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BIO462 BIOCHEMISTRYAS2012BLAB REPORT 1,2,3 & 4TITLE :1. DILUTION AND MOLARITY2. PROTEIN DETERMINATION3. DETERMINATION OF ENZYMES ACTIVITY4. GEL ELECTROPHORESISNAME: DAYANG NOR SHAFIQAH BINTI AWANG SHUKERANSTUDENT ID: 2020994777DATE OF SUBMISSION: 24 DECEMBER 2020PREPARED FOR: MISS SARAH SHAZ...

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BIO462 BIOCHEMISTRY AS2012B1 LAB REPORT 1,2,3 & 4 TITLE : 1. DILUTION AND MOLARITY 2. PROTEIN DETERMINATION 3. DETERMINATION OF ENZYMES ACTIVITY 4. GEL ELECTROPHORESIS

NAME: DAYANG NOR SHAFIQAH BINTI AWANG SHUKERAN STUDENT ID: 2020994777 DATE OF SUBMISSION: 24 DECEMBER 2020

PREPARED FOR: MISS SARAH SHAZWANI ZAKARIA

EXPERIMENT 1: DILUTION AND MOLARITY DILUTION

PROBLEMS

i. How would you prepare: a) 10 mL of a 1:10 dilution

of a 1 M NaCl solution and what would be the final

concentration of NaCl? M1 = 1 M

V1 = ?

M2 = 1:10 x 1 M

V2 = 10 mL

M2 =

1 20

x1

= 0.05 M M1V1 = M2V2 (1)(V1) = (0.1)(10) V1 = 1 mL Hence, 1 mL of stock solution must be added up with 9 mL of solvent.

b) 80 mL of a 1:20 dilution of a 1 M NaCl solution? M1 = 1 M

V1 = ?

M2 = 1:20 x 1 M

V2 = 80 mL

M2 =

1 20

x1

= 0.05 M M1V1 = M2V2 (1)(V1) = (0.05)(80) V1 = 4 mL Hence, 4 mL of stock solution must be added up with 76 mL of solvent.

c) 50 mL of a 1:25 solution of a 1 M NaCl solution? M1 = 1 M

V1 = ?

M2 = 1:25 x 1 M

V2 = 50 mL

1 25

M2 =

x1

= 0.04 M M1V1 = M2V2 (1)(V1) = (0.04)(50) V1 = 2 mL Hence, 2 mL of stock solution must be added up with 48 mL of solvent.

ii. How would you prepare exactly 6 mL of a 1/20 dilution (assume the concentration of your starting solution is ‘1’)? M1 = 1 M

V1 = ?

M2 = 1:20 x 1 M

V2 = 6 mL

1 20

M2 =

x1

= 0.05 M M1V1 = M2V2 (1)(V1) = (0.05)(6) V1 = 0.3 mL Hence, 0.3 mL of stock solution must be added up with 5.7 mL of solvent.

iii. You are prepared with an antibody solution (Ab) that has a concentration of 600 microgram (µg) / microlitre (µL). For lab, it is necessary to make the following dilutions.

a) 10 µL of 600 µg/L Ab + 190 µL of buffer to make a 1:20 dilution at 30 µg/µL. M1 = 600 µg/µL

V1 = 10 µL

M2 = ?

V2 = (190 µL + 10 µL) = 200 µL

M1V1 = M2V2 (600)(10) = M2(200 M2 = 30 µg/µL

b) 20 µL of 1:20 Ab + 40 µL buffer to make a 1:60 dilution at 10 µg/µL. M1 = 1:20 x 600 µg/µL

V1 = 20 µL

M2 = ?

V2 = (20 µL + 40 µL) = 60 µL

1 20

M1 =

x 600 µg/µL

= 30 µg/µL M1V1 = M2V2 (30)(20) = M2 (60) M2 = 10 µg/µL c) 5 µL of 1:60 Ab + 5 µL of buffer to make a 1:120 dilution at 5 µg/µL. M1 = 1:60 x 600 µg/µL

V1 = 5 µL

M2 = ?

V2 = (5 µL + 5 µL) = 10 µL

1 60

M1 =

x 600 µg/µL

= 10 µg/µL

M1V1 = M2V2

dilution factor =

5 600

=

1 120

(10)(5) = M2 (10) M2 = 5 µg/µL

d) 10 µL of 1:60 Ab + 90 µL of buffer to make a 1:600 dilution at 1 µg/µL. M1 = 1:60 x 600 µg/µL

V1 = 10 µL

M2 = ?

V2 = (90 µL + 10 µL) = 100 µL

M1 =

1 60

x 600 µg/µL

= 10 µg/µL

M1V1 = M2V2 (10)(10) = M2 (100) M2 = 1 µg/µL

dilution factor =

1 600

= 1:600

= 1:120

e) 10 µL of 1:60 Ab + 40 µL of buffer to make a 1:300 dilution at 2 µg/µL. M1 = 1:60 x 600 µg/µL

V1 = 10 µL

M2 = ?

V2 = (10 µL + 40 µL) = 50 µL

1 60

M1 =

x 600 µg/µL

= 10 µg/µL

M1V1 = M2V2

dilution factor =

2 600

=

1 300

= 1:300

1 120

= 1:120

(10)(10) = M2 (50) M2 = 2 µg/µL

f) 10 µL of 1:60 Ab + 10 µL of buffer to make a 1:120 dilution at 5 µg/µL. M1 = 1:60 x 600 µg/µL

V1 = 10 µL

M2 = ?

V2 = (10 µL + 10 µL) = 20 µL

M1 =

1 60

x 600 µg/µL

= 10 µg/µL

M1V1 = M2V2

dilution factor =

5 600

=

(10)(10) = M2 (20) M2 = 5 µg/µL

iv. How much 2.0 M NaCl solution would you need to make 250 mL of 0.15 M NaCl solution? M1 = 0.15 M M2 = 2.0 M

V1 = 250 mL V2 = ?

M1V1 = M2V2 (0.15)(250) = (2.0) V2 V2 = 18.75 mL ≈ 19 Ml

v. What would be the concentration of a solution made by diluting 450 mL of 4.2 M KOH to 250 mL? M1 = 4.2 M

V1 = 450 mL

M2 = ?

V2 = 250 mL

M1V1 = M2V2 (4.2)(450) = M2(250) M2 = 0.756 M

vi. What would be the concentration of a solution made by adding 250 mL of water to 45.0 mL of 4.2 M KOH? M1 = 4.2 M

V1 = 45.0 mL

M2 = ?

V2 = (250 + 45.0) mL = 295 mL

M1V1 = M2V2 (4.2)(45.0) = M2 (295) M2 = 0.641 M

vii. How much 0.20 M glucose solution can be made from 50.0 mL of 0.50 M glucose solution? M1 = 0.50 M

V1 = 50.0 mL

M2 = 0.20 M

V2 = ?

M1V1 = M2V2 (0.50)(50.0) = (0.20)V2 V2 = 125 mL

viii. What is the molarity of a solution that has 4.5 mol of solute dissolved in 300 mL of solution? Molarity, M = =

mol volume in L 4.5 mol 0.3 L

= 15 M

ix. What is the molarity of a solution of NaOH that has 0.941 g dissolved in 400 mL of solution? Moles of NaOH =

0.941 g 39.997 g/mol

= 0.01228 mol Molarity, M =

0.01228 mol 0.4 L

= 0.0307 M

x. What is the molarity of a solution prepared by diluting 10.00 mL of a 4.281 M solutions to 50.00 mL? M1 = 4.281 M

V1 = 10.00 mL

M2 = ?

V2 = 50.00 mL

M1V1 = M2V2 (4.281)(10) = M2 (50) M2 = 0.8562 M

EXPERIMENT PROTEIN DETERMINATION

2:

1. Name the dye

used in Bradford Assay.

The dye used in Bradford Assay is Coomassie Brilliant Blue.

2. State the colour change that occurs when proteins combine with the dye reagent. The colour change that occurs when proteins combine with the dye reagent results in a change from brown to blue.

3. In Lowry protein assay, name the bond in protein that binds to copper ions. The bond in protein that binds to copper ions is peptide bond.

4. A Bradford assay was conducted to determine the total protein concentration in a sample. A volume of 2 μL of the original sample was diluted to 100 μL with buffer before performing the assay. The diluted sample gave an absorbance at 595 nm of 0.255. Using the data for the standards below,

Protein concentration (μg/mL)

A595

25

0.008

125

0.087

250

0.113

500

0.197

750

0.295

1000

0.429

1500

0.608

a) Plot a standard curve based on the data given. b) Determine the concentration of protein in the diluted sample. c) Calculate the total protein concentration in the original sample.

a). Standard curve graph

Absorbance vs. Protein Concentration 0.7

Absorbance at 595 nm, A

0.6

f(x) = 0 x + 0.01 R² = 0.99

0.5 0.4 0.3 0.2 0.1 0 0

200

400

600

800

1000

Protein concentration (μg/mL)

b). Protein concentration in diluted sample y = mx + c y = 0.0004x + 0.0117 y = absorbance x = protein concentration

y = 0.0004x + 0.0117 0.255 = 0.0004x + 0.0117 protein concentration in diluted sample, x = 608.25 μg/mL

c). Total protein concentration in original sample 2 μL = 0.002 mL

100 μL = 0.1 mL

M1 = protein concentration in original sample V1 = 0.002 mL M2 = 608.25 μg/mL V2 = 0.1 mL M1V1 = M2V2 M1 (0.002) = (608.25) (0.1) M1 = 30412.5 μg/mL

1200

1400

1600

1.

EXPERIMENT 3: DETERMINATION OF ENZYMES ACTIVITY

Substitute the solution in the video into the

equation below. Protease + Casein

Protease-casein

Tyrosine

2. State the function of the Folin & Ciocalteau’s in the video. It primarily reacts with free tyrosine and then will produce a blue coloured chromophore.

3. In an enzymatic reaction, 1.0 mL of acethylcholinesterase was reacted with 2.0mL of acethylthiocholine iodide. 2.0 mL of phosphate buffer was then been added in. Incubation for 10 minutes took place before 1.0 mL of acetic acid and 1.0 mL of dithiobisnitrobenzoic acid was added up to the mixture. The mixture was further incubated for 10 minutes and 1.0 ml of the sample was placed in cuvette for absorbance reading at 405 nm.

Using the data for the product standards below,

405 nm

Choline (μmol)

0.00

0

0.067

0.05

0.113

0.1

0.197

0.2

0.395

0.4

0.729

0.8

a) Plot a standard curve based on the data given. b) Determine the activity of the enzyme if the absorbance of enzymatic activity gave a reading of 0.552. c) Calculate the specific enzyme activity if the total protein content is 10 mg/mL and the enzyme has been diluted 5X before enzymatic reaction takes place.

a) Standard curve graph

Absorbance (405 nm) vs. Choline (μmol) 0.8 f(x) = 0.9 x + 0.02 R² = 1

Absorbance at 405 nm

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Choline (μmol)

b)

Activity of enzyme y = mx + c

y = absorbance

y = 0.9014x + 0.0173

x = choline

y = 0.9014x + 0.0173 0.552 = 0.9014x + 0.0172 x = 0.6 μmol

Activity of enzyme = =

choline released x total assay time in assay x volume of enzyme x volume in cuvette 0.6 μmol x 7 .0 mL 10 min x 1.0 mL x 1.0 mL

= 0.42

c)

units mL

Specific enzyme activity

units mL mg ÷ 10 mL 0.42

Specific enzyme activity =

= 0.21

units mg

5

0.9

EXPERIMENT 4: GEL ELECTROPHORESIS 1. Define

gel

electrophoresis.

Gel electrophoresis is a method used for separating a mixed population of macromolecules such as DNA in a matrix of agarose. In a simple word, it is a tool used to separate nucleic acids.

2. Describe how 1% agarose gel was prepared. 1) 1 g of agarose is weighed out into a 250ml Erlenmeyer flask. 2) 100ml of 0.5xTBE was added and swirled to mix them. 3) The solution was carefully boiled about 40s by putting it in the microwave to dissolve the agarose. 4) The solution was left to cool down. 5) A magnetic stirrer was put into the flask and the stir plate was turned on for three minutes for that cool down. 6) About 0.8 µl of ethidium bromide stock (10mg/ml) was added into the flask and the solution was swirled. 7) The gel was slowly poured into the tray with the well comb in place. 8) The comb was removed carefully when the gel has cooled down and solidified.

3. State the number of wells in the agarose gel as shown in the first video. There are three wells in an agarose gel.

4. State the function of ethidium bromide. Ethidium bromide is used to visualize DNA bands in agarose gel electrophoresis experiment.

5. State the role of DNA ladder. DNA ladder consists of a set of DNA fragments of different sizes that is used as reference to estimate the size of the unknown fragment by comparing it to the closest band in the ladder lane. 6. Identify which part of DNA molecule contain negative charge. Phosphate groups in the DNA backbone.

7. Explain the function of restriction enzyme. Restriction enzyme is used to cut DNA specifically into fragment that then can be separated by fragment size on an agarose gel....