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PROCTOR VERSION 4.7 B: Regulating Eukaryotic Gene Expression Quiz 1. Scientists have observed that when double-stranded RNA is inserted into cells, genes that produce mRNA transcripts that are similar to the inserted RNA are not expressed. When this happens, the double-stranded RNA is chopped into smaller pieces, called microRNAs or miRNAs, and then a sequence of processes occurs, as shown in the diagram below. miRNAs are also normally transcribed in cells and used to regulate gene expression. Page 1 of 7 PROCTOR VERSION 4.7 B: Regulating Eukaryotic Gene Expression Quiz Which statement best explains how miRNAs prevent gene expression? (A) miRNAs prevent gene expression by binding to the RNA primer, so the process of transcription is blocked and mRNA is not produced. Distractor Rationale: This answer suggests the student may understand that an RNA primer is needed for nucleic acid synthesis, but does not understand that miRNAs binding to the RNA primer would not affect gene expression because the RNA primer is needed for DNA replication, not transcription. (B) miRNAs prevent gene expression by binding to tRNA and preventing it from accepting the amino acids needed for translation. Distractor Rationale: This answer suggests the student may understand that tRNA is important for translation and gene expression, but does not understand that miRNAs are unlikely to bind specifically to tRNAs because, given their structure, miRNAs are better suited to bind to single-stranded RNA transcripts that have not folded upon themselves to form a 3D structure. (C) miRNAs prevent gene expression by binding to and degrading rRNA, which prevents translation in the ribosome. Distractor Rationale: This answer suggests the student may understand that binding to and disabling rRNA could prevent gene expression, but does not understand that this is unlikely because miRNAs bind to single-stranded RNA transcripts, which would most likely be mRNA. (D) miRNAs prevent gene expression by binding to mRNA and disrupting the process that decodes mRNA to produce a polypeptide. Rationale: This answer suggests the student understands that miRNAs could prevent gene expression by binding to mRNA transcripts and preventing translation in the ribosome. Aligned to: LO 4.7 CA 4.7: Represent Mechanisms of Specialization Page 2 of 7 PROCTOR VERSION 4.7 B: Regulating Eukaryotic Gene Expression Quiz 2. The diagram below shows an activator binding to an enhancer region on DNA and then looping the DNA to combine with a transcription factor and RNA polymerase on the promoter region of a gene. Which statement best explains how the process shown in the diagram is involved in the expression of a gene? (A) The enhancer–activator–transcription factor complex cuts out a non-coding region of DNA, enabling RNA polymerase to transcribe the correct sequence of DNA. Distractor Rationale: This answer suggests the student may understand that gene expression involves removal of non-coding regions on the RNA transcript, but does not understand that this process would not involve enhancers, activators, transcription factors, or RNA polymerase and that these structures are involved in the activation of a gene for transcription. (B) The enhancer–activator–transcription factor complex forms a loop to expose the coding sequence of DNA, allowing the RNA polymerase to bind to and transcribe the looped region. Distractor Rationale: This answer suggests the student may understand that gene expression involves binding of the activated enhancer to the promoter, but does not understand that this process would not involve transcription of the looped region because this part of the gene will not be transcribed, since the looped region is located between the enhancer and the promoter and is not part of the coding sequence for the protein. Page 3 of 7 PROCTOR VERSION 4.7 B: Regulating Eukaryotic Gene Expression Quiz (C) The enhancer–activator–transcription factor complex binds to the promoter and alters its configuration, allowing the RNA polymerase to bind to the promoter and initiate transcription of the gene. Rationale: This answer suggests the student understands how an activator and a transcription factor work with enhancer and promoter sequences to recruit RNA polymerase to a gene to initiate transcription. (D) The enhancer–activator–transcription factor complex brings nucleotides to the promoter region of the gene, where the nucleotides are paired to complementary bases on the gene by RNA polymerase during transcription. Distractor Rationale: This answer suggests the student may understand that RNA polymerase pairs complementary RNA nucleotides with bases on the gene during transcription, but does not understand that this process does not involve the enhancer or the activator because the process can only occur after the enhancer-activator activates the promoter region of the gene. Aligned to: LO 4.7 CA 4.7: Represent Mechanisms of Specialization 3. The diagram below shows the transcription of a gene in two different cells in a eukaryotic organism. Which statements best explain the process occurring in step Y in the two cells? Page 4 of 7 PROCTOR VERSION 4.7 B: Regulating Eukaryotic Gene Expression Quiz (A) Non-coding regions are removed and coding regions are spliced together during mRNA processing. Each cell will produce a different polypeptide, depending on which coding regions are spliced together in the final transcript. Rationale: This answer suggests the student understands that mRNA processing can produce different RNA sequences and that this allows for differential gene expression in different cells. (B) RNA is being processed by the attachment of unique 5' caps and poly-A tails. Each cell will produce a unique polypeptide because 5' caps and poly-A tails are different in each cell. Distractor Rationale: This answer suggests the student may understand that post-transcription processing involves the attachment of a 5' cap and a poly-A tail on each end of the mRNA molecule, but does not understand that this does not change the genetic message because all of the codons involved in translation are unaffected. (C) RNA is being processed to produce different promoter regions in the final mRNA of the two cells. Each cell will express the gene in response to different factors, because each cell has different promoters. Distractor Rationale: This answer suggests the student may understand that different promoters respond to different transcription factors and enhancers, but does not understand that this would not occur during mRNA processing because promoters regulate gene expression before transcription, not after transcription. (D) One of the cells has a mutation, causing the RNA to be processed differently in the two cells. The mutation in one cell will cause the cell’s mRNA to contain non-coding sequences that will produce extra or incorrect amino acids in the polypeptide. Distractor Rationale: This answer suggests the student may understand that mutations result in differences in the mRNA sequence, which result in differences in polypeptides, but does not understand that there is no evidence of a mutation in the diagram, because both cells have the same gene and initial mRNA sequences, and differences between the cells appeared only after mRNA processing. Aligned to: LO 4.7 CA 4.7: Represent Mechanisms of Specialization Page 5 of 7 PROCTOR VERSION 4.7 B: Regulating Eukaryotic Gene Expression Quiz 4. The diagram below shows homeotic genes that govern body development in fruit flies (Drosophila melanogaster). Locations governed by certain genes are shown in an adult (below top) and an embryo (below bottom). A mutation occurs that results in the change shown in the diagram below. Page 6 of 7 PROCTOR VERSION 4.7 B: Regulating Eukaryotic Gene Expression Quiz Based on the information provided in the diagrams, which statement describes the most likely outcome in the affected adult fruit fly? (A) The fruit fly would have no wings, because the Antp gene regulates wing development, and the Antp gene would be in the wrong location in the sequence. Distractor Rationale: This answer suggests the student may understand that mutations in homeotic genes can result in entire structures being omitted in organisms, but does not understand that this mutation would not result in missing wings because the gene for wing development (Antp) is not deleted, but is duplicated in another region. (B) The fruit fly would have an extra set of wings but would be missing the third pair of legs, because the Antp gene would be expressed in place of the Ubx gene. Distractor Rationale: This answer suggests the student may understand that mutations in homeotic genes can result in duplications of entire structures, but does not understand that this mutation would not result in missing legs because the duplicated gene governs the production of legs, although the legs would resemble the second pair, not the third pair. (C) The fruit fly would have an extra set of wings but would still have six legs, because the Antp gene would be expressed in place of the Ubx gene. Rationale: This answer suggests the student understands that the duplication of the Antp gene and the insertion of this gene in place of the Ubx gene would result in an extra set of wings because these appendages would replace what was coded for on the Ubx gene. (D) The fruit fly would have an extra set of legs, because the Antp gene is now duplicated in the same region of the fly, and this would result in an extra set of legs. Distractor Rationale: This answer suggests the student may understand that mutations in homeotic genes can result in the duplications of structures, but does not understand that this mutation would not result in extra legs because the gene for wing and leg development (Antp) would be inserted in place of a gene that governs leg development only (Ubx). Aligned to: LO 4.7 CA 4.7: Represent Mechanisms of Specialization Page 7 of 7