enzym dan urin titimetri

8/12/2019 Enzym Dan Urin Titimetri

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4. Test tube

5. Beaker Glass

6. Buret

7. Erlenmeyer

Materials :

1. Glucose 0.5%

2. Mixture of urine with 0.5% glucose (2:1)

3. CuSO 4 solution (35 grams of CuSO 4 + 5 ml concentrated H 2SO 4 per liter)

4. Potassium ferro cyanide 5%

5. Signette solution (150 g K.Na tartrate + 90 grams of NaOH per liter) as the catalyst.

IV. PROCEDURE

A. Titration using 0.5% glucose solution

1. Pipette 10 ml CuSO 4 enter into erlenmeyer

2. Added 10 ml signette

3. Added 5 ml of potassium ferro-cyanide

4. Heat until boiling

5. Titrated with 0.5% glucose solution above the fire

6. Titrated until the color changes from blue to green and then to yellow. Titrated dropwise

until brown color

B. Titration using mixture of urin : 0.5% glucose solution (2:1)

1. Do same steps no 1 to 4

2. Titrated with mixture of urin : 0.5% glucose solution (2:1) above the fire


until brown color


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C. Titration using urine

1. Do same steps no 1 to 4

2. Titrated with urine


until brown color

Example calculation:

Example:

Volume (ml)0,5 % glucose Mixture urin : glucose (2:1)

8.93 11.75

Titration results of weighting glucose = 0.4989 gram

Method 1:

Weight of glucose 0.4989 = 0.4989 gram/100ml x 8.93ml = 0.04455 gram

Weight of glucose in urine glucose mixture (2 : 1)

Volume glucose 0.5% = 1/3 x 11.75 ml

So weight of glucose in urin glucose mixture is

= 1/3 x 11.75 x (0.4989 gram/100ml)

= 0.01954 grams

So urine volume is

= 11.75 ml - (1/3 x 11.75 ml)

= 2/3 x 11.75

= 7.833 ml

levels of glucose in the urine as penitir


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= (Weight of glucose Weight of glucose in urin glucose mixture ) x 100%) /

volume of urine

= (0.04455 - 0.01954) x 100%) / 7.837

= 0.3193%

Method 2:

Level of glucose in urine glucose mixture = level of 0.5% glucose = 0.4989 grams

V1 x N1 = V2 x N2

11.75 x N1 = 8.93 x 0.4989

N1 = 0.3792%

Ratio of urine : glucose 0.4989 = 2 : 1

Levels of glucose in urine glucose mixture = Glucose + glucose in the urine within 0.4989%

0.3792% = 2/3x + (1/3 x 0.4989%)

0.3792 = (2x + 0.4989): 3

3 x 0,3792 = 2x + 0.4989

2x = 1.1376 - 0.4989

x = 0.3194%

Note:

1. Titrating using pure urine is as a controller

2. Do variation mixing of urine and glucose such as 4:3, 3:2; 3:1 etc.


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V. QUESTIONS AND TASKS

1. Draw the diagram for analysis of glucose in urine using titrimetric method.

2. In this analysis, described the end point titration indicator so that the titration can be done.

3. Why titration done during boiling? Explain!


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EXPERIMENT 7

ENZYME

I. OBJECTIVES

1. Understanding enzyme function in the human body.

2. Identify enzyme activities by the sympton and phenomenone which can be observe.

3. be able to do critical experiment of enzyme activities.

II. BASIC THEORY

Food which enter in digestive canal can not be used by the body if food can not be

absorbed through digestive canal wall and carried by blood to all over the body. Food can be

absorbed must micro molecules. Food digestion system from macro molecules to micro

molecules did by digestive canal which supported digestive enzyme. Since food in the mouth,

occured carbohydrate methabolism by tooth and ptyalin enzyme become smaller sakaride

molecule, such as oligosakaride moreover disakaride and monosakaride. whereas protein, fat

and other substance have mechanic methabolism, that are methabolism by tooth. Food which

have methabolism in the mouth by physic or enzymatic, they will entering to stomach.

In normal condition, food stay in the stomach for some hours,while hydrochoric acid

(HCl) and pepsine scatter protein and carbohydrate become oligopeptide and oligosakaride.

Then, digestive process occurs in small intestine by enzymatic digestion. Digestive enzymes

also secreted by pancreas, bile and gland digestion and small intestine that finally it would

become monosakaride, glyserol, fatty acid and amino acid. Futhermore result substance will

be absorbed through small intestine wall then enter to blood circulation and transfered to

other part body which is need together with vitamin and other mineral.


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Based on the reaction, enzyme are grouped become 6 class, they are : (1) oxide-

reductase, (2) isomerase, (3) ligase, (4) liase, (5) Hidrolase, (6) transferase. Enzyme is protein

catalyst of a specific biochemical reaction. one enzyme product become another enzyme

substrate. Enzyme catalytic activities is influenced by pH, temperature, substrate level,

enzyme level and inhibitor.

Enzyme mechanism reaction through forming complex of enzyme substrate (ES).

Therefor obstacle in reaction which use enzyme as catalyst can be occured when substrate

annexation(merge) in active side of enzyme have inhibition. Molecule or ion which obstacle

reaction is called inhibitor. Inhibition by inhibitor as :

1. irreversible inhibition, generally caused by destruction process or modification a

functional group or more.

2. reversiblle inhibition

a. Competitive inhibition is inhibition caused there are molecule which similar with

substrate, so can able to make complexs, that is enzyme inhibitor complex (EI).

b. Non competitive inhibition is inhibition which is not influenced by amount substrate

and inhibitor consentration. example non competitive inhibitor is heavy metals (Cu 2+,

Hg 2+ , Ag +).

III. Equipment and Material :

Equipment :

1. Test tube

2. Filter paper

3. balance

4. Porcelain mortar

5. Drop pipete


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6. Spititus burner

7. Tripod

8. Beaker glass

Material :

1. Saliva sample

2. Amilum solution

3. I2 solution in KI (lugol)

4. Indicator universal

5. gastric juices of animals

6. aquades

7. benedict reagent

8. yeast bread

9. clean sand

10. toluene

11. sodium carbonate solution

12. buffer solution pH = 4

13. sucrose

14. soybean filtrate

15. pp indicator

16. HgCl 2 solution

17. Urea 1%

IV. PROCEDURE

A. Activities Ptialin Test


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1. Enter 3 ml of saliva into a clean and dry test tube

2. Dilute with 6 ml distilled water

3. Shake until homogeneous

4. Filtered to obtain a clean filtrate

5. Prepare 2 pieces of clean and dry test tubes

6. Put 2 mL of starch solution into each test tube

7. Add 2 ml of saliva on the tube 1, and add 2 ml of distilled water in tube 2

8. Shake each test tube

9. Add 1-2 drops of Lugol's solution to both test tube, observe and record the changes

B. Stomach Sap Test

1. Take 2 pieces of tube, first tube with 2 ml of the liquid contents of the stomach while

the second tube filled with distilled water

2. Check the pH of each tube with universal idicator, record the pH obtained

C. sucrase test

1. Prepare enzyme with 1 gram of yeast bread, put in porcelain mortar

2. Add 5 grams of clean and dry sand

3. Add 10 ml of toluene, and then grinding the mixture until yeast bread smooth and

formed homogeneous mixture

4. Add 30 ml of distilled water form a homogeneous suspension

5. Separate the liquid from filtrate

6. Take the supernatant with a pipette

7. Treated supernatant according to the following table:


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No supernatant Buffer

acetate

Sucrose Amylum Aqua Na 2CO 3 Benedict Treatment

1 3 cc 1 cc - - 3 cc 5 drops 5 cc heat

2 3 cc 1 cc 3 cc - - 5 drops 5 cc heat

3 3 cc 1 cc - 3 cc - 5 drops 5 cc heat

4 3 cc

boiling

1 cc 3 cc - - 5 drops 5 cc heat

5 3 cc

boiling

1 cc - 3 cc - 5 drops 5 cc heat

heated in a water bath 10 minutes .

8. Observe the color change.

D. Activity urease test in soybean.

Ureum can analyzed by urease enzyme become CO 2 and NH 3. Because of NH 3, so it can

give red color in pp indicator. Enzyme work can be destroyed or obstructed by hot

temperature or by adding Hg. Reactant that used to urease test is urea solution 1%. Soybean

filtrate, pp indicator and HgCl 2.

1. Prepare 3 clean reaction tube, then fill it with material as:

Tube Sample Adding HgCl 2 Adding

ureum 5%

PP

I 1cc soybean

filtrate

- 5cc 1 drop

II 1cc soybean

filtrate

5 drop 5cc 1 drop


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III 1cc soybean

filtrate that

boiled

- 5cc 1 drop

3. Put 3 reaction tube in water bath at 40 0C during 5 minute.

4. Observe the color change!

V. Question and task

1. Write the function of ptyalin in digestive process and what is enzyme included in

ptyalin?explain it.

2. how to make iodium solution? Explain it.

3. why activity ptyalin test appear the color change?explain it.

4. at gastric juice test , why do not use litmus paper?

5. Why the color change at gastric juice test occurred in one of many reaction tube?

Explain it.

6. What the function of gastric juice in the body?

7. What the enzyme that work in pH of gastric juice?

8. In yeast bread there are enzyme. Mention and explain the function of it!

9. What the function of buffer solution at sucrase test?

10. At sucrase test, why the 3 th tube do not happen the color change as in the 2 nd tube?

Explain it.

11. At activity urease test, what the tube that happen the color change? Explain it.

12. What the function of using HgCl 2 in 2 nd tube and mention the heavy metal other?

13. Based on the reaction,what urease included in enzyme?

14. Why enzyme can not work with the heavy metal?

enzym dan urin titimetri

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