millions of years, we didn't evolve the random mutations which would allow us to do it.

Flying and wings do not contribute directly to the survival of an organism. Being able to obtain energy from a source of food around on the plant for billions of years - does.

Evolution isn't a single-minded pursuit of the perfect life-form for any situation. If you were to personify it, it would be more like a man successively asking ⟊n I still get laid if I do this?' as he gets weirder and weirder. Good enough to breed is good enough.

Meat is denser in calories and usable proteins than forage, so once a species can catch meat, evolution is usually biased toward doing so more often. At some point, a random proto-human lost the ability to digest or taste grass properly or mutated a slightly better meat-digesting colon or even just a taste for more meat, causing him to eat more meat, allowing him to be stronger and breed more. Repeat ad nauseam until we're 'omnivorous'. Which is good for us, because if weɽ kept refining the ability to digest the easy stuff instead, weɽ have probably evolved to be more like the cow -- four times the digestive tract and a quarter of the intelligence.

We can still eat fruits and tubers and such during lean times, which has been, you guessed it, 'good enough' for us. In fact, we can eat field clover, which is pretty damn close to grazing. It's slightly lemony.


The Enzyme Commission is responsible for creating an enzyme classification system based on numbers. The first number describes which class the enzyme belongs to, the second number reference sub-class, the third value specifies the nature of the substrate, and the fourth number is a serial number assigned to enzymes within a subclass. [2] The EC (Enzyme Commission) number of β-galactosidase is β-galactosidase belongs to class 3, which refers to the hydrolases. [3] β-gal belongs to a sub-class of glycosylases with an oxygen substrate nature.

β-galactosidase is an exoglycosidase which hydrolyzes the β-glycosidic bond formed between a galactose and its organic moiety. It may also cleave fucosides and arabinosides but with much lower efficiency. It is an essential enzyme in the human body. Deficiencies in the protein can result in galactosialidosis or Morquio B syndrome. In E. coli, the lacZ gene is the structural gene for β-galactosidase which is present as part of the inducible system lac operon which is activated in the presence of lactose when glucose level is low. β-galactosidase synthesis stops when glucose levels are sufficient. [4]

Beta-galactosidase has many homologues based on similar sequences. A few are evolved beta-galactosidase (EBG), beta-glucosidase, 6-phospho-beta-galactosidase, beta-mannosidase, and lactase-phlorizin hydrolase. Although they may be structurally similar, they all have different functions. [3] Beta-gal is inhibited by L-ribose, non-competitive inhibitor iodine, and competitive inhibitors 2-phenylethyl 1-thio-beta-D-galactopyranoside (PETG), D-galactonolactone, isopropyl thio-beta-D-galactoside (IPTG), and galactose. [5]

β-galactosidase is important for organisms as it is a key provider in the production of energy and a source of carbons through the break down of lactose to galactose and glucose. It is also important for the lactose intolerant community as it is responsible for making lactose-free milk and other dairy products. [6] Many adult humans lack the lactase enzyme, which has the same function of beta-gal, so they are not able to properly digest dairy products. Beta-galactose is used in such dairy products as yogurt, sour cream, and some cheeses which are treated with the enzyme to break down any lactose before human consumption. In recent years, beta-galactosidase has been researched as a potential treatment for lactose intolerance through gene replacement therapy where it could be placed into the human DNA so individuals can break down lactose on their own. [7] [8]

The 1,023 amino acids of E. coli β-galactosidase were accurately sequenced in 1983, [9] and its structure determined twenty-four years later in 1994. The protein is a 464-kDa homotetramer with 2,2,2-point symmetry. [10] Each unit of β-galactosidase consists of five domains domain 1 is a jelly-roll type β-barrel, domain 2 and 4 are fibronectin type III-like barrels, domain 5 a novel β-sandwich, while the central domain 3 is a distorted TIM-type barrel, lacking the fifth helix with a distortion in the sixth strand. [10]

The third domain contains the active site. [11] The active site is made up of elements from two subunits of the tetramer, and disassociation of the tetramer into dimers removes critical elements of the active site. The amino-terminal sequence of β-galactosidase, the α-peptide involved in α-complementation, participates in a subunit interface. Its residues 22-31 help to stabilize a four-helix bundle which forms the major part of that interface, and residue 13 and 15 also contributing to the activating interface. These structural features provide a rationale for the phenomenon of α-complementation, where the deletion of the amino-terminal segment results in the formation of an inactive dimer.

β-galactosidase can catalyze three different reactions in organisms. In one, it can go through a process called transgalactosylation to make allolactose, creating a positive feedback loop for the production of β-gal. Allolactose can also be cleaved to form monosaccharides. It can also hydrolyze lactose into galactose and glucose which will proceed into glycolysis. [3] The active site of β-galactosidase catalyzes the hydrolysis of its disaccharide substrate via "shallow" (nonproductive site) and "deep" (productive site) binding. Galactosides such as PETG and IPTG will bind in the shallow site when the enzyme is in "open" conformation while transition state analogs such as L-ribose and D-galactonolactone will bind in the deep site when the conformation is "closed". [5]

The enzymatic reaction consists of two chemical steps, galactosylation (k2) and degalactosylation (k3). Galactosylation is the first chemical step in the reaction where Glu461 donates a proton to a glycosidic oxygen, resulting in galactose covalently bonding with Glu537. In the second step, degalactosylation, the covalent bond is broken when Glu461 accepts a proton, replacing the galactose with water. Two transition states occur in the deep site of the enzyme during the reaction, once after each step. When water participates in the reaction, galactose is formed, otherwise, when D-glucose acts as the acceptor in the second step, transgalactosylation occurs . [5] It has been kinetically measured that single tetramers of the protein catalyze reactions at a rate of 38,500 ± 900 reactions per minute. [12] Monovalent potassium ions (K + ) as well as divalent magnesium ions (Mg 2+ ) are required for the enzyme's optimal activity. The beta-linkage of the substrate is cleaved by a terminal carboxyl group on the side chain of a glutamic acid.

In E. coli, Glu-461 was thought to be the nucleophile in the substitution reaction. [13] However, it is now known that Glu-461 is an acid catalyst. Instead, Glu-537 is the actual nucleophile, [14] binding to a galactosyl intermediate. In humans, the nucleophile of the hydrolysis reaction is Glu-268. [15] Gly794 is important for β-gal activity. It is responsible for putting the enzyme in a "closed", ligand bounded, conformation or "open" conformation, acting like a "hinge" for the active site loop. The different conformations ensure that only preferential binding occurs in the active site. In the presence of a slow substrate, Gly794 activity increased as well as an increase in galactosylation and decrease in degalactosylation. [5]

The β-galactosidase assay is used frequently in genetics, molecular biology, and other life sciences. [16] An active enzyme may be detected using artificial chromogenic substrate 5-bromo-4-chloro-3-indolyl-β-d-galactopyranoside, X-gal. β-galactosidase will cleave the glycosidic bond in X-gal and form galactose and 5-bromo-4-chloro-3-hydroxyindole which dimerizes and oxidizes to 5,5'-dibromo-4,4'-dichloro-indigo, an intense blue product that is easy to identify and quantify. [17] [18] It is used for example in blue white screen. [19] Its production may be induced by a non-hydrolyzable analog of allolactose, IPTG, which binds and releases the lac repressor from the lac operator, thereby allowing the initiation of transcription to proceed.

It is commonly used in molecular biology as a reporter marker to monitor gene expression. It also exhibits a phenomenon called α-complementation which forms the basis for the blue/white screening of recombinant clones. This enzyme can be split in two peptides, LacZα and LacZΩ, neither of which is active by itself but when both are present together, spontaneously reassemble into a functional enzyme. This property is exploited in many cloning vectors where the presence of the lacZα gene in a plasmid can complement in trans another mutant gene encoding the LacZΩ in specific laboratory strains of E. coli. However, when DNA fragments are inserted in the vector, the production of LacZα is disrupted, the cells therefore show no β-galactosidase activity. The presence or absence of an active β-galactosidase may be detected by X-gal, which produces a characteristic blue dye when cleaved by β-galactosidase, thereby providing an easy means of distinguishing the presence or absence of cloned product in a plasmid. In studies of leukaemia chromosomal translocations, Dobson and colleagues used a fusion protein of LacZ in mice, [20] exploiting β-galactosidase's tendency to oligomerise to suggest a potential role for oligomericity in MLL fusion protein function. [21]

A new isoform for beta-galactosidase with optimum activity at pH 6.0 (Senescence Associated beta-gal or SA-beta-gal) [22] which is specifically expressed in senescence (the irreversible growth arrest of cells). Specific quantitative assays were even developed for its detection. [23] [24] [25] However, it is now known that this is due to an overexpression and accumulation of the lysosomal endogenous beta-galactosidase, [26] and its expression is not required for senescence. Nevertheless, it remains the most widely used biomarker for senescent and aging cells, because it is reliable and easy to detect.

Some species of bacteria, including E. coli, have additional β-galactosidase genes. A second gene, called evolved β-galactosidase (ebgA) gene was discovered when strains with the lacZ gene deleted (but still containing the gene for galactoside permease, lacY), were plated on medium containing lactose (or other 3-galactosides) as sole carbon source. After a time, certain colonies began to grow. However, the EbgA protein is an ineffective lactase and does not allow growth on lactose. Two classes of single point mutations dramatically improve the activity of ebg enzyme toward lactose. [27] [28] and, as a result, the mutant enzyme is able to replace the lacZ β-galactosidase. [29] EbgA and LacZ are 50% identical on the DNA level and 33% identical on the amino acid level. [30] The active ebg enzyme is an aggregate of ebgA -gene and ebgC-gene products in a 1:1 ratio with the active form of ebg enzymes being an α4 β4 hetero-octamer. [31]

Much of the work done on β-galactosidase is derived from E. coli. However β-gal can be found in many plants (especially fruits), mammals, yeast, bacteria, and fungi. [32] β-galactosidase genes can differ in the length of their coding sequence and the length of proteins formed by amino acids. [33] This separates the β-galactosidases into four families: GHF-1, GHF-2, GHF-35, and GHF- 42. [34] E. Coli belongs to GHF-2, all plants belong to GHF-35, and Thermus thermophilus belongs to GHF-42. [34] [33] Various fruits can express multiple β-gal genes. There are at least 7 β-gal genes expressed in tomato fruit development, that have amino acid similarity between 33% and 79%. [35] A study targeted at identifying fruit softening of peaches found 17 different gene expressions of β-galactosidases. [33] The only other known crystal structure of β-gal is from Thermus thermophilus. [34]

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Unit 3. Lesson 1: Carbohydrates

By the end of this section, you will be able to do the following:

1. Discuss the role of carbohydrates in cells and in the extracellular materials of animals and plants

2. Explain carbohydrate classifications

3. List common monosaccharides, disaccharides, and polysaccharides

What’s in a potato? Besides water, which makes up most of the potato’s weight, there’s a little fat, a little protein…and a whole lot of carbohydrate (about 37 grams in a medium potato).

Some of that carbohydrate is in the form of sugars. These provide the potato, and the person eating the potato, with a ready fuel source. A bit more of the potato's carbohydrate is in the form of fiber, including cellulose polymers that give structure to the potato’s cell walls. Most of the carbohydrate, though, is in the form of starch, long chains of linked glucose molecules that are a storage form of fuel. When you eat French fries, potato chips, or a baked potato with all the fixings, enzymes in your digestive tract get to work on the long glucose chains, breaking them down into smaller sugars that your cells can use.

Carbohydrates make up a group of chemical compounds found in plant and animal cells. They have the empirical formula CnH2nOn, or (CH2O)n. An empirical formula tells the atomic composition of the compound, but nothing about structure, size, or what chemical bonds are present. Since this formula is essentially a combination of carbon and water, these materials are called “hydrates of carbon”, or carbohydrates for short.

Carbohydrates are widely available and inexpensive, and are used as an energy source for our bodies and for cell structures. Food carbohydrates include the simple carbohydrates (sugars) and complex carbohydrates (starches and fiber). Before a big race, distance runners and cyclists eat foods containing complex carbohydrates (pasta, pizza, rice and bread) to give them sustained energy. Carbohydrates are divided into monosaccharides, disaccharides, and polysaccharides.

Most of the oxygen atoms in monosaccharides are found in hydroxyl (OH) groups, but one of them is part of a carbonyl (C=O) group. The position of the carbonyl (C=O) group can be used to categorize the sugars:

· If the sugar has an aldehyde group, meaning that the carbonyl C is the last one in the chain, it is known as an aldose.

· If the carbonyl C is internal to the chain, so that there are other carbons on both sides of it, it forms a ketone group and the sugar is called a ketose.

Sugars are also named according to their number of carbons: some of the most common types are trioses (three carbons), pentoses (five carbons), and hexoses (six carbons).

Figure 2.1. Different types of Carbohydrates depending of their number of carbons

Glucose and its isomers

The chemical formula for glucose is C6H12O6. In humans, glucose is an important source of energy. During cellular respiration, energy releases from glucose, and that energy helps make adenosine triphosphate (ATP). Plants synthesize glucose using carbon dioxide and water, and glucose in turn provides energy requirements for the plant. Humans and other animals that feed on plants often store excess glucose as catabolized (cell breakdown of larger molecules) starch.

Galactose (part of lactose, or milk sugar) and fructose (found in sucrose, in fruit) are other common monosaccharides. Although glucose, galactose, and fructose all have the same chemical formula (C6H12O6), they differ structurally and chemically (and are isomers) because of the different arrangement of functional groups around the asymmetric carbon. All these monosaccharides have more than one asymmetric carbon (Figure 2.2).

Figure 2.2 Glucose, galactose, and fructose are all hexoses.

Glucose and galactose are stereoisomers of each other: their atoms are bonded together in the same order, but they have a different 3D organization of atoms around one of their asymmetric carbons. You can see this in the diagram as a switch in the orientation of the hydroxyl (OH) group, marked in red. This small difference is enough for enzymes to tell glucose and galactose apart, picking just one of the sugars to take part in chemical reactions .

Glucose, galactose, and fructose are isomeric monosaccharides (hexoses), meaning they have the same chemical formula but have slightly different structures. Glucose and galactose are aldoses, and fructose is a ketose.

Ring forms of sugars of Monosaccharides

You may have noticed that the sugars we’ve looked at so far are linear molecules (straight chains). That may seem odd because sugars are often drawn as rings. As it turns out both are correct: many five- and six-carbon sugars can exist either as a linear chain or in one or more ring-shaped forms.

Monosaccharides can exist as a linear chain or as ring-shaped molecules. In aqueous solutions they are usually in ring forms (Figure 2.3). Glucose in a ring form can have two different hydroxyl group arrangements (OH) around the anomeric carbon (carbon 1 that becomes asymmetric in the ring formation process). If the hydroxyl group is below carbon number 1 in the sugar, it is in the alpha (α) position, and if it is above the plane, it is in the beta (β) position.

Figure 2.3 Five and six carbon monosaccharides exist in equilibrium between linear and ring forms. When the ring forms, the side chain it closes on locks into an α or β position. Fructose and ribose also form rings, although they form five-membered rings as opposed to the six-membered ring of glucose.

Disaccharides (di- = “two”) form when two monosaccharides join together via a dehydration reaction, also known as a condensation reaction or dehydration synthesis. In this process, the hydroxyl group of one monosaccharide combines with the hydrogen of another, releasing a molecule of water and forming a covalent bond known as a glycosidic linkage.

For instance, the diagram below shows glucose and fructose monomers combining via a dehydration reaction to form sucrose, a disaccharide we know as table sugar. (The reaction also releases a water molecule, not pictured.)

In some cases, it’s important to know which carbons on the two sugar rings are connected by a glycosidic bond. Each carbon atom in a monosaccharide is given a number, starting with the terminal carbon closest to the carbonyl group (when the sugar is in its linear form). This numbering is shown for glucose and fructose, above. In a sucrose molecule, the 1 carbon of glucose is connected to the 2 carbon of fructose, so this bond is called a 1-2 glycosidic linkage.

Common disaccharides include lactose, maltose, and sucrose. Lactose is a disaccharide consisting of glucose and galactose and is found naturally in milk. Many people can't digest lactose as adults, resulting in lactose intolerance (which you or your friends may be all too familiar with). Maltose, or malt sugar, is a disaccharide made up of two glucose molecules. The most common disaccharide is sucrose (table sugar), which is made of glucose and fructose (Figure 2.4).

Figure 2.4 Common disaccharides include maltose (grain sugar), lactose (milk sugar), and sucrose (table sugar).

A long chain of monosaccharides linked by glycosidic bonds is known as a polysaccharide (poly- = “many”). The chain may be branched or unbranched and may contain different types of monosaccharides.

Plants store starch in the form of sugars. In plants, an amylose and amylopectic mixture (both glucose polymers) comprise these sugars. Plants are able to synthesize glucose, and they store the excess glucose, beyond their immediate energy needs, as starch in different plant parts, including roots and seeds. The starch in the seeds provides food for the embryo as it germinates and can also act as a food source for humans and animals. Enzymes break down the starch that humans consume. For example, an amylase present in saliva catalyzes, or breaks down this starch into smaller molecules, such as maltose and glucose. The cells can then absorb the glucose.

Storage polysaccharides

Starch is the stored form of sugars in plants and is made up of a mixture of two polysaccharides, amylose and amylopectin (both polymers of glucose). Plants are able to synthesize glucose using light energy gathered in photosynthesis, and the excess glucose, beyond the plant’s immediate energy needs, is stored as starch in different plant parts, including roots and seeds. The starch in the seeds provides food for the embryo as it germinates and can also serve as a food source for humans and animals, who will break it down into glucose monomers using digestive enzymes.

In starch, the glucose monomers are in the α form (with the hydroxyl group of carbon 1 sticking down below the ring), and they are connected primarily by 1-4 glycosidic linkages (i.e., linkages in which carbon atoms 1 and 4 of the two monomers form a glycosidic bond).

· Amylose consists entirely of unbranched chains of glucose monomers connected by 1-4 linkages.

· Amylopectin is a branched polysaccharide. Although most of its monomers are connected by 1-4 linkages, additional 1-6 linkages occur periodically and result in branch points.

Because of the way the subunits are joined, the glucose chains in amylose and amylopectin typically have a helical structure.

Figure 2.5 Amylose and amylopectin are two different starch forms. Unbranched glucose monomer chains comprise amylose by α 1-4 glycosidic linkages. Unbranched glucose monomer chains comprise amylopectin by α 1-4 and α 1-6 glycosidic linkages .

That’s great for plants, but what about us, humans? Glycogen is the storage form of glucose in humans and other vertebrates. Like starch, glycogen is a polymer of glucose monomers, and it is even more highly branched than amylopectin.

Glycogen is usually stored in liver and muscle cells. Whenever blood glucose levels decrease, glycogen is broken down via hydrolysis to release glucose monomers that cells can absorb and use.

Structural polysaccharides

Although energy storage is one important role for polysaccharides, they are also crucial for another purpose: providing structure. Cellulose, for example, is a major component of plant cell walls, which are rigid structures that enclose the cells (and help make lettuce and other veggies crunchy). Wood and paper are mostly made of cellulose, and cellulose itself is made up of unbranched chains of glucose monomers linked by 1-4 glycosidic bonds.

Unlike amylose, cellulose is made of glucose monomers in their β form, and this gives it very different properties. As shown in the figure, every other glucose monomer in the chain is flipped over in relation to its neighbors, and this results in long, straight, non-helical chains of cellulose. These chains cluster together to form parallel bundles that are held together by hydrogen bonds between hydroxyl groups. This gives cellulose its rigidity and high tensile strength, which are important to plant cells.

The β glycosidic linkages in cellulose can't be broken by human digestive enzymes, so humans are not able to digest cellulose. (That’s not to say that cellulose isn’t found in our diets, it just passes through us as undigested, insoluble fiber.) However, some herbivores, such as cows, koalas, buffalos, and horses, have specialized microbes that help them process cellulose. These microbes live in the digestive tract and break cellulose down into glucose monomers that can be used by the animal. Wood-chewing termites also break down cellulose with the help of microorganisms that live in their guts.

Cellulose is specific to plants, but polysaccharides also play an important structural role in non-plant species. For instance, arthropods (such as insects and crustaceans) have a hard external skeleton, called the exoskeleton, which protects their softer internal body parts. This exoskeleton is made of the macromolecule chitin, which resembles cellulose but is made out of modified glucose units that bear a nitrogen-containing functional group. Chitin is also a major component of the cell walls of fungi, which are neither animals nor plants but form a kingdom of their own.

Figure 2.6 In cellulose, glucose monomers are linked in unbranched chains by β 1-4 glycosidic linkages. Because of the way the glucose subunits are joined, every glucose monomer is flipped relative to the next one resulting in a linear, fibrous structure

CAREER CONNECTION ______________________________________________________________________

Registered Dietitian

Obesity is a worldwide health concern, and many diseases such as diabetes and heart disease are becoming more prevalent because of obesity. This is one of the reasons why people increasingly seek out registered dietitians for advice. Registered dietitians help plan nutrition programs for individuals in various settings. They often work with patients in health care facilities, designing nutrition plans to treat and prevent diseases. For example, dietitians may teach a patient with diabetes how to manage blood sugar levels by eating the correct types and amounts of carbohydrates. Dietitians may also work in nursing homes, schools, and private practices.

To become a registered dietitian, one needs to earn at least a bachelor’s degree in dietetics and nutrition. In addition, registered dietitians must complete a supervised internship program and pass a national exam. Those who pursue careers in dietetics take courses in nutrition, chemistry, biochemistry, biology, microbiology, and human physiology. Dietitians must become experts in the chemistry and physiology (biological functions) of food (proteins, carbohydrates, and fats).

Benefits of Carbohydrates

Are carbohydrates good for you? Some people believe that carbohydrates are bad and they should avoid them. Some diets completely forbid carbohydrate consumption, claiming that a low-carbohydrate diet helps people to lose weight faster. However, carbohydrates have been an important part of the human diet for thousands of years. Artifacts from ancient civilizations show the presence of wheat, rice, and corn in our ancestors’ storage areas.

As part of a well-balanced diet, we should supplement carbohydrates with proteins, vitamins, and fats. Calorie-wise, a gram of carbohydrate provides 4.3 Kcal. For comparison, fats provide 9 Kcal/g, a less desirable ratio. Carbohydrates contain soluble and insoluble elements. The insoluble part, fiber, is mostly cellulose. Fiber has many uses. It promotes regular bowel movement by adding bulk, and it regulates the blood glucose consumption rate. Fiber also helps to remove excess cholesterol from the body. Fiber binds to the cholesterol in the small intestine, then attaches to the cholesterol and prevents the cholesterol particles from entering the bloodstream. Cholesterol then exits the body via the feces. Fiber-rich diets also have a protective role in reducing the occurrence of colon cancer. In addition, a meal containing whole grains and vegetables gives a feeling of fullness. As an immediate source of energy, glucose breaks down during the cellular respiration process, which produces ATP, the cell's energy currency. Without consuming carbohydrates, we reduce the availability of “instant energy”. Eliminating carbohydrates from the diet may be necessary for some people, but such a step may not be healthy for everyone.


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