Chapter 16: The Molecular Basis of Inheritance - ? Chapter 16: The Molecular Basis of Inheritance

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    Name_______________________ Period___________ Chapter 16: The Molecular Basis of Inheritance Concept 16.1 DNA is the genetic material 1. What are the two chemical components of chromosomes? DNA and protein 2. Why did researchers originally think that protein was the genetic material? Until the 1940s, the case for proteins seemed stronger, especially since biochemists had

    identified them as a class of macromolecules with great heterogeneity and specificity of function, essential requirements for the hereditary material. Moreover, little was known about nucleic acids, whose physical and chemical properties seemed far too uniform to account for the multitude of specific inherited traits exhibited by every organism.

    3. Distinguish between the virulent and nonvirulent strains of Streptococcus pneumoniae studied

    by Frederick Griffith. The virulent strains are pathogenic (disease-causing), whereas the nonvirulent strains are non-

    pathogenic (harmless). 4. What was the purpose of Griffiths studies? Griffith was attempting to develop a vaccine against pneumonia. 5. Use this figure to summarize the experiment in which Griffith became aware that hereditary

    information could be transmitted between two organisms in an unusual manner. See page 306 of your text for the labeled figure. Frederick Griffith studied two strains of the bacterium Streptococcus pneumoniae. Bacteria of

    the S (smooth) strain can cause pneumonia in mice; they are pathogenic because they have an outer capsule that protects them from an animals defense system. Bacteria of the R (rough) strain lack a capsule and are nonpathogenic. To test for the trait of pathogenicity, Griffith injected mice with the two strains. Griffith concluded that the living R bacteria had been transformed into pathogenic S bacteria by an unknown, heritable substance from the dead S cells that allowed the R cells to make capsules.

    6. Define transformation.

    A change in genotype and phenotype due to the assimilation of external DNA by a cell. When the external DNA is from a member of a different species, transformation results in horizontal gene transfer.

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    7. What did Oswald Avery determine to be the transforming factor? DNA Explain his

    experimental approach. Avery broke open the heat-killed pathogenic bacteria and extracted the cellular contents. He

    treated each of three samples with an agent that inactivated one type of molecule, and then tested the sample for its ability to transform live nonpathogenic bacteria. Only when DNA was allowed to remain active did transformation occur.

    8. Sketch a T2 bacteriophage and label its head, tail sheath, tail fiber, and DNA. See page 306 of your text for the labeled figure. 9. How does a bacteriophage destroy a bacterial cell? Look ahead to Chapter 19, Figure 19.5, to

    explain this. The T4 phage uses its tail fibers to bind to specific receptor sites on the outer surface of an E.

    coli cell. The sheath of the tail contracts, injecting the phage DNA into the cell and leaving an empty capsid outside. The cells DNA is hydrolyzed. The phage DNA directs production of phage proteins and copies of the phage genome by host and viral enzymes, using components within the cell. Three separate sets of proteins self-assemble to form phage heads, tails, and tail fibers. The phage genome is packaged inside the capsid as the head forms. The phage directs production of an enzyme that damages the bacteria cell wall, allowing fluid to enter. The cell swells, and finally bursts, releasing 100 to 200 phage particles.

    10. How did Hershey and Chase label viral DNA and viral protein so that they could be

    distinguished? Explain why they chose each radioactive tag in light of the chemical composition of DNA and protein.

    Hershey and Chase used radioactive isotopes of sulfur to tag protein in one batch of T2 and a

    radioactive isotope of phosphorus to tag DNA in a second batch. Because proteins, but not DNA, contain sulfur, radioactive sulfur atoms were incorporated only into the protein of the phage.

    11. Describe the means by which Hershey and Chase established that only the DNA of a phage

    enters an E. coli cell. What conclusions did these scientists draw based on these observations? Separate samples of the nonradioactive E. coli cells were allowed to be infected by the protein-

    labeled and DNA-labeled batches of T2. The researchers then tested the two samples shortly after the onset of infection to see which type of moleculeprotein or DNAhad entered the bacterial cells and would therefore be capable of reprogramming them. Hershey and Chase found that the phage of DNA entered the host cells but the phage protein did not. Hershey and Chase concluded that the DNA injected by the phage must be the molecule carrying the genetic information that makes the cells produce new viral DNA and proteins.

    12. What are Chargaffs rules? How did he arrive at them? Chargaffs rules are:

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    1. The base composition varies between species. 2. Within a species, the number of A and T bases are equal and the number of G and C bases are

    equal.

    Chargaff analyzed the base composition of DNA from a number of different organisms. He reported that the base composition of DNA varies from one species to another. He also noticed a peculiar regularity in the ratios of nucleotide bases. In the DNA of each species he studied, the number of adenines approximately equaled the number of thymines, and the number of guanines approximately equaled the number of cytosines.

    13. List the three components of a nucleotide. Phosphate, Sugar (deoxyribose), Nitrogenous base 14. Who are the two men who built the first molecular model of DNA and shared the 1962 Nobel

    Prize for discovery of its structure? James Watson and Francis Crick 15. What was the role of Rosalind Franklin in the discovery of the double helix? Rosalind Franklin, a very accomplished X-ray crystallographer, conducted critical experiments

    resulting in the photograph that allowed Watson and Crick to deduce the double-helical structure of DNA.

    16. Distinguish between the structure of pyrimidines and purines. Explain why adenine bonds only

    to thymine. Pyrimidinescytosine (C), thymine (T), and uracil (U)are characterized by a six-membered

    ring. Purinesadenine (A) and guanine (G)are characterized by a six-membered ring fused to a

    five-membered ring. Adenine bonds only with thymine because adenine is a purine, and thymine is a pyrimidine.

    Always pairing a purine with a pyrimidine results in a uniform diameter in the double helix. Additionally, each base has chemical side groups that can form hydrogen bonds with its appropriate partner. Adenine can form two hydrogen bonds with thymine and only thymine.

    17. How did Watson and Cricks model explain the basis for Chargaffs rules? Watson and Crick began building models of a double helix that would conform to the X-ray

    measurements and what was then known about the chemistry of DNA, including Chargaffs rule of base equivalences. Through trial and error, Watson and Crick deduced that the nitrogenous bases of the double helix are paired in specific combinationsadenine (A) with thymine (T), and guanine (G) with cytosine (C)and they reflected these findings in their model. Whenever one strand of a DNA molecule has an A, the partner strand has a T. And a G in one strand is always paired with a C in the complementary strand. Therefore, in the DNA of any organism, the amount of adenine equals the amount of thymine, and the amount of guanine

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    equals the amount of cytosine. 18. Given that the DNA of a certain fly species consists of 27.3% adenine and 22.5% guanine, use

    Chargaffs rules to deduce the percentages of thymine and cytosine. 27.6% thymine 22.5% cytosine 19. Name the five nitrogenous bases, and put a checkmark in the correct column for each base.

    Also indicate if the base is found in DNA (D), RNA (R), or both (B). Nitrogenous Base Purine Pyrimidine D, R, or B

    Adenine X B

    Guanine X B

    Thymine X D

    Cytosine X B

    Uracil X R

    20. What DNA base is complementary to adenine? Thymine What DNA base is complementary to guanine? Cytosine 21. Describe the structure of DNA relative to each of the following. Indicate the distance in the

    correct location on the figure as well. See page 16.7 for the labeled figure.

    a. distance across molecule 1 nm b. distance between nucleotides 0.34 nm c. distance between turns 3.4 nm d. components of the backbone sugar-phosphate e. components of the rungs A T G C

    22. Explain what is meant by 5' and 3' ends of the nucleotide. The two free ends of the polymer are distinctly different from each other. One end has a

    phosphate attached to a 5' carbon, and the other has a hydroxyl group on the 3' carbon. We refer to these as the 5' end and the 3' end respectively.

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    23. What do we mean when we say the two strands of DNA are antiparallel? Their subunits run in opposite directions. Concept 16.2 Many proteins work together in DNA replication and repair 24. What is the semiconservative model of replication?

    Type of DNA replication in which the replicated double helix consists of one old strand, derived from the parental molecule, and one newly made strand

    25. Who performed the experiments that elucidated the correct mechanism of DNA replication? Matthew Meselson and Franklin Stahl 26. How did Meselson and Stahl create heavy DNA for their experiments? Meselson and Stahl cultured E. coli for several generations in a medium containing nucleotide

    precursors labeled with a heavy isotope of nitrogen, 15N. 27. Use Figure 16.11 to explain how Meselson and Stahl confirmed the semiconservative

    mechanism of DNA replication. See page 312 in your text for the labeled figure. Meselson and Stahl transferred their heavy DNA to a medium with a lighter isotope, 14N. A sample was taken after DNA replicated one; another sample was taken after DNA replicated again. They extracted DNA from the bacteria in the samples and then centrifuged each DNA sample to separate DNA of different densities. Meselson and Stahl compared their results to those predicted by each of the three models (conservative, semiconservative, and dispersive). The first replication in the 14N medium produced a band of hybrid DNA. This result eliminated the conservative model. The second replication produced both light and hybrid DNA, a result that refuted the dispersive model and supported the semiconservative model. They therefore concluded that DNA replication is semiconservative. 28. Define the origins of replication. Site where the replication of a DNA molecule begins, consisting of a specific sequence of

    nucleotides 29. Distinguish between the leading and the lagging strands during DNA replication. The leading strand is the new complementary DNA strand synthesized continuously along the

    template strand toward the replication fork in the mandatory 5' 3' direction. The lagging strand is a discontinuously synthesized DNA strand that elongates by means of

    Okazaki fragments, each synthesized in a 5' 3' direction away from the replication fork.

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    30. What is the direction of synthesis of the new strand? 5' 3' direction 31. What are Okazaki fragments? How are they welded together? Okazaki fragments are short segments of DNA synthesized away from the replication fork on a

    template strand during DNA replication. Many such segments are joined together by the enzyme DNA ligase to make up the lagging strand of newly synthesized DNA.

    32. Which enzyme does each of the following? a. untwists and separates strands helicases

    b. holds DNA strands apart single-strand binding proteins

    c. synthesizes RNA primer primase

    d. adds DNA nucleotides to new strands DNA polymerases

    e. relieves strain caused by unwinding topoisomerase

    f. joins DNA fragments together topoisomerase

    g. removes RNA primer and replaces with DNA DNA polymerases

    33. Label the following figures. Include 3' and 5' strands, RNA primer, primase, SSBP, topoisomerase, helicase, leading strand, lagging strand, DNA pol I, DNA pol III, DNA ligase, parental DNA, and new DNA On the second figure, also add arrows to indicate the direction of synthesis.

    See pages 314-315 in your text for the labeled figures. (Note: See both Figures 16/13 and 16.15.)

    34. Put it all together! Make a detailed list of the steps that occur in the synthesis of a new strand. 1. Helicase unwinds the parental double helix.

    2. Molecules of single-stranded binding protein stabilize the unwound template strands.

    3. The leading strand is synthesized continuously in the 5' 3' direction by DNA polymerase III after being primed by primase.

    4. Primase begins synthesis of the RNA primer for the lagging strand.

    5. DNA polymerase III synthesizes discontinuously the lagging strand in the 5' 3' direction.

    6. DNA polymerase I removes all the RNA primer sections and replaces them with DNA nucleotides.

    7. The replacement of the primer with DNA leaves the new DNA nucleotides with a free 3' end. DNA ligase joins the free 3' end to its adjacent 5' end, forming a continuous and unbroken strand of DNA on both the leading and lagging strands.

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    35. Explain the roles of each of the following enzymes in DNA proofreading and repair.

    36. What is a thymine dimer? How might it occur? How is it repaired? A thymine dimmer is the covalent linking of thymine bases that are adjacent on a DNA strand,

    causing the DNA to buckle and interfere with DNA replication. In order to repair this damage, a nuclease enzyme cuts the damaged DNA strand, and the damaged section is removed. DNA polymerase fills in the missing nucleotides, and DNA ligase seals the free end of the new DNA to the old DNA, making the strand complete.

    37. Make a sketch of a chromosome and label the telomeres. See page 319 in your text for the labeled figure. 38. Explain telomere erosion and the role of telomerase. Telomeres provide their protective function by postponing the erosion of genes located near

    the ends of DNA molecules. Telomeres become shorter during every round of replication. Telomeric DNA tends to be shorter in dividing somatic cells of older individuals and in cultured cells that have divided many times. Importantly, some cell genomes (such as germ cells) must persist virtually unchanged from an organism to its offspring over many generations. In order to accomplish this, an enzyme called telomerase catalyzes the lengthening of telomeres in eukaryotic germ cells, thus restoring their original length and compensating for the shortening that occurs during DNA replication.

    39. Why are cancer cells immortal even though most body cells have a limited life span? Researchers have found telomerase activity in cancerous somatic cells, suggesting that its

    ability to stabilize telomere length may allow these cancer cells to persist.

    Enzyme Role

    DNA polymerase Fills the gap left after nuclease excises damaged segments of the

    DNA strand

    Nuclease DNA-cutting enzyme

    Ligase Joins the sugar-phosphate backbones of all the Okazaki fragments into a continuous DNA strand

    Repair enzymes Proofread and repair DNA

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    Concept 16.3 A chromosome consists of a DNA molecule packed together with proteins 40. On the following diagram, identify the following: 30-nm fiber, metaphase chromosome, double

    helix, histone proteins, nucleosomes, protein scaffold, and looped domains (300-nm fiber). See pages 320-321 in your text for the labeled figure. 41. Distinguish between heterochromatin and euchromatin. Heterochromatin is eukaryotic chromatin that remains highly compacted during interphase and is generally not transcribed. Euchromatin is a less condensed form of eukaryotic chromatin that is available for transcription. Test Your Understanding Answers Now you should be ready to test your knowledge. Place your answers here: 1. c 2. c 3. b 4. b 5. c 6. d 7. b 8. a

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