UNIT 5 EXAM - AP BIOLOGY PART 2
Sickle-cell anemia is associated with a mutation in the gene encoding the beta subunit of hemoglobin that results in a change from glutamic acid to valine at position 6. All other amino acids are identical to a normal hemoglobin molecule. Based on the information above, which of the following mutations is the most likely cause of sickle-cell anemia? Responses A A single base-pair substitution in the gene encoding the beta subunit B A single base-pair insertion in the gene encoding the beta subunit C A single base-pair deletion in the gene encoding the beta subunit D A translocation of DNA from one chromosome to another
A A single base-pair substitution in the gene encoding the beta subunit
Comparison of DNA sequences in Table II suggests that a functional GULO gene in lemurs can have a G, C, or T at position 21 but only a G at position 22. Which of the following pairs of predictions is most helpful in explaining the discrepancy? A A substitution at position 21 would result inA substitution at position 22 would result inNo change to the protein A premature stop codon or an amino acid with differentbiochemical characteristics B A substitution at position 21 would result inA substitution at position 22 would result inA different amino acidA premature stop codon or an amino acid with differentbiochemical characteristics C A substitution at position 21 would result inA substitution at position 22 would result inNo change to the proteinA frame shift producing an inactive protein
A A substitution at position 21 would result inA substitution at position 22 would result inNo change to the protein A premature stop codon or an amino acid with differentbiochemical characteristics
Bacterial cells that contain green fluorescent protein (GFP) will fluoresce under ultraviolet light. Which of the following is the most likely outcome of replacing the lacZ gene in the E. coli lac operon with the gene encoding GFP? Responses A Bacteria growing in the presence of lactose will fluoresce under ultraviolet light. B Beta-galactosidase will be made only when bacteria are cultured under ultraviolet light. C Ultraviolet light will cause a bond to form between glucose and galactose monomers. D Ultraviolet light will cause a duplication of the lac operon.
A Bacteria growing in the presence of lactose will fluoresce under ultraviolet light.
Which of the following correctly compares the two processes shown above? Responses A Both processes increase genetic variation. B Both processes represent aspects of sexual reproduction. C Both processes require partitioning of genetic material and organelles. D The amount of genetic material per cell remains constant in both processes.
A Both processes increase genetic variation.
In which of the following would there NOT be a change in the amino acid sequence of the peptide coded for by this DNA? Responses A Changing 3' AAA 5' to read 3' AAG 5' B Changing 3' TTC 5' to read 3' ATC 5' C Changing 3' CCG 5' to read 3' GGC 5' D Deleting the first A from 3' AAA 5' E Deleting the last triplet
A Changing 3' AAA 5' to read 3' AAG 5'
Which of the following modifications of the DNA would produce the greatest change in the primary structure of the polypeptide chain? Responses A Deleting the first T in the second triplet B Changing the second triplet to read 3' CTC 5' C Changing the third triplet to read 3' AAC 5' D Changing the fourth triplet to read 3' CCA 5' E Deleting the sixth triplet
A Deleting the first T in the second triplet
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? Responses 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 Gene A promotes neuron development; gene B promotes programmed cell death in neuronal precursors.
Based on the codon chart above, which of the following amino acid changes is most likely found in the mutated protein? Responses A Glu→Val B Val→Glu C Glu→Pro D Pro->Val
A Glu→Val
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? Responses 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.
A new mutation that arose in one copy of gene X in a somatic cell resulted in the formation of a tumor. Which of the following pieces of evidence best describes how the new mutation directly caused the tumor? Responses A Protein X normally stimulates cell division, and the mutation created an overactive version of protein X. B Protein X normally activates a growth hormone receptor, and the mutation decreased the stability of protein X. C Protein X normally prevents passage through the cell cycle, and the mutation created an overactive version of protein X. D Protein X normally regulates gene expression, and the mutation created an underactive version of protein X that blocked the cell cycle.
A Protein X normally stimulates cell division, and the mutation created an overactive version of protein X.
Which of the following best explains how the prokaryotic expression of a metabolic protein can be regulated when the protein is already present at a high concentration? Responses A Repressor proteins can be activated and bind to regulatory sequences to block transcription. B Transcription factors can bind to regulatory sequences to increase RNA polymerase binding. C Regulatory proteins can be inactivated to increase gene expression. D Histone modification can prevent transcription of the gene.
A Repressor proteins can be activated and bind to regulatory sequences to block transcription.
The bacterium Vibrio cholerae is harmless unless a lysogenic bacteriophage provides the gene coding for the cholera toxin, which converts the bacterium to the virulent form that causes cholera. Which of the following best explains how the gene encoding cholera toxin becomes part of the bacterial genome? Responses A The bacteriophage inserts the toxin gene into the host cell DNA, and the gene is expressed with the rest of the host cell's genes. B The bacteriophage makes copies of the toxin gene and expresses the copies inside the bacteriophage. C The bacteriophage converts its toxin gene into mRNA, which is then translated by the host cell. D The bacteriophage transforms itself into a self-replicating protein that can survive inside the host cell.
A The bacteriophage inserts the toxin gene into the host cell DNA, and the gene is expressed with the rest of the host cell's genes.
Arctic foxes typically have a white coat in the winter. In summer, when there is no snow on the ground, the foxes typically have a darker coat. Which of the following is most likely responsible for the seasonal change in coat color? Responses A The decrease in the amount of daylight in winter causes a change in gene expression, which results in the foxes growing a lighter-appearing coat. B The diet of the foxes in summer lacks a particular nutrient, which causes the foxes to lose their white coat and grow a darker-colored coat. C Competition for mates in the spring causes each fox to increase its camouflage with the environment by producing a darker-appearing coat. D The lower temperatures in winter denature the pigment molecules in the arctic fox coat, causing the coat to become lighter in color.
A The decrease in the amount of daylight in winter causes a change in gene expression, which results in the foxes growing a lighter-appearing coat.
Which of the following best explains how bacteria can be genetically engineered to produce a desired antigen? Responses 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 mRNA 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.
Which of the following best explains why there is no growth on plate II? Responses A The initial E. coli culture was not ampicillin-resistant. B The transformation procedure killed the bacteria. C Nutrient agar inhibits E. coli growth. D The bacteria on the plate were transformed.
A The initial E. coli culture was not ampicillin-resistant.
The TAS2R38 receptor protein has been detected on the surface of cells from individuals who are homozygous for the nontaster allele of the TAS2R38 gene. Which of the following is the most likely effect of the mutations associated with the nontaster allele on TAS2R38 gene expression? Responses A The mutations change the primary structure of the encoded receptor protein. B The mutations increase the stability of the TAS2R38 mRNA. C The mutations prevent transcription of the TAS2R38 gene. D The mutations prevent translation of the TAS2R38 mRNA.
A The mutations change the primary structure of the encoded receptor protein.
The cells lining the respiratory tract of pigs have receptors for both avian and human influenza viruses. Based on the model above, which of the following best describes the origin of the new strain of human influenza virus? Responses A The new viral strain inherited a mixture of genetic material from both avian influenza virus and human influenza virus. B The new viral strain inherited RNA molecules from the avian influenza virus and packaged them inside the human influenza virus membrane. C The new viral strain inherited a mutant DNA molecule from the pig respiratory tract cell. D The new viral strain inherited an RNA molecule that had recombined with a DNA molecule from the pig respiratory tract cell.
A The new viral strain inherited a mixture of genetic material from both avian influenza virus and human influenza virus.
Which of the following can best be used to justify why the GFP is expressed by E. coli cells after transformation with the plasmid? Responses A The presence of arabinose in the nutrient agar activated the expression of the genes located downstream of the ara operon regulatory sequences. B The combination of ampicillin and arabinose in the nutrient agar inhibited the expression of certain gene products, resulting in the increased expression of the GFP. C The nutrient agar without arabinose but with ampicillin activated the expression of the genes located downstream of the ara operon regulatory sequences. D Both arabinose and ampicillin were required in the nutrient agar to activate the expression of genes located downstream of the ara operon regulatory sequences.
A The presence of arabinose in the nutrient agar activated the expression of the genes located downstream of the ara operon regulatory sequences.
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. Responses 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 ���� 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 ���� 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.
Which of the following is a correct statement about mutations? Responses A They are a source of variation for evolution. B They drive evolution by creating mutation pressures. C They are irreversible. D They occur in germ cells but not in somatic cells. E They are most often beneficial to the organisms in which they occur.
A They are a source of variation for evolution.
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.
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.
Gibberellic acid stimulates the cells of germinating grass seeds to produce mRNA molecules that code for hydrolytic enzymes. In this case the role of gibberellic acid can best be described as that of Responses A a regulator of gene activity B a stimulator of hydrolase secretion C a stimulator of DNA replication D an allosteric activator of hydrolase E an activator of translation
A a regulator of gene activity
Plates I and III were included in the experimental design in order to Responses A demonstrate that the E. coli cultures were viable B demonstrate that the plasmid can lose its ampr gene C demonstrate that the plasmid is needed for E. coli growth D prepare the E. coli for transformation
A demonstrate that the E. coli cultures were viable
Sickle-cell anemia results from a point mutation in the HBB gene. The mutation results in the replacement of an amino acid that has a hydrophilic R-group with an amino acid that has a hydrophobic R-group on the exterior of the hemoglobin protein. Such a mutation would most likely result in altered Responses A properties of the molecule as a result of abnormal interactions between adjacent hemoglobin molecules B DNA structure as a result of abnormal hydrogen bonding between nitrogenous bases C fatty acid structure as a result of changes in ionic interactions between adjacent fatty acid chains D protein secondary structure as a result of abnormal hydrophobic interactions between R-groups in the backbone of the protein
A properties of the molecule as a result of abnormal interactions between adjacent hemoglobin molecules
If the scientist had forgotten to use DNA ligase during the preparation of the recombinant plasmid, bacterial growth would most likely have occurred on which of the following? Responses A 1 and 2 only B 1 and 4 only C 4 and 5 only D 1, 2, and 3 only E 4, 5, and 6 only
B 1 and 4 only
If the scientist used the cultures to perform another experiment as shown above, using medium that contained lactose as the only energy source, growth would most likely occur on which of the following plates? Responses A 10 only B 7 and 8 only C 7 and 9 only D 8 and 10 only E 9 and 10 only
B 7 and 8 only
Which of the following correctly predicts the movement of a mutant allele compared with the movement of the normal allele when analyzed by gel electrophoresis? Responses A An allele with base pair substitutions in an exon will travel farther than the normal allele does. B An allele with a duplication in an exon will travel a shorter distance than the normal allele does. C An allele with base pair deletions in an exon will travel the same distance as the normal allele does. D An allele with a duplication in an exon will travel a longer distance than the normal allele does.
B An allele with a duplication in an exon will travel a shorter distance than the normal allele does.
The processes illustrated in the models depicted above all result in which of the following? Responses A Transcription B An increase in genetic variation C An increase in the chromosome number D Horizontal gene transfer
B An increase in genetic variation
Which of the following best explains why there is a difference in phenotype between BMD and DMD? Responses A BMD can be caused by the deletion of exons, while DMD is caused by the deletion of introns. B BMD can be caused by a deletion that affects a single amino acid, while DMD is caused by a deletion that alters all the amino acids that follow the mutation. C BMD can be caused by a deletion that removes the transcription start site, while DMD is caused by a deletion that removes the transcription stop signal. D BMD can be caused by the inheritance of two alleles containing deletions, while DMD is caused by the inheritance of one allele containing the deletion and one normal allele.
B BMD can be caused by a deletion that affects a single amino acid, while DMD is caused by a deletion that alters all the amino acids that follow the mutation.
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? Responses A Because nondisjunction occurred in anaphase I, all gametes will be normal and the resulting individual will be phenotypically normal. B Because nondisjunction occurred in anaphase I, all gametes will have an abnormal chromosome number and the individual will likely exhibit phenotypic evidence of the nondisjunction event. C Because nondisjunction occurred in anaphase II, all gametes will be normal and the resulting individual will be phenotypically normal. D Because nondisjunction occurred in anaphase II, all gametes will have an abnormal chromosome number and the individual will likely exhibit phenotypic evidence of the nondisjunction event.
B Because nondisjunction occurred in anaphase I, all gametes will have an abnormal chromosome number and the individual will likely exhibit phenotypic evidence of the nondisjunction event.
Which of the following is the most likely consequence of a mutation at the operator locus that prevents binding of the repressor protein? Responses A Expression of the structural genes will be repressed, even in the presence of lactose. B Beta-galactosidase will be produced, even in the absence of lactose. C RNA polymerase will attach at the Plac locus, but transcription will be blocked. D The operator locus will code for a different protein and thereby prevent transcription of the structural gene.
B Beta-galactosidase will be produced, even in the absence of lactose.
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? Responses 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 Expression of the Pitx1 gene is affected by mutations at other genetic loci.
An answer to which of the following questions would provide the most information about the association between the CFTR mutation and the viscous mucus? Responses A Is the mucus also secreted from the cells through the CFTR proteins? B How does the disrupted chloride movement affect the movement of sodium ions and water by the cell? C How does the mutation alter the structure of the CFTR proteins? D What is the change in nucleotide sequence that results in the CFTR mutation?
B How does the disrupted chloride movement affect the movement of sodium ions and water by the cell?
The Trp operon is a coordinately regulated group of genes (trpA-trpE) that are required for tryptophan biosynthesis in E. coli. Based on the figure above, which of the following correctly describes the regulation of the Trp operon? Responses A In the absence of tryptophan, the repressor is active and binds to the Trp operator, preventing RNA polymerase from transcribing the operon. B In the presence of tryptophan, the repressor is active and binds to the Trp operator, preventing RNA polymerase from transcribing the operon. C In the absence of tryptophan, the trpR gene is inactive, preventing the production of the repressor that blocks expression of the operon. D In the presence of tryptophan, the trpR gene is inactive, preventing the production of the repressor that blocks expression of the operon.
B In the presence of tryptophan, the repressor is active and binds to the Trp operator, preventing RNA polymerase from transcribing the operon.
Which of the following questions will most likely allow the scientist to design an experiment to determine whether unregulated tryptophan synthesis in a strain of E. coli is caused by a mutation in the trpR gene or in the operator of the trp operon? Responses A Is normal regulation of tryptophan synthesis restored when the operator sequence of this strain of E. coli is duplicated? B Is normal regulation of tryptophan synthesis restored when a gene encoding a normal repressor protein is introduced into and expressed by this strain of E. coli? C Is normal regulation of tryptophan synthesis restored when a plasmid that contains a normal operator sequence is introduced into this strain of E. coli? D Is normal regulation of tryptophan synthesis restored when the repressor protein gene is deleted in this strain of E. coli?
B Is normal regulation of tryptophan synthesis restored when a gene encoding a normal repressor protein is introduced into and expressed by this strain of E. coli?
Which of the following is the most likely effect of a mutation in the gene coding for a DNA repair enzyme? Responses A The cell containing the mutation will divide more frequently because the cell cycle checkpoints will not function properly. B Mutations will accumulate more quickly because the cell will not be able to fix errors in replication. C The mutated gene will not be transcribed because RNA polymerase cannot transcribe mutated DNA. D The cell will immediately undergo apoptosis so that mutated DNA is not replicated in future rounds of cell division.
B Mutations will accumulate more quickly because the cell will not be able to fix errors in replication.
Which of the following statements best explains why there are fewer colonies on plate IV than on plate III? Responses A Plate IV is the positive control. B Not all E. coli cells are successfully transformed. C The bacteria on plate III did not mutate. D The plasmid inhibits E. coli growth.
B Not all E. coli cells are successfully transformed.
The p53 protein regulates a cellular response to DNA damage. Based on the diagram above, which of the following best describes the role of p53 in the response to DNA damage? Responses A Phosphorylated p53 binds to DNA and repairs the damage. B Phosphorylated p53 stimulates transcription of p21, and the resulting p21 protein suppresses cell division until the DNA damage is repaired. C Phosphorylated p53 binds cyclin−Cdk complexes, and the resulting protein complex repairs the DNA damage. D Phosphorylated p53 activates p21 proteins, and the p21 proteins in turn repair the DNA damage.
B Phosphorylated p53 stimulates transcription of p21, and the resulting p21 protein suppresses cell division until the DNA damage is repaired.
Phenotype is determined, in part, by which genes are expressed. The diagram below illustrates how the product of gene X regulates the expression of other genes. Which of the following statements best explains how protein X regulates gene expression? Responses A Protein X is responsible for processing pre-mRNA. B Protein X is responsible for activating transcription of some genes but not others. C Protein X is a member of some cytoplasmic protein complexes but not others. D Protein X causes specific base-pair changes to produce new alleles.
B Protein X is responsible for activating transcription of some genes but not others.
Which of the following biotechnology approaches could be used to identify ferritin mRNA in a sample of total cellular RNA? Responses A RNA samples could be directly cloned into a DNA plasmid, grown in bacteria, and tested for the ability to bind iron. B RNA samples could be separated by size using agarose gel electrophoresis and incubated with labeled single-stranded DNA molecules that are complementary to the ferritin mRNA. C RNA samples could be converted to protein and subsequently cut with a restriction endonuclease that recognizes DNA sequences. D RNA samples could be examined under a high-power microscope to visually identify the ferritin mRNA.
B RNA samples could be separated by size using agarose gel electrophoresis and incubated with labeled single-stranded DNA molecules that are complementary to the ferritin mRNA.
Which of the following is an ethical question about Tay-Sachs disease that cannot be answered using scientific methods? Responses A Would a difference in the HEXA alleles in subpopulations I and II affect the severity of the neurological symptoms? B Should genetic testing be required before individuals in subpopulation III are advised to not have children? C Could the frequency of Tay-Sachs carriers in subpopulation I be a consequence of a genetic bottleneck? D Should a statistical test be used to evaluate whether the general population is in Hardy-Weinberg equilibrium?
B Should genetic testing be required before individuals in subpopulation III are advised to not have children?
Which of the following best explains why the polymerase from the species T. aquaticus is often used for PCR? Responses A T. aquaticus polymerase has an optimal temperature of 68°C. B T. aquaticus polymerase does not denature at high temperatures. C T. aquaticus polymerase can be used more than once without degrading. D T. aquaticus polymerase adds nucleotides to both the 3′ and 5′ ends of DNA.
B T. aquaticus polymerase does not denature at high temperatures.
Which of the following is the most likely effect of the mutation at nucleotide position 7 in the GULO gene of humans? Responses A The mutation results in the deletion of the GULO gene, so no polypeptide can be translated. B The deletion of the single nucleotide causes a frame shift, changing the primary structure downstream of the mutation and resulting in a nonfunctional protein. C The point mutation causes a substitution of the amino acid isoleucine (Ile)for histidine (His) at position 7, resulting in a protein with higher than normal activity. D The substitution of a single nucleotide in the GULO coding region results in a stop codon. This results in a smaller nonfunctional protein.
B The deletion of the single nucleotide causes a frame shift, changing the primary structure downstream of the mutation and resulting in a nonfunctional protein.
Which of the following most likely explains how the chromosomes circled in Figure 1 could cause a genetic disorder in the person from whom the cells were obtained? Responses A The extra chromosome causes crowding in the nucleus of the cells and blocks RNA polymerase from binding to and transcribing certain genes. B The extra chromosome will affect the levels of RNA transcribed from certain genes and the amount of protein produced from those genes in each cell. C The cells will not divide and enable growth, because the extra chromosome will interfere with the pairing of homologous chromosomes. D The extra chromosome will cause other chromosomes in the cell to become triploid during future rounds of cell division.
B The extra chromosome will affect the levels of RNA transcribed from certain genes and the amount of protein produced from those genes in each cell.
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? Responses 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 The most anterior expression of Hoxc6 is the eighth vertebra in mammals.
Which of the following best explains how the dystrophin protein is synthesized by certain muscle cells of an individual but not by other cell types? Responses A Each cell has different genes as a result of changes that occur during development and cell specialization. B The presence of different transcription factors in different cell types results in differential gene expression. C A cell can only translate mRNA transcripts that have gene-specific sequences that are recognized by the ribosomes. D The promoter region of the gene is necessary for expression to occur, and this region is only present in certain cell types.
B The presence of different transcription factors in different cell types results in differential gene expression.
Which is a scientific claim that is consistent with the information provided and Figure 1 ? Responses 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 When bound to the operator, the repressor protein prevents lactose metabolism in E. coli.
The single colony found within the clear ring in plate I is most likely made up of the descendants of a bacterial cell that Responses A contaminated the agar plate B contained information conferring resistance to Aureo-mycin C changed its response to Aureomycin as a result of being exposed to the antibiotic D was susceptible to both penicillin and Aureomycin E emigrated from another area of the agar plate
B contained information conferring resistance to Aureo-mycin
Figure 1. Domain structure of two proteins Investigators studied the interactions between two different proteins, protein I and protein II. Each protein has two structural domains (Figure 1) that represent different functional parts of the protein. Different combinations of the protein domains were tested for their ability to bind to a known DNA sequence, interact with each other, and activate transcription. The results are shown in Table 1. Which of the following best models the interaction of protein � and protein II that results in the transcription of target genes? A) Pill shape: G,F Rect: L,M straight B) Pill shape: G,F Rect: M,L diagonal C) Pill shape: G,F Rect: M,L parallel D) Pill shape: M,L Rect: F,G diagonal
B) Pill shape: G,F Rect: M,L diagonal
actase is the enzyme needed to digest lactose, the sugar found in milk. Most mammals produce lactase when they are young but stop once nursing ends. In humans however, many people continue to produce lactase into adulthood and are referred to as lactase-persistent. Which of the following mutations is most likely to cause lactase persistence in humans? Responses A A nucleotide substitution in the coding region of the lactase gene that interferes with the interaction between lactase and lactose B A mutation that turns off the expression of transcription factors that activate the expression of lactase C A mutation that increases the binding of transcription factors to the promoter of the lactase gene D The insertion of a single nucleotide into the lactase gene that results in the formation of a stop codon
C A mutation that increases the binding of transcription factors to the promoter of the lactase gene
Some strains of the bacterium Streptococcus pyogenes secrete poisonous substances called exotoxins. The genes encoding the exotoxins are thought to have originated in bacteriophages, which are viruses that infect bacteria. Which of the following is the most likely mechanism by which the S. pyogenes acquired the ability to produce the exotoxins? Responses A Bacteriophages engulfed cellular debris from dead bacteria. B Bacteriophages in the environment activated bacterial cell division. C Bacteriophage DNAbecame integrated in the bacterial chromosome. D Bacteriophage proteins were absorbed into bacteria cells by endocytosis.
C Bacteriophage DNA became integrated in the bacterial chromosome.
Directions: This group of questions consists of five lettered headings followed by a list of phrases or sentences. For each phrase or sentence, select the one heading to which it is most closely related. Each heading may be used once, more than once, or not at all. (A) Transcription(B) Translation(C) Transformation(D) Replication(E) Reverse transcription Process in which naked DNA is taken up by a bacterial or yeast cell Responses A A B B C C D D
C C
Which of the following is most likely to create genetic variation in a population? Responses A RNA polymerase errors during transcription B Helicase failure to unwind DNA during DNA replication C DNA polymerase errors during replication D Misincorporation of amino acids by tRNA during translation
C DNA polymerase errors during replication
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? Responses 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. coli with 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 Enzymes required for the synthesis of tryptophan were continuously produced whether tryptophan was absent or present in large quantities.
Which of the following best explains the genetic variation among primate species? Responses A Food sources on the island of Madagascar were deficient in ascorbic acid, which caused the primates to migrate to the mainland where ascorbic acid-rich foods are more widely available. B Human activity on Madagascar, including poaching, drove the other primates into extinction on the island. C Food sources where non-Madagascar primates lived provided ample ascorbic acid in the diet, which removed the selective pressure for maintaining a functional GULO gene. D Food sources on the island of Madagascar were deficient in ascorbic acid, which caused lemurs to adapt by developing the ability to produce an active GULO protein.
C Food sources where non-Madagascar primates lived provided ample ascorbic acid in the diet, which removed the selective pressure for maintaining a functional GULO gene.
In a second experiment, the plasmid contained the gene for human insulin as well as the ampr gene. Which of the following plates would have the highest percentage of bacteria that are expected to produce insulin? Responses A I only B III only C IV only D I and III
C IV only
In a transformation experiment, a sample of E. coli bacteria was mixed with a plasmid containing the gene for resistance to the antibiotic ampicillin (ampr). Plasmid was not added to a second sample. Samples were plated on nutrient agar plates, some of which were supplemented with the antibiotic ampicillin. The results of E. coli growth are summarized below. The shaded area represents extensive growth of bacteria; dots represent individual colonies of bacteria. Plates that have only ampicillin-resistant bacteria growing include which of the following? Responses A I only B III only C IV only D I and II
C IV only
Plates that have only ampicillin-resistant bacteria growing include which of the following? Responses A I only B III only C IV only D I and II
C IV only
Samples of DNA were isolated from four different individuals and each sample was digested by the same restriction enzymes. Gel electrophoresis was used to separate the resulting DNA fragments and the results are shown above. These data best support which of the following hypotheses? Responses A Individual 1 is the offspring of 2 and 3. B Individual 1 is the offspring of 3 and 4. C Individual 2 is the offspring of 1 and 3. D Individual 2 is the offspring of 1 and 4. E Individual 3 is the offspring of 1 and 4.
C Individual 2 is the offspring of 1 and 3.
Oncogenes are genes that can cause tumor formation as a result of a particular mutation. Which of the following potential therapies would be most effective at preventing the expression of an oncogene? Responses A Reducing the number of ribosomes in the cell to prevent the creation of the oncogene's proteins B Blocking membrane-bound receptors of transcription factors C Introducing a chemical that binds to transcription factors associated with the oncogene's promoter D Producing additional transcription factors for tumor suppressor genes in the cell
C Introducing a chemical that binds to transcription factors associated with the oncogene's promoter
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? Responses 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.
Which of the following best explains the distribution of lactase persistence in the areas shown in Figure 1 ? Responses 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.
Protein X activates gene expression only in cells exposed to a specific signaling molecule. In a study, researchers determined the intracellular location of Protein X in cultured cells both before and after exposing the cells to the signaling molecule. The results of the study are shown in the diagram. Based on the results, which of the following best describes what Protein X is? Responses A Protein X is an RNA splicing enzyme. B Protein X is a cell membrane receptor protein. C Protein X is a transcription factor. D Protein X is a hormone.
C Protein X is a transcription factor.
Which of the following best describes the type of mutation shown in Figure 1 that leads to sickle-cell anemia? Responses A Insertion B Deletion C Substitution D Frameshift
C Substitution
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 ? Responses 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 The activator is a sequence-specific DNA-binding protein that is present in some tissues but not in other tissues.
Which of the following best explains how some cells of an individual produce and secrete a specific enzyme, but other cells of the same individual do not? Responses A The cells contain different genes and therefore do not make the same proteins. B The cells have evolved under different selective pressures, resulting in some cells making proteins that others cannot. C The cells transcribe and translate different combinations of genes, leading to the production of different sets of proteins. D The cells produce different types of ribosomes that enable the translation of different genes.
C The cells transcribe and translate different combinations of genes, leading to the production of different sets of proteins.
A researcher claims that Tay-Sachs carriers are protected against the infectious disease tuberculosis (TB). Which of the following observations about the annual incidence of tuberculosis in subpopulation II could best be used to support the researcher's claim? Responses A The incidence of TB in subpopulation II is roughly equal to the incidence of TB in the general population. B The incidence of TB in subpopulation II is greater than the incidence of TB in the general population. C The incidence of TB in subpopulation II is lower than the incidence of TB in the general population. D The incidence of TB in subpopulation II is roughly equal to the incidence of Tay-Sachs disease in the general population.
C The incidence of TB in subpopulation II is lower than the incidence of TB in the general population.
The banding patterns of the DNA fragments reveal that Responses A child 1 and child 2 cannot be biological siblings B child 1 and child 3 probably look like the mother C the mother cannot be the biological parent of all three children D the mother's DNA has the same DNA sequence as the father's DNA E child 2 and child 3 inherited all of their DNA from the father
C the mother cannot be the biological parent of all three children
Maximal transcription of the lac operon requires Responses A the presence of high levels of glucose B the removal of the operator region of the operon C the presence of cyclic AMP and lactose D the activation of the repressor protein E the presence of cyclin and cyclin-dependent kinase
C the presence of cyclic AMP and lactose
A researcher claims that an individual has protoporphyria, based on a physical exam. Which of the following techniques would most likely confirm the researcher's claim? Responses A Transforming bacteria with the mutant variation of ����2 to measure gene expression B Culturing cells from the individual in the lab and measuring the cells' growth rate C Using light microscopy to examine the individual's chromosomes during metaphase D Determining the nucleotide sequence of the individual's ����2 alleles
D Determining the nucleotide sequence of the individual's ����2 alleles
Which of the following scientific claims is most consistent with the information provided in Figure 1 ? Responses A Gene � codes for a transcription factor required for transcription of gene �. B A single transcription factor regulates transcription similarly, regardless of the specific gene. C Transcription of genes �, �, and � is necessary to transcribe gene �. D Different genes may be regulated by the same transcription factor.
D Different genes may be regulated by the same transcription factor.
Mutations in the ���6 and ���4�3 genes have been associated with a form of hereditary hearing loss in humans. Researchers studying the genes have proposed that ���4�3 encodes a transcription factor that influences the regulation of ���6. Which of the following questions will best help guide the researchers toward a direct test of their proposal? Responses A Have mutations in other genes also been associated with hearing loss? B In what types of cells are the mutant forms of the ���4�3 gene expressed? C Are mutations in the ���6 and ���4�3 genes also found in mice? D Do mutations in the ���4�3 gene affect ���6 mRNA levels in cells?
D Do mutations in the ���4�3 gene affect ���6 mRNA levels in cells?
Living cells typically have biosynthetic pathways to synthesize at least some of the amino acids used in making proteins. Some strains of E. coli, a prokaryote, can synthesize the amino acid tryptophan, while other E. coli strains cannot. Similarly, some strains of the yeast S. cerevisiae, a eukaryote, can synthesize tryptophan, while other S. cerevisiae strains cannot. Which of the following describes the most likely source of genetic variation found in the tryptophan synthesis pathways of both species? Responses A Exchange of genetic information occurs through crossing over. B Viral transmission of genetic information required to synthesize tryptophan occurs. C Random assortment of chromosomes leads to genetic variation. D Errors in DNA replication lead to genetic variation.
D Errors in DNA replication lead to genetic variation.
Histone methyltransferases are a class of enzymes that methylate certain amino acid sequences in histone proteins. A research team found that transcription of gene � decreases when histone methyltransferase activity is inhibited. Which scientific claim is most consistent with these findings? Responses A DNA methylation inhibits transcription of gene �. B Histone modifications of genes are usually not reversible. C Histone methylation condenses the chromatin at gene � so transcription factors cannot bind to DNA. D Histone methylation opens up chromatin at gene � so transcription factors can bind to DNA more easily.
D Histone methylation opens up chromatin at gene � so transcription factors can bind to DNA more easily.
Methicillin-resistant Staphylococcus aureus (MRSA) can be a serious threat to human health. There is evidence that S. aureus infections are common in hospitals and that MRSA have become resistant to other antibiotics besides methicillin. This suggests that the rapid evolution of resistance in the bacteria poses a serious public-health challenge. Which of the following best explains the ability of MRSA to evade existing drug therapies? Responses A MRSA have very long generation times and very large population sizes. B MRSA develop new alleles by intentionally introducing specific mutations that will give them a selective advantage over other bacteria. C MRSA metabolize many drugs in their lysosomes and therefore evolve resistance at a high rate. D MRSA exchange genetic material with other antibiotic-resistant bacteria, which can spread resistance in the S. aureus population.
D MRSA exchange genetic material with other antibiotic-resistant bacteria, which can spread resistance in the S. aureus population.
Which of the following claims is best supported by the data in Table 1 ? Responses A The transformation procedure killed all the bacteria that were added to plate 3. B More bacteria on plates 1 and 2 were successfully transformed than on any other plate. C None of the bacteria on plate 2 were successfully transformed with the kanamycin resistance gene. D Only the bacteria that were successfully transformed with the kanamycin resistance gene grew on plate 4.
D Only the bacteria that were successfully transformed with the kanamycin resistance gene grew on plate 4.
Genetic engineering techniques can be used when analyzing and manipulating DNA and RNA. Scientists used gel electrophoresis to study transcription of gene � and discovered that mRNA strands of three different lengths are consistently produced. Which of the following explanations best accounts for this experimental result? Responses A Gel electrophoresis can only be used with DNA (not mRNA), so experimental results are not interpretable. B RNA polymerase consistently makes the same errors during transcription of gene �. C Gene � is mutated, so RNA polymerase does not always transcribe the correct sequence. D Pre-mRNA of gene � is subject to alternative splicing, so three mRNA sequences are possible.
D Pre-mRNA of gene � is subject to alternative splicing, so three mRNA sequences are possible.
A mutation that affects Pitx1 gene function in all tissue types is most likely to be at which of the following genetic loci? Responses A Hindlimb enhancer B Pituitary enhancer C Jaw enhancer D Promoter
D Promoter
Which of the following is the best explanation for the fragment pattern for individual X ? Responses A She has only one member of this chromosome pair. B She has only one living parent. C She is homozygous for this particular DNA fragment. D She is the mother's child from another marriage. E She is not related to any member of the family being tested.
D She is the mother's child from another marriage.
Based on the information presented, which of the following best explains the difference in phenotype between Tay-Sachs carriers and homozygous recessive individuals? Responses A Tay-Sachs carriers received a vaccination that homozygous recessive individuals did not receive. B Tay-Sachs carriers inherited an extra chromosome that homozygous recessive individuals did not inherit. C Tay-Sachs carriers have access to a critical nutrient that homozygous recessive individuals did not inherit. D Tay-Sachs carriers synthesize an essential enzyme that homozygous recessive individuals cannot synthesize.
D Tay-Sachs carriers synthesize an essential enzyme that homozygous recessive individuals cannot synthesize.
In sticklebacks, which of the following is most likely to occur if the jaw enhancer is disabled instead of the hindlimb enhancer? Responses A The jaw and a pronounced pelvic spine develop normally because the Pitx1 gene is expressed in both developing tissues. B Neither the jaw nor a pronounced pelvic spine develop normally because there is no Pitx1 gene expression in either developing tissue. C The jaw develops normally, but a pronounced pelvic spine does not develop because the Pitx1 gene is expressed in the developing jaw but not in the developing pelvis. D The jaw does not develop normally, but a pronounced pelvic spine does develop because the Pitx1 gene is expressed in the developing pelvis but not in the developing jaw.
D The jaw does not develop normally, but a pronounced pelvic spine does develop because the Pitx1 gene is expressed in the developing pelvis but not in the developing jaw.
The diagram above represents a segment of the E. coli chromosome that contains the lacI gene and part of the lac operon, a coordinately regulated set of genes that are required for the metabolism of lactose. The presence of lactose, which causes the repressor to be released from the operator, results in increased transcription of the lac operon. Which of the following best explains the contribution of the lac operon to the metabolic efficiency of a bacterial cell? Responses A Expression of the lacI gene requires lactose. B RNA polymerase is rapidly degraded by the product of the lacP locus. C The repressor binds to DNA only when the cellular concentration of glucose is low. D The lacZ gene is highly expressed only when lactose is available.
D The lacZ gene is highly expressed only when lactose is available.
Which of the following best supports a claim that an E. coli strain has a mutation in gene E that results in the production of a nonfunctional enzyme? Responses A Cells of this strain synthesize very little of the repressor protein and produce a large amount of tryptophan. B RNA polymerase does not bind to the promoter of the operon in cells of this strain unless tryptophan is added to the nutrient medium. C The operator sequence of the operon always has repressor proteins bound to it, independent of the amount of tryptophan in the nutrient medium. D mRNA is continuously transcribed from the operon, but the amount of tryptophan produced from the cells of this strain remains low.
D mRNA is continuously transcribed from the operon, but the amount of tryptophan produced from the cells of this strain remains low.
If no new mutations occur, it would be most reasonable to expect bacterial growth on which of the following plates? Responses A 1 and 2 only B 3 and 4 only C 5 and 6 only D 4, 5, and 6 only E 1, 2, 3, and 4 only
E 1, 2, 3, and 4 only
Which of the following would most likely be observed in plate II after 24 hours? Responses A A clear ring larger than that around disc A in plate I would appear around disc A only. B A clear ring larger than that around disc A in plate I would appear around disc P only. C A clear ring smaller than that around disc A in plate I would appear around disc P only. D There would be a clear ring around both disc A and disc P. E There would not be a clear ring around either disc A or disc P.
E There would not be a clear ring around either disc A or disc P.
Which of the following statements best describes the results seen in Figure 2 ? Responses A Individuals III-2 and III-3 carry two different alleles of the ��� gene, a mutant allele and a wild-type allele. Individual IV-1 inherited two copies of the mutant allele. B Individuals III-2 and III-3 carry two different alleles of the ��� gene, a mutant allele and a wild-type allele. Individual IV-1 inherited two copies of the wild-type allele. C Individuals III-2 and III-3 both carry two wild-type alleles. Individual IV-1 inherited two copies of the wild-type allele. D Individuals III-2 and III-3 both carry two wild-type alleles, but individual IV-1 inherited one copy of the wild-type allele and one copy of the mutant allele.
Individuals III-2 and III-3 carry two different alleles of the ��� gene, a mutant allele and a wild-type allele. Individual IV-1 inherited two copies of the mutant allele.