Food Structure and physical properties - Appunti studio completi per preparazione esame

Food structure and physical properties

Types of food materials

Eight groups of different foods:

• Viscous fluid ex: Honey
• Plastic or elastic gels ex: Jam, Pectin
• Fibrous material ex: Meat (Hierarchical organization of fiber)
• Water contained in cell, ex: Fruit
• Greasy material (unto, fat). For example, Gorgonzola, where the core can flow, and the external material is like a film. This difference is due to the enzymatic process able to hydrolyze the lipid and protein.
• Dried and crunchy material: for example, biscuits. There are two different structure types (granulose or crystalline). Granulose is made of several small crystals. For example, in fried potatoes, there are several empty bubbles. When bitten, many of them and the thin layers between the bubbles break down and produce the typical sound.
• Glassy materials, for example, candies, flour, pasta
• Aerated material, bread crumb, ice cream, beer foam

Food systems

The food can be a homogeneous or dispersed system. The focus is on the interaction among different compounds rather than their concentration. The difference between Grana Padano and Mozzarella is the addition of enzymes. The rennet breaks down the bond between two parts of casein micelles. It forms curd "CAGLIATA"; in the mozzarella, it isn’t important to break down this bond, so an elastic protein network is formed through the lowering of pH.

The process that involves food doesn’t go through thermodynamic equilibrium, intended as a solution of different compounds. Most food materials can’t reach thermodynamic equilibrium. Food isn’t just the mixing of different compounds. The structure depends on factors like temperature, pH, and Aw of the making process. In most cases, food structure is given from non-covalent bonds like Van der Waals, hydrogen bonds. Meanwhile, the covalent bond breaking requires high energy processes, an example being the roasting of coffee beans, where fibers go through carbonization and release CO₂.

Dimension and complexity of food structure

We can discriminate three sciences:

• Polymer science, about molecular level
• Colloid science, about supermolecules level, like drops, bubbles, fibers, colloids
• Food material structure, about macroscopic level, life suspension, gel, foams, composite

A composite material is like concrete, so a material made of different elements that set down in its firm state while it chills. An example is torrone, or chocolate. First, the melting of fat is needed in order to mix all the compounds. When the temperature decreases, the system is allowed to form crystals responsible for the food structure. Chocolate is made of cocoa powder, sugar, cocoa butter, whey protein, and flavoring. You can recognize the sugar crystal from its sharp edges, and the protein for its irregular shape. The different bond types are notable from the difference in terms of energy.

An example of hierarchical structure is the apple, which consists mainly of a sugar solution with water trapped in wall cells to give the whole solid body of the apple that we know.

Structural definition of food processing

Structure preservation is illustrated with myofibrils and an apple section. Note the empty spaces shown in the microscopic picture that are the result of an artifact because of the necessary removal of water from food material samples.

Yeast-produced enzymes are pectin hydrolysis and cellulose hydrolysis, and these metabolic processes cause an increase in bacterial count. The microbiological proliferation can be inhibited by refrigeration and atmospheric conditions. A bacterial total count like 10³ is considered low-pathogen probable presence, while a bacterial count total like 10¹⁰ is considered dangerous.

The picture shows tomato cells disposed around the vegetal vessels. In the wheat grain picture, you can see the starch granules stored in the vegetal cells. Starch granules are protected by water. The contact of water with the starch crystal contained in the amyloplast will cause expansion in volume due to the extraordinary capacity of starch to absorb water, and eventually germination. These are the reasons behind the air-dry operation "ESSICAZIONE".

The bean picture shows the protein matrix made of albumin. Cooked beans are characterized by gelatinized starch.

Structure disruption

Controlled disruption of food structure is necessary to:

• Release desired compounds
• Simplify food material management
• Produce ingredients

Grounding, homogenization, cutting, slicing, sectioning, and chewing are operations able to disrupt the potato tissue structure. Structure disruption is always associated with higher instability due to the loss of natural compartmentalization and preservation systems.

Structure transformation

The structure of ingredients is transformed into final foods by a sequence of unit operations. For almond cells, they were tightly packed in the fresh sample. Their contents appear contracted after roasting. The minute pits are void spaces developed in the matrix following the extraction of oil bodies. Roasting is conducted at temperatures higher than 200°C. During roasting, there is a volume expansion, and a huge amount of carbon dioxide is produced. In fact, the most quantitative compound produced in the Maillard reaction is CO₂. Produced carbon dioxide pushes the almond structures outward. During warming of food material, until there is a single molecule of water, the temperature is 100°C. When all water is gone, the temperature rises to 250°C.

In the bread crumb shown in the photo, gluten is highlighted in red color and highly hydrated in gel form starch. The starch can make only hydrogen bonds, meanwhile proteins can form hydrogen bonds and disulfide bridges, with the latter being much stronger than H-bonds. For fried onion, you can see cell structures are splashed due to the fact water is entirely replaced by oil during frying.

In cooked meat, you can see the disappearance of muscle fiber and a much more opened structure due to the increase of hydrophilic properties. In the meat, you can recognize the fat globules highlighted in white color. The casein micelles in yogurt and cheese curd respectively are examples of a secondary protein structure that exposes hydrophilic groups to the watery medium. The fat globules in cheese products are protected by protein structure from the aqueous environment. Most proteins have numerous hydrophilic groups so proteins can stay near the fat globules. Between fat globules and protein, there is always an aqueous phase.

Chocolate annealing "CONCAGGIO" is the process operation that guarantees the proper type of fat crystal in the final product, in order to inhibit chocolate blooming. In margarine, you can see the fat medium in a crystalline structure that entraps water droplets. Liquids other than water are the fat fraction with a higher amount of low-boiling fatty acid.

Water in food

Aw is measured by the measurement of %RH of headspace air, which is in the food sample. Aw can assume values from 0 to 1 because it is calculated by a ratio.

Water activity: equilibrium property of water

The amount of water in the sample is a quantitative extensive property that depends on the amount of material. The energy status of water in the system is a qualitative intensive property that doesn’t depend on the amount of material. It's not a driving force, whereas Aw is a driving force. Empirical methods with no standard exist for measuring the amount, while there are known standards (like salt solutions) for measuring water activity.

IMF: Intermediate moisture food like jam has an Aw of 0.5.

Aw is influenced by:

• Solute interactions (colligative effect): freezing point depression, boiling point elevation, osmotic pressure, vapor pressure lowering
• Capillary suction forces
• Surface force interactions

According to Raoult's law, where X is the water molar fraction:

Raoult's law can also be written as follows:

Since P/P°=Aw, Raoult's law is equal to:

Exercises

• Calculate the amount of CaCl₂ (PM 111) to be solubilized in 1L water to get a solution with 0.75 Aw. Compare this amount to that required to get the same using NaCl (PM 58) instead of CaCl₂. N = 1000g (1L)/18 = 55.5 mol H₂O 55.50.75 = 55.5 + CaCl₂ = 0.08 mol = 9.76g For ideal systems, with good approximation for dilute solutions containing low MW species. For real systems where γ is the activity coefficient of water, accounting for deviations from the linearity of Raoult's equation vs X_w, γ is between 0 (max deviation) and 1 (ideal).

Ross' equation

Aw = Aw_i x Aw_j x Aw

Aw and capillarity: The Aw of aqueous solution in the capillarity is always lower than that of the same solution in a larger container because in the capillary the water is retained (TRATTENUTA).

Surface force interactions: They are caused by interactions among water and polar molecules (dipole-dipole forces, ionic bonds, dipolar ionic interactions, van der Waals forces, and hydrogen bonding). The effect is the decreasing of Aw.

Monolayer value: in which each polar and ionic group has a water molecule attached to it, to form the start of a liquid-like phase. The water of the monolayer requires extra energy to turn into vapor, resulting in a reduced Aw.

Effect of temperature on Aw

The effect of temperature on Aw is described by the Clausius-Clapeyron equation where a₁ and a₂ are Aw values at T₁ and T₂ (K) respectively; R is the gas constant (1.987 cal/mole K); Q is the latent heat of evaporation.

The pressure of temperature on Aw is less proportional to temperature rise compared to the pressure of a solution. At sub-freezing temperatures, water can exist as supercooled water (SCW) and ice. Ice and SCW have different vapor pressures.

Water activity of frozen foods depends on the ratio between P_ice and P_scw.

During the temperature fall, the food material records an increase in viscosity. Below the freezing point, the water activity of frozen foods depends on the ratio between P_ice and P_scw. Aw depends on the straight line in the graph and is thus linearly related to the temperature, following the correlation: AW = 0.0088(T°C) + 0.9974.

Exercises

• Calculate the Aw of a food rich in water when stored at -18°C. Through this linear correlation, you can estimate the Aw of frozen food material at the temperature of -18°C. Aw is equal to 0.842.
• The ice cream base made of milk, cream, and sugar has 0.985 Aw at 20°C. What will its Aw be at -14°C? The Aw of the cream is about 0.85 and it will occur a moisture migration to the waffle, making it soggy.
• The same mix is used to produce a frozen dessert containing an osmo-dried fruit filling with Aw 0.90 at 20°C. Will the product stored at 14°C have any issues of water transfer between the two components? You can’t expect very intense moisture migration between dried fruit and ice cream.
• If a biscuit with 0.3 Aw at 20°C was added, are you expecting any water transfers between the biscuit and the ice cream stored at 14°C?
• If yes to the previous point, could the problem be solved by keeping the product at 18°C?

How to measure water activity?

Aw is measured by assessing ERH%, which refers to the relative humidity of the atmosphere surrounding the food. In all Aw-meters, the food is placed in a holder and inserted in a small chamber where the equilibrium between liquid and vapor water should be reached.

Aw meters:

• Electric hygrometer: based on the moisture dependence of electric resistance or polymer capacitance. Changes in electric resistance or polymer capacitance with environmental RH% generate an electric signal that is converted into an Aw value. Advantages/disadvantages: Accurate and reliable, a calibration with salt solutions at known Aw is required, time requiring, not adequate for Aw measurement of foods with volatile substances like alcohol.
• Dew point hygrometers: measure the temperature of the dew point "PUNTO DI RUGIADA" (the temperature at which the air in the headspace becomes saturated with moisture and water droplets condense). The decrease in temperature needed to condense headspace moisture allows the RH% to be calculated. The instrument is equipped with a stainless-steel mirror, a fan to cool the system, and a thermometer. Advantages/disadvantages: Accurate and reliable, very fast. Calibration is not required, periodic cleaning and maintenance are required, not adequate for food with volatile compounds.

Determination or calculation of food Aw

Experimental measure using a hygrometer or calculation using predictive equations:

• Raoult (only with the exact concentrations of solute and their molecular weight).
• Ross equation, the Aw can have different origins: Experimentally measured, calculated with predictive equations, calculated from isotherms, obtained from literature data.
• Lerici’s equation: for the contribution of ethanol.

Calculating Aw based on the freezing point depression

The calculation can be used for single components, especially volatile ones for whom it is not possible to experimentally measure Aw. It is used with ethanol-containing solutions, and the contribution of the volatile ingredient to Aw can be calculated according to the freezing point depression (FPD) with the following empirical equation. The FPD value must be calculated.

Example

• Calculate the Aw of a 6% (w/w) hydroalcoholic solution. The handbook shows that the FPD of a 6% solution is 2.54° and so the freezing point is 270.46K.
• Exercise: Calculate the Aw value of an alcoholic beverage containing: water 100 mL; ethanol 15 g; sucrose 5 g; lemon flavor 0.1%. The lemon flavor’s effect on Aw is negligible. To calculate the contribution of sucrose to Aw, the Raoult equation is used. To calculate ethanol contribution to Aw, the Lerici equation is used: water 100g ethanol 15g total 115g. %EtOH (w/w): 100/18 = 0.997. 100 5( ) + (342)18. Lerici for ethanol: 15:115=x:100. x=13% (p/p). Δcr=6.13°C 266.87K. Aw=0.997*0.939=0.9362.
• Calculate the Aw of a wine containing 11% (w/w) ethanol and 3.5% (w/w) sugar. Assume that all sugars are made of glucose and other wine components (glycerol, salts, etc.) have a negligible effect.

What do we do with Aw?

Stability and food preservation during storage can be obtained.

Moisture vs Aw prediction curve-isotherm

Water sorption isotherm: a sorption isotherm is the relation between food moisture and water activity at constant temperature.

• Zone 1: Water is strongly adsorbed by water-dipole and water-ion interactions. This zone corresponds to the monolayer.
• Zone 2: Water is structured in multilayers interacting with polar molecules and in capillaries. Water-water and water-dipole hydrogen bonding prevail.
• Zone 3: Water is weakly bounded, being only held within the biological structure and shows properties like pure water and easily evaporates.

Sorption and desorption isotherms

Isotherms can be experimentally obtained starting from a wet food (desorption) or dried food (adsorption):

• Desorption: The hydrated food is progressively dried to get the isotherm.
• Adsorption: The dried food is progressively hydrated to get the adsorption isotherm.

Hysteresis effect:

• Stearic rearrangement: In the absence of water, polar molecules may irreversibly bond. The number of functional groups interacting with water decreases. At the same Aw, food is able to adsorb less water upon rehydration.
• Phase transition: Upon drying, molecules may undergo phase transitions (crystallization, glass transition, protein denaturation...) these structures have different water sorption capacity than original ones.
• Capillary water: During desorption, water may remain embedded in food pores and capillaries. At the same Aw, food undergoing desorption has a higher water content.

Isotherm shape

The sigmoidal shape of the isotherms provides information about the nature and intensity of the interactions between water and food:

• Type 1: At low Aw, a moisture increase corresponds to a slight increase in Aw. Food contains highly polar molecules able to strongly interact with water.
• Type 2: Classic behavior of foods interacting with water by both colligative and capillary suction forces. At low Aw, a slight increase in moisture corresponds to a significant increase in Aw. Food has a good capacity of water sorption.
• Type 3: Food has very low capacity for water sorption (sugar crystals, microcrystalline cellulose).

Effect of composition on isotherm shape

• Fat: Essentially no moisture adsorption, isotherm similar on a non-fat basis.

Exercise

Below are the sorption data at 20°C of a food made of starch and sugars. Food formulation is modified by adding 10 g sunflower oil per g dry matter. Draw the sorption isotherm of the reformulated food. At Aw: 0.11, M is 1, so 1 g of food dry matter adsorbs 1 g of H₂O. If 10 g of fat is added, the dry matter becomes 1+10 g =11. So, M becomes 1/11=0.090. For example, at 0.33 M is 3, so 3 g of water per 1 g of food dry matter, with 10 g of fat I have 3/11=0.27.

Proteins

Easy hydration. 3-6 g water/100g protein at Aw 0.26 and 10-12g/100g protein at Aw 0.6, heat denaturation reduces adsorption. In protein, we have type II isotherm.

Carbohydrates

Polymers

More H-bonds than protein but some of them can be intermolecular and thus not easily hydrated. There is a great variety of behaviors: gums are easily hydrated, starch needs more than 30% moisture and heat for rehydration, cellulose is crystalline and has different behavior (type III) than other polysaccharides. Usually, there is a type II isotherm but other types are possible.

Monomers and dimers

When sugars are crystalline, water interacts only at the crystal surface. The crystal size affects the amount of water bound. When the water amount is equal to the solubility limit, sugar dissolves into water (sugar→deliquescence) usually there is type III isotherm.

Exercise

Let’s consider the following ingredients/foods: Chestnut flour, Light ricotta cheese, Pork sausage. Discuss the possible interactions based on their composition and water activity.