Bsc 1010L Ch 16

Front 1

1. In some organisms, the primary function of a gene in a cell is to participate in regulating the body as a whole rather than responding to the cell's immediate environment. These organisms would be
A. multicellular.
B. diploid.
C. bacterial.
D. prokaryotic.

Back 1

A. multicellular.

Front 2

2. The maintenance of a constant environment in a cell is called __________.
A. active transport
B. homeostasis
C. gene expression
D. translation

Back 2

B. homeostasis

Front 3

3. Through control of gene expression, a bacterial cell responds to changing __________ conditions.
A. internal
B. protein
C. environmental
D. genetic

Back 3

C. environmental

Front 4

4. In multicellular organisms, the mechanism most directly responsible for directing development and maintaining homeostasis is gene
A. expression.
B. duplication.
C. deletion.
D. regulation.

Back 4

D. regulation.

Front 5

5. Eukaryotic cell mRNA transcripts can remain in the cell for hours because they are
A. stable.
B. long.
C. isolated.
D. analogous.

Back 5

A. stable.

Front 6

6. As a microbiologist you have been asked to investigate gene regulation in a new bacteria. Given what is known about bacteria, the logical place to begin your investigation is
A. transcriptional control.
B. translational control.
C. promoter control.
D. repressor control.
E. operator control.

Back 6

A. transcriptional control.

Front 7

7. Transcriptional control-proteins increase the rate of transcription by binding to
A. mRNA sequences within the DNA.
B. tRNA sequences within the DNA.
C. operator sequences within the DNA.
D. promoter sequences within the DNA.
E. enhancer sequences within the DNA.

Back 7

E. enhancer sequences within the DNA.

Front 8

8. You have been asked to design a synthetic DNA motif, able to bind proteins to regulate specific genes. The location on this motif that you would necessarily design for protein binding is the
A. minor groove of the DNA double helix.
B. major groove of the DNA double helix.
C. outside groove of the DNA double helix.
D. inside groove of the DNA double helix.
E. hydrogen bonding groove of the DNA double helix.

Back 8

B. major groove of the DNA double helix.

Front 9

9. Vertebrate cells possess a protein that binds to clusters of 5-methylcytosine and ensures the gene will stay in the "off" position. This control of gene regulation is a result of
A. translation.
B. enhancer expression.
C. methylation.
D. promoter expression.
E. operator suppression.

Back 9

C. methylation.

Front 10

10. You lead a research team challenged with the task of creating a regulatory protein able to shut off transcription. You focus your design around a binding site called an operator that is associated with the promotor. The physical location of the operator most likely to affect transcription would be
A. upstream of the gene promoter.
B. downstream of the gene promoter.
C. internal to the gene itself.
D. internal to the gene promoter.

Back 10

B. downstream of the gene promoter.

Front 11

11. Histones that are tightly wound by DNA and are the basic unit of chromatin are called
A. operons.
B. nucleosomes.
C. protein clusters.
D. repressor genes.
E. facilitators sites.

Back 11

B. nucleosomes.

Front 12

12. The hallmark of multicellular organisms is their ability to
A. grow and divide rapidly.
B. adjust quickly to outside environment.
C. maintain homeostasis.
D. quickly synthesize amount and type of enzymes according to available nutrients.
E. respond by gene action to oxygen availability.

Back 12

C. maintain homeostasis.

Front 13

13. Enhancers are the binding sites for the
A. promoters of DNA synthesis.
B. suppressor factors.
C. co-activation factors.
D. mediator factors.
E. specific transcription factors.

Back 13

E. specific transcription factors.

Front 14

14. Within its core a nucleosome contains ____ histones.
A. 2
B. 4
C. 6
D. 8
E. 64

Back 14

D. 8

Front 15

15. The basic tool of genetic regulation is the ability of certain proteins to bind to specific
A. regulatory RNA sequences.
B. regulatory DNA sequences.
C. repressor parts of the gene.
D. promoter parts of the gene.
E. enzymes of the cell.

Back 15

B. regulatory DNA sequences.

Front 16

16. In vertebrates, DNA methylation - the addition of a methyl group to DNA nucleotides - ensures that
A. the gene functions without interruption.
B. no errors will be made during transcription.
C. the nucleosome will quickly form, which assists in mRNA formation.
D. once that gene is transcribed, the mRNA is saved and used over and over again.
E. once a gene is turned off, it will remain off.

Back 16

E. once a gene is turned off, it will remain off.

Front 17

17. You belong to a pharmaceutical company that designs small RNAs, able to control expression of genes that cause chronic illness. The primary focus area of your research should be
A. transcriptional repression.
B. transcriptional activation.
C. translational repression.
D. translational activation.

Back 17

C. translational repression.

Front 18

18. Although the specific mechanism of RNA interference has not been fully defined, it involves
A. double stranded RNA interference with mRNA.
B. double stranded RNA interference with DNA.
C. double stranded DNA interference with mRNA.
D. double stranded mRNA interference with DNA.

Back 18

A. double stranded RNA interference with mRNA.

Front 19

19. The DNA-binding proteins of almost all regulatory proteins use one of a small set of shapes that enable them to fit into the DNA major groove. These shapes are called
A. structural motifs.
B. DNA prints.
C. fingerprints.
D. repressors.
E. transcriptional domains.

Back 19

A. structural motifs.

Front 20

20. Small RNAs can regulate gene expression. One type, called micro RNA (miRNA), acts by binding directly to
A. mRNA to prevent translation.
B. tRNA to prevent transcription.
C. mRNA to prevent transcription.
D. tRNA to prevent translation.

Back 20

A. mRNA to prevent translation.

Front 21

21. The lac operon regulatory system is important to bacteria primarily because lactose
A. cannot be made by bacteria unless the genes are turned on.
B. is the most common source of food; enzymes are needed all the time.
C. is only rarely available; producing enzymes all the time is costly.
D. is incorporated into the nucleic acid of the bacteria.
E. switches the system off and on whether lactose is present or not.

Back 21

C. is only rarely available; producing enzymes all the time is costly.

Front 22

22. In eukaryotes, specific transcription factors have two distinct domains:
A. a DNA-binding domain and a RNA-binding domain.
B. a DNA-binding domain and an activation domain.
C. a DNA-binding domain and a repressor domain.
D. a DNA-binding domain and an enhancer domain.
E. a DNA-binding domain and an operator domain.

Back 22

B. a DNA-binding domain and an activation domain.

Front 23

23. In order for transcription to be initiated
A. DNA polymerase must have access to the DNA double helix and must also be capable of binding to the gene's promoter.
B. RNA polymerase must have access to the DNA double helix and must also be capable of binding to the gene's promoter.
C. DNA polymerase must have access to the RNA and must also be capable of binding to the gene's promoter.
D. RNA ligase must have access to the DNA double helix and must also be capable of binding to the gene's promoter.
E. RNA polymerase must have access to the DNA double helix and also must be capable of binding to the gene's operator.

Back 23

B. RNA polymerase must have access to the DNA double helix and must also be capable of binding to the gene's promoter.

Front 24

24. Certain proteins can bind to specific DNA regulatory sequences by entering
A. the major groove of the DNA and reading the nucleotide base pairs.
B. the minor groove of the DNA and reading the nucleotide base pairs.
C. the major groove of RNA and reading the nucleotide base pairs.
D. DNA's major groove by using DNA polymerase and reading the nucleotide base pairs.
E. DNA's minor groove by using DNA polymerase and reading the nucleotide base pairs.

Back 24

A. the major groove of the DNA and reading the nucleotide base pairs.

Front 25

25. Regulatory proteins can identify specific sequences on the DNA double helix without unwinding the helix. This is accomplished by inserting
A. RNA promoters into either the major groove or the minor groove of the double helix where the edges of the nitrogen bases protrude.
B. DNA-binding motifs into the minor groove of the double helix where the edges of the nitrogen bases protrude.
C. DNA polymerase into the major groove of the double helix where the edges of the nitrogen bases protrude.
D. RNA polymerase into the major groove of the double helix where the edges of the nitrogen bases protrude.
E. DNA-binding motifs into the major groove of the double helix where the edges of the nitrogen bases protrude.

Back 25

E. DNA-binding motifs into the major groove of the double helix where the edges of the nitrogen bases protrude.

Front 26

26. When E. coli cells produce the amino acid tryptophan, a cluster of five genes is transcribed together. This cluster of genes is referred to as the
A. trp transcriptional operator.
B. trp regulator.
C. trp suppressor.
D. trp operon.
E. trp promoter.

Back 26

D. trp operon.

Front 27

27. The proteins necessary for the use of lactose in E. coli are collectively called the
A. lac regulator.
B. lac suppressor.
C. lac operon.
D. lac promoter.
E. lac transcriptional operator.

Back 27

C. lac operon.

Front 28

28. A defining characteristic of eukaryotic organisms is that they
A. have their transcription occurring in the cytoplasm and translation in the nucleus.
B. have their transcription occurring in the nucleus and translation in the cytoplasm.
C. have only operons to assist in gene expression.
D. carry out protein synthesis only in the presence of the cAMP molecule.
E. use the leucine zipper primarily for the production of the amino acid tryptophan.

Back 28

B. have their transcription occurring in the nucleus and translation in the cytoplasm.

Front 29

29. The primary transcripts in eukaryotes are most accurately described as
A. composed of RNA polymerase and associated histones.
B. having the exons removed and the introns retained for translation.
C. a faithful copy of the entire gene including exons and introns.
D. an exact copy of the gene, but the introns have been removed.
E. an RNA copy, but the non-coding exons and introns have been removed.

Back 29

C. a faithful copy of the entire gene including exons and introns.

Front 30

30. One of the DNA-binding motifs in many eukaryotic organisms that contains a nearly identical sequence of 60 amino acids is known as the
A. non-helical zipper.
B. leucine zipper.
C. zinc finger.
D. homeodomain.
E. helix-turn-helix.

Back 30

D. homeodomain.

Front 31

31. The most common DNA-binding motif is the
A. non-helical zipper.
B. leucine zipper.
C. zinc finger.
D. homeodomain.
E. helix-turn-helix.

Back 31

E. helix-turn-helix.

Front 32

32. A protein that regulates transcription by binding to the operator is known as the
A. operon.
B. repressor.
C. promoter.
D. operator.
E. CAP.

Back 32

B. repressor.

Front 33

33. A protein that initiates gene transcription and allows for non-glucose molecules to be used is
A. operon.
B. repressor.
C. promoter.
D. operator.
E. CAP.

Back 33

E. CAP.

Front 34

34. A cluster of functionally-related genes that are regulated together and encoded into a single mRNA molecule is called a(n)
A. operon.
B. repressor.
C. promoter.
D. operator.
E. CAP.

Back 34

A. operon.

Front 35

35. A site of negative genetic regulation where binding by repressor blocks transcription is the
A. operon.
B. repressor.
C. promoter.
D. operator.
E. CAP.

Back 35

D. operator.

Front 36

36. A site at the 5' end of a gene where RNA polymerase attaches to initiate transcription is called a(n).
A. operon.
B. repressor.
C. promoter.
D. operator.
E. CAP.

Back 36

C. promoter.

Front 37

37. The enzyme b-galactosidase acts on lactose to form galactose. In turn, the presence of galactose leads to expression of the enzymes responsible for the metabolism of galactose. In this case, lactose is serving as a carbon source and as a(n)
A. inducer.
B. repressor.
C. effector.
D. operon.

Back 37

A. inducer.

Front 38

38. The lactose analog isopropyl-b-D-thio-galactoside (IPTG) is often used to regulate gene expression systems in bacteria. IPTG does not act as a substrate for b-galactosidase, but can bind to, and inactivate, the repressor. In this case, IPTG serves as a(n)
A. inducer.
B. repressor.
C. effector.
D. operon.

Back 38

A. inducer.

Front 39

39. The main form of glucose repression in the lac operon is
A. induction.
B. repression.
C. inducer exclusion.
D. the CAP/cAMP system.

Back 39

C. inducer exclusion.

Front 40

40. If you were given a sequence with a mutation in the -35 region of the trp operon the most likely effects would be the
A. interference with RNA polymerase binding to the promotor.
B. prevention of the repressor from binding to the operator.
C. depression of the trp operon.
D. enabling of the trp operon to be expressed in the absence of tryptophan.

Back 40

A. interference with RNA polymerase binding to the promotor.

Front 41

41. You are studying regulation of a prokaryotic operon. Experimental results show that expression of the operon is increased when product levels are low. Based on this information, you conclude that the likely mode of regulation is
A. the operon is ON in the absence of its regulatory protein.
B. the operon is OFF in the absence of its regulatory protein.
C. the presence of an inducer will always cause a repressor to bind the operator.
D. the presence of an inducer will always prevent a repressor from binding the operator.

Back 41

B. the operon is OFF in the absence of its regulatory protein.

Front 42

42. The binding of TFIID is followed by the binding of a number of other general transcription factors. The correct order of binding of these factors would be
A. TFIIA, TFIIB, TFIIE, TFIIF, TFIIH
B. TFIIB-TFIIA, TFIIF, TFIIE, TFIIH, TFIIJ
C. TFIIF-TFIIH, TFIIB, TFIIE, TFIIA, TFIIH
D. TFIIE, TFIIB, TFIIA, TFIIH, TFIIF

Back 42

B. TFIIB-TFIIA, TFIIF, TFIIE, TFIIH, TFIIJ

Front 43

43. Your research project involves the characterization of a recently identified transcription factor. As part of your project, you want to determine if this transcription factor binds directly to any of the general transcription factors. Unfortunately, however, you are having trouble expressing and purifying the full length protein. An option that may help you achieve your research goal would be to
A. abandon this experiment and try a vastly different approach to your question, since transcription domains cannot be separated.
B. express large amounts of the general transcription factors.
C. express and purify only the DNA-binding domain, since transcription domains can be separated.
D. try to express and purify only the activation domain, since transcription domains can be separated.

Back 43

D. try to express and purify only the activation domain, since transcription domains can be separated.

Front 44

44. You are working to identify enhancer regions of a particular gene. The best place to begin your search is
A. immediately upstream of the promoter.
B. immediately downstream of the promoter.
C. primarily upstream of the promoter; distance from the promoter does not matter.
D. primarily downstream of the promoter; distance from the promoter does not matter.

Back 44

C. primarily upstream of the promoter; distance from the promoter does not matter.

Front 45

45. The progesterone receptor (PR) is a steroid hormone receptor and transcription factor. The protein SRC-1, which does not bind DNA by itself, can bind to PR and increase expression of genes regulated by PR. Based on this information, SRC-1 is a(n)
A. coactivator.
B. inducer.
C. basal transcription factor.
D. specific transcription factor.

Back 45

A. coactivator.

Front 46

46. The TFIID complex is formed by
A. TATA-binding protein TBP and TAFs.
B. TATA-binding protein TBP, TAFs and RNA pol II.
C. TAFs and the core promoter.
D. TATA-binding protein and activators.

Back 46

A. TATA-binding protein TBP and TAFs.

Front 47

47. The p300/CBP (CREB-binding protein) coactivator proteins are histone acetyltransferases that help regulate the transcription of many genes. Based on this information you can conclude that these proteins are involved in
A. DNA acetylation.
B. chromatin remodeling.
C. recruitment of RNA pol II.
D. formation of the transcription initiation complex.

Back 47

B. chromatin remodeling.

Front 48

48. DNA methylation is the only known natural modification of DNA. It affects
A. adenine bases.
B. guanine bases.
C. cytosine bases.
D. thymine bases.

Back 48

C. cytosine bases.

Front 49

49. If you were able to look very closely at a portion of DNA and find methylated histones, you would
A. be mistaken since only DNA can be methylated, not histones.
B. be looking at a region of active chromatin.
C. be looking at a region of inactive chromatin.
D. be looking at a chromatin remodeling complex.

Back 49

C. be looking at a region of inactive chromatin.

Front 50

50. Elucidation of the histone code might
A. allow us to turn specific genes on or off.
B. allow the determination of nucleosome composition.
C. lead to chromatin remodeling.
D. allow us to control translation.

Back 50

A. allow us to turn specific genes on or off.

Front 51

51. You are studying the function of a recently identified gene in C. elegans. You perform genetic screens for several months in an attempt to isolate loss-of-function gene mutations, but your efforts are unsuccessful. Your advisor suggests you try another approach to eliminate gene function. The best technique to accomplish this goal would be
A. to design a repressor to bind to the operon of this gene.
B. use a histone deacetylase to induce a transcriptionally inactive state.
C. use a C. elegans strain with a homozygous TFIID mutation to prevent the translation initiation complex from forming.
D. use RNA interference to prevent mRNA translation.

Back 51

D. use RNA interference to prevent mRNA translation.

Front 52

52. The enzyme dicer chops dsRNA molecules into small pieces of
A. mRNA and miRNA.
B. miRNA and siRNA.
C. siRNA and rRNA.
D. mRNA and siRNA.

Back 52

B. miRNA and siRNA.

Front 53

53. The gene encoding apolipoprotein B exists in two isoforms, APOB100 and APOB48. These two forms are produced as a result of
A. tissue-specific expression.
B. a gene mutation that results in a stop codon.
C. RNA editing.
D. alternative splicing.

Back 53

C. RNA editing.

Front 54

54. Production of the iron-storing protein ferritin is regulated by aconitase, which binds to a 30-nucleotide sequence at the beginning of the ferritin mRNA and interferes with ribosome binding. Aconitase must be a
A. transcription repressor protein.
B. translation repressor protein.
C. RNA interference protein.
D. translation initiation protein.

Back 54

B. translation repressor protein.

Front 55

55. What is the difference between a gene that is derepressed and one that is induced?
A. A gene that is derepressed is turned on because a repressor molecule is absent. By comparison, a gene that is induced is turned on because an inducer molecule is present.
B. Genes that are derepressed are turned on because an inducer molecule is present. By comparison, a gene that is induced is turned on because a repressor molecule is absent.
C. There is no functional difference between a gene that is derepressed and one that is induced.
D. A derepressed gene is turned off and an induced gene is activated to be expressed.

Back 55

A. A gene that is derepressed is turned on because a repressor molecule is absent. By comparison, a gene that is induced is turned on because an inducer molecule is present.

Front 56

57. You have discovered a way to damage proteins associated with initiating over-active inflammatory responses. You must now begin a study on the pathway that follows the clean up of those damaged proteins. Where would your study be focused?
A. Attachment of ubiquitin to the protein, which would lead to transport of the protein to the lysosome that would engulf the protein and degrade it with lysozymes.
B. Tagging of the protein with ubiquitin and then proteolysis of the protein by proteases in the proteasomes.
C. Tagging of the protein with oligosaccharides first, which would then lead into the transport of the protein to proteasomes for proteolysis via proteases.
D. Oligonucleotides would attach to the protein and transport it to the lysosomes for digestion and recycling into products for energy formation.

Back 56

B. Tagging of the protein with ubiquitin and then proteolysis of the protein by proteases in the proteasomes.

Front 57

58. You are studying the affects of transcription factors on the activation of gene expression. However, when first starting your study you notice that some of the transcription factors bind further away from the DNA that it is aiming to transcribe. What description best explains this?
A. The transcription factor transcribes small RNAs that then bind to the promoter and activate the gene's expression.
B. Signals originate via inducers from the transcription factor to the site of promotion that activate transcription of the gene.
C. DNA looping transports the transcription factor closer to the promoter and initiates gene transcription.
D. The assistance of RNA looping moves the transcription factor closer to the promoter.

Back 57

C. DNA looping transports the transcription factor closer to the promoter and initiates gene transcription.