· Web viewTo obtain accurate pH readings, we must standardize the electrode using buffers of known pH before measuring an unknown. Static electricity can affect pH readings, so always

University of Nizwa CHEM 334 Biochemistry Laboratory Manual.

BIOCHEMISTRY LAB MANUAL

By

Dr. Zakira Naureen

1


RULES FOR WORKING IN A BIOCHEMISTRY LABORATORY

There are two major concerns to consider when working in a biochemistry laboratory. First is safety: this

can never be overemphasized. General guidelines for safety are discussed below. The second is efficiency

in the laboratory work. Although the latter very much depends on the individuals doing the experiments,

there are general rules students are advised to follow:

1. Keep the benches and shelves clean and well-organized

2. Avoid contaminating the chemicals; use only clean glassware and spatulas; label glassware in use,

3. Plan your experiments before starting to carry them out,

4. Pay attention to others in the laboratory.

SAFETY IN THE LABORATORY

Students working in a biochemistry laboratory must always be aware that the chemicals used are

potentially toxic, irritating and flammable. Such chemicals are hazards, however, only when they are

mishandled. Students who come to the laboratory session must have a complete understanding of the

laboratory procedures to carry out and be familiar with both the physical and chemical properties of

chemicals and reagents to be used. Since the carelessness on the part of one student can often cause injury

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to other students, one must have a special concern for the safety of classmates. Students must be familiar

with general safety practices, facilities and emergency action.

I. Safety rules in general

1. Do not work alone in the laboratory.

2. Unauthorized experiments are not allowed.

3. Eating, drinking and smoking in the laboratory are strictly prohibited.

4. Become familiar with the location and the use of standard safety features in the

laboratory. The laboratory is equipped with fire extinguishers, eye washes, safety showers, fume hoods

and first-aid kits. Any question regarding the use of these facilities should be addressed to your instructor.

5. Special care for eye protection is required. Safety glasses must be used when certain procedures are

being carried out. The instructor will call the students' attention to those procedures. The use of contact

lenses is not recommended, since they reduce the rate of self-cleansing of the eye.

ii. Special safety rules

1. While heating a solution one should make sure not to overheat it; be careful when vigorous mixing of

the solution by shaking or stirring is required. The mouth of the glassware containing the solution to be

heated should never be pointed toward anyone.

2. Handling of strong acids and bases requires special attention. When diluting concentrated acids, the

acid should be poured into the water and never the opposite.

3. The pipettes should never be filled with solutions of toxic substances, biological fluids, strong acids

and bases by mouth suction. Use either automatic pipettes or pipette pumps.

3


4. Volatile liquids and solids that are toxic or irritating should be handled under fume hoods.

5. While handling flammable liquids such as ether, alcohols, benzene, naked flame (burners, matches)

must not be in use. The above liquids must not be stored near radiating heat sources, such as the

laboratory oven.

6. Before using electrical appliances, make sure they are grounded.

7. Flasks with flat-bottoms or thin walls should not be desiccated.

8. Before leaving the laboratory, electrical equipment should be turned off, and gas burners extinguished.

No tap water should be left running.

iii. Rules to follow in the case of accidents and injuries

Chemical splatters into the eye. First the eyelid should be opened by using the thumb and the pointing

finger. Then, by using the eye wash kit, the eye should be rinsed with large amounts of water. When an

acid or alkaline solution gets into eye, the eye should be rinsed with 1 % NaHCO3 or 1 % boric acid,

respectively. The victim should be taken to the doctor as soon as possible.

Burning. The burned spot on the skin should not be treated with water; rather, a special bandage should be

used. Consult the doctor if necessary.

Poisoning. Prompt medical treatment should be obtained.

All injuries and accidents must be reported to the instructor.

4


ACIDS, BASES AND THE pH SCALE

INTRODUCTION

PART 1: ACIDS, BASES, AND THE pH SCALE

The nature of acids and bases has been known for quite some time. Chemically speaking, acids are

interesting compounds. But one of the best reasons for studying acids is that a large number of common

household substances are acids or acidic solutions. For example, vinegar contains ethanoic or acetic acid,

cranberries contains malic acid, sauerkraut (fermented cabbage) and yogurt (fermented milk) contain

lactic acid, lemons contain citric acid. Acids cause foods to have a sour taste. On the other hand, many

common household substances are bases (i.e., alkaline). For example, milk of magnesia contains the base

magnesium hydroxide, and household ammonia is a common basic cleaning agent. Bases have a smooth

or slippery feel between the fingertips. NOTE: Do not taste or feel compounds unless you are sure of

what they are due to their caustic nature. Indicator dyes, in addition to litmus, turn various colors

according to the strength of an acid or base that is applied to it.

5


An acid in water solution contains more hydrogen ions than hydroxide ions. Pure water, which is neutral,

exists mostly as H2O molecules. To a very slight extent, it does break into H+ and OH- ions.

HOH H+ (aq) + OH- (aq)

This reaction forms to the extent of 0.0000001 moles of H+ (aq) per liter of water. In scientific notation, it

is written as 1 x 10-7 moles of H+ per liter of solution.

The pH scale was devised to measure the concentration of hydrogen ion in a solution. The term pH

refers to the “power of hydrogen”, the concentration of hydrogen ion in solution. In a neutral solution, the

concentration of H+ (aq) is 1 x 10-7 moles per liter and has a pH of 7. If the concentration is H+ (aq) is 1 x

10-5 moles per liter, its pH is 5; if the concentration is H+ (aq) is 1 x 10-12 moles per liter and its pH is 12.

Please note that the pH scale is a logarithmic scale and each whole number change of the pH represents a

change in hydrogen ion concentration of 10 times. A pH of 8 has 10 times the concentration of hydrogen

ions as a pH of 9 and 100 times that of a pH 10.

Aqueous systems are seldom pure water and other substances affect the pH of the solution. If a substance

increases the concentration of hydrogen ion in a solution, we call it an acid. Acids cause the pH of a

solution to decrease, so the number becomes smaller than 7. If the substance decreases the concentration

of hydrogen ion or increases the hydroxide concentration, we call it a base. Bases increase the pH of the

solution, raising it above a pH of 7.

Methods of measuring pH

6


Color Indicators: An important method of determining pH values involves the use of “indicators”.

These are certain organic substances that have the property of changing color in dilute solutions when the

hydrogen-ion concentration of the solution reaches a definite value. For example, phenolphthalein is a

colorless substance in any aqueous solution of which the hydrogen-ion concentration if greater than 10 -9

M, or the pH is less than 9. In solutions for which the hydrogen-ion concentration is less than 10 -9, the

phenolphthalein imparts a red or pink color to the solution. Substances like phenolphthalein are called

acid-base indicators and they often are used for determining the approximate pH of solutions. Electrical

measurements can determine the pH more precisely.

The pH meter: Where rapid and accurate pH measurements are required, we use an instrument known as

a pH meter. The pH meter is essentially a voltmeter designed to measure the voltage difference between a

reference electrode and a sensory electrode. The reference electrode usually contains silver chloride

solution of known concentration. The sensory electrode is in contact with the solution to be tested. The

pH meter is calibrated so that a certain difference between the voltages of silver chloride and the test

solution reads a certain pH value. To obtain accurate pH readings, we must standardize the electrode

using buffers of known pH before measuring an unknown. Static electricity can affect pH readings, so

always blot the end of the electrode, do not wipe it off, when washing the electrode and moving it to the

next solution. Temperature also affects pH, so make sure that your calibration buffer is at the same

temperature as the test solution. Occasionally, the pH meter has a temperature compensation adjustment

with an internal temperature sensor.

pH METHODS

NOTE: In all activities, replace the cap on the bottles and DO NOT touch the tip of the bottle to the

GLASS PLATE or to the solution in the position. This could contaminate the reagents for the remainder

7


of the students. When placing the droplets, hold the bottle vertically and squeeze gently, dropping only

one drop per well.

Activity 1 – Determining the effect of pH on indicator dyes

Purpose: To determine the color change for each of 3 indicators: methyl orange, bromo thymol blue and

phenolphthalein.

Procedure:

1. Place the GLASS PLATE on a sheet of white paper.

2. Place 1 drop of methyl orange in position #1 and #2.

3. Place 1 drop of brom thymol blue in position #3 and #4.

4. Place 1 drop of phenylphthalein in position #5 and #6.

5. Carefully add 1 drop of pH 1 to position #1, #3 and #5.

6. Carefully add 1 drop of pH 13 to position #2, #4 and #6.

7. Record your observations on your data sheet for this exercise.

8. Rinse the GLASS PLATE with tap water in the sink and dry with a paper towel.

RESULTS: Activity 1– Determining the effect of pH on indicator dyes

Methyl orange solution is what color? ________________________________

Methyl orange changes to ___________________________ in an acid (pH1) solution.

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(color)

Methyl orange changes to ___________________________ in a base (pH13) solution.

(color)

Brom thymol solution is what color? __________________________________

Brom thymol changes to ____________________________ in an acid (pH1) solution.

(color)

Brom thymol changes to ____________________________ in a base (pH13) solution.

(color)

Phenolphthalein solution is what color? ________________________________

Phenolphthalein changes to ____________________________ in an acid (pH1) solution.

(color)

Phenolphthalein changes to ____________________________ in a base (pH13) solution.

(color)

Table 1

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Activity 2 – Determining the pH range with indicator dyes

Purpose: To establish the specific pH range in which the color changes for each indicator dye.

Procedure:

Methyl Orange

1. Place 1 drop of methyl orange in each position numbered 1-7.

2. Carefully add 1 drop of pH1 to position #1, 1 drop of pH3 to position #2, 1 drop of pH5 to

position #3, 1 drop of pH7 to position #4, 1 drop of pH9 to position #5, 1 drop of pH11 to position

#6 and 1 drop of pH13 to position #7.

10


3. Record the color change for methyl orange and the pH on your data sheet.


Brom Thymol Blue

1. Place 1 drop of brom thymol blue in each position numbered 1-7.




3. Record the color change for brom thymol blue and the pH on your data sheet.


Phenolphthalein

1. Place 1 drop of phenolphthalein in each position numbered 1-7.




3. Record the color change for phenolphthalein and the pH on your data sheet.


RESULTS: Activity 2 – Determining the pH range with indicator dyes

Methyl orange changes to ___________________________ at ___________pH.

(color) (number)

11


Brom thymol changes to ____________________________ at ___________pH.

(color) (number)

Phenolphthalein changes to _________________________ at ___________pH.

(color) (number)

Table 2.

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Activity 3 – Determining a color standard using a universal indicator dye

Purpose: To determine a color standard for universal indicator dyes to determine the pH of unknown

samples.

Procedure:

1. Place 1 drop of universal indicator in each position numbered 1-7.




3. Record the color of the indicator for each pH on your data sheet.

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4. DO NOT rinse the GLASS PLATE or mix the reactions in the position.

RESULTS: Activity 3 – Determining a color standard using a universal indicator dye

#1 (pH 1) ___________________

#2 (pH 3) ___________________

#3 (pH 5) ___________________

#4 (pH 7) ___________________

#5 (pH 9) ___________________

#6 (pH 11) __________________

#7 (pH 13) __________________

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Activity 4 – Determining the pH of unknown solutions

Purpose: To identify the pH of an unknown solution by using the universal indicator.

Procedure:

1. Place 1 drop of universal indicator in position #10, #11 and #12.

2. Place 1 drop of unknown I in position #10.

3. Place 1 drop of unknown II in position #11.

4. Place 1 drop of unknown III in position #12.

5. Compare the colors in position #10, #11 and #12 with the colors in position numbered 1-

7 from Activity 3.

6. Record the pH for the unknown samples on your data sheet.


RESULTS: Activity 4 – Determining the pH of unknown solutions

The pH of Unknown I is ________________________

The pH of Unknown II is _______________________

The pH of Unknown III is _______________________

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Activity 5 – Determination of the pH of common household products

Purpose: To use the pH meter to determine the pH of common solutions.

THE INSTRUCTOR WILL DEMONSTRATE THE PROPER USE OF THE pH METER,

WHICH WILL NEED TO BE CALIBRATED USING THREE SOLUTIONS OF KNOWN

pH (i.e., pH 4, pH 7, and pH 10). THE INSTRUCTOR WILL ALSO SHOW YOU HOW TO

CLEAN THE ELECTRODE AND STORE IT SO THAT THE CHANCE OF DAMAGE TO

THIS INSTRUMENT IS MINIMIZED DURING USAGE.

Procedure:

1. Select any 4 beakers of common household solutions from the bench at the front of class.

2. Record the name of your selections on the group worksheet.

3. Insert the probe of the pH meter into each solution and record the pH on the data sheet.

4. List the products in the order of increasing acidity.

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RESULTS: Activity 5 – Determination of the pH of common household products

List the household products in order of increasing acidity:

Activity 6:

Analyze data and discuss your results by providing at least 3 references.

17

Product pH


18


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PART 2: ACIDS, BASES, AND BUFFERS IN LIVING SYSTEMS

Cells “live” in, and contain aqueous, or water-based, solutions. Therefore, water plays an

important role in cellular reactions. Remember that water is neither acidic nor basic, but is

neutral, with a pH of 7.

Acids and bases are damaging to the molecules in a cell, causing dysfunction or even cell death.

It is important to the health of cells that they maintain a pH as close to neutral (or a value of 7) as

possible. If pH changes by even 1 pH unit, the proteins in a cell can denature, leading to changes

in cell structure and, more importantly, to catalysis of cellular reactions. (Review: what does the

term “denature” mean?)

pH

You will remember that water dissociates into hydrogen ion (H+) and hydroxyl ion (OH). This

reaction can be written:

H2O H+ + OH-

Like all reactions, the dissociation of water is an equilibrium reaction, for which an equilibrium

constant (Keq) can be written. The Keq is a measure of the extent of the reaction at

thermodynamic equilibrium. That is, at equilibrium, the value of Keq tells how much product is

formed relative to how much reactant remains. This is calculated by multiplying the

concentrations of products, and dividing that value by the concentrations of reactants. So, for the

dissociation of water, the Keq can be written:

Keq = ([H+][OH-])/[H2O]

The dissociation of water, and its Keq value, is the basis for pH measurement. It is known that

the concentration of pure water ([H2O]) is 55.5 moles/L, and experiments have shown that the

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Keq of water is 1.8 x 10-16 moles/L. Rearranging the Keq equation and using these values, we

find:

Keq x [H2O] = [H+][OH-]

(1.8x10-16) x 55.5 = [H+][OH-]

1.0 x 10-14 = [H+][OH-]

In a neutral solution (neither acidic nor basic), [H+] = [OH-], so

1.0 x 10-7 = [H+], and

1.0 x 10-7 = [OH-]

Let’s consider the hydrogen ion concentration ([H+]): we have just shown that in neutral

solution [H+] is 1.0 x 10-7. Writing concentrations in the form of negative exponents is

cumbersome, so scientists have developed a more convenient method of measurement, the “p”

notation. First, they use logarithms in place of exponents. This would convert the value 1.0 x

10-7 to the whole number -7. To do away with the negative sign, they multiply these values by -1

(so, -7 x -1 = 7). Now the [H+] concentrations can be written as whole, positive numbers (such

as 7) instead of as negative exponents (1.0 x 10-7). This gives us the pH scale.

Acids and Bases

If a solution has a large number of H+ ions (for example 1.0 x 10 -2), it is considered an acid, and,

in this example, has a pH of 2.

An acid is a molecule that releases a hydrogen ion (H+). An acid can be either strong or weak,

depending on how easily the H+ is released from the rest of the molecule. Strong acids easily

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release H+, tending to be fully “dissociated.” An example of a strong acid is HCl. In water,

most HCl molecules exist as H+ and Cl-, with little of the hydrochloric acid (HCl) remaining in

the solution. The HCl dissociation reaction can be written:

HCl H+ + Cl-

(Here, Cl- is called the “conjugate base” of the original acid. (Note the conjugate base holds a

negative charge.)) A strong acid solution would be expected to have a high concentration of

H+.

The HCl dissociation reaction at equilibrium has a characteristic Keq value. The Keq for this

reaction can be written:

Keq = ([H+] [Cl-])/[HCl]

In the case of acids, the Keq is called the acid dissociation constant, or Ka, rather than Keq.

(Remember that if Ka has a large value, the concentrations of the products of the reaction are

much higher than the concentrations of the reactants. So this tells you that at equilibrium, the

reaction has gone far to the right (or much product has been formed, and little reactant remains)).

A large Ka is expected for a strong acid such as HCl, which easily dissociates and goes mostly to

product (individual ions) at equilibrium.

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Most biological acids are not as strong as HCl. That is, they don’t give up their H+ as easily as

does HCl. The weaker the acid, the less likely it is to dissociate into hydrogen ion and the

conjugate base of the acid. The strength of an acid depends on its chemical and electronic

structure, which cause it to want to either “hold on” to or release its H+. A solution of a very

weak acid would have some H+ but most of the acid molecules would remain associated with its

original H.

An example of a weak acid is acetic acid (CH3COOH). The acetic acid dissociation reaction can

be written:

CH3COOH H+ + CH3COO-

(Here, the conjugate base is called acetate ion.)

The Ka for acetic acid would be written:

Ka = ([H+] [CH3COO-])/ [CH3COOH]

Because acetic acid does not give up its H+ as easily as does HCl, you would not expect the

numerator in the Ka equation to be high, while the denominator in the Ka equation would be

large; overall the Ka for acetic acid is low, which is true of all weak acids. Therefore, a solution

of acetic acid would not be expected to have as high a concentration of H+ as would a strong

acid like HCl, and the pH of an acetic acid solution would be higher than the pH of an equimolar

HCl solution.

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pKa

The strength or weakness of an acid is a chemical characteristic of that acid, and can be

determined mathematically using the Ka equation. Using acetic acid as an example:

Ka = ([H+] [CH3COO-])/ [CH3COOH]

Rearranging: [H+] = (Ka x [CH3COOH])/[CH3COO-]

Taking negative log: - log [H+] = -log Ka – log ([CH3COOH]/[CH3COO-])

- log [H+] = -log Ka + log ([CH3COO-]/[CH3COOH])

Use “p” notation: pH = pKa + log ([CH3COO-]/CH3COOH])

This is the Henderson-Hasselbach equation, which relates the pKa of a solution (a measure of

how easily an acid gives up its H+) to the pH of a solution of that acid. Note, however, that the

initial concentrations of the weak acid and its conjugate base are also important to the equation.

Because the Ka is now in “p” notation, a high value denotes a weak acid (one which doesn’t

easily give up its H+; its Ka would be small, but its pKa would be large.) Consider an

example of a very strong acid, like HCl. Its Ka would be large (approximately 1.0 x 10 -2

[H+] in solution); in “p” notation, pKa=2. For a very weak acid, with a Ka that is small

(say 1.0 x 10-6 [H+] in solution), the pKa would be 6.

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The Henderson Hasselbach equation allows experimental determination of pKa. pKa can

be determined by measuring the pH of a solution where the concentration of weak acid =

concentration of conjugate base:

For acetic acid: pH = pKa + log ([CH3COO-]/CH3COOH])

When [acid]=[conj base] pH = pKa + log 1

pH = pKa + 0

pH = pKa

Buffer Systems

Buffers are composed of a weak acid and its conjugate base. Buffers have the ability to resist a

change in pH when either H+ or OH- ions are introduced.

The weak base in a buffer solution is available to react with any added acid (H+), thus neutralizing

the acid and keeping the pH from changing to a great extent. (Would you expect the addition of

added H+ to cause the pH to decrease or increase, if a buffer were not present?) Likewise, if base

(OH-) were added to a buffer solution, the weak acid available could give up its H+ to react with, and

neutralize, the added base (making H2O), again keeping the pH from changing significantly.

However, buffer systems have limits, and their neutralizing effects can be destroyed by the

introduction of an excessive amount of either H+ or OH- ion.

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Acetic acid and its conjugate base (acetate ion) can form acetic acid buffer. The Henderson-

Hasselbach equation can be used to determine how to prepare the buffer, and can calculate the

effect of adding additional acid and base to the existing buffer, by means of a titration

experiment. The scientist starts with a solution of acetic acid of known concentration, and

measures its pH using a pH meter. Hydroxide ion (in the form of 1 M sodium hydroxide

solution) is added in increments, and the pH of the solution is measured after each addition. It is

expected that as OH- is added, the acetic acid will give up its H+, forming H2O with the OH-.

As more and more H+ is dissociated and forms neutral water, producing more and more acetate

ion, the pH will increase (fewer H+ ions will be free in the solution and more acetate will be

available). A plot of amount of NaOH added (x axis) vs. pH (y axis) will yield a curve. Weak

acids that can act as buffers will have a flattened region, at which a relatively large addition of

added base will result in a relatively small pH change in the region of the acid’s pKa. As shown

above, when pH=pKa, the concentration of weak acid = concentration of its conjugate base.

This is the area of highest buffering ability, since there are as many acid molecules available (to

react with any base added to the solution) as there are conjugate base molecules (to react with

any acid added to the solution).

Cells, and all the water in living organisms, maintain active buffer solutions to guard against

extensive changes in pH, should acids or bases be encountered. This allows the cells’ proteins

to maintain their conformation, and their function.

In today’s experiment, students will perform a titration experiment, and will make acetic acid

buffer solutions of varied concentrations. The pH of these solutions will be measured

experimentally, as well as calculated using the Henderson-Hasselbach equation. A buffer will be

26


“tested” by measuring pH after adding acid or base, and the results of this “test” compared to pH

changes in pure water with the addition of acid and base.

PROCEDURES

Standardizing Your pH Meter

Obtaining accurate reading with a pH meter depends on effective standardization, the degree of

static charge, and the temperature, as well as other factors. Particularly important to the proper

use of a pH meter is accurate standardization, as glass electrodes must be carefully calibrated

using buffers of known pH. Generally, it is desirable to calibrate the meter with a standard

buffer of a pH as close as possible to the pH of the solution to be measured. The problem of

static charge on the electrode can be minimized by blotting (not rubbing) the surface of the

electrode dry when transferring from one solution to another. Temperature also affects pH to a

certain extent, but the problem can be easily eliminated by keeping all solutions at the same

temperature.

Standardize the meter at your table for pH 4, since your experiments will be centered on acetic

acid, which has a relatively low pH.

The Titration of Acetic Acid

You will use 0.2 M acetic acid and 1.0 M NaOH for this experiment.

1. Add 100 mL of 0.2 M acetic acid to a 250 mL beaker or container. One person in each group

should gently swirl the solution throughout the experiment.

2. Lower the pH electrode into the solution.

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3. With constant swirling, add NaOH in 1-mL increments, recording the resulting pH after every

addition of base on the data sheet.

4. Continue the titration until the pH is approximately 8.0.

5. Follow the directions on the data sheet regarding plotting your data and finding conclusions.

How to Make an Acetate Buffer

You will use 0.2 M acetic acid and 0.2 M sodium acetate solutions for this experiment.

1. Obtain 5 vials, and label them 0, 1, 2, 3 and 4. Each number represents the volume of acetic

acid you will place in the vial; the total volume in each vial will be 5 mL.

2. In vial 0, measure 5 mL sodium acetate.

3. In vial 1, measure 1 mL acetic acid + 4 mL sodium acetate.




7. Measure the pH of each vial, and record the data on your data sheet. Follow the directions on

your data sheet regarding calculations with your results.

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Effect of Added Acid and Base on Buffers

You will use pure water, 0.2 M acetic acid and 0.2 M sodium acetate solutions for this

experiment. You will also need 0.5 M HCl and 0.5 M NaOH solutions.

1. Obtain 4 vials. Label 2 “buffer” and 2 “water.

2. In the vials labeled “buffer,” add 2.5 mL acetic acid and 2.5 mL sodium acetate in each.

3. In the vials labeled “water,” add 5 mL pure water in each.

4. Measure the pH of one of the buffer vials and record this value on your data sheet.

5. To this vial, with constant swirling, add 0.2 mL HCl. Record the pH.

6. Add another 0.2 mL acid and test the pH again.

7. Continue this process of adding acid at 0.2 mL increments until 1.0 mL acid has been added.

8. Repeat this procedure with a distilled water vial.

9. Use the other buffer and water vials in a similar manner, except add 0.5 M NaOH.

10. Follow the directions on your data sheet regarding calculations with your results.

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DATA SHEET

Titration of Acetic Acid

NaOH Added (mL) pH NaOH Added (mL) pH

_____________ ______________ ______________

_________

_____________ ______________ ______________

_________

_____________ ______________ ______________

_________

_____________ ______________ ______________

_________

_____________ ______________ ______________

_________

_____________ ______________ ______________

_________

_____________ ______________ ______________

_________

_____________ ______________ ______________

_________

_____________ ______________ ______________ _________

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_____________ ______________ ______________ ________

_____________ ______________ ______________ _________

_____________ ______________ ______________ _________

_____________ ______________ ______________ _________

_____________ ______________ ______________ _________

Use your data to plot a graph of NaOH added (x axis) vs. pH (y axis).

From your graph, determine:

the buffering region (Draw a box around this region on your graph)

the pKa of acetic acid ____________________________________

the point where [CH3COO-] = [CH3COOH] _______________________

the chemical formula of acetic acid that predominates

at the beginning of the titration ___________________________

the chemical formula of acetic acid that predominates

at the end of the titration ___________________________

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How to Make an Acetate Buffer

Vial Number [CH3COO-]/[CH3COOH] pH (exper) pH (calc’d)

___________ _____________ _______ _______

___________ _____________ _______ _______

___________ _____________ _______ _______

___________ _____________ _______ _______

___________ _____________ _______ _______

___________ _____________ _______ _______

_acetic acid__ _____________ _______ _______

* This value can be obtained from the first reading of the titration experiment, when 0 mL

NaOH was added.

Use the Henderson-Hasselbach equation to calculate the pH of each buffer solution.

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Effect of Added Acid and Base on Buffers

Additions pH of Buffer pH of Water

None _____________ ______________

0.2 mL HCl _____________ ______________

0.4 mL HCl _____________ ______________

0.6 mL HCl _____________ ______________

0.8 mL HCl _____________ ______________

1.0 mL HCl _____________ ______________

None _____________ ______________

0.2 mL NaOH _____________ ______________

0.4 mL NaOH _____________ ______________

0.6 mL NaOH _____________ ______________

0.8 mL NaOH _____________ ______________

1.0 mL NaOH _____________ ______________

Draw two graphs: one of mL HCl added (x axis) vs. pH (y axis), and one of mL NaOH added (x

axis) vs. pH (y axis).

33


What conclusions can you draw about the effects of a buffer on pH with added acid and base,

compared to adding acid or base to water?

Use your graphs to determine the pKa of acetic acid buffer, when [CH3COO-]=[CH3COOH].

34


QUALITATIVE ANALYSIS OF AMINO ACIDS AND PROTEINS

Amino acids are molecules containing an amine group, a carboxylic acid group and a side

chain that varies betwen different amino acids. Amino acids of the general formula

RCH(NH2)COOH are amphoteric, behaving as amines in some reactions and as carboxylic

acids in others. At a certain pH known as the isoelectric point an amino acid has no overall

charge, since the number of protonated ammonium groups (positive charges) and deprotonated

carboxylate groups (negative charges) are equal. Since the amino acids at their isoelectric points

have both negative and positive charges, they are known as zwitterions.

Amino acids are critical to life. They have particularly important functions like being the

building blocks of proteins and being the intermediates in metabolism.

Amino acids are generally classified by the properties of their side chain into four groups. The

side chain can make an amino acid a weak acid or a weak base, and a hydrophile if the side chain

is polar or a hydrophobe if it is nonpolar.

Proteins (also known as polypeptides) are organic compounds made of amino acids arranged in

a linear chain. The amino acids in a polymer are joined together by the peptide bonds between

the carboxyl and the amino groups of adjacent amino acid residues.

Like other biological macromolecules such as polysaccharides and nucleic acids, proteins are

essential parts of organisms and participate in virtually every process within cells. Proteins are

important in:

- catalyzing biochemical reactions (enzymes)

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- structural and mechanical functions (actin and myosin)

- cell signaling

- immune responses

- cell adhesion

- cell cycle

TESTS ON AMINO ACIDS:

1) Solubility Tests:

The solubility of amino acids and proteins is largely dependent on the solution pH. The structural

changes in an amino acid or protein that take place at different pH values alter the relative

solubility of the molecule. In acidic solutions, both amino and carboxylic groups are protonated.

In basic solutions, both groups are deprotonated.

Amino acids are essentially soluble in water. Their solubilities in water, dilute alkali and dilute

acid vary from one compound to the other depending on the structure of their side chains. Apply

this test to glycine, tyrosine, glutamic acid and cysteine.

Procedure:

- Note the solubility of amino acids in water and alcohol by placing a small amount in a test tube,

adding a few mL of solvent and warming if necessary.

- Determine the amino acid solution is acidic or basic by using a litmus paper while testing the

solubility in water.

- Repeat the solubility test using dilute HCl and dilute NaOH.

2) Ninhydrin Test:

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Ninhydrin (triketohydrindene hydrate) is a chemical used to detect ammoniaor primary and

secondary amines. Amino acids also react with ninhydrin at pH=4. The reduction product

obtained from ninhydrin then reacts with NH3 and excess ninhydrin to yield a blue colored

substance. This reaction provides an extremely sensitive test for amino acids. Apply this test to

any of the amino acids you choose.

WARNING: Avoid spilling ninhydrin solutions on your skin, as the resulting stains are difficult

to remove. (Ninhydrin is the most commonly used method to detect fingerprints, as the terminal

amines or lysine residues in peptides and proteins sloughed off in fingerprints react with

ninhydrin).

Procedure:

- To 1 mL amino acid solution add 5 drops of 0.2% ninhydrine solution in acetone.

- Boil over a water bath for 2 min.

- Allow to cool and observe the blue color formed.

Questions:

Write the reaction(s) involved in Ninhydrin Test.

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3) Stability to Alkali:

Amino acids, unlike amides and volatile amines, do not evolve NH3 or alkaline vapor when

boiled with alkali. This method can be used to differentiate amino acids from amines and amides.

Apply this test to the provided amine or amide and also to glycine.

Procedure:

- Pipette 1 mL 1% glycine and the amide or amine solution into separate test tubes.

- Add 1 mL dilute NaOH to each test tube and boil.

- Test the vapor from each boiling tube with wet litmus paper.

Questions:

What type of reaction is responsible for the evolution of alkaline vapor? Write the reaction and

explain briefly.

4) Specific Reactions for Individual Amino Acids:

WARNING: Please DO NOT use vast amounts of solution for these tests, since most of the

amino acids are very expensive!!

a) Xanthoproteic Test:

Some amino acids contain aromatic groups that are derivatives of benzene. These aromatic

groups can undergo reactions that are characteristics of benzene and benzene derivatives. One

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such reaction is the nitration of a benzene ring with nitric acid. The amino acids that have

activated benzene ring can readily undergo nitration. This nitration reaction, in the presence of

activated benzene ring, forms yellow product. Apply this test to tyrosine, tryptophan,

phenylalanine and glutamic acid.

Procedure:

- To 2 mL amino acid solution in a boiling test tube, add equal volume of concentrated HNO3.

- Heat over a flame for 2 min and observe the color.

- Now COOL THOROUGHLY under the tap and CAUTIOSLY run in sufficient 40% NaOH to

make the solution strongly alkaline.

- Observe the color of the nitro derivativative of aromatic nucleus.

Questions:

Write the reaction(s) involved in Xanthoproteic Test.

Define “activated benzene ring”, briefly.

Do all the amino acids with aromatic side chains give positive result? Why?

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b) Millon’s Test:

Millon’s test is specific to phenol containing structures (tyrosine is the only common phenolic

amino acid). Millon’s reagent is concentrated HNO3, in which mercury is dissolved. As a result

of the reaction a red precipitate or a red solution is considered as positive test. A yellow

precipitate of HgO is NOT a positive reaction but usually indicates that the solution is too

alkaline. Apply this test to tyrosine, phenylalanine, glycine and β-naphtol.

Procedure:

- To 2 mL amino acid solution in a test tube, add 1-2 drops of Millon2s reagent.

- Warm the tube in a boiling water bath for 10 min.

o A brick red color is a positive reaction.

o Note that this is a test for phenols, and the ninhydrin test should also be positive if it is to be

concluded that the substance is a phenolic amino acid.

Questions:

Write the reaction(s) involved in Millon’s Test.

You have phenol, tyrosine, cysteine and β-naphtol in separate test tubes. By using which

test(s) would you find the tyrosine containing test tube?Explain, briefly.

c) Hopkin’s Cole Test:

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The indole group of tryptophan reacts with glyoxylic acid (glacial acetic acid, which has been

exposed to light, always contains glyoxylic acid CHOCOOH as an impurity) in the presence of

concentrated H2SO4 to give a purple color. Apply this test to glycine, tryptophan and tyrosine.

Procedure:

- To a few mL of glacial acetic acid containing glyoxylic acid, add 1-2 drops of the amino acid

solution.

- Pour 1-2 mL H2SO4 down the side of the sloping test tube to form a layer underneath the

acetic acid.

- The development of a purple color at the interface proves a positive reaction.

Questions:

Write the reation(s) involved in Hopkin’s Cole Test.

What is the role of H2SO4 in this test? Explain, briefly.

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d) Lead-Sulfide Test:

When cystine is boiled with 40% NaOH, some of sulfur in its structure is coverted to sodium

sulfide (Na2S). The Na2S can be detected by using sodium plumbate solution which causes the

precipitation of PbS from an alkaline solution. In order to apply this test, first the sodium

plumbate solution should be prepared. Apply this test to cysteine and cystine.

Procedure:

- Sodium Plumbate Solution Preparation:

o Add 5 mL dilute NaOH to 2 mL dilute lead acetate.

o A white precipitate of lead hydroxide forms.

o Boil until the precipitate dissolves with the formation of sodium plumbate.

- Boil 2 mL amino acid solution with a few drops of 40% NaOH for 2 min.

- Cool and add a few drops of the sodium plumbate solution.

- A brown color or precipitate is a positive test for sulfides.

Questions:

Write reation(s) involved in the Lead-Sulfide Test.

Explain what is “plumbate”?

e) Ehrlich Test:

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Aromatic amines and many organic compounds (indole and urea) give a colored complex with

this test. Apply this test to tryptophan, urea and glycine.

Procedure:

- Put 0.5 mL of the amino acid solution to a test tube.

- Add 2 mL Ehrlich reagent and observe the color changes.

- Repeat the test with urea solution.

Questions:

What chemicals are found in Ehrlich’s reagent.

Explain the reaction involved in Ehrlich Test.

Explain your observation for the urea solution when it is tested with Ehrlich’s reagent.

f) Sakaguchi Test:

The Sakaguchi reagent is used to test for a certain amino acid and proteins. The amino acid that

is detected in this test is arginine. Since arginine has a guanidine group in its side chain, it gives a

red color with α-naphthol in the presence of an oxidizing agent like bromine solution. Apply this

test to arginine.

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Procedure:

- 1 mL NaOH and 3 mL arginine solution is mixed and 2 drops of α-naphthol is added.

- Mix thoroughly and add 4-5 drops of bromine solution UNDER THE HOOD!!

- Observe the color change.

Questions:

Define and give the structure of guanidine.

g) Nitroprusside Test:

The nitroprusside test is specific for cysteine, the only amino acid containing sulfhydryl group (-

SH). This group reacts with nitroprusside in the presence of excess ammonia. Apply this test

cysteine, cystine and methionin.

Procedure:

- Put 2 mL amino acid solution into the test tube.

- Add 0.5 mL nitroprusside solution and shake thoroughly.

- Add 0.5 mL ammonium hydroxide.


Questions:

Write the reaction(s) involved in Nitroprusside Test.

Is there any difference in the test results of cystine and cysteine? If there is, explain the

reasons by giving the related structures.

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5) Tests for Proteins:

a) Biuret Test:

The Biuret Test positively identifies the presence of proteins (not less than two peptides). The

reaction in this test involves the complex formation of the proteins with Cu2+ ions in a strongly

alkaline solution. Apply this test to gelatin, casein and albumin.

Procedure:

- To 2 mL protein solution, add 5-6 drops of dilute CuSO4 (Fehling’s solution A diluted 1/10

with water)

- Add 3 mL 40% NaOH solution.


If the protein tested is insoluble in water, then apply the procedure given below:

- Measure 3 mL acetone and 1.5 mL water into a test tube.

- Add 1 drop of dilute NaOH and a little piece of protein to be tested.

- Boil continuously over a small flame for 2 min and cool.

- Add 0.5 mL 40% NaOH and 2 drops of a 1/10 diluted Fehling’s solution A.


Questions:

Write the reaction(s) involved in Biuret’s Test.

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b) Ninhydrin Test:

This test is given by only amino acids and proteins which contain free –NH2 groups in their

structure. Apply this test for all the proteins provided.

c) Test for Amino Acids:

Perform the tests for individual amino acids on the provided proteins.

Xanthoproteic Test, Millon’s Test , Hopkin’s Cole Test, and Lead Sulphite Test.

Questions:

According to your test results, indicate which amino acids are found on the protein structures

that are tested.

d) Precipitation of Proteins:

The precipitation of a protein occurs in a stepwise process. The addition of a precipitating agent

and steady mixing sestabilizes the protein solution. Mixing causes the precipitant and the target

product to collide. Enough mixing time is required for molecules to diffuse accross the fluid.

I. By Neutral Salts:

The precipitation of a protein by neutral salt is commonly known as salting-out method. Addition

of a neutral salt, such as ammonium sulfate, compresses the solvation layer and increases the

protein-protein interaction. As the salt concentration of a solution is increased, more of the bulk

water becomes associated with the ions. As a result, less water is available to take part in the

46


solvation layer around the protein, which exposes hydrophobic parts on the protein surface.

Therefore, proteins can aggregate and form precipitates from the solution. The amount of neutral

salt required to cause protein precipitation varies with the nature of the protein and the pH of the

solution. Apply this test to all the proteins provided.

Procedure:

- Add solid ammonium sulfate to about 5 mL of protein solution in a test tube (the salt should be

added in quantities of approximately 1 g at a time)

- Agitate the solution gently after each addition to dissolve the ammonium sulfate.

Questions:

The salting-out process occurs spontaneously. Can you explain the reason for this spontaneity

with free energy, enthalpy and entropy concepts.

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II. By salts of Heavy Metals:

Heavy metal salts usually contain Hg2+, Pb2+, Ag1+, Tl1+, Cd2+ and other metals with high

atomic weights. Since salts are ionic, they disrupt salt bridges in proteins. The reaction of a

heavy metal salt with a protein usually leads to an insoluble metal protein salt. Apply this test to

all the proteins provided.

Procedure:

- Treat 3 mL of the protein solution provided with a few drops of mercuric nitrate.

- A white precipitate formation should be observed.

Questions:

What would you expect to happen when you add mercuric nitrate on the solution of cystine

amino acid? Explain, briefly.

III. By Acid Reagents:

The precipitation of a protein in the presence of acid reagents is probably due to the formation of

insoluble salts between the acid anions and the positively charged protein particles. These

precipitants are only effective in acid solutions. Apply this test to all the proteins provided.

Procedure:

- Treat 3 mL of protein solution provided with a few drops of trichloroacetic acid solution.

- Note the protein precipitate formed.

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Questions:

What could be the reason of using trichloroacetic acid as an acid reagent instead of commonly

used ones?

6) Char test:

Procedure

Place about a pea-sized bit of casein in an evaporating dish and ignite it with the Bunsen

burner.

Question:

What does the odour/ smell reminds you of?

What do you think it smells like?

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7) Unknown Part:

- Take an unknown solid from your assistants and please DO NOT forget to write your

unknown number in your lab reports.

- Carry out the amino acid and protein tests in a reasonable sequence to determine your unknown

solid (Please DO NOT trust on your solubility observations and physical appearances of

your unknown and complete all the tests).

50


EXPERIMENT 1- QUALITATIVE ANALYSIS OF CARBOHYDRATES

A carbohydrate is an organic compound with the general formula Cm(H2O)n, that is, consists

only of carbon, hydrogen and oxygen, with the last two in the 2:1 atom ratio. Carbohydrates

make up the bulk of organic substances on earth and perform numerous roles in living things.

The carbohydrates (saccharides) are divided into four chemical groups: monosaccharides,

disaccharides, oligosaccharides and polysaccharides. Polysaccharides serve for the storage of

energy (e.g., starch in plants and glycogen in animals) and as structural components (e.g.,

cellulose in plants and chitin in arthropods). Structural polysaccharides are frequently found in

combination with proteins (glycoproteins or mucoproteins) or lipids (lipopolysaccharides). The

5-carbon monosaccharide ribose is an important component of coenzymes (e.g., ATP, FAD and

NAD) and the backbone of the genetic molecule known as RNA. The related deoxyribose is a

component of DNA. Saccharides and their derivatives include many other important

biomolecules that play key roles in the immune system, fertilization, preventing pathogenesis,

blood clotting and development [1].

This experiment aims to introduce you with the identification of unknown carbohydrates. To

gain maximum benefit, observations should be related, as far as possible, to the structure of

the substances examined.

Some important points:

1. Most of the tests and reactions described are not quantitative and volumes are approximate,

despite these facts some tests do not work if quantities greatly in excess of those stated are used.

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2. DO NOT place your pipettes in reagent bottles as this leads to contamination.

3. In most tests, it is important to apply a control test using water instead of the solution under

examination. If you are in doubt about the result of a test, perform the reaction with a suitable

known compound.

4. In this experiment, sugar samples are given in their solid state. To perform each procedure,

you should prepare your own sugar solution by taking very small amounts of solid sugars.

5. When you need to boil your sample in a test tube, prepare a hot water in a large beaker and put

your test tube inside the beaker. DO NOT forget to put boiling chips in the beaker.

TESTS ON CARBOHYDRATES:

1) Molisch’s Test:

Molisch’s Test is a sensitive chemical test for all carbohydrates, and some compounds containing

carbohydrates in a combined form, based on the dehydration of the carbohydrate by sulfuric acid

to produce an aldehyde (either furfural or a derivative), which then condenses with the phenolic

structure resulting in a red or purple-colored compound.

Procedure:

- Apply this test two different carbohydrate solutions of your own choice , preferably to one

monosaccharide and one polysaccharide.

- Place 2 mL of a known carbohydrate solution in a test tube, add 1 drop of Molisch’s reagent

(10% α-naphthol in ethanol).

- Pour 1-2 mL of conc. H2SO4 down the side of the test tube, so that it forms a layer at the

bottom of the tube.

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- Observe the color at the interface between two layers and compare your result with a control

test.

A brown color due to charring must be ignored and the test should be repeated with a

more dilute sugar solution.

Questions:

Write the reaction step(s) involved in this test?

Give an example of a protein structure that would give positive test with Molisch’s Reagent.

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2) Solubility Tests:

- Apply this test to all carbohydrates provided.

- Observe the solubility of the carbohydrates both in water and ethanol.

DO NOT depend on your solubility observations during identification of your unknown

compound.

3) Carbohydrates as Reducing Sugars:

A reducing sugar is any sugar that, in a solution, has an aldehyde or a ketone group. The

enolization of sugars under alkaline conditions is an important consideration in reduction tests.

The ability of a sugar to reduce alkaline test reagents depends on the availability of an aldehyde

or keto group for reduction reactions. A number of sugars especially disaccharides or

polysaccharides have glycosidic linkages which involve bonding a carbohydrate (sugar)

molecule to another one, and hence there is no reducing group on the sugar; like in the case of

sucrose, glycogen, starch and dextrin. In the case

54


of reducing sugars, the presence of alkali causes extensive enolization especially at high pH and

temperature. This leads to a higher susceptibility to oxidation reactions than at neutral or acidic

pH. These sugars, therefore, become potential agents capable of reducing Cu+2 to Cu+, Ag+ to

Ag and so fort. Most commonly used tests for detection of reducing sugars are Fehling’s Test,

Benedict’s Test and Barfoed’s Test.

a) Fehling’s Test:

Fehling’s Solution (deep blue colored) is used to determine the presence of reducing sugars and

aldehydes. Perform this test with fructose, glucose, maltose and sucrose.

Procedure:

- To 1 mL of Fehling’s solution A (aqueous solution of CuSO4) add 1 mL of Fehling solution B

(solution of potassium tartrate).

- Add 2 mL of the sugar solution, mix well and boil.

Try to see the red precipitate of cuprous oxide that forms at the end of the reaction.

Questions:

Write the reaction(s) involved in Fehling’s Test.

What is the function of tartrate?

Some disaccharides such as maltose are reducing agents, whereas others, such as sucrose are

not. Explain briefly by incluiding the structures of the sugars.

55


56


b) Barfoed’s Test:

Barfoed’s reagent, cupric acetate in acetic acid, is slightly acidic and is balanced so that is can

only be reduced by monosaccharides but not less powerful reducing sugars. Disaccharides may

also react with this reagent, but the reaction is much slower when compared to monosaccharides.

Perform this test with glucose, maltose and sucrose.

Procedure:

- To 1-2 mL of Barfoed’s reagent, add an equal volume of sugar solution.

- Boil for 5 min. in a water bath and allow to stand.

You will observe a brick-red cuprous oxide precipitate if reduction has taken place.

Questions:

Write the reaction(s) involved in the Barfoed’s Test.

When you test starch with Barfoed’s reagent, what would be the answer, positive or negative?

Explain your answer by giving reasons and structures.

57


58


c) Seliwanoff’s Test:

Seliwanoff’s Test distinguishes between aldose and ketose sugars. Ketoses are distinguished

from aldoses via their ketone/aldehyde functionality. If the sugar contains a ketone group, it is a

ketose and if it contains an aldehyde group, it is an aldose. This test is based on the fact that,

when heated, ketoses are more rapidly dehydrated than aldoses. Perform this test with glucose,

fructose, maltose and sucrose.

Procedure:

- Heat 1 mL of sugar solution with 3 mL Seliwanoff’s reagent (0.5 g resorcinol per liter 10%

HCl) in boiling water.

In less than 30 seconds, a red color must appear for ketoses.

Upon prolonged heating, glucose will also give an appreciable color.

Questions:

Write the reaction(s) involved in Seliwanoff’s Test.

What is the funtion of resorcinol?

What is the aim of using a strong acid?

What is the result of testing sucrose with Seliwanoff’s reagent? Explain your answers by

giving reasons and structures.

59


60


d) Bial’s Test:

Bial’s Test is to determine the presence of pentoses (5C sugars). The components of this reagent

are resorcinol, HCl, and ferric chloride. In this test, the pentose is dehydrated to form furfural

and the solution turns bluish and a precipitate may form. Perform this test with ribose and

glucose.

Procedure:

- To 5 mL of Bial’s reagent, add 2-3 drops of sugar solution and boil.

Upon boiling, note the green-blue color formed.

Questions:

Write the reaction(s) involved in Bial’s Test.

Is it possible to distinguish DNA and RNA structures by using Bial’s Test?

The boiling step is common for each test for the reducing sugars. Why boiling is necessary for

the reduction to take place?

61


62


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4) Action of Alkali on Sugars:

Procedure:

- Heat 1 mL glucose solution with 1 mL 40% NaOH for 1 min.

- Cool and apply test for reducing sugars (e.g.; Fehling’s Test).

- Apply a control test with glucose solution to observe the difference.

Questions:

Explain the reaction of glucose in the alkali medium by giving your reasons and related

structures.

Explain the difference between glucose solution and alkali treated glucose solution when a test

for reducing sugars is applied.

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65


5) The Inversion of Sucrose:

Sucrose is a disaccharide, which means that it is a molecule that is derived from two simple

sugars (monosaccharides). In the case of sucrose, these simple sugars are glucose and fructose.

Inverted sugar is a mixture of glucose and fructose. It is obtained by splitting sucrose into these

two components. The splitting of sucrose is a hydrolysis reaction which can be induced simply

by heating an aqueous solution of sucrose. Acid also accelerates the conversion of sucrose to

invert.

Procedure:

- Add 5 mL of sucrose solution to two test tubes.

- Add 5 drops of conc. HCl to one test tube.

- Heat both tubes in boiling water bath for 10 min.

- Cool and neutralize with diluted NaOH (use litmus paper).

- Test both solutions for the presence of reducing sugar with Fehling’s Test.

Questions:

Explain the result by giving the reasons and related reactions and structures.

What would you expect from a similar reation with starch?

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67


6) Iodine Test:

Iodine test is an indicator for the presence of starch. Iodine solution (iodine dissolved in an

aqueous solution of potassium iodide) reacts with starch producing a blue-black color. Apply

this test to all the polysaccharides provided.

Procedure:

- To 2-3 mL of polysaccharide solution, add 1-2 drops of iodine solution.

- Observe the different colors obtained for each of the polysaccharide solutions.

Questions:

Explain the reaction between iodine solution and polysaccharides by giving the structures of

related compounds.

Each polysaccharide tested gives different color results with the iodine test. Explain the reason

briefly.

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7) Unknown Part:

- Take an unknown solid from your assistants and please DO NOT forget to write your

unknown number in your lab reports.

- Carry out the carbohydrate tests in a reasonable sequence to determine your unknown (Please

DO NOT trust on your solubility observations and physical appearances of your unknown).

69


EXPERIMENT 1- QUALITATIVE ANALYSIS OF LIPIDS

Fats and oils are the principle stored forms of energy in many organisms. They are highly

reduced compounds and are derivatives of fatty acids. Fatty acids are carboxylic acids with

hydrocarbon chains of 4 to 36 carbons; they can be saturated or unsaturated. The simplest lipids

constructed from fatty acids are triacylglycerols or triglycerides. Triacylglycerols are composed

of three fatty acids each in ester linkage with a single glycerol. Since the polar hydroxyls of

glycerol and the polar carboxylates of the fatty acids are bound in ester linkages, triacyl glycerols

are non-polar, hydrophobic molecules, which are insoluble in water.

TESTS ON LIPIDS:

1-Solubility test:

The test is based on the property of solubility of lipids in organic solvents and insolubility in

water. The oil being insoluble in water will float on water because of lesser specific gravity.

Procedure

Take 3ml of solvents (water, ethanol, chloroform/ ether) in seperate test tubes and add 5 drops of

sample. For water and ethanol, it is insoluble and for chloroform and ether, it is soluble and

hence the given sample is lipid.

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2-Transparency test:

All the lipids are greasy in nature. Therefore the test may be taken as group test for lipids.

The oil does not wet the paper rather it will make a greasy spot on it.

Procedure

Take 3ml of ether in a test tube and dissolve 5 drops of oil in it. Put a drop of the solution on the

filter paper and let it dry. A translucent spot on the filter paper will be observed and this indicates

the greasy character of the lipid.

3- Emulsification test:

When oil and water, which are immiscible, are shaken together, the oil is broken up into very

tiny droplets which are dispersed in water. This is known as oil in water emulsion. The water

molecule due to the high surface tensions has a tendency to come together and form a separate

layer. This is why the oil and water emulsion is unstable in the presence of substances that lower

the surface tension of water. E.g.: Sodium carbonate, soap, bile salts etc. The tendency of the

water molecule to coalesce is decreased and the emulsion becomes stable. Since bile salts cause

the greatest decrease in surface tension they are best emulsifying agents.

Procedure

1. Add 1 mL of the food sample to 2 mL of ethanol

2. Shake well

3. Allow to settle in a test tube rack for two minutes

4. Empty this liquid into a test tube containing 2mL of distilled water

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Observations

If the mixture remains clear there are no fats present in the sample

However, if the mixture becomes cloudy there are fats present in the sample

4- Saponification test

Saponification is the hydrolysis of fats or oils under basic conditions to afford glycerol and the

salt of the corresponding fatty acid. Saponification literally means "soap making". It is important

to the industrial user to know the amount of free fatty acid present, since this determines in large

measure the refining loss. The amount of free fatty acid is estimated by determining the quantity

of alkali that must be added to the fat to render it neutral. This is done by warming a known

amount of the fat with strong aqueous caustic soda solution, which converts the free fatty acid

into soap. This soap is then removed and the amount of fat remaining is then determined. The

loss is estimated by subtracting this amount from the amount of fat originally taken for the test.

The saponification number is the number of milligrams of potassium hydroxide required to

neutralize the fatty acids resulting from the complete hydrolysis of 1g of fat. It gives information

concerning the character of the fatty acids of the fat- the longer the carbon chain, the less acid is

liberated per gram of fat hydrolysed. It is also considered as a measure of the average molecular

weight (or chain length) of all the fatty acids present. The long chain fatty acids found in fats

72


have low saponification value because they have a relatively fewer number of carboxylic

functional groups per unit mass of the fat and therefore high molecular weight.

Principle:

Fats (triglycerides)

upon alkaline

hydrolysis

(either with KOH or NaOH ) yield glycerol and potassium or sodium salts of fatty acids (soap) .

Procedure

1) Weigh 1g of fat in a tarred beaker and dissolve in about 3mL of the fat solvent [ethanol

/ether mixture].

2) Quantitatively transfer the contents of the beaker three times with a further 7ml of the

solvent.

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3) Add 25ml of 0.5N alcoholic KOH and mix well, attach this to a reflux condenser.

4) Set up another reflux condenser as the blank with all other reagents present except the fat.

5) Place both the flask on a boiling water bath for 30 minutes.

6) Cool the flasks to room temperature.

7) Now add phenolphthalein indicator to both the flasks and titrate with 0.5N HCl .

8) Note down the endpoint of blank and test.

9) The difference between the blank and test reading gives the number of milliliters of

0.5N KOH required to saponify 1g of fat.

10) Calculate the saponification value using the formula:

Saponification value or number of fat = mg of KOH consumed by 1g of fat.

Weight of KOH = Normality of KOH * Equivalent weight* volume of KOH in litres

Volume of KOH consumed by 1g fat = [Blank – test] ml

5- Test for unsaturation:

The unsaturated fatty acids absorb iodine at the double bonds until all the double bonds are

saturated with iodine. Hence the amount of iodine required to impart its color to the solution is a

measure of the degree of the fatty acids.

Procedure

Take 1ml of chloroform and add a drop of methanol and one drop of oil. To this add 1drop of

iodine. Chloroform dissolved sample give red color which decolorizes the iodine giving brown

color. The intensity of color indicates the presence of unsaturated fatty acids.

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6- Sudan IV test for the presence of lipids

Sudan IV is a red β-naphthol diazo dye used for staining triglycerides, lipids and lipoproteins in

cells and tissues.

Like lipids, the chemical Sudan IV is not soluble in water; it is, however, soluble in lipids.

Therefore to test for the presence of lipids in a solution you will use a Sudan IV Test. In this test,

if lipids are present the Sudan IV will stain (by “sticking to” or binding with) them reddish-

orange, giving a positive test result.

NOTE: Sudan IV is mildly irritating and will stain skin and clothing. Handle with care!

Procedure:

1. Mark pieces of filter paper as shown below (use PENCIL):

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2. Fill in testing solutions column of data char given below. Use water as negative and oil

as positive control .

3. Using a clean pipette for each sample, place a drop of each solution in the appropriately

labeled area. Repeat when the first drops are dry.

4. Allow the paper to dry for 30 minutes.

a. Fill in prediction column of data chart using the key described in step #9.

5. In a Petri dish, soak the paper in Sudan IV solution for 3 minutes.

6. Fill a second Petri dish with distilled water.

7. Use forceps to transfer the filter paper from the Sudan IV Petri dish to the Petri dish with

water.

8. Rinse for one minute and allow to dry on a paper towel.

9. Compare the intensity of the red coloring. Record your results in the data table.

Negative for lipids = 0

Faint red: +

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Definite red: ++

7- Grease Spot Test for Fats

You perform this test every time you buy muffins or doughnuts in a paper bag. Lipids make

unglazed paper (brown paper, writing paper) translucent.

1. Fill in grease spot prediction column. Predict whether or not the testing sample will

leave a spot and how translucent it will be (a little, very, etc…)

2. Draw and label several circles on the paper (same as on filter paper).

3. Put a drop of each sample on a piece of unglazed paper.

4. Write the name of the sample in pencil next to the spot.

5. Allow all spots to dry thoroughly.

6. Hold the paper in front of a light source and observe the spots.

7. Record your observations in data table.

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Data Table:

Testing

Solution

Prediction

Sudan IV

Sudan IV

Solution

Prediction

Grease Spot

Grease Spot

Test

Water

Oil

Sample 1

Sample 2

Sample 3

Sample 4

Sample 5

Sample 6

Questions:

1. Compare your predictions to your results. How many correct predictions did you have?

______ How many incorrect predictions? _______

2. Were there any foods that did NOT contain fat that surprised you? Which ones and why did it

surprise you?

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__________________________________________________________________

3. Were there any foods that DID contain fat that surprised you? Which ones and why did it

surprise you?

__________________________________________________________________

4. Compare the results of the Sudan IV solution and the grease spot test.

__________________________________________________________________

__________________________________________________________________

5. Do the results of your lab correspond to information you have learned about lipids? EXPLAIN

HOW.

__________________________________________________________________

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__________________________________________________________________

__________________________________________________________________

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EXPERIMENT: QUANTITATIVE ESTIMATION OF CHOLESTEROL USING

SPECTROPHOTOMETRIC METHOD

Cholesterol is a member of a large group of substances called steroid, which include vitamin D.

Cholesterol is an essential component of cell membrane, brain and nerve cells, and bile, which

helps the body absorb fats and fat soluble vitamins. The body uses cholesterol to make Vitamin

D and various hormones, such as estrogen, testosterone, and cortisol. The body can produce all

the cholesterol that it is needs, but it also obtains cholesterol from food.

Lieberman-burchard test: The Lieberman-Burchard or acetic anhydride test is used for the

determination of cholesterol. The formation of a green or green-blue color after a few minutes is

positive. Lieberman-Burchard is a reagent used in a colorimetric test to detect cholesterol, which

gives a deep green color. This color begins as a purplish, pink color and progresses through to a

light green then very dark green color. The color is due to the hydroxyl group (-OH) of

cholesterol reacting with the reagents and increasing the conjugation of the un-saturation in the

adjacent fused ring. Because this test uses acetic anhydride and sulfuric acid as reagents caution

must be exercised so as not to receive severe burns.

Material:

1- Cholesterol reagent (20 mL Acetic anhydride + 1 mL Conc. sulphuric acid)

2- Sulphuric acid 95- 97 %

3-Standard cholesterol (300 mg/dl)

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4-Samples (any vegetable oil like olive oil, olive pomace oil, ghee etc)

5-Test tubes

6-Pipettes

7-Cuvettes

8-Spectrophotometer

9-Water bath

Procedure: Label 7 test tubes as (A, B) for test 1, (C, D) for test 2, (E, F) for the standard, and

(G) for the blank, the following will be added to each labeled test tube and OD recorded at

610nm:

1 2 Standard Blank

Contents (A,B) (C,D) (E,F) (G) Absorbance

at 610 nm

Sample 1 0.1 ml

Sample 2 0.1 ml

Cholesterol

standard

0.1 ml

DH2O (blank) 0.1 ml

Cholesterol

Reagent

4 ml 4 ml 4 ml 4 ml

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Then the tubes will be incubated at room temperature for 20 minutes. Then 1.0 ml of sulphuric

acid will be added to each tube. Incubate in a water bath at room temperature for 5 minutes.

Remove from the water bath and shake vigorously. Measure the absorbance after 10 minutes.

The absorbance will be measured for the samples against the blank at 610 nm.

Calculation:

(Absorbance of sample\Absorbance of standard) × 300

Sample1:

Sample 2:

Activity:

What is the reason of getting different absorbance in different oil samples?

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EXPERIMENT: ISOELECTRIC PRECIPITATION OF CASEIN PRESENT IN MILK.

Principle:

Milk is a mixture of many types of protein, most of them present in very small amounts. Milk

proteins are classified into three main groups of proteins on the basis of their widely different

behavior and form of existence. They are caseins (80%), whey proteins and minor proteins.

Casein is a heterogeneous mixture of phosphorous containing proteins in milk. Casein is present

in milk as calcium salt, calcium caseinate. It is a mixture of alpha, beta and kappa caseins to

form a cluster called micelle. These micelles were responsible for the white opaque appearance

of milk.

The casein, as proteins, is made up of many hundreds of individual amino acids, each of which

may have a positive or a negative charge, depending on the pH of the [milk] system. At some pH

value, all the positive charges and all the negative charges on the [casein] protein will be in

balance, so that the net charge on the protein will be zero. That pH value is known as the

isoelectric point (IEP) of the protein and is generally the pH at which the protein is least soluble.

For casein, the IEP is approximately 4.6 and it is the pH value at which acid casein is

precipitated. In milk, which has a pH of about 6.6, the casein micelles have a net negative charge

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and are quite stable. During the addition of acid to milk, the negative charges on the outer

surface of the micelle are neutralized (the phosphate groups are protonated), and the neutral

protein precipitates.

The same principle applies when milk is fermented to curd. The lactic acid bacillus produces

lactic acid as the major metabolic end-product of carbohydrate [lactose in milk] fermentation.

The lactic acid production lowers the pH of milk to the IEP of casein. At this pH, casein

precipitates.

Materials required:

1) Raw milk - 100ml

2) 0.2N HCl - 50ml

3) Diethyl ether - 50ml

4) 50% Ethanol - 50ml

5) Whatman No 1 filter paper strip (Size 25×50mm) - 2 no.

Procedure:

1. Measure 100ml of milk in a measuring cylinder and transfer aliquots of 25ml of milk to

four centrifuge tubes.

2. Centrifuge the milk in a centrifuge at 4000 rpm at room temperature (25- 30 oC) for 20

minutes. This is done to remove the fats and lipids from the mixture.

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3. After centrifugation, carefully remove the fats and lipids from the surface of the milk

with a spatula.

4. Then transfer the milk from all the tubes into a beaker and add equal volume of distilled

water and stir well. Now check the pH.

5. Start adding 0.2N HCl drop by drop into the milk mixture and stir well.

6. Note the PH at which precipitation (white curdy substances) appears. The pH should be

4.6.

7. Take the curdy precipitate and allow it to sediment.

8. Now decant the supernatant using a filter paper and funnel and wash the precipitate with

distilled water to remove the salts, then wash with diethyl ether and ethanol.

9. Dry the precipitate and take the weight of the casein and record it.

Activity:

What happens to a protein if the pH is lowered than the isoelectric point of that protein?

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EXPERIMENT: QUANTIFICATION OF THE AMOUNT OF AMINO ACIDS BY

USING NINHYDRIN REACTION.

Theory:

Amino acids are known as the building blocks of all proteins. There are 20 different amino acids

commonly found in proteins. Amino acids comprise of a carboxyl group and an amino group

attached to the same carbon atom (the α carbon). They vary in size, structure, electric charge and

solubility in water because of the variation in their side chains ( R groups). Thus detection,

quantification and identification of amino acids in any sample constitute an important steps in the

study of proteins.

The general structure of an amino acid is shown below:

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Alpha amino acids react with Ninhydrin involved in the development of color which is explained

by the following five steps.

1. alpha-amino acid + Ninhydrin ---> Reduced ninhydrin +Alpha amino acid +H2O

This is an oxidative deamination reaction that elicit two hydrogen from the alpha amino acid to

produce an alpha – imino acid. Also the ninhydrin reduced and loses an oxygen atom with the

formation of water molecule.

2. alpha-amino acid + H2O ---> alpha-keto acid +NH3

The rapid hydrolysis of NH group in the alpha – imino acid will cause the formation of an alpha-

keto acid with an ammonia molecule. This alpha-keto acid further involved in the

decarboxylation reaction of step.

3. alpha-keto acid + NH3 ---> aldehyde + CO2

Under a heated condition to form an aldehyde that has one less carbon atom than the original

amino acid. A carbon dioxide molecule is produced along with aldehyde. These first three steps

produce the reduced ninhydrin and ammonia that are required for the production of color .The

overall reaction for the above reactions is simply explained in Reaction (4) as follows:

4. alpha-amino acid + 2 ninhydrin ---> CO2 + aldehyde + final complex (BLUE) + 3H2O

In summary, ninhydrin, which is originally yellow, reacts with amino acid and turns deep purple.

It is this purple color that is detected in this method. Ninhydrin will react with a free alpha-amino

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group, NH2-C-COOH. This group is present in all amino acids, proteins or peptides. Whereas,

the decarboxylation reaction will proceed for a free amino acid, it will not happen for peptides

and proteins. Theoretically only amino acids produce color with ninhydrin reagent. However,

one should always check out the possible interference from peptides and proteins by performing

blank tests especially when such solutions are readily available. For example, one can simply add

the ninhydrin reagent to a solution of only proteins and see if there is any color development.

There is no excuse for failing to perform such a vital test when the sample mixture contains both

proteins and amino acids. There are also reports that chemical compounds other than amino acids

also respond positively to this reaction.

The ninhydrin reaction, one of the most important methods of detecting amino acids, both

technically and historically, has been conventionally used to detect their microgram amounts.

When amino acids with free alpha amino groups are treated with an excess of ninhydrin, they

yield a purple colored product. Under appropriate conditions, the color intensity produced is

proportional to the amino acid concentration.

The primary amino groups react with ninhydrin to form the purple colour dye now called

Ruhemann's purple (RP) was discovered by Siegfried Ruhemann in 1910. Iminoacids like

proline, the guanidino group of arginine, the amide groups of asparagine, the indole ring of

tryptophan, the sulfhydryl group of cysteine, amino groups of cytosine and guanine, and cyanide

ions also react with ninhydrin to form various chromophores that can be analyzed.

The overall reaction can be written as follows:

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Primary amines also react with ninhydrin, but do not liberate of CO2.

[ Caution : Ninhydrin is a very reactive oxidizing agent, so should be handled with care].

Several other convenient reagents are available which can react with the alpha amino group to

form colored or fluorescent derivatives. These include fluorescamine, dansyl chloride, dabsyl

chloride etc used in the detection of trace amounts of amino acids at the nanogram level.

In the quantitative estimation of amino acid using Ninhydrin reagent, the absorbance of the

Ruhemann's purple formed by the reaction at 570nm is measured. For imino acids, the

absorbance is done at 440nm. The principle behind the colorimetric estimation is given below:

Principle of Colorimeter:

Measurement of Absorbance (A):

Unknown compounds may be identified by their characteristic absorption spectra in the

ultraviolet, visible or infrared regions. Enzyme-catalysed reactions frequently can be followed by

measuring spectrophotometrically the appearance of a product or disappearance of a substrate. A

spectrophotometer /colorimeter is an instrument for measuring the absorbance of a solution by

measuring the amount of light of a given wavelength that is transmitted by a sample.

Light and Spectrum profile:

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Light can be categorized according to its wavelength. Figure 1 shows the relationship between

the wavelength of light and the common types of electromagnetic radiation. Light in the short

wavelengths of 200 to 400 nm is referred to as ultraviolet (UV). Light in the longer wavelengths

of 700 to 900 nm is referred to as near infrared (near IR).

Visible light falls between the wavelengths of 400 and 700 nm. All the colors visible to human

eye fall under this wavelength range. Any solution that contains a compound that absorbs light in

the visible region will appear colored to the eye. The solution is colored because specific

wavelengths of light are absorbed as they pass through the solution. Then, the only light that the

eye will perceive are the wavelengths of light that are transmitted (not absorbed).

Beer-Lambert law:

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The Beer-Lambert law states that the amount of light absorbed is proportional to the number of

molecules of absorbing substance in the light path, ie., absorption is proportional both to the

concentration of the sample solution and to the length of the light path through the solution. This

relationship can be

expressed as

follows:

Absorbance, A= ε

x c x l c =

concentration of

the sample (in Moles/liter),

l = length of the light path through the solution (in cm) and ε = molar extinction coefficient

To determine the absolute concentration of a pure substance, a standard curve is constructed

from the known concentrations and using that standard curve, the absorbance reading of the

unknown concentration was determined. The determination of unknown concentration from the

standard curve is done by drawing a line parallel to the X- axis from the point on the Y axis that

corresponds to the absorbance of the unknown. This line will be made to intersect the standard

curve drawn, and is extended vertically such that it meets the X-axis and the concentration of

unknown is read from the X-axis. A typical standard curve is depicted in the figure.

Materials Required:

Reagents Required:

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1. Standard amino acid stock solution (150 micrograms of Standard amino acid stock

solution (150µg/ml)).

2. 0.2M Acetate buffer (pH=5.5).

3. 8% w/v of Ninhydrin reagent [Preparation: Weigh 8g of ninhydrin and dissolve in 100ml

of acetone].

4. 50% v/v ethanol.

5. Distilled water.

6. Any sample like milk, beef broth as unknown solution.

Apparatus and Glassware Required:

1. Test / Boiling tubes.

2. Pipettes [glass / micropipette].

3. Water bath.

4. Colorimeter.

Procedure:

1. Pipette out different volumes (0.1ml-1ml) of standard amino acid solution to the

respective labeled test tubes.

2. Add distilled water in all the test tubes to make up the volume to 4ml.

3. Add 4ml of distilled water to the test tube labeled Blank.

4. Now add 1ml of ninhydrin reagent to all the test tubes including the test tubes labeled

'blank' and 'unknown'.

5. Mix the contents of the tubes by vortexing /shaking the tubes.

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6. Put a few marble chips in each tube.

7. Cover the mouth of the tubes with aluminum foil.

8. Place all the test tubes in boiling water bath for 15 minutes.

9. Cool the test tubes in cold water and add 1ml of ethanol to each test tube and mix well.

10. Now record the absorbance at 570 nm of each solution using a colorimeter.

How to set the colorimeter:

1. Switch on the colorimeter and set the wavelength at which the absorbance is to be

measured [this should be done about 30 min before taking the readings].

2. Rinse the cuvette with the blank solution; drain off the solution by inverting the cuvette

and touching its mouth on to the filter paper.

3. Fill the cuvette with the blank solution up to the appropriate mark and place it in the hole

provided on the colorimeter.

4. Care should be taken that the cuvette is properly inserted into the hole.

5. Read the OD value and tare the instrument to zero.

6. Take out the cuvette and drain off the blank solution.

7. Now measure the OD values from the lower to the higher concentration.

Activity

Plot the values of OD of each sample along Y-axis and concentration of sample along X-axis to

make a standard curve. Find the concentration of unknown sample from the standard curve.

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Documents

· Web viewTo obtain accurate pH readings, we must standardize the electrode using buffers of known pH before measuring an unknown. Static electricity can affect pH readings, so always