Assignment 3 PDF

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Summary

Assignment 3 for BIOC2000...

Description

Course Code

BIOC2000

Course Title

Biochemistry and Molecular Biology

Course Coordinator

Associate Professor Susan Rowland

Due Date

28/05/18

Assignment Title

Report 3: Practical Report

Word Count

1309

Date Submitted

28/05/2018

Extension applied for Yes

/ No

Revised Date

Student Number

Surname

First Name

4484268

Andrews

Myles

extracts were given accurately by the enzyme assay

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Section 3:

a. 1 Marker

2 SDS1

3 SDS2

4 SDS3

5 SDS4

6 SDS5

7 Marker

8 SDS1

9 SDS2

10 SDS3

11 SDS4

12 SDS5

b. Marker: Should see multiple bands (ladder-like) as it marks the levels as a comparison SDS1: Should see multiple bands as there are many proteins (as it is induced cell lysate) SDS2: Should see most the same as SDS1 however more dilute (as it is only the flow through of 1), and without anything at approximately 50 dKa (shouldn’t see GST-GFP) SDS3: One band should be seen (the GST-GFP) at around 50kDa SDS4: One band should be seen (purified GFP) at around 25kDa SDS5: Should see multiple bands (same as SDS1), however again less obvious, due to naturally degraded cell membrane (non induced cell lysate)

c. Refer to photo d. Refer to Photo: GFP-GST - * GFP - ^ e. Protein Name GST-GFPS65T GFPS65T

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kDa 52.4 26.9

Molecular Weight 52400 g/mol 26900 g/mol

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GST

25.5

25500 g/mol

Section 4: a. Lane 11: 75% Purity of GFP b. Based on lane 11, there are 4 very faint lines near the top of the gel at about ¼ of the intensity of the GFP band, meaning it is not 100% pure, however close to it as the other bands are very faint and minimal. Section 5: a. b. c. d. e.

Coefficient = 22015 (L/mol*cm) Didn’t have any dilutions (already too weak) ^^^ A=0.134

C= C=

A εL 0.134 −1

−1

22015 M ∙c m × 1 cm C=6.087 ×10−6 M =6.087 × 10−6 mol ∙ L−1 GFP =26.9 kDa=26900 Da=26900 g ∙ mo l−1 C=6.087 ×10−6 mol ∙ L−1 ×26900 g ∙ L−1 C=0.1637 mg/ml f. Elute = 2.4mL g. 2.4 mL × 0.1637 mg∙ mL−1=0.3929 mg of GFPS65T in 2.4mL of eluate

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Section 6: a. 2.5

2

A562

1.5 Replicate 1 Replicate 2 Replicate 3

1

0.5

0

0

500

1000

1500

2000

2500

BSA Concentration (ug/mL)

Figure 1 - BCA assay standard curve showing all three replicates of A562 vs BCA concentration (ug/mL)

b. 2.5

f(x) = 0 x + 0.05 R² = 1

2

A562

1.5

1

0.5

0

0

500

1000

1500

2000

2500

BSA Concentration (ug/mL)

Figure 2 - BCA assay standard curve showing average A562 of 3 replicates vs BSA concentration (ug/mL) with linear trendline

Equation – y = 0.001x + 0.0488

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Section 7: 1 Dilutio n of Sample 1/1 1/1 1/1 1/3 1/3 1/3 1/5 1/5 1/5

2 3 4 5 6 A562 for [GFPS65T] in Actual [GFPS65T] in Vol. of the Total mg of GFPS65T each diluted sample undiluted sample GFPS65T in entire sample sample (ug/mL) (mg/mL) eluate (ml) volume 0.299 250.2 250.2 * 10-3 3 0.7506 0.287 238.2 238.2 * 10-3 3 0.7146 0.3 251.2 251.2 * 10-3 3 0.7536 0.115 66.2 198.6 * 10-3 3 0.5958 0.112 63.2 189.6 * 10-3 3 0.5688 -3 0.113 64.2 192.6 * 10 3 0.5778 0.075 26.2 131 * 10-3 3 0.393 -3 0.076 27.2 136 * 10 3 0.408 0.081 32.2 161 * 10-3 3 0.483 Amount of GFPS65T in the sample (mg) (averaged from samples) 0.5828

Section 8: The calculated amount of GFPS65T recovered by Alex should be less than the amount of GSTGFPS65T in the initial lysate sample before purification. This is because before purification, the protein sample will be contaminated by other proteins of similar size. The starting protein sample is also very crude. These contaminating proteins will be removed (mostly) during purification, and the pure GFPS65T will be present (at a lower amount than originally, as multiple other proteins were removed during purification). Therefore, less mg of protein = less weight. Another reason is because the of the molar mass of each compound. The MM of GFPS65T alone is only 26.9kDa, however GST-GFPS65T is 52.4kDa. This means that once GST-GFP has been purified into only GFP, the MM of the measured compound is less, meaning the overall weight is less again. One thing to note, however, is that the mols of the protein will stay the same. Moles is the measure of the amount of a substance expressed at its MM. During purification, only the unwanted contaminants are removed, therefore the mols will not change as GFPS65T remains unchanged.

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Section 9: a.

Assay Curves of GST-GFPS65T Amounts

A340

0.6 0.5

f(x) = 0.09 x + 0.05

0.4

f(x) = 0.08 x + 0.02

0.3

0μg(GST-GFP) Linear (0μg(GST-GFP)) 0.5μg(GST-GFP) Linear (0.5μg(GST-GFP)) 1.0μg(GST-GFP) Linear (1.0μg(GST-GFP)) 2.0μg(GST-GFP) Linear (2.0μg(GST-GFP)) 3.0μg(GST-GFP) Linear (3.0μg(GST-GFP)) 4.0μg(GST-GFP) Linear (4.0μg(GST-GFP))

f(x) = 0.06 x + 0.01

0.2 f(x) = 0.03 x + 0.01

0.1

f(x) = 0.02 x + 0.01 f(x) = 0.01 x + 0.02 1 1.5 2 2.5

0 0.5

3

3.5

4

4.5

5

5.5

Time (min)

b. Rate was found using M (y=mx+c) 0ug 0.0055

0.5ug 0.0161

1ug 0.0282

2ug 0.0588

3ug 0.0797

4ug 0.0935

c. 0.1 f(x) = 0.02 x + 0.01

0.09 0.08

A340/min

0.07 0.06 0.05 0.04 0.03 0.02 0.01 0

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

GST-GFP (ug)

Figure 3 - standard curve activity of GST-GFPS65T for CDNB Assay. Dotted line shows linear equation

d. Refer to above figure

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Section 10: a. . 0.6

0.5

A340

0.4

0.3

A340 for 1/8 Dilution of Lysate A340 for 1/4 Dilution of Lysate A340 for 1/2 Dilution of Lysate

0.2

0.1

0

0

1

2

3

4

5

6

Time (min)

Figure 4 - A340 plotted against time for each dilution

b. . 0.4 0.35 0.3 A340 for 1/8 Dilution of Lysate Linear (A340 for 1/8 Dilution of Lysate) A340 for 1/4 Dilution of Lysate Linear (A340 for 1/4 Dilution of Lysate) A340 for 1/2 Dilution of Lysate Linear (A340 for 1/2 Dilution of Lysate)

A340

0.25 0.2 0.15 f(x) = − 0.03 x + 0.14 0.1 0.05

f(x) = − 0.02 x + 0.08 f(x) = − 0.01 x + 0.04

0

0

1

2

3

4

5

6

Time (min)

Figure 5 - A340 plotted against time for each dilution, removing all except two linear data points to gather slope for rate of reaction

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c. Refer to Figure 5 for computerized calculation, and below for manual (between 0 minutes and 1 for vmax) Rate = Rise/Run 1/8 Dilution (Using minutes 0 and 5) Rate = 0.11-0/1-0 = 0.11 A340/minute 1/4 Dilution (Using minutes 4 and 5) Rate = 0.2-0/1-0 = 0.2 A340/minute 1/2 Dilution (Using minutes 2 and 5) Rate = 0.38-0/1-0 = 0.38 A340/minute

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Section 11: a. 9c equation: y = 0.0231x + 0.0066 y = A340/minute, x=GST-GFP(μg) 1/8 Dilution (Using minutes 4 and 5) 0.11 A340/minute = 0.0231x + 0.0066 X = 4.48 μg SAMPLE CALCULATION Rate(ΔA340/min) = 0.0231 (minutes) x amount of GST-GFP(μg) + 0.0066(ΔA340) 0.11(ΔA340/min) = 0.0231 (minutes) x amount of GST-GFP(μg) + 0.0066(ΔA340) (0.11(ΔA340/min)-0.0066(ΔA340))/0.0231(min) = amount of GST-GFP (μg) 4.48 = amount of GST-GFP(μg)

1/4 Dilution (Using minutes 4 and 5) 0.2 A340/minute = 0.0231x + 0.0066 X = 8.37 μg 1/2 Dilution (Using minutes 4 and 5) 0.38 A340/minute = 0.0231x + 0.0066 X = 16.16 μg b. . Concentration of Amount of GSTGFP (μg) in 40 μl GST-GFP in of diluted lysate undiluted lysate (μg/ml) from standard curve 1/8 5 Minutes 0.11 A340/minute 4.48 896 1/4 1 Minute 0.07 A340/minute 8.37 837 1/2 3 Minutes 0.04 A340/minute 16.16 808 Average GST-GFP concentration in lysate from samples giving linear rate = 847 Standard Deviation = 36.615 c. Refer to table above d. Assuming Alex had 5ml of lysate at the beginning, approximately 4.235 mg of GST-GFP will be used (based on samples giving linear rate). Dilution

Linear for how long?

Rate

WORKING: Average GST-GFP concentration = 847 μg/ml 5mL of lysate GST-GFP = 847 μg/ml x 5mL = 4235 μg = 4.235mg

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Section 12: a. The rate of product output from the dilution became non-linear very quickly. This is because originally there were many proteins that needed binding. As time goes on, the rate of reaction decreases and is less linear as there are less proteins that need binding (as many became bound quickly in the beginning). b. SDS-PAGE provides an estimate of the concentration of protein in the sample to be calculated, while the assay is advantageous, allowing for an accurate concentration to be calculated (as well as targeting only one protein, as in a soup there are protein bands everywhere) c. GST tagged proteins can be separated and purified from other cellular proteins in the resin, while GFPS65T cannot. Section 13: a. GST-GFP 0.004235 g 52.4 molar mass

GFP 0.0005828 g 26.9 molar mass

Mols = mass/mm GST-GFP Mols = 0.004235g / 52400g/mol = 8.08 x10-8 Mols GFP Mols = 0.0005828g / 26900g/mol = 2.17 x10-8 Mols % Yield = GFP Mols / GST-GFP Mols x 100 = (2.17x10-8 mols/8.08 x10-8 mols) x100 = 26.8% Yield

b. Yes. The 26.8% yield is quite reasonable, however could be better. This is as even though 26.8% is not a lot, it is still a considerable amount and could be much worse (still over a quarter).

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