Biochemistry Lab Report

BiochemistryLab Report

Purpose

Toidentify unknown amino acids present in sample A, and B using theknown standards.

Reagents

Thereagents provided were as follows the standards concentrations ofleucine (Leu), proline (Pro), glutamic acid (Glu), cysteine (Cys)tyrosine (Tyr), arginine (Arg), glycine (Gly), alanine (Ala) andaspartic acid (Asp). Hydrolyzed casein sample was among the providedreagents. Beside two unknown samples – sample A and B – wereprovided. Then, a straight line was drawn about 2 cm from the oneedge of the chromatography paper. On the line, 1 cm spaced marks weremade. They were then labeled in agreement with all provided samplesto be tested (spots for nine amino acids standards, two unknownsamples and hydrolyzed casein sample) On the baseline, a singlespot of each standard amino acids and that of the unknown sample Aand B were made. Similarly, a single spot of the hydrolyzed caseinsample was placed on the designated respective spots on thechromatography paper.

Discussion

Responseto Question 5 (a) to (g)

StandardRelative to Front (Rf) Values

Distancemoved by the front = 12.6cm

Distancesmoved by the standards Leu 11.5cm Pro 8.9cm Glu 8.5cm Cys 3.7Tyr 9.2cm Arg 6.6cm Gly 7.5cm Ala 8.9cm and Asp 7.5cm whereasCasein constituents were not clearly defined.

Rf=

LeuRf value= = 0.9127

Figure1:Leucine Structure

ProRf value= = 0.7063

Figure 2: ProlineStructure

GluRf value= = 0.6746

Figure 3: GlutamicAcid Structure

CysRf value= = 0.2936

Figure4: Cysteine Structure

TyrRf value= = 0.7302

Figure 5: TyrosineStructure

ArgRf value= = 0.5238

Figure 6: ArginineStructure

GlyRf value= = 0.5952

Figure 7:Glycine Structure

AlaRf value= = 0.7063

Figure 8:Alanine Structure

AspRf value= = 0.5952

Figure 9: AsparticAcid Structure

RfsValues for Unknown Components

UnknownSample A

ComponentA1 distance moved by the front = 9.5 cm.

A1Rf value= = 0.7540

ElementA1 (R= 0.7540) was possibly tyrosine (Rf = 0.7302).

ComponentA2: distance moved by the front = 7.4 cm.

A2Rf value= = 0.5873

ElementA2 (Rf= 0.5873) were/was possibly aspartic acid (Rf = 0.5952) andGlycine (Rf= 0.5952).

Therefore,unknown amino acid residues in sample A were:

  • Tyrosine

  • Aspartic acid and

  • Glycine

UnknownSample B

ComponentB1: distance moved by the front = 11.6cm.

B1Rf value= = 0.9206

ElementBI (Rf= 0.9206) was possibly leucine (Rf = 0.9127).

ComponentB2: distance moved by the front = 9.3cm.

B2Rf value= = 0.7381

ElementB2 (Rf= 0.7381) was possibly tyrosine (Rf = 0.7302). Otherpredictable candidates include proline (Rf= 0.7063) or alanine (Rf=0.7063).

ComponentB3: distance moved by the front = 8.3cm.

B3Rf value= = 0.6587

ElementB3 (Rf= 0.6587) was possibly glutamic acid (Rf = 0.6746).

Therefore,unknown amino acid residues in sample B were:

  • Leucine

  • Tyrosine

  • Proline or alanine and

  • Glutamic acid.

Responseto Question 6

Answerto 6(d)

Challengesand limitations

  • The spots did not move perpendicularly to the baseline.

  • The spots were too large thus making it difficult to determine the migration distance from the baseline.

  • Some spots failed to separate clearly especially that of sample B, casein and cysteine.

Answerto 6 (b)

  • The baseline should be set perpendicularly to the solvent surface line, perhaps on a flat table.

  • The amount of the sample or standard placed at the baseline should be reduced.

  • Small sample or standard and taller chromatography paper may aid better separation of spots.

  • The hydrolyzed casein samples should be diluted further.

Answerto 6 (c)

Gaschromatography is an excellent method for fractional identificationseparation of amino acids. However, mass spectrometry will berequired to help identify gas chromatography eluted peaks. Therefore,gas chromatography – mass spectrometry (GC-MS) will be therecommended method. The polar properties of amino acids, primarilydue to carboxylic acid and amino groups, reduce their gas pressures.Consequently, the amino acids should be treated to reduce theirpolarity before passage through a GS column (Poole, 2012).

Answerto 6 (d)

Val-Gly-Ser-Ala.

Answerto 6 (e)

V-G-S-A.

Answerto 6 (f)

2-L-valine-glycine-2-serine-3-L-alanine

Answerto 6 (g)

CH(CH3)2CHNH3+CONHCH2CONHCH(CH2OH)CONHCH(CH3)COOH

Answerto 6 (h)

CH(CH3)2CHNH3+CONHCH2CONHCH(CH2OH)CONHCH(CH3)CO2

Answerto 6 (i)

CH(CH3)2CHNH2CONHCH2CONHCH(CH2OH)CONHCH(CH3)CO2

Answerto 6 (j)

Alanineresidue

Answerto 6 (k)

Valineresidue

Answerto 6 (l)

Lysine(Lys)

Answerto 6 (m)

pHRange pH 10.0 (pKa2)to pH 10.5 (pKa3)([email protected],n.d.).

References

[email protected].(n.d). Titration of Lysine with Hydroxide. Retrieved on 17 September2016 from,http://faculty.une.edu/com/courses/bionut/distbio/obj-512/Chap6-titration-lysine.html

Poole,C. (2012). GasChromatography. NewYork: Elsevier.