Food is composed of organic and inorganic substances.

• Organic substances :
1. Carbohydrates
2. Proteins
3. Lipids
4. Vitamins

• Inorganic substances : H2O , mineral salts.

 Tests for Inorganic Substances

Test for water Test for mineral salts

• Put a piece of food(bread) in • Put a piece of bread in a
a test tube test tube.
• Heat the test tube • Add drops of silver nitrate
• Drops of water(mist) (AgNO₃) or ammonium
appears on the presence of oxalate.
water in bread. • White precipitate appears
in the presence of salt.
 Tests for Organic Substances
1. Test for Carbohydrates

Test for starch Test for reducing sugar

(glucose,maltose,fructose,lactose, galactose)
- Iodine solution test - Benedict’s/Fehling solution test
brown orange (original color) (blue color) requires heating
+ present + present
dark blue (final color) Brick red color
Note: Sucrose the white sugar we eat is not a reducing sugar
 2. Test for proteins
• Biuret test (copper sulfate + sodium hydroxide)
(Blue color)
+ presence of protein
Violet color

• Coagulation test (requires heating) is used to indicate the presence of

certain type of protein albumin or globulin.
Coagulation/White small clots appears in presence of albumin or globulin.
Probing page # 11

Iodine water

Dark blue

Salts, water, reducing sugar, starch, proteins

 3. Test for Lipids
• Translucent spot test

Rub a piece of food or add drop of oil on a paper, if lipids

present a
a translucent spot appears.

• Sudan III test

Add sudan III solution to oil a red color appears

indicating the presence of lipids.
Title : Table showing the tests for simple food
Probing page 11

The bread is a complex food since it is made

up of several constituents : water, mineral
salts, carbohydrates, proteins and lipids

.
Milk has reducing sugar and proteins, since positive tests
appear upon the Fehling which turned brick red and the Biuret
tests which turned violet, indicating the presence of reducing
sugars and proteins respectively.
Grape contains reducing sugar, since the Fehling test changes
to brick red / positive test indicating the presence of reducing
sugar.
Meat and eggs contains proteins, since positive test appears
only upon Biuret test indicating the presence of proteins.

Banana contains starch and reducing sugar, since

upon applying iodine water and Fehling solution,
tests are positive ,indicating the presence of starch
and reducing sugar respectively
Describe:
• State the details of an observation (exp., graph, doc.) using correct scientific language.
• No acquired knowledge, no justification (NO since, because…)
• Use: First, then, followed by, finally, later, at the end
• Use a legend/key if given

Concept 4 number 5 p: 8. Describe the test used to prove that:

a. Milk contains lactose.
First put in a tube some milk and add a few drops of blue Fehling solution. Heat gently for
few minutes. The color will change to brick red.
b. Egg albumin contains proteins.
Add into a test tube egg. Then add a few drops of blue copper sulfate (CuSO4 ) , and a few
drops of colorless sodium hydroxide. The color will change to violet.
a. What is the color of the Biuret solution , the iodine solution, and the Fehling
solution before performing any test?
The color of the Biuret solution is blue
The color of the iodine solution is brown orange
The color of the Fehling solution is blue
b. Which test requires heat ? Describe how this test is performed.
Fehling solution test
Put in a tube some bread filtrate. Add a few drops of blue Fehling
solution. Then heat gently for few minutes , the color will change to
brick red.
c. Determine the contents of each tube. Justify your answer.
In Tube A , positive test shows only upon Biuret test. This indicates the presence of proteins. Thus,
tube A has egg albumin.
While Tube B is only positive upon Fehling test . This indicates the presence of a reducing sugar. So
Tube B has fructose solution
Tube C is only positive with translucent spot test . This indicates the presence of lipids. So Tube C
has sunflower oil.
Tube D is only positive with Iodine solution test . This indicates the presence of starch. This shows
that Tube D has starch paste.
 Digestion
Mechanical Digestion Chemical Digestion

• Physical Breakdown of complex food • Using enzymes to break down of a into

into smaller piece without changing of food to a simple food molecule.
the identity of the food.
• It changes the identity of the food
• It doesn’t need enzymes
• Enzymes are involved
Example: grinding, cutting, peristalsis
Enzymes:
- are proteins that speed up the chemical reaction
- are biological catalysts. ( biological , since it is produced by body of
living things) ( catalyst, since they speed up chemical reactions)
- are needed in small amounts in the body since one enzyme can work
on several reactions of the same type.
- Remains unchanged/intact at the end of the reaction
- Enzymes work better on small pieces of food
Note:
• **Mechanical digestion facilitates the action of an enzyme**
• ** enzymes are necessary for chemical digestion of food **
The starch milk is a raw starch and the starch paste is a cooked starch
• In vitro experiments are done in test tubes.
• In vivo experiments are directly tested in the body.

• Digestion of starch ( refer to doc a and b page 12 )

Title: Table showing the experimental conditions of the in vitro digestion of
starch
The three tubes contain the same quantity of starch paste and are placed under the
same temperature of 37˚C for the same time 20 min. But they differ by the
presence of distilled water in the tube A, the commercial enzyme (1) in tubes B and
C and of the commercial enzyme (2) in the tube C
˚

At the beginning of the experiment, the aspect of the three tubes is the same
turbid. Whereas at the end of the experiment (after 20 min), this aspect remained
turbid only in tube A, different than tubes B and C that are clear.
Tube A is a control tube used as a standard for comparison with other
experimental tubes.
Under the same conditions, at temperature 37 ˚C, having cooked starch in tubes A, B and C. A t=0min,
positive results appear in the 3 tubes upon adding iodine water while negative results appear upon
adding Fehling solution. While after 20min, the same results appear upon adding iodine and Fehling
water to tube A containing distilled water only. However, negative results appear upon adding iodine
water and positive results upon adding Fehling solution in both tubes B and C that contain enzyme 1
and enzyme 1 +2 respectively instead of water.

This indicates that no chemical transformation of starch takes place in tube A (starch is not digested in
tube A)while in tubes B and C starch paste was transformed/digested into sugar under the action of
enzymes 1 and 2.
• Fresh saliva juice ( in the mouth secreted by salivary gland ) contains
salivary amylase enzyme that digests cooked starch into reducing sugar
Maltose

• Pancreatic juice ( in the small intestine secreted by the pancreas and

released into the small intestine ) contains pancreatic amylase enzyme
that transform cooked starch into a reducing sugar ( maltose)

• substrate: is a substance acted upon by an enzyme.

• Salivary amylase and pancreatic amylase are the enzymes for cooked
starch, therefore the substrate for salivary amylase and pancreatic
amylase is starch paste.
Dialysis experiment ( refer to doc c)

Both the negative iodine water test and the positive Fehling solution test indicate
the presence of glucose and the absence of starch in the water of the container.
Thus the glucose is a small molecule which crosses the membrane of cellophane
while the starch is a macromolecule which does not cross this membrane.
• Molecular simplification of starch molecules into sugar/ hydrolysis of
starch .(p.13 doc d)
Starch has to be broken down into simple sugars in two steps.
Starch maltose glucose ( simple sugar)

• ***Progressive molecular simplification : the breakdown of large

molecules into simpler molecules (nutrients) in more than one step.

• Hydrolysis reaction : the breakdown of large molecules into simpler

ones in the presence of water.
Schematize the progressive molecular simplification of cooked starch.
The macromolecule of starch is first hydrolyzed into maltose by amylase enzyme
and then maltose is hydrolyzed into simpler molecules glucose( nutrient) by
maltase enzyme. Therefore, starch is simplified into a nutrient by more than one
enzyme

The commercial enzyme 1 is amylase

The commercial enzyme 2 is maltase

Saliva and pancreatic juice contains both amylase.

Maltase enzyme is secreted by the small intestine
Essentials #7 page 9
The document below shows the progressive molecular simplification of
a macromolecule (A) first into (B) And then into small molecules (C)

a. Molecule A can be identified by the iodine solution test. Name this

molecule.
starch
b. Molecules E and F represent enzymes. What is the role of an
enzyme?
enzymes speed up a chemical reaction without being consumed.

c. Are E and F the same enzyme? Justify your answer

No, because each enzyme works on a specific substrate. Enzyme E
works on molecule A and enzyme F works on molecule B.
d. Identify molecules B and C .
Molecule B is a disaccharide ( maltose) resulting from the breakdown of starch.
Molecule C is a monosaccharide ( glucose ) resulting from the breakdown of
maltose.

e. Explain why the expression “ progressive molecular identification” is used to

describe the chemical reaction in the document above.
Since starch ( complex molecule) is broken down into glucose( simpler molecule) in
more than one step
Essentials #8 page 10
Wheat starch is a complex molecule constituted of many molecules of
glucose. In the course of this digestion, starch reacts with water and is
degraded into very small sized molecules. This chemical reaction is
called hydrolysis. The enzymes, contained in saliva and pancreatic juice,
activate the hydrolytic reaction and favor the molecular simplification
of starch. These enzymes remain intact at the end of this reaction.

1. In reference to the text:

a. Indicate the constituent molecules of wheat starch.
glucose molecules
b. Define “hydrlolysis”
Break down of complex molecules into simpler ones in the presence of
water.
Wheat starch is a complex molecule constituted of many molecules of
glucose. In the course of this digestion, starch reacts with water and is
degraded into very small sized molecules. This chemical reaction is
called hydrolysis. The enzymes, contained in saliva and pancreatic juice,
activate the hydrolytic reaction and favor the molecular simplification
of starch. These enzymes remain intact at the end of this reaction

c. Pick put the sentence that shows the role of enzymes.

Enzymes activate the hydrolytic reaction and favor the molecular
simplification.

2. Name the enzyme responsible for the molecular simplification of

wheat starch.
amylase, maltase
3. Schematize the molecular simplification of starch ( cooked starch) in
the presence of saliva.
Title: Table showing the experimental conditions of the in vitro digestion of
proteins

Key: ( + ) : presence ( - ) : absence

Under the same conditions, at PH8 and Temperature 37˚C with eggalbumin in tubes A and B, at
the beginning of the experiment, positive result appear with Biuret test in both tubes A and B.
However, at the end of the experiment, the results remain the same (constant) for tube A in
presence of distilled water, While in tube B negative results appear in the presence of pancreatin
instead of water.
This indicates that chemical transformation of proteins took place only in tube B by commercial
pancreatin.

(**pancreatic extract is same as pancreatic juice and has lipase, trypsin and amylase )
The pancreatin has enzyme that digests proteins.
The favorable experimental conditions are temperature of 37⁰C, a
basic medium pH = 8 , and duration of 1 hr
The P1 protease transforms the macromolecule of protein into peptides and
the P2 protease transforms peptides into amino acids
A protein differs from other one by : the number, the type, and the
sequence of amino acids chain . After the digestion, the protein is
transformed into nutrients ( amino acids) . Therefore the sequence of
chain is lost. Hence, the specificity is lost.
Schematize the progressive molecular simplification of protein
Note:
( pepsin and trypsin hydrolyze proteins only, thus enzymes are specific )
( digestion of proteins start in the stomach and end in the small
intestine )
Digestion of lipids
A test tube containing: Oil + water + bile + pancreatic extract
- is placed at 37 oC ( body temperature)
- using bile (substance produced in the liver and stored in the gall
bladder)

**Lipids accumulate to form globules. In the presence of bile, the lipids are
emulsified into droplets to facilitate the action of lipase. Lipase acts on lipid
molecules to break them down into fatty acids and glycerol.
**Note : Fat digestion takes a lot of time, that’s why you feel tired after eating
lipids since it needs a long process and all energy will be spent in digestion.
Molecular simplification of lipids
Since lipid ( a complex molecule ) is broken down into fatty acids and
glycerol ( simpler molecules )in more than one step

The digestion of food is a molecular simplification where the

macromolecules are transformed, under the effect of the specific
enzymes, into simpler molecules.
Experiments to come up with the characteristics of enzymes
1. The effect of pH on enzyme activity ( refer to doc a pg 16)
Under the same conditions, at T=37⁰C, with fresh saliva and starch paste in tubes
A,B and C. Negative result appear upon adding iodine water and positive results
upon adding Fehling solution in tube A where the pH is 7. On the contrary, positive
result appears upon adding iodine water while negative result appears upon adding
Fehling solution in tubes B and C where the pH is 3 and 10 respectively,
This indicates that digestion of starch took place in tube A only where the pH is 7
while no digestion took place in tubes B and C where pH is 3 and 10 respectively.
Salivary amylase works best at an optimum pH=7
Characteristic # 1 of enzymes: enzymes work best at optimum pH.
2. The effect of temperature on enzymes ( refer to doc b page 16)
The optimum temperature is 37 ⁰C , because the negative iodine water
test and the positive Fehling solution test indicate that digestion of
starch paste into a reducing sugar took place only in tubes B and in
tube A after being placed at 37 ⁰C
The previously boiled enzyme at 60 ⁰C in tube C did not facilitate the digestion of
starch paste. When the tube was placed at 37 ⁰C, the positive iodine water test and
the negative Fehling solution test indicate the presence of starch and the absence of a
reducing sugar respectively. Thus the boiled enzyme became denatured at 60 ⁰C and
was not able to regain its activity at 37 ⁰C
The enzyme did not facilitate the digestion of starch paste in tube A placed at 0 ⁰C.
When this tube was replaced at 37⁰ C, digestion took place and this is confirmed by
the negative iodine water test and the positive Fehling solution test which indicate
the absence of starch paste and the presence of a reducing sugar respectively.
Accordingly, the enzyme was inactive at 0 ⁰C and it regained its activity at 37 C
Characteristic # 2 of enzymes: enzymes work best at optimum
temperature
3. Specificity of enzymes ( doc c)
What hypothesis is tested?
Hypothesis : An enzyme can act on one substrate
Fresh saliva facilitates the chemical transformation of starch paste only
but has no action on the milk starch and on egg albumin
Enzymes are specific since the enzyme of saliva facilitated the digestion
of the starch paste but is not able to digest the egg albumin and the
milk starch since this enzyme acts on a specific food and has no action
on another one.
Characteristic # 3 of enzymes: enzymes are specific to a substrate
4. Duration for enzyme activity ( refer to doc d )
The suitable duration to the transformation of egg albumin is 1 hr
because only tube B where the duration is 1 hr , has a negative Biuret
test indicating the absence of proteins, thus the transformation of
proteins into amino acids.
Favorable conditions for the activity of enzymes found in saliva and
pancreatin
Characteristic # 4 of enzymes: Activity of enzymes is optimum after
favorable duration
 Properties of Enzymes
• Enzymes act on specific substrate
Example: Salivary amylase acts on cooked starch only.
• Enzymes act on Specific pH (medium)
Example: Salivary amylase acts at pH 7 (neutral medium)
• Enzymes act on specific temperature 37℃
At cold temperature, enzymes become inactive and regain their activity
at temperature 37℃.
At High temperature, enzymes are denatured and do not regain its activity
• Enzymes act on specific duration
Example: Cooked starch needs 20minutes to be digested by salivary amylase
The salivary amylase in saliva facilitated the transformation of starch
paste in 20 min. However, outside the body and without enzymes this
transformation needs few days. So enzymes are biological catalysts
which accelerate the chemical reactions in the body
Essentials # 9 page 10
In a series of 5 test tubes, we introduce in each 2g of coagulated and fragmented
egg albumin (a protein), pepsin and hydrochloric acid (HCl). The tubes are placed at
different temperatures. After one hour, the quantity of egg albumin is measured in
each tube. The results are presented in the table below.

Tube Contents Temperature Quantity of egg albumin

measured after one hour

A 2g egg albumin + pepsin + HCl 0°C 2g

B 2g egg albumin + pepsin + HCl 20°C 1.7g
C 2g egg albumin + pepsin + HCl 37°C 0g

D 2g egg albumin + pepsin + HCl 45°C 1.7g

E 2g egg albumin + pepsin + HCl 60°C 2g

1. In which tube is the egg albumin totally digested ? Justify the answer by
referring to the obtained results only.
In tube C because after 1 hr, the quantity of albumin measured is 0 gr which means
that albumin has been totally digested in this tube
 Tube Contents Temperature Quantity of egg albumin
measured after one hour
A 2g egg albumin + pepsin + HCl 0°C 2g
B 2g egg albumin + pepsin + HCl 20°C 1.7g
C 2g egg albumin + pepsin + HCl 37°C 0g
D 2g egg albumin + pepsin + HCl 45°C 1.7g
E 2g egg albumin + pepsin + HCl 60°C 2g

b. Analyze the obtained results. What can you deduce concerning the enzyme
activity?
Having the same contents, In tubes A and E, with temperatures 0℃ and 60℃ respectively the
quantity of egg albumin remained 2 gr after 1 hr. In tubes B and D placed at temperature 20 and 45
℃ respectively the quantity decreased from 2 gr to 1.7 gr after 1 hr. In tube C, that is at
temperature 37℃, 2gr of egg albumin are not present at all ( 0gr ) after 1 hr.
Enzymatic activity is highest at 37 ℃, inactive at 0 ℃ and 60 ℃, and low at 20 ℃ and 45 ℃
 Tube Contents Temperature Quantity of egg albumin
measured after one hour
A 2g egg albumin + pepsin + HCl 0°C 2g
B 2g egg albumin + pepsin + HCl 20°C 1.7g
C 2g egg albumin + pepsin + HCl 37°C 0g
D 2g egg albumin + pepsin + HCl 45°C 1.7g
E 2g egg albumin + pepsin + HCl 60°C 2g

c. We place tubes A and E at 37°C for a duration of one hour. What

result can be obtained in each tube? Justify.

In tube A, digestion of albumin took place and the quantity of egg albumin became
o gr. At O⁰ C the enzyme was inactive and It regained its activity at 37⁰C. In tube E,
no digestion took place and the quantity of egg albumin remained 2 gr. At 60⁰C the
enzyme was denatured and cant regain its activity at 37⁰C
Essential # 10 page 11

a. What is the importance of tube 1

Tube 1 is a control tube used as a standard for comparison with other
experimental tubes
b. Which tube(s) show digestion of starch? Justify using information
from the table.
tube 2 shows digestion of starch since the iodine test was negative result indicating
the absence of starch and the Fehling test was positive indicating the presence of a
reducing sugar
c. Indicate the name of the products revealed by the iodine test and
the Fehling test.
A positive iodine test reveals the presence of starch and a positive
fehling test indicates the presence of a reducing sugar.
d. Explain why digestion didn’t occur in the other tubes.
In tube 1 there is no digestion due to the absence of a substrate.
In tube 3 there is no digestion because the enzyme in boiled saliva is
denatured.
In tube 4 there is no digestion because the enzyme is inactive at 0 oC.
In tube 5 there is no digestion because the enzyme in saliva doesn’t work in
an acidic medium ( pH=2).
In tube 6 there is no digestion because the enzyme in saliva does not act
with uncooked starch.
e. Which enzyme properties are signified by tubes 3,5, and 6?
Tube 3 shows that enzymes work best at an optimum temperature.
Tube 5 shows shows that enzymes work best at an optimum pH.
Tube 6 shows that enzymes are specific to substrate.
f. Indicate what will happen if we place tube 4 in a water bath at a
temperature of 37 C for 20 minutes.
the enzyme will regain its activity and digestion will occur. Iodine
solution test will be negative and the Fehling test will be positive.
digestion takes place in tube A. A brown orange color in tube A indicates the
absence of starch.
b. Explain why digestion didn’t take place in the remaining tubes.
In tube B no digestion took place , because starch paste does not work in an acidic
medium. In tube C no digestion took place because at temperature 70 oC , the
enzyme salivary amylase is denatured. In tube D, no digestion took place because
salivary amylase does not work on starch suspension.
Essentials # 12 page 12
Table showing different experimental conditions for each tube
Tube At the Medium Temperature Duration At the end of
beginning of the
the experiment
experiment

A Protein pH=8 37 oC 60 min Amino acid+

+water+ water +
trypsin trypsin

B Starch pH=8 37 oC 20 min Maltose +

paste+water water +
+amylase amylase

C Protein + pH=8 37 oC 60 min Protein +

water + water +
amylase amylase
D Starch paste+ pH=8 37 oC 20 min Starch paste
water+ trypsin + water +
trypsin
 Contents Tube A Tube B Tube C Tube D
At the Proteins Starch paste Proteins Starch paste
beginning of
Water Water Water Water
the experiment
Trypsin Amylase Amylase Trypsin
At the end of Amino acids Maltose Proteins Starch paste
the experiment
Water Water Water Water
Trypsin Amylase Amylase Trypsin

b. In which tubes does chemical transformation occur? Justify the answer by referring to
the table.
Chemical transformation occurs in tubes A and B. At the end of the experiment, the
proteins are transformed into amino acids in tube A and the starch paste is transformed
into maltose in tube B.
c. What property of enzyme activity is revealed in these experiments?
an enzyme is specific to a substance.
d. "An enzyme remains unchanged during a chemical transformation." Explain this
statement by referring to the table only.
AT the end of the experiment, we find that the enzyme in tube A( trypsin) and the
enzyme in tube B ( amylase) remained unchanged even when digestion occurred.
The molecules which undergo chemical transformation are the proteins, starch and
lipids.
The molecules that do not undergo chemical transformation are the amino acids,
mineral ions, glucose, wastes, fatty acids, glycerol, vitamins, water and fibers
Enzymes that help in the simplification of food and the
resulting nutrient

Food Enzymes Nutrients

Starch Amylase, maltase Glucose

Maltose Maltase Glucose

Lactose Lactase Glucose + galactose

Sucrose Sucrase Glucose + fructose

Proteins Pepsin, trypsin, peptidase Amino acids

Lipids lipase Fatty acids + glycerol

The small intestine, since the majority of enzymes are at this level ( pancreatic juice
and intestinal enzymes )
Enzymes are specific. Each enzyme acts on a specific food and has no action on the
other food.
Hypothesis:The undigested cellulose is eliminated with the fecal matter outside the
body
The small intestine:
- is folded to increase surface area for absorption.
- every big fold is called ridge (
- ridge has many folds called villi ( singular villus)
- villus is the functional unit of the small intestine site of absorption.
 function : site of absorption of nutrients
• Function of small intestine:
1. complete digestion of organic molecules into nutrients
2.Absorption of nutrients from the lumen of the small intestine into the
blood or lymph.

• Nutrients : are small and soluble molecules (in the small intestine),
that result from complete digestion of food.
Example of nutrients : glucose, fructose, amino acids, water, vitamins,
mineral salts, fatty acids, glycerol.
The different structures are folds or valves, the villi, the microvilli,
blood vessels, and lymphatic vessels
The pathway followed by nutrients :
intestinal lumen microvilli  villi folds  blood
The characteristics of small intestine :
1. Thin walls
2. Highly vascularized
3. Large surface area of absorption
Just after the digestion of a meal, the content of nutrients in the blood increases
from 1 to 1.8 g/L for glucose, from 0.5 g/L to 15 g/L for amino acids. The increase in
nutrients’ rate in blood is due to the fact that food is digested, transformed into
nutrients and then absorbed by the blood
After absorption , the concentration of water and mineral ions increased in
both the blood and the lymph. However, the concentration of glucose and
amino acids increased only in blood. Whereas, the concentration of fatty
acids and glycerol increased in lymph only. Thus, water and mineral salts
pass from the small intestine into blood and lymph while glucose and amino
acids pass from the small intestine into the blood only however, fatty acids
and glycerol pass from the small intestine into lymphatic vessels only
Glucose is a nutrient that is taken by cells as it is while starch is a
complex sugar that has to be broken down before cells can use it
Yes, the hypothesis is validated since having tubes A and B under the same conditions with the same quantity of
sucrose at time 0min, the quantity of sucrose was 5ml in both tubes, as time increases to 30 minutes,
the quantity of sucrose remains constant in tube B that has no sucrase while it decreases gradually in tube A
that has sucrose to become 0ml.
This shows that sucrase permits the digestion of sucrose