# Critical Thinking Questions

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 changed morphology and physiology.
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-inactivated 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.

Explain why radioactive sulfur and phosphorous were used to label bacteriophages in the Hershey and Chase experiments.

1. Protein was labeled with radioactive sulfur and DNA was labeled with radioactive phosphorous. Phosphorous is found in DNA, so it will be tagged by radioactive phosphorous.
2. Protein was labeled with radioactive phosphorous and DNA was labeled with radioactive sulfur. Phosphorous is found in DNA, so it will be tagged by radioactive phosphorous.
3. Protein was labeled with radioactive sulfur and DNA was labeled with radioactive phosphorous. Phosphorous is found in DNA, so DNA will be tagged by radioactive sulfur.
4. Protein was labeled with radioactive phosphorous and DNA was labeled with radioactive sulfur. Phosphorous is found in DNA, so DNA will be tagged by radioactive sulfur.
36.

How can Chargaff’s rules be used to identify different species?

1. The amount of adenine, thymine, guanine, and cytosine varies from species to species and is not found in equal quantities. They do not vary between individuals of the same species and can be used to identify different species.
2. The amount of adenine, thymine, guanine, and cytosine varies from species to species and is found in equal quantities. They do not vary between individuals of the same species and can be used to identify different species.
3. The amount of adenine and thymine is equal to guanine and cytosine and is found in equal quantities. They do not vary between individuals of the same species and can be used to identify different species.
4. The amount of adenine, thymine, guanine, and cytosine varies from species to species and is not found in equal quantities. They vary between individuals of the same species and can be used to identify different species.
37.

In the Hershey-Chase experiments, what conclusion would the scientists have drawn if bacteria containing both radioactive phosphorus and sulfur were found in the final pellets?

38.

Describe the structure and complementary base pairing of DNA.

1. DNA is made up of two strands that are twisted around each other to form a helix. Adenine pairs up with thymine and cytosine pairs with guanine. The two strands are anti-parallel in nature; that is, the 3' end of one strand faces the 5' end of other strand. Sugar, phosphate and nitrogenous bases contribute to the DNA structure.
2. DNA is made up of two strands that are twisted around each other to form a helix. Adenine pairs up with cytosine and thymine pairs with guanine. The two strands are anti-parallel in nature; that is, the 3' end of one strand faces the 5' end of other strand. Sugar, phosphate and nitrogenous bases contribute to the DNA structure.
3. DNA is made up of two strands that are twisted around each other to form a helix. Adenine pairs up with thymine and cytosine pairs with guanine. The two strands are parallel in nature; that is, the 3' end of one strand faces the 3' end of other strand. Sugar, phosphate and nitrogenous bases contribute to the DNA structure.
4. DNA is made up of two strands that are twisted around each other to form a helix. Adenine pairs up with thymine and cytosine pairs with guanine. The two strands are anti-parallel in nature; that is, the 3' end of one strand faces the 5' end of other strand. Only sugar contributes to the DNA structure.
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 create heterochromatin and euchromatin, 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 create heterochromatin and euchromatin, 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 create heterochromatin and euchromatin, 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 semi- conservative fashion.

1. Meselson and Stahl experimented with E. coli. DNA grown in 15N was heavier than DNA grown in 14N . When DNA in 15N was switched to 14N media, DNA sedimented halfway between the 15N and 14N levels after one round of cell division, indicating 50 percent presence of 14N . This supports the semiconservative replication model.
2. Meselson and Stahl experimented with S. pneumonia. DNA grown in 15N was heavier than DNA grown in 14N . When DNA in 15N was switched to 14N media, DNA sedimented halfway between the 15N and 14N levels after one round of cell division, indicating 50 percent presence of 14N . This supports the semiconservative replication model.
3. Meselson and Stahl experimented with E. coli. DNA grown in 14N was heavier than DNA grown in 15N . When DNA in 15N was switched to 14N media, DNA sedimented halfway between the 15N and 14N levels after one round of cell division, indicating 50 percent presence of 14N . This supports the semiconservative replication model.
4. Meselson and Stahl experimented with S. pneumonia. DNA grown in 15N was heavier than DNA grown in 14N . When DNA in 15N was switched to 14N media, DNA sedimented halfway between the 15N and 14N levels after one round of cell division, indicating complete presence of 14N . This supports the semiconservative 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.
Explain the events taking place at the replication fork. If the gene for helicase is mutated, what part of replication will be affected?
1. Helicase separates the DNA strands at the origin of replication. Topoisomerase breaks and reforms DNA’s phosphate backbone ahead of the replication fork, thereby relieving the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes RNA primer which is used by DNA polymerase to form a daughter strand. If helicase is mutated, the DNA strands will not be separated at the beginning of replication.
2. Helicase joins the DNA strands together at the origin of replication. Topoisomerase breaks and reforms DNA’s phosphate backbone after the replication fork, thereby relieving the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes RNA primer which is used by DNA polymerase to form a daughter strand. If helicase is mutated, the DNA strands will not be joined together at the beginning of replication.
3. Helicase separates the DNA strands at the origin of replication. Topoisomerase breaks and reforms DNA’s sugar backbone ahead of the replication fork, thereby increasing the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes DNA primer which is used by DNA polymerase to form a daughter strand. If helicase is mutated, the DNA strands will be separated at the beginning of replication.
4. Helicase separates the DNA strands at the origin of replication. Topoisomerase breaks and reforms DNA’s sugar backbone ahead of the replication fork, thereby relieving the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes DNA primer which is used by RNA polymerase to form a parent strand. If helicase is mutated, the DNA strands will be separated at the beginning of replication.
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 make two copies of a 1.2 million base pair genome?

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.
What would be the consequence of a mutation in a mismatch repair enzyme? How will this affect the function of a gene?
1. Mismatch repair corrects the errors after the replication is completed by excising the incorrectly added nucleotide and adding the correct base. Any mutation in a mismatch repair enzyme would lead to more permanent damage.
2. Mismatch repair corrects the errors during the replication by excising the incorrectly added nucleotide and adding the correct base. Any mutation in the mismatch repair enzyme would lead to more permanent damage.
3. Mismatch repair corrects the errors after the replication is completed by excising the added nucleotides and adding more bases. Any mutation in the mismatch repair enzyme would lead to more permanent damage.
4. Mismatch repair corrects the errors after the replication is completed by excising the incorrectly added nucleotide and adding the correct base. Any mutation in the mismatch repair enzyme would lead to more temporary damage.
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 nonfunctional 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 nonfunctional 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 nonfunctional 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.