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Critical Thinking Questions

Biology for AP® CoursesCritical Thinking Questions
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34 .
Explain Griffith’s transformation experiments. What did he conclude from them?
  1. Two strains of S. pneumoniae were used for the experiment. Griffith injected a mouse with heat- inactivated S strain (pathogenic) and R strain (non-pathogenic). The mouse died and S strain was recovered from the dead mouse. He concluded that external DNA is taken up by a cell that changed morphology and physiology.
  2. Two strains of Vibrio cholerae were used for the experiment. Griffith injected a mouse with heat-inactivated S strain (pathogenic) and R strain (non-pathogenic). The mouse died and S strain was recovered from the dead mouse. He concluded that external DNA is taken up by a cell that reversed the effects of heat inactivation.
  3. Two strains of S. pneumoniae were used for the experiment. Griffith injected a mouse with heat-inactivated S strain (pathogenic) and R strain (non-pathogenic). The mouse died and R strain was recovered from the dead mouse. He concluded that external DNA is taken up by a cell that changed morphology and physiology.
  4. Two strains of S. pneumoniae were used for the experiment. Griffith injected a mouse with heat-activated S strain (pathogenic) and R strain (non-pathogenic). The mouse died and S strain was recovered from the dead mouse. He concluded that mutation occurred in the DNA of the cell that changed morphology and physiology.
35 .
(credit: modification of work from The Regents of the University of California)

Hershey and Chase were running experiments to understand which molecule transfered genes. They were suspecting DNA and proteins as the possible molecules.

Hershey and Chase marked some viruses with radioactive sulfur and phosphorous. Then they let the viruses infect some bacteria. They then centrifuged the mixture. The heavier bacteria, also contiaining viral DNA, would sink to the bottom. The remaining parts of the viruses floated to the top. The graph shows what Hershey and Chase detected at the top.

Based on the graph, what can be said about DNA and proteins?

  1. Proteins contain higher amounts sulfur. DNA contains higher amounts of phosphorous.
  2. Proteins contain higher amounts phosphorous. DNA contains higher amounts of sulfur.
  3. Both proteins and DNA contain high amounts of sulfur.
  4. Both proteins and DNA contain high amounts of phosphorous.
36 .
How can Chargaff’s rules be used to identify different species?
  1. The base pairs are not found in equal ratios. The ratios vary between species, but do not vary by any significant amount among different individuals of the same species.
  2. The amount of adenine is equal to the amount of thymine. The amount of cytosine varies between species, but does not vary by any significant amount among different individuals of the same species.
  3. The base pairs are found in equal quantities. Different species have different molecular weight of DNA.
  4. the amount of guanine is equal to the amount of thymine. The amount of cytosine varies between species, but does not vary by any significant amount among different individuals of the same species.
37.

In the Avery, Macleod, and McCarty experiments, what conclusion would the scientists have drawn if the use of proteases prevented the transformation of R strain bacteria?

38 .
(credit: modification of work by Kostas Kampourakis/Cambridge University Press)
Identify the 3' and 5' ends of the given DNA strand.
  1. A and B are the 5' ends; C and D are the 3' ends.
  2. A, B and C are the 5' ends; D is the 3' end.
  3. A and D are the 5' ends; B and C are the 3' ends.
  4. A is the 5' end, B, C and D the 3' ends.
39 .
Provide a brief summary of the Sanger sequencing method.
  1. Frederick Sanger’s sequencing is a chain termination method that is used to generate DNA fragments that terminate at different points using dye-labeled dideoxynucleotides. DNA is separated by electrophoresis on the basis of size. The DNA sequence can be read out on an electropherogram generated by a laser scanner.
  2. Frederick Sanger’s sequencing is a chain elongation method that is used to generate DNA fragments that elongate at different points using dye-labeled dideoxynucleotides. DNA is separated by electrophoresis on the basis of size. The DNA sequence can be read out on an electropherogram generated by a laser scanner.
  3. Frederick Sanger’s sequencing is a chain termination method that is used to generate DNA fragments that terminate at different points using dye-labeled dideoxynucleotides. DNA is joined together by electrophoresis on the basis of size. The DNA sequence can be read out on an electropherogram generated by a laser scanner.
  4. Frederick Sanger’s sequencing is a chain termination method that is used to generate DNA fragments that terminate at different points using dye-labeled dideoxynucleotides. DNA is separated by electrophoresis on the basis of size. The DNA sequence can be read out on an electropherogram generated by a magnetic scanner.
40 .
Compare and contrast the similarities and differences between eukaryotic and prokaryotic DNA.
  1. Eukaryotes have a single, circular chromosome, while prokaryotes have multiple, linear chromosomes. Prokaryotes pack their chromosomes by super coiling, managed by DNA gyrase. Eukaryote chromosomes are wrapped around histone proteins that create heterochromatin and euchromatin, which is not present in prokaryotes.
  2. Prokaryotes have a single, circular chromosome, while eukaryotes have multiple, linear chromosomes. Prokaryotes pack their chromosomes by super coiling, managed by DNA gyrase. Eukaryote chromosomes are wrapped around histone proteins that could form heterochromatin, which is not present in prokaryotes.
  3. Prokaryotes have a single, circular chromosome, while eukaryotes have multiple, linear chromosomes. Eukaryotes pack their chromosomes by super coiling, managed by DNA gyrase. Prokaryotes chromosomes are wrapped around histone proteins that could form heterochromatin, which is not present in eukaryotes.
  4. Prokaryotes have a single, circular chromosome, while eukaryotes have multiple, linear chromosomes. Prokaryotes pack their chromosomes by super coiling, managed by DNA gyrase. Eukaryote chromosomes are wrapped around histone proteins that could form heterochromatin, which is present in prokaryotes.
41 .
DNA replication is bidirectional and discontinuous; explain your understanding of those concepts.
  1. DNA polymerase reads the template strand in the 3’ to 5’ direction and adds nucleotides only in the 5’ to 3’ direction. The leading strand is synthesized in the direction of the replication fork. Replication on the lagging strand occurs in the direction away from the replication fork in short stretches of DNA called Okazaki fragments.
  2. DNA polymerase reads the template strand in the 5’ to 3’ direction and adds nucleotides only in the 5’ to 3’ direction. The leading strand is synthesized in the direction of the replication fork. Replication on the lagging strand occurs in the direction away from the replication fork in short stretches of DNA called Okazaki fragments.
  3. DNA polymerase reads the template strand in the 3’ to 5’ direction and adds nucleotides only in the 5’ to 3’ direction. The leading strand is synthesized in the direction away from the replication fork. Replication on the lagging strand occurs in the direction of the replication fork in short stretches of DNA called Okazaki fragments.
  4. DNA polymerase reads the template strand in the 5’ to 3’ direction and adds nucleotides only in the 3’ to 5’ direction. The leading strand is synthesized in the direction of the replication fork. Replication on the lagging strand occurs in the direction away from the replication fork in long stretches of DNA called Okazaki fragments.
42 .
Discuss how the scientific community learned that DNA replication takes place in a semiconservative fashion.
  1. Meselson and Stahl experimented with E. coli. DNA grown in 15 N was heavier than DNA grown in 14 N . When DNA in 15 N was switched to 14 N media, DNA sedimented halfway between the 15 N and 14 N levels after one round of cell division, indicating fifty percent presence of 14 N . This supports the semi-conservative replication model.
  2. Meselson and Stahl experimented with S. pneumonia. DNA grown in 15 N was heavier than DNA grown in 14 N . When DNA in 15 N was switched to 14 N media, DNA sedimented halfway between the 15 N and 14 N levels after one round of cell division, indicating fifty percent presence of 14 N . This supports the semi-conservative replication model.
  3. Meselson and Stahl experimented with E. coli. DNA grown in 14 N was heavier than DNA grown in 15 N . When DNA in 15 N was switched to 14 N media, DNA sedimented halfway between the 15 N and 14 N levels after one round of cell division, indicating fifty percent presence of 14 N . This supports the semi-conservative replication model.
  4. Meselson and Stahl experimented with S. pneumonia. DNA grown in 15 N was heavier than DNA grown in 14 N . When DNA in 15 N was switched to 14 N media, DNA sedimented halfway between the 15 N and 14 N levels after one round of cell division, indicating complete presence of 14 N . This supports the semi-conservative replication model.
43 .
Explain why half of DNA is replicated in a discontinuous fashion.
  1. Replication of the lagging strand occurs in the direction away from the replication fork in short stretches of DNA, since access to the DNA is always from the 5’ end. This results in pieces of DNA being replicated in a discontinuous fashion.
  2. Replication of the leading strand occurs in the direction away from the replication fork in short stretches of DNA, since access to the DNA is always from the 5’ end. This results in pieces of DNA being replicated in a discontinuous fashion.
  3. Replication of the lagging strand occurs in the direction of the replication fork in short stretches of DNA, since access to the DNA is always from the 5’ end. This results in pieces of DNA being replicated in a discontinuous fashion.
  4. Replication of the lagging strand occurs in the direction away from the replication fork in short stretches of DNA, since access to the DNA is always from the 3’ end. This results in pieces of DNA being replicated in a discontinuous fashion.
44 .
(credit: modification of work by Alexander M Makhov/ResearchGate)

The image shows electron microscope images of the DNA replication process. The strings in the image are DNA strands. The blob-like shapes are enzymes. The two DNA strands marked with arrows are single, not double. The enzymes are splitting the DNA into two strands.

Based on this information, what type of enzyme is seen here?

  1. DNA polymerase.
  2. Helicase.
  3. DNA ligase.
  4. Topoisomerase.
45 .
What are Okazaki fragments and how they are formed?
  1. Okazaki fragments are short stretches of DNA on the lagging strand, which is synthesized in the direction away from the replication fork.
  2. Okazaki fragments are long stretches of DNA on the lagging strand, which is synthesized in the direction of the replication fork.
  3. Okazaki fragments are long stretches of DNA on the leading strand, which is synthesized in the direction away from the replication fork.
  4. Okazaki fragments are short stretches of DNA on the leading strand, which is synthesized in the direction of the replication fork.
46 .
Compare and contrast the roles of DNA polymerase I and DNA ligase in DNA replication.
  1. DNA polymerase I removes the RNA primers from the developing copy of DNA. DNA ligase seals the ends of the new segment, especially the Okazaki fragments.
  2. DNA polymerase I adds the RNA primers to the already developing copy of DNA. DNA ligase separates the ends of the new segment, especially the Okazaki fragments.
  3. DNA polymerase I seals the ends of the new segment, especially the Okazaki fragments. DNA ligase removes the RNA primers from the developing copy of DNA.
  4. DNA polymerase I removes the enzyme primase from the developing copy of DNA. DNA ligase seals the ends of the old segment, especially the Okazaki fragments.
47 .
If the rate of replication in a particular prokaryote is 900 nucleotides per second, how long would it take to replicate a 1.2 million base pair genome twice?
  1. 22.2 minutes
  2. 44.4 minutes
  3. 45.4 minutes
  4. 54.4 minutes
48 .
How do the linear chromosomes in eukaryotes ensure that their ends are replicated completely?
  1. The ends of the linear chromosomes are maintained by the activity of the telomerase enzyme.
  2. The ends of the linear chromosomes are maintained by the formation of a replication fork.
  3. The ends of the linear chromosomes are maintained by the continuous joining of Okazaki fragments.
  4. The ends of the linear chromosomes are maintained by the action of the polymerase enzyme.
49 .
Compare and contrast prokaryotic and eukaryotic DNA replication.
  1. A prokaryotic organism’s rate of replication is ten times faster than that of eukaryotes. Prokaryotes have a single origin of replication and use five types of polymerases, while eukaryotes have multiple sites of origin and use fourteen polymerases. Telomerase is absent in prokaryotes. DNA pol I is the primer remover in prokaryotes, while in eukaryotes it is RNase H. DNA pol III performs strand elongation in prokaryotes and pol δ and pol ε do the same in eukaryotes.
  2. A prokaryotic organism’s rate of replication is ten times slower than that of eukaryotes. Prokaryotes have a single origin of replication and use five types of polymerases, while eukaryotes have multiple sites of origin and use fourteen polymerases. Telomerase is absent in eukaryotes. DNA pol I is the primer remover in prokaryotes, while in eukaryotes it is RNase H. DNA pol III performs strand elongation in prokaryotes and pol δ and pol ε do the same in eukaryotes.
  3. A prokaryotic organism’s rate of replication is ten times faster than that of eukaryotes. Prokaryotes have five origins of replication and use a single type of polymerase, while eukaryotes have a single site of origin and use fourteen polymerases. Telomerase is absent in prokaryotes. DNA pol I is the primer remover in prokaryotes, while in eukaryotes it is RNase H. DNA pol III performs strand elongation in prokaryotes and pol δ and pol ε do the same in eukaryotes.
  4. A prokaryotic organism’s rate of replication is ten times slower than that of eukaryotes. Prokaryotes have a single origin of replication and use five types of polymerases, while eukaryotes have multiple sites of origin and use fourteen polymerases. Telomerase is absent in prokaryotes. DNA pol I is the primer remover in eukaryotes, while in prokaryotes it is RNase H. DNA pol III performs strand elongation in prokaryotes and pol δ and pol ε do the same in eukaryotes.
50 .
(credit: modification of work by Subhadip Mondal/Biology Discussion)

The table shows which codons code for which amino acids in the DNA.

Due to a mutation, the nucleotide sequence CCT-TAT-GGA becomes CCT-TAA-GGA on a gene.

What type of mutation would this be?

  1. Silent mutation.
  2. Missense mutation.
  3. Nonesense mutation.
  4. Frameshift mutation.
51 .
A mutation has occurred in the DNA and in the mRNA for a gene. Discuss which would have a more significant effect on gene expression. Why?
  1. Both will result in the production of defective proteins. The DNA mutation, if not corrected, is permanent, while the mRNA mutation will only affect proteins made from that mRNA strand. Production of defective protein ceases when the mRNA strand deteriorates.
  2. Both will result in the production of defective proteins. The DNA mutation, if not corrected, is permanent, while the mRNA mutation will not affect proteins made from that mRNA strand. Production of defective protein continues when the mRNA strand deteriorates.
  3. Only DNA will result in the production of defective proteins. The DNA mutation, if not corrected, is permanent. Production of defective protein ceases when the DNA strand deteriorates.
  4. Only mRNA will result in the production of defective proteins. The mRNA mutation will only affect proteins made from that mRNA strand. Production of defective protein ceases when the mRNA strand deteriorates.
52 .
Discuss the effects of point mutations on a DNA strand.
  1. Mutations can cause a single change in an amino acid. A nonsense mutation can stop the replication or reading of that strand. Insertion or deletion mutations can cause a frame shift. This can result in non-functional proteins.
  2. Mutations can cause a single change in amino acid. A missense mutation can stop the replication or reading of that strand. Insertion or deletion mutations can cause a frame shift. This can result in non-functional proteins.
  3. Mutations can cause a single change in amino acid. A nonsense mutation can stop the replication or reading of that strand. Substitution mutations can cause a frame shift. This can result in non-functional proteins.
  4. Mutations can cause a single change in amino acid. A nonsense mutation can stop the replication or reading of that strand. Insertion or deletion mutations can cause a frame shift. This can result in functional proteins.
53 .
Discuss the significance of mutations in tRNA and rRNA.
  1. Mutations in tRNA and rRNA would lead to the production of defective proteins or no protein production.
  2. Mutations in tRNA and rRNA would lead to changes in the semi-conservative mode of replication of DNA.
  3. Mutations in tRNA and rRNA would lead to production of a DNA strand with a mutated single strand and normal other strand.
  4. Mutations in tRNA and rRNA would lead to skin cancer in patients of xeroderma pigmentosa.
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