Download Enzymatic Action in Digestion

Enzymatic Action in Digestion
Procedural Inquiries
Preparation
1. What is hydrolysis?
2. Which major molecules must be
hydrolyzed in digestion?
3. How do we simulate body conditions
in setting up digestive experiments?
Experimentation
4. How do we test for starch?
5. What is the principle behind the
starch test?
6. How do we test for reducing sugars?
7. What is the principle behind the
reducing-sugar test?
8. How do we test for proteins?
9. What is the principle behind the
protein test?
10. How do we test for lipids?
11. What is the principle behind the
lipid test?
12.
13.
14.
Additional Inquiries
What is digestion?
What is the role of enzymes in the
digestive process?
What are bile salts?
Key Terms
Amylase
Benedict’s Test
Bile
Biuret’s Test
Carbohydrate
Hydrolysis
Lipid
Lugol’s Iodine
Micelles
Pancreatin
Protein
Reducing Sugar
Starch
Materials Needed
Water bath - 37°C
Water bath - 100°C
Test Tubes
Test Tube Rack
Test Tube Clamp
Pipets
Spot Plate
Solutions
1% Starch
2% Maltose
2% Glucose
Water (pH8)
Iodine
Benedict’s Solution
Albumin
5% Pancreatin (pH8)
Biuret’s Solution
Cream (or Half and Half)
Bile Salts
Litmus Solution
The foods we eat are composed primarily of carbohydrates, proteins, and lipids.
In foods, however, these building blocks are often not in their most usable form and
therefore must be digested before they can be transformed into whatever substance the
body may need. Digestion is the process of breaking foods down into that usable form.
The digestive process is a series of enzyme-catalyzed reactions. Recall that an
enzyme is an organic catalyst, a molecule at least part protein whose function is to change
the rate of a chemical reaction.
Enzymes do not do anything that does not happen naturally. For instance, if you
had had a steak for dinner last night, you really would not have needed those digestive
enzymes for the digestive processes to have taken place. The only problem is that you
would have starved to death waiting for digestion! Without digestive enzymes, what
normally occurs in a few hours (digesting the steak) would take about 50 years to happen.
Time is the justifying factor for enzyme catalysis. Enzymes, by today’s
definition, increase the rate of biochemical reactions without being changed themselves.
In this laboratory exercise we will demonstrate the principles of enzymatic action.
Preparation
I. Introduction
A. Hydrolysis
Hydrolysis, the basic digestive breakdown mechanism, involves the splitting of a
biological molecule by the addition of part of a water molecule.
 Study Figure 59-1 so that you understand exactly what is happening to the
molecules being hydrolyzed. If necessary, go back to your lecture text or to your
chemistry book to review the molecular intricacies of hydrolysis.
Dehydration
synthesis
+
+
H2O
(a) During dehydration synthesis two molecules are joined by
the removal of a water molecule.
Dehydration
synthesis
+
Dehydration
synthesis
+
Hydrolysis
H2O
(b) Hydrolysis reverses the steps of dehydration synthesis: a complex
molecule is broken down
by the addition of a
water molecule.
Figure Figure
59-1b, 59-1a
c, d
(b) Carbohydrate
Hydrolysis
Starch Molecule
(c) Lipid Hydrolysis
(d) Dipeptide Hydrolysis
Hydrolysis
+ H2O
2
Disaccharide
(maltose)
Two monosaccharides
(glucose)
(b)
Hydrolysis
+
+ H2O
Monoglyceride
Fatty acid
(lauric acid)
Glycerol
(c)
O
C
O
H
OH
C
C
C
H
N
NH2
H
H
C
H
H
O
C
C
N
H
H
H
O
C
C
C
H
N
H
Hydrolysis
O
H
+
H
O
H
H
H
H
H
Dipeptide
Alanine
Glycine
Two Amino Acids
(d)
 What are the major molecules that must undergo hydrolysis in the digestive
processes? ________________________________________________________
_________________________________________________________________
 In digestive hydrolysis, what happens to the water molecule?
B. Overview of the Experiments
Specific enzymes act on each of the major biochemical molecules so that those
molecules can be broken down into usable components. We will be demonstrating
this biochemical action. Because each enzyme has its own optimum temperature and
pH, we will be performing these experiments at 37°C to simulate the conditions of the
body, and at pH 8.0 to simulate the specific conditions in the small intestine where
most digestion takes place.
For each of the following tests, we will:
 Demonstrate or identify the substance.
 Simulate digestion under specific conditions.
 Demonstrate the substance is no longer present after digestion.
Experimentation
Be certain you have all your equipment ready for a particular experiment before
beginning that part of the exercise.
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II. Carbohydrates
A. Background
Starch, a basic storage molecule of plants, is a long chain of glucose molecules
that occurs in two forms, amylase and amylopectin, the latter being more highly branched
and actually more common in the everyday diet. (As a comparison, glycogen is the
animal storage molecule. Glycogen is a highly branched glucose chain.)
The same enzyme, amylase, is required to break down either of these starches.
Amylase is found in the saliva as salivary amylase and in the small intestine as
pancreatic amylase. Pancreatic amylase, which is produced in the pancreas, is virtually
identical to salivary amylase.
 Refer back to Figure 59-1. Note that the principle breakdown
products of amylase action are disaccharide maltose and certain
other short-chain glucose molecules.
B. Iodine Test for Starch Breakdown
If you wish to test a substance for starch, add Lugol’s iodine (IKI). If starch is
present, the geometric pattern of the starch molecules will interact with the iodine and
give a purplish-black color. If no starch is present, no geometric interaction can occur
and the substance in question will not become dark.
1 → Obtain a beaker of starch solution. (Technically you could use any
starch or starch-like substance, including a hamburger bun or a piece
of paper. However, for scientific validity, you will want to use the
same starch supply for all your starch experiments. Be careful of
your clothes because this experiment even works accidentally on
cotton shirts!)
2 → Also obtain: maltose solution, glucose solution, water (pH 8), Lugol’s
iodine, and a spot plate. (A clean microscope slide or chemistry
watch glass can be substituted for a spot plate.)
3 → On the spot plate put separate drops of starch, maltose, glucose, and
water. To each spot add a drop of Lugol’s iodine. What happened?
Starch ___Black_________________
Maltose __No Change____________
Glucose ___No Change___________
Water ____No Change___________
Neither water, maltose, nor glucose has the geometric configuration to affect a color
change. Only the starch drop should have changed color.
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C. Benedict’s Test
Benedict’s solution tests for the presence of reducing sugars. Glucose, maltose,
and fructose are reducing sugars. Sucrose is not a reducing sugar. Neither is starch. Go
back to the molecular structures of these carbohydrates to see if you can figure out why
certain molecules are reducing sugars while other molecules are not.
________________________________________________________________________
________________________________________________________________________
Reducing sugars react with Benedict’s solution to form the insoluble red
precipitat, cuprous (Copper I) oxide. If only small amounts of a reducing sugar are
present, the solution will be green. With more and more reducing sugar, the solution will
be yellow, orange, or red.
1 → Obtain four test tubes, labeled #1, #2, #3, #4.
2 → Also obtain: Glucose solution, maltose solution, starch solution,
water (pH 8), and Benedict’s solution.
3 → Into test tube #1:
Test tube #2:
Test tube #3:
Test tube #4:
10 drops glucose, 10 drops Benedict’s solution.
10 drops maltose, 10 drops Benedict’s solution.
10 drops starch, 10 drops Benedict’s solution.
10 drops water, 10 drops Benedict’s solution.
4 → Place the tubes in the boiling water bath for 30 seconds. Remove and
read your results.
#1 __Orange (+)___________ #3 ___Clear__(-)_________
#2 __Yellow_(+)__________ #4 ___Clear___(-)________
Glucose and maltose should have given you a positive Benedict’s test. Starch and
water should have yielded a negative test. Why? Glucose and Maltose are reducing
sugars, Starch is not a Reducing sugar and water is hydrogen and oxygen only
D. Experiment
In this experiment we are demonstrating the hydrolysis of starch. As starch is
hydrolyzed, more and more maltose and other short-chain glucose molecules should
be present.
1→ Obtain two test tubes, labeled #1 and #2.
2→ Also obtain: starch solution, water (pH8), iodine solution, Benedict’s
solution, and salivary amylase.
3→ Into test tube #1: 10 drops starch, 10 drops water
Test tube #2: 10 drops starch, 10 drops saliva (salivary amylase!)
4→ Place these tubes in the 37° water bath for 30 minutes. Remove from
the water bath.
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5→ Place 1 drop from each tube onto your spot plate.
Record your results.
#1 Black indicating starch present
#2 Yellow indicating starch is being digested
Add iodine.
6→ Place 9 drops of Benedict’s solution in each tube and place the tubes
in the boiling water for 30 seconds. Remove and read your results.
#1 No reaction indicating no reducing sugars present
#2 Orange indicating amylase digested the starch leaving reducing
sugars
E. Explanation
If starch was hydrolyzed, you should get no color change upon adding iodine
because no starch should be present.
If starch was hydrolyzed, you should get a positive Benedict’s test because
maltose is a breakdown product of starch and maltose is a reducing sugar. Glucose
and other short-chain glucose molecules that may be present in your hydrolyzed
starch solution are also reducing sugars.
Do your results match the predictions? If they do not, can you figure out what
happened?
________________________________________________________________________
________________________________________________________________________
(Some people do not produce salivary amylase. If you are such a person, you will
still have starch present and thus will have a positive iodine test and a negative
Benedict’s test. The inability to produce salivary amylase is not usually a problem
because most amylase action comes from the pancreatic amylase. Before assuming
you did not produce salivary amylase, check to make certain you did not mix up your
test tubes.)
III. Proteins
A. Background
Proteins are long chains of amino acids held together by peptide bonds. Pepsin,
secreted by the stomach, usually only begins the process of protein digestion by splitting
the long amino acid chains into shorter amino acid chains.
 Refer back to Figure 59-1. Most protein digestion takes place in the small
intestine under the influence of the pancreatic enzymes – trypsin, chymotrypsin,
and carboxypolypeptidase. Trypsin and chymotrypsin split the amino acid chains
into short polypeptides. Carboxypolypeptidase cleaves individual amino acids
from the polypeptide. (Several additional enzymes can also be found along the
brush border of the epithelium.)
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B. Biuret’s Test
Biuret’s solution is a mixture of sodium hydroxide (NaOH) and copper sulfate
(CuSO4). In the presence of protein, the copper sulfate reacts with the peptide bonds of
the protein, causing the mixture to turn a deep violet color. If protein was never present,
or if protein has been completely digested, no color change will occur.
1. → Obtain two test tubes, labeled #1 and #2.
2. → Also obtain: protein solution (albumin), water (pH 8.0), pancreatin (a
mixture of pancreatic enzymes), and Biuret’s solution.
3. → Into test tube #1: 10 drops of protein solution
Into test tube #2: 10 drops water (pH 8.0)
4. → Add Biuret’s solution to both tubes, drop by drop, until you observe a
definite, deep color change. Did the protein solution in tube #1 turn
violet? _Yes____ Tube #2 should have remained light blue.
C. Experiment
1. → Obtain two fresh test tubes, labeled #1 and #2.
2. → Also obtain: protein (albumin) solution, water (pH 8.0), Biuret’s
solution, and pancreatin.
3. → Into test tube #1: 10 drops protein solution, 10 drops water (pH 8.0).
Into test tube #2: 10 drops protein solution, 10 drops pancreatin.
4. → Place these tubes in the 37°C water bath for 30 minutes. Remove from
the water bath.
5. → Add about 5 drops of Biuret’s solution and gently shake the tubes.
Record the color: #1 Dark Purple indicating protein is present
#2 Light Purple indicating pancreatin digested the protein
D. Explanation
Test tube #1 should have changed color because no protein was digested: the color
changed because protein was still present.
Test tube #2 may have changed color. If the color change was pronounced, your
proteins were not digested. Head back to the drawing board and try to figure out what
happened! If you had no color change, your proteins were completely digestested. If you
had some color change (say to a light pink or a slight purple), you had some protein
digestion but you still have some protein present.
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IV. Lipids
A. Background
Triglycerides are the primary lipids in the average diet. If necessary, review the
chemistry of the triglycerides so that you have a picture of what we are ding in this
experiment.
 Refer back to Figure 59-1 for lipid breakdown.
Although an insignificant amount of lipid is digested in the mouth by lingual lipase
and in the stomach by gastric lipase, virtually all fat digestion takes place in the small
intestine.
The digestive enzymes involved in fat breakdown are water-soluble. Unfortunately,
fats themselves are not water-soluble. Therefore, if the fat traversed the intestine in
“glob” form, very little of it would be digested before the “glob” reached the ileocecal
valve.
Enter bile, a secretory product of the liver. Bile does not contain any enzymes so it
cannot function as an organic catalyst. Bile is an emulsifier. In other words, bile
physically separates parts of the lipid into smaller and smaller pieces. Bile increases the
surface area of the lipid, thus enhancing the possibility of lipase action.
The polar ionic and carboxyl parts of the bile salt molecule are highly water-soluble,
while the sterol portion of the molecule is highly lipid-soluble. The lipid-soluble portion
of the bile dissolves the fat, while the polar portion extends into the water-based fluid.
This lowers the interfacial tension of the lipid, making the molecules far less cohesive. In
other words the lipid becomes more soluble in water. The agitation of the small intestine
fragments the lipid, thus facilitating digestion.
As stated, the lipases are water-soluble and can only catalyze reactions on the surface
of the fat globule. Since bile greatly increases the surface area of the fat, fat digestion is
accelerated.
Bile salts also form aggregates of 20 to 40 molecules called micelles. In a micelle, all
hydrophobic sterol units adhere at a central point while the hydrophilic tails extend out
into the fluid.
Triglycerides are split into free fatty acids and monoglycerides, which are quickly
dissolved into the micelle aggregates. The micelles ferry these products to the brush
border of the intestinal epithelium.
B. Experiment
1. → Obtain three test tubes, labeled #1, #2, #3.
2. → Also obtain: cream, water (pH 8.0), litmus solution, and bile salts.
3. → Into test tube #1: 2ml cream, 2ml water (pH 8.0), pinch of bile salts, 15
drops litmus solution.
Into test tube #2: 2ml cream, 2ml pancreatin(pH 8.0), pinch of bile
salts, 15 drops litmus solution.
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Into test tube #3: 2ml cream, 2ml water (pH 8.0), 15 drops litmus
solution.
If you are using litmus paper instead of a litmus solution, ignore the directions
for adding litmus. When the experiment is complete, dip your litmus paper in the
solution and read the color.
4. → If the above solutions are not purple, add more litmus.
5. → Place these tubes in the 37°C water bath for 30 minutes. Remove from
the water bath. Record the color changes.
#1 light purple color indicating the fat was emulsified
#2 Light Purple to pink indicating fat was emulsified and digested by
pancreatin
#3 Light purple to pink, indicating pancreatin digested small amount of fat.
C. Explanation
Test tube #1 should have shown no change. Bile salts do not contain digestive
enzymes. Although your bile salts emulsified the fats, the fats were not digested.
Test tube #2 should have turned a nice shade of pink. The bile salts emulsified the
cream, and the enzymes in the pancreatin hydrolyzed the fats. You created an acidic
environment in the process.
Test tube #3 should have turned a color somewhere between #1 and #2. You put the
enzymes in but you did not add the bile salts. Thus, the digestive process was proceeding
but at a much slower rate.
D. Fatty Acids
Fatty acids without carbon-carbon double bonds are classified as saturated, whereas
those with at least one carbon-carbon double bond are classified as unsaturated.
Saturated fatty acids are solids at room temperature. Grease, lard, and butter are
examples of fats containing saturated fatty acids. Unsaturated fatty acids are less dense
than saturated fatty acids, have lower melting point, and tend to be liquids at room
temperature. Fats composed of unsaturated fatty acids are referred to as oils, such as corn
oil.
E. Fatty Acid Experiment
1. → Add 10 drops of vegetable oil to a test tube half filled with water.
2. → Shake the test tube and observe that the two liquids do not mix. The oil
molecules are hydrophobic; they are insoluble in water.
3. → Add 5 drops of Sudan IV and shake the test tube. Observe in which of
the two liquids the dye is soluble. What can you deduce from this
observation? Sudan IV is lipid soluable
4. → Prepare another test tube with oil, water, and Sudan IV. To this test tube
add several droppers full of diluted detergent water. Shake the test tube.
Observe the change that has taken place in the two liquids. Detergent is
an emulsifier. It surrounds the oil droplets and allows them to stay
suspended in the water.
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