AP Biology Unit 6

In a transformation experiment, a sample of E. coli bacteria was mixed with a plasmid containing the gene for resistance to the antibiotic ampicillin (amp r ). 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?A) I only B III only C )IV only D) I and II

C) IV only

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

The initial E. coli culture was not ampicillin-resistant.

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 amp r 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

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 Not all E. coli cells are successfully transformed.

In a second experiment, the plasmid contained the gene for human insulin as well as the amp r 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 IV only

A scientist is using an ampicillin-sensitive strain of bacteria that cannot use lactose because it has a nonfunctional gene in the lac operon. She has two plasmids. One contains a functional copy of the affected gene of the lac operon, and the other contains the gene for ampicillin resistance. Using restrictions enzymes and DNA ligase, she forms a recombinant plasmid containing both genes. She then adds a high concentration of the plasmid to a tube of the bacteria in a medium for bacterial growth that contains glucose as the only energy source. This tube (+) and a control tube (-) with similar bacteria but no plasmid are both incubated under the appropriate conditions for growth and plasmid uptake. The scientist then spreads a sample of each bacterial culture (+ and -) on each of the three types of plates indicated below. If no new mutations occur, it would be most reasonable to expect bacterial growth on which of the following plates?

A1 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

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?

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?

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

A sterile agar plate, I, is streaked with a pure culture of bacteria by means of aseptic techniques. Paper discs treated with the antibiotics Aureomycin (A) and penicillin (P) are placed at opposite sides of the plate, as shown in the diagram above. The plate is examined after a 24-hour incubation period, and a clear ring is discovered around disc A, but not around disc P. Within the clear ring around disc A, a single bacterial colony with physical characteristics like those of the pure culture is observed. A second sterile agar plate, II, is then streaked with this single colony and also incubated with antibiotics. The single colony found within the clear ring in plate I is most likely made up of the descendants of a bacterial cell that

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

Which of the following would most likely be observed in plate II after 24 hours?

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.

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 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.

The regulatory sequences of the operon controlling arabinose metabolism (ara operon) were studied to determine whether bacteria can respond to changes in nutrient availability. It is predicted that if those regulatory sequences are functioning properly, the bacteria will produce the enzymes involved in arabinose metabolism (structural genes B, A, and D) in the presence of arabinose. If a gene that encodes a green fluorescent protein (GFP) is substituted for the structural genes of the operon, activation of the regulatory sequences can be assayed by GFP expression. A culture of E. coli cells underwent a transformation procedure with a plasmid containing the regulatory sequences of the ara operon directly upstream of the gene encoding the GFP. The plasmid also confers ampicillin resistance to bacteria. Samples were then plated on different types of culture media. (Note: The GFP fluoresces only under UV light, not under white light.) The table below shows the results. 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 A 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 A the ara operon regulatory sequences.

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?

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 the 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.

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.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 A 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 A in both H. arsenicoxydans and O. tritici. In contrast, E. coli has only one operon devoted to arsenic removal.

Ascorbic acid (vitamin C) is an organic molecule necessary for the health of plants and animals. The majority of animals, including most mammals, synthesize ascorbic acid from organic precursors, but some primates are unable to synthesize ascorbic acid and must instead acquire it from dietary sources, such as certain fruits and vegetables. The L-gulonolactone oxidase (GULO) gene encodes an enzyme that catalyzes a required step in the biosynthesis of ascorbic acid. Most mammals carry a functional copy of the GULO gene, but some primates carry only a GULO pseudogene, which is a nonfunctional variant.

A comparison of GULO genes and GULO pseudogenes from different animals can provide insight into the evolutionary relatedness of the animals. In Table I, selected members of some mammalian groups are listed, along with an indication of their ability to synthesize ascorbic acid. Table II shows an alignment of amino acid coding sequences from homologous regions of the GULO genes and GULO pseudogenes of the organisms listed in Table I. Figure 1 represents the universal genetic code.

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 No change to the protein ;A substitution at position 22 would result in A premature stop codon or an amino acid with different biochemical 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 different biochemical 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)A substitution at position 21 would result No change to the protein ;A substitution at position 22 would result in A premature stop codon or an amino acid with different biochemical characteristics

Lemurs are primates that live on the island of Madagascar off the coast of Africa. Lemurs have a functional GULO gene and are able to produce ascorbic acid. However, primates that live in other places (e.g., humans, chimpanzees, and orangutans) have a GULO pseudogene and are unable to produce ascorbic acid. Which of the following best explains the genetic variation among primate species?

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.

Tay-Sachs disease is a rare inherited disorder caused by an autosomal recessive allele of the HEXA gene. Affected individuals exhibit severe neurological symptoms and do not survive to reproductive age. Individuals who inherit one copy of the allele (Tay-Sachs carriers) typically show no symptoms of the disorder. The frequencies of TaySachs carriers in the general population of North America and in three different subpopulations are presented in the table.

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)Tay-Sachs carriers synthesize an essential enzyme that homozygous recessive individuals cannot synthesize

A model that represents a process occurring in a cell of a particular organism is shown in Figure 1. 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 mRNA of 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.

The table below describes the action of two genes involved in the regulation of nervous system development in the nematode C. elegans.

Which of the following claims is best supported by the data?

A)Gene A promotes neuron development; gene B promotes programmed cell death in neuronal precursors.

B)Gene A promotes programmed cell death in neuronal precursors; gene B promotes neuron development.

C)Gene B must be active before gene A can function

D)Gene B must be inactive before gene A can function

A)Gene A promotes neuron development; gene B promotes programmed cell death in neuronal precursors.

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?

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 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)The reading frame of the sequence would shift, causing a change in the amino acid sequence after that point

The mRNA transcribed from the DNA would read

a)5’ TAG TTC AAA CCG CGT AAC AAT 3’

b)5’ ATC AAG TTT GGC GCA TTG TAA 3’

c)5’ AUC AAG UUU GGC GCA UUG UAA 3’

d)5’ AAU CAA UGC GCC AAA CUU GAU 3’

e)5’ AUU GUU ACG CGG UUU GAA CUA 3’

c)5’ AUC AAG UUU GGC GCA UUG UAA 3’

Which of the following modifications of the DNA would produce the greatest change in the primary structure of the polypeptide chain?

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

In which of the following would there NOT be a change in the amino acid sequence of the peptide coded for by this DNA?

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’

The processes illustrated in the models depicted above all result in which of the following?

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

Cystic fibrosis is a recessively inherited disorder that results from a mutation in the gene encoding CFTR chloride ion channels located on the surface of many epithelial cells. As shown in the figure, the mutation prevents the normal movement of chloride ions from the cytosol of the cell to the extracellular fluid. As a consequence of the mutation, the mucus layer that is normally present on the surface of the cells becomes exceptionally dehydrated and viscous. An answer to which of the following questions would provide the most information about the association between the CFTR mutation and the viscous mucus?

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?

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’ to 5’ direction from their DNA templates.

b)New DNA strands are being synthesized in the 5’ to 3’ direction from their DNA templates

c)A new RNA strand is being synthesized in the 3’ to 5’ end from its DNA template

d)Two new RNA strands are being synthesized in both directions from their DNA templates.

b)New DNA strands are being synthesized in the 5’ to 3’ direction from their DNA templates

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 (dTMP ) 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’ to 3’ direction?

a)The two strands need to be antiparallel to bond properly.

b)Thymine and adenine would not bond properly if the strand grew from 3’ to 5’.

c)The translation of mRNA occurs in the 5’ to 3’ direction; therefore, the growing DNA strand must also grow in the 5’ to 3’ direction.

d)The phosphate group, attached to the 5’ carbon of the dTMP , forms a covalent bond with the oxygen atom attached to the 3’carbon of the growing strand.

d)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.

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)Noncoding sequences are removed from a newly synthesized RNA molecule.

The diagram below illustrates the results of electrophoresis of DNA sequences obtained from a family of two adults and three children, and amplified using PCR. The bands represent short repeating sequences of variable length. Results for another female (X) are included for comparison.The banding patterns of the DNA fragments reveal that

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

Which of the following is the best explanation for the fragment pattern for individual X ?

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.

Histone methyltransferases are a class of enzymes that methylate certain amino acid sequences in histone proteins. A research team found that transcription of gene R decreases when histone methyltransferase activity is inhibited. Which scientific claim is most consistent with these findings?

a)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.

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?

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.

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 Bare positively charged, and moleculeA 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.

. . . glutamine-glutamine-glutamine . . . . . . serine-serine-serine . . . Which of the following messenger RNA sequences could code for both of the two amino acid sequences above, simply by a shift in the reading frame?

a). . . AGCAGCAGCAGC . . .

b). . AGUAGUAGUAGU . . .

c). . . CAACAACAACAA . . .

d). . .GCUGCUGCUGCU . . .

e). . . GCAAGCGCAAGC . . .

a). . . AGCAGCAGCAGC . . .

Genetic engineering techniques can be used when analyzing and manipulating DNA and DNA. 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) RNA polymerase consistently makes the same errors during transcription of gene .

c)Gene L is mutated, so RNA polymerase does not always transcribe the correct sequence

d)Pre-mRNA of gene L is subject to alternative splicing, so three mRNA sequences are possible.

d)Pre-mRNA of gene L is subject to alternative splicing, so three mRNA sequences are possible.

Iron is an essential nutrient that is acquired by organisms from the environment. When intracellular levels of iron are relatively high, living cells synthesize an iron-storage protein called ferritin. The induction of ferritin synthesis by iron was investigated in rats. Figure 1 shows the results of an experiment in which cellular levels of ferritin protein were measured in the presence or absence of iron and actinomycin D, a drug that inhibits transcription. Figure 2 shows the results of an experiment in which cellular levels of ferritin protein were measured in the presence or absence of iron and cycloheximide, a drug that inhibits translation.. 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)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.

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)Iron increases ribosome binding to ferritin mRNA.

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.

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)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.

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.

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.

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 w , all gametes will be normal and the resulting individual will be phenotypically normal.

d)Because nondisjunction occurred in anaphase w, 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 1 , all gametes will have an abnormal chromosome number and the individual will likely exhibit phenotypic evidence of the nondisjunction event

a)The presence of excess lactose blocks the functioning 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.

a)Phosphorylated 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 complexes, and the resulting protein complex repairs the damage.

d)Phosphorylated p 53 activates 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.

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

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.

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.

a)The synthesis of 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

a)GeneX codes for a transcription factor required for transcription of geneD .

b)A single transcription factor regulates transcription similarly, regardless of the specific gene.

c)Transcription of genes A,B C, and 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.

A)Insertion

b)Deletion

c)Substitution

d)Frameshift

c)Substitution

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

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

a)Bacteriophages engulfed cellular debris from dead bacteria.

b)Bacteriophages in the environment activated bacterial cell division

c)Bacteriophage DNA became integrated in the bacterial chromosome

d)Bacteriophage proteins were absorbed into bacteria cells by endocytosis.

c)Bacteriophage DNA became integrated in the bacterial chromosome

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.

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 GTPcap

c) The antisense strand, because it is serving as a template

a) Asp-Arg-met-val-thr-lys-phe-gly-his

b)met-arg-stop-his-gly-phe-lys-thr-val

c)met-valt-thr-lys-phe-gly-his

d)val-thr-lys-phe-gly-his

c)met-valt-thr-lys-phe-gly-his

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.

a) Sample 1

b) sample 2

c) sample 3

d) sample 4

a) sample 1

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

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.

a)Prokaryotic cells are preparing for cell division, because prokaryotes must double the quantity of histone proteins each time they replicate their DNA.

b)Eukaryotic cells are preparing for cell division, because eukaryotes must double the quantity of histone proteins each time they replicate their DNA.

c)Prokaryotic cells are preparing for protein synthesis, because prokaryotes must double the quantity of histone proteins when they manufacture proteins.

d)Eukaryotic cells are preparing for protein synthesis, because eukaryotes must double the quantity of histone proteins when they manufacture proteins.

b)Eukaryotic cells are preparing for cell division, because eukaryotes must double the quantity of histone proteins each time they replicate their DNA.

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.

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.

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 molecules DNA, 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 DNA molecules that provide genes that may aid in survival of the prokaryotic cell.

a)The rate in E. coli is one-tenth the rate in S. cerevisiae

b)The rate in E. coli is half the rate in S. cerevisiae.

c)The rate in E. coli is one-and-one-half times the rate in S. cerevisiae

d)The rate in E. coli is twice the rate in S. cerevisiae.

d)The rate in E. coli is twice the rate in S. cerevisiae.

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.

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

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

(A) Cell signaling to increase gene expression

(B) Translation of an mRNA to produce a protein

(C) Regulation of gene expression by microRNA

(D) Replication of a plasmid

(D) Replication of a plasmid

(A) The host-cell ribosomes translate the viral RNA genome that enters the cell upon initial viral infection.

(B) The viral RNA polymerase that transcribes host genes has a high error rate.

(C) The RNA viral genome is reverse transcribed into DNA that integrates into the host genome.

(D) The RNA viral genome integrates into the host genome

(C) The RNA viral genome is reverse transcribed into DNA that integrates into the host genome.

(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.

(A) The synthesis of 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

A) The chemical prevents the formation of RNA primers.

(B) The chemical inhibits DNA ligase.

(C) The chemical blocks DNA polymerase.

(D) The chemical disrupts hydrogen bonding.

(B) The chemical inhibits DNA ligase.

(A) The cell’s ability to transport the amino acids needed for translation will be reduced.

(B) The cell’s ability to transcribe RNA transcripts that will be translated will be reduced.

(C) The cell’s ability to properly assemble ribosomes and initiate translation will be reduced.

(D) The cell’s ability to modify proteins after they have been assembled will be reduced.

(C) The cell’s ability to properly assemble ribosomes and initiate translation will be reduced.

(A) Reversible changes in DNA he 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.

(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.

(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) Amino acid synthesis will be inhibited.

(B) No RNA 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.

(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.

(A) Normal expression of trpR causes the operon to be transcribed regardless of tryptophan levels.

(B) When the operator sequence is mutated, the trpR 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.

(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.

(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?

(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.

(A) Prokaryotic genes do not contain introns.

(B) Only prokaryotes can translate an mRNA while the mRNA is being transcribed.

(C) Eukaryotes require more amino acids than prokaryotes do and therefore do not shut down production of tryptophan.

(D) Eukaryotes do not have groups of genes that are regulated together.

(B) Only prokaryotes can translate an mRNA while the mRNA is being transcribed.