Question

Kornberg and his colleagues incubated soluble extracts of E.coli. with a mixture of dATP, dTTP, dGTP, and dCTP, all labeled with 32P in the α phosphate group. After a time, the incubation mixture was treated with trichloroacetic acid which precipitates the DNA but not the nucleotide precursors. The precipitate was collected and the extent of precursor incorporation into DNA was determined from the amount of radioactivity present in the precipitate.

a. If any one of the 4 nucleotide precursors were omitted from the incubation mixture, would radioactivity be found in the precipitate? Explain

b. Would 32P be incorporated into the DNA if only dTTP were labeled? Explain.

c. Would radioactivity be found in the precipitate if 32P labeled the βor Υ phosphate rather than the α phosphate of the deoxyribonucleotides? Explain.

Answer 1

a) For synthesis of a new DNA strand, all the four dNTPs are required. Hence, if one of the four nucleotides is removed from the incubation mixture, there would be no radioactivity present in the precipitate.

b) During synthesis of new strands of DNA, all the four nucleotides must be present in the form of dNTPs. Even if one of the nucleotide is tagged with fluorescence, there will be incorporation of 32P. So, yes, there will be radioactivity even if only the dTTP was labelled.

c) The synthesis of a phosphodiester bond occurs with the α-phosphate of the triphosphate. Hence , the radioactivity is seen only if the label is in the α-phosphate . The β and Υ phosphate is cleaved off as pyrophosphate(PPi) and hence tagging or labelling them with 32P will not incorporate the radioactivity in the synthesized DNA.