The correct answer is d A. Answer a is incorrect. The lethality of smooth-coat bacteria was known before Griffith started the experiment.

Transcription

1 1. What was the key finding from Griffith s experiments using live and heat-killed pathogenic bacteria? a. Bacteria with a smooth coat could kill mice. b. Bacteria with a rough coat are not lethal. c. Heat-killed smooth-coat bacteria would not cause death in mice. d. Heat-killed smooth-coat bacteria could transform the nonlethal live bacteria. A. Answer a is incorrect. The lethality of smooth-coat bacteria was known before Griffith started the experiment. B. Answer b is incorrect. This observation was not a result of this experiment. C. Answer c is incorrect. This is a result from this experiment; however, it is not the important result. Heat-killed smooth-coat bacteria could transform the nonlethal live bacteria. D. Answer d is correct. Transformation of nonlethal cells to lethal cells was the critical result that demonstrated the ability of genetic material to be transferred between cells. 2. When Hershey and Chase differentially tagged the DNA and proteins of bacteriophages and allowed them to infect bacteria, what did the viruses transfer to the bacteria? a. Radioactive phosphorous and sulfur b. Radioactive sulfur c. DNA d. Both b and c A. Answer a is incorrect. Radioactive phosphorous labeled the DNA and radioactive sulfur labeled protein. If both phosphorous and sulfur were transferred to the bacteria that would mean that both DNA and protein are transferred from the virus. B. Answer b is incorrect. Radioactive sulfur labeled proteins. Proteins were not transferred to the bacteria. DNA C. Answer c is correct. The experiment demonstrated that DNA is transferred from the virus to the bacteria. D. Answer d is incorrect. Radioactive sulfur labeled proteins. Proteins were not transferred to the bacteria. 3. Which of the following is NOT a component of DNA? a. Pyrimidines such as uracil

2 b. Five-carbon sugars c. Purines such as adenine d. Phosphate groups Pyrimidines such as uracil A. Answer a is correct. Uracil is a pyrimidine that is found in RNA. B. Answer b is incorrect. DNA is composed of ribose a five-carbon sugar. C. Answer c is incorrect. The purine adenine is present in DNA. D. Answer d is incorrect. All the nucleotides in DNA are composed of a phosphate group. 4. What type of chemical bond allows DNA or RNA to form a long polymer? a. Hydrogen bonds b. Peptide bonds c. Ionic bonds d. Phosphodiester bonds A. Answer a is incorrect. Hydrogen bonds are important for the formation complementary base pairing. B. Answer b is incorrect. Peptide bonds are involved in forming proteins. C. Answer c is incorrect. Ionic bonds are not involved in the formation of nucleic acid polymers. Phosphodiester bonds D. Answer d is correct. Phosphodiester bonds are the covalent bonds formed between the sugar and phosphate groups of nucleic acids. 5. What is Chargaff s rule? a. The number of phosphate groups always equals the number of five-carbon sugars. b. The proportions of A equal that of C and G equals T. c. The proportions of A equal that of T and G equals C. d. Purines binds to pyrimidines A. Answer a is incorrect. Chargaff s rule refers to the numbers of purines and pyrimidines. Since every nucleic acid is composed of a sugar and phosphate group there would be no variation.

3 B. Answer b is incorrect. Chargaff s rule is based on complementary base-pairing. Adenine does not base pair with cytosine. The proportions of A equal that of T and G equals C. C. Answer c is correct. Chargaff s rule is based on complementary base-pairing. Adenine base-pairs with thymine, and guanine base-pairs with cytosine. D. Answer d is incorrect. Chargaff s rule is related to the relative numbers of purines and pyrimidines. It does not address the ability of the nucleotides to form hydrogen bonds. 6. The bonds that hold two complementary strands of DNA together are a. Hydrogen bonds b. Peptide bonds c. Ionic bonds d. Phosphodiester bonds Hydrogen bonds A. Answer a is correct. Hydrogen bonds are formed between the nitrogenous bases of the nucleotides. B. Answer b is incorrect. Peptide bonds are found in proteins, not nucleic acids. C. Answer c is incorrect. Ionic bonds are not found in nucleic acids. D. Answer d is incorrect. Phosphodiester bonds make up the sugar phosphate backbone of the nucleic acid polymer. 7. If one strand of a DNA molecule has the sequence ATTGCAT, then the complementary strand will have the sequence a. ATTGCAT b. TACGTTA c. TAACGTA d. GCCTAGC A. Answer a is incorrect. This is an identical sequence. Adenine will not form a hydrogen bond with itself. B. Answer b is incorrect. This cannot be the complementary sequence since the base-pairing is not correct. A binds to T and G binds to C. TAACGTA C. Answer c is correct. The base-pairing rule is that A binds to T and G binds to C. This sequence meets this rule.

4 D. Answer d is incorrect. The base-pairing rule is that A binds to T and G binds to C. This sequence does not meet this rule. 8. Which of the following is NOT part of the Watson Crick model of the structure of DNA? a. DNA is composed of two strands. b. The two DNA strands are oriented in parallel (5 to 3 ). c. Purines bind to pyrimidines. d. DNA forms a double helix. A. Answer a is incorrect. The Watson Crick model proposes that DNA is composed of two strands. The two DNA strands are oriented in parallel (5 to 3 ). B. Answer b is correct. The Watson Crick model proposes that DNA is composed of antiparallel strands one running 5 to 3 and the other 3 to 5. C. Answer c is incorrect. Base-pairing between purine and pyrimidines is a part of the Watson Crick model of DNA. D. Answer d is incorrect. The double-helix structure of DNA is a central feature of the Watson Crick model. 9. Meselson and Stahl demonstrated that a. DNA replication occurs in bacteria b. DNA replication is dispersive c. DNA replication is conservative d. DNA replication is semiconservative A. Answer a is incorrect. Meselson and Stahl examined the mechanism of DNA replication, not what type of cells can replicate DNA. B. Answer b is incorrect. Dispersive replication would lead to the formation of strands of DNA that were a mix of parental and new DNA. This did not occur. C. Answer c is incorrect. Conservative replication would lead to the formation of separate strands of parental and new DNA. This did not occur. DNA replication is semiconservative D. Answer d is correct. The newly synthesized DNA is a combination of one parental strand and one new strand.

5 10. Which of the following steps in DNA replication involves the formation of new phosphodiester bonds? a. Initiation at an origin of replication b. Elongation by a DNA polymerase c. Unwinding of the double helix d. Termination A. Answer a is incorrect. Initiation involves the opening up of the DNA helix. Elongation by a DNA polymerase B. Answer b is correct. Elongation involves the formation of phosphodiester bonds between the new nucleotides through the activity of DNA polymerases. C. Answer c is incorrect. Unwinding of the helix is important for replication; however, this step does not involve bond formation. D. Answer d is incorrect. Termination is the point where the formation of new phosphodiester bond formation stops. 11. The difference in leading- versus lagging-strand synthesis is a consequence of a. the antiparallel configuration of DNA b. DNA polymerase III synthesizing only 5 to 3 c. the activity of DNA gyrase d. both a and b A. Answer a is incorrect. The antiparallel configuration contributes to the difference between leading- and lagging-strand synthesis, but it is not the only factor. B. Answer b is incorrect. The 5 to 3 activity (true of all DNA polymerases) of DNA polymerase III contributes to the differences between leading- and lagging-strand synthesis, but is not the only factor. C. Answer c is incorrect. The activity of DNA gyrase does not contribute to the difference between leading- and lagging-strand synthesis. both a and b D. Answer d is correct. DNA polymerase III synthesizes DNA only in a 5 to 3 direction; however, because DNA has an antiparallel orientation there is a difference in the rate of synthesis between the two strands. 12. Okazaki fragments are a. synthesized in the 3 to 5 direction b. found on the lagging strand

6 c. found on the leading strand d. made of RNA A. Answer a is incorrect. An Okazaki fragment is a segment of DNA and DNA is synthesized in the 5 to 3 direction. found on the lagging strand B. Answer b is correct. Okazaki fragments are only found in the lagging strand as a consequence of discontinuous synthesis. C. Answer c is incorrect. The leading strand undergoes continuous synthesis and therefore does not generate Okazaki fragments. D. Answer d is incorrect. Okazaki fragments are made of DNA. The primers are composed of RNA. 13. Successful DNA synthesis requires all of the following except a. helicase b. endonuclease c. DNA primase d. DNA ligase A. Answer a is incorrect. The helicase is required to unwind the DNA, making it possible for the polymerase to bind to the replication fork. endonuclease B. Answer b is correct. Endonuclease function is important for DNA repair. DNA polymerases contain exonuclease activity. C. Answer c is incorrect. Primase generates the RNA primer required to start DNA polymerase III synthesis. D. Answer d is incorrect. Ligase activity is required to generate phosphodiester bonds to link Okazaki fragments. 14. What is a telomere? a. An A-T rich region of DNA b. The point of DNA termination on a bacterial chromosome c. Regions of repeated sequences of DNA on the ends of eukaryotic chromosomes d. The sequence of RNA found on a replicating molecule of DNA

7 A. Answer a is incorrect. An origin of replication is characterized by the high proportion of adenine and thymine. B. Answer b is incorrect. Bacteria possess circular chromosomes. They do not have telomeres. Regions of repeated sequences of DNA on the ends of eukaryotic chromosomes C. Answer c is correct. Telomeres are found at the end of eukaryotic chromosomes. They are characterized by repeated sequences of DNA. D. Answer d is incorrect. The sequence of RNA used in DNA replication is known as a primer. 15. Which type of enzyme is involved in excision repair? a. photolyase b. DNA polymerase III c. endonuclease d. telomerase A. Answer a is incorrect. A photolyase is the enzyme involved in repair of a thymine thymine dimer. B. Answer b is incorrect. DNA polymerase III is involved in the normal synthesis of DNA. endonuclease C. Answer c is correct. An endonuclease can break the phosphodiester bonds between the nucleotides in a damaged region of DNA. D. Answer d is incorrect. A telomerase is the enzyme that catalyzes the synthesis of DNA at the ends of eukaryotic chromosomes. Challenge Questions 1. The work by Griffith provided the first indication that DNA was the genetic material. Review the four experiments outline in Figure Predict the likely outcome for the following variations on this classic research. a. Heat-killed pathogenic and heat-killed nonpathogenic b. Heat-killed pathogenic and live nonpathogenic in the presence of an enzyme that digests proteins (proteases) c. Heat-killed pathogenic and live nonpathogenic in the presence of an enzyme that digests DNA (endonuclease)

8 Answer The key finding from the Griffith experiments is that heat-killed pathogenic bacteria can transform living, nonpathogenic bacteria. It was later discovered that the transforming factor was DNA. a. If both bacteria are heat-killed, then the transfer of DNA will have no effect since pathogenicity requires the production of proteins encoded by the DNA. Protein synthesis will not occur in a dead cell. b. The nonpathogenic cells will be transformed to pathogenic cells. Loss of proteins will not alter DNA. c. The nonpathogenic cells remain nonpathogenic. If the DNA is digested, it will not be transferred and no transformation will occur. 2. Imagine that you have identified the sequence 5 -TTATAAAGCAATAGT-3 in a eukaryotic chromosome. Could this region of the chromosome function as an origin of replication? Predict the sequence of an RNA primer that would be formed in association with this sequence. Answer The region could be an origin of replication. Origins of replication are adenine- and thymine-rich regions since only these nucleotides form two hydrogen bonds versus the three hydrogen bonds formed between guanine and cytosine, making it easier to separate the two strands of DNA. The RNA primer sequences would be 5 -ACUAUUGCUUUAUAA-3. The sequence is antiparallel to the DNA sequence (review Figure 14.16) meaning that the 5 end of the RNA is matching up with the 3 end of the DNA. It is also important to remember that in RNA the thymine nucleotide is replaced by uracil (U). Therefore, the adenine in DNA will form a complementary base-pair with uracil. 3. Enzyme function is critically important for the proper replication of DNA. Predict the consequence of a loss of function for each of the following enzymes. a. DNA gyrase b. DNA polymerase III c. DNA ligase d. DNA polymerase I Answer Consider the role of each enzyme in DNA replication to get an insight into what a loss of function might mean for the cell. a. DNA gyrase functions to relieve torsional strain on the DNA. If DNA gyrase were not functioning, the DNA molecule would undergo supercoiling, causing the DNA to wind up on itself, preventing the continued binding of the polymerases necessary for replication. b. DNA polymerase III is the primary polymerase involved in the addition of new nucleotides to the growing polymer and in the formation of the phosphodiester bonds that make up the sugar phosphate backbone. If this enzyme were not functioning, then no new DNA strand would be synthesized and there would be no replication. c. DNA ligase is involved in the formation of phosphodiester bonds between Okazaki fragments. If this enzyme was not functioning, then the fragments would remain disconnected and would be more susceptible to digestion by nucleases. d. DNA polymerase I functions to remove and replace the RNA primers that are required for DNA polymerase III function. If DNA polymerase I was not available, then the

9 RNA primers would remain and the replicated DNA would become a mix of DNA and RNA.