AP Biology

1. AP Biology Molecular Genetics Unit Chapters 16 & 17

2. Chapter 16 Objectives • Explain how Griffith and Avery’s experiment with pneumococcus bacteria contributed to the understanding of molecular genetics. • Define transformation. Use Griffith & Avery's results to explain how it occurs. • Using Hershey & Chase's procedure, use given hypotheses to predict results. Evaluate each hypothesis based upon the actual results of the experiments. • Identify the component parts of nucleic acids. Describe their organization into DNA and RNA. • Explain how Chargaff's rule supports the base pairing rule. • Explain how the two strands of a DNA molecule are complementary. • Explain how DNA replication is semiconservative. • Outline the process of DNA replication. Include the names and roles of any enzymes involved in the process. • Explain why one strand of DNA is replicated continuously, while the other strand is discontinuous. • Define Telomere. Explain why the mechanism of replication results in the shortening of telomeres. • Explain how DNA condenses into chromosomes in eukaryotic cells.

3. What part of the cell controls heredity? • Actual traits are not inherited. Traits are developed. Inheritance must come from something that is in the gametes and in the resulting zygote • Since the gametes and zygote are cells, the control center of heredity must have some cellular origin • Cells have 3 fundamental regions • Cell membrane • Cytoplasm • Nucleus • One of these 3 regions must control heredity & determine how traits will develop in the organism

4. What part of the cell controls heredity? • It follows logically that hereditary control is centered either in the membrane, the cytoplasm, or the nucleus. • We will assume two alternative hypotheses: • The nucleus controls heredity • The cytoplasm controls heredity • An experiment to test these must allow for simple manipulation of cell parts and simple observation of developing traits • Manipulating cell parts would be simplified by using a single celled organism, but it must be large enough to work with and observe easily, and must have some distinct inherited trait.

5. 1943 - Hammerling’s Experiment • The giant unicellular algae of the genus Acetabularia are an ideal experimental subject. • They are large, easy to manipulate, and easy to observe

6. Hammerling’s Experiment • All Acetabularia species have a base (which contains the nucleus), a stalk, and a cap. • The base and stalk are similar between species, but the cap varies greatly

7. Hammerling’s Experiment • A. mediterranea has an umbrella shaped cap. • A. crenulata has a crenulated (fringed) cap. • If you cut off the cap, it will regenerate. The regeneration is under hereditary control

8. Experimental Design: • Cut off the cap of each alga • Remove a section of the stalk (which contains cytoplasm, but not the nucleus) • Place a stalk from A. crenulata on a base (containing a nucleus) of A. mediterannea (Experiment 1) • Place a stalk from A. mediterannea on a base of A. crenulata (Experiment 2) • Observe the regeneration of the cap on each “hybrid” alga

9. Exp. #1 – Stalk C on Base M • If the cytoplasm controls heredity . . . • The cap that regenerates will follow instructions from the cytoplasm in the stalk. • The cap should resemble type C • If the nucleus controls heredity . . . • The cap that regenerates will follow instructions from the nucleus in the base • The cap should resemble type M

10. Exp. #1 – Stalk C on Base M • The algae regenerated type M caps, suggesting the nucleus in the base of the cell controlled the development of the cap

11. Exp. #2 – Stalk M on Base C • Similarly, algae with a nucleus from type C regenerated a crenulated cap. • In both experiments, the nucleus controlled regeneration

12. The Nucleus as Control Center • If the nucleus is the center of hereditary control, there must be something inside of it that actually directs the development of traits. • The nucleus consists almost entirely of: • Proteins • Nucleic Acids • DNA • RNA

13. What material in the nucleus controls heredity? • Nucleic Acids? • DNA and RNA • Complex polymer made of nucleotides • Nitrogenous bases are variable • 4 types (AGCT) • Protein? • Complex polymer made of amino acids • Amino acid side chains are variable • 20 different types of essential amino acids commonly found in protein • With more variability, protein was the favored hypothesis

14. What material in the nucleus controls heredity? • To test these hypotheses, we need to separate the proteins from the nucleic acids, treat a test subject with each, and observe the results. • Viruses are made entirely out of nucleic acids and proteins. • Viruses inject their genetic material into a host cell, take over the hereditary machinery of the host, and use it to reproduce itself. • We can use viruses as a test subject if we can find a way to easily identify or manipulate the proteins and the nucleic acids.

15. Tobacco Mosaic Virus & Holmes Ribgrass Virus • Tobacco Mosaic Virus (TMV) was the first virus discovered and isolated. • TMV infects plants, and forms characteristic lesions that are identifiable as tobacco mosaic disease. Holmes ribgrass virus (HRV) also infects plants. The lesions formed in Holmes Ribgrass disease are distinctly different from TMV lesions • Viruses consist of a protein coat surrounding a core of nucleic acids (RNA in the case of these 2 viruses) • Viruses inject their host with their genetic material, and take over the host for the purpose of replication

16. Heinz Fraenkel-Conrat 1955 • In the first of a series of experiments, Fraenkel-Conrat enzymatically digested the protein coat from TMV viruses and isolated the RNA core (and vice versa) • He then infected tobacco plants with only the protein he derived from the virus. The plants did not develop tobacco mosaic disease • He also infected tobacco plants with only the RNA from his viruses. These plants developed tobacco mosaic lesions • These results suggest that nucleic acids control the heredity of the virus and the characteristics associated with the viral disease • Remember, protein was the favored hypothesis for the hereditary material (due to its greater structural variability)!

17. Fraenkel-Conrat; Hybrid Viruses • In this experiment, he created hybrid viruses consisting of the protein coat from HRV and an RNA core from TMV • He infected tobacco plants with the hybrid viruses

18. Fraenkel-Conrat; Hybrid Viruses • If protein was the hereditary material, the plants should exhibit the symptoms of HRV • If the RNA is the heredity material, the plants should form lesions characteristic of TMV

19. Results • Not only did the plants show disease symptoms characteristic of tobacco mosaic disease, but viruses collected from the infected plants were fully formed tobacco mosaic viruses. • The hereditary material from the TMV core caused the disease symptoms, was replicated in the host cells, and directed the formation of TMV protein coats in the offspring viruses. • That core material was RNA, not protein as hypothesized

20. Hershey and Chase 1952 • In another classic experiment to identify the hereditary material, Alfred Hershey and Martha Chase used radioactively tagged viruses (bacteriophages – viruses that infect bacteria) • Both protein and DNA contain carbon, hydrogen, oxygen and nitrogen, but each contains one element that the other does not. • Protein contains sulfur, while DNA contains phosphorus. • Viruses grown in a culture containing radioactive sulfur will tag the proteins in a way that can be identified and tracked in the lab. • If viruses are grown in a culture containing radioactive phosphorus, their nucleic acids can be tracked

21. Hershey/Chase Experiment • Produce radioactively tagged viruses • Allow them to infect bacteria • Agitate, wash, and centrifuge the cultures • Test the wash solution and the bacterial cells for radioactive residue

22. Predictions: • If protein is the hereditary material, then the wash solution should contain radioactive phosphorus residue and radioactive sulfur should be detectable in the bacterial cultures • If DNA is the hereditary material, the wash solution should contain sulfur, while the bacterial culture should contain the tagged phosphorus

23. Results and conclusions: • The cells that were infected contained residues of radioactive tagged phosphorus. • DNA was injected into the host cells • DNA is the hereditary material

24. Frederic Griffith 1927 • Pneumonia is a deadly disease. There are various causes, including both bacterial and viral types • The causes and treatment of bacterial pneumonia has been of enormous importance to medical science for generations • To determine the pathogen responsible for infectious disease, a researcher will adhere to a series of logical concepts called Koch’s Postulates: • The bacteria must be present in every case of the disease. • The bacteria must be isolated from the host with the disease and grown in pure culture. • The specific disease must be reproduced when a pure culture of the bacteria is inoculated into a healthy susceptible host. • The bacteria must be recoverable from the experimentally infected host.

25. Smooth vs. Rough Strains Streptococcus pneumoniae exists in two distinct strains. One which forms rough textured pinpoint colonies and another which forms smooth, spreading colonies

26. Smooth = Capsulated We now know that the smooth strains are smooth due to a slime layer, or “capsule” which surrounds the outside of the cell wall

27. Experiment #1 • Inject rough type pneumococci into healthy mice • Result: • Mice remain healthy • No pneumonia symptoms are observed • No bacteria are recovered from blood • Inject smooth type pneumococci into healthy mice • Result: • Mice exhibit pneumonia symptoms • Smooth type bacteria are recovered from the blood of the infected mice

28. Follow-Up; Experiment #2 • Only the smooth type cause the disease, but is it a result of the action of the cells themselves or is it a result of some poison in the capsule? • Mice were injected with heat killed smooth type bacteria • Results: • Mice remain healthy • No pneumonia symptoms are observed • No bacteria are recovered from blood

29. OK, so what now? • Clearly it is the cells themselves, and not the capsule that cause pneumonia, but the capsule must have some significance • The third experiment in the series involved taking living rough type cultures and growing them in a medium containing the remains of heat killed smooth type bacteria

30. Experiment #3 - Predictions • Live rough type hadn’t caused pneumonia in the 1st experiment • Heat killed smooth didn’t cause pneumonia in the 2nd Experiment • The only logical assumption would be that since neither caused pneumonia by itself, there would be no reason to expect them to cause pneumonia together

31. Experiment #3 - Results • Combined live rough/dead smooth cultures are injected into healthy mice • Mice develop pneumonia symptoms • Smooth type bacteria are recovered from the blood of the infected animals

32. How do we account for that? • So not only did we get pneumonia from 2 things that were proven not to cause pneumonia, but it sure looks like the smooth type cells came back from the dead like slime covered microscopic pneumonia zombies

33. Not Likely • OK, so scrap the zombie hypothesis • What appears to have happened is that hereditary material from the heat killed smooth cells was absorbed by the living rough cells • Even though the cells were dead, the hereditary material was still operational, and could cause the rough cells to become “transformed” • The transformed cells expressed the “smooth” genes they picked up from the dead cells. They gained the ability to make a capsule, and because of that they were able to cause pneumonia in the mice • It turns out that the capsule doesn’t harm the mice, but it does protect the bacteria from the mouse’s immune system, allowing them to stay alive, multiply, and cause the disease

34. Avery 1943 https://www.youtube.com/watch?v=RWFc8Iqz4Jg

35. Frederic Griffith and Oswald Avery

36. DNA Structure • DNA is a polymer consisting of nucleotide subunits • Nucleotides have 3 parts: • Phosphoric Acid (phosphate) • Pentose sugar (deoxyribose) • Nitrogenous Base • In DNA nucleotides, the phosphate and the deoxyribose are constants • There are 2 categories of nitrogenous base: • Purines • Adenine and Guanine • Pyrimidines • Cytosine and Thymine

37. Pentoses, Purines, and Pyrimidines

38. DNA Polymer • There are 2 fundamental structural components of a DNA strand, the Sugar/Phosphate backbone and the Nitrogen Base side-chains

39. Some Vocabulary Clarification • The deoxyribose is a 5 carbon sugar (pentose) • Structural positions are designated by numbering the carbons • The nitrogen base is at position 1 (1’, read as “1 prime”) • The phosphate is at the 5’ position • The bottom corner (3’) will become the point of attachment of the next nucleotide in the polymer

40. Rosalind Franklin - 1951 • Franklin’s x-ray crystallography photographs of DNA demonstrated that DNA consisted of 2 strands twisted in a “double helix”

41. Erwin Chargaff 1950 Chargaff enzymatically digested DNA from a variety of organisms and determined the relative proportion of each base. With great regularity, the proportions of A and T are equal to each other, as are G and C

42. Base Pairing • Chargaff’s rule (#A = #T; #G = #C) is explained if the 2 DNA strands align with bases paired across between the strands, A to T and G to C • Note the formation of hydrogen bonds between the base pairs

43. The Double Helix Illustrated • The polymer is formed by the sugar-phosphate backbone • The “phosphodiester” links are covalent • The 2 strands are “antiparallel” • The strands are joined by base pairing • The hydrogen bonds are weak • They can separate and rejoin easily

44. Watson and Crick - 1953 • James Watson and Francis Crick won the Nobel Prize for publishing the structure of DNA • Neither of them is particularly good looking

45. Bozeman Videos • DNA and RNA part 1 • http://www.youtube.com/watch?v=qoERVSWKmGk&list=PLFCE4D99C4124A27A&index=34 • DNA and RNA part 2 • http://www.youtube.com/watch?v=W4mYwsr9gGE&list=PLFCE4D99C4124A27A • Gene Regulation • http://www.youtube.com/watch?v=3S3ZOmleAj0&list=PLFCE4D99C4124A27A • Signal Transmission and Gene Expression • http://www.youtube.com/watch?v=D-usAds_-lU&list=PLFCE4D99C4124A27A

46. MIT OpenCourseWare Videos • DNA Structure and Classic Experiments, excerpt 1 • https://www.youtube.com/watch?v=P-Ry4rRdDbk • DNA Structure and Classic Experiments, excerpt 2 • https://www.youtube.com/watch?v=YCeKtM6Hnmc • DNA Replication: Fundamentals of Biology • https://www.youtube.com/watch?v=DRBREvFL19g • Transcription and Translation, excerpt 1 • https://www.youtube.com/watch?v=tMr9XH64rtM • Transcription and Translation, excerpt 2 • https://www.youtube.com/watch?v=uBRdfsz_YB4

47. Replication • “Replica” = exact copy. Replication is copying of the DNA • DNA is copied in preparation for cell division • (mitosis/meiosis/binary fission) • Replication occurs in the S (synthesis) phase of Interphase • In replication, the entire genome is copied

48. Replication Process Sequence

49. Unzipping the strands • The area that unzips forms a “replication fork” • Unzipping is a more complex process than it might seem, largely because of the helical nature of DNA • The two major players in unzipping are the Helix Destabilizing Protein and the enzyme DNA Helicase • The Helix Destabilizing Protein does exactly what its name suggests. It interferes with the twisted shape of the double helix, creating some physical stress between the base pairs • This creates space for the binding of the DNA Helicase, which will act like a wedge to separate the base pairs from each other • Remember, the sugar-phosphate backbone is permanently linked together with covalent phosphodiester bonds, but the base pairs are only attracted to each other by hydrogen bonding. Hydrogen bonding is strong as polar attractions go, but weak in comparison to covalent bonds.

50. Resolving complications • Once unzipping begins, the action of the single strand binding proteins and topoisomerase become necessary • Single strand binding proteins hold the strands apart so that they don’t immediately rejoin. Remember, the base pairs are held together by hydrogen bonding. If they come back into contact they will stick back together • Remember that DNA is a double helix. Disrupting the helix at the replication fork creates stress further down on the molecule. Picture uncoiling an extension cord. Every coil you unwrap compresses a coil you already unwrapped, and pretty soon you have a tangled mess. Add to that the fact that you are turning one double helix into 2 double helices. The enzyme topoisomerase resolves these stresses by cutting the DNA every once in a while to relieve the stress and then bonding it back together afterwards • Don’t try that with your extension cords. They don’t go back together.