Quiz 7: Anatomy and Function of a Gene: Dissection Through Mutation

In a segment of DNA, an A base is changed to a C and the error is copied during replication.Under which circumstances is that change considered a mutation? (Select all that apply. ) A)The change alters the phenotype of the organism's offspring. B)The change results in production of a novel protein. C)The change is in a somatic cell, such as a skin cell. D)The change is in a gene, but has no effect on a gene product or phenotype.

Replacing a thymine nucleotide with a guanine is an example of a A)translocation. B)transition. C)transversion. D)forward mutation. E)reversion or reverse mutation

Replacing an adenine nucleotide with a guanine is an example of a A)translocation. B)transition. C)transversion. D)forward mutation. E)reversion or reverse mutation

Assume that a wild-type sequence is 5' AGCCTAC 3'.Which sequence could be produced by a transversion? A)5' AGTCTAC 3' B)5' AGCCGCCGCCGCCTAC 3' C)5' AGCCCAC 3' D)5' ATCCTAC 3' E)5' AGCCTGC 3'

Assume that the mutation rate for a given gene is 5 × 10⁻⁶ mutations per generation.For that gene, how many mutations would be expected if 10 million sperm are examined? A)none B)5 × 10⁻⁶ C)5 D)50 E)500

Which type of mutation is least likely to revert? A)deletion B)transition C)transversion D)insertion E)All are equally likely.

When Luria and Delbruck assayed bacterial growth on the petri plates, what did they find? A)All plates had some bacterial colonies; some plates had more bacterial colonies than others. B)Some plates had no bacterial colonies; others had varied numbers of colonies. C)All plates had a similar number of bacterial colonies. D)Some plates had no bacterial colonies; all of the other plates had a similar number of bacterial colonies. E)No bacterial colonies formed on any of the plates.

In the Luria Delbruck experiment, what did the appearance of a bacterial colony on a petri plate indicate? A)The bacterial colony was derived from one initial cell that was resistant to infection by the bacteriophage. B)The bacterial colony was made up of many cells that became resistant to bacteriophage infection independently. C)A colony formed when a bacterial cell was not resistant to bacteriophage infection. D)The bacterial cells that made up a colony were resistant to antibiotic.

What did the results of the Luria Delbruck fluctuation experiment indicate? A)All bacteria are naturally resistant to infection by phage. B)Some individuals in any bacterial population are resistant to infection by phage. C)Exposure to phage induces some bacteria to mutate and become resistant to phage. D)Resistance to phage in bacteria is caused by random spontaneous mutation. E)Some phage mutated to produce large plaques with sharp edges.

A researcher planned to duplicate the Luria Delbruck fluctuation experiment.This researcher inoculated twenty samples of liquid nutrient media with bacteria from the same colony and let them grow overnight.The next morning the researcher noticed that all but one of the flasks had come open and were ruined.Not wishing to redo the experiment, the researcher spread samples of bacteria from the one intact liquid culture on twenty petri plates that had nutrient agar infused with bacteriophage.What results do you expect the researcher obtained from this experiment? A)All plates had some bacterial colonies; some plates had more bacterial colonies than others. B)Some plates had no bacterial colonies; others had varied numbers of colonies. C)All plates had a similar number of bacterial colonies. D)Some plates had no bacterial colonies; all of the other plates had a similar number of bacterial colonies. E)No bacterial colonies formed on any of the plates.

What does the hydrolysis of a purine base from the deoxyribose-phosphate backbone leave behind? A)an apurinic site B)a modified base C)a double-strand break D)a mutation E)a deletion

Thymine dimers are caused by A)X-rays. B)active oxygen species. C)a mutagen such as ethylmethane sulfonate (EMS). D)depurination. E)UV light.

Exposure to UV light from the sun or tanning beds causes A)depurination. B)deamination. C)alkylation. D)thymine dimers. E)oxidation.

The heritable disorder fragile X syndrome, a major cause of intellectual disability, is caused by A)production of enzymes that break the phosphate backbone. B)UV light. C)X-rays. D)the presence of an extra X chromosome in the sperm or egg. E)the expansion of a region of trinucleotide repeats.

If a man has one premutation allele for fragile X syndrome, what is the probability that he will pass the premutation allele on to his son? A)100% B)75% C)50% D)25% E)0%

The base analog 5-bromouracil can tautomerize.One tautomer behaves like thymine and the other behaves like cytosine.If 5-bromouracil is used as a mutagen, what is the minimum number of times replication must occur to mutate the codon for proline (CCC)into the codon for alanine (GCC)? A)one generation B)two generations C)three generations D)5-bromouracil cannot induce this mutation

The base analog 5-bromouracil can tautomerize.One tautomer behaves like thymine and the other behaves like cytosine.If 5-bromouracil is used as a mutagen, what is the minimum number of times replication must occur to mutate the codon for proline (CCC)into the codon for serine (TCC)? A)one generation B)two generations C)three generations D)5-bromouracil cannot induce this mutation

Intercalating agents, such as acridine orange, are mutagens because they A)promote transitions. B)remove amine groups. C)attach to purines causing distortions. D)add ethyl or methyl groups. E)fit between stacked bases and disrupt replication.

Alkylating agents, such as ethylmethane sulfate (EMS), are mutagens because they A)promote deletions and insertions. B)remove amine groups. C)add hydroxyl groups to bases. D)add ethyl or methyl groups to bases. E)fit between stacked bases and disrupt replication.

In a test tube, a particular DNA polymerase has an error rate of 1 mistake in every 106 bases copied.Why is the mutation rate in a cell that uses this DNA polymerase much lower? A)The polymerase is more careful in replicating regions where genes exist. B)Repair mechanisms correct errors made by the polymerase. C)Not all mutations can be detected easily. D)The DNA polymerase has no proofreading function. E)Mutations do not occur if mutagens are not present.

The first step in base excision repair is A)removal of a double-stranded fragment of damaged DNA. B)nicking the backbone of one DNA strand. C)recognizing and excising the incorrect base from a nucleotide. D)removing extraneous groups such as methyl or oxygen added by mutagens. E)replacing an A-T base pair with a C-G base pair.

The genetic condition xeroderma pigmentosum, which can lead to skin cancer, results from the inability to A)correct UV-induced thymine dimers. B)process the amino acid phenylalanine. C)produce functional hemoglobin. D)correct transitions. E)correct breaks in the X chromosome.

A bacterial mismatch repair system is able to correct replication errors that insert an incorrect nucleotide.How is this repair system able to determine which mismatched base is incorrect? A)The incorrect base results in a bulge on only the new strand. B)The incorrect base is in the parental strand which is modified with methyl groups. C)Some bases of the parental strand are methylated, while none of the new strand are. D)A methyl group is attached to the incorrect base in the newly synthesized DNA strand. E)DNA polymerase is bound to the new strand at the location of the incorrect base.

What is the consequence to a bacterial cell of a mutation that inactivates the enzyme responsible for methylating the A in the DNA sequence 5′ GATC? A)Replication of the lagging strand cannot be completed. B)Thymine dimers cannot be repaired. C)Parental and new DNA strands cannot be distinguished during mismatch repair. D)Expression of certain metabolic genes is decreased. E)The mutation rate of all genes will be lower.

In the Ames test for mutagenicity, A)reverse mutations convert auxotrophs to prototrophs that survive. B)colonies are due to forward mutations that convert prototrophs to auxotrophs. C)cells are treated with mutagen and only cells with no mutations survive. D)cells are treated with excess amino acids and cells with mutations are killed. E)rat liver enzymes are present to protect cells from mutation.

How is the Ames test for mutagenicity useful for identifying potential carcinogens? A)Bacteria do not get cancer so they can survive lethal carcinogens. B)Substances that mutate bacterial DNA are likely to mutate human DNA. C)Bacteria thrive on substances that could cause cancer in humans. D)The same genes that cause cancer in humans can be mutated in bacteria. E)Liver enzymes alter the bacteria so they will behave like human cells.

What is true about the mutations within a complementation group? A)They produce the same mutant phenotype. B)They complement each other and are in the same gene. C)They do not complement each other and are in the same gene. D)They complement each other and are in different genes. E)They do not complement each other and are in different genes.

Which statement(s)about the rII- strain of T4 bacteriophage that Benzer studied is (are)true? (Select all that apply. ) A)It grows in E.coli B, but not in E.coli K(λ). B)It produces larger plaques than wild-type. C)It grows in E.coli K(λ), but not in E.coli B. D)It produces smaller plaques than wild-type.

A plaque is A)a colony of bacteria growing on a plate. B)a clump of bacterial cells that contain phage within them. C)a region on a plate where bacteria that have survived phage infection are living. D)an area on a plate that has live phage-resistant bacteria. E)an area on a plate that has phage and dead bacteria.

One strain of rII- phage has a deletion in the rII region, another has a point mutation in the rII region.When E.coli K(λ)cells are infected with either rII- phage (not both)no plaques form.When E.coli K(λ)cells are infected with both rII- phage simultaneously plaques do form.Why do plaques form when E.coli K(λ)cells are infected with both types of rII- phage simultaneously? A)Recombination between the two phage genomes resulted in a wild-type phage. B)The two phage have mutations in different rII genes and their genomes complement each other. C)A protein made by the deletion phage can repair the DNA of the phage with the point mutation. D)Bacteria can survive infection with one phage, but they cannot survive infection with two phages.

Shown below are the maps of a series of rII- deletion strains (1-5).The deleted region is indicated as (...... )and the intact region as ______. 1 ___________(........... )_______________ 2 _________________(........... )_________ 3 (..................... )_____________________ 4 ________________________(................ ) 5 _____(.......... )______________________ rII- phage strains A-E have point mutations in the rII region.E.coli K(λ)cells are coinfected with one phage that has a deletion and one phage that has a point mutation.The presence of wild-type progeny phage is assessed by the presence (+)or absence (o)of plaques. -Experiment 1 depends on recombination.

Shown below are the maps of a series of rII- deletion strains (1-5).The deleted region is indicated as (...... )and the intact region as ______. 1 ___________(........... )_______________ 2 _________________(........... )_________ 3 (..................... )_____________________ 4 ________________________(................ ) 5 _____(.......... )______________________ rII- phage strains A-E have point mutations in the rII region.E.coli K(λ)cells are coinfected with one phage that has a deletion and one phage that has a point mutation.The presence of wild-type progeny phage is assessed by the presence (+)or absence (o)of plaques. -Experiment 2 depends on recombination.

What statement about biochemical pathways is correct? A)All enzymes in the pathway catalyze the same reaction. B)If an enzyme in a pathway is inactive, adding excessive amounts of its substrate will restore the normal phenotype. C)If an enzyme in a pathway is inactive, adding excessive amounts of its product will restore the normal phenotype. D)If the enzyme that catalyzes the final step in a pathway is inactive, all the other enzymes will be inactivated as well. E)If the first enzyme in a pathway is inactivated, adding the final product will not restore the normal phenotype.

Fruit flies normally have red eyes.Seven different true-breeding strains of fly with white eyes have been identified (A-H).In each strain, the white eye trait is due to an autosomal recessive allele.It is possible that all seven strains have mutations in the same gene.Alternatively, some or all of the strains may have mutations in different genes.To determine how many genes are involved in eye color in these flies, all possible pairwise crosses are performed.The offspring phenotypes resulting from each cross are observed.(+ = wild-type; − = mutant) -Which strains' mutations represent a complementation group? A)C, D, F B)A, B C)E D)A, B, E, G, H

Based on these crosses, how many different eye color genes are mutant in all of these strains collectively? A)1 B)2 C)3 D)4 E)8

In the human genetic disorder alkaptonuria, urine turns black because of the presence of homogentisic acid in individuals with the trait.This is due to A)the presence of large amounts of homogentisic acid in the diet. B)failure of individuals with alkaptonuria to manufacture enzymes involved in the synthesis of homogentisic acid. C)failure of wild-type individuals to manufacture enzymes involved in the synthesis of homogentisic acid. D)failure of the kidneys to remove homogentisic acid from the urine. E)failure of individuals with alkaptonuria to manufacture enzymes involved in the breakdown of homogentisic acid.

Consider the pathway for the synthesis of the amino acid arginine in Neurospora: Mutant strains of Neurospora may carry one or more mutations.Neurospora mutant strain a is grown on minimal media plus supplements as shown.Growth is shown by (+)and no growth is shown by (o).Growing strain a cells accumulate citrulline. Which statement is true about strain a? A)It has one mutation in ARG-H. B)It has two mutations, one in ARG-F and one in ARG-H. C)It has two mutations, one in ARG-E and one in ARG-H. D)It has two mutations, one in ARG-E and one in ARG-F. E)It has mutations in ARG-E, ARG-F, and ARG-H.

Consider the pathway for the synthesis of the amino acid arginine in Neurospora: Mutant strains of Neurospora may carry one or more mutations.Neurospora mutant strain b is grown on minimal media plus supplements as shown.Growth is shown by (+)and no growth is shown by (o). What can you conclude from these data? A)Strain a has only one mutation and it is in ARG-E. B)Strain b has only one mutation and it is in ARG-H. C)Strain a has a mutation in ARG-F and strain b has a mutation in ARG-E. D)Strain b has mutations in ARG-E, ARG-F, and ARG-H.

Which statement about amino acids is false? A)Every amino acid contains a carboxyl group. B)The side chain or R group is the same in each amino acid. C)In a polypeptide, amino acids are joined by peptide bonds. D)One end of a polypeptide, the N terminus, contains a free amino group.

Which statement about amino acids is false? A)A polypeptide is made up of several amino acids attached by peptide bonds. B)Amino acids are linked by peptide bonds that form between two amino groups. C)The C terminus of a polypeptide chain contains a free carboxylic acid group. D)Each type of amino acid has a unique R group.

Which condition does not involve a defect in an enzyme pathway? A)Alkaptonuria B)Albinism C)Sickle cell anemia D)Phenylketonuria (PKU)

Which interaction is not involved in maintaining tertiary structure in protein molecules? A)covalent bonds B)hydrogen bonds C)hydrophobic interactions D)electrostatic interactions E)All of the choices may be involved in maintaining tertiary structure in proteins.

The condition sickle cell anemia is due to A)the insertion of an amino acid. B)the deletion of an amino acid. C)substitution of an amino acid. D)failure to synthesize a hemoglobin molecule. E)deletion of the β-chain hemoglobin gene.

Which statement about sickle cell anemia is false? A)Individuals who are heterozygous for the sickle cell allele cannot make hemoglobin. B)The sickle cell hemoglobin molecule contains an amino acid substitution. C)The hemoglobin molecules of an individual with sickle cell anemia clump together. D)The red blood cells of an individual with sickle cell anemia distort and elongate.

In a polypeptide, what level of structure refers to a localized region that takes on a particular geometry? A)primary structure B)secondary structure C)tertiary structure D)quaternary structure E)both tertiary and quaternary structures

Interaction between two distinct polypeptide chains is what type of structure in a protein? A)primary structure B)secondary structure C)tertiary structure D)quaternary structure E)both primary and secondary structure

What level(s)of protein structure could be affected by a nucleotide substitution that results in an amino acid change in a polypeptide? (Select all that apply. ) A)primary structure B)secondary structure C)tertiary structure D)quaternary structure E)There will be no structural changes.

The photoreceptor protein rhodopsin A)is found in cone cells and is sensitive to weak light at many wavelengths. B)is found in rod cells and is sensitive to weak light at many wavelengths. C)is found in cone cells and is responsible for blue color vision. D)is found in rod cells and is responsible for red color vision. E)is missing in individuals who exhibit red-green color blindness.

Examination of the rhodopsin gene family provides evidence for gene evolution by A)duplication and divergence. B)accumulation of random mutations. C)convergent evolution. D)spontaneous generation.

Red-green color blindness is more common in males than females because A)the red pigment gene is on the X chromosome, the green is on an autosome. B)the green pigment gene is on the X chromosome, the red is on an autosome. C)the rhodopsin gene is on the X chromosome. D)both the red and the green pigment genes are on the X chromosome. E)both the red and the green pigment genes are on an autosome.

Assume that a series of compounds has been discovered in Neurospora.Compounds A-F appear to be intermediates in a biochemical pathway.Conversion of one intermediate to the next is controlled by enzymes that are encoded by genes.Several mutations in these genes have been identified and Neurospora strains 1-4 each contain a single mutation.Strains 1-4 are grown on minimal media supplemented with one of the compounds A-F.The ability of each strain to grow when supplemented with different compounds is shown in the table (+ = growth; o = no growth). Which biochemical pathway fits the data presented? Media supplement A)A → B → C → D → E → F B)A → B → C → F → D → E C)F → B → C → D → A → E D)A → B → C → D → F → E E)A → B → F → E → C → D