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, while 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 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

Answer 3(a) molecular weight of 75 nucleotide long mRNA can be calculated by the formula

M.W = number of AMPx mol wt. of AMP + number of GMPx mol. wt. of GMP + number of CMP x mol wt of CMP + number of UMP x mol. wt. of UMP + mol wt of 5' triphosphate(which is 159)

As given that each nucleotide is in equal proportion means 18.75 because total nucleotides were 75 that can be considered as 19.

= 19x 329.2 + 19 x 306.2 + 19 x 305.2 + 19 x 345.2 + 159

= 24589.2 D or 24.5 KD

3(b) As we know that 1 unit of absorbance corresponds to 40 microgm/ml of RNA.

therefore absorbance of 0.0126 will be = .0126 x 40 microgm/ml = 0.504 microgm/ml

RNA conc. is 0.504 microgm/m which cna be converted in to molarity by using mol. wt that we calculated in first part

M= Conc. / mol. wt

= 0.020 micromoles

extinction coeficient cand be calculated by the formula

A = ECL

E = A/CL

= 0.0126/0.504 microgm/ml x 1 cm

= 2.5 x 104

3 (C) and 3 (D) is not clear what do you want to know

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.