IB/AP Biology Unit 6 Gene Expression and Regulation

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 tRNA to mRNA in 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 mRNA will be transcribed from DNA. (C) Posttranslational modifications will be prevented. (D) Synthesis of polypeptides will be inhibited.

(D) Synthesis of polypeptides will be inhibited.

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 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) Liver cells possess transcriptional activators that are different from those of lens cells.

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? Responses (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 R so transcription factors can bind to DNA more easily.

(D) Histone methylation opens up chromatin at gene R so transcription factors can bind to DNA more easily.

A model that represents a process occurring in a cell of a particular organism is shown in Figure 1. Figure 1. Process occurring in a cell 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) Initiation of transcription is occurring because a strand of RNA is being produced from a DNA template strand.

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

(B) New DNA strands are being synthesized in the 5' to 3' direction from their DNA templates.

Which of the following best explains how the pattern of DNADNA arrangement in chromosomes could be used, in most cases, to determine if an organism was a prokaryote or a eukaryote? (A) Prokaryotic DNA: Single circular chromosome Eukaryotic DNA: Multiple circular chromosomes (B) Prokaryotic DNA: Multiple chromosomes Eukaryotic DNA: Single chromosome (C) Prokaryotic DNA: Single linear chromosome Eukaryotic DNA: Multiple linear chromosomes (D) Prokaryotic DNA: Single circular chromosome Eukaryotic DNA: Multiple linear chromosomes

(D) Prokaryotic DNA: Single circular chromosome Eukaryotic DNA: Multiple linear chromosomes

Eukaryotes transcribe RNA from DNA that contains introns and exons. Alternative splicing is one posttranscriptional modification that can create distinct mature mRNA molecules that lead to the production of different proteins from the same gene. Figure 1 shows a gene and the RNA produced after transcription and after alternative splicing. Figure 1. Model of posttranscriptional alternate splicing of mRNA 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 A mRNA. (D) The cell will remain healthy because enzyme-substrate interactions are nonspecific and enzyme C will eventually metabolize the substrate.

(A) The cell will die because it is unable to metabolize the substrate without enzyme A, which is structurally specific for the substrate shown.

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 DNA of the antigen has to be transcribed in order for the m R N A 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) The gene coding for the antigen can be inserted into plasmids that can be used to transform the bacteria.

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. Table 1. Comparison of Selected Features of Transcription and Translation 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) The two processes will occur simultaneously in prokaryotes but not eukaryotes

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 R773 is found in environments with low arsenic levels. Herminiimonas arsenicoxydans and Ochrobactrum tritici are both found in arsenic‑rich environments. Figure 1. Operons found in three selected bacteria for arsenic removal 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) 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.

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. Figure 1. Phytochrome response to exposure to red light Figure 2. Phytochrome response to exposure to far‑red light 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, MYBMYB is transcribed, initiating transcription of various other proteins in the cell.

(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.

In mammals, the dark color of skin, hair, and eyes is due to a pigment called melanin. Melanin is produced by specialized skin cells called melanocytes. The melanin is then transferred to other skin cells called keratinocytes. Melanocytes synthesize melanin in a multistep metabolic pathway (Figure 1). The amount of melanin produced is dependent on the amount of the enzymes TYR, TRP2, and TRP1 present inside melanocytes. Figure 1. Melanin synthesis pathway The peptide hormone α-melanocyte-stimulating hormone (α-MSH) activates a signal transduction pathway leading to the activation of MITF. MITF is a transcription factor that increases the expression of the TYR, TRP2, and TRP1 genes (Figure 2). Figure 2. Activation of melanin synthesis genes in melanocytes Some mammals increase melanin production in response to ultraviolet (UV)(UV) radiation. The UV radiation causes damage to DNA in keratinocytes, which activates the p53 protein. p53 increases the expression of the POMC gene. The POMC protein is then cleaved to produce α-MSH. The keratinocytes secrete α-MSH, which signals nearby melanocytes. The increased melanin absorbs UV radiation, reducing further DNA damage. Figure 3. Production of α-MSH in keratinocytes in response to UV radiation Mice have melanocytes in the skin on their ears and show a tanning response to UV radiation. Researchers were studying a mutant population of mice that do not show a tanning response. Genetic testing of these mutant mice showed that the pathway causing the production of α-MSH by keratinocytes in response to UV radiation was fully functional. Thus, the researchers claimed that the lack of tanning response was due a nonfunctional MC1R. 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) When researchers applied a drug that activates adenylyl cyclase to the mutant mice's ears, the level of melanin increased.

Question 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. Figure 1. Model of a nondisjunction event 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 1 , all gametes will be normal and the resulting individual will be phenotypically normal. (B) Because nondisjunction occurred in anaphase 1 , 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 2 , 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) 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.

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) Plasmids are circular, double-stranded DNADNA molecules that provide genes that may aid in survival of the prokaryotic cell.

Huntington's disease, an autosomal dominant disorder, is caused by a mutation in the ��� gene. The ��� gene contains multiple repeats of the nucleotide sequence CAG. A person with fewer than 35 CAG repeats in the ��� 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? (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) She inherited an allele with more than 40 CAG repeats in the HTT gene from her father.

Erwin Chargaff investigated the nucleotide composition of DNA. He analyzed DNA from various organisms and measured the relative amounts of adenine (AA), guanine (GG), cytosine (CC), and thymine (TT) present in the DNA of each organism. Table 1 contains a selected data set of his results. Table 1. Nucleotide composition of sample DNA from selected organisms 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-AA 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) 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.

Figure 1 represents a portion of a process that occurs during protein synthesis. Figure 1. Model of selected features of DNA transcription 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) Some sequences of DNA can interact with regulatory proteins that control transcription.

In mammals, the dark color of skin, hair, and eyes is due to a pigment called melanin. Melanin is produced by specialized skin cells called melanocytes. The melanin is then transferred to other skin cells called keratinocytes. Melanocytes synthesize melanin in a multistep metabolic pathway (Figure 1). The amount of melanin produced is dependent on the amount of the enzymes TYR, TRP2, and TRP1 present inside melanocytes. Figure 1. Melanin synthesis pathway The peptide hormone α-melanocyte-stimulating hormone (α-MSH) activates a signal transduction pathway leading to the activation of MITF. MITF is a transcription factor that increases the expression of the TYR, TRP2, and TRP1 genes (Figure 2). Figure 2. Activation of melanin synthesis genes in melanocytes Some mammals increase melanin production in response to ultraviolet (UV)(UV) radiation. The UV radiation causes damage to DNA in keratinocytes, which activates the p53 protein. p53 increases the expression of the POMC gene. The POMC protein is then cleaved to produce α-MSH. The keratinocytes secrete α-MSH, which signals nearby melanocytes. The increased melanin absorbs UV radiation, reducing further DNA damage. Figure 3. Production of α-MSH in keratinocytes in response to UV radiation 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) The TYR, TRP2, and TRP1 genes may be located on different chromosomes but are activated by the same transcription factor.

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. DNADNA analysis indicates that the genes for antibiotic resistance are not normally present in bacterial chromosomal DNA. 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) The genes for antibiotic resistance are located on a plasmid that can be passed to neighboring bacteria.

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? (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) The individual has an increased risk of developing colon cancer.

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. 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 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) When bound to the operator, the repressor protein prevents lactose metabolism in E. coli.

Figure 1 illustrates a model of the molecules involved in DNA replication and their placement relative to each other. Figure 1. Model including molecules involved in DNA replication 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) DNA replication will be initiated immediately to the left of the RNA , since DNA polymerase requires an RNA primer.

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? (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) Deletion, because a thymine is missing, which changes the reading frame.

Scientists conducted a transformation experiment using E. coli bacteria and the pTru���� plasmid. Samples of the pTru���� plasmid (lane AA) and the chromosomal DNADNA from two different E. coli strains that the scientists attempted to transform (lane BB and lane CC) were compared using gel electrophoresis. The results are shown in Figure 1. Figure 1. Results of E. coli transformation with pTru���� plasmid Which of the following statements best explains the experimental results observed in Figure 1 ? (A) E. coli in both lanes BB and CC have been successfully transformed and contain additional genetic information. (B) E. coli in lane BB have been successfully transformed and contain additional genetic information. (C) E. coli in lane CC have been successfully transformed and contain additional genetic information. (D) Which E. coli have been transformed cannot be determined from this gel.

(C) E. coli in lane CC have been successfully transformed and contain additional genetic information.

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) Hydrogen bonds join a double-ringed structure to a single-ringed structure in each pair.

Figure 1 depicts a simplified model of a replication bubble. Figure 1. DNA replication bubble. 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′ 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) I is synthesized continuously in the 5′ to 3′ direction, and III is synthesized in segments in the 5′ to 3′ direction.

In mammals, the dark color of skin, hair, and eyes is due to a pigment called melanin. Melanin is produced by specialized skin cells called melanocytes. The melanin is then transferred to other skin cells called keratinocytes. Melanocytes synthesize melanin in a multistep metabolic pathway (Figure 1). The amount of melanin produced is dependent on the amount of the enzymes TYR, TRP2, and TRP1 present inside melanocytes. Figure 1. Melanin synthesis pathway The peptide hormone α-melanocyte-stimulating hormone (α-MSH) activates a signal transduction pathway leading to the activation of MITF. MITF is a transcription factor that increases the expression of the TYR, TRP2, and TRP1 genes (Figure 2). Figure 2. Activation of melanin synthesis genes in melanocytes Some mammals increase melanin production in response to ultraviolet (UV)(UV) radiation. The UV radiation causes damage to DNA in keratinocytes, which activates the p53 protein. p53 increases the expression of the POMC gene. The POMC protein is then cleaved to produce α-MSH. The keratinocytes secrete α-MSH, which signals nearby melanocytes. The increased melanin absorbs UV radiation, reducing further DNA damage. Figure 3. Production of α-MSH in keratinocytes in response to UV radiation 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) Keratinocytes do not express the MITF gene.

Students subjected three samples of five different molecules to gel electrophoresis as shown in Figure 1. Figure 1. Gel electrophoresis of three prepared samples 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) Molecules A and B are negatively charged, and molecule A is smaller than molecule B.

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. Figure 1. Distribution of lactase persistence in Europe, North Africa, and parts of Asia 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) Mutations conferring lactase persistence likely arose independently in different geographic areas and offered a selective advantage.

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 ? (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) Sections of the pre-rRNA are removed, and the mature rRNA molecules are available to combine with proteins to form the ribosomal subunits.

An evolutionary biologist hypothesizes that two morphologically similar plant species are not closely related. To test the hypothesis, the biologist collects DNA samples from each of the two plant species and then uses restriction enzymes to cut the DNA samples into fragments, which are then subjected to gel electrophoresis. The results are shown in Figure 1. Figure 1. DNA analysis of two species 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) Species B has more short fragments of DNA than species A does.

Sickle-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. Partial sequence for normal hemoglobin: Partial sequence for sickle-cell hemoglobin: Figure 1. Comparison of partial DNADNA sequences for normal hemoglobin and hemoglobin with a sickle-cell mutation 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) Substitution

Figure 1 represents part of a process essential to gene expression. Figure 1. Model of process involved in gene expression. Which of the following best explains what strand X represents? Responses (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) The antisense strand, because it is serving as a template

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? (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 reverse transcriptase will produce DNA from the viral RNA, which can be incorporated into the host's genome and then transcribed and translated.

In mammals, the dark color of skin, hair, and eyes is due to a pigment called melanin. Melanin is produced by specialized skin cells called melanocytes. The melanin is then transferred to other skin cells called keratinocytes. Melanocytes synthesize melanin in a multistep metabolic pathway (Figure 1). The amount of melanin produced is dependent on the amount of the enzymes TYR, TRP2, and TRP1 present inside melanocytes. Figure 1. Melanin synthesis pathway The peptide hormone α-melanocyte-stimulating hormone (α-MSH) activates a signal transduction pathway leading to the activation of MITF. MITF is a transcription factor that increases the expression of the TYR, TRP2, and TRP1 genes (Figure 2). Figure 2. Activation of melanin synthesis genes in melanocytes Some mammals increase melanin production in response to ultraviolet (UV)(UV) radiation. The UV radiation causes damage to DNA in keratinocytes, which activates the p53 protein. p53 increases the expression of the POMC gene. The POMC protein is then cleaved to produce α-MSH. The keratinocytes secrete α-MSH, which signals nearby melanocytes. The increased melanin absorbs UV radiation, reducing further DNA damage. Figure 3. Production of α-MSH in keratinocytes in response to UV radiation 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) A poly‑A tail is added to RNA.

Figure 1 illustrates processes related to control of transcription and translation in a cell. Figure 1. Model of a relationship between a transcription factor and selected genes 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) Different genes may be regulated by the same transcription factor.

Labeled nucleotides were supplied to a cell culture before the cells began DNA replication. A simplified representation of the process for a short segment of DNA is shown in Figure 1. Labeled DNA bases are indicated with an asterisk (*). Figure 1. A simplified representation of the DNA replication process 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 molecule.

(D) Each newly synthesized strand remains associated with its template strand to form two copies of the original DNA molecule.

A simplified model of a DNA replication fork is represented in Figure 1. The protein labeled Enzyme 1 carries out a specific role in the DNA replication process. Figure 1. A simplified model of a DNA replication fork 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) Enzyme 1 is a topoisomerase that relieves tension in the overwound DNA in front of a replication fork.

Cycloheximide (CHX) is a eukaryote protein synthesis inhibitor. It is used in biomedical research to inhibit protein synthesis in eukaryotic cells studied in vitro. Its effects are rapidly reversed by simply removing it from the culture medium. In a translation experiment using a fungus culture, radiolabeled amino acids were added to the culture, allowing the researchers to measure the growth of a single polypeptide chain by measuring counts per minute (CPM). As the chain grew, the CPM increased. After a certain amount of time, CHX was added to the mixture, and the experiment continued. After an additional amount of time, the CHX was removed from the culture medium. Which of the following graphs best predicts the data collected during the experiment? (A) Graph A (B) Graph B (C) Graph C (D) Graph D

(D) Graph D

Genetic engineering techniques can be used when analyzing and manipulating DNA and RNA. Scientists used gel electrophoresis to study transcription of gene L and discovered that mRNA 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 DNA (not mRNA ), so experimental results are not interpretable. (B) R N A polymerase consistently makes the same errors during transcription of gene L. (C) Gene L is mutated, so R N A 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) Pre-mRNA of gene L is subject to alternative splicing, so three mRNA sequences are possible.

In mammals, the dark color of skin, hair, and eyes is due to a pigment called melanin. Melanin is produced by specialized skin cells called melanocytes. The melanin is then transferred to other skin cells called keratinocytes. Melanocytes synthesize melanin in a multistep metabolic pathway (Figure 1). The amount of melanin produced is dependent on the amount of the enzymes TYR, TRP2, and TRP1 present inside melanocytes. Figure 1. Melanin synthesis pathway The peptide hormone α-melanocyte-stimulating hormone (α-MSH) activates a signal transduction pathway leading to the activation of MITF. MITF is a transcription factor that increases the expression of the TYR, TRP2, and TRP1 genes (Figure 2). Figure 2. Activation of melanin synthesis genes in melanocytes Some mammals increase melanin production in response to ultraviolet (UV)(UV) radiation. The UV radiation causes damage to DNA in keratinocytes, which activates the p53 protein. p53 increases the expression of the POMC gene. The POMC protein is then cleaved to produce α-MSH. The keratinocytes secrete α-MSH, which signals nearby melanocytes. The increased melanin absorbs UV radiation, reducing further DNA damage. Figure 3. Production of α-MSH in keratinocytes in response to UV radiation 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? (A) DNA damage due to UV radiation will be strongly inhibited, resulting in 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) Skin pigmentation will not be able to change, resulting in a negative selection pressure.

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) The enzyme-regulated processing of pre‑mRNA into mature mRNA

In mammals, the dark color of skin, hair, and eyes is due to a pigment called melanin. Melanin is produced by specialized skin cells called melanocytes. The melanin is then transferred to other skin cells called keratinocytes. Melanocytes synthesize melanin in a multistep metabolic pathway (Figure 1). The amount of melanin produced is dependent on the amount of the enzymes TYR, TRP2, and TRP1 present inside melanocytes. Figure 1. Melanin synthesis pathway The peptide hormone α-melanocyte-stimulating hormone (α-MSH) activates a signal transduction pathway leading to the activation of MITF. MITF is a transcription factor that increases the expression of the TYR, TRP2, and TRP1 genes (Figure 2). Figure 2. Activation of melanin synthesis genes in melanocytes Some mammals increase melanin production in response to ultraviolet (UV)(UV) radiation. The UV radiation causes damage to DNA in keratinocytes, which activates the p53 protein. p53 increases the expression of the POMC gene. The POMC protein is then cleaved to produce α-MSH. The keratinocytes secrete α-MSH, which signals nearby melanocytes. The increased melanin absorbs UV radiation, reducing further DNA damage. Figure 3. Production of α-MSH in keratinocytes in response to UV radiation Researchers discovered a mutant form of the TYR gene with a deletion of a single guanine nucleotide in the beginning of the coding sequence. 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 a nonfunctional TYR protein. Since the TRP1���1 and TRP2 genes were not affected, the TRP1 and TRP2 proteins will fill the role of the TYRTYR 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 UVUV 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 UVUV radiation.

Figure 1 shows some relevant details of a model of how a deoxynucleotide, in this case dTMP, is added to a growing strand of DNA. Figure 1. Model showing details of adding a deoxythymidine monophosphate (dTMPdTMP) nucleotide 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 mRNA occurs in the 5′5′ to 3′3′ direction; therefore, the growing DNA strand must also grow in the 5′5′ to 3′3′ direction. (D) The phosphate group, attached to the 5′5′ carbon of the dTMP, forms a covalent bond with the oxygen atom attached to the 3′ carbon of the growing strand.

(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.

Exposure to ultraviolet (UV) radiation is the leading cause of skin cancer in humans. Figure 1 shows a model of how UV exposure damages DNA. Figure 1. Model of damage to DNA caused by UV exposure 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) UV photons cause dimers to form, leading to misshapen DNA, which results in replication and transcription errors.

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? (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) When the miRNA binding site sequence is altered, translation of Q mRNA occurs in the presence of miRNA-delta.

The trp operon in E. coli is an example of a repressible operon that consists of genes coding for enzymes used to synthesize tryptophan. When tryptophan levels are high, the operon is turned off and these genes are not transcribed. However, it is also known that tryptophan does not bind directly to the operator DNA sequence. A regulatory gene called trpR has also been discovered although it is not part of the trp operon. The proposed model of how tryptophan acts as a corepressor is shown in Figure 1. Figure 1. Model of proposed regulation of the trp operon by corepressors trp repressor and tryptophan 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) When trpR is mutated, the trp operon is transcribed regardless of tryptophan levels.

Nuclear pores regulate the passage of substances into and out of the nucleus. Antibodies such as mAb414 have been used to inhibit the movement of substances through the nuclear pores of rat liver cells. Scientists cultured rat liver cells (eukaryotic) and bacteria cells (prokaryotic) in separate dishes with radioactively labeled amino acids. A specific gene in both cell types was engineered to synthesize identical polypeptide chains, and translation of this gene was measured. The procedure was repeated with mAb414 (the inhibitor) added to each of the two cell cultures, and translation was monitored again. Which of the following sets of graphs best summarizes the results of the experiments? CONTROL (No inhibitor) / EXPERIMENTAL (With inhibitor) (A) Graph 1 (B) Graph 2 (C) Graph 3 (D) Graph 4

Graph A