C2005/F2401 '09 -- Hints & Answers to Recitation Problems #6

 

1. Hints: A & B. In the experiment described here, you have used conditions of hydrolysis (probably acid) that break most, but not all covalent bonds.  Will you break weak bonds?

    B. Consider the dinucleotide AT. What is the complementary dinucleotide in the "other" DNA strand? Be sure you have correctly labeled the 5' ends of the two dinucleotides.

    Remember that you have many, many, molecules. Some are cut one way, some another. On average, the more times a mono- or di-nucleotide occurs in the original sequence, the more often it will turn up in the hydrolysate. For example, the more times TT 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.

 

1. Answers:
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 'Remember' in hints above.) 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').

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. Hints:
    A-1. This is not a trick question -- it's as simple as it looks. It's included just to be sure you know the terminology.

    A-2. Is replication semi-conservative or conservative? Draw an original, unlabeled DNA molecule (double stranded), and diagram out what will you get after one round of replication in G*. Then take whatever you get, and replicate it again.

    A-3. When you replicate DNA 3 times, how many molecules will you have? Do you still have the molecules from the 1st and 2nd replications?

    B-1. Diagram out the experiment: Make a picture or flow chart of what happens to the starting G*, how you get G*-DNA etc. You need to be able to follow the G* through all the steps. Then consider what would happen in each of the possibilities given  -- if the DNA is degraded to bases (or nucleosides, or nucleotides), will radioactivity end up in both RNA and DNA? Why or why not?

    B-2. When the G*-DNA is broken down into nucleotides, what do you get? G*MP? G*DP? Something else? When you reuse the nucleotides, to make new DNA chains, what form do you use? G*MP? G*DP? Something else?

    C. What would it take to get colonies here? What is required for successful transformation?


2. Answers:
    A-1. G is a purine and a base.

    A-2. It will take 2 generations. After one replication, all the DNA will contain radioactivity in one strand, and none in the other. After a second replication, half the DNA will have radioactivity in both strands, and half will have radioactivity in one strand only.

    A-3. 3/4.  There should be a total of 8 double stranded molecules at the end. (Did you get 15? You forgot that when you replicate a molecule, you don't have the original any more.) You go from  1 molecule → 2 → 4 → 8. Two of the molecules contain one new strand (labeled) and one of the original old strands (unlabeled). The other 6 molecules contain two new (labeled) strands.

    B-1. 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.

    B-2. 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.

    C. 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.