BSCI222 – Lecture 2 (9/5/13)- Chapter 10. DNA: The Chemical Nature of the Gene- 3 classic experiments to demonstrate that DNA was genetic material:o Simple, repetitive, couldn’t possibly carry much information. Proteins were much more interesting because more complex, more different shapes and sizes. People kept assuming that the protein part of the chromosomes carried the information. It’s really just the simple DNA structure though.o Griffith: provided an assay for identifying which component of the chromosome was carrying the info. Assay for transformation. Worked with 2 types of pneumococcus, labeled Rough (mouse develops immune response, survives) and Smooth (has polysaccharide coating, protects it from the immune system, thereby virulent, will kill the mouse). Heat-kill the smooth cells (boil them) and inject them into a mouse -> mouse lives. Then mixed Rough cells with heat-killed Smooth cells and injected them -> mouse died (even though neither by themselvescould kill it). Something from the heat-killed cells is transforming the rough cells into a virulent strain, some “component” is mixing with the benign cells and genetically transforming them. (found living Smooth cells in the dead mouse’s heart)o Avery, Macleod and McCarty: developed purified extracts of the chromosomes, demonstrate that it was the DNA component. Not everybody believed this experiment. Started to fractionate the heat-killed Smooth cells into components. Wash boiled virulent cells 3 times with saline. Then, add proteinase (remove all proteins). Finally, add Rough cells and see if any transform into virulent cells. They did and that means that protein is not the genetic material. The reason it’s not completely convincing is that you can never be sure that the proteinase chewed up all the protein. If all protein gone, that leaves only nucleic acid as possibilities (RNA and DNA). Chewed up RNA -> still got the transformation into the virulent cells. Only when added deoxyribonuclease, ridding it of all the DNA, did no transformation happen, no Smooth cells appeared. This is essentiallystill how you extract DNA. Characteristics of “the transforming principle”: high molecular weight, highly negatively charged, UV absorption matches that of DNA, nitrogen/phosphorus ratio was like DNA (DNA has phosphorus, protein does not). o Hershey and Chase: used radioactive compounds, did a convincing experiment that finally settled the argument. Took advantage of a bacteriophage (bacterial virus for E. coli). DNA is packed in the head (looks like the lunar module), everything else is protein, very simply organism. Binds to receptors on the surfaceof E. coli, injects its genome, which then begins to be expressed and then a varietyof proteins start to chop up the bacteria’s chromosome and turn it into a phage factory, and finally lyses the bacterial wall and releases all the new phages. The researchers used 35_Sulfur and 32_Phosphorus. Used the Sulfur to grow E. coli,then grew phage on it, which became labeled with the 35_S (in the proteins). In separate vial did the same thing with the Phosphorus, which labeled 32_P only in the nucleic acid (because DNA has phosphorus, proteins don’t). Then took unlabeled E. coli and infected separately with each, and after a few minute put the2 mixes in 2 blenders to separate the phages from the E. coli cells. After getting rid of the phages, the cells that were taken over by Sulfur phages should have no traces of it left, but those infected by (radioactive) Phosphorus phages should. Most of the bacteria got infected, almost all of the S_35 wound up in the superfluous medium, as well as a little P (not all phages bound). 75% of the P stayed with the E. coli cells, 75% of the S wound up in the extracellular solution.o After this, race was on to understand DNA structure and how it carries information.- Chemical components of DNA:o Composed of nucleotides. Each nucleotide has a phosphate group (PO4), a sugar (ribose (OH on 2’ and 3’) or deoxyribose (lost the O’s on 2’ and 3’), 5 Carbons labeled with prime numbers), and a nitrogenous base (purines (double ring structures, A and G) and pyrimidines (single ring structures, C and T (and U instead of T in RNA, different because of a methyl group)). Don’t need to know structure of the bases. Do need to remember that A and G are purines, and that C, T, and U are pyrimidines.o Watson and Crick:  First clues they used: data from Chargaff, ratios of A, G, C, and T in various DNA sources (E. coli, yeast, sea urchin, rat, and human). No matter what the actual frequency was, the ratio of A/T was always about 1,and G/C same. Also had preliminary x-ray crystallography data from Rosalind Franklin, studied arrangement of atoms in a crystal. Generate a bunch of x-rays through a lead screen, shine at crystal (DNA) atoms diffract the beam, and detect it on X-ray film as a series of bands and dots (a diffraction pattern). The pattern that Rosalind Franklin was able to generate (very difficult to do, “spaghetti of DNA”). Told us that: cross angle -> indicates helix spacing, triangles -> indication of helix diameter, and varying intensities in the X in the middle -> arise from major and minor grooves.  First model: the phosphate groups have all this negative charge (in the backbone of the DNA), which would repel each other unless they can stabilize themselves somehow. So the first idea was that the bases are on the outside of the DNA (info thus accessible), with positive ions on the inside to balance the negative ones. Seemed perfect; total opposite of whatRosalind Franklin’s images indicated. Linus Pauling suggested a triple helix model, with bases sticking out of a helix made of 3 strands. Also completely wrong, although he didn’t have access to Franklin’s x-rays. Completely guessed the structure. Found hydrogen bonds where like bases could pair and the two helices were running in the same direction. The DNA strand has a polarity, arisingfrom the sugar. Polarity runs from the 5’ (sugar) to the 3’ (when drawing it, it’s an arrow from 5’ to 3’). The model was wrong because they were both going the same way. (Textbook had sugar in the enol form, but in reality it’s in the keto form in the cell. Once they knew that, knew the H bonds were totally different, only took a few hours to understand what the correct pattern must be. This is why the ratios were around 1, because for each C need a G,