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10.01 What evidence suggested that the transforming principle in Griffith's experiments was really a gene or genes?
10.01 What evidence suggested that the transforming principle in Griffith's experiments was really a gene or genes?
Answer: When living R cells and extract of S cells were injected into a mouse, the mouse died. On examination of the dead mouse, living S cells were found. If these cells were isolated, they would only produce more S cells. This indicated that a permanent genetics change had taken place.
Answer: When living R cells and extract of S cells were injected into a mouse, the mouse died. On examination of the dead mouse, living S cells were found. If these cells were isolated, they would only produce more S cells. This indicated that a permanent genetics change had taken place.
10.02 Would 14C and 3H have been suitable radioactive tracers for the Hershey-Chase experiment Why, or why not?
10.02 Would 14C and 3H have been suitable radioactive tracers for the Hershey-Chase experiment Why, or why not?
Answer: No, these isotopes would not have been suitable tracers because both proteins and DNA would have been labeled at the same time. Both hydrogen and carbon are present in both DNA and protein.
Answer: No, these isotopes would not have been suitable tracers because both proteins and DNA would have been labeled at the same time. Both hydrogen and carbon are present in both DNA and protein.
10.03 Was the model of DNA proposed by Watson and Crick consistent or inconsistent with Chargaff's rules? Explain.
10.03 Was the model of DNA proposed by Watson and Crick consistent or inconsistent with Chargaff's rules? Explain.
Answer: The Watson-Crick model was consistent with Chargaff’s rules. Chargaff found that the quantities of adenine and thymine bases were the same in any particular DNA sample. Also the amounts of guanine and cytosine were the same. Watson and Crick’s model proposed that the As and Ts paired and the Gs and Cs paired.
Answer: The Watson-Crick model was consistent with Chargaff’s rules. Chargaff found that the quantities of adenine and thymine bases were the same in any particular DNA sample. Also the amounts of guanine and cytosine were the same. Watson and Crick’s model proposed that the As and Ts paired and the Gs and Cs paired.
10.04 A particular double-stranded DNA sample has a C+G ratio of 44%. What is the fraction of all four bases in this sample?
10.04 A particular double-stranded DNA sample has a C+G ratio of 44%. What is the fraction of all four bases in this sample?
Answer: Cytosine percentage must equal to Guanine. Adenine percentage must be equal to Thymine. This DNA is 22% cytosine, 22% guanine, 28% adenine, and 28% thymine.
Answer: Cytosine percentage must equal to Guanine. Adenine percentage must be equal to Thymine. This DNA is 22% cytosine, 22% guanine, 28% adenine, and 28% thymine.
10.05 An certain virus contains a single chromosome of double-stranded DNA with 120,000 base pairs. The percentage of cytosine bases is 18%.
10.05 An certain virus contains a single chromosome of double-stranded DNA with 120,000 base pairs. The percentage of cytosine bases is 18%.
(a) Approximately how many helical turns does each chromosome contain?
(a) Approximately how many helical turns does each chromosome contain?
Answer: Each turn is about 10 base pairs, so 120,000/10 = 12,000 turns
Answer: Each turn is about 10 base pairs, so 120,000/10 = 12,000 turns
(b) How long is the chromosome in micrometers?
(b) How long is the chromosome in micrometers?
Answer: The distance between base pairs is 0.34 nm, so (120,000 )(0.34 nm)(1 µm/1000 nm) = 40.8 µm
Answer: The distance between base pairs is 0.34 nm, so (120,000 )(0.34 nm)(1 µm/1000 nm) = 40.8 µm
(c) What is the approximate molecular weight of this DNA? (The average molecular of a nucleotide is approximately 330)
(c) What is the approximate molecular weight of this DNA? (The average molecular of a nucleotide is approximately 330)
Answer: Each base pair has an average molecular weight of about 660, so the approximate molecular weight is 120,000 x 660 = 79.2 x 106.
Answer: Each base pair has an average molecular weight of about 660, so the approximate molecular weight is 120,000 x 660 = 79.2 x 106.
(d) Approximately how many phosphorous atoms does a single chromosome contain?
(d) Approximately how many phosphorous atoms does a single chromosome contain?
Answer: Each nucleotide has one phosphorus atom, so 120,000 base pairs x 2 = 240,000 phosphorus atoms.
Answer: Each nucleotide has one phosphorus atom, so 120,000 base pairs x 2 = 240,000 phosphorus atoms.
(e) Approximately how many molecules of adenine, cytosine, guanine, and thymine would a single chromosome contain?
(e) Approximately how many molecules of adenine, cytosine, guanine, and thymine would a single chromosome contain?
Answer: The C and G percentages must be the same (18%). So 2 x 18 = 36% of the base pairs are C-G pairs. The rest at A-T pairs, 100 - 36 = 64%. The percentage of A and T must each be 1/2 of 64 (32%). The total number of bases is 120,000 x 2 = 240,000. So multiplying each percentage times this total give the following numbers of bases: adenine = 76,800, thymine = 76,800, guanine = 43,200, cytosine = 43,200.
Answer: The C and G percentages must be the same (18%). So 2 x 18 = 36% of the base pairs are C-G pairs. The rest at A-T pairs, 100 - 36 = 64%. The percentage of A and T must each be 1/2 of 64 (32%). The total number of bases is 120,000 x 2 = 240,000. So multiplying each percentage times this total give the following numbers of bases: adenine = 76,800, thymine = 76,800, guanine = 43,200, cytosine = 43,200.
10.06 Draw a short segment of DNA with at least 3 base pairs. Use the following abbreviations in your diagram: D=deoxyribose, P=phosphate, A=adenine, G=guanine, T=thymine, C=cytosine. You don't need to draw the complete structures of the nucleotides, but clearly show how the components listed above are bound together. Use solid lines for covalent bonds; dotted lines for hydrogen bonds. Also label the 3' and 5' ends of each nucleotide strand.
10.06 Draw a short segment of DNA with at least 3 base pairs. Use the following abbreviations in your diagram: D=deoxyribose, P=phosphate, A=adenine, G=guanine, T=thymine, C=cytosine. You don't need to draw the complete structures of the nucleotides, but clearly show how the components listed above are bound together. Use solid lines for covalent bonds; dotted lines for hydrogen bonds. Also label the 3' and 5' ends of each nucleotide strand.
Answer:
Answer:
10.07 The graphs below show the results of the analysis of two different samples of DNA (labelled A and B in the graphs). The left graph shows the change in absorbance of UV light as the DNA samples are slowly heated. The right graph shows how UV light absorbance drops when the DNA samples are first heated above their melting points and then allowed to cool. What can you conclude about theses two samples of DNA? Be sure to explain how the data support your conclusions.
10.07 The graphs below show the results of the analysis of two different samples of DNA (labelled A and B in the graphs). The left graph shows the change in absorbance of UV light as the DNA samples are slowly heated. The right graph shows how UV light absorbance drops when the DNA samples are first heated above their melting points and then allowed to cool. What can you conclude about theses two samples of DNA? Be sure to explain how the data support your conclusions.
Answer: From the first graph, we see that DNA sample B has the greater melting point. This would suggest that it has a greater percentage of C-G pairs than does sample A. G-C base pairs are held together by three hydrogen bonds; A-T pairs have only two hydrogen bonds. From the second graph, we see that sample A has the greater Cot1/2 value. It takes longer for the dissociated DNA strands to rejoin. This could mean that sample A is more complex (has more genes), or it might be that the DNA segments of sample B are shorter than those of sample A, making it easier for the rejoining strands to "find" complementary matches.
Answer: From the first graph, we see that DNA sample B has the greater melting point. This would suggest that it has a greater percentage of C-G pairs than does sample A. G-C base pairs are held together by three hydrogen bonds; A-T pairs have only two hydrogen bonds. From the second graph, we see that sample A has the greater Cot1/2 value. It takes longer for the dissociated DNA strands to rejoin. This could mean that sample A is more complex (has more genes), or it might be that the DNA segments of sample B are shorter than those of sample A, making it easier for the rejoining strands to "find" complementary matches.
10.08 Draw diagrams of the DNA molecules as hypothesized by the Meselson-Stahl experiment after three generations in 14N medium assuming: (a) conservative, and (b) semiconservative replication. Use solid lines for 15N-labeled DNA (old strands) and dotted lines for 14N-labeled DNA (new strands). Also diagram the number and positions of the bands of DNA as they would be expected to appear in the centrifuge tube that would arise from both possible replication methods.
10.08 Draw diagrams of the DNA molecules as hypothesized by the Meselson-Stahl experiment after three generations in 14N medium assuming: (a) conservative, and (b) semiconservative replication. Use solid lines for 15N-labeled DNA (old strands) and dotted lines for 14N-labeled DNA (new strands). Also diagram the number and positions of the bands of DNA as they would be expected to appear in the centrifuge tube that would arise from both possible replication methods.
Answer:
Answer:
10.09 In the DNA replication studies of Meselson and Stahl, DNA was labelled with nucleotides containing an isotope of nitrogen that is more dense than typical nitrogen. What if they had labelled the DNA with radioactive nucleotides instead. Would this have worked? Explain why or why not?
10.09 In the DNA replication studies of Meselson and Stahl, DNA was labelled with nucleotides containing an isotope of nitrogen that is more dense than typical nitrogen. What if they had labelled the DNA with radioactive nucleotides instead. Would this have worked? Explain why or why not?
Answer: Just based on the detecting Labeling the DNA strands with radioactive isotopes would not have worked using density gradient centrifugation unless the nucleotides also differed in density. However, since radioactive isotopes differ in number of neutrons, and neutrons contribute to the atomic mass of an atom, radioactive isotopes would have different densities. Another possibility is that a technique that detects radioactive emissions could have been used to distinguish between old and new strands.
Answer: Just based on the detecting Labeling the DNA strands with radioactive isotopes would not have worked using density gradient centrifugation unless the nucleotides also differed in density. However, since radioactive isotopes differ in number of neutrons, and neutrons contribute to the atomic mass of an atom, radioactive isotopes would have different densities. Another possibility is that a technique that detects radioactive emissions could have been used to distinguish between old and new strands.
10.10 Make a diagram showing how bidirectional DNA replication occurs. Mark the 5' and 3' ends of each strand of nucleotides. Also label: leading strand, lagging strand, Okasaki fragment, and RNA primer.
10.10 Make a diagram showing how bidirectional DNA replication occurs. Mark the 5' and 3' ends of each strand of nucleotides. Also label: leading strand, lagging strand, Okasaki fragment, and RNA primer.
Answer:
Answer:
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