Thursday, September 25, 2014

SOALSUBSTANSI GENETIKA GRADE 3


1) The role of a metabolite that controls a repressible operon is to
A) bind to the promoter region and decrease the affinity of RNA polymerase for the promoter.
B) bind to the operator region and block the attachment of RNA polymerase to the promoter.
C) increase the production of inactive repressor proteins.
D) bind to the repressor protein and inactivate it.
E) bind to the repressor protein and activate it.
Answer: E

2) The tryptophan operon is a repressible operon that is
A) permanently turned on.
B) turned on only when tryptophan is present in the growth medium.
C) turned off only when glucose is present in the growth medium.
D) turned on only when glucose is present in the growth medium.
E) turned off whenever tryptophan is added to the growth medium.
Answer: E

3) Which of the following is a protein produced by a regulatory gene?
A) operon
B) inducer
C) promoter
D) repressor
E) corepressor
Answer: D

4) A lack of which molecule would result in the cell's inability to "turn off" genes?
A) operon
B) inducer
C) promoter
D) ubiquitin
E) corepressor
Answer: E

5) Which of the following, when taken up by the cell, binds to the repressor so that the repressor no longer binds to the operator?
A) ubiquitin
B) inducer
C) promoter
D) repressor
E) corepressor
Answer: B

6) Most repressor proteins are allosteric. Which of the following binds with the repressor to alter its conformation?
A) inducer
B) promoter
C) RNA polymerase
D) transcription factor
E) cAMP
Answer: A

7) A mutation that inactivates the regulatory gene of a repressible operon in an E. coli cell would result in
A) continuous transcription of the structural gene controlled by that regulator.
B) complete inhibition of transcription of the structural gene controlled by that regulator.
C) irreversible binding of the repressor to the operator.
D) inactivation of RNA polymerase by alteration of its active site.
E) continuous translation of the mRNA because of alteration of its structure.
Answer: A

8) The lactose operon is likely to be transcribed when
A) there is more glucose in the cell than lactose.
B) the cyclic AMP levels are low.
C) there is glucose but no lactose in the cell.
D) the cyclic AMP and lactose levels are both high within the cell.
E) the cAMP level is high and the lactose level is low.
Answer: D

9) Transcription of the structural genes in an inducible operon
A) occurs continuously in the cell.
B) starts when the pathway's substrate is present.
C) starts when the pathway's product is present.
D) stops when the pathway's product is present.
E) does not result in the production of enzymes.
Answer: B

10) For a repressible operon to be transcribed, which of the following must occur?
A) A corepressor must be present.
B) RNA polymerase and the active repressor must be present.
C) RNA polymerase must bind to the promoter, and the repressor must be inactive.
D) RNA polymerase cannot be present, and the repressor must be inactive.
E) RNA polymerase must not occupy the promoter, and the repressor must be inactive.
Answer: C

11) Allolactose, an isomer of lactose, is formed in small amounts from lactose. An E. coli cell is presented for the first time with the sugar lactose (containing allolactose) as a potential food source. Which of the following occurs when the lactose enters the cell?
A) The repressor protein attaches to the regulator.
B) Allolactose binds to the repressor protein.
C) Allolactose binds to the regulator gene.
D) The repressor protein and allolactose bind to RNA polymerase.
E) RNA polymerase attaches to the regulator.
Answer: B

12) Altering patterns of gene expression in prokaryotes would most likely serve the organism's survival in which of the following ways?
A) organizing gene expression so that genes are expressed in a given order
B) allowing each gene to be expressed an equal number of times
C) allowing the organism to adjust to changes in environmental conditions
D) allowing young organisms to respond differently from more mature organisms
E) allowing environmental changes to alter the prokaryote's genome
Answer: C

13) In response to chemical signals, prokaryotes can do which of the following?
A) turn off translation of their mRNA
B) alter the level of production of various enzymes
C) increase the number and responsiveness of their ribosomes
D) inactivate their mRNA molecules
E) alter the sequence of amino acids in certain proteins
Answer: B

14) If glucose is available in the environment of E. coli, the cell responds with a very low concentration of cAMP. When the cAMP increases in concentration, it binds to CAP. Which of the following would you expect to be a measurable effect?
A) decreased concentration of the lac enzymes
B) increased concentration of the trp enzymes
C) decreased binding of the RNA polymerase to sugar metabolism-related promoters
D) decreased concentration of alternative sugars in the cell
E) increased concentrations of sugars such as arabinose in the cell
Answer: E

15) In positive control of several sugar-metabolism-related operons, the catabolite activator protein (CAP) binds to DNA to stimulate transcription. What causes an increase in CAP?
A) increase in glucose and increase in cAMP
B) decrease in glucose and increase in cAMP
C) increase in glucose and decrease in cAMP
D) decrease in glucose and increase in repressor
E) decrease in glucose and decrease in repressor
Answer: B

16) There is a mutation in the repressor that results in a molecule known as a super-repressor because it represses the lac operon permanently. Which of these would characterize such a mutant?
A) It cannot bind to the operator.
B) It cannot make a functional repressor.
C) It cannot bind to the inducer.
D) It makes molecules that bind to one another.
E) It makes a repressor that binds CAP.
Answer: C

17) Which of the following mechanisms is (are) used to coordinate the expression of multiple, related genes in eukaryotic cells?
A) Genes are organized into clusters, with local chromatin structures influencing the expression of all the genes at once.
B) The genes share a common intragenic sequence, and allow several activators to turn on their transcription, regardless of location.
C) The genes are organized into large operons, allowing them to be transcribed as a single unit.
D) A single repressor is able to turn off several related genes.
E) Environmental signals enter the cell and bind directly to promoters.
Answer: A

18) If you were to observe the activity of methylated DNA, you would expect it to
A) be replicating nearly continuously.
B) be unwinding in preparation for protein synthesis.
C) have turned off or slowed down the process of transcription.
D) be very actively transcribed and translated.
E) induce protein synthesis by not allowing repressors to bind to it.
Answer: C

19) Genomic imprinting, DNA methylation, and histone acetylation are all examples of
A) genetic mutation.
B) chromosomal rearrangements.
C) karyotypes.
D) epigenetic phenomena.
E) translocation.
Answer: D

20) When DNA is compacted by histones into 10-nm and 30-nm fibers, the DNA is unable to interact with proteins required for gene expression. Therefore, to allow for these proteins to act, the chromatin must constantly alter its structure. Which processes contribute to this dynamic activity?
A) DNA supercoiling at or around H1
B) methylation and phosphorylation of histone tails
C) hydrolysis of DNA molecules where they are wrapped around the nucleosome core
D) accessibility of heterochromatin to phosphorylating enzymes
E) nucleotide excision and reconstruction
Answer: B

21) Two potential devices that eukaryotic cells use to regulate transcription are
A) DNA methylation and histone amplification.
B) DNA amplification and histone methylation.
C) DNA acetylation and methylation.
D) DNA methylation and histone modification.
E) histone amplification and DNA acetylation.
Answer: D

22) During DNA replication,
A) all methylation of the DNA is lost at the first round of replication.
B) DNA polymerase is blocked by methyl groups, and methylated regions of the genome are therefore left uncopied.
C) methylation of the DNA is maintained because methylation enzymes act at DNA sites where one strand is already methylated and thus correctly methylates daughter strands after replication.
D) methylation of the DNA is maintained because DNA polymerase directly incorporates methylated nucleotides into the new strand opposite any methylated nucleotides in the template.
E) methylated DNA is copied in the cytoplasm, and unmethylated DNA is copied in the nucleus.
Answer: C

23) In eukaryotes, general transcription factors
A) are required for the expression of specific protein-encoding genes.
B) bind to other proteins or to a sequence element within the promoter called the TATA box.
C) inhibit RNA polymerase binding to the promoter and begin transcribing.
D) usually lead to a high level of transcription even without additional specific transcription factors.
E) bind to sequences just after the start site of transcription.
Answer: B

24) Steroid hormones produce their effects in cells by
A) activating key enzymes in metabolic pathways.
B) activating translation of certain mRNAs.
C) promoting the degradation of specific mRNAs.
D) binding to intracellular receptors and promoting transcription of specific genes.
E) promoting the formation of looped domains in certain regions of DNA.
Answer: D

25) Transcription factors in eukaryotes usually have DNA binding domains as well as other domains that are also specific for binding. In general, which of the following would you expect many of them to be able to bind?
A) repressors
B) ATP
C) protein-based hormones
D) other transcription factors
E) tRNA
Answer: D

26) Gene expression might be altered at the level of post-transcriptional processing in eukaryotes rather than prokaryotes because of which of the following?
A) Eukaryotic mRNAs get 5' caps and 3' tails.
B) Prokaryotic genes are expressed as mRNA, which is more stable in the cell.
C) Eukaryotic exons may be spliced in alternative patterns.
D) Prokaryotes use ribosomes of different structure and size.
E) Eukaryotic coded polypeptides often require cleaving of signal sequences before localization.
Answer: C

27) Which of the following experimental procedures is most likely to hasten mRNA degradation in a eukaryotic cell?
A) enzymatic shortening of the poly-A tail
B) removal of the 5' cap
C) methylation of C nucleotides
D) methylation of histones
E) removal of one or more exons
Answer: B

28) Which of the following is most likely to have a small protein called ubiquitin attached to it?
A) a cyclin that usually acts in G₁, now that the cell is in G₂
B) a cell surface protein that requires transport from the ER
C) an mRNA that is leaving the nucleus to be translated
D) a regulatory protein that requires sugar residues to be attached
E) an mRNA produced by an egg cell that will be retained until after fertilization
Answer: A

29) In prophase I of meiosis in female Drosophila, studies have shown that there is phosphorylation of an amino acid in the tails of histones of gametes. A mutation in flies that interferes with this process results in sterility. Which of the following is the most likely hypothesis?
A) These oocytes have no histones.
B) Any mutation during oogenesis results in sterility.
C) All proteins in the cell must be phosphorylated.
D) Histone tail phosphorylation prohibits chromosome condensation.
E) Histone tails must be removed from the rest of the histones.
Answer: D

30) The phenomenon in which RNA molecules in a cell are destroyed if they have a sequence complementary to an introduced double-stranded RNA is called
A) RNA interference.
B) RNA obstruction.
C) RNA blocking.
D) RNA targeting.
E) RNA disposal.
Answer: A

31) At the beginning of this century there was a general announcement regarding the sequencing of the human genome and the genomes of many other multicellular eukaryotes. There was surprise expressed by many that the number of protein-coding sequences was much smaller than they had expected. Which of the following could account for most of the rest?
A) "junk" DNA that serves no possible purpose
B) rRNA and tRNA coding sequences
C) DNA that is translated directly without being transcribed
D) non-protein-coding DNA that is transcribed into several kinds of small RNAs with biological function
E) non-protein-coding DNA that is transcribed into several kinds of small RNAs without biological function
Answer: D

32) Among the newly discovered small noncoding RNAs, one type reestablishes methylation patterns during gamete formation and block expression of some transposons. These are known as
A) miRNA.
B) piRNA.
C) snRNA.
D) siRNA.
E) RNAi.
Answer: B

33) Which of the following best describes siRNA?
A) a short double-stranded RNA, one of whose strands can complement and inactivate a sequence of mRNA
B) a single-stranded RNA that can, where it has internal complementary base pairs, fold into cloverleaf patterns
C) a double-stranded RNA that is formed by cleavage of hairpin loops in a larger precursor
D) a portion of rRNA that allows it to bind to several ribosomal proteins in forming large or small subunits
E) a molecule, known as Dicer, that can degrade other mRNA sequences
Answer: A

34) One way scientists hope to use the recent knowledge gained about noncoding RNAs lies with the possibilities for their use in medicine. Of the following scenarios for future research, which would you expect to gain most from RNAs?
A) exploring a way to turn on the expression of pseudogenes
B) targeting siRNAs to disable the expression of an allele associated with autosomal recessive disease
C) targeting siRNAs to disable the expression of an allele associated with autosomal dominant disease
D) creating knock-out organisms that can be useful for pharmaceutical drug design
E) looking for a way to prevent viral DNA from causing infection in humans
Answer: C

35) Which of the following describes the function of an enzyme known as Dicer?
A) It degrades single-stranded DNA.
B) It degrades single-stranded mRNA.
C) It degrades mRNA with no poly-A tail.
D) It trims small double-stranded RNAs into molecules that can block translation.
E) It chops up single-stranded DNAs from infecting viruses.
Answer: D

36) In a series of experiments, the enzyme Dicer has been inactivated in cells from various vertebrates so that the centromere is abnormally formed from chromatin. Which of the following is most likely to occur?
A) The usual mRNAs transcribed from centromeric DNA will be missing from the cells.
B) Tetrads will no longer be able to form during meiosis I.
C) Centromeres will be euchromatic rather than heterochromatic and the cells will soon die in culture.
D) The cells will no longer be able to resist bacterial contamination.
E) The DNA of the centromeres will no longer be able to replicate.
Answer: C

37) Since Watson and Crick described DNA in 1953, which of the following might best explain why the function of small RNAs is still being explained?
A) As RNAs have evolved since that time, they have taken on new functions.
B) Watson and Crick described DNA but did not predict any function for RNA.
C) The functions of small RNAs could not be approached until the entire human genome was sequenced.
D) Ethical considerations prevented scientists from exploring this material until recently.
E) Changes in technology as well as our ability to determine how much of the DNA is expressed have now made this possible.
Answer: E

38) You are given an experimental problem involving control of a gene's expression in the embryo of a particular species. One of your first questions is whether the gene's expression is controlled at the level of transcription or translation. Which of the following might best give you an answer?
A) You explore whether there has been alternative splicing by examining amino acid sequences of very similar proteins.
B) You measure the quantity of the appropriate pre-mRNA in various cell types and find they are all the same.
C) You assess the position and sequence of the promoter and enhancer for this gene.
D) An analysis of amino acid production by the cell shows you that there is an increase at this stage of embryonic life.
E) You use an antibiotic known to prevent translation.
Answer: B

39) In humans, the embryonic and fetal forms of hemoglobin have a higher affinity for oxygen than that of adults. This is due to
A) nonidentical genes that produce different versions of globins during development.
B) identical genes that generate many copies of the ribosomes needed for fetal globin production.
C) pseudogenes, which interfere with gene expression in adults.
D) the attachment of methyl groups to cytosine following birth, which changes the type of hemoglobin produced.
E) histone proteins changing shape during embryonic development.
Answer: A

40) The fact that plants can be cloned from somatic cells demonstrates that
A) differentiated cells retain all the genes of the zygote.
B) genes are lost during differentiation.
C) the differentiated state is normally very unstable.
D) differentiated cells contain masked mRNA.
E) differentiation does not occur in plants.
Answer: A

41) In animals, embryonic stem cells differ from adult stem cells in that
A) embryonic stem cells are totipotent, and adult stem cells are pluripotent.
B) embryonic stem cells are pluripotent, and adult stem cells are totipotent.
C) embryonic stem cells have more genes than adult stem cells.
D) embryonic stem cells have fewer genes than adult stem cells.
E) embryonic stem cells are localized to specific sites within the embryo, whereas adult stem cells are spread throughout the body.
Answer: A

42) What is considered to be the first evidence of differentiation in the cells of an embryo?
A) cell division occurring after fertilization
B) the occurrence of mRNAs for the production of tissue-specific proteins
C) determination of specific cells for certain functions
D) changes in the size and shape of the cell
E) changes resulting from induction
Answer: B

43) Embryonic lethal mutations result in
A) phenotypes that prevent fertilization.
B) failure to express maternal effect genes.
C) death during pupation.
D) phenotypes that are never born/hatched.
E) homeotic phenotype changes.
Answer: D

44) Your brother has just purchased a new plastic model airplane. He places all the parts on the table in approximately the positions in which they will be located when the model is complete. His actions are analogous to which process in development?
A) morphogenesis
B) determination
C) induction
D) differentiation
E) pattern formation
Answer: E

45) The product of the bicoid gene in Drosophila provides essential information about
A) lethal genes.
B) the dorsal-ventral axis.
C) the left-right axis.
D) segmentation.
E) the anterior-posterior axis.
Answer: E

46) If a Drosophila female has a homozygous mutation for a maternal effect gene,
A) she will not develop past the early embryonic stage.
B) all of her offspring will show the mutant phenotype, regardless of their genotype.
C) only her male offspring will show the mutant phenotype.
D) her offspring will show the mutant phenotype only if they are also homozygous for the mutation.
E) only her female offspring will show the mutant phenotype.
Answer: B

47) Mutations in which of the following genes lead to transformations in the identity of entire body parts?
A) morphogens
B) segmentation genes
C) egg-polarity genes
D) homeotic genes
E) inducers
Answer: D

48) Which of the following genes map out the basic subdivisions along the anterior-posterior axis of the Drosophila embryo?
A) homeotic genes
B) segmentation genes
C) egg-polarity genes
D) morphogens
E) inducers
Answer: B

49) Gap genes and pair-rule genes fall into which of the following categories?
A) homeotic genes
B) segmentation genes
C) egg-polarity genes
D) morphogens
E) inducers
Answer: B

50) The bicoid gene product is normally localized to the anterior end of the embryo. If large amounts of the product were injected into the posterior end as well, which of the following would occur?
A) The embryo would grow to an unusually large size.
B) The embryo would grow extra wings and legs.
C) The embryo would probably show no anterior development and die.
D) Anterior structures would form in both sides of the embryo.
E) The embryo would develop normally.
Answer: D

51) What do gap genes, pair-rule genes, segment polarity genes, and homeotic genes all have in common?
A) Their products act as transcription factors.
B) They have no counterparts in animals other than Drosophila.
C) Their products are all synthesized prior to fertilization.
D) They act independently of other positional information.
E) They apparently can be activated and inactivated at any time of the fly's life.
Answer: A

52) Which of the following statements describes proto-oncogenes?
A) Their normal function is to suppress tumor growth.
B) They are introduced to a cell initially by retroviruses.
C) They are produced by somatic mutations induced by carcinogenic substances.
D) They can code for proteins associated with cell growth.
E) They are underexpressed in cancer cells.
Answer: D

53) Which of the following is characteristic of the product of the p53 gene?
A) It is an activator for other genes.
B) It speeds up the cell cycle.
C) It causes cell death via apoptosis.
D) It allows cells to pass on mutations due to DNA damage.
E) It slows down the rate of DNA replication by interfering with the binding of DNA polymerase.
Answer: A

54) Tumor-suppressor genes
A) are frequently overexpressed in cancerous cells.
B) are cancer-causing genes introduced into cells by viruses.
C) can encode proteins that promote DNA repair or cell-cell adhesion.
D) often encode proteins that stimulate the cell cycle.
E) do all of the above.
Answer: C

55) BRCA1 and BRCA2 are considered to be tumor-suppressor genes because
A) they prevent infection by retroviruses that cause cancer.
B) their normal products participate in repair of DNA damage.
C) the mutant forms of either one of these promote breast cancer.
D) the normal genes make estrogen receptors.
E) they block penetration of breast cells by chemical carcinogens.
Answer: B

56) The cancer-causing forms of the Ras protein are involved in which of the following processes?
A) relaying a signal from a growth factor receptor
B) DNA replication
C) DNA repair
D) cell-cell adhesion
E) cell division
Answer: A

57) Forms of the Ras protein found in tumors usually cause which of the following?
A) DNA replication to stop
B) DNA replication to be hyperactive
C) cell-to-cell adhesion to be nonfunctional
D) cell division to cease
E) growth factor signaling to be hyperactive
Answer: E

58) A genetic test to detect predisposition to cancer would likely examine the APC gene for involvement in which type(s) of cancer?
A) colorectal only
B) lung and breast
C) small intestinal and esophageal
D) lung only
E) lung and prostate
Answer: A


59. In Drosophila after ~100 minutes postfertilization, the embryo looks like the following diagram, with all nuclei having moved to the periphery and, subsequently, four of the nuclei being sequestered at the posterior end
.
At this point, the embryo is characterized as
A) a first-stage larva.
B) nuclei in the cortex that has not undergone cytokinesis.
C) nuclei in the cortex forming a single-cell layer over the surface.
D) an embryo with segmentation beginning to be apparent.
Answer: B

60. In Drosophila after ~100 minutes postfertilization, the embryo looks like the following diagram, with all nuclei having moved to the periphery and, subsequently, four of the nuclei being sequestered at the posterior end.
The four sequestered cells at one end are most probably destined to become
A) the legs of the adult fly.
B) the germ cells of the adult.
C) mouthparts.
D) antennae.
E) wing primordial.
Answer: B

61. In Drosophila after ~100 minutes postfertilization, the embryo looks like the following diagram, with all nuclei having moved to the periphery and, subsequently, four of the nuclei being sequestered at the posterior end.
Formation of the pole cells (the four sequestered cells) demonstrates the role of
A) segmentation genes.
B) homeotic genes.
C) maternal effect genes.
D) zygotic genes.
E) all of the above.
Answer: C

62. In Drosophila after ~100 minutes postfertilization, the embryo looks like the following diagram, with all nuclei having moved to the periphery and, subsequently, four of the nuclei being sequestered at the posterior end.
The next step after the embryo is formed would be
A) division of the embryo into five broad regions.
B) use of pair-rule genes to divide the embryo into stripes, each of which will become two segments.
C) use of zygotic segment polarity genes to divide each segment into anterior and posterior halves.
D) enclosure of the nuclei in membranes, forming a single layer over the surface.
E) separation of head, thoracic, and abdominal segments of the embryo.
Answer: D

63. In Drosophila after ~100 minutes postfertilization, the embryo looks like the following diagram, with all nuclei having moved to the periphery and, subsequently, four of the nuclei being sequestered at the posterior end.
The developmental stages described for Drosophila illustrate
A) a hierarchy of gene expression.
B) homeotic developmental control.
C) the blockage of cell-to-cell communication.
D) homeotic developmental control and the blockage of cell-to-cell communication.
E) a hierarchy of gene expression and the blockage of cell-to-cell communication.
Answer: A

64. Suppose an experimenter becomes proficient with a technique that allows her to move DNA sequences within a prokaryotic genome
If she moves the promoter for the lac operon to the region between the beta galactosidase gene and the permease gene, which of the following would be likely?
A) Three structural genes will no longer be expressed.
B) RNA polymerase will no longer transcribe permease.
C) The operon will no longer be inducible.
D) Beta galactosidase will be produced.
E) The cell will continue to metabolize but more slowly.
Answer: D

65. Suppose an experimenter becomes proficient with a technique that allows her to move DNA sequences within a prokaryotic genome
If she moves the operator to the far end of the operon (past the transacetylase gene), which of the following would likely occur when the cell is exposed to lactose?
A) The inducer will no longer bind to the repressor.
B) The repressor will no longer bind to the operator.
C) The operon will never be transcribed.
D) The structural genes will be transcribed continuously.
E) The repressor protein will no longer be produced.
Answer: D

66. Suppose an experimenter becomes proficient with a technique that allows her to move DNA sequences within a prokaryotic genome
If she moves the repressor gene (lac I), along with its promoter, to a position at some several thousand base pairs away from its normal position, which will you expect to occur?
A) The repressor will no longer be made.
B) The repressor will no longer bind to the operator.
C) The repressor will no longer bind to the inducer.
D) The lac operon will be expressed continuously.
E) The lac operon will function normally.
Answer: E

67. Suppose an experimenter becomes proficient with a technique that allows her to move DNA sequences within a prokaryotic genome
If she moves the operator to a position upstream from the promoter, what would occur?
A) The lac operon will function normally.
B) The lac operon will be expressed continuously.
C) The repressor will not be able to bind to the operator.
D) The repressor will bind to the promoter.
E) The repressor will no longer be made.
Answer: B

68. A geneticist introduces a transgene into yeast cells and isolates five independent cell lines in which the transgene has integrated into the yeast genome. In four of the lines, the transgene is expressed strongly, but in the fifth there is no expression at all.
Which of the following is a likely explanation for the lack of transgene expression in the fifth cell line?
A) A transgene integrated into a heterochromatic region of the genome.
B) A transgene integrated into a euchromatic region of the genome.
C) The transgene was mutated during the process of integration into the host cell genome.
D) The host cell lacks the enzymes necessary to express the transgene.
E) A transgene integrated into a region of the genome characterized by high histone acetylation.
Answer: A

69. A geneticist introduces a transgene into yeast cells and isolates five independent cell lines in which the transgene has integrated into the yeast genome. In four of the lines, the transgene is expressed strongly, but in the fifth there is no expression at all.
Of the lines that express the transgene, one is transcribed but not translated. Which of the following is a likely explanation?
A) no promoter
B) no AUG in any frame
C) no compatible ribosome
D) high histone acetylation
E) missing transcription factor
Answer: B

70. A researcher found a method she could use to manipulate and quantify phosphorylation and methylation in embryonic cells in culture.
In one set of experiments using this procedure in Drosophila, she was readily successful in increasing phosphorylation of amino acids adjacent to methylated amino acids in histone tails. Which of the following results would she most likely see?
A) increased chromatin condensation
B) decreased chromatin condensation
C) abnormalities of mouse embryos
D) decreased binding of transcription factors
E) inactivation of the selected genes
Answer: B

71. A researcher found a method she could use to manipulate and quantify phosphorylation and methylation in embryonic cells in culture.
In one set of experiments she succeeded in decreasing methylation of histone tails. Which of the following results would she most likely see?
A) increased chromatin condensation
B) decreased chromatin condensation
C) abnormalities of mouse embryos
D) decreased binding of transcription factors
E) inactivation of the selected genes
Answer: A

72. A researcher found a method she could use to manipulate and quantify phosphorylation and methylation in embryonic cells in culture.
One of her colleagues suggested she try increased methylation of C nucleotides in a mammalian system. Which of the following results would she most likely see?
A) increased chromatin condensation
B) decreased chromatin condensation
C) abnormalities of mouse embryos
D) decreased binding of transcription factors
E) inactivation of the selected genes
Answer: E

73. A researcher found a method she could use to manipulate and quantify phosphorylation and methylation in embryonic cells in culture.
She tried decreasing the amount of methylation enzymes in the embryonic stem cells and then allowed the cells to further differentiate. Which of the following results would she most likely see?
A) increased chromatin condensation
B) decreased chromatin condensation
C) abnormalities of mouse embryos
D) decreased binding of transcription factors
E) inactivation of the selected genes
Answer: C

74. A researcher introduces double-stranded RNA into a culture of mammalian cells, and can identify its location or that of its smaller subsections experimentally, using a fluorescent probe.
Within the first quarter hour, the researcher sees that the intact RNA is found in the cells. After 3 hours, she is not surprised to find that
A) Dicer enzyme has reduced it to smaller double-stranded pieces.
B) the RNA is degraded by 5' and 3' exonucleases.
C) the double-stranded RNA replicates itself.
D) the double-stranded RNA binds to mRNAs to prevent translation.
E) the double-stranded RNA binds to tRNAs to prevent translation.
Answer: A

75. A researcher introduces double-stranded RNA into a culture of mammalian cells, and can identify its location or that of its smaller subsections experimentally, using a fluorescent probe.
Some time later, she finds that the introduced strand separates into single-stranded RNAs, one of which is degraded. What does this enable the remaining strand to do?
A) attach to histones in the chromatin
B) bind to complementary regions of target mRNAs
C) bind to Dicer enzymes to destroy other RNAs
D) activate other siRNAs in the cell
E) bind to noncomplementary RNA sequences
Answer: B

76. A researcher introduces double-stranded RNA into a culture of mammalian cells, and can identify its location or that of its smaller subsections experimentally, using a fluorescent probe.
In addition, she finds what other evidence of this single-stranded RNA piece's activity?
A) She can measure the degradation rate of the remaining single strand.
B) She can measure the decrease in the concentration of Dicer.
C) The rate of accumulation of the polypeptide to be translated from the target mRNA is reduced.
D) The amount of miRNA is multiplied by its replication.
E) The cell's translation ability is entirely shut down.
Answer: C

77. A researcher has arrived at a method to prevent gene expression from Drosophila embryonic genes. The following questions assume that he is using this method.
The researcher in question measures the amount of new polypeptide production in embryos from 2—8 hours following fertilization and the results show a steady and significant rise in polypeptide concentration over that time. The researcher concludes that
A) his measurement skills must be faulty.
B) the results are due to building new cell membranes to compartmentalize dividing nuclei.
C) the resulting new polypeptides are due to translation of maternal mRNAs.
D) the new polypeptides were inactive and not measurable until fertilization.
E) polypeptides were attached to egg membranes until this time.
Answer: C

78. A researcher has arrived at a method to prevent gene expression from Drosophila embryonic genes. The following questions assume that he is using this method.
The researcher continues to study the reactions of the embryo to these new proteins and you hypothesize that he is most likely to see which of the following (while embryonic genes are still not being expressed)?
A) The cells begin to differentiate.
B) The proteins are evenly distributed throughout the embryo.
C) Larval features begin to make their appearance.
D) Spatial axes (anterior → posterior, etc.) begin to be determined.
E) The embryo begins to lose cells due to apoptosis from no further gene expression.
Answer: D

79. The researcher measures the concentration of the polypeptides from different regions in the early embryo and finds the following pattern (darker shading = greater concentration): SEE IMAGE



Which of the following would be his most logical assumption?
A) The substance has moved quickly from region 5 to region 1.
B) Some other material in the embryo is causing accumulation in region 1 due to differential binding.
C) The cytosol is in constant movement, dispersing the polypeptide.
D) The substance is produced in region 1 and diffuses toward region 5.
E) The substance must have entered the embryo from the environment near region 1.
Answer: D

80. One hereditary disease in humans, called xeroderma pigmentosum (XP), makes homozygous individuals exceptionally susceptible to UV-induced mutation damage in the cells of exposed tissue, especially skin. Without extraordinary avoidance of sunlight exposure, patients soon succumb to numerous skin cancers.
Which of the following best describes this phenomenon?
A) inherited cancer taking a few years to be expressed
B) embryonic or fetal cancer
C) inherited predisposition to mutation
D) inherited inability to repair UV-induced mutation
E) susceptibility to chemical carcinogens
Answer: D

81. One hereditary disease in humans, called xeroderma pigmentosum (XP), makes homozygous individuals exceptionally susceptible to UV-induced mutation damage in the cells of exposed tissue, especially skin. Without extraordinary avoidance of sunlight exposure, patients soon succumb to numerous skin cancers.
Given the damage caused by UV, the kind of gene affected in those with XP is one whose product is involved with
A) mending of double-strand breaks in the DNA backbone.
B) breakage of cross-strand covalent bonds.
C) the ability to excise single-strand damage and replace it.
D) the removal of double-strand damaged areas.
E) causing affected skin cells to undergo apoptosis.
Answer: C

82. A few decades ago, Knudsen and colleagues proposed a theory that, for a normal cell to become a cancer cell, a minimum of two genetic changes had to occur in that cell. Knudsen was studying retinoblastoma, a childhood cancer of the eye.
Two children are born from the same parents. Child one inherits a predisposition to retinoblastoma (one of the mutations) and child two does not. However, both children develop the retinoblastoma. Which of the following would you expect?
A) an earlier age of onset in child one
B) a history of exposure to mutagens in child one but not in child two
C) a more severe cancer in child one
D) increased levels of apoptosis in both children
E) decreased levels of DNA repair in child one
Answer: A

83. A few decades ago, Knudsen and colleagues proposed a theory that, for a normal cell to become a cancer cell, a minimum of two genetic changes had to occur in that cell. Knudsen was studying retinoblastoma, a childhood cancer of the eye.
In colorectal cancer, several genes must be mutated in order to make a cell a cancer cell, supporting Knudsen's hypothesis. Which of the following kinds of genes would you expect to be mutated?
A) genes coding for enzymes that act in the colon
B) genes involved in control of the cell cycle
C) genes that are especially susceptible to mutation
D) the same genes that Knudsen identified as associated with retinoblastoma
E) the genes of the bacteria that are abundant in the colon
Answer: B

84. A few decades ago, Knudsen and colleagues proposed a theory that, for a normal cell to become a cancer cell, a minimum of two genetic changes had to occur in that cell. Knudsen was studying retinoblastoma, a childhood cancer of the eye.
Knudsen and colleagues also noted that persons with hereditary retinoblastoma that had been treated successfully lived on but then had a higher frequency of developing osteosarcomas (bone cancers) later in life. This provided further evidence of their theory because
A) osteosarcoma cells express the same genes as retinal cells.
B) p53 gene mutations are common to both tumors.
C) both kinds of cancer involve overproliferation of cells.
D) one of the mutations involved in retinoblastoma is also one of the changes involved in osteosarcoma.
E) retinoblastoma is a prerequisite for the formation of osteosarcoma later in life.
Answer: D

85. A few decades ago, Knudsen and colleagues proposed a theory that, for a normal cell to become a cancer cell, a minimum of two genetic changes had to occur in that cell. Knudsen was studying retinoblastoma, a childhood cancer of the eye.
One of the human leukemias, called CML (chronic myelogenous leukemia), is associated with a chromosomal translocation between chromosomes 9 and 22 in somatic cells of bone marrow. Which of the following allows CML to provide further evidence of this multistep nature of cancer?
A) CML usually occurs in more elderly persons (late age of onset).
B) The resulting chromosome 22 is abnormally short; it is then known as the Philadelphia chromosome.
C) The translocation requires breaks in both chromosomes 9 and 22, followed by fusion between the reciprocal pieces.
D) CML involves a proto-oncogene known as abl.
E) CML can usually be treated by chemotherapy.
Answer: C

86. Epstein Bar Virus (EBV) causes most of us to have an episode of sore throat and swollen glands during early childhood. If we first become exposed to the virus during our teen years, however, EBV causes the syndrome we know as mononucleosis. However, in special circumstances, the same virus can be carcinogenic.
In areas of the world in which malaria is endemic, notably in sub-Saharan Africa, EBV can cause Burkitt's lymphoma in children, which is usually associated with large tumors of the jaw. Which of the following is consistent with these findings?
A) EBV infection makes the malarial parasite able to produce lymphoma.
B) Malaria's strain on the immune system makes EBV infection worse.
C) Malaria occurs more frequently in those infected with EBV.
D) Malarial response of the immune system prevents an individual from making EBV antibodies.
E) A cell infected with the malarial parasite is more resistant to the virus.
Answer: D

87. Epstein Bar Virus (EBV) causes most of us to have an episode of sore throat and swollen glands during early childhood. If we first become exposed to the virus during our teen years, however, EBV causes the syndrome we know as mononucleosis. However, in special circumstances, the same virus can be carcinogenic.
In a different part of the world, namely in parts of southeast Asia, the same virus is associated with a different kind of cancer of the throat. Which of the following is most probable?
A) Viral infection is correlated with a different immunological reaction.
B) The virus infects the people via different routes.
C) The virus only infects the elderly.
D) The virus mutates more frequently in the Asian population.
E) Malaria is also found in this region.
Answer: A

88. Epstein Bar Virus (EBV) causes most of us to have an episode of sore throat and swollen glands during early childhood. If we first become exposed to the virus during our teen years, however, EBV causes the syndrome we know as mononucleosis. However, in special circumstances, the same virus can be carcinogenic.
A very rare human allele of a gene called XLP, or X-linked lymphoproliferative syndrome, causes a small number of people from many different parts of the world to get cancer following even childhood exposure to EBV. Given the previous information, what might be going on?
A) The people must have previously had malaria.
B) Their ancestors must be from sub-Saharan Africa or southeast Asia.
C) They must be unable to mount an immune response to EBV.
D) They must have severe combined immune deficiency (SCID).
E) Their whole immune system must be overreplicating.
Answer: C

89. Epstein Bar Virus (EBV) causes most of us to have an episode of sore throat and swollen glands during early childhood. If we first become exposed to the virus during our teen years, however, EBV causes the syndrome we know as mononucleosis. However, in special circumstances, the same virus can be carcinogenic.
What must characterize the XLP population?
A) They must have severe immunological problems starting at birth.
B) They must all be males with affected male relatives.
C) They must all be males with affected female relatives.
D) They must all inherit this syndrome from their fathers.
E) They must live in sub-Saharan Africa.
Answer: C

90). If a particular operon encodes enzymes for making an essential amino acid and is regulated like the trp operon, then
A) the amino acid inactivates the repressor.
B) the enzymes produced are called inducible enzymes.
C) the repressor is active in the absence of the amino acid.
D) the amino acid acts as a corepressor.
E) the amino acid turns on transcription of the operon.
Answer: D

91) Muscle cells differ from nerve cells mainly because they
A) express different genes.
B) contain different genes.
C) use different genetic codes.
D) have unique ribosomes.
E) have different chromosomes.
Answer: A

92) The functioning of enhancers is an example of
A) transcriptional control of gene expression.
B) a post-transcriptional mechanism to regulate mRNA.
C) the stimulation of translation by initiation factors.
D) post-translational control that activates certain proteins.
E) a eukaryotic equivalent of prokaryotic promoter functioning.
Answer: A

93) Cell differentiation always involves
A) the production of tissue-specific proteins, such as muscle actin.
B) the movement of cells.
C) the transcription of the myoD gene.
D) the selective loss of certain genes from the genome.
E) the cell's sensitivity to environmental cues, such as light or heat.
Answer: A

94) Which of the following is an example of post-transcriptional control of gene expression?
A) the addition of methyl groups to cytosine bases of DNA
B) the binding of transcription factors to a promoter
C) the removal of introns and alternative splicing of exons
D) gene amplification contributing to cancer
E) the folding of DNA to form heterochromatin
Answer: C

95) What would occur if the repressor of an inducible operon were mutated so it could not bind the operator?
A) irreversible binding of the repressor to the promoter
B) reduced transcription of the operon's genes
C) buildup of a substrate for the pathway controlled by the operon
D) continuous transcription of the operon's genes
E) overproduction of catabolite activator protein (CAP)
Answer: D

96) Absence of bicoid mRNA from a Drosophila egg leads to the absence of anterior larval body parts and mirror-image duplication of posterior parts. This is evidence that the product of the bicoid gene
A) is transcribed in the early embryo.
B) normally leads to formation of tail structures.
C) normally leads to formation of head structures.
D) is a protein present in all head structures.
E) leads to programmed cell death.
Answer: C

97) Which of the following statements about the DNA in one of your brain cells is true?
A) Most of the DNA codes for protein.
B) The majority of genes are likely to be transcribed.
C) Each gene lies immediately adjacent to an enhancer.
D) Many genes are grouped into operon-like clusters.
E) It is the same as the DNA in one of your heart cells.
Answer: E

98) Within a cell, the amount of protein made using a given mRNA molecule depends partly on
A) the degree of DNA methylation.
B) the rate at which the mRNA is degraded.
C) the presence of certain transcription factors.
D) the number of introns present in the mRNA.
E) the types of ribosomes present in the cytoplasm.
Answer: B

99) Proto-oncogenes can change into oncogenes that cause cancer. Which of the following best explains the presence of these potential time bombs in eukaryotic cells?
A) Proto-oncogenes first arose from viral infections.
B) Proto-oncogenes normally help regulate cell division.
C) Proto-oncogenes are genetic "junk."
D) Proto-oncogenes are mutant versions of normal genes.
E) Cells produce proto-oncogenes as they age.

Answer: B

SOAL SUBSTANSI GENETIKA DASAR GRADE 2

1) Which of the following variations on translation would be most disadvantageous for a cell?
A) translating polypeptides directly from DNA
B) using fewer kinds of tRNA
C) having only one stop codon
D) lengthening the half-life of mRNA
E) having a second codon (besides AUG) as a start codon
Answer: A

2) Garrod hypothesized that "inborn errors of metabolism" such as alkaptonuria occur because
A) metabolic enzymes require vitamin cofactors, and affected individuals have significant nutritional deficiencies.
B) enzymes are made of DNA, and affected individuals lack DNA polymerase.
C) many metabolic enzymes use DNA as a cofactor, and affected individuals have mutations that prevent their enzymes from interacting efficiently with DNA.
D) certain metabolic reactions are carried out by ribozymes, and affected individuals lack key splicing factors.
E) genes dictate the production of specific enzymes, and affected individuals have genetic defects that cause them to lack certain enzymes.
Answer: A

3) Garrod's information about the enzyme alteration resulting in alkaptonuria led to further elucidation of the same pathway in humans. Phenylketonuria (PKU) occurs when another enzyme in the pathway is altered or missing, resulting in a failure of phenylalanine (phe) to be metabolized to another amino acid: tyrosine. Tyrosine is an earlier substrate in the pathway altered in alkaptonuria. How might PKU affect the presence or absence of alkaptonuria?
A) It would have no effect, because PKU occurs several steps away in the pathway.
B) It would have no effect, because tyrosine is also available from the diet.
C) Anyone with PKU must also have alkaptonuria.
D) Anyone with PKU is born with a predisposition to later alkaptonuria.
E) Anyone with PKU has mild symptoms of alkaptonuria.
Answer: B

4) The nitrogenous base adenine is found in all members of which group?
A) proteins, triglycerides, and testosterone
B) proteins, ATP, and DNA
C) ATP, RNA, and DNA
D) α glucose, ATP, and DNA
E) proteins, carbohydrates, and ATP
Answer: C

5) A particular triplet of bases in the template strand of DNA is 5' AGT 3'. The corresponding codon for the mRNA transcribed is
A) 3' UCA 5'.
B) 3' UGA 5'.
C) 5' TCA 3'.
D) 3' ACU 5'.
E) either UCA or TCA, depending on wobble in the first base.
Answer: A

6) The genetic code is essentially the same for all organisms. From this, one can logically assume which of the following?
A) A gene from an organism can theoretically be expressed by any other organism.
B) All organisms have experienced convergent evolution.
C) DNA was the first genetic material.
D) The same codons in different organisms translate into the different amino acids.
E) Different organisms have different numbers of different types of amino acids.
Answer: A

7) The "universal" genetic code is now known to have exceptions. Evidence for this can be found if which of the following is true?
A) If UGA, usually a stop codon, is found to code for an amino acid such as tryptophan (usually coded for by UGG only).
B) If one stop codon, such as UGA, is found to have a different effect on translation than another stop codon, such as UAA.
C) If prokaryotic organisms are able to translate a eukaryotic mRNA and produce the same polypeptide.
D) If several codons are found to translate to the same amino acid, such as serine.
E) If a single mRNA molecule is found to translate to more than one polypeptide when there are two or more AUG sites.
Answer: A

8) Which of the following nucleotide triplets best represents a codon?
A) a triplet separated spatially from other triplets
B) a triplet that has no corresponding amino acid
C) a triplet at the opposite end of tRNA from the attachment site of the amino acid
D) a triplet in the same reading frame as an upstream AUG
E) a sequence in tRNA at the 3' end
Answer: D

9) Which of the following provides some evidence that RNA probably evolved before DNA?
A) RNA polymerase uses DNA as a template.
B) RNA polymerase makes a single-stranded molecule.
C) RNA polymerase does not require localized unwinding of the DNA.
D) DNA polymerase uses primer, usually made of RNA.
E) DNA polymerase has proofreading function.
Answer: D

10) Which of the following statements best describes the termination of transcription in prokaryotes?
A) RNA polymerase transcribes through the polyadenylation signal, causing proteins to associate with the transcript and cut it free from the polymerase.
B) RNA polymerase transcribes through the terminator sequence, causing the polymerase to separate from the DNA and release the transcript.
C) RNA polymerase transcribes through an intron, and the snRNPs cause the polymerase to let go of the transcript.
D) Once transcription has initiated, RNA polymerase transcribes until it reaches the end of the chromosome.
E) RNA polymerase transcribes through a stop codon, causing the polymerase to stop advancing through the gene and release the mRNA.
Answer: B

11) Which of the following does not occur in prokaryotic eukaryotic gene expression, but does in eukaryotic gene expression?
A) mRNA, tRNA, and rRNA are transcribed.
B) RNA polymerase binds to the promoter.
C) A poly-A tail is added to the 3' end of an mRNA and a cap is added to the 5' end.
D) Transcription can begin as soon as translation has begun even a little.
E) RNA polymerase requires a primer to elongate the molecule.
Answer: C

12) RNA polymerase in a prokaryote is composed of several subunits. Most of these subunits are the same for the transcription of any gene, but one, known as sigma, varies considerably. Which of the following is the most probable advantage for the organism of such sigma switching?
A) It might allow the transcription process to vary from one cell to another.
B) It might allow the polymerase to recognize different promoters under certain environmental conditions.
C) It could allow the polymerase to react differently to each stop codon.
D) It could allow ribosomal subunits to assemble at faster rates.
E) It could alter the rate of translation and of exon splicing.
Answer: B

13) Which of the following is a function of a poly-A signal sequence?
A) It adds the poly-A tail to the 3' end of the mRNA.
B) It codes for a sequence in eukaryotic transcripts that signals enzymatic cleavage ~1035 nucleotides away.
C) It allows the 3' end of the mRNA to attach to the ribosome.
D) It is a sequence that codes for the hydrolysis of the RNA polymerase.
E) It adds a 7-methylguanosine cap to the 3' end of the mRNA.
Answer: B

14) In eukaryotes there are several different types of RNA polymerase. Which type is involved in transcription of mRNA for a globin protein?
A) ligase
B) RNA polymerase I
C) RNA polymerase II
D) RNA polymerase III
E) primase
Answer: C

15) Transcription in eukaryotes requires which of the following in addition to RNA polymerase?
A) the protein product of the promoter
B) start and stop codons
C) ribosomes and tRNA
D) several transcription factors (TFs)
E) aminoacyl synthetase
Answer: D

16) A part of the promoter, called the TATA box, is said to be highly conserved in evolution. Which of the following might this illustrate?
A) The sequence evolves very rapidly.
B) The sequence does not mutate.
C) Any mutation in the sequence is selected against.
D) The sequence is found in many but not all promoters.
E) The sequence is transcribed at the start of every gene.
Answer: C

17) The TATA sequence is found only several nucleotides away from the start site of transcription. This most probably relates to which of the following?
A) the number of hydrogen bonds between A and T in DNA
B) the triplet nature of the codon
C) the ability of this sequence to bind to the start site
D) the supercoiling of the DNA near the start site
E) the 3-D shape of a DNA molecule
Answer: A

18) What is a ribozyme?
A) an enzyme that uses RNA as a substrate
B) an RNA with enzymatic activity
C) an enzyme that catalyzes the association between the large and small ribosomal subunits
D) an enzyme that synthesizes RNA as part of the transcription process
E) an enzyme that synthesizes RNA primers during DNA replication
Answer: B

19) A transcription unit that is 8,000 nucleotides long may use 1,200 nucleotides to make a protein consisting of approximately 400 amino acids. This is best explained by the fact that
A) many noncoding stretches of nucleotides are present in mRNA.
B) there is redundancy and ambiguity in the genetic code.
C) many nucleotides are needed to code for each amino acid.
D) nucleotides break off and are lost during the transcription process.
E) there are termination exons near the beginning of mRNA.
Answer: A

20) During splicing, which molecular component of the spliceosome catalyzes the excision reaction?
A) protein
B) DNA
C) RNA
D) lipid
E) sugar
Answer: C

21) Alternative RNA splicing
A) is a mechanism for increasing the rate of transcription.
B) can allow the production of proteins of different sizes from a single mRNA.
C) can allow the production of similar proteins from different RNAs.
D) increases the rate of transcription.
E) is due to the presence or absence of particular snRNPs.
Answer: B

22) In the structural organization of many eukaryotic genes, individual exons may be related to which of the following?
A) the sequence of the intron that immediately precedes each exon
B) the number of polypeptides making up the functional protein
C) the various domains of the polypeptide product
D) the number of restriction enzyme cutting sites
E) the number of start sites for transcription
Answer: C

23) In an experimental situation, a student researcher inserts an mRNA molecule into a eukaryotic cell after he has removed its 5' cap and poly-A tail. Which of the following would you expect him to find?
A) The mRNA could not exit the nucleus to be translated.
B) The cell recognizes the absence of the tail and polyadenylates the mRNA.
C) The molecule is digested by restriction enzymes in the nucleus.
D) The molecule is digested by exonucleases since it is no longer protected at the 5' end.
E) The molecule attaches to a ribosome and is translated, but more slowly.
Answer: D

24. Use the following model of a eukaryotic transcript to answer the next few questions.

5' UTR E₁ I₁ E₂ I₂ E₃ I₃ E₄ UTR 3'

Which components of the previous molecule will also be found in mRNA in the cytosol?
A) 5' UTR I₁ I₂ I₃ UTR 3'
B) 5' E₁ E₂ E₃ E₄ 3'
C) 5' UTR E₁ E₂ E₃ E₄ UTR 3'
D) 5' I₁ I₂ I₃ 3'
E) 5' E₁ I₁ E₂ I₂ E₃ I₃ E₄ 3'
Answer: C

25. Use the following model of a eukaryotic transcript to answer the next few questions.

5' UTR E₁ I₁ E₂ I₂ E₃ I₃ E₄ UTR 3'

When the spliceosome binds to elements of this structure, where can it attach?
A) to the exons
B) to the 5' UTR
C) to the 3' UTR
D) to an adjacent intron and exon
E) to the end of an intron
Answer: E

26. Use the following model of a eukaryotic transcript to answer the next few questions.

5' UTR E₁ I₁ E₂ I₂ E₃ I₃ E₄ UTR 3'

Which of the following is a useful feature of introns for this model?
A) They are translated into small polypeptides.
B) They become parts of snRNPs.
C) Each intron has enzymatic properties.
D) Introns allow exon shuffling.
E) Introns protect exon structure.
Answer: D

27. Use the following model of a eukaryotic transcript to answer the next few questions.

5' UTR E₁ I₁ E₂ I₂ E₃ I₃ E₄ UTR 3'

Suppose that exposure to a chemical mutagen results in a change in the sequence that alters the 5' end of intron 1 (I₁). What might occur?
A) loss of the gene product
B) loss of E₁
C) premature stop to the mRNA
D) inclusion of I₁ in the mRNA
E) exclusion of E₂
Answer: D

28. Use the following model of a eukaryotic transcript to answer the next few questions.

5' UTR E₁ I₁ E₂ I₂ E₃ I₃ E₄ UTR 3'

Suppose that an induced mutation removes most of the 5' end of the 5' UTR. What might result?
A) Removal of the 5' UTR has no effect because the exons are still maintained.
B) Removal of the 5' UTR also removes the 5' cap and the mRNA will quickly degrade.
C) The 3' UTR will duplicate and one copy will replace the 5' end.
D) The first exon will not be read because I₁ will now serve as the UTR.
E) Removal of the 5' UTR will result in the strand not binding to tRNAs.
Answer: B

29) A particular triplet of bases in the coding sequence of DNA is AAA. The anticodon on the tRNA that binds the mRNA codon is
A) TTT.
B) UUA.
C) UUU.
D) AAA.
E) either UAA or TAA, depending on first base wobble.
Answer: C

30) Accuracy in the translation of mRNA into the primary structure of a polypeptide depends on specificity in the
A) binding of ribosomes to mRNA.
B) shape of the A and P sites of ribosomes.
C) bonding of the anticodon to the codon.
D) attachment of amino acids to tRNAs.
E) bonding of the anticodon to the codon and the attachment of amino acids to tRNAs.
Answer: E

31) What is the function of GTP in translation?
A) GTP energizes the formation of the initiation complex, using initiation factors.
B) GTP hydrolyzes to provide phosphate groups for tRNA binding.
C) GTP hydrolyzes to provide energy for making peptide bonds.
D) GTP supplies phosphates and energy to make ATP from ADP.
E) GTP separates the small and large subunits of the ribosome at the stop codon.
Answer: A

32) A mutant bacterial cell has a defective aminoacyl synthetase that attaches a lysine to tRNAs with the anticodon AAA instead of the normal phenylalanine. The consequence of this for the cell will be that
A) none of the proteins in the cell will contain phenylalanine.
B) proteins in the cell will include lysine instead of phenylalanine at amino acid positions specified by the codon UUU.
C) the cell will compensate for the defect by attaching phenylalanine to tRNAs with lysine-specifying anticodons.
D) the ribosome will skip a codon every time a UUU is encountered.
E) none of the options will occur; the cell will recognize the error and destroy the tRNA.
Answer: B

33) There are 61 mRNA codons that specify an amino acid, but only 45 tRNAs. This is best explained by the fact that
A) some tRNAs have anticodons that recognize four or more different codons.
B) the rules for base pairing between the third base of a codon and tRNA are flexible.
C) many codons are never used, so the tRNAs that recognize them are dispensable.
D) the DNA codes for all 61 tRNAs but some are then destroyed.
E) competitive exclusion forces some tRNAs to be destroyed by nucleases.
Answer: B

34) Which of the following is the first event to take place in translation in eukaryotes?
A) elongation of the polypeptide
B) base pairing of activated methionine-tRNA to AUG of the messenger RNA
C) binding of the larger ribosomal subunit to smaller ribosomal subunits
D) covalent bonding between the first two amino acids
E) the small subunit of the ribosome recognizes and attaches to the 5' cap of mRNA
Answer: E

35) Which of the following is a function of a signal peptide?
A) to direct an mRNA molecule into the cisternal space of the ER
B) to bind RNA polymerase to DNA and initiate transcription
C) to terminate translation of the messenger RNA
D) to translocate polypeptides across the ER membrane
E) to signal the initiation of transcription
Answer: D

36) When translating secretory or membrane proteins, ribosomes are directed to the ER membrane by
A) a specific characteristic of the ribosome itself, which distinguishes free ribosomes from bound ribosomes.
B) a signal-recognition particle that brings ribosomes to a receptor protein in the ER membrane.
C) moving through a specialized channel of the nucleus.
D) a chemical signal given off by the ER.
E) a signal sequence of RNA that precedes the start codon of the message.
Answer: B

37) An experimenter has altered the 3' end of the tRNA corresponding to the amino acid methionine in such a way as to remove the 3' AC. Which of the following hypotheses describes the most likely result?
A) tRNA will not form a cloverleaf.
B) The nearby stem end will pair improperly.
C) The amino acid methionine will not bind.
D) The anticodon will not bind with the mRNA codon.
E) The aminoacylsynthetase will not be formed.
Answer: C

38) The process of translation, whether in prokaryotes or eukaryotes, requires tRNAs, amino acids, ribosomal subunits, and which of the following?
A) polypeptide factors plus ATP
B) polypeptide factors plus GTP
C) polymerases plus GTP
D) SRP plus chaperones
E) signal peptides plus release factor
Answer: B

39) When the ribosome reaches a stop codon on the mRNA, no corresponding tRNA enters the A site. If the translation reaction were to be experimentally stopped at this point, which of the following would you be able to isolate?
A) an assembled ribosome with a polypeptide attached to the tRNA in the P site
B) separated ribosomal subunits, a polypeptide, and free tRNA
C) an assembled ribosome with a separated polypeptide
D) separated ribosomal subunits with a polypeptide attached to the tRNA
E) a cell with fewer ribosomes
Answer: A

40) What is the function of the release factor (RF)?
A) It separates tRNA in the A site from the growing polypeptide.
B) It binds to the stop codon in the A site in place of a tRNA.
C) It releases the amino acid from its tRNA to allow the amino acid to form a peptide bond.
D) It supplies a source of energy for termination of translation.
E) It releases the ribosome from the ER to allow polypeptides into the cytosol.
Answer: B

41) When the function of the newly made polypeptide is to be secreted from the cell where it has been made, what must occur?
A) It must be translated by a ribosome that remains free of attachment to the ER.
B) Its signal sequence must target it to the ER, from which it goes to the Golgi.
C) It has a signal sequence that must be cleaved off before it can enter the ER.
D) It has a signal sequence that targets it to the cell's plasma membrane where it causes exocytosis.
E) Its signal sequence causes it to be encased in a vesicle as soon as it is translated.
Answer: B

42) Suppose that a mutation alters the formation of a tRNA such that it still attaches to the same amino acid (phe) but its anticodon loop has the sequence AAU that binds to the mRNA codon UUA (that usually specifies leucine leu).
A) The modified tRNA will cause this mRNA to make only nonfunctioning product.
B) The tRNA-leu will not be able to enter the site of the ribosome to bind to the UUA.
C) One mutated tRNA molecule will be relatively inconsequential because it will compete with many "normal" ones.
D) The tRNA will be so unstable that it will not participate in translation.
E) The mutated tRNA will result in an amino acid variant in all copies of the protein.
Answer: C

43) Why might a point mutation in DNA make a difference in the level of protein's activity?
A) It might result in a chromosomal translocation.
B) It might exchange one stop codon for another stop codon.
C) It might exchange one serine codon for a different serine codon.
D) It might substitute an amino acid in the active site.
E) It might substitute the N-terminus of the polypeptide for the C-terminus.
Answer: D

44) In the 1920s Muller discovered that X-rays caused mutation in Drosophila. In a related series of experiments in the 1940s, Charlotte Auerbach discovered that chemicals–she used nitrogen mustards–have a similar effect. A new chemical food additive is developed by a cereal manufacturer. Why do we test for its ability to induce mutation?
A) We worry that it might cause mutation in cereal grain plants.
B) We want to make sure that it does not emit radiation.
C) We want to be sure that it increases the rate of mutation sufficiently.
D) We want to prevent any increase in mutation frequency.
E) We worry about its ability to cause infection.
Answer: D

45) Which of the following types of mutation, resulting in an error in the mRNA just after the AUG start of translation, is likely to have the most serious effect on the polypeptide product?
A) a deletion of a codon
B) a deletion of two nucleotides
C) a substitution of the third nucleotide in an ACC codon
D) a substitution of the first nucleotide of a GGG codon
E) an insertion of a codon
Answer: B

46) What is the effect of a nonsense mutation in a gene?
A) It changes an amino acid in the encoded protein.
B) It has no effect on the amino acid sequence of the encoded protein.
C) It introduces a premature stop codon into the mRNA.
D) It alters the reading frame of the mRNA.
E) It prevents introns from being excised.
Answer: C

47) A frameshift mutation could result from
A) a base insertion only.
B) a base deletion only.
C) a base substitution only.
D) deletion of three consecutive bases.
E) either an insertion or a deletion of a base.
Answer: E

48) Which of the following DNA mutations is the most likely to be damaging to the protein it specifies?
A) a base-pair deletion
B) a codon substitution
C) a substitution in the last base of a codon
D) a codon deletion
E) a point mutation
Answer: A

49) Which small-scale mutation would be most likely to have a catastrophic effect on the functioning of a protein?
A) a base substitution
B) a base deletion near the start of a gene
C) a base deletion near the end of the coding sequence, but not in the terminator codon
D) deletion of three bases near the start of the coding sequence, but not in the initiator codon
E) a base insertion near the end of the coding sequence, but not in the terminator codon
Answer: B

50) The most commonly occurring mutation in people with cystic fibrosis is a deletion of a single codon. This results in
A) a base-pair substitution.
B) a nucleotide mismatch.
C) a frameshift mutation.
D) a polypeptide missing an amino acid.
E) a nonsense mutation.
Answer: D

51) Which of the following mutations is most likely to cause a phenotypic change?
A) a duplication of all or most introns
B) a large inversion whose ends are each in intergenic regions
C) a nucleotide substitution in an exon coding for a transmembrane domain
D) a single nucleotide deletion in an exon coding for an active site
E) a frameshift mutation one codon away from the 3' end of the nontemplate strand
Answer: D

52) If a protein is coded for by a single gene and this protein has six clearly defined domains, which number of exons below is the gene likely to have?
A) 1
B) 5
C) 8
D) 12
E) 14
Answer: C

53) Which of the following statements is true about protein synthesis in prokaryotes?
A) Extensive RNA processing is required before prokaryotic transcripts can be translated.
B) Translation can begin while transcription is still in progress.
C) Prokaryotic cells have complicated mechanisms for targeting proteins to the appropriate cellular organelles.
D) Translation requires antibiotic activity.
E) Unlike eukaryotes, prokaryotes require no initiation or elongation factors.
Answer: B

54) Of the following, which is the most current description of a gene?
A) a unit of heredity that causes formation of a phenotypic characteristic
B) a DNA subunit that codes for a single complete protein
C) a DNA sequence that is expressed to form a functional product: either RNA or polypeptide
D) a DNA—RNA sequence combination that results in an enzymatic product
E) a discrete unit of hereditary information that consists of a sequence of amino acids
Answer: C

55) Gene expression in the domain Archaea in part resembles that of bacteria and in part that of the domain Eukarya. In which way is it most like the domain Eukarya?
A) Domain Archaea have numerous transcription factors.
B) Initiation of translation is like that of domain Eukarya.
C) There is only one RNA polymerase.
D) Transcription termination often involves attenuation.
E) Post-transcriptional splicing is like that of Eukarya.
Answer: A

56) Which of the following is true of transcription in domain Archaea?
A) It is regulated in the same way as in domain Bacteria.
B) There is only one kind of RNA polymerase.
C) It is roughly simultaneous with translation.
D) Promoters are identical to those in domain Eukarya.
E) It terminates in a manner similar to bacteria.
Answer: C

57) In comparing DNA replication with RNA transcription in the same cell, which of the following is true only of replication?
A) It uses RNA polymerase.
B) It makes a new molecule from its 5' end to its 3' end.
C) The process is extremely fast once it is initiated.
D) The process occurs in the nucleus of a eukaryotic cell.
E) The entire template molecule is represented in the product.
Answer: E

58) In order for a eukaryotic gene to be engineered into a bacterial colony to be expressed, what must be included in addition to the coding exons of the gene?
A) the introns
B) eukaryotic polymerases
C) a bacterial promoter sequence
D) eukaryotic ribosomal subunits
E) eukaryotic tRNAs
Answer: C

59) When the genome of a particular species is said to include 20,000 protein-coding regions, what does this imply?
A) There are 20,000 genes.
B) Each gene codes for one protein.
C) Any other regions are "junk" DNA.
D) There are also genes for RNAs other than mRNA.
E) The species is highly evolved.
Answer: D


60) According to Beadle and Tatum's hypothesis, how many genes are necessary for this pathway?



A) 0
B) 1
C) 2
D) 3
E) It cannot be determined from the pathway.
Answer: C

61) A mutation results in a defective enzyme A. Which of the following would be a consequence of that mutation?
A) an accumulation of A and no production of B and C
B) an accumulation of A and B and no production of C
C) an accumulation of B and no production of A and C
D) an accumulation of B and C and no production of A
E) an accumulation of C and no production of A and B
Answer: A

62) If A, B, and C are all required for growth, a strain that is mutant for the gene-encoding enzyme A would be able to grow on which of the following media?
A) minimal medium
B) minimal medium supplemented with nutrient A only
C) minimal medium supplemented with nutrient B only
D) minimal medium supplemented with nutrient C only
E) minimal medium supplemented with nutrients A and C
Answer: C

63) If A, B, and C are all required for growth, a strain mutant for the gene-encoding enzyme B would be capable of growing on which of the following media?
A) minimal medium
B) minimal medium supplemented with A only
C) minimal medium supplemented with B only
D) minimal medium supplemented with C only
E) minimal medium supplemented with nutrients A and B
Answer: D


64) A possible sequence of nucleotides in the template strand of DNA that would code for the polypeptide sequence phe-leu-ile-val would be



A) 5' TTG-CTA-CAG-TAG 3'.
B) 3' AAC-GAC-GUC-AUA 5'.
C) 5' AUG-CTG-CAG-TAT 3'.
D) 3' AAA-AAT-ATA-ACA 5'.
E) 3' AAA-GAA-TAA-CAA 5'.
Answer: E


65) What amino acid sequence will be generated, based on the following mRNA codon sequence?


5' AUG-UCU-UCG-UUA-UCC-UUG 3'
A) met-arg-glu-arg-glu-arg
B) met-glu-arg-arg-glu-leu
C) met-ser-leu-ser-leu-ser
D) met-ser-ser-leu-ser-leu
E) met-leu-phe-arg-glu-glu
Answer: D


66) A peptide has the sequence NH2-phe-pro-lys-gly-phe-pro-COOH. Which of the following sequences in the coding strand of the DNA could code for this peptide?


A) 3' UUU-CCC-AAA-GGG-UUU-CCC
B) 3' AUG-AAA-GGG-TTT-CCC-AAA-GGG
C) 5' TTT-CCC-AAA-GGG-TTT-CCC
D) 5' GGG-AAA-TTT-AAA-CCC-ACT-GGG
E) 5' ACT-TAC-CAT-AAA-CAT-TAC-UGA
Answer: C

DNA template strand
5' ____________________________ 3'

DNA complementary strand
3' ____________________________ 5'

67) Given the locally unwound double strand above, in which direction does the RNA polymerase move?
A) 3' → 5' along the template strand
B) 5' → 3' along the template strand
C) 3' → 5' along the complementary strand
D) 5' → 3' along the complementary strand
E) 5' → 3' along the double-stranded DNA
Answer: A
             68
DNA template strand
5' ____________________________ 3'

DNA complementary strand
3' ____________________________ 5'

68) In the transcription event of the previous DNA, where would the promoter be located?
A) at the 3' end of the newly made RNA
B) to the right of the template strand
C) to the left of the template strand
D) to the right of the sense strand
E) to the left of the sense strand
Answer: B


69. A part of an mRNA molecule with the following sequence is being read by a ribosome: 5' CCG-ACG 3' (mRNA). The following charged transfer RNA molecules (with their anticodons shown in the 3' to 5' direction) are available. Two of them can correctly match the mRNA so that a dipeptide can form.

The dipeptide that will form will be
A) cysteine-alanine.
B) proline-threonine.
C) glycine-cysteine.
D) alanine-alanine.
E) threonine-glycine.
Answer: B

70. A part of an mRNA molecule with the following sequence is being read by a ribosome: 5' CCG-ACG 3' (mRNA). The following charged transfer RNA molecules (with their anticodons shown in the 3' to 5' direction) are available. Two of them can correctly match the mRNA so that a dipeptide can form.

The anticodon loop of the first tRNA that will complement this mRNA is
A) 3' GGC 5'
B) 5' GGC 3'
C) 5' ACG 3'
D) 5' UGC 3'
E) 3' UGC 5'
Answer: A

71) What type of bonding is responsible for maintaining the shape of the tRNA molecule?



A) covalent bonding between sulfur atoms
B) ionic bonding between phosphates
C) hydrogen bonding between base pairs
D) van der Waals interactions between hydrogen atoms
E) peptide bonding between amino acids
Answer: C

72) The figure represents tRNA that recognizes and binds a particular amino acid (in this instance, phenylalanine). Which codon on the mRNA strand codes for this amino acid?
A) UGG
B) GUG
C) GUA
D) UUC
E) CAU
Answer: D

73) The tRNA shown in the figure has its 3' end projecting beyond its 5' end. What will occur at this 3' end?

A) The codon and anticodon complement one another.
B) The amino acid binds covalently.
C) The excess nucleotides (ACCA) will be cleaved off at the ribosome.
D) The small and large subunits of the ribosome will attach to it.
E) The 5' cap of the mRNA will become covalently bound.
Answer: B

74. The enzyme polynucleotide phosphorylase randomly assembles nucleotides into a polynucleotide polymer.
You add polynucleotide phosphorylase to a solution of adenosine triphosphate and guanosine triphosphate. How many artificial mRNA 3 nucleotide codons would be possible?

A) 3
B) 4
C) 8
D) 16
E) 64
Answer: C

75. The enzyme polynucleotide phosphorylase randomly assembles nucleotides into a polynucleotide polymer.
You add polynucleotide phosphorylase to a solution of ATP, GTP, and UTP. How many artificial mRNA 3 nucleotide codons would be possible?
A) 3
B) 6
C) 9
D) 27
E) 81
Answer: D

76. A transfer RNA (#1) attached to the amino acid lysine enters the ribosome. The lysine binds to the growing polypeptide on the other tRNA (#2) in the ribosome already.
Where does tRNA #2 move to after this bonding of lysine to the polypeptide?
A) A site
B) P site
C) E site
D) exit tunnel
E) directly to the cytosol
Answer: C

77. A transfer RNA (#1) attached to the amino acid lysine enters the ribosome. The lysine binds to the growing polypeptide on the other tRNA (#2) in the ribosome already.
Which component of the complex described enters the exit tunnel through the large subunit of the ribosome?
A) tRNA with attached lysine (#1)
B) tRNA with polypeptide (#2)
C) tRNA that no longer has attached amino acid
D) newly formed polypeptide
E) initiation and elongation factors
Answer: D

78) In eukaryotic cells, transcription cannot begin until
A) the two DNA strands have completely separated and exposed the promoter.
B) several transcription factors have bound to the promoter.
C) the 5' caps are removed from the mRNA.
D) the DNA introns are removed from the template.
E) DNA nucleases have isolated the transcription unit.
Answer: B

79) Which of the following is not true of a codon?
A) It consists of three nucleotides.
B) It may code for the same amino acid as another codon.
C) It never codes for more than one amino acid.
D) It extends from one end of a tRNA molecule.
E) It is the basic unit of the genetic code.
Answer: D

80) The anticodon of a particular tRNA molecule is
A) complementary to the corresponding mRNA codon.
B) complementary to the corresponding triplet in rRNA.
C) the part of tRNA that bonds to a specific amino acid.
D) changeable, depending on the amino acid that attaches to the tRNA.
E) catalytic, making the tRNA a ribozyme.
Answer: A

81) Which of the following is not true of RNA processing?
A) Exons are cut out before mRNA leaves the nucleus.
B) Nucleotides may be added at both ends of the RNA.
C) Ribozymes may function in RNA splicing.
D) RNA splicing can be catalyzed by spliceosomes.
E) A primary transcript is often much longer than the final RNA molecule that leaves the nucleus.
Answer: A

82) Which component is not directly involved in translation?
A) mRNA
B) DNA
C) tRNA
D) ribosomes
E) GTP
Answer: B

83) Which of the following mutations would be most likely to have a harmful effect on an organism?
A) a nucleotide-pair substitution
B) a deletion of three nucleotides near the middle of a gene
C) a single nucleotide deletion in the middle of an intron
D) a single nucleotide deletion near the end of the coding sequence
E) a single nucleotide insertion downstream of, and close to, the start of the coding sequence
Answer : E

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