roblem #3 (25 points). You run a biochemical analysis lab you have isolated a nu
ID: 91263 • Letter: R
Question
roblem #3 (25 points). You run a biochemical analysis lab you have isolated a nucleotide piece of rom an E. coli sample you were provided. You have determined that 76 percent this RNA is part of a coding of region, while the remaining percentage is the very end of a gene product and hence is non-coding An HPLC analysis of an the RNase-treated aliquot of the mRNA reveals that it contains about equal fractions G,C,A, and U. What is your best guess for molecular mass of this mRNA? Provide a reasonable extinction coefficient for this RNA for UV absorption at 260 nm and use it to calculate the micromolar concentration of your original sample if a 1/1000 dilution gives an OD,60 of 0.0126 in an Eppendorf Biophonomer with a 1 cm cuvette. Without any further analysis, how much information content (in bits) exists in the nucleotide sequence of the coding region? Similarly, how much information content (in bits) exists in the amino acid sequence generated from the coding region? Describe in as much detail as you can the methods you would use to determine the exact nucleotide sequence of this mRNA based on procedures described in class.Explanation / Answer
a) Molar mass of single stranded mRNA = (Number of nucleotides x 320.5) + 159
320.5 is the average molecular weight of four nucleotides
159 is the molecular weight of 5' triphosphate.
Approximate molar mass of single stranded mRNA = (75 x 320.5) + 159 = 24196.5 g/mol
b)
Standard Coefficient = extinction coefficient for ss RNA = 0.025 (ug/ml) cm-1
Nucleic Acid Concentration = (OD260 / Pathlength) x Standard Coefficient x Sample Dilution
= (0.0126 / 1) x 0.025 x 1/1000 = 0.000000315 = 3.15 x 10 -7 ug/ml
c) Information content : mRNA
Total = 75 nucloetides
76 % coding region
100 % = 75
76% = 57
Number of coding nucleotides = 57 bits
d) Information content : Amino acid
Three nucleotide = 1 codon
57/3 = 19 bits
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