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Food Chemistry

by: Andreanne Collier

Food Chemistry FD SC 400

Andreanne Collier
Penn State
GPA 3.58


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This 0 page Class Notes was uploaded by Andreanne Collier on Sunday November 1, 2015. The Class Notes belongs to FD SC 400 at Pennsylvania State University taught by Staff in Fall. Since its upload, it has received 24 views. For similar materials see /class/233076/fd-sc-400-pennsylvania-state-university in Food Science & Technology at Pennsylvania State University.

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Date Created: 11/01/15
4 Food Proteins Proteins are synthesized by plants and animals to play a role in their physiology The functions a protein has from the organism s point of view include communication e g insulin structural e g collagen in skin or keratin in hair biochemical catalysts eg enzymes transportation e g hemoglobin to transport oxygen in the blood defense e g antibodies and storage e g globulins in seeds Very rarely do our needs from the protein match those of the plantanimal we are going to eat Our needs are firstly nutritional 7 of the 20 amino acids common in nature we can synthesize ll of them and must gain the rest from dietary sources and secondly functional 7 we rely on proteins to change the texture of our food for example by forming gels or by stabilizing foams and emulsions The only area where the proteins evolved function overlaps with our needs is when we use proteins as enzymes in bioprocessing e g making corn syrup from the action of amylase enzymes on starch We are also interested in how proteins interact with other food components eg avor binding or have toxic or antinutritional properties e g botulinum toxin is a protein In this section we will start by brie y reviewing the biochemical structure of proteins We will then examine how they can be modified postmortem by processing and through interactions with other ingredients and how this leads to some functional roles We will finally look at some examples ofthe uses of enzymes in food processing 1 Protein structure A protein is a linear sequence of amino acids linked together by O I peptide bonds There are 20 amino acids prevalent in nature and the CCNH2 sequence in which the cell assembles them is set in the DNA code H ie the primary structure of the protein The 20 amino acids all have distinct structures and unique properties but with a couple of important exceptions we can classify them as o Chargeduncharged Some amino acids have functional groups eg carboxylic acid amine phenol thiol that can carry a charge depending on the pH Hydrophobichydrophilic Amino acids may be more or less water soluble depending on the polarity of their structure Charged amino acids tend to be more watersoluble Some important example amino acids include The side chain Phenylalanine has 0 O 0 1k of glutamrc ac1d a bulky nonpolar HO I OH is a very polar benzene ring as S carboxylic acid part of its side H NH which may carry chain a negative charge depending on the pH SH quotI CIH HSN M s 1 3 x Cysteine contains a functional thiol group Thiols can dimerize under oxidizing conditions to form disulfide bonds a di cysteine is called cystine During their RNA mediated polymerization process the amine group of one amino acid reacts with the carboxylic acid group of another and water is eliminated The dipeptide formed still has a reactive amino and carboxylic acid group which can continue to react to form sequentially larger polypeptides and eventually a complete protein The number of amino acids required to form a protein varies Lysine contains an amine group that is an unusually strong ya N Hi nucleoph11e The 1 reactivity of this amine with reducing sugars 7 quot in the Maillard 393 reaction is particularly Hams 39x DH strong and so lysine is quickly destroyed during nonenzymatic browning The suquunur 0m 2H Exam aw side chain Libe pnum39y smmuma 395 ml 113 Rmnaimngmaqliwc wads allow Further mlymcrimiliml the DNA 03357 39 R t R quot 0391 I 1 oil xc B m 39 R 39C g NH I 0 l 0 p s llh D T HO s lhzldrlu IRNIIEI It39s39 39ll hiS H 47 391 into chainl mcklmnelo quot 3 limit exibility 039 qcadmn 39 C E N H widely e g lysozyme is egg white is a simple globular protein with 129 residues while collagen is a triple helix of chains each containing more than 1400 residues A freshly synthesized polypeptide is no real use to the organism It must first spontaneously fold into a characteristic shape necessary for it to function properly Protein folding is a spontaneous process but incredibly precise process that remains one of the wonders of the natural world The first step in understanding protein folding is to think about the preferred conformation of a polymer A exible chain will not stand straight in solution the random conformation of the movable bond angles means instead it will take on a random coil conformation In the case of a protein there are a number of hydrophobic groups on the chain that would prefer to remove themselves from the aqueous cell matrix by being buried in the core of the protein This hydrophobic effect will tend to fold the random coil somewhat more tightly however folding too tightly would have an entropy cost Incmas almi I I emery r If Increase s elmamt 11111ij Onc nailm stain Hang denatured slams Another way of looking at protein folding is as a competition between the chain entropy and the solvent entropy The highest entropy a chain can take is to form a random coil Each bond angle is set randomly and can ex randomly Note that a random coil does not imply a single structure for the protein instead a number of interchanging shapes all based on the principle that the bond angles are random However a random coil protein still has many hydrophobic amino acid residues exposed to the aqueous solvent Water molecules organize themselves into a structured clathrate cage around hydrophobic groups ie lowering their entropy The lowered solvent entropy is a problem that can be overcome by more tightly folding the protein but that would create a problem by increasing the degree of ordering of the chain The basis of protein folding then is a balance of forces argument between chain entropy seek to unfold to maximize the disorder of the polymer itself and solvent entropy seek to fold up to remove the ordering effects of hydrophobic amino acid residues on water This simple argument allows us to image a protein in solution spontaneously taking on a tightly folded conformation but the details distinguishing proteins lead to other structural intricacies A whole range of other noncovalent interactions are important in supporting and refining the shapes taken Important amongst these are i steric restraints 7 the chains exibility is restricted to movement about the acarbon and may be further restricted by bulky side chains ii strong interchain hydrogen bonding may strengthen certain configurations and iii permanent charges on the chain will lead to interchain electrostatic I repulsion or attraction these forces will be l strongly pH dependent Finally covalent disulfide bonds between two cysteine residues are very f K a H I I I quotI39 Rf faquot strong and can read11y stablllze structure There are some common structural features that occur across a wide range of protein types These are the secondary structures 7 a local folding of the polypeptide chain over a part of its length A protein may contain several types of different secondary structure or it may contain none Examples of secondary structure include the xhelix and sheet T An a helix is a coil formed over a region of the quotTN 01 e tide chain There are stron h dro en 9 i hydrmphol c 01111 between the nitrogen on the Eha in angd the oxygen on the forth proceeding amino acid in the helix However probably more important is the radial positioning of the side chains sticking out at angles from the helix One side of the helix tends to accumulate hydrophobic residues and one side hydrophilic The overall helix is therefore amphiphilic and can easily be used to build tertiary 39 I structure by hiding the hydrophobic half in the core of the protein L3 W In a B sheet polypeptides line up either parallel or antiparallel with 392 t i one another There are strong In L I hydrogen bonds between the hydrophobic nitrogens on one chain and the 5 in Q l oxygen on a second chain However probably a more 399quot 55 I J F 1 important driving force is half a hi dl ophllw u quoti the side chains line up above the plane of the sheet and half below The half above tend to be hydrophobic and the half below hydrophilic The overall structure is amphipillic and held together by hydrophobicity r9 1 m J 39 k U r J FJ Ciwgpm I g Q 714 939 Tertiary structure is the bulk folding of the chain to make a defined threedimensional structure As we have seen the primary driving force for protein folding is solvent entropy Interchain hydrogen bonding electrostatic interactions and disulfide bonds may reinforce the structure formed Most proteins are at least approximately spherical blobs in solution as this allows them to minimize their surface to volume ratio and hide more hydrophobic amino acids in the core away from water More hydrophilic polymers will tend to be more open and more hydrophobic more densely packed Nonspherical proteins e g collagen need a strong secondary structure to maintain an extended shape normally for structural reasons 2 Protein Denaturation The native structure of a protein is the energetic minimum under physiological protein Any change in conformation away from this shape will represent an energy cost The protein is not completely static 7 it will ex and bend in response to thermal energy or binding ligands but will always tend to return to the same shape When the protein is moved out of the physiological state the balance of forces acting are changed and the protein may respond to change shape to minimize its energy under the new conditions For example a globular protein folds up tightly in water to hide its hydrophobic residues If the same molecule was moved into an organic solvent it would wan to unfold to hide its more hydrophilic groups in the core away from the nonpolar solvent The most common way to denature proteins is through heat Consider how the energy balance holding the protein together change with temperature for the reaction of native protein transforming to denatured protein PNlt gtPD The reaction will proceed spontaneously if the Gibbs free energy for the reaction is negative The two major contributions to Gibbs free energy here are solvent entropy opposes unfolding and chain entropy favors unfolding AG ACTsolvent ACkhain 39TAssolvent39 TAS chain If a protein unfolds the chain becomes less organized so its entropy increases so iTDS becomes negative and the reaction is favored On the other hand if the chain unfolds more clathrate water is formed the solvent becomes more organized so its entropy AS decreases so iTAS becomes positive and the reaction is opposed Under physiological conditions these factors balance each other with AGmal slightly positive and the reaction stays in the native state As temperature increases the T multiplier on both entropy terms increases and the value of AGED th and AGchain increase in parallel AGED th becomes increasingly positive and AGchain increasingly negative Their sum is unchanged and the protein remains stable However at about 700C the structure of the clathrate water starts to break down and the hydrophobic opposition to chain entropy stops being able to keep up and their sum the free energy for denaturation becomes negative and the protein unfolds In practice the various other interactions e g interchain hydrogen bonds electrostatic interactions disulfide bonds etc will oppose or favor denaturation to different extents depending on their prevalence and strength in a given protein and the denaturation temperature will vary between types of protein Other factors can also cause denaturation Any factor or combination of factors that increase the forces forcing a protein apart or reduces those holding them together may cause denaturation Examples include 0 Changing the pH so the protein is highly charged positive below the luau pm pI or negative above can favor 3 protein denaturation because the 1 like charges on the chain repel one another and favor expansion of the ml Pm Phase fulded strnetnre Makrng the sulventmurenunrpular e g by addrng eertan a1enhn1s rednees the energy enst ufexpusmg thehydrnphnhre armnu aerds tn snlyent and ean favur en there rs a surface present 3 g a eshly prepared fuam nr emulsmn 1 an aqueuus prntern ml 1 adsnrh at the rnterfaee tn all nw any hydrnphnhre surface atehes tn rnnye nnt quhe aqueuus enyrrnnrnent The surface prntern may then znn n1d at the snrfaeetn a1an rnnre ufthehydmphubm armnu ands tn rnnye rntn the nunrpular enyrrnnrnent prntern wru try tn get back rntn rts energy rnrnrrnnrn whreh underphysmlugmal prense sequence nfstepsneeded tn fuld the prntern rs hard tn repeat xtrs very easy fur e prntern rs equen y stnek seenndary energy rnrnrrnnrn and never regenerates urface These ammu aerds wuuld rdeanyrepaekrntn the enre quhe prntern hnt equendy ear bele stranded atthe surface The surface hydrnphnhrepatehes ean 1ead tn prnterns aggregatang tn hrde 39nm water Summag anrnt Denatum nn 4 a The nataye prntern rs the rnnst In e lb 3 rm 11 d denatured fnrrnsrnnst re n1d m remsely the sarne sequence tn rehndd therdea1 natrye strnetnre 3 Protein Functionality Protein structure describes the chemical and physical shape of a protein Functionality describes what the protein does from our point of view As noted above an important functional role of protein that depends on their native state functionality ie biological functionality depends on their behavior as enzymes Enzymes are biological catalysts that increase the rate of a chemical reaction by providing a lowerenergy pathway between starting conditions and end point Frequently the reaction accelerated by the enzyme may be not apparent in its absence and it may seem as if a completely new pathway is being opened up but in fact the only effect is catalytic Enzymes are unusually specific catalysts in that they will work on a very defrned starting material to make a single often chirally pure form Because the catalyst is an enzyme any factor leading to denaturation may quickly destroy its activity A detailed treatment of enzyme structurefunctionality kinetics and associated biotechnology is beyond the scope of this course and we shall instead focus on some illustrative examples 0 Polyphenoloxidase catalyzes the oxidation of a wide range of phenolic compounds to diphenols and hence to quinones him i mml Quinones spontaneously ie nonenzymicaly H m polymerize to brown melanin pigments 7quot i o eno oxi ase nee s 0 en an a co er I 39 1 quot Plyph 1 d d Xyg d pp quot 39 cofactor to function CH cu Polyphenoloxidase activity is a plant defense mechanism 1 W designed to turn natural plant phenols into quoti l j 397quot 3 HAHN5 antimicrobial products in the case of 39 39 M tissue injury In plant processing this dimich cmmle is a problem because freshly cut fruit and vegetables will spontaneously brown quickly unless steps are taken to i denature the enzyme ie blanching ii remove the oxygen e g store the vegetables under water or add ascorbic acid which reacts preferentially with the oxygen or iii inhibit the enzyme eg drop the surface pH with acid Sulfur dioxide is a very effective inhibitor of enzymatic browning by binding to the quinnones and preventing the nonenzymatic polymerization step that leads to melanins The products of polyphenoloxidase are nontoxic and avorless but extensive browning quickly leads to perceived spoilage Lipoxygenase catalyzes the formation of lipid peroxides from polyunsaturated fatty acids Lipid peroxides are a crucial step in the oxidation of fats to form rancid avors and peroxide activity can quickly lead to off avors e g if the germ is not adequately separated before milling grain In other cases lipoxygenase is added to facilitate avor formation The radicals forms as the peroxides break down radicals are formed that can cooxidize pigments and destroy vitamins Soy lipoxygenase is often added to bleach flour Amylase is an important starchdegrading enzyme Amylase enzymes are used to produce deXtrins small glucose oligimers and glucose from suspensions of starch Initially a heatsensitive endoamylase ccamylase is added to a hot suspension of starch to rapidly reduce the average molecular weight The starch is then cooled but is less viscous because ofthe lower MW and a more heat sensitive exoamylase glucoamylase is added to further reduce the starch polymers to smaller units that can be partially puri ed and used as food ingredients see carbohydrates section below The other major functional roles of proteins depend more on their properties as polymers rather than their evolved biological functionality and frequently denatured proteins can be as or more functional than the native form While there is a wide range of these functions we will categorize them as hydrodynamic and surface properties Hydrodynamic functionality depends on the physical size and shape of the protein as an object suspended in a uid In this case the protein can be seen as a very small colloidal particle or neutral buoyancy It will build viscosity by forcing uid streamlines to de ect around it and may aggregate to form a gel The analogy between a polymer molecule and a colloidal particle is pursued further in the polysaccharides section below and at this stage we can settle for the general rule that interprotein cross links cf occulation in emulsions tend to lead to rst an increase in viscosity and second the formation of a gel The mechanisms of interchain bond formation include disulfide bonds but the most common are the attractions between the hydrophobic groups on the surface of denatured proteins Hydrophobic attraction can be mediated by strong electrostatic repulsion cf the DVLO potential if the protein is far from its isoelectric point and the salt concentration not too high Another important interaction is the strong affinity between certain amino acid residues and calcium If two separate proteins attempt to bind the same calcium it can form a strong link between the Calciummediated aggregation is particularly important in forming a tofu gel from soy proteins or setting caseinate gels The second group of functionalities important in proteins depends on their surface hydrophobicity Because a native protein can o en not fold adequately to protect all of their hydrophobic amino acids there are frequently hydrophobic regions on their surfaces A denatured protein has even more exposed surface hydrophobicity Either of these cases can allow a protein to adsorb at the surface of an emulsion droplet or a foam bubble and protect them against occulation and coalescence In addition small hydrophobic molecules can bind to the hydrophobic regions on the protein Importantly a protein can bind up avor molecules and stop them becoming volatile and thus perceived Water in Foods 1 The structure of water and solutions a BulkPropem39es of Solutions Probably the main bulk property of a solution we can readily vary in foods is pH log10H The pH ofmost foods is in the range 37 The pH is important as it alters the structure of molecules present by either protonating or deprotonating the ionizable groups The main protonationdeprotonation reactions we are interested in involve carboxylic acid and amine groups RCOOH RCOO39 H R NH 4 RNH2 H As the reaction proceeds the charge on the carboxylic acid will change from zero to l and the charge on the amine from 1 to zero Qualitatively we can see that as the pH drops H increases the reactions will be forced to the left hand side However different functional groups have different affinity for protons that can be expressed quantitatively in terms of the equilibrium constant of the reaction So for a general protonation reaction AH 4 A39 H K A39HAH A391AH H1 Taking negative logs of both sides we get the Henderson Hasselbach equation pH 3Ka log A39 HA The Henderson Hasselbach equation can tell us the ratio of protonated to deprotonated groups ie COOH and 7COO39 for a carboxylic acid or 7NH3 and 7NH2 for an amine as a function of pH and the pK of that particular functional group Plotting the concentration of both molecules as a function of pH we get a characteristic sigmoidal shape with protonateddeprotonated at pHpK The concentration ratio changes rapidly around the pK but more than 2 pH units above and below changes in pH will not affect the ratio 100 E Protonated form 2 E Deprotonated 0 form 2 O 50 a O m E a 8 a n 0 l i l i i i i i i 0 2 4 6 8 10 12 pH Figure 1 Characteristic acidbase dissociation for a pK5 salt calculated using the HendersonHasselbach equation Example Vinegar is acetic acid solution It deprotonates according to the following reaction with a pK of 4 74 CH3 COOH CHg COO39 H If we made a meat sauce containing acetic acid and ended up with a 1 M solution at pH 5 what proportion of species would be present Calculate the ratio of species using the Henderson H asselbach equation 54 74 log A39HAA39HA1003926 181 Calculate the sum of the deprotonated and protonated forms equal to the total amount of acetic acid present Acetic Acid A39HA1 Solve for the concentration of each species 181 HA HA 1 HA1281 035 35 so A3965 Question 1 The structure of glutamic acid is shown below i Calculate the proportion of each functional group that is charged at pH 4 ii indicate which functional group would change the most if the pH was decrease by a further 01 unit a 5 LD lJH pK188 b Microscopic Properties of Solutions The unique properties of water are essential to the biochemistry responsible for all life Indeed most recent efforts to find life on Mars or other planets focus on a search for present or past water eg httpnew bbc co 11k quot 39 39 39 2846897stm The basics of water molecular structure bond polarization the hydrogen bond and the interconnected structure are available in detail elsewhere Please look at one or more of these sites until you are con dent on these points 39 I 0 http sbc rd 39 wtmtm html 0 httpwww lsbn ac 11k wate 39 39 html 0 httpwww lsl m ac uk 39 39 J html 0 httpwwwlsbuaculdwaterabstrcthtml We are mainly interested in how water molecules can interact with the solutes present Conceptually we can write this as a reaction Solute dissolved Unmodified water Dissolved solute modified water The reaction will proceed spontaneously if the Gibbs free energy is negative AGAHTAS Where AG is the change in Gibbs free energy AH the change in enthalpy making and breaking bonds and AS the change in entropy how disordered the system is T is absolute temperature The change in enthalpy will depend on the strength of the bonds formed between the water and the solute many strong bonds will make the term negative and favor the reaction The solute is always more disordered after dissolution AS positive TDS negative favoring the reaction but the water molecules may sometimes be ordered around the solute so Aswm is negative and the reaction is opposed For an ionic solute there is a strong electrostatic attraction between the permanent charge on the ion and dipole on the water molecule The 5 hydrogens will align towards the cations present and the 5 oxygens towards the anions Figure 2a The binding of water by solutes in this way is not a permanent bond and many times weaker than a covalent linkage An individual water molecule will diffuse in and out of proximity to the solute many times a second but there is a net accumulation as shown in the Figure There is a limited amount of water that can bind to any charge present but there may also be a secondary alignment of water molecules bonded to the orientated dipoles of the first shell water The iondipole attractions for an ionic solvent are very strong and allow most charged materials to readily interact with and dissolve in water For a polar but uncharged solute the interactions are weaker They may include hydrogen bonding to the water or dipoledipole interactions Figure 2b Although relatively weak these interactions are usually enough to allow polar materials to dissolve in water HH r Q I 9 Pee 1 o 39 I e5 7 A i o c is 3 a H Ho 5 Figure 2a Water molecules orientated in Figure 2b Water molecules bound to an multilayers around a positive ion uncharged polar solvent by hydrogen bonds and dipoledipole interactions Nonpolar molecules can only weakly interact with water through dipoleinduced dipole bonds The water instead forms a tightly hydrogenbonded cage a clathrate cage around the nonpolar molecule bonding tightly to itself to avoid contact The important net effect of dissolving a nonpolar molecule is to increase the organization of water molecules This is an entropy cost for the water and means the dissolution reaction does not proceed The cost of bringing a nonpolar molecule into solution is proportional to the number of water molecules ordered around it and hence the solute surface area For example small alcohols eg ethanol are very water soluble because the strong hydrogen bonding with the 70H group offsets the cost of forming a clathrate cage around the hydrocarbon portion of the molecule while larger alcohols eg decanol are water insoluble because many more water molecules must orientate around the long nonpolar portion of the molecule The drive to minimize the area of contact between nonpolar materials and water means they will tend to attract one another in solution and aggregate Some examples are illustrated in Figure 3 a 11 rHu N j s 4 H M hrydmphilic hydrophobic a i Figure 3 Examples of consequences of the hydrophobic effect not to scale 7 a typical surfactant is in the order of 01 nm a globular protein 23 nm an oil droplet 1000 nm a Oil droplets in water are spherical minimize surface area and will tend to coalesce if not prevented b A protein is a chain or more or less polar amino acids It will coil in solution to hide the less polar groups in the core out of contact with water An amphiphile is a molecule with part of its structure polar and part non polar e g lecithin many proteins c Some small molecules can form supramolecular assemblies in water eg micelles or membranes d Most amphiphiles will tend to align at surfaces eg oilwater airwater again to bring their nonpolar portions out of contact with water 2 Water activity and moisture sorption In most solutions water is present to excess and only an in nitesimally small fraction is bound to the surface of solute molecules However as the solution becomes more concentrated e g by drying freezing or adding more solutes the fraction of bound water increases Bound water is less able to take part in chemical reactions so the stability of the food should also increase Imagine drying down a simple sugar solution How dif cult is it to remove a water molecule The majority of water present has no association with the sugar molecules so can be removed with very little effort However the last few molecules become increasingly dif cult and it is necessary to increase the force applied ie by lowering the humidity of the headspace above the food Ifwe plot the amount of residual moisture on a dry weight basis as a function of the force applied ie relative humidity or PPo where P is the partial pressure of water vapor above the sample and P the partial pressure of water vapor above pure water or aW water activity we get a moisture sorption isotherm For almost all foods these appear Jshaped with almost all of the moisture removed if the humidity in the air is lowered slightly below saturation Figure 4a However if the lower part of the isotherm is magni ed we can see the detailed effects of the solute on the ease of water molecules and the differences between foods Solutes with very strong af nity for water can maintain a larger amount of bound moisture at lower humidities than solutes with less af nity I 2 I 5 2 JH hsu arJ bnd i E Equot 3 3 t E 5 39 s I a a d E al mc W a n i 5 u Figure 4a Moisture sorption isotherm Figure 4b Magni cation of the lower Almost all gtgt90 of the moisture in most portion of Figure 4a showing the different foods is driven off by a slight lowering of sorption properties of a typical polymeric relative humidity but a small fraction of or highsugar food bound water is retained to lower moisture contents Some example sorption isotherms are shown in Figure 4b The typical behavior for a sugar or other small molecule rich food is Jshaped isotherm with very strong binding of water above a critical moisture content This is typically seen in table sugar or hard candies which remain stable in all but the most humid conditions then will rapidly cake or even dissolve if the moisture content is slightly increased Polymeric foods on the other hand tend to have a sigmoidal moisture sorption isotherm Lowering the humidity slightly reduces the moisture content to a low level but after that the humidity needs to be drastically lowered to drive off the last part of the water This is probably due to there being different populations of bound water A physical basis for Figure 4b polymeric can be gained from an understanding of the interactions between water and a solute Figure 5 As the water activity of the environment is lowered first the unbound Zone 3 then the loosely bound Zone 2 then the tightly bound water Zone 3 is removed The more bound water is not able to act as a solvent for chemical reactions so under these conditions the rate is drastically lower The rate of various deterioration reactions has been measured as a function of water activity and some common patterns can readily be seen see Figure 6 The parameter that governs reactivity in this theory is the amount of water available as a solvent If we define Zone 1 water as that bound closely to food components we can say the moisture content at the boundary between Zone 1 and Zone 2 is the monolayer value 7 a layer of water one molecule thick around the binding sites If Zone 1 water is unavailable for reaction then the difference in water activity between the current environment and the monolayer aW is a measure of food stability Figure 5 Zones of moisture in a typical isotherm Zone 3 Unbound water driven off by only a slight drop in aw The bulk of the water in most foods is indistinguishable from bulk water Zone 2 Intermediately bound water Loosely bound water perhaps in the secondarily binding sites on food Moi 511m comment cllwh Zone 1 Tightly bound water To remove it from direct proximity to binding sites on the food requires lowering the aw to very low levels lag RATE Moisture content dank aw Figure 6 Food Stability Map after Labuza University of Minnesota The rate of most reactions decreases rapidly with decreasing water activity Some important points in Figure 6 l 2 3 4 5 6 The rate of reaction is expressed logarithmically on the yaxis so a small change on the graph means a very large change in reality The rate of most chemical reaction decreases rapidly with aW and often reaches zero at or before the monolayer value The argument of this is that as there is less water able to act as a solute for the reaction the rate is slowed In some cases there is a decrease in rate at very high aW due to a dilution of reagents Bacterial growth is also governed by chemical reaction and rapidly tends to zero as aW is decreased The rate of lipid oxidation may increase at very low water activities Lipid oxidation is obviously not a conventionally aqueous chemical reaction so it is surprising that it is as affected as it is The decrease in rate is believed to be due to a lack of solvent for the aqueous components of the reaction eg catalysts and the increase at low aW to the dehydration of metal catalysts to make them more effective Changes in physical properties often follow changes in aw For example baked goods can undergo a crisp to soft transition as water activity and powders can begin to clump Another value of knowing the water activity of a food is it can be used to predict the direction of water diffusion Remember water activity is a measure of the ease that which a water molecule can leave the food If there are two foods with different water activities sealed together in a closed system water molecules will continually leave and return to the surface of each by the normal processes of molecular diffusion However because it is slightly harder to leave the low aW food there will be a net accumulation of moisture there until the water activities of the two foods become equal In summary 7 water diffusion is driven by differences in water activity not by differences in water 20 content Water diffusion in foods can limit the stability of packaged composite foods For example moisture can diffuse between 0 a lettuce leaf and make bread in a sandwich soft 0 a frozen pizza topping into the crust 0 The ice in ice cream and the crispy cone making it soft Moisture diffusion can be limited by either matching the water actiVities of the components of the food by adding humectants to lower the water actiVity of a moist component eg glycerol A good humectant should lower water actiVity without impacting avor or safety of the product It should be readily soluble and should not crystallize Alternatively the diffusion path can be blocked by packaging the food components separately or adding a moisture impermeable barrier between the components e g cheese in a sandwich a chocolate layer in a frozen ice cream cone Question 2 Moisture Sorption The moisture sorption data for pasta and pasta sauce are listed in the following table Both materials were dried to 7 moisture and packed together equal masses of each ingredient 0 Plot a moisture sorption isotherm for both materials on the same axis 0 Would there be any moisture diffusion in this system In which direction Using your sorption isotherm estimate the final moisture contents and water actiVities of each component RelativeHumidity 0 0112 0226 0438 0518 0577 0660 0753 0843 0903 Pasta moisture 0 40 60 70 75 80 90 110 140 180 Sauce mix moisture 0 10 30 140 180 230 280 21 3 Phase transitions We can also use an understanding of the properties of solutions to de ne their limits 7 to construct a phase diagram useful phase diagram for a solution would map the different events that can occur when temperature and concentration are altered The Figure 7 below shows such a phase diagram for a sucrose solution taken from Crystallization in Foods by Richard Hartel The annotations on the diagram are explained below Temperaluro quot0 ICE summon Compositlon wt 79 Figure 7 Phase diagram for sucrose from Hartel Important points in Figure 7 a b C e yaxis shows temperature the Xaxis the composition of the solution p e The freezing point of pure water is 0 C cooling below this will lead to freezing Heating above 100 C will cause the water to boil to steam The melting and freezing points mark the limits of the liquid water phase When a solute is added the freezing point of water decreases Ideally the extent of the freezing point depression should be proportional to the number of added molecules only ie a colligative property although reality o en shows some deviation If the solution is cooled below that line the water present can still freeze and the solution will be further concentrated the phase diagram only refers to the liquid solution component As more ice is formed the solution will be progressively concentrated so it will need to be cooled even further to induce more ice formation The liquid solution will follow the freezing line as it is freeze concentrated d Similarly when a solution is boiled only water will leave the system as pure steam The residual solution will be progressively concentrated and the boiling point will further increase again a colligative property As a solution is boiled its temperature will increase as it is concentrated e Eventually both the boiling and freezing line will intersect with the saturation limit line of the solution At a given temperature the solubility of the solute reaches a limit and the solute will begin to crystallize As more water is removed from the system by boiling or freezing the solute will simultaneously crystallize out of the solution and the net composition will not change Generally more solute will dissolve at higher temperatures so there is a slope in the saturation line f The intersection of the freezing and saturation lines is the eutectic point g The region bounded by the boiling freezing and saturation lines defines the compositions possible for that solutesolvent mixture However in practice crystallization of either water or solute takes a finite time1 and it is possible on cooling or boiling to overshoot into the supercooled or supersaturated regions for extended periods of time before a phase transition begins To form a crystal the molecules in a liquid must arrange into a crystal lattice shape This takes time particularly if the solution is particularly viscous or there isn t much water present to act as a solvent Furthermore the small crystals formed easily redissolved and it can take a long time to get one large enough that it can grow and act as a nucleus for complete crystallization So when a solution is supersaturated or supercooled the further it is away below or to the right from the line bounding the solution region in the phase diagram 7 the greater the thermodynamic pressure to crystallize However the cooler the solution is the slower the molecules move and the time taken to form crystals is increased similarly the more concentrated the solution the more viscous and again the slower the molecules move and the slower the crystallization rate In principle in very cool concentrated solutions the molecular mobility would tend to zero and therefore the rate of crystallization and chemical reaction would also tend to zero If the coolingconcentrating rate is relatively fast it is possible to get a solution into this region where reactivity is effectively nil without crystallization occurring When molecular mobility lviscosity of a solution approaches zero we say it has become glassy The viscous supersaturated region prior to glass formation is the rubber state The glassy and rubbery states are together described as amorphous and are not crystalline Material in the rubbery state may crystallize if sufficient time is available but glassy material is too viscous to rearrange in to a crystalline form The glass transition line is also shown in Figure 7 To form a glass it is necessary to either concentrate a solution to very high levels by removing water as ice or steam or by cooling a moderately concentrated solution to very low temperatures In either case the concentration must often be quite fast to get to the glassy state before crystallization occurs 1 The reasons for this are explained further in the lipids section 23 Example As an illustration consider how a phase diagram can be used to describe a freezing The food starts as a true solution A which is cooled until it reaches the melting point line B At this temperature a little less than 00 C crystals of pure ice form leaving the food material more concentrated Further cooling concentrates the food more and more with little change in temperature until it reaches TE the eutectic point Thermodynamically the food solute would like to start crystallizing at this point but cannot Instead the food continues along the line into the supersaturated region a rubbery state forms and mobility is rapidly reduced At the glass transition temperature the solute finally locks up and forms a very low mobility glass and most diffusion limited food chemical reactions e ectively stop Cooling beyond this point causes no further change in composition as even the crystallization process of water occurs too slowly in the viscous glass Temp 0C 0 Cone Solute 100 At the bulk scale we see a glassy material as hard and brittle and a rubbery material is soft and pliable While a glass is almost completely unreactive and biochemically stable there can be some reactivity in a rubbery material The key aspect limiting reactivity in both of these cases is the amount of molecular mobility available One measurement of this is the amount a rubbery food must be cooled before it becomes a glass ie TTg A food can be made more stable by cooling so T is lowered to closer to Tg or drying so Tg is raised closer to T Question 3 Glass Transition Use your knowledge of glass transition theory to brie y explain the following phenomenon Wherever possible illustrate your answer with an annotated state diagram a When you boil pasta it goes from brittle to soft Why doesn t pasta soften with dry heat in an oven or harden on cooling after cooking b Hard candies left in a moist environment become sticky on the outside but stay crunchy in the center c Freshly baked bread packed immediately in a plastic bag will develop a soft crust but left in an open container will develop a crispy crust Hint Freshly baked bread is still steaming d To prepare aqueous samples for visualization in an electron microscope it is necessary to freeze them solid This is often done by dripping the sample into a liquid nitrogen slush and cracking open the solid balls formed for imaging Why does this process known as vitrification give better samples than simply putting them in a freezer overnight 24 e Cotton candy is made by melting sucrose then spinning the liquid through narrow nozzles into thin strings What state is the sucrose in cotton candy How and why does it change when left in a moist environment for even a relatively short period of time Close up of the spinning head in a cotton candy machine The center is lled with granular sugar which is heated to melt temperature then spun to force the melt out through narrow holes f Dry corn is a hard brittle material about 14 moisture but it will pop when heated above a critical temperature What phase transitions are important in the formation of popcorn 25 5 Carb0hydrates Carbohydrates are essential in foods as an energy source starch is the main source of human calories a avoring simple sugars are usually sweet and as a functional ingredient sucrose allows ice cream to be soft in the freezer xanthan gum thickens a lowfat salad dressing As with all our approaches to food ingredients constituents we will first examine the structure of carbohydrates and then try to elucidate how their structures allow them to function as they do 1 Carbohydrate Structure As their name suggests carbohydrates basically made up from sugar and water ie CxHzOy although this ratio is often not strictly true and occasionally other atoms may be present The carbons are arranges in a chain most often 56 atoms functionalized with alcohol groups The terminal carbon either carries either an aldehyde or a ketone functional group O Fructose is a ketose sugar ie Glucose has the same atoms as O with a ketone functional group fructose but it is an aldose sugar 0 It has five alcohol groups ie with an aldehyde functional 0 no ie fructose is a polyol group 0 Note that in both of these diagrams 39 J and carbons are 39 39 quot J Each of these sugars contains several chiral carbons so there are many mirror image versions of the same functional groups Fructose has three chiral carbons so 23 different version Glucose has four so 24 different versions All of the different sugars have distinct properties The bond angles around each carbon atom are tetrahedral so each carbon in a sugar has a fixed three dimensional configuration of groups bonded two it If all four of the bonding groups are different then there are two distinct arrangements that cannot be superimposed ie the carbon is chiral The two different versions have the same chemical formula but can have very different properties Chiral carbons are three dimensional structures that are hard to represent on paper so various conventions have been developed to convey the shape information The most common of these is the Fisher projection where the carbon is arranged as the center point of a cross with the up out of plane groups left and right and the down out of plane groups top and bottom A further simplification of the Fisher projection commonly used in sugar chemistry is the Rosanoff projection where the cross shape is retained along with its implications of configuration but only alcohol groups are drawn in as a straight line and aldehydeketone groups as a circle These conventions are illustrated below with glyceraldehydes a threecarbon molecule that is the basis of aldose sugars IIIOCIIEIIOHl IIQOH Dglyeemldeh dc nglycamldehyde 9H0 QHO 39 Ho L H CHZOH g CHO g quotg H OH HO H E quotg a Rustinmull projection Dglyceraldehyde is the starting point for all of the Dseries sugars the most important Three carbon sugars are not common in our diet but we can imagine them as the starting point of a family of molecules created by adding a CHOH between the aldehyde and the rst chiral carbon Each new carbon added is also chiral so there are two alternative con gurations The following diagram shows the Dseries of sugars with D glyceraldehyde in the center The outer ring contains the hexosuloses they have four chiral carbons so 16 enantiomers but only 8 of these are in the Dseries Some important Dsugars are marked on the diagram I 39 I l rnrr l39J39 m I Dxylose I 7 d l r j 39 1L 4 1 I D 1 I 393 i Dglyceraldehyde I ga actose I i quot Dglucose 1 4 39 c3 39139 39 Dm annose The chain forms of sugars shown above are not prevalent in nature The carbonyl carbon part of the aldehyde or ketone functional group can react with an alcohol to form an ether link In sugars the bond angles conspire to make the alcohol on carbon 4 or 5 of the same molecule particularly reactive in this respect yielding a 5 139 e furanose or 6 139 e pyranose 39 A ring I 39y The 39 J alcoholcarbonether structure produced is a hemiacetal C4 H 0 l V I H o t IIO E l O 9 gquot flOH f H Hap4 1 3 39 CH 039 As shown in the diagram the alignment of the hemiacetal alcohol formed from the carbonyl oxygen can either lie axially OLform or equatorially Bform to the ring depending on the way the alcohol initially attacked the carbonyl The new alcohol has made carbon number one chiral so there are two new forms known as the 0L and B anomers Monomeric sugars can therefore from ve possible structures 7 linear open 0 pyranose Bpyranose OLfuranose and Bfuranose The ring forms are by far the most prevalent with the linear form typically present at about 002 at any instant However the system is very dynamic and individual molecules are constantly transforming from ringform to ringform through the linear intermediate The linear from is the only one with a free carbonyl that allows the sugar to take part in important reactions An excellent demonstration of sugar structure is available from the University of Hertfordshire httpWWW herts 20 11k nntsci P 39 39 Rim se 39 htm Oxidation and Reduction An aldehyde can readily be oxidized to a mixture of a carboxylic acid and other products while ketones cannot The oxidation must be coupled with the reduction of another group and this is commonly exploited in a chemical test for reducing sugars where CuII is reduced to CuI and forms a brickred precipitate Ketose sugars can interconvert with aldose sugars and give a positive reducing sugar test and so are also referred to as reducing sugars The interconversion isomerisation is catalyzed by enzyme eg glucose isomerase or high pH Aldehyde groups can be reduced typically by catalytic hydrogenation to their corresponding alcohols ie glucose to glycitol or xylose to xylitol Sugar alcohols are sometimes used as reduced calorie sweeteners particularly in chewing gums and diabetic foods They often give a mild cooling sensation in the mouth Disaccharides A hemiacetal can be attacked by another alcohol to loose water and form a full acetal The attacking alcohol can come from any of the alcohols on a second sugar to form a disaccharide This type of link a glycosidic bond is described as an 0L or 5 depending on the orientation of the hemiacetal alcohol 1 because the hemiacetal carbon is number 1 of the ring where is the number of the carbon on the second bonding sugar Some examples include Maltose or malt sugar is a formed from starch by enzymatic digestion It is used for its mild sweetening taste 7 I and characteristic avor a f Structurally maltose is two 1 glucose pyranose rings joined i by an 0L l4 glycosidic 4 39 O 1 linkage The 1 carbon on the A second ring is unreacted so 3 DH 3 1 u m 5 can open isomerize and close to form 0L and 5 versions of maltose It can also be oxidized so maltose behaves as a reducing sugar H0 Lactose is a major component of the nonfat solids of milk about 5 by weight in uid milk It is a dimer of two different monosaccharides glucose and galactose joined by a 14 B glycosidic link The galactose can still open to form an oxdizable aldehyde therefore lactose is a reducing sugar Lactose is vulnerable to acid hydrolysis and the enzyme lactase is required to split the glycosidic link for digestion Lactase is absent in many adults so the sugar can accumulate in the gut where it is fermented by bacteria to form acid and gas Lactose can crystallize in two types of crystal the OLhydrate and the Banhydrous The 0L form is much less soluble than the 5 form Commercial lactose is extracted from whey and crystallized as alactose Crystals larger than about 10 mm can be perceived and as alactose does not melt well its crystals can lead to quotsandinessquot defect in certain dairy products importantly those with a high solids content and or cold storage condensed milk ice cream To avoid the formation of large crystals it is often useful to add seed crystals to prevent too much supercooling Sucrose is a disaccharide formed by the reaction of OLglucopyranose and B fructofuranose Because the acetal formed is a 11 linkage neither ring can open and expose reducing groups Sucrose is table sugar Trehalose was for years a scientific novelty Certain classes of animal and plant are known as anhydrobiotic ie they can survive drying and freezing eg frogs desert plants and insects It was noted that they accumulate a lot of simple sugars in their tissues before and during this process importantly including mm m n ghaapymmse mmrwxysmhrm malmse Insbellzved that h kebabs malecule ms mm h W h pulymex 5mm mmhy m hd mm h maessaryhydmgenband uhhs that Suppmquot pm bacan Cammzxcu mhalnsus 2mg mu preserve a andfmnn funds 2 Ttha hldRun nn Mallardxeactmns chmhahhg 5 hahhm gmnps m farm bmwn pigments hanmhchmwhha mm hsaccurs Wham mp1 5 hahhm gmnps mm Th Tasman a hhdxyghm pmcee Th nacnan mamas hth Ascm mc mm c can 51m bmwnma whamm mmxbmh th andmthz presence afammn campmmds h m Tasman xespuns nlz rmh manye ects nh mm c h hhhh arha tho aquama ofsz 2h 2 cg w h m oftha ammo W 0 amino 1 1 WW 1 glycoadn hydmxylyoup of sugars 2h aquama 2mmth mfgth ofbmwn mtroganouspolymavsov malanomms fmm yhhh aMhThhp1s hm Chzmlsh39y Th hth behzvedtn hasme a 1 Funnw39nnufan ghuunlninz m m h 1 leaningwizh quota 9 Evans hmyhhhhmghh mac39s wnh h hhh Th hhh hhh mmhm Th m mph pmhem Insfxequzndy lysmz 5 h gm gmnp h m macaw mth acids 2 Rnrnmgunznluhl glytunminzvi a Anudnximuhan39lm Th pmdvcns ahyuma hd h can 21th cychu a farm ghlmsammz ax lsamznn m farm h mmrmms dzth samznmes callzd 1h Am ma 5 Degradation of Amadori product The Amadori product is quite unstable and will 39 1s omerize through several intermediates to form deoxyhexosulose The DH can react further with another amino group or can cyclize to form hydroxy methyl furfural HMF HMF has a mildly sweet caramel avor 4 Condensation and polymerization The HMF and o er intermediates can react with each other and more amino compounds to form a complex mix of high molecular weight polymers The actual reaction mechanism 1s complicated and the structure of the producw only understood up to a point wdicarbonyl e g DH T C O T i O HZNiCiCOOH 1 L amino and esp lysine ii VHZO R l 00 R izNiliCOOH i l 1L we R l c i RiCHO NH 0 l R aldehyde The reactants are largely colorless but the larger and greater num er o polymers formed the darker brown the solution food becomes There is frequently a lag time in the browning reaction before any color is seen during which the intermediates accumulate and the polymers grow to produce color Different starting ingredients can be used to produce different color pigments and this can be exploited to some extent in the controlled formation of process colors Reaction of Maillard Products with Proteins Several Maillard producw contain the highly reactive 0Ldicarbonyl structure e g 3deoxy hexosulose which can react with an amino group in particular the eamino of lysine The adduct breaks down to form an aldehyde amongst other products Aldehydes have strong distinctive aromas and their formation is important in many quotcookedquot avors eg meaty nutty toasted The reaction shown is the Strecker degradation Cross linking of proteins as part of Maillard product formation can lead to some insolubilization and loss of protein functionality Amino acids most importantly lysine are destroyed as a result of the Maillard reaction Several of the heterocyclic compounds formed have been shown to be carcinogenic and mutagenic Some Maillard products are antioxidants and have been claimed to have some health benefits Control of the Maillard Reaction The Maillard reaction proceeds much faster at high temperatures doubling or tripling in rate about every 10 C so keeping a reaction mixture food cool can limit the browning In fact Maillard browning is usually only associated with cooked foods or very long term storage Reactions sometime show a maximum rate at reduced water activities before dropping to zero at or near the monolayer moisture value This effect is particularly strong for Maillard chemistry and intermediate moisture foods brown many times faster than their fully hydrated counterparts The drying process itself is particularly difficult as the foods have to pass through the intermediate moisture dangerzone at elevated temperatures The reaction is slower under acid conditions and is inhibited by the additive sulfur dioxide under acid conditions amine groups may be protonated and less reactive Question 1 Browning Problems 1 Why do roasted and fried foods taste richer and better than boiled or steamed foods 2 Why don t these sugar cookie go brown Ingredients 2 12 cupsflour 1 12 tsp baking powder 34 tsp salt 1 tsp cinnamon 1 cup sugar 34 cup vegetable oil 2 eggs 1 tsp vanilla sugar Directions Sift together flour baking powder salt and cinnamon In a separate bowl combine sugar and oil Add to the second mixture the eggs and vanilla Add the flour mixture all at once andbeat well Shape the dough into 12 inch balls Flatten the balls as thin as you can between lightly floured hands To give a corrugated e ect score them in parallel lines with a fork dipped in flour Sprinkle with granulated sugar Bake about 1012 minutes on a lightly greased cookie sheet Hydrolyzed proteins are often added to bakedroasted products to improve the avor What is protein hydrolysis Why would you expect a hydrolyzed protein to produce more avor on roasting than a whole protein native or denatured Milk is sometimes painted on the surface of cakes and bread to help form a brown crust Why Beer is made by allowing the grains to germinate slightly malting The amylase enzymes are expressed and start to convert starch to glucose After a limited amount of malting the grains are dried and roasted kilning Lager grains to make light beer are roasted to approximately 79 C and stout grains to make dark beer to approximately 105 C What aspects of this are important to the Maillard reaction and the quality of the beer Which ingredients are involved in browning these pretzels 1 cup warm water 1 package dry active yeast 1 12 cupsflour 2 tbs vegetable oil 12 tsp Salt 1 14 cupflour 4 cups water 2 tbs baking soda 2 tbs coarse salt Dissolve the yeast in the warm water and let stand for 10 minutes Add the vegetable oil salt and 12 cups flour Stir together until thoroughly combined Add remaining flour and knead dough for 5 minutes Let the dough rest for hour Divide the dough into 12 equal shapes and reform them into small balls Let them restfor 15 minutes Roll them into 18 lengths and form them into pretzel shapes or cut each length in half to make sticks Preheat oven to 475 degrees In a large pot place the baking soda and water to a boil Let the pretzels rise for a 12 hour Add the pretzels to the boiling water for minute Remove and place on a greased sheetpan Sprinkle with coarse salt and bake for 12 minutes From Freemantle M 2000 Chemical and Engineering News 78 49 p 82 The essential ingredient in all chocolate is cocoa which is made from the creamcoloredbeans that grow in pods on a tree with the botanical name Theobroma cacao The cocoa or cacao tree as it is commonly known is a native of the tropical regions of South and Central America Nowadays it is also cultivated in WestAfrican and SoutheastAsian countries that have humid tropical climates and lie within 20 degrees ofthe equator Afterharvesting the beans are removed from the pods and piled in heaps The growers allow the beans to ferment for several days in order to develop the chemical precursors of the chocolate flavor The beans are then dried and transported to chocolate factories At the factory the cured beans are sorted and impurities such as sand and plant materials are removed The beans are then roasted This process makes the bean shells brittle darkens the color of the beans and converts the beans flavor precursors into the aldehydes esters lactones pyrazines and other groups of compounds that give chocolate its distinctive flavor and aroma The next step is to break up the roasted beans into pieces called nibs and remove the thin shells by blowing air through the beans in a process known as winnowing The nibs are then ground into chocolate liquora thickbrown liquid that solidi es at about room temperature Approximately 55 ofthe liquor is cocoa butter afat consisting ofvarious triglycerides Each triglyceride has three fatty acids attached to a glycerol backbone Oleic acid stearic acid andpalmitic acid accountfor more than 95 ofthefatty acids in cocoa butter The concentration offat in the liquor is too highfor making cocoa powder and too lowfor making socalledeating chocolate The trick is to remove about half of the cocoa butter from the liquor using heavyduty presses and use the butter for making eating chocolate The solid block of cocoa that remains is pulverized The powder is used to manufacture drinking chocolate and cocoa Dairies bakeries and confectionery manufacturers also use the powder as a flavoring ingredient Which chemical and biochemical processes are required to develop chocolate avor Where in this process do they occur 3 Fund Pulymer FuncLiunzlity contort ofpolysaccharldes could equally apply to protelns see above polymerwlll take on an extended randorn coll conflguratlon In solutlon The greater the length ofthe chaln the larger the coll mdlus Any feature that favors polymerrpolymer lnteractlons e g a hydrophoblc backbone wlll tend to collapse the coll to a smaller man strong Slmhm lnteractlons e g averyhydrophlllc chaln wlll favora rnore expanded Coll each coll behaves as an lsolated dd 49 E R E a 8 E 5 8E Dllulc Scull lull Cunccmmmd VISCOM IV Bearlnmln a the spheres we are 1 W h u m l S E 2 5 E E elements responslble for vrscosrty Conccmmllon git 3 rnass actlon of gum 1 can provlde a large dlspersed phase volume fractron and vrscosrty The slrnple StokeerlnsLeln approach Eventually n The Note that solvent present to allow the polymer to be considered a solution and not an amorphous solid A viscous polymer solution will still ow albeit slowly If a force is applied to it for example by tipping a container the uid will immediately respond by owing at a constant rate The higher the viscosity the lower the rate of ow To gel it must form some form of interchain bonds between polymer molecules The interactions between molecules if they extend over the entire container allow it to instantaneously transmit mechanical forces and behave as a solid The two basic pictures of a gel are a particle gel left where the structural objects e g globular proteins fat crystals emulsion droplets are linked together in a network of discrete pieces and a polymer gel right e g mostly polysaccharides where local regions Particle gel of the chains interact to form cross P01 Imer gel lmks but other regions do not If a force is applied to a solid it will stretch but not ow The amount of deformation will increase with the magnitude of the force applied and decrease with the elastic modulus of the gel When the force is removed the gel will spring back into its original shape and be unchanged by the experience We can picture the mechanism of elasticity in terms of the conformation of a polymer chain between two cross linking points In its normal state a the polymer can take on many conformations ie entropy but as it is stretched b the slack in the chain is taken up and eventually drawn tight The stretched polymer has very little exibility between the xed 1 b points ie low entropy When the c H applied force is removed the polymer 1 39 L T L 7 7 will spring back to its original shape to A L Fr A A allow the chain to regain some entropy In fact the behavior of many polymeric foods cannot be readily described in terms of either a viscous liquid eg salad dressing or an elastic gel eg JellO but lie somewhere in between e g bread dough We can investigate the simultaneously solid and liquid like behavior of foods by measuring a property known as mechanical creep In a creep study a force is applied to a food sample for a period of time then removed and the subsequent change in shape monitored mass Du Fnica a 39 rumxv may albumin 3 395 L I w y 7528 M c 11 mo 39 A solid will In a liquid the food Many real foods are instantaneously deform begins to deform at a viscoelastic They will a fixed amount when the constant rate when the partially but not force is applied and will force is applied and completely recover their instantaneously and when the force is original shape and will completely recover removed will stop and do so fairly slowly when it is removed not recover We can understand viscoelasticity by If 7 i T retuming to the conceptual model of A 1 LI T quot 1 quot A elasticity Now instead of imagining Pnh mar has relaxed 1mmutable chem1cal 11nks between the F 39 p L 1 Iquot 1 llmn mg through cha1ns picture the fixed pomts are in 39 r A UN in mum merely tangles of polymer Now when stretched and released the polymer can either 1 pull the two tangles back to closer to each other as if they were fixed points in an elastic solid or 2 allow the chains to slide and worm their way through the tangle and relieve the tension that way In practice both phenomenal will occur simultaneously and the material will partially recover its shape by mechanism 1 and partially dissipate the applied energy as frictional heat and not recover by mechanism 2 In any case the more cross links and the stronger they are the more elastic and less viscous will be a given polymer system In proteins we have seen crosslinks form in the form of disulfide bonds and hydrophobic interactions In polysaccharides the cross linking is more often the formation of multichain helices of polymer supported by hydrogen bonding andor 39 J r 39 39 39 or 39 quot binding of specific ions usually calcium by two different chains Page on hydrocolloid rheology httpwwwlsbuaculdwaterhyrhehtml 4 Starch Starch is tremendously important to plants as their main fuel store and to the animals that eat them From the plant s point of view the goal is to store as much reduced carbon in a small amount of space as possible without disrupting cell function The plant must be able to access the store when needed Given these objectives what design options are open to the plant 0 Small molecule sugars would create a huge osmotic pressure if stored in sufficient quantities to be useful It is necessary to polymerize the sugar to reduce the number of molecules present and hence the osmotic effects A large concentration of free polymers would be too viscous to allow the cell to function probably It is necessary to pack the polymers together in some dense function These needs are met by storing the reduced carbon in starch granules Starch is comprised oftwo types ofpolymeric glucose Amylose is linked via 0Ll4 bonds It typically has a degree of polymerization DP of approximately 180320 units and a molecular weight of about 1000000 About 1 in 200 links there may be a branch but it is basically a linear polymer The axialequatorial nature of the link forces the polymer into a helical structure The core of the helix is slightly hydrophobic so amylose can form inclusion complexes with certain lipids and iodine Amylopectin is a much larger molecule molecular weight 100000000 It is also has about one an 114 chain but approximately every 20 residues there is an XI6 link branching off the chain Amylopectin has one reducing end a glucose that can unfold and expose the aldehyde functional group The 114 chain that ows from that is the Cchain Branches from the Cchain are B chains and branches from that are Achains The overall structure of the molecules is believed to look like a branching tree The dense helices of the Cchains are close to each other and can form a double helix In a typical starch amylose and amylopectin are present at a ratio of about 14 However this ratio will vary according to the biological source of the starch and certain genetic mutants are commercially available with arti cially modified rations Waxy starches are effectively pure amylopectin with a few percent amylose and sugary mutants have elevated levels of amylose e g sumaize A starch granule is a lensshaped object a few micrometers to a few tens of micrometers in diameter Under polarized light it appears as a bright Maltese cross Polarizing light microscopy shows only the crystalline structures as bright but the presence of a bright Maltese cross does not mean just those regions of the granule are crystalline Instead the cross shape is characteristic of radial symmetry about the intersection point of the cross There must be bands of alternating crystalline and noncrystalline ie amorphous materials present in a granule Further evidence for this model comes from electron microscopy studies of granules that have been sliced open Ifthe granule is sliced than etched with acid to digest out the noncrystalline portions a series of raise ie not etched therefore crystalline bands are seen The crystalline regions are believed to be due to the formation of double helices between Achains of the amylopectin molecule The molecule s C chain and single free aldehyde group is orientated towards the center of the granule and radiates out towards the edge The B and A chains radiate from that but only the spacing of the branch points allows the relatively short Achains to naturally lie close together and form their double helices The double helices pack together like stacks of pipes to form the crystalline lamellae seen in electron microscopy and inferred from the polarized light microscopy The location of amylose is not clear in this model but is believed to be non crystalline and lie in the amorphous portions of the granule alongside the B chains of the amylopectin C Char 4 Chain Starch granules are insoluble although they will swell as they take on a little water However as they are heated in the presence of water at a critical temperature they will suddenly loose their Maltese crosses and swell to many times their original diameter in a process known as gelatinizationl In fact complete swelling of granules is often limited due to a lack of available water The swollen granules readily leak amylose out of their structure and can be disrupted by shear forces acting cde mlwounhndmsmm mldocmnmn The need for both heart and moisture to gelatinize starch granules provides a strong suggestion that a glass transition must be involved in starch gm m2 gelatinization and indeed the rst step seems to m t g Z 5 a w 1nvolve the noncrystalllne portlon of the granule As E g the temperature is increased the crystalline regions A f will begin to melt or dissolve in the water but in this case they are physically locked into place by the concentrated fairly dry polymer in the amorphous regions Heat and moisture eventually softens the amorphous region and induces a glass to rubbery transition The softer rubber cannot contain the crystalline regions and they rapidly melt and the granule swells and soaks up a huge amount of water as the structures holding it together are disrupted The amylose is not covalently linked into the granule structure so can diffuse out into the water The network of mainly amylopectin ion the swollen granule is still linked together but can be disrupted under shear to leave granule fragments known as ghosts Indeed some people have suggested that starch granules contain only a single huge amylopectin molecule Time mins 1 Frequently confused with gelation These are different processes Starch gelatinization is associated with a sudden increase in solution viscosity because the hydrodynamic radius of the swollen granule is so many times larger than the native granule However if the swollen granules are intensively mixed or allowed to heat for a prolonged period particularly under acid conditions the viscosity will decrease as the swollen granules are ruptured If a suspension of swollen granules is allowed to cool it will gelz As we have seen in our general discussion of macromolecular functionality gelation requires the formation of some form of interchain linkage to support solidlike behavior In starch the links are initially the formation of double helices between amylose chains ie retrogradation The amylose molecules are smaller and are therefore able to diffuse faster but over the next days and weeks the larger amylopectin molecules may also begin to retrograde and increase the gel strength The natural gelatinization and retrogradation behavior of starch are important in controlling the texture of many foods but natural granules are not sufficiently robust or are otherwise inconvenient for many food processing operations Various chemically and physically modified starch ingredients are available to improve on nature Some specific defects and their solutions include 0 Starch must be cooked to act as a thickening agent Pregelatinized starch is made by quickly cooking a starch paste between two hot rollers and drying the product Pregelatinized starch rapidly rehydrates without further cooking and is a useful thickening agent in dried sauces and salad dressings that require no further cooking Starch suspensions are not stable to heating Swollen granules rapidly break down in hot stirred and acid conditions and loose their viscosity A classic example of this problem is when cans of fruit pie filling mix being retorted The heat acid and very high temperatures rapidly depolymerize the swollen granules and the finished product would be fruit pieces sitting at the bottom of can of colored sugar syrup Chemical cross links can be added to the alcohol groups of the starch chain ie cross linked starch to stabilize the molecule and prevent breakdown Cross links also make a granule slightly more resistant to gelatinization Typical cross links include acetic anhydride esters and they are used at levels of DS3000202 Cross linked starch must be labeled as modified starch Starch gels change their properties during storage The slow retrogradation of amylopectin means the texture of a starch gel will change and frequently show some syneresis Monofunctional reagents commonly acetate or phosphate DS0 1 can be chemically added to the starch molecules These modifications do not fit well into the double helical arrangement and thus inhibit retrogradation and allow the formation of a stable gel The stabilized starches formed must also be labeled as modified starch As well as an ingredient in its own right starch is a useful feedstock to make other reagents by partially or completely depolymerizing them Depolymerization can involve 2 Frequently confused with gelatinization These are different processes 3 Defined as Degree of Substitution the number of free alcohols per residue containing a modification The maximum DS is therefore 3 either acid or enzymatic hydrolysis of the polymers The more extensive the hydrolysis the lower the molecular weight of the product The size of glucose oligimers made from starch hydrolysis is usually expressed as a dextrose ie glucose equivalents reducing power of the sample relative to that of glucose In general the lower the molecular weight of the product or higher the DE the lower the viscosity of the product the sweeter the avor the greater the tendency towards nonenzymatic browning and the easier it is for the products to crystallize Some examples of starch gelatinization products include dextrins daltodextrin corn syrup DE2060 a few glucose units 5 Other Polysaccharides See also httpwwwlsbuaculdwaterhydrohtml The great advantage of starch is the tremendous efficiency of midwestem corn farmers means it is incredibly cheap However limitations to the functionality of even modified starch in combination with the costs and label declarations associated with modification mean other polysaccharides have useful roles as food ingredients typically as gelling and thickening agents Cellulose and Cellulose Derivatives Cellulose is the main polymeric content of plants and is mainly located as a structural material in cell walls Chemically cellulose molecules are like very large amylose molecules but the glycosidic links are 514 as opposed to 0Ll4 in amylase which are indigestible by human enzymes and so cellulose behaves as dietary fiber The 514 links in cellulose allow the molecule to lie as a at ribbon while 514 links put a step into the chain conformation that facilitated helix formation in amylose In the cell wall the cellulose chains lie parallel to one another and form crystalline regions supported by strong interchain hydrogen bonds However the very long chains are hard to line up to form a perfect crystal over their entire length and the crystalline regions are interspersed with noncrystalline amorphous regions Cellulose fibers form into bundles Tquot connected at intervals along their length by the crystalline regions They are very IMO if q rigid and strong and are completely 39 water insoluble They are resistant to m our digestive enzymes and make up the bulk of the dietary fiber we consume Cellulose can be modified to make a number of useful products Most simply the amorphous regions can be hydrolyzed out to form short rods of microcrystalline cellulose This can be dried to form an insoluble powder useful both as a fiber additive and as a dusting form shredded cheese to prevent it sticking The microcrystalline cellulose bundles can be disrupted by intense shear and acid The fibers formed are then derivatized to form carboxymethyl cellulose methyl cellulose or hydroxypropyl methyl cellulose Cellulose gums tend to be highly viscous and can be used to form gels Xanthan Gum Xanthan gum is an extracellular polysaccharide produced by the bacterium xanthomonis campesm39s The xanthan gum backbone is very similar to cellulose Bl4 polyglucose but does not crystallize because alternate residues have an XI6 linked 3sugar residue attached specifically 3 l0Llinked DmannopyranoseZ l BD glucuronic acid 4 lBD mannopyranose The side chains contain carboxylic acid residues glucuronic acid which are negatively charged at all but the lowest pH and their intrachain electrostatic repulsion helps keep the molecule relatively linear in solutions The extended conformation of a xanthan molecule contributes to its high viscosity The molecules also form weak transient cross links as double helices between two chains Xanthan gum is a non gelling polysaccharide but forms very viscous temperaturestable solutions Alginate Alginate is extracted from seaweed It is a linear polysaccharide containing two types of residue ie a co polymer BD mannopyranosyluronic acid and M xLgulopyrasonic acid G The structure tends to be either sequences of lV V V V VlM GGGGGG or MGMGMGMGM The MG ratio is important in determining the overall quality of the product The polyguluronic acid regions Gblock can bind calcium with the negative charge on the carboxylic acid residues to form the eggbox structure shown Alginate solutions are viscous and will gel in the presence of calcium or acid to form soft temperaturestable gels Proplylene glycol can be reacted with the carboxylic acid residues to form propylene glycol alginate PGA that has a reduced sensitivity to calcium Pectin Pectin is extracted from fruit particularly citrus pulp Chemically it is largely a linear polymer of polygalacturonic acid with varying degrees of methyl esterification If more than 50 of the IMP 311 acid groups are present as methyl esters ie DEgt50 it is quot classified as a highmethoxy pectin otherwise it is a low methoxy pectin It also contains hairy regions with considerable side chains of the main backbone High I Gal with VHFquot s 1 It iul 1 ill milth amt lml u llsl methoxy pectin will gel in the presence of acid and high J 7 sugar concentrations and low methoxy pectin will gel in the a in am presence of calcium but there is a continuum of properties um rpm15 between these two extremes Gelation of pectin requires the formation of crosslinks between two different polymer chains The charged groups on the galacturonic acid residues repel one another and this can only be overcome for highmethoxy pectin by acidifying to protonate and remove charge and adding sugar to compete for the hydration water Low methoxy pectin carries too much charge to aggregate under these conditions and instead calcium is The calcium ions complex with the negative charges from the galacturinic acid to form local polymerpolymer bonds The hairy regions never cross link and allow the pectin to form a wellhydrated gel Question 2 Polysaccharide Applications to be completed in class Work as a group of 35 students to investigate the applications of a given polysaccharide as a food ingredient The first class meeting for this exercise will be held in the computer lab and you will search the internet for suppliers of your ingredient Prepare a a list of the major applications of this product b a list of about 5 of the major suppliers c for a selected important supplier d a table summarizing the major variants of this product available details of how the products are chemically or physically distinct and the suggested applications of the variation Finally go to a supermarket and try to purchase a real representative example of your applications I will reimburse receipts up to 15 per group Organize your material in the form of a PowerPoint presentation under the title Applications of YOUR GUM and email it to me before noon of the day of the following class meeting In the next possible two class sessions I will provide a brief lecture on the structure and functionality of each of the example ingredients I will then ask a representative of each group or more than one person 7 whatever you prefer to give a brief talk on applications Your talk should cover the material in your PowerPoint presentation and should be illustrated with the examples you have purchased


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