unit 6 bio
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? 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
Based on the codon chart above, which of the following amino acid changes is most likely found in the mutated protein? A Glu→Val B Val→Glu C Glu→Pro D Pro→Val
A
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 in No change to the protein A substitution at position 22 would result in A premature stop codon or an amino acid with differentbiochemical characteristics B A substitution at position 21 would result in A different amino acid A substitution at position 22 would result in A premature stop codon or an amino acid with differentbiochemical characteristics C A substitution at position 21 would result in No change to the protein A substitution at position 22 would result in A frame shift producing an inactive protein D A substitution at position 21 would result in An amino acid with different biochemical characteristics A substitution at position 22 would result in No transcription of the gene
A
Plates I and III were included in the experimental design in order to 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
The data can best be used to support which of the following claims about the mechanism for regulating ferritin gene expression? A Iron increases ribosome binding to ferritin mRNA. B Iron causes the ferritin mRNA to be degraded. C Iron increases the effect of actinomycin D on RNA polymerase. D Iron decreases the ability of ferritin mRNA to bind to ribosomes.
A
Which of the following best explains how continuity of genetic information in cells is ensured across generations? A Replication uses a parental strand of DNA as a template to create a new strand of DNA. B DNA molecules are shaped like a double helix with a constant diameter throughout. C Transcription copies the information in DNA into an RNA transcript. D Cells contain different polymerases for DNA replication and transcription.
A
Which of the following best explains how the prokaryotic expression of a metabolic protein can be regulated when the protein is already present at a high concentration? A Repressor proteins can be activated and bind to regulatory sequences to block transcription. B Transcription factors can bind to regulatory sequences to increase RNA polymerase binding. C Regulatory proteins can be inactivated to increase gene expression. D Histone modification can prevent transcription of the gene.
A
Which of the following best explains why there is no growth on plate II? 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
Which of the following can best be used to justify why the GFP is expressed by E. coli cells after transformation with the plasmid? 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
Which of the following samples most likely contains a double-stranded RNA virus? A Sample 1 B Sample 2 C Sample 3 D Sample 4
A
Which of the following statements best describes the results seen in Figure 2 ? A Individuals III-2 and III-3 carry two different alleles of the FXN 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 FXN 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.
A
A karyotype is a representation of all the metaphase chromosomes in a sample of cells from a particular individual (Figure 1). Which of the following most likely explains how the chromosomes circled in Figure 1 could cause a genetic disorder in the person from whom the cells were obtained? A The extra chromosome causes crowding in the nucleus of the cells and blocks RNARNA 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
A scientist adds a chemical to a culture of dividing cells in order to disrupt DNA replication. The replicated DNA produced by the cells is double-stranded, but sections of it lack covalent bonds between adjacent nucleotides (Figure 1). Which of the following claims is best supported by the data? A The chemical prevents the formation of RNA primers. B The chemical inhibits DNA ligase. C The chemical blocks DNA polymerase. D The chemical disrupts hydrogen bonding.
B
As represented in Figure 1 and Figure 2, the phenotypic difference between marine and freshwater sticklebacks involves Pitx1, a gene that influences the formation of the jaw, pituitary gland, and pelvic spine. Enhancer sequences upstream of the Pitx1 genetic locus regulate expression of the Pitx1 gene at the appropriate times and in the appropriate tissues during development. Previous studies have found that a mutation in the hindlimb enhancer interferes with the formation of a pronounced pelvic spine. Which of the following best describes how sticklebacks in the same population with identical copies of the Pitx1 gene can still show phenotypic variation in the pelvic spine character? A The Pitx1 gene is carried on different chromosomes in different individuals. B Expression of the Pitx1 gene is affected by mutations at other genetic loci. C The genetic code of the Pitx1 gene is translated differently in males and females. D The subcellular location of the Pitx1 gene changes when individuals move to a new environment.
B
One mutation in ALAS2 that is associated with protoporphyria is a four-nucleotide deletion. The protein expressed from the mutant allele is 2020 amino acids shorter than the wild-type protein. Which of the following best explains why a shortened protein is produced? A The mutation disrupts the start codon, preventing the ribosome from beginning translation. B The mutation introduces a premature stop codon, causing translation to end early. C The mutation changes the gene's regulatory region, causing unregulated gene expression. D The mutation affects posttranscriptional modifications by preventing the removal of introns.
B
Which of the following best explains why the polymerase from the species T. aquaticus is often used for PCR? 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
Which of the following biotechnology approaches could be used to identify ferritin mRNA in a sample of total cellular RNA? 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
Which of the following is the most likely consequence of a mutation at the operator locus that prevents binding of the repressor protein? 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
Which of the following is the most likely effect of a mutation in the gene coding for a DNA repair enzyme? A The cell containing the mutation will divide more frequently because the cell cycle checkpoints will not function properly. B Mutations will accumulate more quickly because the cell will not be able to fix errors in replication. C The mutated gene will not be transcribed because RNA polymerase cannot transcribe mutated DNA. D The cell will immediately undergo apoptosis so that mutated DNA is not replicated in future rounds of cell division.
B
Which of the following is the most likely effect of the mutation at nucleotide position 7 in the GULO gene of humans? 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
Which of the following statements best explains how protein X regulates gene expression? 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
Which of the following statements best explains why there are fewer colonies on plate IV than on plate III? 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
Based on the results, which of the following best describes what Protein X is? A Protein X is an RNA splicing enzyme. B Protein X is a cell membrane receptor protein. C Protein X is a transcription factor. D Protein X is a hormone.
C
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? A I only B III only C IV only D I and III
C
Lactase 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? 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
Plates that have only ampicillin-resistant bacteria growing include which of the following? A I only B III only C IV only D I and II
C
Which of the following best describes a characteristic of the process shown in Figure 1 that is unique to prokaryotes? A The mRNA is synthesized in a 5′ to 3′ direction. B A single strand of the DNA is being used as a template for the transcription of the mRNA. C The translation of the mRNA is occurring while the mRNA is still being transcribed. D The enzyme that is transcribing the mRNA is RNA polymerase.
C
Which of the following best describes the most likely effect of the formation of a triplex DNA structure (Figure 3) on the synthesis of the frataxin protein? A The binding of the ribosome to the mRNA is prevented, resulting in a decrease in frataxin translation. B The DNA will not degrade in the cytoplasm, leading to an increase in frataxin translation. C RNA polymerase is prevented from binding to the DNA, resulting in a decrease in frataxin mRNA transcription. D The protein will include extra amino acids, resulting in a protein with an altered secondary structure.
C
Which of the following best explains how the expression of a eukaryotic gene encoding a protein will differ if the gene is expressed in a prokaryotic cell instead of in a eukaryotic cell? A No transcript will be made, because eukaryotic DNA cannot be transcribed by prokaryotic RNA polymerase. B The protein will have a different sequence of amino acids, because prokaryotes use a different genetic code. C The protein will be made but will not function, because prokaryotes cannot remove introns. D The protein will not be made, because prokaryotes lack the ribosomes necessary for translation.
C
Oncogenes are genes that can cause tumor formation as a result of a particular mutation. Which of the following potential therapies would be most effective at preventing the expression of an oncogene? A Reducing the number of ribosomes in the cell to prevent the creation of the oncogene's proteins B Blocking membrane-bound receptors of transcription factors C Introducing a chemical that binds to transcription factors associated with the oncogene's promoter D Producing additional transcription factors for tumor suppressor genes in the cell
CWhich of the following statements best explains how protein X regulates gene expression? 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.
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? A Transforming bacteria with the mutant variation of ALAS2ALAS2 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 ALAS2ALAS2 alleles
D
Based on the information presented, which of the following best explains the difference in phenotype between Tay-Sachs carriers and homozygous recessive individuals? 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
The following the DNA sequence is a small part of the coding (nontemplate) strand from the open reading frame of β-hemoglobin gene. Given the codon chart listed below, what would be the effect of a mutation that deletes the G at the beginning of the DNA sequence? A The mutation precedes the gene, so no changes would occur. B Lysine (lys) would replace glutamine (gln), but there would be no other changes. C The first amino acid would be missing, but there would be no other change to the protein. D The reading frame of the sequence would shift, causing a change in the amino acid sequence after that point.
D
Which of the following best describes a characteristic of DNA that makes it useful as hereditary material? A There are many different types of nucleotide bases that can be incorporated into DNA. B The nucleotide bases can also be used to provide the energy needed for reproduction. C Nucleotide bases can be randomly replaced with different nucleotide bases to increase variation. D Nucleotide bases in one strand can only be paired with specific bases in the other strand.
D
Which of the following best describes an event during step 2 in the simplified model above? A A new RNA molecule is synthesized using a DNA template. B A new polypeptide is synthesized using an RNA template. C Thymine nucleotides in an RNA molecule are replaced with uracil nucleotides. D Noncoding sequences are removed from a newly synthesized RNA molecule.
D
Which of the following claims is best supported by the data in Table 1 ? 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
