Chat with us, powered by LiveChat An Investigation of ABO Group Antigen Structure The ABO blood group antigen epitopes are carbohydrates present on a cell membrane protein. The basic core carbohydrate (known as H) is present in group | Writedemy

An Investigation of ABO Group Antigen Structure The ABO blood group antigen epitopes are carbohydrates present on a cell membrane protein. The basic core carbohydrate (known as H) is present in group

An Investigation of ABO Group Antigen Structure The ABO blood group antigen epitopes are carbohydrates present on a cell membrane protein. The basic core carbohydrate (known as H) is present in group

uestion
Antibody Isolation

Practical write-up report

Student name:

Student number:

Introduction

Introduce what a polyclonal antibody is, and why and how you might want to purify a particular immunoglobulin isotype away from the remaining serum components

(maximum 300 words)

Methods

Present an extremely brief overview of the methods used

(maximum 200 words)

Results

There is no maximum word count for any of the results section 1-5.

Section 1:

Present the results of your ouchterlony and describe which column fractions were pooled to form fraction 5 (F5)

Section 2:

Calculate total protein of all your fractions F1-5. Present all values as:

A. protein concentration in mg/ml

B. Total protein in your sample having corrected your value for any dilution factor and the volume of the sample

A calibration curve must be completed and attached to this report. Sample values collected from the graph must be clearly labeled

Section 3:

Calculate total IgG contained within all of your fractions F1-5. Present all values as:

A. IgG concentration in mg/ml of each sample

B. Total IgG in each sample having corrected your value for any dilution factor and the volume of the sample

A calibration curve must be completed and attached to this report. Sample values collected from the graph must be clearly labeled

Section 4:

Calculate percentage yield of IgG from the starting sample

((total IgG/ total protein) x 100)

Section 5:

Present the results of your immune-electrophoresis clearly labeling where your samples were loaded onto the gel and which antiserum was placed into which trough. Indicate where you observed lines of precipitation.

Discussion:

Discuss the results from section 1-5 in terms of quantity yield and quality of your samples. Critically review the relative advantages and disadvantages of both purification methods

(maximum 300 words)

Results

Results for BSA (first week)

Concentration of BSA (mg/ml)

Absorbance at 280nm

1.0

0.646

0.5

0.328

Results for Fractions( 1-4)

Fractions

Absorbance at 280nm

F1

0.621

F2

0.504

F3

0.242

F4

0.440

The fractions F1-F4 were diluted at 1/50

Calibration curve were plotted from the BSA absorbance and concentration in order to find the concentration of these fractions 1-4.

BSA and fractions 1-4 calibration curve

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Fraction

Concentration mg/ml

F1

0.96

F2

0.78

F3

0.38

F4

0.68

Results for BSA (second week)

Concentration of BSA (mg/ml)

Absorbance @ 280nm

1

0.628

0.5

0.311

Fractions

Absorbance at 280nm

F5

0.340

F5 diluted at 1/50

Calibration curve were plotted from the BSA absorbance and fraction 5 absorbance from the second weekin order to find the concentration of fraction 5.

BSA and fractions 1-4 calibration curve

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Fractions

Concentration mg/ml

F5

0.55

Section 1:

Present the results of your ouchterlony and describe which column fractions were pooled to form fraction 5 (F5)

Ouchterlony

Precipitation line was relatively visible for samples 2, 3 and 5. These samples were pooled to make fraction 5.

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6

2

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3

5

4

Section 2:

Calculate total protein of all your fractions F1-5. Present all values as:

A. protein concentration in mg/ml

B. Total protein in your sample having corrected your value for any dilution factor and the volume of the sample

A calibration curve must be completed and attached to this report. Sample values collected from the graph must be clearly labeled

Total Protein = Concentration (mg/ml) x dilutions x volume

F1: 0.96 x 50 x 0.5 = 24

F2: 0.78 x 50 x 0.5 = 19.5

F3: 0.38 x 50 x 0.5 = 9.5

F4: 0.68 x 50 x 1= 34

F5: 0.55 x 50 x 3 = 82.5

Fractions

Concentration (mg/ml)

Dilution

Volume (ml)

Total concentration (mg)

F1

0.96

50

0.5

24

F2

0.78

50

0.5

19.5

F3

0.38

50

0.5

9.5

F4

0.68

50

1.0

34

F5

0.55

50

3.0

82.5

Section 3:

Calculate total IgG contained within all of your fractions F1-5. Present all values as:

A. IgG concentration in mg/ml of each sample

B. Total IgG in each sample having corrected your value for any dilution factor and the volume of the sample

A calibration curve must be completed and attached to this report. Sample values collected from the graph must be clearly labeled

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1 0.5 0.2 0.1

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At dilution 1/20, fractions F1, F2, F3, F5 circle formed. Fraction F4 didn’t form any circle.

Standards

Diameter (mm)

Diameter(mm2)

1

13.32

177.4

0.5

11.21

125.7

0.2

8.90

79.21

0.1

7.11

50.55

Fractions

Diameter (mm)

Diameter (mm2)

F1

8.5

72.25

F2

8.8

77.44

F3

7.1

50.41

F4

0.0

0.00

F5

7.5

56.25

The diameter (mm2) for the IgG standard were plotted on a graph and the diameter (mm2) for each of the fractions that produced precipitation circles were plotted against the line of best fit to determine the IgG concentration in each fraction.

Single radial immunodiffusion of IgG

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Calculate total IgG contained within all of your fractions F1-5. Present all values as:

Following calculation was used to calculate the total IgG in each fraction

Total IgG concentration = concentration x dilution x volume

F1: 0.20 x 20 x 0.5 = 2

F2: 0.23 x 20 x 0.5 =2.3

F3: 0.04 x 20 x 0.5 = 0.4

F4: didn’t form any circle.

F5: 0.08 x 20x 3 = 4.8

Concentration of IgG example of F1: look up diameter of circle which for F1 is 72.25 and look across line of best fit and go down which reads 0.20.

Fractions

Concentration of IgG

Dilution

Volume (ml)

Total IgG (mg)

F1

0.20

20

0.5

2.0

F2

0.23

20

0.5

2.3

F3

0.04

20

0.5

0.4

F4

0.00

0

0.0

0.0

F5

0.08

20

3.0

4.8

Section 4:

Calculate percentage yield of IgG from the starting sample

((total IgG/ total protein) x 100)

F1 1:20 (2/24) x 100 = 8.3%

F2 1:20 (2.3/19.5) x 100 = 11.8%

F3 1:20 (0.4/9.5) x 100 = 4.21%

F5 1:100 (24/78) x 100 = 6.2%

Section 5:

Present the results of your immune-electrophoresis clearly labeling where your samples were loaded onto the gel and which antiserum was placed into which trough. Indicate where you observed lines of precipitation.

Plate 1

Fraction three with a-IgG produced a single precipitation line, indicating no other proteins present and pure IgG is present. Fraction 1 with a-IgG produced multiple precipitation lines indicating fraction 1 contained other proteins. Therefore fraction 1 was not pure IgG. Moreover, fraction 2 with a-whs also produced multiple precipitation lines. lines indicating fraction 1 contained other proteins. Therefore fraction 1 was not pure IgG

Plate 2

Fraction 3 with a-whs showed three precipitation lines so it’s not pure IgG. Fraction 4 showed no precipitin lines. Fraction 5 with a-whs showed many precipitation lines indicating clearly other proteins were present therefore, fraction 5 was relatively contaminated.Antibody Isolation 1

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