You run a biochemical analysis lab and you have isolated a 75-nucleotide piece o

Question

You run a biochemical analysis lab and you have isolated a 75-nucleotide piece of mRNA from an E. coli sample you were provided. You have determined that 76 percent of this RNA is part of a coding region, white the remaining percentage is the very 5' end of a gene product and hence is non-coding. An HPLC analysis of an RNase-treated aliquot of the mRNA reveals that it contains about equal fractions G, C, A, and U. What is your best guess for the 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_260 of 0.0126 in an Eppendorf Bioptome with a 1 cm cuvette. Without any further analysis, how much information content (in bits) exists in the nucleotide sequence of the coadding region? Similarly, how much information content (in bits) exists in the amino acid sequence generated from the coadding 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

As said is question, there are 76% coding region and remaining noncoding region. This means 26% part shall be rich in adenine. Therefore, calculation should be done with approximation only.

Let’s assume

75 nucleotides are equally shared. Therefore, 75/4, and assume 18 numbers of each nucleotide and gives 3 additional number to adenine.

Thus, total molecular weight of chain should be equal to 21A+18U+18+G+18U

=21*347.2+18*323.2+18*363.2+18*324.2

=25482

Absorbance at given wavelength can be correlated as follows:

A = c L

Where, c= molar concentration; L=length of path travelled by light in cm; = molar absorptivity

Usually, A260 for the ssRNA = 40 µg mL-1

Nucleic Acid Concentration          = (OD260/Path length) Standard Coefficient*Sample Dilution

                                                                = (0.0126/1)*40*1/1000

                                                                =0.000504 µg/µl

A = c L; or                                        = A/cL

                                                                =0.0126/0.000504*1

Thus, the reasonable extinction coefficient will be =25

Since, the no of coding nucleotides is 75 so, the amino acid will be 24 amino acid in length by excluding one termination codon.

For the sequencing of RNA direct sequencing of RNA can be done by application of reverse transcription. this is followed by the manipulation and ligation.

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