Genetics - Analysis and Principles

Biology

Quiz 12 :

Gene Transcription and Rna Modification

Quiz 12 :

Gene Transcription and Rna Modification

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What is the subunit composition of bacterial RNA polymerase holoenzyme What are the functional roles of the different subunits
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The subunit composition of the holoenzyme of the enzyme RNA polymerase in bacteria is:
• 2 subunits
• 1 subunit

img subunit
• 1 subunit
• subunit
The functional roles of the various subunits given above are:
The two subunits have a role of binding to DNA and also are instrumental for the proper assembling of the holoenzyme.
The two subunits are also needed along with the two subunits for binding to DNA. These subunits have an additional function of the catalytic synthesis of RNA.
The subunit is important for the assembly of the core enzyme.
The subunit is the most important subunit of all. It is instrumental in the recognition of the - 10 and the - 35 promoter sequences in DNA.
The holoenzyme is required for the initiation of transcription.

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At the molecular level, describe how factor recognizes a bacterial promoter. Be specific about the structure of factor and the type of chemical bonding.
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The factor is instrumental in the recognition of the - 10 and the - 35 promoter sequences in DNA.
The recognition and binding of the factor to DNA on the molecular level can be explained in this way:
After attachment of the holoenzyme of RNA polymerase, the whole enzyme slides along the DNA strand. The factor can slide along the major groove of the DNA where the - 10 and the - 35 promoter sequences are present.
While sliding, the factor is able to recognize the bases in the major groove of DNA. In this way, it recognizes the promoter sequences in the major groove. When the factor meets a promoter sequence, hydrogen bonding occurs.
Hydrogen bonding occurs between the factor and the nucleotide bases of the promoter sequences. This interaction is specific and thus the binding is very tight.

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What is the consensus sequence of the following six DNA sequences GGCATTGACT GCCATTGTCA CGCATAGTCA GGAAATGGGA GGCTTTGTCA GGCATAGTCA
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The different DNA sequences are given below:
• GGCATTGACT
• GCCATTGTCA
• CGCATAGTCA
• GGAAATGGGA
• GGCTTTGTCA
• GGCATAGTCA
The term consensus sequence indicates that the sequences have some conserved bases and some variable bases. The sequences are seen in almost all organisms without much variation.
The consensus sequence for the above DNA sequence is GGCATTGTCA. This consensus sequence can be determined by finding out the most common nucleotides for each position of the sequence.
There are slight variations in each organism but the sequence usually remains the same.

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A gel retardation assay can be used to study the binding of proteins to a segment of DNA. This method is described in Chapter 20. When a protein binds to a segment of DNA, it retards the movement of the DNA through a gel, so the DNA appears at a higher point in the gel (see the following). img Lane 1: 900-bp fragment alone Lane 2: 900-bp fragment plus a protein that binds to the 900-bp fragment In this example, the segment of DNA is 900 bp in length, and the binding of a protein causes the DNA to appear at a higher point in the gel. If this 900-bp fragment of DNA contains a eukaryotic promoter for a protein-encoding gene, draw a gel that shows the relative locations of the 900-bp fragment under the following conditions: Lane 1: 900 bp plus TFIID Lane 2: 900 bp plus TFIIB Lane 3: 900 bp plus TFIID and TFIIB Lane 4: 900 bp plus TFIIB and RNA polymerase II Lane 5: 900 bp plus TFIID, TFIIB, and RNA polymerase II/ TFIIF
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In bacteria, what event marks the end of the initiation stage of transcription
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According to the examples shown in Figure 12.5, which positions of the 35 sequence (i.e., first, second, third, fourth, fifth, or sixth) are more tolerant of changes Do you think these positions play a more or less important role in the binding of factor Explain why. FIGURE 12.5 Examples of 35 and 10 sequences within several E. coli promoters. This figure shows the 35 and 10 sequences for one DNA strand found in seven different E. coli promoters. The consensus sequence is shown at the bottom. The spacer regions contain the designated number of nucleotides between the 35 and 10 region or between the 10 region and the transcriptional start site. For example, N 17 means there are 17 nucleotides between the end of the 35 region and the beginning of the 10 region. img
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What is an R loop In an R loop experiment, to which strand of DNA does the mRNA bind, the coding strand or the template strand
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What is the meaning of the term consensus sequence Give an example. Describe the locations of consensus sequences within bacterial promoters. What are their functions
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If a gene contains three introns, draw what it would look like in an R loop experiment.
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Based on your knowledge of introns and pre-mRNA splicing, discuss whether or not you think alternative splicing fully explains the existence of introns. Can you think of other possible reasons to explain the existence of introns
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Discuss the types of RNA transcripts and the functional roles they play. Why do you think some RNAs form complexes with protein subunits
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In Chapter 9, we considered the dimensions of the double helix (see Figure 9.15). In an helix of a protein, there are 3.6 amino acids per complete turn. Each amino acid advances the helix by 0.15 nm; a complete turn of an helix is 0.54 nm in length. As shown in Figure 12.6, two helices of a transcription factor occupy the major groove of the DNA. According to Figure 12.6, estimate the number of amino acids that bind to this region. How many complete turns of the helices occupy the major groove of DNA FIGURE 12.6 The binding of -factor protein to the promoter region. The -factor protein contains two helices connected by a turn, termed a helix-turn-helix motif. Two helices of the protein fit within the major groove of the DNA. Amino acids within the helices form hydrogen bonds with the bases in the DNA. The DNA strands are shown in red and green. img FIGURE 9.15 Key features of the structure of the double helix. Note: In the drawing on the left and in the inset, the planes of the bases and sugars are shown parallel to each other in order to depict the hydrogen bonding between the bases. In an actual DNA molecule, the bases would be rotated about 90° so the planes of the bases would be facing each other, as shown in Figure 9.16a. img FIGURE 9.16 Two models of the double helix. (a) Ball-and-stick model of the double helix. The deoxyribose-phosphate backbone is shown in detail, whereas the bases are depicted as flattened rectangles. img
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A mutation within a gene sequence changes the start codon to a stop codon. How will this mutation affect the transcription of this gene
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Genes may be structural genes that encode polypeptides, or they may be nonstructural genes. A. Describe three examples of genes that are not structural genes. B. For structural genes, one DNA strand is called the template strand, and the complementary strand is called the coding strand. Are these two terms appropriate for nonstructural genes Explain.
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A research group has sequenced the cDNA and genomic DNA from a particular gene. The cDNA is derived from mRNA, so it does not contain introns. Here are the DNA sequences. cDNA: 5 -ATTGCATCCAGCGTATACTATCTCGGGCCCAATTAATGCCA-GCGGCCAGACTATCACCCAACTCGGTTACCTACTAGTATATC-CCATATACTAGCATATATTTTACCCATAATTTGTGTGTGGGTATA-CAGTATAATCATATA-3 Genomic DNA (contains one intron): 5 -ATTGCATCCAGCGTATACTATCTCGGGCCCAATTAATGCCAGCGGCCAGACTATCACCCAACTCGGCCCACCCCCCAGGTTTA-CACAGTCATACCATACATACAAAAATCGCAGTTACTTATCCCA-AAAAAACCTAGATACCCCACATACTATTAACTCTTTCTTTCTAG-GTTACCTACTAGTATATCCCATATACTAGCATATATTTTAC-CCATAATTTGTGTGTGGGTATACAGTATAATCATATA-3 Indicate where the intron is located. Does the intron contain the normal consensus splice site sequences based on those described in Figure 12.21 Underline the splice site sequences, and indicate whether or not they fit the consensus sequence. FIGURE 12.21 Consensus sequences for pre-mRNA splicing in complex eukaryotes. Consensus sequences exist at the intronexon boundaries and at a branch site found within the intron itself. The adenine nucleotide shown in blue in this figure corresponds to the adenine nucleotide at the branch site in Figure 12.22. The nucleotides shown in bold are highly conserved. Designations: A/C = A or C, Pu = purine, Py = pyrimidine, N = any of the four bases. img FIGURE 12.22 Splicing of pre-mRNA via a spliceosome. img
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Chapter 18 describes a technique known as Northern blotting that can be used to detect RNA transcribed from a particular gene. In this method, a specific RNA is detected using a short segment of cloned DNA as a probe. The DNA probe, which is radioactive, is complementary to the RNA that the researcher wishes to detect. After the radioactive probe DNA binds to the RNA, the RNA is visualized as a dark (radioactive) band on an X-ray film. As shown here, the method of Northern blotting can be used to determine the amount of a particular RNA transcribed in a given cell type. If one type of cell produces twice as much of a particular mRNA as occurs in another cell, the band will appear twice as intense. Also, the method can distinguish if alternative RNA splicing has occurred to produce an RNA that has a different molecular mass. img Lane 1 is a sample of RNA isolated from nerve cells. Lane 2 is a sample of RNA isolated from kidney cells. Nerve cells produce twice as much of this RNA as do kidney cells. Lane 3 is a sample of RNA isolated from spleen cells. Spleen cells produce an alternatively spliced version of this RNA that is about 200 nucleotides longer than the RNA produced in nerve and kidney cells. Let's suppose a researcher was interested in the effects of mutations on the expression of a particular structural gene in eukaryotes. The gene has one intron that is 450 nucleotides long. After this intron is removed from the pre-mRNA, the mRNA transcript is 1100 nucleotides in length. Diploid somatic cells have two copies of this gene. Make a drawing that shows the expected results of a Northern blot using mRNA from the cytosol of somatic cells, which were obtained from the following individuals: Lane 1: A normal individual Lane 2: An individual homozygous for a deletion that removes the -50 to -100 region of the gene that encodes this mRNA Lane 3: An individual heterozygous in which one gene is normal and the other gene had a deletion that removes the -50 to -100 region Lane 4: An individual homozygous for a mutation that introduces an early stop codon into the middle of the coding sequence of the gene Lane 5: An individual homozygous for a three-nucleotide deletion that removes the AG sequence at the 3 splice site
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The technique of DNA footprinting is described in Chapter 20. If a protein binds over a region of DNA, it will protect the DNA in that region from digestion by DNase I. To carry out a DNA foot-printing experiment, a researcher has a sample of a cloned DNA fragment. The fragments are exposed to DNase I in the presence and absence of a DNA-binding protein. Regions of the DNA fragment not covered by the DNA-binding protein will be digested by DNase I, producing a series of bands on a gel. Regions of the DNA fragment not digested by DNase I (because a DNA-binding protein is preventing DNase I from gaining access to the DNA) will be revealed, because a region of the gel will not contain any bands. In the DNA footprinting experiment shown here, a researcher began with a sample of cloned DNA 300 bp in length. This DNA contained a eukaryotic promoter for RNA polymerase II. For the sample loaded in lane 1, no proteins were added. For the sample loaded in lane 2, the 300-bp fragment was mixed with RNA polymerase II plus TFIID and TFIIB. img A. How long of a region of DNA is "covered up" by the binding of RNA polymerase II and the transcription factors B. Describe how this binding would occur if the DNA was within a nucleosome structure. (Note: The structure of nucleosomes is described in Chapter 10.) Do you think that the DNA is in a nucleosome structure when RNA polymerase and transcription factors are bound to the promoter Explain why or why not.
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Mutations in bacterial promoters may increase or decrease the rate of gene transcription. Promoter mutations that increase transcription are termed up-promoter mutations, and those that decrease transcription are termed down-promoter mutations. As shown in Figure 12.5, the sequence of the 10 region of the promoter for the lac operon is TATGTT. Would you expect the following mutations to be up-promoter or down-promoter mutations A. TATGTT to TATATT B. TATGTT to TTTGTT C. TATGTT to TATGAT FIGURE 12.5 Examples of 35 and 10 sequences within several E. coli promoters. This figure shows the 35 and 10 sequences for one DNA strand found in seven different E. coli promoters. The consensus sequence is shown at the bottom. The spacer regions contain the designated number of nucleotides between the 35 and 10 region or between the 10 region and the transcriptional start site. For example, N 17 means there are 17 nucleotides between the end of the 35 region and the beginning of the 10 region. img
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As described in Table 12.1, several different types of RNA are made, especially in eukaryotic cells. Researchers are sometimes interested in focusing their attention on the transcription of structural genes in eukaryotes. Such researchers want to study mRNA. One method that is used to isolate mRNA is column chromatography. (Note: See the Appendix for a general description of chromatography.) Researchers can covalently attach short pieces of DNA that contain stretches of thymine (i.e., TTTTTTTTTTTT) to the column matrix. This is called a poly-dT column. When a cell extract is poured over the column, mRNA binds to the column, but other types of RNA do not. A. Explain how you would use a poly-dT column to obtain a purified preparation of mRNA from eukaryotic cells. In your description, explain why mRNA binds to this column and what you would do to release the mRNA from the column. B. Can you think of ways to purify other types of RNA, such as tRNA or rRNA
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As described in Chapter 20 and in experimental question E5, a gel retardation assay can be used to determine if a protein binds to DNA. This method can also determine if a protein binds to RNA. In the combinations described here, would you expect the migration of the RNA to be retarded due to the binding of a protein A. mRNA from a gene that is terminated in a -independent manner plus protein B. mRNA from a gene that is terminated in a -dependent manner plus protein C. Pre-mRNA from a protein-encoding gene that contains two introns plus the snRNP called U1 D. Mature mRNA from a protein-encoding gene that contains two introns plus the snRNP called U1 Question E5 A gel retardation assay can be used to study the binding of proteins to a segment of DNA. This method is described in Chapter 20. When a protein binds to a segment of DNA, it retards the movement of the DNA through a gel, so the DNA appears at a higher point in the gel (see the following). img Lane 1: 900-bp fragment alone Lane 2: 900-bp fragment plus a protein that binds to the 900-bp fragment In this example, the segment of DNA is 900 bp in length, and the binding of a protein causes the DNA to appear at a higher point in the gel. If this 900-bp fragment of DNA contains a eukaryotic promoter for a protein-encoding gene, draw a gel that shows the relative locations of the 900-bp fragment under the following conditions: Lane 1: 900 bp plus TFIID Lane 2: 900 bp plus TFIIB Lane 3: 900 bp plus TFIID and TFIIB Lane 4: 900 bp plus TFIIB and RNA polymerase II Lane 5: 900 bp plus TFIID, TFIIB, and RNA polymerase II/TFIIF
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