C2005/F2401 '06 -- Answers to Recitation Problems #6

1. A. A = T; B. AA = TT. These are the proportions in the original double-stranded molecules and they will be the proportions found in the hydrolysate. AT does not equal TA because the strands are antiparallel.

Detailed Explanation: The conditions of hydrolysis are sufficiently harsh to break all weak bonds (although the weak bonds are not "hydrolyzed" -- they are not broken by addition of water across the bond). Therefore you have no base pairs, only single nucleotides and dinucleotides connected by covalent bonds. The DNA was broken at random spots, so you can't say where any individual molecule was broken, but overall, in the total hydrolysate, the proportions of bases and dinucleotides will be the same as the proportions in the original molecules. (See * below.) In the original double stranded molecules, A = T and AA = TT but AT is not = TA. Why? For every sequence of AA in one strand there is a TT in the other. For every 5'AT3' in one strand there is a 3'TA5' (or AT) in the other. So in the original molecule, AA == TT but AT = AT not TA (if everything is written 5' to 3').

* Note: You have many, many, molecules. Some are cut one way, some another. On average, the more times TT, for example, appears in the original sequence, the more often TT will turn up in the collection of dinucleotides, even though not every TT is preserved in the dinucleotides.

If the strands of DNA were parallel, then you would find AT = TA. An experiment like this was actually done to confirm that the strands were antiparallel. (This was back in the ancient days, in the 60's, when there was no way to sequence DNA.)

2. A. G is a purine and a base.

B. G* DNA is NOT degraded to bases. The sugar (deoxyribose) must remain attached to the G* if the radioactivity can be recycled to DNA but not RNA. If the sugar were removed, and the DNA was degraded to bases, the *G could be re-incorporated into both DNA and RNA. Some of the G* could be attached to deoxyribose and incorporated into DNA; some could be attached to ribose and incorporated into RNA. But this does not happen – radioactivity from G*-DNA breakdown does not end up in RNA. So DNA must be broken down to nucleosides or nucleoside monophosphates. Then the G*-deoxyribose or G*-deoxyribose-phosphate is incorporated only into DNA.

C. Phosphate/G should be = 0; Moles of phosphate released should be 2. When the G*-DNA is broken down to mononucleotides, there is one phosphate per G*. The phosphate is not radioactive (but the G* is). Two phosphates must be added from ATP to convert the dG*MP to dG*TP, before the G* can be incorporated into DNA. The two added phosphates may be radioactive, but they are released when the dG* (really d*GMP) is added to the growing DNA chain. The only phosphate incorporated into the DNA is NOT radioactive.

D. No colonies. Under the conditions described here, the DNA that is taken up by the recipient is degraded to nucleotides, so it cannot transfer any genetic information. So the recipients will all be unable to degrade compound X and will not form colonies.

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