Deck 7: Gene Expression: the Flow of Information From Dna to Rna to Protein
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Deck 7: Gene Expression: the Flow of Information From Dna to Rna to Protein
1
If a man shows the premutation allele for fragile X syndrome,what is the probability that he will pass it on to his son?
A) 100%
B) 75%
C) 50%
D) 25%
E) 0%
A) 100%
B) 75%
C) 50%
D) 25%
E) 0%
E
2
Assume that a researcher set out to duplicate the Luria-Delbruck fluctuation experiment.This researcher planted twenty small flasks 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 took bacteria samples from the one remaining intact flask and placed them on twenty phage plates.What results would you expect to see when the twenty phage plates are examined,and how would these results compare with those of the original Luria-Delbruck fluctuation experiment?
A) Identical to the Luria-Delbruck results,namely different numbers of resistant colonies.
B) Identical to the Luria-Delbruck results,namely identical numbers of resistant colonies.
C) Not like the Luria-Delbruck results,namely different numbers of resistant colonies.
D) Not like the Luria-Delbruck results,namely identical numbers of resistant colonies.
E) Identical to the Luria-Delbruck results,namely no resistant colonies.
A) Identical to the Luria-Delbruck results,namely different numbers of resistant colonies.
B) Identical to the Luria-Delbruck results,namely identical numbers of resistant colonies.
C) Not like the Luria-Delbruck results,namely different numbers of resistant colonies.
D) Not like the Luria-Delbruck results,namely identical numbers of resistant colonies.
E) Identical to the Luria-Delbruck results,namely no resistant colonies.
D
3
The term mutation refers to
A) only changes in the DNA that result in new phenotypes.
B) only changes in the DNA that result in novel proteins.
C) any change in the DNA of a cell.
D) a heritable change in the DNA of a cell.
E) any change in the cell that changes its survival chances.
A) only changes in the DNA that result in new phenotypes.
B) only changes in the DNA that result in novel proteins.
C) any change in the DNA of a cell.
D) a heritable change in the DNA of a cell.
E) any change in the cell that changes its survival chances.
D
4
The results of the Luria-Delbruck fluctuation experiment indicated that
A) bacteria are naturally resistant to phage.
B) a low level of any bacterial population is naturally resistant to phage.
C) bacteria become resistant to phage by mutation when exposed to phage.
D) bacteria become resistant to phage by random spontaneous mutation.
E) the phage mutate to produce large plaques with sharp edges.
A) bacteria are naturally resistant to phage.
B) a low level of any bacterial population is naturally resistant to phage.
C) bacteria become resistant to phage by mutation when exposed to phage.
D) bacteria become resistant to phage by random spontaneous mutation.
E) the phage mutate to produce large plaques with sharp edges.
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5
The heritable disorder fragile X syndrome,a major cause of mental retardation,is caused by
A) production of enzymes that break the phosphate backbone.
B) UV light.
C) X-rays.
D) presence of an extra X chromosome in the sperm or egg
E) duplication of multiple three-nucleotide repeats.
A) production of enzymes that break the phosphate backbone.
B) UV light.
C) X-rays.
D) presence of an extra X chromosome in the sperm or egg
E) duplication of multiple three-nucleotide repeats.
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6
A heritable change in DNA base sequence is called a
A) forward mutation.
B) reversion.
C) substitution.
D) deletion.
E) mutation.
A) forward mutation.
B) reversion.
C) substitution.
D) deletion.
E) mutation.
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7
A mutation in which parts of two nonhomologous chromosomes change places is called a(n)
A) reciprocal translocation.
B) transition.
C) transversion.
D) insertion.
E) deletion.
A) reciprocal translocation.
B) transition.
C) transversion.
D) insertion.
E) deletion.
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8
UV light is a mutagen that can cause
A) depurination.
B) deamination.
C) alkylation.
D) thymine dimers.
E) oxidation.
A) depurination.
B) deamination.
C) alkylation.
D) thymine dimers.
E) oxidation.
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9
Unequal crossing over results in
A) an exchange between nonhomologous chromosomes.
B) a loss of genetic material.
C) a repair of UV-induced damage.
D) a production of eggs containing Y chromosomes.
E) a creation of deletions and duplications.
A) an exchange between nonhomologous chromosomes.
B) a loss of genetic material.
C) a repair of UV-induced damage.
D) a production of eggs containing Y chromosomes.
E) a creation of deletions and duplications.
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10
Consider the following results.When 50 million sperms were examined for a specific mutation,100 mutations were found.Indicate the mutation rate for that gene.
A) 5 × 10-6
B) 50 × 10-6
C) 2 × 10-6
D) 2 × 10-5
E) 5 × 10-5
A) 5 × 10-6
B) 50 × 10-6
C) 2 × 10-6
D) 2 × 10-5
E) 5 × 10-5
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11
The hydrolysis of a purine base from the deoxyribose-phosphate backbone is called
A) depurination.
B) deamination.
C) replica plating.
D) excision repair.
E) deletion.
A) depurination.
B) deamination.
C) replica plating.
D) excision repair.
E) deletion.
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12
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.
A) translocation.
B) transition.
C) transversion.
D) forward mutation.
E) reversion or reverse mutation.
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13
Thymine dimers are caused by
A) X-rays.
B) free radicals such as oxygen.
C) EMS or NSG.
D) depurination.
E) UV light.
A) X-rays.
B) free radicals such as oxygen.
C) EMS or NSG.
D) depurination.
E) UV light.
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14
The duplication of the triplet sequence CGG resulting in elongation or breakage of the X chromosome is termed
A) bar-eyed.
B) Huntington's disease.
C) unequal crossing over.
D) fragile X syndrome.
E) Rhys syndrome.
A) bar-eyed.
B) Huntington's disease.
C) unequal crossing over.
D) fragile X syndrome.
E) Rhys syndrome.
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15
Assume that a wild-type sequence is 5'AGCCTAC3'.Indicate the sequence that might be produced by a transversion.
A) 5'AGTCTAC3'
B) 5'AGCCGCCGCCGCCTAC3'
C) 5'AGCCCAC3'
D) 5'ATCCTAC3'
E) 5'AGCCTGC3'
A) 5'AGTCTAC3'
B) 5'AGCCGCCGCCGCCTAC3'
C) 5'AGCCCAC3'
D) 5'ATCCTAC3'
E) 5'AGCCTGC3'
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16
Which type of mutation is least likely to revert?
A) deletion
B) transition
C) transversion
D) insertion
E) all are equally likely
A) deletion
B) transition
C) transversion
D) insertion
E) all are equally likely
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17
Assume that the mutation rate for a given gene is 5 × 10-6 mutations per gene per generation.For that gene how many mutations would be expected if 10 million sperms are examined?
A) none
B) 5 × 10-6
C) 5
D) 50
E) 500
A) none
B) 5 × 10-6
C) 5
D) 50
E) 500
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18
Genes on the X chromosome of mammals and Drosophila are particularly suitable for genetic study because
A) males have only one X and most genes behave as haploids.
B) females have only one X and most genes behave as haploids.
C) the X chromosome is large and many more genes are located there.
D) when present as Barr bodies they are exposed for electron microscopic examination.
E) they behave as diploids in females.
A) males have only one X and most genes behave as haploids.
B) females have only one X and most genes behave as haploids.
C) the X chromosome is large and many more genes are located there.
D) when present as Barr bodies they are exposed for electron microscopic examination.
E) they behave as diploids in females.
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19
In the Luria-Delbruck fluctuation experiment,the bacteria + phage plates showed
A) all plates had some resistant colonies,some had very many.
B) some plates had no resistant colonies,a few plates had very many resistant colonies.
C) all plates had the same number of resistant colonies.
D) some plates had no resistant colonies;the plates that had resistant colonies all had the same number of resistant colonies.
E) phage caused mutations to occur in some of the plates but not in others.
A) all plates had some resistant colonies,some had very many.
B) some plates had no resistant colonies,a few plates had very many resistant colonies.
C) all plates had the same number of resistant colonies.
D) some plates had no resistant colonies;the plates that had resistant colonies all had the same number of resistant colonies.
E) phage caused mutations to occur in some of the plates but not in others.
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20
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.
A) translocation.
B) transition.
C) transversion.
D) forward mutation.
E) reversion or reverse mutation.
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21
Intercalating agents such as ethidium bromide function as mutagens to
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.
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.
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22
Assume that a researcher is studying coat colour in voles.Three strains of white vole have been isolated: milky,blanc,and weiss.White is a recessive trait in each strain.Homozygous white voles are obtained for each strain.Consider the following crosses:
Milky × blanc = all white progeny
Milky × weiss = all brown (wild-type vole colour)
Blanc × weiss = all brown (wild-type vole colour)
The conclusion most consistent with these results is
A) all three strains have mutations in the same gene.
B) all three strains have mutations in different genes.
C) milky and blanc have mutations on the same gene,weiss has a mutation in a different gene.
D) milky and weiss have mutations on the same gene,blanc has a mutation in a different gene.
E) weiss and blanc have mutations on the same gene,milky has a mutation in a different gene.
Milky × blanc = all white progeny
Milky × weiss = all brown (wild-type vole colour)
Blanc × weiss = all brown (wild-type vole colour)
The conclusion most consistent with these results is
A) all three strains have mutations in the same gene.
B) all three strains have mutations in different genes.
C) milky and blanc have mutations on the same gene,weiss has a mutation in a different gene.
D) milky and weiss have mutations on the same gene,blanc has a mutation in a different gene.
E) weiss and blanc have mutations on the same gene,milky has a mutation in a different gene.
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23
Excision repair corrects DNA by
A) removing a double-stranded fragment of damaged DNA.
B) detecting,removing,and replacing damaged or incorrect nucleotides in a single strand of DNA.
C) excising the incorrect base from a nucleotide.
D) removing extraneous groups such as methyl or oxygen added by mutagens.
E) correcting A = T to C = G transitions.
A) removing a double-stranded fragment of damaged DNA.
B) detecting,removing,and replacing damaged or incorrect nucleotides in a single strand of DNA.
C) excising the incorrect base from a nucleotide.
D) removing extraneous groups such as methyl or oxygen added by mutagens.
E) correcting A = T to C = G transitions.
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24
If a base analogue such as 5-bromouracil is used as a mutagen,how many generations will be required to mutate the codon for proline (CCC)into the codon for alanine (GCC)?
A) one generation
B) two generations
C) three generations
D) at least two,but perhaps more due to chance
E) it will not occur
A) one generation
B) two generations
C) three generations
D) at least two,but perhaps more due to chance
E) it will not occur
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25
The bacterial repair system that corrects mismatched bases after polymerization is able to discriminate between the old and newly made DNA strands because
A) the new strand will contain the incorrect base if a mismatch occurs.
B) the older DNA is more likely to contain errors.
C) the older DNA contains methyl groups at specific sequences.
D) the newer DNA contains methyl groups at specific sequences.
E) the DNA polymerase is attached to the new strand.
A) the new strand will contain the incorrect base if a mismatch occurs.
B) the older DNA is more likely to contain errors.
C) the older DNA contains methyl groups at specific sequences.
D) the newer DNA contains methyl groups at specific sequences.
E) the DNA polymerase is attached to the new strand.
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26
Alkylating agents such as ethylmethane sulfonate (EMS)function as mutagens to
A) promote deletions and insertions.
B) remove amine groups.
C) add oxygen free radicals to bases.
D) add ethyl or methyl groups.
E) fit between stacked bases and disrupt replication.
A) promote deletions and insertions.
B) remove amine groups.
C) add oxygen free radicals to bases.
D) add ethyl or methyl groups.
E) fit between stacked bases and disrupt replication.
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27
In the Ames test for mutagenicity
A) auxotrophic bacteria are converted to prototrophs that survive.
B) prototrophic bacteria are converted to auxotrophs that survive.
C) cells are treated with mutagen and only those with no mutations survive.
D) cells are treated with excess amino acids,killing cells that carry mutations.
E) rat liver enzymes protect cells from mutation.
A) auxotrophic bacteria are converted to prototrophs that survive.
B) prototrophic bacteria are converted to auxotrophs that survive.
C) cells are treated with mutagen and only those with no mutations survive.
D) cells are treated with excess amino acids,killing cells that carry mutations.
E) rat liver enzymes protect cells from mutation.
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28
A complementation group is a group of mutations
A) that produce the same phenotype.
B) that are in the same gene and complement each other.
C) that are in the same gene and do not complement each other.
D) in two different genes that complement each other.
E) in two different genes that do not complement each other.
A) that produce the same phenotype.
B) that are in the same gene and complement each other.
C) that are in the same gene and do not complement each other.
D) in two different genes that complement each other.
E) in two different genes that do not complement each other.
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29
Assume eight different strains of fly have been isolated,each shows a recessive white eye trait.Crosses are performed as follows (w)indicates white-eyed progeny, (R)indicates wild-type red eyes.
Based on these crosses,how many different genes are present and what strains have mutations in the same gene as does strain A?
A) 2,B,E,and H
B) 3,B,and C
C) 3,B,C,and H
D) 3,B,E,and H
E) 4,B,and H
Based on these crosses,how many different genes are present and what strains have mutations in the same gene as does strain A?
A) 2,B,E,and H
B) 3,B,and C
C) 3,B,C,and H
D) 3,B,E,and H
E) 4,B,and H
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30
Assume that in the organism under study,the DNA polymerase has an error rate of 1 mistake in every 106 bases copied.However,the overall mutation rate is much lower.This is most likely because
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.
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.
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31
Assume that a new low-calorie sweetener is developed.The structure is novel and is tested with the Ames test for mutagenicity.The following results are obtained:
What conclusion is most consistent with these data?
A) The sweetener is not mutagenic.
B) Rat liver enzymes are highly mutagenic.
C) The sweetener is not mutagenic but can be converted into strong mutagens.
D) The sweetener is mutagenic and can be converted into strong mutagens.
E) The sweetener and its conversion products are equally mutagenic.
What conclusion is most consistent with these data?
A) The sweetener is not mutagenic.
B) Rat liver enzymes are highly mutagenic.
C) The sweetener is not mutagenic but can be converted into strong mutagens.
D) The sweetener is mutagenic and can be converted into strong mutagens.
E) The sweetener and its conversion products are equally mutagenic.
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32
The genetic condition Xeroderma pigmentosum,which can lead to skin cancer,results from
A) inability to correct UV induced dimers.
B) inability to process phenylalanine.
C) inability to produce functional haemoglobin.
D) inability to correct transitions.
E) breaks in the X chromosome.
A) inability to correct UV induced dimers.
B) inability to process phenylalanine.
C) inability to produce functional haemoglobin.
D) inability to correct transitions.
E) breaks in the X chromosome.
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33
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 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.
A) the presence of large amounts of homogentisic acid in the diet.
B) failure 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.
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34
Assume seven different strains of fly have been isolated,each shows a recessive white eye trait.Crosses are performed as follows (w)indicates white-eyed progeny, (R)indicates wild-type red eyes.
Based on these crosses,how many different genes are present?
A) only one gene with several different alleles
B) 2
C) 3
D) 4
E) 7
Based on these crosses,how many different genes are present?
A) only one gene with several different alleles
B) 2
C) 3
D) 4
E) 7
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35
The Ames test for mutagenicity is useful to identify potential carcinogens because
A) bacteria do not get cancer they can survive lethal carcinogens.
B) mutagens that affect bacterial DNA are likely to cause human mutation.
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 mammal cells.
A) bacteria do not get cancer they can survive lethal carcinogens.
B) mutagens that affect bacterial DNA are likely to cause human mutation.
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 mammal cells.
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36
In the Ames test for mutagenicity,rat liver enzymes are included with the compound under test because
A) bacterial cell walls must be treated to permit uptake of the compounds.
B) rat liver enzymes increase the sensitivity of the bacteria to mutagens.
C) rat liver enzymes kill mutant cells and allow colonies to form.
D) rat liver enzymes may modify or break down some harmless compounds to produce mutagens.
E) the mutant strain of bacteria requires rat liver enzymes to digest nutrients for growth.
A) bacterial cell walls must be treated to permit uptake of the compounds.
B) rat liver enzymes increase the sensitivity of the bacteria to mutagens.
C) rat liver enzymes kill mutant cells and allow colonies to form.
D) rat liver enzymes may modify or break down some harmless compounds to produce mutagens.
E) the mutant strain of bacteria requires rat liver enzymes to digest nutrients for growth.
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37
The consequence to a bacterial cell of a mutation that inactivated the enzyme that methylates the A of the sequence GATC in newly made DNA would be
A) failure to carry out replication.
B) failure to correct thymine dimers.
C) failure to distinguish old and new DNA during mismatch repair.
D) inactivation of certain metabolic genes.
E) decrease in the mutation rate.
A) failure to carry out replication.
B) failure to correct thymine dimers.
C) failure to distinguish old and new DNA during mismatch repair.
D) inactivation of certain metabolic genes.
E) decrease in the mutation rate.
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38
Base analogues differ from other classes of mutagen in that they
A) only alter bases.
B) can only cause transversions.
C) only work during DNA replication or repair.
D) can only cause forward mutations,nor reversions.
E) will not function in bacterial cells.
A) only alter bases.
B) can only cause transversions.
C) only work during DNA replication or repair.
D) can only cause forward mutations,nor reversions.
E) will not function in bacterial cells.
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39
Choose the statement that is most accurate concerning biochemical pathways.
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.
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.
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40
Consider the pathway for the synthesis of the amino acid arginine in Neurospora:
Mutant strains of Neurospora are grown in minimal media supplements as follows.Each mutant strain carries only a single mutation.Growth is shown by (+),no growth is shown by (o).
Indicate the correct pairing: mutant strain defective gene.
A) strain a ARG-F
B) strain a ARG-H
C) strain b ARG-F
D) strain c ARG-E
E) strain c ARG-F
Mutant strains of Neurospora are grown in minimal media supplements as follows.Each mutant strain carries only a single mutation.Growth is shown by (+),no growth is shown by (o).
Indicate the correct pairing: mutant strain defective gene.
A) strain a ARG-F
B) strain a ARG-H
C) strain b ARG-F
D) strain c ARG-E
E) strain c ARG-F
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41
Assume that a series of compounds has been discovered in Neurospora.Compounds A-F appear to be members of an enzyme pathway.Several mutations have been identified and each of strains 1-4 contains a single mutation.Shown below are five possible pathways.Choose the pathway that best fits the data presented.[Growth in minimal media with supplements is shown by (+),no growth is shown by (o).]
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
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
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42
Choose the interaction listed below that is not involved in maintaining tertiary structure in protein molecules.
A) covalent bond
B) hydrogen bond
C) hydrophobic/hydrophilic interactions
D) ionic interactions
E) halogen bond
A) covalent bond
B) hydrogen bond
C) hydrophobic/hydrophilic interactions
D) ionic interactions
E) halogen bond
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43
Indicate the false statement regarding amino acids.
A) Several amino acids linked together are termed an oligopeptide.
B) Amino acids are linked by peptide bonds that join two amino groups together.
C) The C terminus of a polypeptide chain contains a free carboxylic acid group.
D) Two amino acids joined together are termed a dipeptide.
E) Amino acids contain carbon.
A) Several amino acids linked together are termed an oligopeptide.
B) Amino acids are linked by peptide bonds that join two amino groups together.
C) The C terminus of a polypeptide chain contains a free carboxylic acid group.
D) Two amino acids joined together are termed a dipeptide.
E) Amino acids contain carbon.
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44
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 and green colour vision.
D) is found in rod cells and is responsible for blue and green colour vision.
E) is missing in individuals who exhibit red green colour blindness.
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 and green colour vision.
D) is found in rod cells and is responsible for blue and green colour vision.
E) is missing in individuals who exhibit red green colour blindness.
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45
Choose the condition that involves a defect in an enzyme pathway.
A) photolyase
B) white eyes in Drosophila
C) tritanopia
D) retinitis pigmentosa
E) xeroderma pigmentosum
A) photolyase
B) white eyes in Drosophila
C) tritanopia
D) retinitis pigmentosa
E) xeroderma pigmentosum
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46
Consider the pathway for the synthesis of the amino acid arginine in Neurospora:
Mutant strains of Neurospora are grown in minimal media supplements as follows.The strains may carry more than one mutation.Growth is shown by (+)and no growth is shown by (o).
Strain a accumulates citrulline,strain b does not.Indicate the statement that is most correct regarding these two strains.
A) Strain a has a mutation in ARG-E only.
B) Strain b has only one mutation.
C) Strain a has mutations in ARG-F and ARG-H.
D) Strain a has mutations in ARG-E,ARG-F and ARG-H.
E) Strain a has a mutation in ARG-H only.
Mutant strains of Neurospora are grown in minimal media supplements as follows.The strains may carry more than one mutation.Growth is shown by (+)and no growth is shown by (o).
Strain a accumulates citrulline,strain b does not.Indicate the statement that is most correct regarding these two strains.
A) Strain a has a mutation in ARG-E only.
B) Strain b has only one mutation.
C) Strain a has mutations in ARG-F and ARG-H.
D) Strain a has mutations in ARG-E,ARG-F and ARG-H.
E) Strain a has a mutation in ARG-H only.
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47
The structure of a protein that involves the interaction between two distinct polypeptide chains is
A) primary structure.
B) secondary structure.
C) tertiary structure.
D) quaternary structure.
E) both primary and secondary structure.
A) primary structure.
B) secondary structure.
C) tertiary structure.
D) quaternary structure.
E) both primary and secondary structure.
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48
Assume that a series of compounds has been discovered in Neurospora.Compounds A-F appear to be members of an enzyme pathway.Several mutations have been identified and strains 1-4 each contain a single mutation.Shown below are five possible pathways.Choose the pathway that best fits the data presented.[Growth in minimal media with supplements is shown by (+),no growth is shown by (o).]
A) A ? B ? C ? D ? E ? F
B) A ? B ? C ? F ? E ? D
C)
D)
E)
A) A ? B ? C ? D ? E ? F
B) A ? B ? C ? F ? E ? D
C)
D)
E)
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49
The structure of a polypeptide that is characterized by a three-dimensional shape with a characteristic geometry at local regions maintained by hydrogen bonds is
A) primary structure.
B) secondary structure.
C) tertiary structure.
D) quaternary structure.
E) both tertiary and quaternary structures.
A) primary structure.
B) secondary structure.
C) tertiary structure.
D) quaternary structure.
E) both tertiary and quaternary structures.
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50
The condition sickle-cell anaemia 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 haemoglobin molecule.
E) unequal recombination resulting in the deletion of the β chain haemoglobin gene.
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 haemoglobin molecule.
E) unequal recombination resulting in the deletion of the β chain haemoglobin gene.
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51
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.
E) drift.
A) duplication and divergence.
B) accumulation of random mutations.
C) convergent evolution.
D) spontaneous generation.
E) drift.
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52
Assume that a transition mutation results in an amino acid substitution in the resulting polypeptide.What level of protein structure might be affected,as a result?
A) primary structure
B) secondary structure
C) tertiary structure
D) quaternary structure
E) All levels might be affected by a single amino acid substitution.
A) primary structure
B) secondary structure
C) tertiary structure
D) quaternary structure
E) All levels might be affected by a single amino acid substitution.
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53
Indicate the false statement regarding amino acids.
A) Every amino acid contains a carboxyl group.
B) The side chain or R group differs for each amino acid.
C) Amino acids are joined together by peptide bonds.
D) The end of the polypeptide termed the N terminus contains a free amino group.
E) Amino acids contain nitrogen.
A) Every amino acid contains a carboxyl group.
B) The side chain or R group differs for each amino acid.
C) Amino acids are joined together by peptide bonds.
D) The end of the polypeptide termed the N terminus contains a free amino group.
E) Amino acids contain nitrogen.
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54
Red green colour 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.
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.
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55
Choose the statement below that is not true regarding sickle-cell anaemia.
A) Individuals who are heterozygous for the sickle-cell allele cannot make haemoglobin.
B) The sickle-cell haemoglobin molecule contains an amino acid substitution.
C) The haemoglobin molecules of an individual with sickle-cell anaemia clump together.
D) The red blood cells of an individual with sickle-cell anaemia distort and elongate.
E) Sickle-cell anaemia may be lethal.
A) Individuals who are heterozygous for the sickle-cell allele cannot make haemoglobin.
B) The sickle-cell haemoglobin molecule contains an amino acid substitution.
C) The haemoglobin molecules of an individual with sickle-cell anaemia clump together.
D) The red blood cells of an individual with sickle-cell anaemia distort and elongate.
E) Sickle-cell anaemia may be lethal.
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56
Though sickle-cell anaemia is frequently lethal for individuals who are homozygous for the sickle-cell allele,natural selection seems to have maintained that allele in certain geographic locations.A likely explanation for this observation is that
A) the forward mutation rate to sickle-cell is much higher in those regions.
B) individuals with sickle-cell anaemia live longer and have more children.
C) reversion from sickle-cell to wild-type is prevented in some populations.
D) individuals who are heterozygous for the sickle-cell allele are protected from malaria.
E) only certain populations have been tested for the presence of the sickle-cell allele.
A) the forward mutation rate to sickle-cell is much higher in those regions.
B) individuals with sickle-cell anaemia live longer and have more children.
C) reversion from sickle-cell to wild-type is prevented in some populations.
D) individuals who are heterozygous for the sickle-cell allele are protected from malaria.
E) only certain populations have been tested for the presence of the sickle-cell allele.
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57
Which of the following is used in site-directed mutagenesis?
A) alkylating agent
B) hydroxylating agent
C) deaminating agent
D) synthetic oligonucleotide
E) intercalating agent
A) alkylating agent
B) hydroxylating agent
C) deaminating agent
D) synthetic oligonucleotide
E) intercalating agent
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58
How are somatic mutations different from germline mutations?
A) Only somatic mutations are spontaneous.
B) Only germline mutations are heritable.
C) Alkylating agents only induce germline mutations.
D) Only somatic mutations are inherited by the next generation.
E) Alkylating agents only induce somtaic mutations.
A) Only somatic mutations are spontaneous.
B) Only germline mutations are heritable.
C) Alkylating agents only induce germline mutations.
D) Only somatic mutations are inherited by the next generation.
E) Alkylating agents only induce somtaic mutations.
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59
Base analogues similar to the normal nitrogenous bases of DNA can be incorporated during replication,often causing substitutions in the next round of replication.
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60
What would be the consequence of a loss of 3'-to-5' exonuclease activity?
A) A 100-fold increase in apyrimidinic sites.
B) A 100-fold increase in spontaneous mutation frequency.
C) A 100-fold decrease in base excision repair.
D) A 100-fold increase in induced mutation frequency.
E) A 100-fold increase in trinucleotide repeats.
A) A 100-fold increase in apyrimidinic sites.
B) A 100-fold increase in spontaneous mutation frequency.
C) A 100-fold decrease in base excision repair.
D) A 100-fold increase in induced mutation frequency.
E) A 100-fold increase in trinucleotide repeats.
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61
A "hot spot" is a region of DNA resistant to mutations.
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62
The secondary structure of a protein is its overall three-dimensional structure.
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63
What two hypotheses are tested in the Luria-Delbruck fluctuation test?
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64
How is DNA altered by hydrolysis,radiation,UV light,and oxidation respectively?
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65
Chemical X has just been screened using the Ames test.A total of 5,000 bacteria were tested against 0.001 M,1 M,0.1M,and 1M concentrations for which 4,1,0,and 200 colonies grew respectively.Control plate of minimal media supplemented with histidine had 5,000 colonies,while minimal media alone had only two.Interpret these data.
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66
What technique would you use to test the hypothesis that multiple drug resistant bacterium exists in a heterogeneous population?
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67
A missense mutation is one that creates a premature stop codon.
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68
The local pet store received several shipments of albino ferrets.You choose two males and two females as pets one breeding pair from the same litter,one from two different litters.When your ferrets' litters are born,one litter has normally pigmented offspring.State which offspring are albino and which are pigmented and explain why.
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69
What are hot spots?
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70
A mutation in a single gene that codes for a protein in a biochemical pathway often leads to auxotrophy for the end-product of that pathway.
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71
Only multimeric proteins have quaternary structure.
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72
How could you use the mutagenesis technique co-developed by Michael Smith to demonstrate that a specific amino acid is critical for the function of a particular enzyme?
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73
Geneticists have detected mutations affecting vision that result in blue cone monochromacy.What is this and how does it occur?
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74
The Food and Drug Administration assesses whether an agent is mutagenic by first screening it in an Ames test and then subsequently in rodents.
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75
What mutation occurs in the genetic disease xeroderma pigmentosum and what are the consequences?
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