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Biochemistry Exercise 6
Department of Biochemistry
Second Faculty of Medicine with the English Division and the Physiotherapy Division 1
Student’s name: _____________________________
Index numer: _________________________
Date: __________________
Points:
/3
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EXERCISE 6
DIGESTIVE SYSTEM.
Digestive enzymes are substances produced by our bodies that help us to digest the foods we eat. These
enzymes are secreted by the various parts of our digestive system and they help to break down food
components such as proteins, carbohydrates, and fats. This breakdown allows the nutrients from foods to be
absorbed into our bloodstreams so that they can support the functioning of all of the cells in our bodies.
The release of digestive enzymes
Digestive enzymes are released in both anticipation of food and in response to food. This means that just
thinking about or looking at food is enough to get your juices flowing. As we smell and eventually taste our
food, the amount of enzymes that are being secreted increases.
Enzymes are secreted from our salivary glands, and then from the cells lining our stomach, pancreas, and large
and small intestines. Different types of enzymes are secreted depending on the types of foods that we eat.
Types of digestive enzymes
Our bodies produce many different types of digestive enzymes to help our bodies to take advantage of the
various nutrients found within the foods we consume. Here are examples of some of the more prevalent types
of enzymes:
Amylase is a digestive enzyme essential for our digestion of carbohydrates, as amylase breaks down
starches into sugars.
Amylase is secreted by both our salivary glands and from our pancreas. The measurement of amylase
levels in the blood is sometimes used as an aid in diagnosing various pancreas or other digestive tract
diseases.
Lactase is a type of enzyme that breaks down the sugar, lactose, found in dairy products.
Supplemental lactase may be used to assist people who are lactose intolerant to digest dairy products.
Lipase is the enzyme responsible for the breakdown of fats that we consume. Specifically, lipase breaks
fats into fatty acids and glycerol. Within your body, lipase is produced in small amounts by your mouth
and stomach, and in larger amounts by your pancreas.
Biochemistry Exercise 6
Department of Biochemistry
Second Faculty of Medicine with the English Division and the Physiotherapy Division 2
Maltase is secreted by the small intestine and is responsible for breaking down maltose (malt sugar).
Proteases are digestive enzymes that break down proteins into amino acids. Here are the major types of
proteases found within the human digestive tract:
Carboxypeptidase
Aminopeptidase
Chymotrypsin
Pepsin
Trypsin
Sucrase is secreted by the small intestine where it breaks down sucrose into the simpler sugars of fructose
and glucose.
Digestive enzyme supplements
There are a wide variety of health conditions that interfere with the secretion of sufficient amounts of digestive
enzymes for full digestion of foods. In such a case a person might benefit from taking a digestive enzyme
supplement. One of the areas in which digestive enzyme supplements are most helpful is for people who have
chronic pancreatitis. For these individuals, digestive enzyme supplements not only help with digestion, but also
can help to ease the pain associated with the disease.
Summary of human digestive enzymes:
Biochemistry Exercise 6
Department of Biochemistry
Second Faculty of Medicine with the English Division and the Physiotherapy Division 3
Experiment 1
Amylase
Saliva contains hydrolitic enzyme -amylase (EC 3.2.1.1). It belongs to the hydrolase class of enzymes (EC 3).
Amylase optimal pH is 6.6-6.8 but it also needs Ca2+
which keeps proper conformation of the protein and Cl-
(enzyme’s activator).
Salivary amylase acts in mouth
(buccal cavity) and it produces:
dextrins (fragmented starch molecules,
most commonly containing one
-1,6-glycosidic bond)
maltose (isomaltose) molecules
maltotriose molecules
Aim:
The aim of this investigation is to detect the digestive properties of saliva.
Method used:
Iodine is the indicator for starch because it forms an intense blue
complex. Starch is not a redox indicator; it responds specifically
to the presence of I2, not to a change in redox potential. The
active fraction of starch is amylose, a polymer of the sugar α-D-
glucose. In the presence of starch, iodine forms I6 chains inside
the amylose helix and the color turns dark blue.
Procedure:
Take a test tube (large) and add:
1 ml of phosphate buffer (pH 6.6)
2.5 ml of 0.9% NaCl
2.5 ml of 1% starch solution
Mix the content and place the test tube into water bath (37°C) and leave it until the end of investigation.
Biochemistry Exercise 6
Department of Biochemistry
Second Faculty of Medicine with the English Division and the Physiotherapy Division 4
Prepare 6 test tubes (labelled 1-6) and add to each of them 1 drop of iodine solution and 1 drop of 2M HCl.
The determination of activity of amylase is performed in 5x diluted saliva. Take test tube and make 5x diluted
saliva solution (take your own saliva, split into test tube and add approx. 4 volumes of water). Mix it well to
obtain homogeneous solution.
1. Take 1 ml of diluted saliva and add to the incubation mixture in water bath
2. Immediately mix the content, take 2 drops using a pipette and transfer it to the first test tube containing the
iodine indicator (test tube 1 = time 0)
3. Take another samples of incubation mixture with saliva every 1 min and transfer them to labelled test tubes
4. If the starch is still present in the last test tube you must extend the experiment and take more test tubes
with iodine solution.
Observations and conclusions:
Biochemistry Exercise 6
Department of Biochemistry
Second Faculty of Medicine with the English Division and the Physiotherapy Division 5
Experiment 2
Pepsin
Pepsin is an endopeptidase found in gastric juice. It catalyzes the hydrolysis of peptide bonds within
polypeptide chain: X┼Tyr; X┼Phe; X┼Trp; X┼Leu; X┼Glu; X┼Gln.
Aim:
The aim of this investigation is to determine the influence of pH on the pepsin activity.
Procedure:
Prepare 4 test tubes and label them (1-4).
Reagents (mL) 1 2 3 4 (control)
0.2M HCl 0.5 - - -
Buffer (pH 6) - 0.5 - -
0.2M NaOH - - 0.5 -
Buffer (pH 7) - - - 0.5
Pepsin 0.5 0.5 0.5 0.5
a piece of chicken egg + + + +
Note observations after 5, 15, 30, 60 minutes.
Observations and conclusions:
Biochemistry Exercise 6
Department of Biochemistry
Second Faculty of Medicine with the English Division and the Physiotherapy Division 6
Experiment 3
Trypsin
The pancreatic juice after mixing with a food content from the stomach neutralizes its pH in duodenum. The
same effect has a bile produced in the liver and released from gallbladder. In such conditions different enzymes
which act in the intestines can be active: trypsin, chymotrypsin, elastase, carboxypeptidases, aminopeptidases,
pancreatic amylase, nucleases, maltase, lactase or pancreatic lipase. After digestion all products can be
absorbed.
Trypsin is an endopeptidase which works the best in pH 7-9. It breaks down peptide bonds between: Lys┼X;
Arg┼X.
Aim:
The aim of this investigation is to detect the activity of trypsin.
Method used:
In protein hydrolysate formalin blocks the amino groups of short peptide chains. Free carboxylic groups are
then titrated with the sodium alkaline in the presence of formaldehyde (Søerensen method1). The amount of
base is used to determine the level of trypsin activity.
Procedure:
Prepare 4 Erlenmeyer flasks and label them (1-4).
1. In flask 1 make an incubation solution: add 10 ml of gelatin solution and place it into water bath (37°C)
2. Add 3 ml of formalin with phenolphthalein to flasks 2, 3 and 4 and leave them at room temperature.
3. After 5 min add 3 ml of trypsin buffer solution (pH 8) to the flask 1, mix the content quickly (without
taking the flask out of the water). Take 2 ml of mixture and transfer it into flask 2 (time 0).
4. Titrate the content of flask 2 with the 0.02M NaOH solution until the pinkish colour appears.
You’ve got 5 minutes to do the titration (as the next sample of mixture must be taken from flask 1 after 5 minutes).
5. After 5 minutes take another sample of mixture (2 ml) and transfer it into flask 3. Do the titration.
6. After another 5 minutes take another sample of mixture (2 ml) and transfer it into flask 4. Do the titration.
Calculate the amount of free carboxylic groups formed during protein digestion. Remember, that the amount of
NaOH used correlates with the amount of carboxylic groups due to the proportion:
1 ml 0.02 M NaOH ↔ 20 µmol of carboxylic groups
1 details not needed.
Biochemistry Exercise 6
Department of Biochemistry
Second Faculty of Medicine with the English Division and the Physiotherapy Division 7
Calculations :
Draw a line graph showing the correlation between the time and the amount of hydrolysis product.
Biochemistry Exercise 6
Department of Biochemistry
Second Faculty of Medicine with the English Division and the Physiotherapy Division 8
Experiment 4
Vitamin C
Ascorbic acid (vitamin C) is a derivative of carbohydrates.
It is a coenzyme of oxidoreductases and acts as a donor of protons, e.g. during the synthesis of collagen,
adrenalin or bile acids). It is also needed for Fe3+
ions absorption in the intestines.
Biochemistry Exercise 6
Department of Biochemistry
Second Faculty of Medicine with the English Division and the Physiotherapy Division 9
Aim:
The aim of this investigation is to quantitative determination of vitamin C content in a fruit homogenate.
Method used:
DCPIP (2,6-Dichlorophenolindophenol) is a common indicator for vitamin C.
The amount of ascorbic acid is determined using the titration method.
Biochemistry Exercise 6
Department of Biochemistry
Second Faculty of Medicine with the English Division and the Physiotherapy Division 10
Procedure:
Take 3 test tubes and label them “C” for control test, “S” for standardization test and “E” for experimental test.
Reagents (mL) “C” ”S” “E”
Phosphoric(V) acid in acetic acid 1 - -
Standard solution of vitamin C - 1 -
Fruit homogenate - - 1
Titrate all three solutions with a DCPIP solution until decolourisation (until pinkish colour persists).
Calculate the amount of vitamin C (in g/100g of fruit).
1 ml of vitamin C standard solution contains 0.12 mg of ascorbic acid.
1 ml of homogenate contains 0.2 g of fruit.
Calculations:
Biochemistry Exercise 6
Department of Biochemistry
Second Faculty of Medicine with the English Division and the Physiotherapy Division 11
Experiment 5
Lactase
Lactose consists of galactose and glucose molecules connected with -1,4-glycosidic bond.
Lactose intolerance results due to decreased activity of lactase and may be caused by:
genetics
damage of cells of intestine lining as a result of different drugs taken by the patient or due to the protein
deficiency
The signs and symptoms of lactose intolerance usually begin 30 minutes to two hours after eating or drinking
foods that contain lactose. Common signs and symptoms include: diarrhoea (because of water retention in the
intestines), nausea (sometimes vomiting), abdominal cramps, bloating, gases production (H2, CO2, CH4).
lactose intolerance
Biochemistry Exercise 6
Department of Biochemistry
Second Faculty of Medicine with the English Division and the Physiotherapy Division 12
Aim:
The aim of this investigation is to confirm the hydrolytic properties of lactase.
Procedure:
Take 2 test tubes and label them (1-2).
Reagents (mL) 1(control) 2
4% glucose solution 1 -
2% lactose solution - 1
Lactase in buffer (pH 7) 1 1
Mix the content and incubate (37°C) for 15 minutes. Use the glucose-detecting strips to test for the presence of
products of lactose digestion.
Observations and conclusions: