F 24 Exam 4 Study Answers (3) (3) PDF

**[Regulation of Gene Expression II (Chapter 17)]** 1\. List some important differences between bacterial and eukaryotic cells that affect the way in which genes are regulated. a\. Bacterial genes are frequently organized into operons with coordinate regulation, and genes with operons can be trans...

**[Regulation of Gene Expression II (Chapter 17)]** 1\. List some important differences between bacterial and eukaryotic cells that affect the way in which genes are regulated. a\. Bacterial genes are frequently organized into operons with coordinate regulation, and genes with operons can be transcribed as on a single long mRNA. Eukaryotic genes are not organized into operons and are singly transcribed from their own promoters. b\. In eukaryotic cells, nucleosome structure of the DNA is remodeled prior to transcription occurring. Essentially, the chromatin must assume a more open configuration state, allowing for access by transcription-associated factors. c\. Activator and repressor molecules function in both eukaryotic and bacterial cells. However, in eukaryotic cells activators appear to be more common than in bacterial cells. d\. In bacteria, transcription and translation can occur concurrently. In eukaryotes, the nuclear membrane separates transcription from translation both physically and temporally. This separation results in a greater diversity of regulatory mechanisms that can occur at different points during gene expression. 2\. What changes take place in chromatin structure, and what role do these changes play in eukaryotic gene regulation? Changes in chromatin structure can result in repression or stimulation of gene expression. As genes become more transcriptionally active, DNA shows increased sensitivity to DNase I digestion, suggesting that the chromatin structure is more open. Acetylation of histone proteins by acteyltransferase proteins results in the destabilization of the nucleosome structure and increases transcription as well as hypersensitivity to DNase I. The reverse reaction by deacetylases stabilizes nucleosome structure and lessens DNase I sensitivity. Other transcription factors and regulatory proteins, called chromatin remodeling complexes, bind directly to the DNA-altering chromatin structure without acetylating histone proteins. The chromatin remodeling complexes allow for transcription to be initiated by increasing accessibility to the promoters by transcription factors. DNA methylation is also associated with decreased transcription. Methylated DNA sequences stimulate histone deacetylases to remove acetyl groups from the histone proteins, thus stabilizing the nucleosome and repressing transcription. Demethylation of DNA sequences is often followed by increased transcription, which may be related to the deacetylation of the histone proteins. 3\. Briefly explain how transcriptional activator and repressor proteins affect the level of transcription of eukaryotic genes. Transcriptional activator proteins stimulate transcription by binding DNA at specific base sequences such as an enhancer or regulatory promoter and attracting or stabilizing the basal transcription factor apparatus. Repressor proteins bind to silencer sequences or promoter regulator sequences. These proteins may inhibit transcription by either blocking access to the enhancer sequence by the activator protein, preventing the activator from interacting with the basal transcription apparatus, or preventing the basal transcription factor from being assembled. 4\. What is an enhancer? How does it affect the transcription of distant genes? Enhancers are DNA sequences that are the binding sites of transcriptional activator proteins. Transcription at a distant gene is affected when the DNA sequence located between the gene's promoter and the enhancer is looped out, allowing for the interaction of the enhancer-bound proteins with proteins needed at the promoter, which in turn stimulates transcription. Additionally, the transcription of short enhancer (e)RNA molecules from an enhancer template may be involved in transcriptional activation, but a precise mechanism for such activation has not been determined. 5\. What is an insulator? An insulator or boundary element is a sequence of DNA that inhibits the action of regulatory elements called enhancers in a position dependent manner. 6\. What is a response element? How do response elements bring about the coordinated expression of eukaryotic genes? Response elements are regulatory DNA sequences consisting of short specific sequences located at various distances from the genes that they regulate. Under conditions of stress, a transcription activator protein binds to the response element and stimulates transcription. If the same response element sequence is located in the control regions of different genes, then these genes will be activated by the same stimuli, thus producing a coordinated response. 7\. Outline the role of alternative splicing in the control of sex differentiation in Drosophila. See **Figure 17.11**. Sex development in fruit flies depends on alternative splicing as well as a cascade of genetic regulation. Early in the development of female fruit flies, a female-specific promoter is activated stimulating transcription of the sex-lethal (*Sxl*) gene. Splicing of the pre-mRNA of the transformer (*tra*) gene is regulated by the Sxl protein. The mature mRNA produces the Tra protein. In conjunction with another protein, the Tra protein stimulates splicing of the pre-mRNA from the doublesex (*dsx*) gene. The resulting Dsx protein is required for the embryo to develop female characteristics. Male fruit flies do not produce the Sxl protein, which results in the *tra* pre-mRNA in male fruit flies being spliced at an alternate location. The alternate Tra protein is not functional, resulting in the *dsx* pre-mRNA splicing at a different location as well. Protein synthesis from this mRNA produces a male-specific doublesex protein, which causes development of male-specific traits. 8\. What role does RNA stability play in gene regulation? What controls RNA stability in eukaryotic cells? The total amount of protein synthesized is dependent on how much mRNA is available for translation. The amount of mRNA present is dependent on the rates of mRNA synthesis and degradation. Less-stable mRNAs will be degraded faster so there will be fewer copies available to serve as templates for translation. The presence of the 5′ cap, 3\' poly(A) tail, the 5\' UTR, 3\' UTR, and the coding region in the mRNA molecule are features that can affect the stability of the mRNA molecule. Poly(A) binding proteins (PABP) bind at the 3\' poly(A) tail. These proteins contribute to the stability of the tail and protect the 5\' cap through direct interaction. Once a critical number of adenine nucleotides have been removed from the tail, the protection is lost and the 5\' cap is removed. The removal of the 5\' cap allows for 5\' to 3\' nucleases to degrade the mRNA. AU-rich sequence elements in the 3ʹ UTR can also increase degradation of the mRNA. 9\. Briefly list some of the ways in which siRNAs and miRNAs regulate genes. \(1) Through cleavage of mRNA sequences through "slicer activity": The binding of RISCs containing either siRNA or miRNA to complementary sequences in mRNA molecules stimulates cleavage of the mRNA through "slicer activity." This is followed by further degradation of the cleaved mRNA. \(2) Through binding of complementary regions with the mRNA molecule by miRNAs to prevent translation: The miRNAs as part of RISC bind to complementary mRNA sequences preventing either translation initiation or elongation, which results in premature termination. \(3) Through transcriptional silencing due to methylation of either histone proteins or DNA sequences: The siRNA binds to complementary DNA sequences within the nucleus and stimulates methylation of histone proteins. Methylated histones bind DNA more tightly preventing transcription factors from binding the DNA. The miRNA molecules bind to complementary DNA sequences and stimulate DNA methylases to directly methylate the DNA sequences, which results in transcriptional silencing. \(4) Through slicer-independent mRNA degradation stimulated by miRNA binding to complementary regions in the 3\' UTR of the mRNA: A miRNA binds to the AU rich element in the 3\' UTR of the mRNA stimulating degradation using RISC and dicer. 10\. How does bacterial gene regulation differ from eukaryotic gene regulation? How are they similar? Similarities: Bacterial and eukaryotic gene regulation involves the action of protein repressors and protein activators. Cascades of gene regulation in which the activation of one set of genes affects another set of genes take place in both eukaryotes and bacteria. Regulation of gene expression at the transcriptional level is also common in both types of cells. Both have coordinated expression, although, through different mechanisms. Bacterial genes are often clustered in operons and are coordinately expressed through the synthesis of a single polygenic mRNA. Eukaryotic genes are typically separate, with each containing its own promoter and transcribed on individual mRNAs. Coordinate expression of multiple genes is accomplished through the presence of common \[shared\] response elements. Genes sharing the same response element will be regulated by the same regulatory factors. Differences: In eukaryotic cells, gene-coding regions are interrupted by introns, which are generally longer than exons. An individual intron may be much longer than the entire coding region. Gene expression requires the proper splicing of the pre-mRNA to remove these noncoding regions. In prokaryotic cells, gene-coding regions are usually not interrupted. In eukaryotic cells, chromatin structure plays a role in gene regulation. Chromatin that is condensed inhibits transcription. Therefore, for expression to occur, the chromatin must be altered to allow for changes in structure. Acetylation of histone proteins and DNA methylation are important in these changes. At the level of transcription initiation, the process is more complex in eukaryotic cells. In eukaryotes, initiation requires a complex machine involving RNA polymerase, general transcription factors, and transcriptional activators. Bacterial RNA polymerase is either blocked or stimulated by the actions of regulatory proteins. Finally, in eukaryotes the action of activator proteins binding to enhancers may take place at a great distance from the promoter and structural gene. These distant enhancers occur much less frequently in bacterial cells. 11\. What would be the most likely effect of deleting flowering locus D (FLD) in Arabidopsis thaliana? Explain how this related to the function of FLC. It is likely that flowering will not occur if the *flowering locus D* is deleted. The protein encoded by *FLD* is a deacetylase enzyme. This deacetylase enzyme normally removes acetyl groups from histones surrounding the *flowering locus C* (*FLC*). Once the acetyl groups are removed, the chromatin structure within this region is restored to be transcriptionally repressed. The restored chromatin inhibits transcription from the *FLC* locus. *FLC* codes for a transcriptional activator whose expression activates other genes that suppress flowering. If *FLC* transcription is active due to a deletion in *FLD,* then flowering will not occur. 12\. How do repressors that bind to silencers in eukaryotes differ from repressors that bind to operators in bacteria? In bacteria, repressors that bind to the operator block RNA polymerase from binding to the promoter, and thus, directly block transcription. On the other hand, repressors that bind to silencers in eukaryotes block transcriptional activator proteins from binding at an activator site, thus eliminating transcriptional activation. **[Chapter 18 Gene Mutations and DNA Repair]** 1. What is the difference between a transition and a transversion? Which type of base substitution is usually more common? Transition (more common) purine -\> purine or pyr -\> pyr; transversion (less common) pyr-\>purine or purine -\> pyr 2. Briefly describe expanding nucleotide repeats. How do they account for the phenomenon of anticipation? 3 base insertion/trinucleotide repeats and gets worse with generation; more copies very bad; add more copies of amino acid to make faulty proteins 3. What is the difference between a missense mutation and a nonsense mutation? Between a silent mutation and a neutral mutation? Missense -\> diff codon/amino acid Nonsense -\> changes to stop codon Silent -\> changes but same result/amino acid Neutral -\> codes for new amino acid but function doesn't change 4. How do insertions and deletions arise? How do base analogs lead to mutations? Strand slippage -- formation of small loops: new -\> insertion on new; template -\> deletion on new Unequal crossing over -- deletion in one, insertion on other Base analogues -\> mismatched at higher frequencies 5. What is the purpose of the Ames test? How are *his*-- bacteria used in this test? To test mutagenesis or carcinogenicity of compounds; his- strain is plated so only the mutants that revert the change to be his+ can grow (no his on plate) 6. List the four different types of DNA repair and briefly explain how each is carried out. Base excision -- remove base, create AP site, remove sugar and phosphate, DNA poly/ligase Direct -- revert changed nucleotides back to normal structure; photolyase 7. Hemoglobin is a complex protein that contains four polypeptide chains. The normal hemoglobin found in adults---called adult hemoglobin---consists of two alpha and two beta polypeptide chains, which are encoded by different loci. Sickle-cell hemoglobin, which causes sickle-cell anemia, arises from a mutation in the beta chain of adult hemoglobin. Adult hemoglobin and sickle-cell hemoglobin differ in a single amino acid: the sixth amino acid from one end in adult hemoglobin is glutamic acid, whereas sickle-cell hemoglobin has valine at this position. After consulting the genetic code provided in **Figure 15.10**, indicate the type and location of the mutation that gave rise to sickle-cell anemia. There are two possible codons for glutamic acid, GAA and GAG. Single-base substitutions at the second position in both codons can produce codons that encode valine: GAA\-\-\-\-\-\-\--\> GUA (Val) GAG\-\-\-\-\-\-\--\> GUG (Val) Both substitutions are transversions. However, in the gene encoding the β chain of hemoglobin, the GAG codon is the wild-type codon and the mutated GUG codon results in the sickle-cell phenotype. 8. In many eukaryotic organisms, a significant proportion of cytosine bases are naturally methylated to 5-methylcytosine. Through evolutionary time, the proportion of AT base pairs in the DNA of these organisms increases. Can you suggest a possible mechanism for this increase? Spontaneous deamination of 5-methylcytosine produces thymine. If the subsequent repair of the GT mispairing is repaired incorrectly or, more likely, not repaired at all because the thymine is a normal base, then a GC to AT transition will result. Over time, the incorrect repairs will lead to an increase in the number of AT base pairs. 9. What conclusion would you make if the number of bacterial colonies in **Figure 18.22** were the same on the control plate and the treatment plate? Explain your reasoning. A screenshot of a cell phone Description automatically generated The chemical tested is not mutagenic and likely not carcinogenic. The number of colonies on each plate represents the number of bacterial cells that underwent a mutation. Because the number of cells undergoing mutation on the control plate, without the tested chemical, was the same as the number undergoing mutation on the plate treated with the chemical, there is no evidence that the chemical elevates the mutation rate or is potentially carcinogenic. 10. A genetics instructor designs a laboratory experiment to study the effects of UV radiation on mutation in bacteria. In the experiment, the students expose bacteria plated on petri plates to UV light for different lengths of time, place the plates in an incubator for 48 hours, and then count the number of colonies that appear on each plate. The plates that have received more UV radiation should have more pyrimidine dimers, which block replication; thus, fewer colonies should appear on the plates exposed to UV light for longer periods of time. Before the students carry out the experiment, the instructor warns them that while the bacteria are in the incubator, the students must not open the incubator door unless the room is darkened. Why should the bacteria not be exposed to light? Exposure of DNA to UV light results in the formation of pyrimidine dimers in the DNA molecule. Often the repair of these dimers leads to mutations. Because the SOS repair system is error-prone and leads to an increased accumulation of mutations, UV light produces more mutations in bacteria when the SOS repair system is activated to repair the damage caused by the UV light. However, many species of bacteria have a direct DNA repair system that can repair pyrimidine dimers by breaking the covalent linkages between the pyrimidines that form the dimer. The enzyme that repairs the DNA is called photolyase and is activated and energized by light. The photolyase is a very efficient repair enzyme and typically makes accurate repairs of the damage. If the bacteria in the UV radiation experiment are exposed to light, then the photolyase will be activated to repair the damage, resulting in fewer mutations in the irradiated bacteria. 11. What is a somatic cell? What is a germ line cell? Contrast the consequences of mutations in germ line and somatic cells. Somatic arise in somatic tissues which do not produce gametes. When a somatic cell with a mutation divides by mitosis, the mutation is passed on to the daughter cells leading to a population of genetically identical cells. Germ line mutations- arise in cells that ultimately produce gametes. A germ line mutation can be passed to future generations producing offspring that carry the mutation in there somatic and germ line cells. 12. What is anticipation? ***Anticipation*** is the increase in the severity of a disease over subsequent generations. In diseases influenced by expansion of polynucleotide repeats, the numbers of repeats increase in subsequent generations, leading to increases in the severity of the diseases in those progenies. *For the rest of the review, please revisit the slides/research on your own in order to answer the comparison and True/False questions!*