In the Kjeldahl method, crude protein is usually calculated by multiplying the nitrogen content by 6.25. What is the basis of this factor?

Food proteins contain on average about 16% nitrogen by mass, and 100/16 = 6.25

Kjeldahl digestion recovers only about 16% of the nitrogen, so the result is scaled up by 6.25

The distilled ammonia reacts with the standard acid in a 1 : 6.25 ratio during the final titration

The factor converts the mass of ammonium sulphate formed in digestion into the mass of ammonia

Which of the following is an example of intelligent rather than active packaging?

A label that changes colour progressively with the cumulative time and temperature a chilled pack has experienced

An iron-powder sachet placed inside a sealed packet of salted nuts that absorbs the residual oxygen left in its headspace

A sachet of potassium permanganate on a porous carrier that removes ethylene from a box of mangoes

A label that changes colour progressively with the cumulative time and temperature a chilled pack has experienced

A wrapping film that slowly releases an antimicrobial agent onto the surface of sliced cheese

Which enzyme is most commonly used to judge whether vegetables have been blanched adequately before freezing, and why?

Peroxidase, because it is among the most heat-stable vegetable enzymes, so its loss implies the others are gone

Alkaline phosphatase, because its inactivation coincides with the destruction of Coxiella burnetii in vegetables

Peroxidase, because it is among the most heat-stable vegetable enzymes, so its loss implies the others are gone

Polyphenol oxidase, because it is the most heat-labile enzyme present and so the first one destroyed on heating

Amylase, because its activity is the main reason frozen vegetables lose their sweetness during long storage

Which of the following statements about the thermal processing of canned foods are correct?

A food with pH above 4.6 and water activity above 0.85 must receive a process designed against spores of Clostridium botulinum

A 12D process at a given temperature lasts twelve times the D-value of the target spores at that temperature

A food with pH above 4.6 and water activity above 0.85 must receive a process designed against spores of Clostridium botulinum

A commercially sterile can contains no living microbial cell or spore of any kind

A 12D process at a given temperature lasts twelve times the D-value of the target spores at that temperature

Tomatoes acidified to pH 4.2 need the same F₀ as unacidified green peas, because both are vegetables

UHT milk processed at about 140 °C for a few seconds tastes much less cooked than milk sterilised in the bottle at about 115 °C for 15–20 min, although both are commercially sterile. Which explanation is correct?

Spore destruction has a much smaller z-value (about 10 °C) than browning and flavour reactions (25 °C or more), so higher temperature cuts the time needed far more than it speeds quality loss

In UHT plants the spores are destroyed mainly by the high pressure of the steam rather than by the temperature, so much less heat reaches the milk overall

Browning and flavour reactions have a much smaller z-value than spore destruction, so at the higher temperature they slow down markedly while the spores die far faster than they did at 115 °C

At 140 °C the Maillard reaction cannot proceed at all, because the lactose is converted entirely to lactulose before it is able to react with milk proteins

Spore destruction has a much smaller z-value (about 10 °C) than browning and flavour reactions (25 °C or more), so higher temperature cuts the time needed far more than it speeds quality loss

A batch of soft dried apricots with a water activity of 0.70 is stored at room temperature in a sealed pack. Which group of organisms is the most likely cause of eventual spoilage?

Xerophilic moulds and osmophilic yeasts, which can grow at water activities around 0.6–0.7

Clostridium botulinum, because the sealed pack creates anaerobic conditions in which spores germinate

Pseudomonas species, which are the main spoilage bacteria of moist foods kept at room temperature

Lactic acid bacteria, which are favoured by the sugars that leach from the fruit during storage

Xerophilic moulds and osmophilic yeasts, which can grow at water activities around 0.6–0.7

Strawberries frozen slowly in a domestic freezer release much more drip on thawing than strawberries frozen in a liquid-nitrogen tunnel. What is the main reason?

Slow freezing forms a few large ice crystals outside the cells that draw water out and rupture cell walls

Slow freezing leaves a larger fraction of the water permanently unfrozen, and this water drains away as drip

Slow freezing forms a few large ice crystals outside the cells that draw water out and rupture cell walls

Slow freezing inactivates the pectin methylesterase that would otherwise firm the fruit during thawing

Fast freezing removes water from the berries by sublimation, so less water remains to form drip later

Sodium benzoate preserves a lemon squash of pH 2.8 effectively but protects a sauce of pH 6.5 poorly at the same concentration. Taking the pKa of benzoic acid as 4.2, what explains this?

At pH 2.8 most of it is undissociated benzoic acid, the uncharged form that crosses microbial membranes and acidifies the cytoplasm

At pH 6.5 the benzoate ion reacts with the sugars of the sauce and is converted into an inactive ester

Benzoic acid is destroyed by heat at neutral pH during the hot filling of the sauce, although it remains fully heat-stable at acid pH

At pH 2.8 the benzoate ion binds calcium in the squash, and this calcium complex is the actual antimicrobial agent

Which of the following statements about intermediate-moisture foods and hurdle technology are correct?

Humectants such as glycerol and sorbitol lower water activity while the food stays soft and moist to the touch

At a water activity of 0.80 a mould inhibitor is still useful, because many moulds can grow at that level

Humectants such as glycerol and sorbitol lower water activity while the food stays soft and moist to the touch

At a water activity of 0.80 a mould inhibitor is still useful, because many moulds can grow at that level

Each hurdle in a hurdle-preserved food is set at a level that would by itself be lethal to the target microbes

Intermediate-moisture foods must be rehydrated before they can be eaten, as dried vegetables must

When a frozen block of minced meat is thawed in a microwave oven at full power, the edges begin to cook while the centre is still frozen. Which explanation is correct?

Liquid water has a far higher dielectric loss factor than ice, so regions that thaw first absorb most of the energy and heat ever faster

Ice has a far higher dielectric loss factor than liquid water, so the frozen centre absorbs the energy and reflects it out towards the edges

Microwaves heat only the outer surface, as infrared does, so the centre warms solely by slow conduction from the edges

Microwave energy is absorbed mainly by the fat in the meat, and in a minced block the fat collects at the edges

Which of the following statements about non-thermal processing technologies are correct?

In high-pressure processing, pressure acts uniformly and almost instantly throughout a packed food, whatever its size or shape

Cold plasma inactivates microbes mainly on surfaces, through reactive oxygen and nitrogen species and ultraviolet light

In high-pressure processing, pressure acts uniformly and almost instantly throughout a packed food, whatever its size or shape

Pulsed electric field treatment suits solid foods packed with air pockets, because the trapped air conducts the pulses

Cold plasma inactivates microbes mainly on surfaces, through reactive oxygen and nitrogen species and ultraviolet light

A pressure of 600 MPa at room temperature reliably destroys bacterial spores, so treated low-acid foods need no refrigeration

In a multilayer squeeze bottle for ketchup, the ethylene–vinyl alcohol (EVOH) layer is placed between layers of polypropylene rather than on the inner or outer surface. Why?

EVOH is an excellent oxygen barrier only when dry, and the polypropylene layers keep moisture away from it

EVOH is a poor oxygen barrier, and it is the polypropylene layers that provide the oxygen resistance of the bottle

EVOH dissolves in acetic acid, so it has to be kept away from the vinegar in the ketchup at all times

EVOH is an excellent oxygen barrier only when dry, and the polypropylene layers keep moisture away from it

EVOH cannot be printed on, so it is hidden inside while the outer polypropylene layer carries the label

A tray of fresh chicken pieces is gas-flushed with 100% CO₂ and sealed with a high-barrier film. After a day in chill storage the film is sucked tightly onto the meat. Which change to the gas mixture would best prevent this while keeping the antimicrobial effect?

Replace part of the CO₂ with nitrogen, an inert and poorly soluble filler that maintains the pack volume

Replace part of the CO₂ with oxygen, which is less soluble and also brightens the colour of the chicken

Use a film of much higher CO₂ permeability, so that the CO₂ escapes before it can dissolve in the meat

Replace all the CO₂ with nitrogen, which is just as bacteriostatic as CO₂ and does not dissolve in the meat

Replace part of the CO₂ with nitrogen, an inert and poorly soluble filler that maintains the pack volume

Food Processing Principles: Quality Analysis, Thermal and Non-Thermal Preservation, Packaging and Storage

This chapter opens Section XL5.3, “Food Products Technology”, of XL5 Food Technology, one of the five optional sections of the Life Sciences (XL) paper (the Engineering Sciences (XE) paper sets the same four headings as its section XE6 Food Technology, one of nine, so this chapter serves a candidate of either paper). It answers the first three items of that heading: “Processing Principles”, from food characterisation and quality analysis through thermal processing, chilling, freezing, dehydration, preservatives and additives, microwave, radiowave and infrared processing, irradiation, fermentation and hurdle technology to intermediate moisture foods; “Non thermal technologies: Ultrasonication, HPP, Pulse Electric Field, Cold plasma”; and “Food packaging and storage”, including active, intelligent and aseptic packaging, controlled and modified atmosphere storage and microplastics in food. The numericals are moisture on wet and dry basis, Kjeldahl protein, F-value accumulation, irradiation dose and time, and shelf life from Q₁₀.

1. Characterising foods and analysing their quality

Proximate analysis divides a food into moisture, protein, fat, ash, crude fibre and carbohydrate. Moisture is the mass lost in a hot-air oven at about 105 °C to constant weight; sugar-rich or heat-sensitive foods go to a vacuum oven at lower temperature, very dry foods to Karl Fischer titration, and spices to distillation with toluene so that volatile oil is not counted as water. Protein is found by Kjeldahl: digestion in concentrated H₂SO₄ with a catalyst converts organic nitrogen to (NH₄)₂SO₄, NaOH releases NH₃, which is distilled into boric acid and titrated with standard acid. Protein = N × 6.25, because proteins average 16% nitrogen (milk uses 6.38, wheat 5.7). Fat is extracted with petroleum ether or hexane in a Soxhlet apparatus (milk fat by the Gerber method). Ash is the residue after burning in a muffle furnace near 550 °C, crude fibre the residue after boiling in dilute acid and then dilute alkali, and carbohydrate is taken by difference, so it absorbs the errors of every other fraction.

Instruments quantify single components (HPLC for sugars and vitamins, gas chromatography for fatty acids and volatiles, atomic absorption for minerals). Quality analysis of raw and processed foods measures what the buyer judges. Physical: colour as CIE Lab* (L* lightness, a* red–green, b* yellow–blue), texture by texture profile analysis (a two-bite compression giving hardness, cohesiveness, springiness, chewiness), viscosity, and total soluble solids by refractometer in °Brix. Chemical: pH, titratable acidity, peroxide value and free fatty acids. Microbiological: total plate count, coliforms, yeasts and moulds. Sensory: difference tests such as the triangle test, and liking on the 9-point hedonic scale.

⚠️ Exam trap

Kjeldahl measures total nitrogen, not protein. Non-protein nitrogen such as urea or melamine added to milk or feed raises the apparent “protein”, which is exactly how such adulteration deceives a nitrogen-based test.

2. Thermal processing: blanching, pasteurisation and sterilisation

Heat must deliver enough lethality with the least loss of colour, flavour and nutrients. Blanching is a brief exposure to hot water or steam given to vegetables before freezing, drying or canning. It inactivates enzymes (peroxidase, catalase, lipoxygenase, polyphenol oxidase) that would cause off-flavours and discoloration in storage, expels tissue air, wilts leaves for filling and lowers surface microbes. Peroxidase is the usual index of adequate blanching because it is among the most heat-stable vegetable enzymes: if it is gone, the others are too. Water blanching leaches more vitamin C and minerals than steam blanching.

Pasteurisation is a mild heat treatment that destroys vegetative pathogens and reduces spoilage organisms, so the product still needs refrigeration or acidity. Milk is held at 63 °C for 30 min (LTLT, batch) or 72 °C for 15 s (HTST, continuous plate heat exchanger), a schedule set against Coxiella burnetii, the most heat-resistant non-spore-forming pathogen of milk. A negative alkaline phosphatase test confirms it, because the enzyme is slightly more heat-resistant than those pathogens. Sterilisation of food means commercial sterility: nothing survives that could grow under normal storage, though dormant thermophilic spores may remain. Low-acid foods (pH above 4.6, aw above 0.85) must be processed against spores of Clostridium botulinum, which cannot grow and make toxin at pH 4.6 or below; acid foods such as most fruits need only a boiling-water or hot-fill process.

Microbial death is first order, so log N falls linearly with time. The D-value is the time at a stated temperature for a one-log (90%) reduction; the z-value is the temperature rise that cuts D tenfold. A process is stated as its F-value, the equivalent minutes at a reference temperature; F₀ uses 121.1 °C and z = 10 °C. As the cold point of a container heats and cools slowly, F is accumulated: F₀ = Σ 10^((T − 121.1)/10) Δt, each minute at T counting as its lethal rate 10^((T − 121.1)/10). Thus 5 min at 111.1 °C (lethal rate 0.1) adds only 0.5 min. The 12D or botulinum cook demands twelve decimal reductions of C. botulinum spores; with D₁₂₁.₁ ≈ 0.21 min that is about 2.5 min, and commercial low-acid processes usually deliver F₀ of 3 min or more.

🎯 Why UHT and HTST work

Spore death speeds up tenfold per about 10 °C (z ≈ 10 °C), but browning and vitamin loss have z of roughly 25 °C or more. Higher temperature therefore cuts the time needed for sterility far more than it speeds quality loss, so UHT (about 135–150 °C for a few seconds) gives sterile milk far less cooked in taste than in-bottle sterilisation.

3. Chilling, freezing, dehydration and water activity

Chilling (about 0–5 °C) slows microbial growth, respiration and enzyme action without stopping them; banana, mango and other tropical produce suffer chilling injury (pitting, browning, failure to ripen) below roughly 10–13 °C, so they are stored warmer. The temperature sensitivity of shelf life is summarised by Q₁₀, the factor by which a deterioration rate rises for a 10 °C increase: shelf life at T₂ = shelf life at T₁ ÷ Q₁₀^((T₂ − T₁)/10). With Q₁₀ = 2, a snack keeping 12 months at 20 °C keeps 6 months at 30 °C; accelerated tests run warm are extrapolated back this way. Freezing concentrates the solutes as ice forms, so some water stays unfrozen even at −18 °C. Slow freezing forms a few large extracellular crystals that draw water out of cells and rupture walls, giving drip loss on thawing; quick freezing (air blast, plate, IQF, liquid nitrogen) passes rapidly through the zone of maximum crystal formation (about −1 to −5 °C) and forms many small crystals. Fluctuating storage temperature causes recrystallisation; sublimation from bare surfaces causes freezer burn.

Dehydration preserves by lowering water activity, aw = p/p₀, the vapour pressure of water over the food divided by that of pure water at the same temperature (equilibrium relative humidity ÷ 100). Approximate lower limits for growth are 0.91 for most bacteria, 0.88 for most yeasts, 0.80 for most moulds, 0.75 for halophilic bacteria and about 0.6 for xerophilic moulds and osmophilic yeasts. Lipid oxidation is slowest near monolayer moisture (aw about 0.2–0.3) and browning fastest at intermediate aw. In hot-air drying a constant-rate period removes surface water as from a wet bulb; the falling-rate period, controlled by internal diffusion, begins at the critical moisture, and drying too fast at first gives case hardening, a dry impervious skin. Freeze drying sublimes ice under vacuum below the triple point of water (611 Pa, 0.01 °C) and best keeps shape, aroma and rehydration.

🧠 Wet basis or dry basis?

Wet-basis moisture is water ÷ total mass (never above 100%); dry-basis moisture is water ÷ dry solids (often above 100%), and M_db = M_wb/(1 − M_wb). Do drying balances on the dry solids, which do not change: 100 kg of carrot at 88% (wb) holds 12 kg solids, so dried to 6% (wb) it weighs 12/0.94 = 12.77 kg and 87.2 kg of water has gone.

4. Preservatives, additives, fermentation, hurdles and intermediate-moisture foods

Weak-acid preservatives act mainly in the undissociated form, which crosses the cell membrane and releases protons inside. Benzoic acid (as sodium benzoate, pKa about 4.2) therefore works in acid foods such as squashes, juices and ketchup but is weak at neutral pH; sorbic acid (as potassium sorbate, pKa about 4.8) is chiefly antifungal, for cheese and bakery goods; propionates check mould and rope in bread. Sulphur dioxide, usually from potassium metabisulphite (KMS), is antimicrobial, antioxidant and anti-browning, but it bleaches anthocyanins, so it is avoided in naturally red or purple products, and it destroys thiamine. Nitrite fixes cured-meat colour and restrains C. botulinum. Indian rules have long classed salt, sugar, vinegar, spices, honey and edible oil as Class I preservatives and chemical ones such as benzoate and SO₂ as Class II. Other additives include antioxidants (BHA, BHT, tocopherols), emulsifiers (lecithin, monoglycerides), stabilisers (pectin, gums, CMC), humectants, acidulants, colours and sweeteners, each allowed only in listed foods at listed levels.

Fermentation preserves by turning sugars into acids or alcohol: lactic acid bacteria lower the pH of curd, sauerkraut and brined pickles, yeasts make ethanol and acetic acid bacteria make vinegar. Hurdle technology, the concept developed by Leistner, combines several mild barriers (reduced aw, low pH, mild heat, preservatives, low redox potential, competitive flora), none lethal alone, so that together they overwhelm the microbes’ homeostasis while the food stays close to fresh. Intermediate-moisture foods (IMF) are the classic case: with aw roughly 0.6–0.85 (texts differ on the upper limit) and about 20–50% moisture, they are soft and eaten without rehydration, yet keep at room temperature because humectants (sugar, salt, glycerol, sorbitol) hold aw below the bacterial limit, while a mould inhibitor such as sorbate and a lowered pH cover the moulds and yeasts that can still grow.

5. Microwave, radio-frequency, infrared and ionising radiation

Microwave (domestic ovens at 2450 MHz; industry also uses lower frequencies near 915 MHz) and radio-frequency heating (13.56, 27.12 and 40.68 MHz) generate heat inside the food: polar water molecules rotate as they try to follow the alternating field (dipolar rotation) and dissolved ions oscillate (ionic conduction). Heating is volumetric, and penetration depth increases as frequency falls, so RF heats thick blocks more evenly (thawing, disinfesting grain, post-bake drying). Microwave heating is fast but uneven: corners overheat, salt confines heating to the surface, and in frozen food thawed water absorbs far more strongly than ice, so thawed spots run away while ice stays frozen, which is why defrost cycles pulse the power. Infrared (about 0.75–1000 µm) is absorbed close to the surface and suits roasting, toasting, peeling, drying thin layers and browning crusts.

Irradiation exposes food to ionising radiation: gamma rays from cobalt-60 (or caesium-137), electron beams of up to 10 MeV or X-rays of up to 5 MeV (the Codex limits), energies too low to make the food radioactive. The main target is DNA, hit directly or through radicals from the radiolysis of water, so larger, more complex organisms need lower doses: insects and parasites die at low doses, vegetative bacteria need more, spores more still, while viruses and preformed toxins are hardly affected. Dose is measured in gray (1 Gy = 1 J absorbed per kg). Low doses (below 1 kGy) inhibit sprouting of potato and onion, disinfest grain and delay ripening; medium doses (1–10 kGy) give radurisation (fewer spoilage organisms) and radicidation (elimination of non-spore-forming pathogens such as Salmonella in spices and frozen seafood); doses above 10 kGy give radappertisation, commercial sterility for special diets. Microbial kill follows D₁₀, the dose for one log reduction. Since 1 kGy deposits only 1 kJ per kg, even 3 kGy warms water-like food by under 1 °C, hence “cold process”. Irradiated food carries the Radura symbol.

6. Non-thermal technologies: HPP, PEF, ultrasonication and cold plasma

Non-thermal methods inactivate vegetative microbes and some enzymes while the food stays near room or chill temperature, so fresh flavour, colour and heat-labile vitamins survive. None reliably kills bacterial spores on its own, so treated low-acid products still need refrigeration or a combination with heat.

Four non-thermal technologies compared
Technology	Principle	Typical uses	Limitations
High-pressure processing (HPP)	300–600 MPa for minutes, transmitted isostatically (uniformly and at once, whatever the pack size or shape); disrupts membranes and non-covalent bonds, not covalent ones	Packed juices, avocado products, ready meals, shucking oysters	Spores and some enzymes resist; batch process; packs must be flexible and water-tight
Pulsed electric field (PEF)	Microsecond high-voltage pulses across a pumpable food open pores in cell membranes (electroporation); about 20–80 kV/cm to kill microbes	Juices, milk, liquid egg; at about 1 kV/cm, softening potatoes for fries and aiding extraction	Air bubbles and large particles cause arcing; spores resist
Ultrasonication	Power ultrasound (20–100 kHz) causes cavitation: collapsing bubbles create local hot spots, shear and free radicals	Extraction, emulsification, cleaning; kills best with heat (thermosonication) or heat and pressure (manothermosonication)	Alone it kills slowly; radicals can cause off-flavours
Cold plasma	Partly ionised gas near room temperature from an electric discharge; reactive oxygen and nitrogen species, UV and charged particles attack cell surfaces	Decontaminating seeds, spices, fresh produce and packaging, even inside a sealed pack; degrading pesticide residues	Acts on surfaces only; may oxidise lipids and pigments

7. Packaging, storage atmospheres and microplastics

Common packaging materials
Material	Strengths	Weaknesses
Glass	Inert, transparent, total gas and moisture barrier	Heavy, brittle, sensitive to thermal shock
Metal (tinplate, aluminium)	Strong, total barrier to light, gas and moisture, withstands retorting	Needs an internal lacquer against corrosion
Paper and paperboard	Cheap, printable, renewable	Poor barrier unless coated or laminated
Polyolefins (PE, PP)	Good water-vapour barrier, heat-sealable; PP survives retorting	Poor oxygen barrier
Polyethylene terephthalate (PET)	Clear, strong, good gas barrier; bottles	Less heat-resistant than glass
EVOH, PVDC	Excellent oxygen barrier	EVOH loses its barrier when wet, so it is buried between polyolefin layers

No single plastic does everything, so films are laminated or co-extruded (a retort pouch is typically PET / aluminium foil / polypropylene). In aseptic packaging the product and the package are sterilised separately and united in a sterile zone: the liquid is UHT-treated and cooled, the packaging material is sterilised (commonly with hot hydrogen peroxide) and filling and sealing take place under sterile air, so light paperboard–polyethylene–foil cartons keep milk or juice for months unrefrigerated. Active packaging changes conditions inside the pack: oxygen scavengers (iron-powder sachets), CO₂ emitters or absorbers, ethylene absorbers (potassium permanganate on a porous carrier), moisture absorbers, antimicrobial films. Intelligent packaging only monitors and communicates: time–temperature indicators, freshness indicators that respond to amines or CO₂ from spoilage, leak indicators, RFID tags.

Controlled-atmosphere storage holds produce in a gas-tight store whose O₂ and CO₂ are measured and corrected continuously; low O₂ (a few per cent) and raised CO₂ slow respiration, ethylene action and ripening, which is how apples keep for months. Modified-atmosphere packaging (MAP) sets the gas once, at sealing, after which respiration and permeation change it: CO₂ is bacteriostatic but dissolves in moist food, N₂ is an inert filler that stops the pack collapsing, and red meat gets high O₂ to keep bright red oxymyoglobin. For respiring produce the film is chosen so that O₂ entering and CO₂ leaving balance respiration; if O₂ falls too low, anaerobic respiration gives ethanol and off-flavours.

Microplastics are plastic particles smaller than 5 mm (below about 1 µm they are commonly called nanoplastics). They reach food from the environment (seafood, especially bivalves eaten whole, sea salt, drinking water) and from processing and packaging (abrasion of plastic equipment and chopping boards, release from bottles, tea bags and containers, more so when heated). Microscopy with FTIR or Raman spectroscopy identifies them and pyrolysis–GC–MS quantifies them. The concern is physical (particle uptake), chemical (additives such as phthalates and bisphenols, and pollutants adsorbed on the particles) and microbial (biofilms); health effects and safe levels are still under research.

Key takeaways

Proximate analysis: oven moisture, Kjeldahl protein (N × 6.25), Soxhlet fat, ash near 550 °C, crude fibre by acid then alkali, carbohydrate by difference. Kjeldahl sees total nitrogen, so non-protein nitrogen fakes protein.
Peroxidase indexes blanching, alkaline phosphatase milk pasteurisation (63 °C/30 min or 72 °C/15 s). Low-acid foods (pH > 4.6, aw > 0.85) need at least a 12D botulinum cook; F₀ = Σ 10^((T − 121.1)/10) Δt.
Shelf life falls by Q₁₀ for each 10 °C rise; quick freezing gives small crystals and little drip; aw limits are about 0.91 for bacteria, 0.88 yeasts, 0.80 moulds, 0.6 xerophiles. Do drying balances on dry solids.
Weak-acid preservatives act as the undissociated acid, so benzoate needs acid foods. Hurdle technology stacks sub-lethal barriers; intermediate-moisture foods rely on humectants plus a mould inhibitor.
Microwaves and RF heat volumetrically, infrared heats surfaces. Irradiation dose is in gray (J/kg): below 1 kGy for sprouting and insects, 1–10 kGy for radurisation and radicidation, above 10 kGy for radappertisation; dose = log reductions × D₁₀.
HPP, PEF, ultrasound and cold plasma kill vegetative cells, not spores. Active packs act, intelligent packs report; controlled atmospheres are corrected continuously, MAP is set once; microplastics come from the environment and from packaging.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

In the Kjeldahl method, crude protein is usually calculated by multiplying the nitrogen content by 6.25. What is the basis of this factor?
1. Food proteins contain on average about 16% nitrogen by mass, and 100/16 = 6.25
2. Kjeldahl digestion recovers only about 16% of the nitrogen, so the result is scaled up by 6.25
3. The distilled ammonia reacts with the standard acid in a 1 : 6.25 ratio during the final titration
4. The factor converts the mass of ammonium sulphate formed in digestion into the mass of ammonia
Show answer
Answer: A — Food proteins contain on average about 16% nitrogen by mass, and 100/16 = 6.25
Average food proteins hold about 16 g N per 100 g, so protein ≈ N × 100/16 = N × 6.25; foods whose proteins differ use other factors (6.38 for milk, 5.7 for wheat). The digestion is designed to recover essentially all organic nitrogen, not 16% of it, and the titration is a simple 1 : 1 neutralisation of ammonia (as borate) by acid, so neither explains the factor.
Which of the following is an example of intelligent rather than active packaging?
1. An iron-powder sachet placed inside a sealed packet of salted nuts that absorbs the residual oxygen left in its headspace
2. A sachet of potassium permanganate on a porous carrier that removes ethylene from a box of mangoes
3. A label that changes colour progressively with the cumulative time and temperature a chilled pack has experienced
4. A wrapping film that slowly releases an antimicrobial agent onto the surface of sliced cheese
Show answer
Answer: C — A label that changes colour progressively with the cumulative time and temperature a chilled pack has experienced
A time–temperature indicator monitors and communicates the product’s history without changing the food or the headspace, which is what defines intelligent packaging. Oxygen scavengers, ethylene absorbers and antimicrobial-releasing films alter conditions in the pack, so they are active packaging; the oxygen-scavenger sachet is tempting because it looks like an indicator insert, but it removes a gas.
A 0.500 g sample of a pulse flour is analysed by the Kjeldahl method. The ammonia distilled into boric acid requires 8.40 mL of 0.100 N HCl, and a reagent blank requires 0.20 mL. Taking the atomic mass of nitrogen as 14.0 and a protein factor of 6.25, the crude protein content of the sample (%, to two decimal places) is ____.
Numerical answer — type the value.
Show answer
Answer: 14.35
Net titre = 8.40 − 0.20 = 8.20 mL. Nitrogen = 8.20 mL × 0.100 meq/mL × 14.0 mg/meq = 11.48 mg. N = 11.48/500 × 100 = 2.296%. Protein = 2.296 × 6.25 = 14.35%. Forgetting the blank gives 14.70%.
Which enzyme is most commonly used to judge whether vegetables have been blanched adequately before freezing, and why?
1. Alkaline phosphatase, because its inactivation coincides with the destruction of Coxiella burnetii in vegetables
2. Peroxidase, because it is among the most heat-stable vegetable enzymes, so its loss implies the others are gone
3. Polyphenol oxidase, because it is the most heat-labile enzyme present and so the first one destroyed on heating
4. Amylase, because its activity is the main reason frozen vegetables lose their sweetness during long storage
Show answer
Answer: B — Peroxidase, because it is among the most heat-stable vegetable enzymes, so its loss implies the others are gone
Peroxidase is one of the most heat-resistant enzymes of vegetables, so a negative peroxidase test shows that blanching also inactivated the less stable enzymes. Alkaline phosphatase is the index of milk pasteurisation, not of blanching, and a heat-labile enzyme would be a poor indicator because its loss says nothing about tougher enzymes.
Which of the following statements about the thermal processing of canned foods are correct?
1. A food with pH above 4.6 and water activity above 0.85 must receive a process designed against spores of Clostridium botulinum
2. A commercially sterile can contains no living microbial cell or spore of any kind
3. A 12D process at a given temperature lasts twelve times the D-value of the target spores at that temperature
4. Tomatoes acidified to pH 4.2 need the same F₀ as unacidified green peas, because both are vegetables
Show answer
Answer: A — A food with pH above 4.6 and water activity above 0.85 must receive a process designed against spores of Clostridium botulinum; C — A 12D process at a given temperature lasts twelve times the D-value of the target spores at that temperature
Low-acid foods (pH > 4.6, aw > 0.85) can support growth and toxin formation by C. botulinum, so they receive at least a 12D botulinum cook, which by definition lasts twelve D-values at the process temperature. Commercial sterility tolerates dormant spores (often thermophilic) that cannot grow under normal storage, so “no living cell of any kind” is wrong. Tomatoes at pH 4.2 are an acid food in which C. botulinum cannot grow, so they need a far milder process than low-acid peas.
The temperature at the cold point of a can during processing is approximated as 6 min at 113.1 °C, then 10 min at 118.1 °C, then 4 min at 120.1 °C; lethality outside these periods is negligible. Taking Tref = 121.1 °C and z = 10 °C, the F₀ value delivered (min, to one decimal place) is ____.
Numerical answer — type the value.
Show answer
Answer: 9.1
Lethal rate L = 10^((T − 121.1)/10). At 113.1 °C, L = 10^(−0.8) = 0.1585, contributing 6 × 0.1585 = 0.951 min. At 118.1 °C, L = 10^(−0.3) = 0.5012, contributing 5.012 min. At 120.1 °C, L = 10^(−0.1) = 0.7943, contributing 3.177 min. F₀ = 0.951 + 5.012 + 3.177 = 9.14 ≈ 9.1 min. Simply adding the 20 min of heating would grossly overstate the lethality.
UHT milk processed at about 140 °C for a few seconds tastes much less cooked than milk sterilised in the bottle at about 115 °C for 15–20 min, although both are commercially sterile. Which explanation is correct?
1. In UHT plants the spores are destroyed mainly by the high pressure of the steam rather than by the temperature, so much less heat reaches the milk overall
2. Browning and flavour reactions have a much smaller z-value than spore destruction, so at the higher temperature they slow down markedly while the spores die far faster than they did at 115 °C
3. At 140 °C the Maillard reaction cannot proceed at all, because the lactose is converted entirely to lactulose before it is able to react with milk proteins
4. Spore destruction has a much smaller z-value (about 10 °C) than browning and flavour reactions (25 °C or more), so higher temperature cuts the time needed far more than it speeds quality loss
Show answer
Answer: D — Spore destruction has a much smaller z-value (about 10 °C) than browning and flavour reactions (25 °C or more), so higher temperature cuts the time needed far more than it speeds quality loss
Spore lethality rises tenfold every ~10 °C, while cooked flavour, browning and vitamin loss rise tenfold only every ~25 °C or more. At high temperature the time needed for the same F₀ shrinks far more than the quality reactions speed up, so their total extent falls. The reversed statement gets the z-values the wrong way round (no reaction slows down on heating); pressure in a UHT plant only keeps water liquid, and Maillard browning does occur in UHT milk, just to a smaller extent.
A vegetable purée contains 80% moisture on a wet basis. It is dried until its moisture content is 0.10 kg water per kg dry solids. The mass of water removed per 100 kg of purée (kg) is ____.
Numerical answer — type the value.
Show answer
Answer: 78
100 kg of purée holds 20 kg dry solids and 80 kg water. The dried product holds 0.10 × 20 = 2 kg water (the moisture is on dry basis, i.e. per kg of solids). Water removed = 80 − 2 = 78 kg. Reading 0.10 as a wet-basis fraction would give a product of 20/0.90 = 22.2 kg and 77.8 kg removed.
In an accelerated shelf-life test, a packaged snack reaches its rejection point after 30 days at 40 °C. If the Q₁₀ of the deterioration reaction is 2.5 and stays constant, its expected shelf life at 25 °C (days, to the nearest whole number) is ____.
Numerical answer — type the value.
Show answer
Answer: 119
Shelf life varies inversely with rate: θ(25 °C) = θ(40 °C) × Q₁₀^((40 − 25)/10) = 30 × 2.5^1.5 = 30 × 3.953 = 118.6 ≈ 119 days. Multiplying by 2.5 × 1.5 instead of raising to the power 1.5 gives 112.5 days, a common slip.
A batch of soft dried apricots with a water activity of 0.70 is stored at room temperature in a sealed pack. Which group of organisms is the most likely cause of eventual spoilage?
1. Clostridium botulinum, because the sealed pack creates anaerobic conditions in which spores germinate
2. Pseudomonas species, which are the main spoilage bacteria of moist foods kept at room temperature
3. Lactic acid bacteria, which are favoured by the sugars that leach from the fruit during storage
4. Xerophilic moulds and osmophilic yeasts, which can grow at water activities around 0.6–0.7
Show answer
Answer: D — Xerophilic moulds and osmophilic yeasts, which can grow at water activities around 0.6–0.7
Most bacteria need aw of about 0.91 and most yeasts about 0.88, so at 0.70 only xerophilic moulds and osmophilic yeasts (limits near 0.6) can grow. C. botulinum is tempting because the pack is sealed, but it needs aw well above 0.9, which anaerobiosis cannot make up for; Pseudomonas and lactic acid bacteria are likewise excluded by the low aw.
Strawberries frozen slowly in a domestic freezer release much more drip on thawing than strawberries frozen in a liquid-nitrogen tunnel. What is the main reason?
1. Slow freezing leaves a larger fraction of the water permanently unfrozen, and this water drains away as drip
2. Slow freezing forms a few large ice crystals outside the cells that draw water out and rupture cell walls
3. Slow freezing inactivates the pectin methylesterase that would otherwise firm the fruit during thawing
4. Fast freezing removes water from the berries by sublimation, so less water remains to form drip later
Show answer
Answer: B — Slow freezing forms a few large ice crystals outside the cells that draw water out and rupture cell walls
Slow passage through the zone of maximum crystal formation gives a few large extracellular crystals; water migrates out of the cells to them, and the crystals damage walls and membranes, so cell water escapes as drip on thawing. The unfrozen fraction at a given storage temperature depends on solute concentration and temperature, not on the freezing rate, and rapid freezing does not dry the fruit by sublimation.
Sodium benzoate preserves a lemon squash of pH 2.8 effectively but protects a sauce of pH 6.5 poorly at the same concentration. Taking the pKa of benzoic acid as 4.2, what explains this?
1. At pH 2.8 most of it is undissociated benzoic acid, the uncharged form that crosses microbial membranes and acidifies the cytoplasm
2. At pH 6.5 the benzoate ion reacts with the sugars of the sauce and is converted into an inactive ester
3. Benzoic acid is destroyed by heat at neutral pH during the hot filling of the sauce, although it remains fully heat-stable at acid pH
4. At pH 2.8 the benzoate ion binds calcium in the squash, and this calcium complex is the actual antimicrobial agent
Show answer
Answer: A — At pH 2.8 most of it is undissociated benzoic acid, the uncharged form that crosses microbial membranes and acidifies the cytoplasm
By Henderson–Hasselbalch, at pH 2.8 the ratio benzoate ion : acid is 10^(2.8 − 4.2) ≈ 0.04, so about 96% is undissociated; at pH 6.5 the ratio is 10^2.3 ≈ 200 and under 1% is undissociated. Only the uncharged acid diffuses through the membrane and releases protons inside the cell. No ester with sugars or calcium complex is involved, and benzoic acid is not destroyed by hot filling.
Which of the following statements about intermediate-moisture foods and hurdle technology are correct?
1. Humectants such as glycerol and sorbitol lower water activity while the food stays soft and moist to the touch
2. At a water activity of 0.80 a mould inhibitor is still useful, because many moulds can grow at that level
3. Each hurdle in a hurdle-preserved food is set at a level that would by itself be lethal to the target microbes
4. Intermediate-moisture foods must be rehydrated before they can be eaten, as dried vegetables must
Show answer
Answer: A — Humectants such as glycerol and sorbitol lower water activity while the food stays soft and moist to the touch; B — At a water activity of 0.80 a mould inhibitor is still useful, because many moulds can grow at that level
Humectants bind water and lower aw without making the product hard and dry, which is the basis of IMF. Most moulds grow down to about aw 0.80 (xerophiles lower), so IMF need a mould inhibitor such as sorbate. Hurdles are deliberately sub-lethal individually and work in combination, and IMF are eaten as they are, without rehydration.
Frozen shrimp are to be irradiated to achieve a 5-log reduction of Salmonella, whose D₁₀ in this product is 0.70 kGy. If the minimum-dose point of the load receives 6.0 kGy per hour, the minimum exposure time (min) is ____.
Numerical answer — type the value.
Show answer
Answer: 35
Required dose = 5 × D₁₀ = 5 × 0.70 = 3.5 kGy. Time = 3.5 kGy ÷ 6.0 kGy h⁻¹ = 0.583 h = 35 min. The time is set by the minimum-dose point; points nearer the source receive more.
When a frozen block of minced meat is thawed in a microwave oven at full power, the edges begin to cook while the centre is still frozen. Which explanation is correct?
1. Liquid water has a far higher dielectric loss factor than ice, so regions that thaw first absorb most of the energy and heat ever faster
2. Ice has a far higher dielectric loss factor than liquid water, so the frozen centre absorbs the energy and reflects it out towards the edges
3. Microwaves heat only the outer surface, as infrared does, so the centre warms solely by slow conduction from the edges
4. Microwave energy is absorbed mainly by the fat in the meat, and in a minced block the fat collects at the edges
Show answer
Answer: A — Liquid water has a far higher dielectric loss factor than ice, so regions that thaw first absorb most of the energy and heat ever faster
Once a region thaws, its liquid water couples strongly with the field and heats rapidly, which raises its absorption further (runaway heating); the remaining ice absorbs weakly and stays frozen. The reversed loss-factor statement is wrong, and microwaves do penetrate some centimetres, unlike infrared. Defrost settings pulse low power so that heat can conduct into the ice between bursts.
Which of the following statements about non-thermal processing technologies are correct?
1. In high-pressure processing, pressure acts uniformly and almost instantly throughout a packed food, whatever its size or shape
2. Pulsed electric field treatment suits solid foods packed with air pockets, because the trapped air conducts the pulses
3. Cold plasma inactivates microbes mainly on surfaces, through reactive oxygen and nitrogen species and ultraviolet light
4. A pressure of 600 MPa at room temperature reliably destroys bacterial spores, so treated low-acid foods need no refrigeration
Show answer
Answer: A — In high-pressure processing, pressure acts uniformly and almost instantly throughout a packed food, whatever its size or shape; C — Cold plasma inactivates microbes mainly on surfaces, through reactive oxygen and nitrogen species and ultraviolet light
Pressure is transmitted isostatically, so HPP treats every part of a pack alike. Cold plasma’s reactive species and UV act at surfaces, which is why it is used for seeds, produce and packaging. PEF needs a pumpable liquid free of air bubbles, which cause dielectric breakdown and arcing, and bacterial spores survive HPP at room temperature, so low-acid HPP foods must be kept chilled.
In a multilayer squeeze bottle for ketchup, the ethylene–vinyl alcohol (EVOH) layer is placed between layers of polypropylene rather than on the inner or outer surface. Why?
1. EVOH is a poor oxygen barrier, and it is the polypropylene layers that provide the oxygen resistance of the bottle
2. EVOH dissolves in acetic acid, so it has to be kept away from the vinegar in the ketchup at all times
3. EVOH is an excellent oxygen barrier only when dry, and the polypropylene layers keep moisture away from it
4. EVOH cannot be printed on, so it is hidden inside while the outer polypropylene layer carries the label
Show answer
Answer: C — EVOH is an excellent oxygen barrier only when dry, and the polypropylene layers keep moisture away from it
EVOH has one of the lowest oxygen permeabilities of food plastics but is hydrophilic: absorbed moisture plasticises it and its barrier falls sharply. Burying it between moisture-resistant polyolefin layers keeps it dry. EVOH is the oxygen barrier of the structure, so the claim that polypropylene provides the oxygen resistance is backwards; polypropylene is a poor oxygen barrier.
A tray of fresh chicken pieces is gas-flushed with 100% CO₂ and sealed with a high-barrier film. After a day in chill storage the film is sucked tightly onto the meat. Which change to the gas mixture would best prevent this while keeping the antimicrobial effect?
1. Replace part of the CO₂ with oxygen, which is less soluble and also brightens the colour of the chicken
2. Use a film of much higher CO₂ permeability, so that the CO₂ escapes before it can dissolve in the meat
3. Replace all the CO₂ with nitrogen, which is just as bacteriostatic as CO₂ and does not dissolve in the meat
4. Replace part of the CO₂ with nitrogen, an inert and poorly soluble filler that maintains the pack volume
Show answer
Answer: D — Replace part of the CO₂ with nitrogen, an inert and poorly soluble filler that maintains the pack volume
CO₂ is bacteriostatic but highly soluble in the water and fat of meat, so the headspace shrinks as it dissolves and the pack collapses. Nitrogen is inert and poorly soluble, so a CO₂/N₂ mixture keeps the volume while retaining enough CO₂ to inhibit spoilage bacteria. Nitrogen itself is not bacteriostatic, so replacing all the CO₂ loses the antimicrobial effect; high oxygen is used for the oxymyoglobin colour of red meat, which poultry does not need, and it favours aerobic spoilage; a permeable film simply loses the CO₂.