AP Bio Unit 6 ALL Questions

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? Responses 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 A 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 universal genetic code, which of the following represents the correct polypeptide that will result from translation of the mRNA A Asp−Arg−Met−Val−Thr−Lys−Phe−Gly−His B Met−Arg−Asp−Stop−His−Gly−Phe−Lys−Thr−Val C Met−Val−Thr−Lys−Phe−Gly−His D Val−Thr−Lys−Phe−Gly−His

C

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

C

Histone methyltransferases are a class of enzymes that methylate certain amino acid sequences in histone proteins. A research team found that transcription of gene R� decreases when histone methyltransferase activity is inhibited. Which scientific claim is most consistent with these findings? A DNADNA 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 DNADNA. D Histone methylation opens up chromatin at gene R� so transcription factors can bind to DNADNA more easily.

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? Responses 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 D N A molecule. D Each newly synthesized strand remains associated with its template strand to form two copies of the original DNA molecule.

D

Which of the following best explains how the pattern of DNA arrangement in chromosomes could be used, in most cases, to determine if an organism was a prokaryote or a eukaryote?

(C)Prokaryotic DNA - Single circular chromosome Eukaryotic DNA - Multiple linear chromosomes

Which of the following claims best explains why keratinocytes do not produce melanin? Responses 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

Genetic engineering techniques can be used when analyzing and manipulating DNADNA and RNARNA. Scientists used gel electrophoresis to study transcription of gene L� and discovered that mRNAmRNA strands of three different lengths are consistently produced. Which of the following explanations best accounts for this experimental result? A Gel electrophoresis can only be used with DNADNA (not mRNAmRNA), so experimental results are not interpretable. B RNARNA polymerase consistently makes the same errors during transcription of gene L�. C Gene L� is mutated, so RNARNA polymerase does not always transcribe the correct sequence. D Pre-mRNAmRNA of gene L� is subject to alternative splicing, so three mRNAmRNA sequences are possible.

D

Small single-stranded RNA molecules called microRNAs (miRNAs) are capable of base pairing with specific binding sites in the 3′ untranslated region of many mRNA transcripts. Transcription of gene Q yields an mRNA transcript that contains such an miRNA binding site, which can associate with miRNA‑delta, a specific miRNA molecule. Which of the following best supports the claim that binding of miRNA‑delta to the miRNA binding site inhibits translation of gene Q mRNA? Responses 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

Which of the following evidence best supports a claim that tryptophan functions as a corepressor? Responses 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

Both liver cells and lens cells have the genes for making the proteins albumin and crystalline. However, only liver cells express the blood protein albumin and only lens cells express crystalline, the main protein in the lens of the eye. Both of these genes have enhancer sequences associated with them. 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 RNARNA polymerase enzymes to transcribe DNADNA. C Liver cells and lens cells possess the same transcriptional activators. D Liver cells and lens cells possess different general transcription factors.

A

Which of the following pieces of evidence would best support the researchers' claim above? AWhen researchers applied a drug that activates adenylyl cyclase to the mutant mice's ears, the level of melanin increased. BWhen researchers viewed sections of mutant mouse ears under the microscope, they found melanocyte numbers comparable to nonmutant mice. CWhen 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

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

C

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

Which of the following graphs best predicts the data collected during the experiment?

D. Up, Medium, Up (GRAPH)

Lynch syndrome is an inherited condition associated with an increased risk for colon cancer, as well as certain other cancers. Mutations in one of several genes involved in DNA repair during DNA replication have been associated with Lynch syndrome. DNA sequencing was performed for an individual. The results indicated that the individual carries one of the dominant alleles that has been associated with Lynch syndrome. Which of the following best explains how the results should be interpreted? Responses 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

A scientist claimed that an E. coli strain had either a mutated trpR gene or a mutated operator. Which of the following observations most likely supports the claim? A Transcription from the operon occurred only in the presence of abundant tryptophan. B The strain of E. coli required more tryptophan for its metabolic processes than does a strain of E. coliwith typical tryptophan regulatory controls. C Enzymes required for the synthesis of tryptophan were continuously produced whether tryptophan was absent or present in large quantities. D The cells died when they were grown in nutrient medium that lacked tryptophan.

C

ickle-cell anemia is an inherited blood disorder in which one of the hemoglobin subunits is replaced with a different form of hemoglobin. Partial DNADNA sequences of the HBB��� gene for normal hemoglobin and for sickle-cell hemoglobin are shown in Figure 1. 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 the process represented by Figure 1 ? A The synthesis of mRNAmRNAin the 5′5′ to 3′3′direction from DNADNA B The modification of a protein to produce a functional form of that protein C The translation of an mRNAmRNA molecule into a polypeptide D The enzyme-regulated processing of pre‑mRNAmRNA into mature mRNA

D

Antigens are foreign proteins that invade the systems of organisms. Vaccines function by stimulating an organism's immune system to develop antibodies against a particular antigen. Developing a vaccine involves producing an antigen that can be introduced into the organism being vaccinated and which will trigger an immune response without causing the disease associated with the antigen. Certain strains of bacteria can be used to produce antigens used in vaccines. 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 DNADNA of the antigen has to be transcribed in order for the mRNAmRNA produced to be inserted into the bacteria. D The mRNAmRNA of the antigen has to be translated in order for the protein to be inserted into the bacteria.

A

Arsenic is a toxic element found in both aquatic and terrestrial environments. Scientists have found genes that allow bacteria to remove arsenic from their cytoplasm. Arsenic enters cells as arsenate that must be converted to arsenite to leave cells. Figure 1 provides a summary of the arsenic resistance genes found in the operons of three different bacteria. E. coli R773R773 is found in environments with low arsenic levels. Herminiimonas arsenicoxydans and Ochrobactrum tritici are both found in arsenic‑rich environments 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

Molecular biologists are studying the processes of transcription and translation and have found that they are very similar in prokaryotes and eukaryotes, as summarized in Table 1. 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 mRNAmRNA is shorter than eukaryotic mRNAmRNA. C Eukaryotic mRNAmRNA contains more coding regions than prokaryotic DNADNA. D The processing of mRNAmRNA by eukaryotes is required for the mRNAmRNA to leave the nucleus.

A

Phytochromes are molecules that change light stimuli into chemical signals, and they are thought to target light-activated genes in plants. A study was conducted to determine how certain cell proteins were made in a plant cell using a phytochrome. Figures 1 and 2 represent findings from the study. 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 PrPr is activated by red light to become phytochrome PfrPfr, it is transported into the nucleus where it binds to the transcription factor PIF3PIF3 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 PIF3PIF3. B Far-red light activates phytochrome PrPr, causing it to travel to the nucleus where it binds to PIF3PIF3 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 PIF3PIF3. C MYBMYB, and not PfrPfr, is activated by red light, causing it to bind to the promoter and stimulate transcription and translation of cellular proteins. D PIF3PIF3 binds to the promoter only in the presence of red light and PfrPfr. Any time PIF3PIF3 is bound to the promoter, MYBMYB is transcribed, initiating transcription of various other proteins in the cell.

A

The table below describes the action of two genes involved in the regulation of nervous system development in the nematode C. elegans. Which of the following claims is best supported by the data? A Gene A promotes neuron development; gene B promotes programmed cell death in neuronal precursors. B Gene A promotes programmed cell death in neuronal precursors; gene B promotes neuron development. C Gene B must be active before gene A can function. D Gene B must be inactive before gene A can function.

A

Antibiotics can be used to kill the specific pathogenic bacterium, Mycobacterium tuberculosis, that causes tuberculosis. The appearance of antibiotic-resistant strains has made it more difficult to cure M. tuberculosis infections. These antibiotic-resistant bacteria survive and pass on the genes to their offspring, making the resistant phenotype more common in the population. DNADNAanalysis indicates that the genes for antibiotic resistance are not normally present in bacterial chromosomal DNADNA. Which of the following statements best explains how the genes for antibiotic resistance can be transmitted between bacteria without the exchange of bacterial chromosomal DNADNA? 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 DNADNAto neutralize the antibiotics. D The antibiotic alters the bacterial genome of each bacterium, which results in an antibiotic-resistant population.

B

Erwin Chargaff investigated the nucleotide composition of DNADNA. He analyzed DNADNA from various organisms and measured the relative amounts of adenine (AA), guanine (GG), cytosine (CC), and thymine (TT) present in the DNADNA of each organism. Table 1 contains a selected data set of his results. Which of the following statements best explains the data set? A Since the %A%A and the %G%G add up to approximately 50 percent in each sample, adenine and guanine molecules must pair up in a double-stranded DNADNA molecule. B Since the %A%A and the %T%T are approximately the same in each sample, adenine and thymine molecules must pair up in a double-stranded DNADNA molecule. C Since the %(A+T)%(A+T) is greater than the %(G+C)%(G+C) in each sample, DNADNA molecules must have a poly-AA tail at one end. D Since the %C%C and the %T%T add up to approximately 50 percent in each sample, cytosine and thymine molecules must both contain a single ring.

B

Figure 1 represents a metabolic process involving the regulation of lactose metabolism by E. coli bacteria. Lactose is utilized for energy by E. coli when glucose is not present. Allolactose is an isomer of lactose that is in the environment of these bacteria when lactose is present. The CAP site prevents the binding of RNA polymerase when glucose is present in the environment. The lacZ, lacY, and lacA genes code for proteins needed for lactose metabolism. Figure 1. Model of lac operon, comparing repressed and active states 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

Figure 1 represents a metabolic process involving the regulation of lactose metabolism by E. coli bacteria. Lactose is utilized for energy by E. coli when glucose is not present. Allolactose is an isomer of lactose that is in the environment of these bacteria when lactose is present. The CAPCAP site prevents the binding of RNARNA polymerase when glucose is present in the environment. The lacZ, lacY, and lacA genes code for proteins needed for lactose metabolism. 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 RNARNA 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

Huntington's disease, an autosomal dominant disorder, is caused by a mutation in the HTT gene. The HTT gene contains multiple repeats of the nucleotide sequence CAG. A person with fewer than 35 CAG repeats in the HTT gene is unlikely to show the neurological symptoms of Huntington's disease. A person with 40 or more CAG repeats almost always becomes symptomatic. Due to errors in meiosis, an individual without symptoms of Huntington's disease can produce gametes with a larger number of CAG repeats than there are in their somatic cells. 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? Responses 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

In animals, the hox genes encode a family of transcription factors that are important for proper development of embryonic segments and are widely conserved in organisms. The figure below shows the embryonic segments in which one such gene, Hoxc6, is expressed in the embryo of a mouse, a chick, and a goose. Embryonic segments are counted from the anterior end. During the formation of vertebrae, the most anterior embryonic segment that expresses Hoxc6 marks the end of the cervical (neck) vertebrae and the beginning of the thoracic (rib) vertebrae. All mammals have seven cervical vertebrae. Which of the following statements is most likely to be true? A The chick and the goose have the same number of thoracic vertebrae. B The most anterior expression of Hoxc6 is the eighth vertebra in mammals. C Hoxc6 is expressed in the same embryonic segments in birds and mammals. D Hoxc6 is expressed in the same vertebra at the anterior end of all bird embryos.

B

Nondisjunction during meiosis can negatively affect gamete formation. A model showing a possible nondisjunction event and its impact on gamete formation is shown in Figure 1. 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 II, all gametes will be normal and the resulting individual will be phenotypically normal. B 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. C Because nondisjunction occurred in anaphase IIII, all gametes will be normal and the resulting individual will be phenotypically normal. D Because nondisjunction occurred in anaphase IIII, all gametes will have an abnormal chromosome number and the individual will likely exhibit phenotypic evidence of the nondisjunction event.

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

The three-spined stickleback (Gasterosteus aculeatus) is a small fish found in both marine and freshwater environments. Marine stickleback populations consist mostly of individuals with pronounced pelvic spines, as shown in Figure 1. Individuals in freshwater stickleback populations, on the other hand, typically have reduced pelvic spines, as shown in Figure 2. As represented in Figure 1 and Figure 2, the phenotypic difference between marine and freshwater sticklebacks involves Pitx1, a gene that influences the formation of the jaw, pituitary gland, and pelvic spine. Enhancer sequences upstream of the Pitx1 genetic locus regulate expression of the Pitx1 gene at the appropriate times and in the appropriate tissues during development. Previous studies have found that a mutation in the hindlimb enhancer interferes with the formation of a pronounced pelvic spine. Which of the following best describes how sticklebacks in the same population with identical copies of the Pitx1 gene can still show phenotypic variation in the pelvic spine character? A The Pitx1 gene is carried on different chromosomes in different individuals. B Expression of the Pitx1 gene is affected by mutations at other genetic loci. C The genetic code of the Pitx1 gene is translated differently in males and females. D The subcellular location of the Pitx1 gene changes when individuals move to a new environment.

B

Which claim is most consistent with the information provided by the diagram and current scientific understanding of gene regulation and expression? Responses 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 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 best explains what process is represented in Figure 1 ? A New DNADNAstrands are being synthesized in the 3'3′ to 5'5′ direction from their DNADNA templates. B New DNADNA strands are being synthesized in the 5'5′ to 3'3′ direction from their DNADNA templates. C A new RNARNAstrand is being synthesized in the 3'3′ to 5'5′ end from its DNADNA template. D Two new RNARNA strands are being synthesized in both directions from their DNADNA templates.

B

Which of the following claims about the TYR, TRP2, and TRP1 mammalian genes is most likely to be accurate? 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 DNADNAreplication 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 RNARNAis being produced from a DNADNAtemplate strand. C Translation is occurring because the two strands have separated and a new strand is being produced. D Alternative splicing of mRNAmRNA is occurring because the mRNAmRNA strand is being synthesized from only one strand of DNADNA.

B

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

B

All cells must transcribe rRNA in order to construct a functioning ribosome. Scientists have isolated and identified rRNA genes that contribute to ribosomal structure for both prokaryotes and eukaryotes. Figure 1 compares the transcription and processing of prokaryotic and eukaryotic rRNA. Figure 1. Comparison of rRNA processing in prokaryotes and eukaryotes Which of the following statements provides the best explanation of the processes illustrated in Figure 1 ? Responses 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

Cystic fibrosis (CF) is a progressive genetic disease that causes persistent lung infections and affects the ability to breathe. CF is inherited in an autosomal recessive manner, caused by the presence of mutations in both copies of the gene for the cystic fibrosis transmembrane conductance regulator (CFTR) protein. Partial nucleotide sequences and the corresponding amino acid sequences for an unaffected individual and an affected individual are modeled in Figure 1. Figure 1. CFTR protein sequences in unaffected and affected individuals 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? Responses 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

Figure 1 illustrates a model of the molecules involved in DNADNAreplication and their placement relative to each other. Which of the following correctly explains where DNADNAreplication will begin on the strand oriented 5'→3'5′→3′, reading from left to right? A DNADNAreplication will be randomly initiated along the unwound portion of the DNADNA strand since base pairing will occur. B DNADNAreplication cannot occur since there is already RNARNA base pairing with the template strand. C DNADNAreplication will be initiated immediately to the left of the RNARNA, since DNADNApolymerase requires an RNARNAprimer. D DNADNA replication will be initiated at the site of the topoisomerase since that is where DNADNA begins to uncoil.

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? Responses 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

Retroviruses such as HIV and hepatitis B virus use RNA as their genetic material rather than DNA. In addition, they contain molecules of reverse transcriptase, an enzyme that uses an RNA template to synthesize complementary DNA. Which of the following best predicts what will happen when a normal cell is exposed to a retrovirus? Responses 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

The enzyme lactase aids in the digestion of lactose, a sugar found in the milk of most mammals. In most mammal species, adults do not produce lactase. Continuing to produce lactase into adulthood in people is called lactase persistence. A number of different alleles have been identified that result in lactase persistence. Figure 1 shows the percentage of people in different geographic areas parts of the Old World that exhibit lactase persistence. 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

The figure above depicts the DNA-protein complex that is assembled at the transcriptional start site of gene X when the expression of gene X is activated in liver cells. Previous studies have shown that gene X is never expressed in nerve cells. Based on the diagram, which of the following most likely contributes to the specific expression pattern of gene X ? A Expression of gene X produces large amounts of tRNA but undetectable amounts of mRNA. B The general transcription factors inhibit the activation of gene X in liver cells by blocking the activator from binding to RNA polymerase II. C The activator is a sequence-specific DNA-binding protein that is present in some tissues but not in other tissues. D The enhancer is a unique DNA segment that is added to the nuclear DNA of some cells of an organism during the process of mitotic cell division but not other cells.

C

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

C

Which of the following statements best explains the experimental results observed in Figure 1 ? Responses 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 AA and BB? A Molecules AA and BB are positively charged, and molecule AA is smaller than molecule BB. B Molecules AA and BB are positively charged, and molecule AA is larger than molecule BB. C Molecules AA and BB are negatively charged, and molecule AA is smaller than molecule BB. D Molecules AA and BB are negatively charged, and molecule AA is larger than molecule BB.

C

Antibiotics interfere with prokaryotic cell functions. Streptomycin is an antibiotic that affects the small ribosomal subunit in prokaryotes. Specifically, streptomycin interferes with the proper binding of tRNAtRNA to mRNAmRNAin prokaryotic ribosomes. 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 mRNAmRNA will be transcribed from DNADNA. C Posttranslational modifications will be prevented. D Synthesis of polypeptides will be inhibited.

D

Based on the information provided in Figure 1 and Figure 2, which of the following best explains the effects of a mutation in the promoter of the TYR gene that prevents it from being transcribed? Responses 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. DSkin pigmentation will not be able to change, resulting in a negative selection pressure.

D

Figure 1 represents part of a process that occurs in eukaryotic cells. There are untranslated regions (UTR) in this sequence. Figure 1. Cellular process involving nucleic acids 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

Figure 1 shows some relevant details of a model of how a deoxynucleotide, in this case dTMPdTMP, is added to a growing strand of DNA The features of this model provide evidence for which explanation of why all growing strands are synthesized in a 5′5′to 3′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′3′to 5′5′. C The translation of mRNAmRNAoccurs in the 5′5′to 3′3′direction; therefore, the growing DNADNAstrand must also grow in the 5′5′to 3′3′direction. D The phosphate group, attached to the 5′5′ carbon of the dTMPdTMP, forms a covalent bond with the oxygen atom attached to the 3′3′ carbon of the growing strand.

D

The following the DNA sequence is a small part of the coding (nontemplate) strand from the open reading frame of β-hemoglobingene. Given the codon chart listed below, what would be the effect of a mutation that deletes the G at the beginning of the DNA sequence? 5'- GTT TGT CTG TGG TAC CAC GTG GAC TGA - 3' A The mutation precedes the gene, so no changes would occur. B Lysine (lys) would replace glutamine (gln), but there would be no other changes. C The first amino acid would be missing, but there would be no other change to the protein. D The reading frame of the sequence would shift, causing a change in the amino acid sequence after that point.

D

Which of the following best explains a process occurring between point 1 and point 2 in Figure 3 ? Responses 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 predicts the phenotype of an individual who is homozygous for this TYR mutation? Responses 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. 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. 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. 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 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? Responses 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 ? Responses 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