Lab Report number one PNP assay, easy PDF

Preview

Summary

lab report number one for biochemistry with tables and introduction. Well explained with a 23/25 grade hope u...

Description

Gueye Cheikh

BioChem Lab

Lab Report 1 Introduction This lab puts in theory and pratique how a substance reacts to light at different wavelengths, with the likes of absorbing, reflecting, or transmitting, causing the sight of colors using a spectrophotometer. A spectrophotometer detects the presence of light absorbing particles that are dissolved in a solution and measures the concentration in them. Photons get absorbed when they have the same amount of energy and wavelength that matches the electron orbital energy leading to the electron getting excited causing the absorbtion. Absorbtion can be defined as the capacity a substance has to absorb light depending on the wavelength. It is a little bit similar with the Extinction coefficient up to the point where light is absorbed at a particular wavelength and a very specific concentration. With the use of the BeerLambert Law, one can see the relation between absorbance and concentration as well as wavelength. As an equation the Beer-Lambert Law is A= Ecl with A as absorbance, E as Extinction coefficient, c as concentration of the chemical and l as the length of the light through the chemical. A graph will be plotted from the information given as means to deriving a standard curve, a graph on a sample that is known which will be used to determine the samples of the unknown through a best line fit, to prove the pratique and theory of the relationship between absorbance and a chemical’s concentration with para-nitrophenol also known as PNP used as the substance. PNP is an organic molecule that has different functional groups like a hydroxyl group, a benzene ring and a nitro group. It has the prefix para because the hydroxyl group and nitro group are in adjacent position on the benzene ring; it also has a pka of 7.08 pH. When we have a pka lower than 7.08 pH, the protonation probability is really high causing the PNP to be colorless which means that at pH 4, there is no absorbtion. At pH 10, PNP is deprotonated and is able to absorb longer wavelengths due to the electron excitations. We have 4 aims in this lab which are: drawing a PNP standard curve, finding the unknown concentration using the best line fit of the standard curve, making an absorbtion spectrum using the Beer-Lambert Law and finding the extinction coefficient. Experimental procedure As Razib highlighted on the video, a 96 well plate will be used to hold our sample labeled A1 to A10 and B1 to B10. The B wells will have the same thing as the A wells. A1 is our control. All the wells that have sodium carbonate will remain 150 ul and starting from wells A2 to A6, the Buffer B and PNP will be added in different amounts. We have an unknown from A7 to A10 that needs to be calculated with the help of our standard curve. After loading the samples, the 96 well plate will be placed in Epoch 2 from Biotech from which we will derive the absorbance result which we will use to get the standard curve. We will follow that with PNP absorbtion spectrum by adding 150ul of buffer B and 150ul of sodium carbonate with a set wavelength. Results of the Experiment

At a pH of 4 which is lower than 7.08, as highlighted above, PNP is clear and there is no absorbtion or very little of it. At a pH of 10, we have a yellowish color and high absorbance. As seen on the worksheet 1 table, the absorbance peak is 400nm. Wells

Volume of 0.12 mole/ml PNP (ls)

Volume of Unknown #____ (ls)

Volume of pH 4 Buffer B (ls) 150l

Volume of 0.1M, pH 11, Na carbonate (ls) 150l

Absorbance (at 405 nm)

Amount of PNP in tube (moles)

A1 (Blank) B1 (Blank) A2

0.0

XXXX

0.063

0

0.0

XXXX

150l

150l

0.056

0

25l

XXXX

125l

150l

0.218

0.003

B2

25l

XXXX

125l

150l

0.238

0.003

A3

50l

XXXX

100l

150l

0.306

0.006

B3

50l

XXXX

100l

150l

0.283

0.006

A4

75l

XXXX

75l

150l

0.441

0.009

B4

75l

XXXX

75l

150l

0.462

0.009

A5

100l

XXXX

50l

150l

0.569

0.0012

B5

100l

XXXX

50l

150l

0.575

0.012

A6

125l

XXXX

25l

150l

0.710

0.015

B6

125l

XXXX

25l

150l

0.770

0.015

A7

XXXX

25l

125l

150l

0.236

0.004

B7

XXXX

25l

125l

150l

0.220

0.004

A8

XXXX

50l

100l

150l

0.394

0.008

B8

XXXX

50l

100l

150l

0.392

0.008

A9

XXXX

75l

75l

150l

0.531

0.011

B9

XXXX

75l

75l

150l

0.561

0.011

A10

XXXX

100l

50l

150l

0.687

0.0014

B10

XXXX

100l

50l

150l

0.701

0.014

Calculations using the standard curve’s best line fit For unknowns of A7 and B7 0.236 + 0.220 = 0.456/2 = 0.228 0.228 = 43.7x + 0.063  0.228 – 0.063 =43.7x  x = 0.165 / 43.7 = 0.004 For unknowns of A8 and B8 0.394 + 0.392 = 0.786/2 = 0.393 0.393 = 43.7x + 0.063  0.393 – 0.063 = 43.7x  x = 0.330 / 43.7 = 0.008 For unknowns of A9 and B9 0.531 + 0.561 = 1.092/2 = 0.546 0.546 = 43.7x + 0.063  0.546 – 0.063 = 43.7x  x = 0.483 / 43.7 = 0.011 For unknowns of A10 and B10 0.687 + 0.701 = 1.388/2 = 0.694 0.694 = 43.7x + 0.063  0.694 – 0.063 = 43.7x  x = 0.4631 / 43.7 = 0.014

Worksheet #2: p-Nitrophenol Absorption Spectrum wavelength

Absorbance

wavelength

Absorbance

340 345

0.391 0.415 0.447 0.482 0.520 0.563 0.616 0.671 0.717 0.760 0.795 0.814 0.817 0.798 0.755 0.692 0.620 0.541 0.458 0.375 0.301 0.235 0.184 0.142 0.111 0.091 0.076 0.067 0.062 0.058 0.057 0.056 0.056

510 520

0.056

350 355 360 365 370 375 380 385 390 395 400 405 410 415 420 425 430 435 440 445 450 455 460 465 470 475 480 485 490 495 500

530 540 550 560 570 580 590 600 610 620 630 640 650 660 670 680 690 700

0.054 0.054 0.054 0.053 0.054 0.053 0.053 0.053 0.053 0.053 0.053 0.053 0.052 0.052 0.052 0.052 0.051 0.051 0.051

Calculation of concentration of A and B from 7 to 10 For A7 and B7:(0.004 umole) / (0.025ml )= 0.16 umole / ml For A8 and B8: (0.008 umole) / (0.050ml) = 0.16 umole / ml For A9 and B10: (0.011 umole) / (0.075ml) = 0.15 umole / ml For A10 and B10: (0.014 umole) / (0.100ml) = 0.14 umole / ml Average  [(0.16 umole / ml) + (0.16 umole/ml) + (0.15 umole / ml) + (0.14 umole/ml)] / 4 = (0.1525 umole/ml) x (1000ml / 1L) = 152.5 umol/L

Extinction Coefficient calculation A = Ecl  E=A/cl A=0.5 C = 0.01 umol / 300ml = (3.33 x 10-5 umol/ml) x (1000ml / 1l) = 3.33 x 10-8 mol/L l= 1cm E= 0.5 / ( 3.33 x 10-8 M cm) = 1.5x107/M cm

Discussion When we increased the pH from 4 to 10, PNP turned yellow and became more basic with more deprotonation occurring leading to PNP being able to absorb more wavelengths up to 405nm which is a visible spectrum; that is why we could see the color eventually. According to Beer-Lambert’s Law, absorption of light and concentration are directly proportional. In other words, when plotted in a graph, there should be a straight line with all points fitting on a standard curve. We have a consistent data; however, it may be inaccurate; that is why we had to come up with the best line fit using the average of the A wells and B wells. We used the equation derived from the best line fit to determine the concentration of the PNP unknown; from there, the answers of the unknowns are averaged, and we found 152.5 uM for the PNP unknown proving the consistency of the data. We used the Beer-Lambert’s equation as a mean to find the coefficient of extinction. After a series of calculation with numbers derived from the standard curve and the use of averages from both experiments, we came up with the numbers that replaced A= Ecl with a as the absorbance = 0.5, c= 3.33x10-8 mol/L and l, the pathlength was given as 1cm. Our calculation gave us 1.5x107/M cm as an answer for the extinction coefficient, E....