AP Bio unit 6 test

Which of the following samples most likely contains a double-stranded RNA virus? A Sample 1 B Sample 2 C Sample 3 D Sample 4

A

Which of the following statements best explains the data set? A Since the %A and the %G add up to approximately 50 percent in each sample, adenine and guanine molecules must pair up in a double-stranded DNA molecule. B Since the %A and the %T are approximately the same in each sample, adenine and thymine molecules must pair up in a double-stranded DNA molecule. C Since the %(A+T) is greater than the %(G+C) in each sample, DNA molecules must have a poly-A tail at one end. D Since the %C and the %T add up to approximately 50 percent in each sample, cytosine and thymine molecules must both contain a single ring.

B

The features of this model provide evidence for which explanation of why all growing strands are synthesized in a 5′ to 3′ direction? A The two strands need to be antiparallel to bond properly. B Thymine and adenine would not bond properly if the strand grew from 3′ to 5′. C The translation of mRNA occurs in the 5′ to 3′ direction; therefore, the growing DNA strand must also grow in the 5′ to 3′ direction. D The phosphate group, attached to the 5′ carbon of the dTMP, forms a covalent bond with the oxygen atom attached to the 3′ carbon of the growing strand.

D

Which of the following best explains the process represented by Figure 1? A The synthesis of mRNA in the 5' to 3′ direction from DNA B The modification of a protein to produce a functional form of that protein C The translation of an mRNA molecule into a polypeptide D The enzyme-regulated processing of pre‑mRNA into mature mRNA

D

Which of the following best supports the claim that binding of miRNA‑delta to the miRNA binding site inhibits translation of gene Q mRNA? A When the promoter for gene Q is altered, transcription is inhibited. B Translation of Q mRNA is inhibited regardless of whether the miRNA binding site sequence is altered. C Translation of Q mRNA is inhibited in the absence of miRNA‑delta. D When the miRNA binding site sequence is altered, translation of Q mRNA occurs in the presence of miRNA-delta.

D

A cell needs to metabolize the substrate illustrated in Figure 1 for a vital cellular function. Which of the following best explains the long-term effect on the cell of splicing that yields only enzyme C mRNA? A The cell will die because it is unable to metabolize the substrate without enzyme A, which is structurally specific for the substrate shown. B The cell will remain healthy because all three of the above enzymes can metabolize the substrate, as they are from the same gene. C The cell will remain healthy because the enzyme C mRNA will undergo alternative splicing again until it transformed into enzyme AA mRNA. D The cell will remain healthy because enzyme-substrate interactions are nonspecific and enzyme C will eventually metabolize the substrate.

A

Based on the codon chart above, which of the following amino acid changes is most likely found in the mutated protein? A Glu→Val B Val→Glu C Glu→Pro D Pro→Val

A

Based on the information in Table 1, which of the following best predicts a key difference in prokaryotes and eukaryotes with regard to transcription and translation? A The two processes will occur simultaneously in prokaryotes but not eukaryotes. B Prokaryotic mRNA is shorter than eukaryotic mRNA. C Eukaryotic mRNA contains more coding regions than prokaryotic DNA. D The processing of mRNA by eukaryotes is required for the mRNA to leave the nucleus.

A

Researchers claim that bacteria that live in environments heavily contaminated with arsenic are more efficient at processing arsenic into arsenite and removing this toxin from their cells. Justify this claim based on the evidence shown in Figure 1. A There are multiple operons controlling the production of proteins that process and remove arsenite from cells in both H. arsenicoxydans and O. tritici. In contrast, E. coli has only one operon devoted to arsenic removal. B Both H. arsenicoxydans and O. tritici contain the arsR gene that codes for a repressor that turns on the operon to eliminate arsenite from the cell. C Both O. tritici and E. coli contain the arsD gene, which codes for a protein that helps remove arsenite from the cell. D Both H. arsenicoxydans and O. tritici. have more arsenic resistance genes than has E. coli.

A

The claim that gene regulation results in differential gene expression and influences cellular products (albumin or crystalline) is best supported by evidence in which of the following statements? A Liver cells possess transcriptional activators that are different from those of lens cells. B Liver cells and lens cells use different RNA polymerase enzymes to transcribe DNA. C Liver cells and lens cells possess the same transcriptional activators. D Liver cells and lens cells possess different general transcription factors.

A

Use the response models shown in Figures 1 and 2 to justify the claim that phytochromes regulate the transcription of genes leading to the production of certain cellular proteins. A When inactive phytochrome Pr is activated by red light to become phytochrome Pfr, it is transported into the nucleus where it binds to the transcription factor PIF3 at the promoter. This stimulates transcription, ultimately leading to protein production. Far-red light inactivates the phytochrome, which will turn transcription off by not binding to PIF3. B Far-red light activates phytochrome Pr, causing it to travel to the nucleus where it binds to PIF3 at the promoter. This stimulates transcription, ultimately leading to protein production. Red light inactivates the phytochrome, which will turn transcription off by not binding to PIF3. C MYB, and not Pfr, is activated by red light, causing it to bind to the promoter and stimulate transcription and translation of cellular proteins. D PIF3 binds to the promoter only in the presence of red light and Pfr. Any time PIF3 is bound to the promoter, MYB is transcribed, initiating transcription of various other proteins in the cell.

A

Which of the following best explains how bacteria can be genetically engineered to produce a desired antigen? A The gene coding for the antigen can be inserted into plasmids that can be used to transform the bacteria. B The bacteria need to be exposed to the antigen so they can produce the antibodies. C The DNA of the antigen has to be transcribed in order for the mRNAmRNA produced to be inserted into the bacteria. D The mRNA of the antigen has to be translated in order for the protein to be inserted into the bacteria

A

Which of the following best explains how continuity of genetic information in cells is ensured across generations? A Replication uses a parental strand of DNA as a template to create a new strand of DNA. B DNA molecules are shaped like a double helix with a constant diameter throughout. C Transcription copies the information in DNA into an RNA transcript. D Cells contain different polymerases for DNA replication and transcription.

A

Which of the following best explains how the prokaryotic expression of a metabolic protein can be regulated when the protein is already present at a high concentration? A Repressor proteins can be activated and bind to regulatory sequences to block transcription. B Transcription factors can bind to regulatory sequences to increase RNA polymerase binding. C Regulatory proteins can be inactivated to increase gene expression. D Histone modification can prevent transcription of the gene

A

Which of the following best explains why ligase is required for DNA replication? A The lagging strand cannot be replicated continuously, and ligase is needed to join the fragments. B Ligase forms the hydrogen bonds between complementary bases in the two strands of DNA. C Ligase facilitates the binding of RNA polymerase to the promoter region. D Ligase enables the newly synthesized DNA to twist into a double helix.

A

Which of the following pieces of evidence would best support the researchers' claim above? A When researchers applied a drug that activates adenylyl cyclase to the mutant mice's ears, the level of melanin increased. B When researchers viewed sections of mutant mouse ears under the microscope, they found melanocyte numbers comparable to nonmutant mice. C When researchers exposed the mutant mice to UV radiation, the amount of POMC mRNA in keratinocytes did not change. D When researchers exposed the mutant mice to UV radiation, the level of melanin production did not change.

A

A woman develops Huntington's disease. Her father had the disorder. Her mother did not, and there is no history of the disorder in the mother's family. Which of the following best explains how the woman inherited Huntington's disease? A She inherited an allele with fewer than 40 CAG repeats in the HTT gene because her mother did not have Huntington's disease. B She inherited an allele with more than 40 CAG repeats in the HTT gene from her father. C Her mother produced eggs that all have more than 40 repeats in the HTT gene. D Her mother produced eggs that all have fewer than 40 CAG repeats in the HTT gene.

B

Figure 1. Replicated DNA produced after a chemical is introduced Which of the following claims is best supported by the data? A The chemical prevents the formation of RNA primers. B The chemical inhibits DNA ligase. C The chemical blocks DNA polymerase. D The chemical disrupts hydrogen bonding.

B

The human TPM1 gene encodes members of the tropomyosin family of cytoskeletal proteins. Which of the following best explains how different proteins can be made in different cell types from the one TPM1 gene? A Different introns are selectively converted to exons. B Different exons are retained or spliced out of the primary transcript. C The GTP cap is selectively added to and activates different exons. D Different portions of the primary transcript remain bound to the template DNA.

B

Which claim is most consistent with the information provided by the diagram and current scientific understanding of gene regulation and expression? A Reversible changes in the DNA sequence may influence how a gene is expressed in a cell. B Some sequences of DNA can interact with regulatory proteins that control transcription. C This is an inducible operon controlled by several regulatory factors. D The transcription factor may produce mutations in the binding site at the promoter sequence inhibiting the synthesis of the protein.

B

Which is a scientific claim that is consistent with the information provided and Figure 1 ? A The presence of excess lactose blocks the functioning of RNA polymerase in this operon. B When bound to the operator, the repressor protein prevents lactose metabolism in E. coli. C The binding of the repressor protein to the operator enables E. coli to metabolize lactose. D Allolactose acts as an inducer that binds to the operator, allowing E. coli to metabolize lactose.

B

Which of the following best describes the most likely impact on an individual produced from fertilization between one of the daughter cells shown and a normal gamete? A Because nondisjunction occurred in anaphase I, all gametes will be normal and the resulting individual will be phenotypically normal. B Because nondisjunction occurred in anaphase I, all gametes will have an abnormal chromosome number and the individual will likely exhibit phenotypic evidence of the nondisjunction event. C Because nondisjunction occurred in anaphase II, all gametes will be normal and the resulting individual will be phenotypically normal. D Because nondisjunction occurred in anaphase II, all gametes will have an abnormal chromosome number and the individual will likely exhibit phenotypic evidence of the nondisjunction event.

B

Which of the following best explains how the results should be interpreted? A The individual does not have an increased risk of developing cancer because one dominant allele is insufficient to cause the disease. B The individual has an increased risk of developing colon cancer. C Because the person's DNA has the mutation, other family members must have cancer. D Results cannot be interpreted until testing determines if additional mutated alleles are present.

B

Which of the following best explains what process is represented in Figure 1 ? A New DNA strands are being synthesized in the 3' to 5' direction from their DNA templates. B New DNA strands are being synthesized in the 5' to 3' direction from their DNA templates. C A new RNA strand is being synthesized in the 3' to 5' end from its DNA template. D Two new RNA strands are being synthesized in both directions from their DNA templates.

B

Which of the following claims about the TYR, TRP2, and TRP1 mammalian genes is most likely to be accurate? A The TYR, TRP2, and TRP1 genes are located next to each other on a single chromosome and are organized into an operon. B The TYR, TRP2, and TRP1 genes may be located on different chromosomes but are activated by the same transcription factor. C The TYR, TRP2, and TRP1 genes are identical genes since they are activated by the same transcription factor. D The TYR, TRP2, and TRP1 genes may be located on different chromosomes but with identical operator sequences.

B

Which of the following correctly explains the process shown in Figure 1 ? A DNA replication is occurring because replication is semi-conservative and the new strand is a copy of the template strand. B Initiation of transcription is occurring because a strand of RNA is being produced from a DNA template strand. C Translation is occurring because the two strands have separated and a new strand is being produced. D Alternative splicing of mRNA is occurring because the mRNA strand is being synthesized from only one strand of DNA.

B

Which of the following is the most likely effect of a mutation in the gene coding for a DNA repair enzyme? A The cell containing the mutation will divide more frequently because the cell cycle checkpoints will not function properly. B Mutations will accumulate more quickly because the cell will not be able to fix errors in replication. C The mutated gene will not be transcribed because RNA polymerase cannot transcribe mutated DNA. D The cell will immediately undergo apoptosis so that mutated DNA is not replicated in future rounds of cell division.

B

Which of the following most likely explains how the chromosomes circled in Figure 1 could cause a genetic disorder in the person from whom the cells were obtained? A The extra chromosome causes crowding in the nucleus of the cells and blocks RNA polymerase from binding to and transcribing certain genes. B The extra chromosome will affect the levels of RNA transcribed from certain genes and the amount of protein produced from those genes in each cell. C The cells will not divide and enable growth, because the extra chromosome will interfere with the pairing of homologous chromosomes. D The extra chromosome will cause other chromosomes in the cell to become triploid during future rounds of cell division.

B

Which of the following statements best explains how the genes for antibiotic resistance can be transmitted between bacteria without the exchange of bacterial chromosomal DNA? A The antibiotic-resistant bacteria release a hormone that signals neighboring bacteria to become resistant. B The genes for antibiotic resistance are located on a plasmid that can be passed to neighboring bacteria. C The antibiotic-resistant bacteria are the result of bacteria that specifically modify their own chromosomal DNA to neutralize the antibiotics. D The antibiotic alters the bacterial genome of each bacterium, which results in an antibiotic-resistant population.

B

Which of the following statements best explains the structure and importance of plasmids to prokaryotes? A Plasmids are circular, single-stranded RNA molecules that transfer information from the prokaryotic chromosome to the ribosomes during protein synthesis. B Plasmids are circular, double-stranded DNA molecules that provide genes that may aid in survival of the prokaryotic cell. C Plasmids are single-stranded DNA molecules, which are replicated from the prokaryotic chromosome, that prevent viral reproduction within the prokaryotic cell. D Plasmids are double-stranded RNA molecules that are transmitted by conjugation that enable other prokaryotic cells to acquire useful genes.

B

A The reverse transcriptase will cut the host DNA into fragments, destroying the host cell. B The reverse transcriptase will insert the viral RNA into the host's genome so it can be transcribed and translated. C The reverse transcriptase will produce DNA from the viral RNA, which can be incorporated into the host's genome and then transcribed and translated. D The reverse transcriptase will force the host ribosomes to translate the viral RNA prior to polypeptide assembly.

C

Based on the information in Figure 1, which type of mutation explains the nature of the change in DNA that resulted in cystic fibrosis in the affected individual? A Substitution, because the amino acid tryptophan is replaced with glycine. B Insertion, because an extra guanine is present, which changes the reading frame. C Deletion, because a thymine is missing, which changes the reading frame. D Duplication, because the amino acid leucine occurs twice, which changes the reading frame.

C

Based on the results, which of the following best describes what Protein X is? A Protein X is an RNA splicing enzyme. B Protein X is a cell membrane receptor protein. C Protein X is a transcription factor. D Protein X is a hormone.

C

Figure 1. Protein synthesis in a prokaryotic cell Which of the following best describes a characteristic of the process shown in Figure 1 that is unique to prokaryotes? A The mRNA is synthesized in a 5' to 3' direction. B A single strand of the DNA is being used as a template for the transcription of the mRNA. C The translation of the mRNA is occurring while the mRNA is still being transcribed. D The enzyme that is transcribing the mRNA is RNA polymerase.

C

Given the results shown in Figure 1, which of the following correctly describes a relationship between the two species? A Species B is the ancestor of species A because it has fewer bands. B Species A is more complex than species B because it has more bands. C Species B has more short fragments of DNA than species A does. D Species A has more short fragments of DNA than species B does.

C

Nucleotide base pairing in DNA is universal across organisms. Each pair (T−A; C−G) consists of a purine and a pyrimidine. Which of the following best explains how the base pairs form? A Ionic bonds join two double-ringed structures in each pair. B Hydrogen bonds join two single-ringed structures in each pair. C Hydrogen bonds join a double-ringed structure to a single-ringed structure in each pair. D Covalent bonds join a double-ringed structure to a single-ringed structure in each pair.

C

Oncogenes are genes that can cause tumor formation as a result of a particular mutation. Which of the following potential therapies would be most effective at preventing the expression of an oncogene? A Reducing the number of ribosomes in the cell to prevent the creation of the oncogene's proteins B Blocking membrane-bound receptors of transcription factors C Introducing a chemical that binds to transcription factors associated with the oncogene's promoter D Producing additional transcription factors for tumor suppressor genes in the cell

C

Which of the following best describes the type of mutation shown in Figure 1 that leads to sickle-cell anemia? A Insertion B Deletion C Substitution D Frameshift

C

Which of the following best explains how some cells of an individual produce and secrete a specific enzyme, but other cells of the same individual do not? A The cells contain different genes and therefore do not make the same proteins. B The cells have evolved under different selective pressures, resulting in some cells making proteins that others cannot. C The cells transcribe and translate different combinations of genes, leading to the production of different sets of proteins. D The cells produce different types of ribosomes that enable the translation of different genes.

C

Which of the following best explains how the expression of a eukaryotic gene encoding a protein will differ if the gene is expressed in a prokaryotic cell instead of in a eukaryotic cell? A No transcript will be made, because eukaryotic DNA cannot be transcribed by prokaryotic RNA polymerase. B The protein will have a different sequence of amino acids, because prokaryotes use a different genetic code. C The protein will be made but will not function, because prokaryotes cannot remove introns. D The protein will not be made, because prokaryotes lack the ribosomes necessary for translation.

C

Which of the following best explains how this model illustrates DNA replication of both strands as a replication fork moves? A I and IV are synthesized continuously in the 5′5′ to 3′3′ direction. B II and III are synthesized in segments in the 3'3′ to 5'5′ direction. C I is synthesized continuously in the 5′5′ to 3′3′ direction, and III is synthesized in segments in the 5′5′ to 3′3′ direction. D II is synthesized in segments after DNA polymerase is released from synthesizing strand IV.

C

Which of the following best explains the distribution of lactase persistence in the areas shown in Figure 1 ? A Lactase persistence developed because people were malnourished in Europe. B Lactase persistence alleles are present in all human populations and are expressed when lactose is consumed. C Mutations conferring lactase persistence likely arose independently in different geographic areas and offered a selective advantage. D The mutations that cause lactase persistence are detrimental to humans and will eventually disappear from the gene pool.

C

Which of the following best explains what strand X represents? A A complementary RNA sequence, because it contains thymine B The coding strand in this process, because it is being read 3′ to 5′ C The antisense strand, because it is serving as a template D The pre‑mRNA, because it does not yet have a GTP cap

C

Which of the following claims best explains why keratinocytes do not produce melanin? A Keratinocytes do not contain the TYR, TRP2, and TRP1 genes. B Keratinocytes do not contain the MC1R gene. C Keratinocytes do not express the MITF gene. D Keratinocytes do not express the POMC gene.

C

Which of the following correctly explains where DNA replication will begin on the strand oriented 5'→3', reading from left to right? A DNA replication will be randomly initiated along the unwound portion of the DNA strand since base pairing will occur. B DNA replication cannot occur since there is already RNA base pairing with the template strand. C DNA replication will be initiated immediately to the left of the RNA, since DNA polymerase requires an RNA primer. D DNA replication will be initiated at the site of the topoisomerase since that is where DNA begins to uncoil.

C

Which of the following describes the most direct effect of a mutation in the DNA that encodes a cell's rRNA? A The cell's ability to transport the amino acids needed for translation will be reduced. B The cell's ability to transcribe RNA transcripts that will be translated will be reduced. C The cell's ability to properly assemble ribosomes and initiate translation will be reduced. D The cell's ability to modify proteins after they have been assembled will be reduced.

C

Which of the following mutations is most likely to cause lactase persistence in humans? A A nucleotide substitution in the coding region of the lactase gene that interferes with the interaction between lactase and lactose B A mutation that turns off the expression of transcription factors that activate the expression of lactase C A mutation that increases the binding of transcription factors to the promoter of the lactase gene D The insertion of a single nucleotide into the lactase gene that results in the formation of a stop codon

C

Which of the following statements best explains the experimental results observed in Figure 1? A E. coli in both lanes B and C have been successfully transformed and contain additional genetic information. B E. coli in lane B have been successfully transformed and contain additional genetic information. C E. coli in lane C have been successfully transformed and contain additional genetic information. D Which E. coli have been transformed cannot be determined from this gel.

C

Which of the following statements best explains the pattern seen on the gel with regard to the size and charge of molecules A and B? A Molecules A and B are positively charged, and molecule A is smaller than molecule B. B Molecules A and B are positively charged, and molecule A is larger than molecule B. C Molecules A and B are negatively charged, and molecule A is smaller than molecule B. D Molecules A and B are negatively charged, and molecule A is larger than molecule B.

C

Which of the following statements provides the best explanation of the processes illustrated in Figure 1 ? A Introns are removed from the pre-rRNA, and the mature rRNA molecules are joined and then translated to produce the protein portion of the ribosome. B Introns are removed from the pre-rRNA, and each mature rRNA molecule is translated to produce the proteins that make up the ribosomal subunits. C Sections of the pre-rRNA are removed, and the mature rRNA molecules are available to combine with proteins to form the ribosomal subunits. D Sections of the pre-rRNA are removed, and the mature rRNA molecules are available to bring different amino acids to the ribosome.

C

Based on the information provided in Figure 1 and Figure 2, which of the following best predicts the effects of a mutation in the promoter of the TYR gene that prevents it from being transcribed? A DNA damage due to UV radiation will be strongly inhibited, resulting in a positive selection pressure. B DNA damage due to UV radiation will be strongly inhibited, resulting in a negative selection pressure. C Skin pigmentation will not be able to change, resulting in a positive selection pressure. D Skin pigmentation will not be able to change, resulting in a negative selection pressure.

D

Which of the following best describes a characteristic of DNA that makes it useful as hereditary material? A There are many different types of nucleotide bases that can be incorporated into DNA. B The nucleotide bases can also be used to provide the energy needed for reproduction. C Nucleotide bases can be randomly replaced with different nucleotide bases to increase variation. D Nucleotide bases in one strand can only be paired with specific bases in the other strand.

D

Which of the following best describes an event during step 2 in the simplified model above? A A new RNA molecule is synthesized using a DNA template. B A new polypeptide is synthesized using an RNA template. C Thymine nucleotides in an RNA molecule are replaced with uracil nucleotides. D Noncoding sequences are removed from a newly synthesized RNA molecule.

D

Which of the following best explains a process occurring between point 1 and point 2 in Figure 3 ? A α-MSH is produced. B The TYR gene is transcribed. C Polypeptides are removed from a protein. D A poly‑A tail is added to RNA.

D

Which of the following best helps explain how the process represented in Figure 1 produces DNA molecules that are hybrids of the original and the newly synthesized strands? A Each template strand is broken down into nucleotides, which are then used to synthesize both strands of a new DNA molecule. B Each template strand is broken into multiple fragments, which are randomly assembled into two different DNA molecules. C Each newly synthesized strand is associated with another newly synthesized strand to form a new DNA molecule. D Each newly synthesized strand remains associated with its template strand to form two copies of the original DNA model.

D

Which of the following best predicts the most direct effect of exposing prokaryotic cells to streptomycin? A Amino acid synthesis will be inhibited. B No mRNA will be transcribed from DNA. C Posttranslational modifications will be prevented. D Synthesis of polypeptides will be inhibited.

D

Which of the following best predicts the phenotype of an individual who is homozygous for this TYR mutation? A The mutation will cause a single amino acid change in the TYR protein, which will not be enough to disrupt its function. Therefore, those with this mutation will produce melanin in the hair, skin, and eyes and tan in response to UV radiation. B The mutation will cause a single amino acid change in the TYR protein, leading to a nonfunctional TYR protein. Therefore, those with this mutation will lack melanin in the hair, skin, and eyes and will not tan in response to UV radiation. C The mutation will change all subsequent amino acids in the TYR protein, leading to nonfunctional TYR protein. Since the TRP1 and TRP2 genes were not affected, the TRP1 and TRP2 proteins will fill the role of the TYR protein. Therefore, those with this mutation will produce melanin in the hair, skin, and eyes in response to UV radiation. D The mutation will change all subsequent amino acids in the TYR protein, leading to nonfunctional TYR protein. Individuals with this mutation will lack melanin in their hair, skin, and eyes and will not tan in response to UV radiation.

D

Which of the following evidence best supports a claim that tryptophan functions as a corepressor? A Normal expression of trpR causes the trp operon to be transcribed regardless of tryptophan levels. B When the operator sequence is mutated, the trp operon is not transcribed. C The trpR gene codes for a repressor protein that has a DNA binding domain. D When trpR is mutated, the trp operon is transcribed regardless of tryptophan levels.

D

Which of the following explanations best accounts for this experimental result? A Gel electrophoresis can only be used with DNA (not mRNA), so experimental results are not interpretable. B RNA polymerase consistently makes the same errors during transcription of gene L. C Gene L is mutated, so RNA polymerase does not always transcribe the correct sequence. D Pre-mRNA of gene L is subject to alternative splicing, so three mRNA sequences are possible.

D

Which of the following scientific claims is most consistent with the information provided in Figure 1 ? A Gene X codes for a transcription factor required for transcription of gene D. B A single transcription factor regulates transcription similarly, regardless of the specific gene. C Transcription of genes A, B, and C is necessary to transcribe gene E. D Different genes may be regulated by the same transcription factor.

D

Which of the following statements best explains the role of Enzyme 1 in the DNA replication process? A Enzyme 1 is a DNA ligase that joins together the DNA fragments at a replication fork to form continuous strands. B Enzyme 1 is a DNA primase that catalyzes the synthesis of RNA primers on the lagging strand of a replication fork. C Enzyme 1 is a DNA polymerase that synthesizes new DNA by using the leading and lagging strands of a replication fork as templates. D Enzyme 1 is a topoisomerase that relieves tension in the overwound DNA in front of a replication fork.

D

Which of the following statements best explains what is shown in Figure 1 ? A UV exposure triggers DNA replication, which results in rapid cell proliferation. B Naturally occurring dimers are removed by the UV photons, causing misshapen DNA, which results in replication errors. C The hydrogen bonds between base pairs absorb the UV photons, separating the two DNA strands, which results in rapid DNA replication. D UV photons cause dimers to form, leading to misshapen DNA, which results in replication and transcription errors.

D

Which scientific claim is most consistent with these findings? A DNA methylation inhibits transcription of gene R. B Histone modifications of genes are usually not reversible. C Histone methylation condenses the chromatin at gene R so transcription factors cannot bind to DNA. D Histone methylation opens up chromatin at gene RR so transcription factors can bind to DNA more easily.

D