Biochemistry Laboratory Manual: Practical Methods and Quantitative Assays

Determination of Iodine Value of Lipids

• Aim:

  • To determine the iodine value of lipids.

• Principle:

  • The iodine value (or iodine number) of a lipid is defined as the number of grams of iodine absorbed per $100\,\text{g}$ of lipid.
  • A solution containing the lipid dissolved in chloroform is mixed with a solution of iodine and an equal quantity of bromine in acetic acid (Hanus reagent).
  • The mixture is allowed to stand for $30\,\text{minutes}$ in the dark to allow complete addition across double bonds.
  • Potassium iodide ($\text{KI}$) is added, reacting with unreacted halogen to liberate free iodine ($\text{I}_2$).
  • The liberated iodine is determined quantitatively by titrating against a standardized solution of sodium thiosulphate ($\text{Na}_2\text{S}_2\text{O}_3$).

• Chemical Reactions:

  • Addition across double bond:

$\text{R-CH=CH-R} + \text{IBr} \rightarrow \text{R-CH(I)-CH(Br)-R}$

• Reaction of excess Hanus reagent with potassium iodide:

$\text{IBr} + \text{KI} \rightarrow \text{KBr} + \text{I}_2$

• Titration of liberated iodine with sodium thiosulphate:

$\text{I}_2 + 2\text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2\text{NaI}$

• Reagents Required:

  • $0.1\,\text{N}\text{ K}_2\text{Cr}_2\text{O}_7$ (Potassium dichromate standard solution)
  • $0.1\,\text{N}\text{ Na}_2\text{S}_2\text{O}_3$ (Sodium thiosulphate solution)
  • $10\%\,\text{KI}$ (Potassium iodide solution)
  • $1\,\text{N}\text{ HCl}$ (Hydrochloric acid)
  • $1\%\text{ starch solution}$ (Indicator)
  • Chloroform
  • Oil sample
  • Hanus Reagent: Prepared by dissolving $6.8\,\text{g}$ of iodine in $500\,\text{mL}$ of glacial acetic acid with heating. After cooling, $1.5\,\text{mL}$ of bromine water is added.

• Procedure:

  • Part A: Standardization of Sodium Thiosulphate Solution:

  • $0.1\,\text{N}$ Potassium dichromate is prepared by dissolving $0.49\,\text{g}$ of $K_2Cr_2O_7$ in $100\,\text{mL}$ of distilled water in a standard volumetric flask.

  • Pipette out $10\,\text{mL}$ of this $K_2Cr_2O_7$ solution into a clean conical flask.

  • Add $10\,\text{mL}$ of $10\%\,\text{KI}$ solution and $10\,\text{mL}$ of $1\,\text{N}\text{ HCl}$.

  • Titrate this mixture against the given sodium thiosulphate solution taken in the burette using $1\%\text{ starch solution}$ as an indicator.

  • End Point: Disappearance of the blue colour.

  • Standardization Titration Data (Part A):

  • Burette Solution: Sodium thiosulphate ($\text{Na}_2\text{S}_2\text{O}_3$)

  • Conical Flask Solution: $10\,\text{mL}\text{ K}_2\text{Cr}_2\text{O}_7 + 10\,\text{mL } 10\%\,\text{KI} + 10\,\text{mL } 1\,\text{N}\text{ HCl}$

  • Indicator: $1\%\text{ starch solution}$

  • End Point: Disappearance of blue colour

• Standardization Calculations:

$V_1 N_1 (K_2Cr_2O_7) = V_2 N_2 (Na_2S_2O_3)$

$N_2 = \frac{V_1 N_1}{V_2}$

$N_2 = \frac{10 \times 0.1}{10.1} = 0.099\,\text{N}$

• Normality of $\text{Na}_2\text{S}_2\text{O}_3$ = $0.099\,\text{N}$

• Part B: Determination of Iodine Value of Lipid Sample:

  • Weigh accurately about $0.5\,\text{g}$ of the lipid into a $250\,\text{mL}$ iodine flask.
  • Add $25\,\text{mL}$ of chloroform to dissolve the lipid completely.
  • Add $25\,\text{mL}$ of Hanus reagent into the flask containing the lipid solution, stopper the flask, shake thoroughly, and keep it in the dark for $30\,\text{minutes}$.
  • At the end of $30\,\text{minutes}$, add $10\,\text{mL}$ of $10\%\,\text{KI}$ solution followed by $25\,\text{mL}$ of distilled water.
  • Titrate the liberated iodine immediately against standardized sodium thiosulphate solution until a pale yellow colour is obtained.
  • Add $1\%\text{ starch solution}$ (a blue colour forms) and continue titrating until the blue colour completely disappears.
  • Repeat the exact same procedure without adding oil to perform a blank titration.

• Sample Titration Data (Part B):

  • Burette Solution: Standardized sodium thiosulphate ($0.099\,\text{N}$)
  • Conical Flask Solution: $0.5\,\text{g}$ oil + $25\,\text{mL}$ chloroform + $25\,\text{mL}$ Hanus reagent (kept in dark for $30\,\text{min}$) + $10\,\text{mL } 10\%\,\text{KI} + 25\,\text{mL}$ distilled water
  • Indicator: $1\%\text{ starch solution}$
  • End Point: Disappearance of blue colour

• Sample Calculations for Iodine Value:
  • Blank Titre Value ($V_b$) = $26.7\,\text{mL}$
  • Sample Titre Value ($V_s$) = $18.6\,\text{mL}$
  • Weight of Oil Taken = $0.5\,\text{g}$
  • Normality of $\text{Na}_2\text{S}_2\text{O}_3$ = $0.099\,\text{N}$
  • Equivalent Weight of Iodine = $12.7$

$\text{Iodine Value} = \frac{(\text{Blank Titre Value} - \text{Sample Titre Value}) \times \text{Normality of } Na_2S_2O_3 \times \text{Equivalent Weight of Iodine} \times 100}{1000 \times \text{Weight of Oil Taken}}$

$\text{Iodine Value} = \frac{(\text{Blank} - \text{Sample Titre Value}) \times \text{Normality of } Na_2S_2O_3 \times 12.7}{\text{Weight of Oil Taken}}$

$\text{Iodine Value} = \frac{(26.7 - 18.6) \times 0.099 \times 12.7}{0.5}$

$\text{Iodine Value} = \frac{8.1 \times 0.099 \times 12.7}{0.5} = 20.368$

• Result:

  • The iodine value of the given oil sample is $20.368$.

• Discussion:

  • The iodine value (or iodine number) provides a direct quantitative measure of the degree of unsaturation in an oil, fat, or wax.
  • Fully saturated oils, fats, and waxes contain no carbon-carbon double bonds and take up no iodine, yielding an iodine value of $0$.
  • Unsaturated oils, fats, and waxes contain double bonds that readily react with and take up iodine; higher iodine values correspond to higher degrees of unsaturation.

Estimation of Total Reducing Sugar in Honey by DNS Method

• Aim:

  • Estimation of total reducing sugar in honey by the DNS (3,5-Dinitrosalicylic acid) method, using a standard glucose solution containing $1.8\,\text{mg/mL}$ of glucose.

• Principle:

  • When an alkaline solution of 3,5-dinitrosalicylic acid (DNS) is heated with a reducing sugar, the nitro group of DNS is reduced to an amino group, producing 3-amino-5-nitrosalicylic acid, which has an intense red colour.
  • This reaction demonstrates the reduction of a non-metallic organic compound by a reducing sugar.
  • While not typically used as a routine qualitative test, this reaction forms the basis of a reliable quantitative spectrophotometric method for measuring reducing sugar concentration at an absorption wavelength of $540\,\text{nm}$.

• Procedure:

  • Pipette out aliquots of standard glucose solution ($1.8\,\text{mg/mL}$) ranging from $0.2\,\text{mL}$ to $1.0\,\text{mL}$ into different test tubes.
  • Dilute the given unknown sugar solution in a standard flask up to the mark with distilled water and mix thoroughly.
  • Pipette out $1.0\,\text{mL}$ of this made-up unknown solution into each of two separate test tubes (Tubes 7 and 8).
  • Adjust the liquid volume in each tube to $2.0\,\text{mL}$ using distilled water. For the blank tube (Tube 1), use $2.0\,\text{mL}$ of distilled water.
  • Add $2.0\,\text{mL}$ of 3,5-dinitrosalicylic acid reagent to every test tube.
  • Heat all tubes in a boiling water bath for exactly $15\,\text{minutes}$.
  • Cool the tubes to room temperature, then add $16.0\,\text{mL}$ of distilled water to each test tube.
  • Mix the tube contents thoroughly and measure the optical absorbance at $540\,\text{nm}$ against the blank.

• Experimental Protocol Table:

• Calculations:
  • Amount of reducing sugar in $1.0\,\text{mL}$ of made-up solution (determined from standard graph) = $X\,\text{mg}$
  • Amount of reducing sugar in the total given volume ($50\,\text{mL}$ solution) =

$\text{Total Sugar} = X \times 50\,\text{mg}$

• Precautions:
  • The spectrophotometer/colorimeter must be set to $540\,\text{nm}$ before taking absorbance measurements. Calibrate zero optical density ($0.00\text{ OD}$) or $100\%$ transmittance using the reagent blank tube prior to testing sample tubes.
  • Rinse the cuvette properly, fill it with solution from the subsequent tube, and dry its exterior completely using clean tissue paper before insertion into the cell holder.
  • Align the position mark on the cuvette precisely with the etched reference mark on the instrument holder during every measurement to prevent path-length errors caused by micro-variations in glass thickness.

Estimation of Protein by Biuret Method

• Aim:

  • To estimate the amount of protein present in the given sample using the Biuret method.

• Principle:

  • Proteins consist of amino acid polymers joined together through peptide bonds.
  • Under alkaline conditions, molecules containing two or more peptide bonds react with cupric ions ($\text{Cu}^{2+}$) present in the biuret reagent.
  • This coordination complex forms between one cupric ion and 4 peptide bond nitrogen atoms, generating a violet-coloured coordination compound.
  • The intensity of the violet color is directly proportional to protein concentration and exhibits maximum optical absorption at $540\,\text{nm}$.

• Reagents Required:

  • Standard Protein Solution: Prepared by dissolving $200\,\text{mg}$ of Bovine Serum Albumin (BSA) in $100\,\text{mL}$ of distilled water ($2\,\text{mg/mL}$).
  • Biuret Reagent: Dissolve $3\,\text{g}$ of copper sulphate ($\text{CuSO}_4$) and $9\,\text{g}$ of sodium potassium tartrate in $500\,\text{mL}$ of $0.2\,\text{N}\text{ NaOH}$. Add $5\,\text{g}$ of potassium iodide ($\text{KI}$) and dilute to a final total volume of $1\,\text{L}$ using $0.2\,\text{N}\text{ NaOH}$.

• Procedure:

  • Pipette out standard protein solution ($2\,\text{mg/mL}$ BSA) in volumes ranging from $0.5\,\text{mL}$ to $2.5\,\text{mL}$ into clean, labeled test tubes.
  • Adjust the liquid volume in each test tube to $2.5\,\text{mL}$ using distilled water.
  • Add $2.5\,\text{mL}$ of Biuret reagent to all tubes and mix thoroughly.
  • Incubate test tubes in a water bath kept at $37\,^\circ\text{C}$ (or $34\,^\circ\text{C}$) for $15\,\text{minutes}$.
  • Allow tubes to cool and read optical absorbance at $540\,\text{nm}$ against the reagent blank.
  • Plot a standard calibration curve of absorbance ($A_{540}$) versus protein concentration ($\text{mg}$).
  • For the unknown sample, pipette $1.0\,\text{mL}$ of unknown protein solution, add $1.5\,\text{mL}$ of distilled water, followed by $2.5\,\text{mL}$ of Biuret reagent. Mix well, incubate at $37\,^\circ\text{C}$ for $15\,\text{minutes}$, and measure absorbance at $540\,\text{nm}$.
  • Calculate total protein content contained in $50\,\text{mL}$ of the unknown sample by interpolation from the standard curve.

• Experimental Protocol Table:

• Calculations:
  • Case 1: At $0.08\text{ OD}$, $1.0\,\text{mL}$ of unknown solution corresponds to $2.65\,\text{mg}$ of protein.

$\text{Total Protein in } 50\,\text{mL} = 2.65 \times 50 = 132.5\,\text{mg}$

• Case 2: At $0.09\text{ OD}$, $1.0\,\text{mL}$ of unknown solution corresponds to $3.0\,\text{mg}$ of protein.

$\text{Total Protein in } 50\,\text{mL} = 3.0 \times 50 = 150\,\text{mg}$

• Result:
  • The amount of protein present in $50\,\text{mL}$ of given sample was found to be $132.5\,\text{mg}$ (or $150\,\text{mg}$ at $0.09\text{ OD}$).

Estimation of Ascorbic Acid in a Sample

• Aim:

  • To estimate the amount of ascorbic acid (Vitamin C) present in the given sample.

• Principle:

  • This assay is based on the strong reducing property of ascorbic acid.
  • The redox dye 2,6-dichlorophenol indophenol acts as an oxidizing agent and is reduced by ascorbic acid to a completely colorless leucobase compound.
  • Because Vitamin C is unstable in neutral or basic conditions, an acidic stabilizing medium composed of an oxalic acid-acetic acid mixture is required.
  • The dye acts as a self-indicator; at the titration endpoint, excess unreacted dye turns the acidic solution pale pink.

• Chemical Reaction:

$\text{L-ascorbic acid} + \text{2,6-dichlorophenol indophenol (oxidized form, blue/pink)} \rightarrow \text{L-dehydroascorbic acid} + \text{Leucobase (reduced form, colorless)}$

• Reagents Required:

  • Standard Ascorbic Acid Solution: Dissolve $100\,\text{mg}$ of pure ascorbic acid in oxalic acid-acetic acid mixture and dilute up to $100\,\text{mL}$ with the same mixture ($1\,\text{mg/mL}$).
  • Unknown Sample Preparation: Extract from $1/2$ lemon diluted to $50\,\text{mL}$ using oxalic acid-acetic acid mixture.
  • Indole Dye Preparation (2,6-Dichlorophenol indophenol): Dissolve $21\,\text{mg}$ of sodium bicarbonate ($\text{NaHCO}_3$) and $26\,\text{mg}$ of 2,6-dichlorophenol indophenol dye in distilled water, making up to $100\,\text{mL}$. Filter and store in a refrigerator for no longer than $3\,\text{days}$.
  • Oxalic Acid - Acetic Acid Mixture: Dissolve $0.5\,\text{g}$ of oxalic acid in $50\,\text{mL}$ of glacial acetic acid, add $450\,\text{mL}$ of distilled water (total $500\,\text{mL}$), filter, and store refrigerated for up to $10\,\text{days}$.

• Procedure:

  • Part I: Standardization of Indophenol Solution:

  • Pipette $1.0\,\text{mL}$ of standard ascorbic acid solution ($1.0\,\text{mg}$ ascorbic acid) into a conical flask.

  • Add $10\,\text{mL}$ of oxalic acid-acetic acid mixture.

  • Titrate against indophenol dye solution filled in the burette until a distinct pale pink color persists for $15\,\text{seconds}$.

  • Standardization Titration Data (Part I):

  • Burette Solution: Indole dye (2,6-dichlorophenol indophenol)

  • Conical Flask Solution: $1\,\text{mL}$ standard ascorbic acid + $10\,\text{mL}$ oxalic acid-acetic acid mixture

  • Indicator: Self-indicator

  • End Point: Appearance of pale pink colour

• Standardization Factor:

  • $8.2\,\text{mL}$ of indophenol dye reacts completely with $1.0\,\text{mg}$ of standard ascorbic acid.

• Part II: Determination of Ascorbic Acid in Unknown Sample:

  • Pipette $1.0\,\text{mL}$ of unknown lemon extract sample into a conical flask.
  • Add $10\,\text{mL}$ of oxalic acid-acetic acid mixture.
  • Titrate against indophenol dye in the burette until a permanent pale pink colour appears.

• Sample Titration Data (Part II):

  • Burette Solution: Indole dye solution
  • Conical Flask Solution: $1\,\text{mL}$ unknown sample + $10\,\text{mL}$ oxalic acid-acetic acid mixture
  • Indicator: Self-indicator
  • End Point: Appearance of pale pink colour

• Calculations:
  • Volume of indophenol dye consumed by $1\,\text{mL}$ of unknown sample = $0.4\,\text{mL}$
  • Since $8.2\,\text{mL}$ of dye corresponds to $1\,\text{mg}$ of ascorbic acid:

$\text{Ascorbic acid in } 1\,\text{mL} \text{ sample} = \frac{0.4}{8.2} = 0.04878\,\text{mg} \approx 0.049\,\text{mg}$

• Ascorbic acid in total $50\,\text{mL}$ sample volume:

$\text{Total Ascorbic Acid} = 0.04878 \times 50 = 2.439\,\text{mg} \approx 2.45\,\text{mg}$

• Result:
  • The amount of ascorbic acid present in $50\,\text{mL}$ of the given sample is $2.45\,\text{mg}$.

Estimation of Amino Acids by Formal Titration

• Aim:

  • To estimate the mass of amino acid present in one litre of the given sample solution.

• Principle:

  • Amino acids are amphoteric compounds containing both basic amino ($-\text{NH}_2$) and acidic carboxyl ($-\text{COOH}$) groups.
  • Direct titration of the carboxyl group in aqueous solution using standard alkali is difficult because the protonated amino group competes with added hydroxyl ions for protons.
  • Formaldehyde solution ($\text{HCOH}$) is added to eliminate this amino group interference.
  • Formaldehyde reacts with uncharged amino groups to form mono- and di-hydroxyl methyl derivatives ($\text{N-hydroxyl methyl}$ and $\text{N,N-dihydroxyl methyl}$ derivatives).
  • These methyl derivatives are vastly weaker bases than the original free amino groups, removing proton competition and permitting direct, sharp titration of carboxyl groups using standard $\text{NaOH}$ solution with phenolphthalein indicator.

• Reaction Scheme:

$\text{R-CH(NH}_3^+\text{)-COO}^- \rightleftharpoons \text{R-CH(NH}_2\text{)-COO}^- + \text{H}^+$

$\text{R-CH(NH}_2\text{)-COO}^- + \text{HCOH} \rightarrow \text{R-CH(NH-CH}_2\text{OH)-COO}^- \quad (\text{N-hydroxyl methyl derivative})$

$\text{R-CH(NH-CH}_2\text{OH)-COO}^- + \text{HCOH} \rightarrow \text{R-CH(N(CH}_2\text{OH)}_2\text{)-COO}^- \quad (\text{N,N-dihydroxyl methyl derivative})$

• Procedure:

  • Part I: Standardization of Sodium Hydroxide Solution:

  • Pipette $10\,\text{mL}$ of standard potassium hydrogen phthalate (KHP) into a clean conical flask.

  • Add 2 drops of phenolphthalein indicator.

  • Titrate against sodium hydroxide ($\text{NaOH}$) solution taken in the burette until a permanent pale pink color appears.

  • Repeat titration to obtain concordant burette values.

  • Part I Titration Table Template:

  • Burette Solution: Sodium hydroxide ($\text{NaOH}$)

  • Conical Flask Solution: $10\,\text{mL}$ Potassium hydrogen phthalate

  • Indicator: Phenolphthalein

  • End Point: Colorless to permanent pale pink

• Standardization Formulas (Part I):
  • Volume of KHP ($V_1$) = $10\,\text{mL}$
  • Normality of KHP ($N_1$)
  • Volume of $\text{NaOH}$ used ($V_2$)

$V_1 N_1 = V_2 N_2$

$N_2 (\text{Normality of } NaOH) = \frac{V_1 N_1}{V_2}$

$\text{Molarity of } NaOH (X) = \text{Normality } (N_2) \times 1$

• Part II: Estimation of Amino Acid:

  • Pipette $10\,\text{mL}$ of the unknown amino acid solution into a clean conical flask.
  • Add 4 drops of phenolphthalein indicator and $10\,\text{mL}$ of neutralized formaldehyde solution.
  • Titrate the resulting amino acid-formaldehyde mixture against standard $\text{NaOH}$ solution taken in the burette until a permanent pale pink color is produced.
  • Record the volume of $\text{NaOH}$ consumed ($Y\,\text{mL}$) and repeat for concordant values.

• Part II Titration Table Template:

  • Burette Solution: Standardized $\text{NaOH}$ ($X\,\text{M}$)
  • Conical Flask Solution: $10\,\text{mL}$ amino acid + $10\,\text{mL}$ neutralized formaldehyde
  • Indicator: Phenolphthalein
  • End Point: Colorless to permanent pale pink

• Calculations:
  • Molarity of $\text{NaOH}$ = $X\,\text{M}$
  • Mean Volume of $\text{NaOH}$ used = $Y\,\text{mL}$
  • Millimoles of alkali reacting with amino acid in $10\,\text{mL}$ sample:

$\text{Millimoles} (Z) = \text{Volume of } NaOH (Y) \times \text{Molarity of } NaOH (X)$

• Moles of amino acid in $10\,\text{mL}$ = $Z \times 10^{-3}\,\text{moles}$
• Moles of amino acid present in $1\,\text{L}$ ($1000\,\text{mL}$):

$\text{Moles/L} = \frac{Z \times 10^{-3} \times 1000}{10} = \frac{Z}{10}\,\text{moles}$

• Mass of amino acid (glycine) present in $1\,\text{L}$ solution:

$\text{Mass of Glycine (g/L)} = \text{Moles/L} \times \text{Molecular Weight of Glycine} (75\,\text{g/mol})$

$\text{Mass of Glycine (g/L)} = \left(\frac{Z}{10}\right) \times 75\,\text{g}$