critical thinking questions dna replication

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Critical Thinking Questions

28 . Compare and contrast a human somatic cell to a human gamete.

29 . What is the relationship between a genome, chromosomes, and genes?

30 . Eukaryotic chromosomes are thousands of times longer than a typical cell. Explain how chromosomes can fit inside a eukaryotic nucleus.

31 . Briefly describe the events that occur in each phase of interphase.

32 . Chemotherapy drugs such as vincristine (derived from Madagascar periwinkle plants) and colchicine (derived from autumn crocus plants) disrupt mitosis by binding to tubulin (the subunit of microtubules) and interfering with microtubule assembly and disassembly. Exactly what mitotic structure is targeted by these drugs and what effect would that have on cell division?

33 . Describe the similarities and differences between the cytokinesis mechanisms found in animal cells versus those in plant cells.

34 . List some reasons why a cell that has just completed cytokinesis might enter the G 0 phase instead of the G 1 phase.

35 . What cell-cycle events will be affected in a cell that produces mutated (non-functional) cohesin protein?

36 . Describe the general conditions that must be met at each of the three main cell-cycle checkpoints.

37 . Compare and contrast the roles of the positive cell-cycle regulators negative regulators.

38 . What steps are necessary for Cdk to become fully active?

39 . Rb is a negative regulator that blocks the cell cycle at the G 1 checkpoint until the cell achieves a requisite size. What molecular mechanism does Rb employ to halt the cell cycle?

40 . Outline the steps that lead to a cell becoming cancerous.

41 . Explain the difference between a proto-oncogene and a tumor-suppressor gene.

42 . List the regulatory mechanisms that might be lost in a cell producing faulty p53.

43 . p53 can trigger apoptosis if certain cell-cycle events fail. How does this regulatory outcome benefit a multicellular organism?

44 . Name the common components of eukaryotic cell division and binary fission.

45 . Describe how the duplicated bacterial chromosomes are distributed into new daughter cells without the direction of the mitotic spindle.

Biology 2e for Biol 111 and Biol 112 Copyright © 2023 by Mary Ann Clark; Jung Choi; and Matthew Douglas is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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Critical Thinking Questions

• No, they cannot be identical because the T nucleotide in DNA is replaced with U nucleotide in RNA and AUG is the start codon.
• No, they cannot be identical because the T nucleotide in RNA is replaced with U nucleotide in DNA.
• They can be identical if methylation of the U nucleotide in RNA occurs and gives T nucleotide.
• They can be identical if de-methylation of the U nucleotide in RNA occurs and gives T nucleotide.
• 2 because the minimum length of an exon is 500 base pairs. In order to fit all 200 amino acids onto the minimum exon the maximum codon length is 2.5 (500 divided by 200). However codons length must be a whole number.
• 3 because by the law of degeneracy there is currently 20 times fewer amino acids than are possible as most codons are redundant. There could be up to 400 amino acids with the current codon length of 3.
• 4 because 4 to the 4th power is 256. 4 to the 3rd power is 64; not enough combinations
• 5 because 4 to the 4th power is only 256. This is not enough combinations because by the law of degeneracy every amino acid must have at least one redundant codon. With 5 codons 1,024 combinations is more than enough.
• The flow of information in HIV is from RNA to DNA, then back to RNA to proteins. Influenza viruses never go through DNA.
• The flow of information is from protein to RNA in HIV virus, while the influenza virus converts DNA to RNA.
• The flow of information is similar, but nucleic acids are synthesized as a result of translation in HIV and influenza viruses.
• The flow of information is from RNA to protein. This protein is used to synthesize the DNA of the viruses in HIV and influenza.

Suppose a gene has the sequence ATGCGTTATCGGGAGTAG. A point mutation changes the gene to read ATGCGTTATGGGGAGTAG. How would the polypeptide product of this gene change?

• In prokaryotes the polymerase is composed of five polypeptide subunits, two of which are identical. Four of these subunits, denoted α , α , β , and β ’, comprise the polymerase core enzyme. The fifth subunit, σ , is involved only in transcription initiation. The polymerase comprised of all five subunits is called the holoenzyme.
• In prokaryotes the polymerase is composed of four polypeptide subunits, two of which are identical. These subunits, denoted α , α , β , and β ’, comprise the polymerase core enzyme. There is a fifth subunit that is involved in translation initiation. The polymerase comprised of all four subunits is called the holoenzyme.
• In prokaryotes the polymerase is composed of five polypeptide subunits, two of which are identical. Four of these subunits, denoted α , α , β , and β ’, comprise the polymerase holoenzyme. The fifth subunit, σ , is involved only in transcription initiation. The polymerase comprised of all five subunits is called the core enzyme.
• In prokaryotes the polymerase is composed of five polypeptide subunits, two of which are identical. Four of these subunits, denoted α , α α, β , and β ’, comprise the polymerase core enzyme. The fifth subunit, σ , is involved only in termination. The polymerase comprised of all five subunits is called the holoenzyme.
• Rho-dependent termination is controlled by rho protein and the polymerase stalls near the end of the gene at a run of G nucleotides on the DNA template. In rho-independent termination, when the polymerase encounters a region rich in C-G nucleotides the mRNA folds into a hairpin loop that causes the polymerase to stall.
• Rho-independent termination is controlled by rho protein and the polymerase stalls near the end of the gene at a run of G nucleotides on the DNA template. In rho-dependent termination, when the polymerase encounters a region rich in C-G nucleotides, the mRNA folds into a hairpin loop that causes polymerase to stall.
• Rho-dependent termination is controlled by rho protein and the polymerase begins near the end of the gene at a run of G nucleotides on the DNA template. In rho-independent termination, when the polymerase encounters a region rich in C-G nucleotides, the mRNA creates a hairpin loop that causes polymerase to stall.
• Rho-dependent termination is controlled by rho protein and the polymerase stalls near the end of the gene at a run of G nucleotides on the DNA template. In rho-independent termination, when the polymerase encounters a region rich in A-T nucleotides, the mRNA creates a hairpin loop that causes polymerase to stall.
• Rho-independent termination involves the formation of a hairpin.
• Rho-dependent termination involves the formation of a hairpin.
• Rho-dependent termination stalls when the polymerase begins to transcribe a region rich in A-T nucleotides.
• Rho-independent termination stalls when the polymerase begins to transcribe a region rich in G nucleotides.
• The initiation step in eukaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different as histones are involved.
• The initiation step in prokaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different as histones are involved.
• The elongation step in eukaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different as histones are involved.
• The initiation step in eukaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different as histones are not involved.
• No, because they have the same α -amanitin sensitivity in all products.
• No, quantitative analysis of products is done to determine the type of polymerase.
• Yes, they can be determined as they differ in α -amanitin sensitivity.
• Yes, they can be determined by the number of molecules that bind to DNA.
• No, alternative splicing can lead to the synthesis of several proteins from a single gene.
• Yes, alternative splicing can lead to the synthesis of several forms of mRNA from a single gene, building more complex proteins.
• No, alternative splicing can lead to the synthesis of several forms of codons from a set of genes.
• Yes, alternative splicing can lead to the synthesis of several forms of ribosomes from a set of genes, but only one protein per gene.
• exporting the mRNA across the nuclear membrane
• importing the mRNA across the nuclear membrane
• the mRNA staying inside the nuclear membrane
• the mRNA translating into proteins within seconds
• The transcript would degrade when the mRNA moves out of the nucleus to the cytoplasm.
• The mRNA molecule would stabilize and start the process of translation within the nucleus of the cell.
• The mRNA molecule would move out of the nucleus and create more copies of the mRNA molecule.
• The mRNA molecule would not be able to add the poly-A tail on its strand at the 5’ end.
• The mRNA would be 5’-AUGGCCGGUUAUUAAGCA-3’ and the protein will be MAGY.
• The mRNA would be 3’-AUGGCCGGUUAUUAAGCA-5’ and the protein will be MAGY.
• The mRNA would be 5’-ATGGCCGGTTATTAAGCA-3’ and the protein will be MAGY.
• The mRNA would be 5’-AUGGCCGGUUAUUAAGCA-3’ and the protein will be MACY.
• rRNA has catalytic properties in the large subunit and it assembles proteins.
• rRNA is a protein molecule that helps in the synthesis of other proteins.
• rRNA is essential for the transcription process.
• rRNA plays a major role in post-translational processes.
• The anticodon will match the codon in mRNA.
• The anticodon will match with the modified amino acid it carries.
• The anticodon will lose the specificity for the tRNA molecule.
• The enzyme amino acyl tRNA synthetase would lose control over the amino acid.

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141 Critical Thinking Questions

21. If a purine were substituted for a pyrimidine at a single position in one strand of a DNA double helix, what would happen?

22. The DNA double helix looks like a twisted ladder. What makes up each rung of the ladder? What holds the rungs together at the sides?

23. Is there mostly empty space between the atoms in a DNA double helix?

24. In a DNA double helix, why doesn’t an A or T form two hydrogen bonds (out of the three possible) with G or C?

Biology Part I Copyright © 2022 by LOUIS: The Louisiana Library Network is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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Critical Thinking Questions

Explain Griffith’s transformation experiments. What did he conclude from them?

• 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.
• 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.
• 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.
• 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.

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

• 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.
• 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.
• 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.
• 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.

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

• 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.
• 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.
• 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.
• 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.

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?

Describe the structure and complementary base pairing of DNA.

• 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.
• 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.
• 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.
• 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.
• 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.
• 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.
• 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.
• 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.
• 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.
• 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.
• 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.
• 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.

DNA replication is bidirectional and discontinuous; explain your understanding of those concepts.

• 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.
• 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.
• 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.
• 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.

Discuss how the scientific community learned that DNA replication takes place in a semi- conservative fashion.

• 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 50 percent presence of 14 N . This supports the semiconservative replication model.
• 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 50 percent presence of 14 N . This supports the semiconservative replication model.
• 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 50 percent presence of 14 N . This supports the semiconservative replication model.
• 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 semiconservative replication model.

Explain why half of DNA is replicated in a discontinuous fashion.

• 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.
• 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.
• 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.
• 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.
• 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.
• 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.
• 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.
• 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.
• Okazaki fragments are short stretches of DNA on the lagging strand, which is synthesized in the direction away from the replication fork.
• Okazaki fragments are long stretches of DNA on the lagging strand, which is synthesized in the direction of the replication fork.
• Okazaki fragments are long stretches of DNA on the leading strand, which is synthesized in the direction away from the replication fork.
• Okazaki fragments are short stretches of DNA on the leading strand, which is synthesized in the direction of the replication fork.
• 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.
• 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.
• 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.
• 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.

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?

• 22.2 minutes
• 44.4 minutes
• 45.4 minutes
• 54.4 minutes
• The ends of the linear chromosomes are maintained by the activity of the telomerase enzyme.
• The ends of the linear chromosomes are maintained by the formation of a replication fork.
• The ends of the linear chromosomes are maintained by the continuous joining of Okazaki fragments.
• The ends of the linear chromosomes are maintained by the action of the polymerase enzyme.
• 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.
• 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.
• 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.
• 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.
• 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.
• 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.
• 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.
• 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.
• 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.
• 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.
• 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.
• 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.

Discuss the effects of point mutations on a DNA strand.

• 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.
• 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.
• 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.
• 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.
• Mutations in tRNA and rRNA would lead to the production of defective proteins or no protein production.
• Mutations in tRNA and rRNA would lead to changes in the semi-conservative mode of replication of DNA.
• Mutations in tRNA and rRNA would lead to production of a DNA strand with a mutated single strand and normal other strand.
• Mutations in tRNA and rRNA would lead to skin cancer in patients of xeroderma pigmentosa.

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