Slide content - Labs Flashcards

1
Q

What is the mechanism of the BCA assay?

A
  • Peptide bonds in protein reduce Cu2+ to Cu1+ in the presence of a base
  • BCA reagent is alkaline (~pH 11)
  • Temperature-dependent assay!!
  • Every assay needs a new standard curve
  • The amount of Cu2+ reduced is proportional to the amount of protein in the sample
  • BCA reagent chelates the Cu1+ from the protein to form a complex
  • 2 moles of BCA to 1 mole of Cu1+
  • Not a reaction, no bonds are formed
  • Creates a colored product that can be quantified at λ = 562 nm
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2
Q

The BCA assay involves a reaction between the BCA reagent and copper.
True or False?

A

False.
No bonds formed.
BCA reagent chelates the Cu+1 (reduced from Cu2+ by peptide bonds in protein) to form a complex

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3
Q

The BCA assay is temperature-dependent.
True or False?

A

True.
Every assay needs a new standard curve.

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4
Q

The BCA agent is acidic.
True or False?

A

False.
It is alkaline.

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5
Q

The BCA reagent is alkaline.
True or False?

A

True.

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6
Q

Why dry foods? [3]

A
  • Microbial activity
    • Not enough water available for bacterial, yeast, or mould growth
    • Does not necessarily kill microbes unless temp >60°C used
  • Enzymatic activity
    • Produces conformational changes in the enzyme, affecting its catalytic activity
  • Chemical reactions
    • Reactants cannot move as freely in the food and thus reaction rates are lower; however, oxidation rates may increase
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7
Q

What are advantages of air-drying?

A

Relatively inexpensive equipment
Can modify drying conditions (temp, air flow, humidity, time)

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8
Q

What are disadvantages of air drying?

A
  • case hardening
  • poor rehydration
  • loss of bioactive compounds
  • slow - especially in later stages
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9
Q

What are the 3 stages of freeze drying?

A
  1. Freezing phase – Freeze product to below it’s triple point so that sublimation rather than melting will occur
  2. Sublimation phase – Pressure is lowered and “heat” (-20°C) is added to shelves, causing water to sublimate; 95% of water is removed
  3. Adsorption phase – Ionically-bound water molecules are removed by raising the temperature (~ 0 - 30°C or above) to break the bonds between water and the materials; 1-5% residual moisture
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10
Q

Describe the drying phases in air drying.

A
  • Heating phase (Product is brought up to drying temperature
  • Constant rate zone (heat is added at the same rate as heat is loss due to evaporation of water; most water is removed
  • Falling rate zone (Last 10% of water is removed; Rate of removal is slow due to bound water and case hardening)
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11
Q

What are advantages of freeze-drying? [4]

A
  • no/little shrinkage
  • no case hardening
  • excellent rehydration
  • good retention of bioactive compounds
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12
Q

What are disadvantages of freeze drying? [4]

A
  • expensive process
  • Slow
  • Cannot achieve partial drying - all or nothing
  • Porous structure; susceptible to oxidation and moisture uptake
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13
Q

What type of drying takes the longest?

A

Freeze drying

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14
Q

What type of drying produces the least dense product?

A

Freeze drying

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15
Q

What type of drying has the best rehydration?

A

Freeze-drying

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16
Q

What does the phenol sulfuric acid method measure?

A
  • Total carbohydrate (using a standard curve)
  • Other food components can interfere with absorbance if they absorb light at the same wavelength (e.g., flavonoids)
17
Q

What does the enzymatic method measure?

A

Free glucose + fructose content (via NADPH)
Less subject to interference by food components due to specificity of enzyme active site.

18
Q

Describe the phenol sulfuric assay mechanism.

A
  1. Sulfuric acid breaks down polysaccharides to monomers and dehydrates them to furfural derivatives
  2. Furfural derivatives react with phenol to produce yellow-gold colour (food dilution should be clear initially)
  3. Absorptivity of solution is measured at 490 nm; colour is stable for several hours
  4. Standard curve of representative sugar is used for quantification; the reaction is not stoichiometric
19
Q

What wavelength is measured in the phenol sulfuric acid assay?

A

490 nm

20
Q

What are advantages of the phenol-sulfuric acid assay? [5]

A
  • Accuracy is within ±2% under proper conditions
  • Simple
  • Rapid
  • Inexpensive
  • Reagents are stable
21
Q

What are disadvantages of the phenol sulfuric acid assay? [3]

A
  • Harsh chemicals
  • Subject to interference by colour and other food components
  • Accuracy depends on how well standard solution represents sugar composition in food
22
Q

What is the principle of the enzymatic method for glucose determination?

A
  • Glucose is converted to glucose-6-phosphate (G6P) by hexokinase (HK) + ATP
  • Glucose-6-phosphate dehydrogenase (G6P-DH) + NADP+ oxidizes G6P to gluconate-6-phosphate and forms NADPH which is then measured at 340 nm and = [glucose]
23
Q

What is the principle of the enzymatic method for fructose determination?

A
  • Fructose is converted to fructose-6-phosphate (F6P) by hexokinase (HK) + ATP
  • F6P converted to G6P by phosphoglucose isomerase (PGI)
  • G6P then again converted to gluconate-6-phosphate and NADP+ to NADPH
  • NADPH measured at 340 nm and = [glucose + fructose]
24
Q

What wavelength is measured in the enzymatic method?

A

340 nm

25
Q

What are advantages of the enzymatic method? [4]

A
  • Simple
  • Rapid
  • Allows for quantification of individual sugars
  • Less subject to interference
26
Q

What are disadvantages of the enzymatic assay? [3]

A
  • Expensive
  • Reagents not stable
  • Total carbohydrates cannot be quantified
27
Q

Yeasts and moulds commonly spoil foods that: [3]

A

Have low pH and low water activity and are stored at refrigerated temperatures

28
Q

Why are yeasts and moulds undesirable?

A
  • Some moulds can produce dangerous toxins (aflatoxins)
  • Yeasts can cause sugar to ferment (ethanol and carbon dioxide production) or degrade in other ways
29
Q

Describe the HGMF method.

A
  • Can examine large sample volumes
  • Used to concentrate microbes onto a grid membrane
  • Membrane is placed on agar surface and colonies grow
  • Hydrophobicity of grid confines microbial growth to within squares, allowing for up to 1600 “colonies” to be counted on one plate
30
Q

What is potato dextrose agar?

A
  • Most widely used medium for growing fungi and bacteria
  • Contains a potato infusion (potatoes + water, then filtered) to promote the growth of yeasts and moulds
  • Relatively low pH (~5.5) partially inhibits the growth of background microflora (bacteria)
  • Trypan blue makes yeast colonies visible
  • Chloramphenicol suppresses bacterial growth
  • Colonies will appear:
  • Yeast = blue
  • Mould = blue-grey & fuzzy
31
Q

What is the D-value?
How is it obtained?

A
  • The D-value is the time is takes for a 1 log reduction in microbial cells (or 90% of a population) at a specific temperature
  • Obtained by plotting log(surviving population) vs. time, followed by calculating -1/slope
32
Q

What is the z-value?
How is it obtained?

A
  • Z-value is the temperature change required for a 1 log reduction in the D-value (described as the thermal resistance of a microorganism)
  • Obtained by plotting log(D-values) vs. temperature, followed by calculating -1/slope
33
Q

How does milk turn into yogurt?

A
  • Milk is inoculated with lactic acid bacteria starter cultures (Usually Streptococcus thermophilus and Lactobacillus delbrueckii subsp. bulgaricus
34
Q

Describe the effect of temperature on pH.

A
  • pH decreases with an increase in temperature. As the temperature increases, molecular vibrations increase, which causes water to ionise and form more H+
35
Q

What is the role of phenolpththalein in titration?

A

Secondary indicator that changes from colourless to pink at the equivalence point.

36
Q

How is equivalent weight determined?

A
  • First determine molecular weight of the acid
  • Then divide by the number of carboxylic acid groups on the molecule – this equals the number of titratable hydrogens
37
Q

How is texture analysed (not viscosity)?

A

Puncture force (N)

38
Q

Desribe what L, a, and b values correspond to in colourimetry.

A

-a = green
+a = red
-b = blue
+b = yellow
L = 0 = black
L = 100 = white
L = 50 = grey

39
Q

What are the megazyme K-FRUGL assay kit components?

A
  1. Buffer (pH 7.6)
  2. NADP+ and ATP
  3. Hexokinase + glucose-6-phosphate dehydrogenase
  4. Phosphoglucose isomerase
  5. D-glucose and D-fructose standards (0.2 mg/mL)