Genetics Ch. 12

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?

A consensus sequence is the most common nucleotide sequence that is found within a group of related sequences. An example is the -35 and -10 consensus sequences found in bacterial promoters. At -35, it is TTGACA, but it can differ by one or two nucleotides and still function efficiently as a promoter. In the consensus sequences within bacterial promoters, the -35 site is primarily for recognition by sigma factor. The -10 site, also known as the Pribnow box, is the site where the double-stranded DNA will begin to unwind to allow transcription to occur.

What does it mean to say that gene expression is colinear?

A gene is colinear when the sequence of bases in the coding strand of the DNA (i.e., the DNA strand that is complementary to the template strand for RNA synthesis) corresponds to the sequence of bases in the mRNA. Most prokaryotic genes and many eukaryotic genes are colinear. Therefore, you can look at the gene sequence in the DNA and predict the amino acid sequence in the polypeptide. Many eukaryotic genes, however, are not colinear. They contain introns that are spliced out of the pre-mRNA.

What is the unique feature of ribozyme function? Give two examples described in this chapter.

A ribozyme has a catalytic part that is composed of RNA. Examples are RNase P and self-splicing group I and II introns. It is thought that the spliceosome may contain catalytic RNAs as well.

Mutations that occur at the end of a gene may alter the sequence of the gene and prevent transcriptional termination. A. What types of mutations would prevent ρ-independent termination? B. What types of mutations would prevent ρ-dependent termination? C. If a mutation prevented transcriptional termination at the end of a gene, where would gene transcription end? Or would it end?

A. Mutations that alter the uracil-rich region by introducing guanines and cytosines and mutations that prevent the formation of the stem-loop structure. B. Mutations that alter the termination sequence, and mutations that alter the ρ-protein recognition site. C. Eventually, somewhere downstream from the gene, another transcriptional termination sequence would be found, and transcription would terminate there. This second termination sequence might be found randomly or it might be at the end of an adjacent gene.

For each of the following transcription factors, explain how eukaryotic transcriptional initiation would be affected if it were missing. A. TFIIB B. TFIID C. TFIIH

A. RNA polymerase would not be bound to the core promoter. B. TFIID contains the TATA-binding protein. If it were missing, RNA polymerase would not bind to the TATA box. C. The formation of the open complex would not take place.

If the following RNA polymerases were missing from a eukaryotic cell, what types of genes would not be transcribed? A. RNA polymerase I B. RNA polymerase II C. RNA polymerase III

A. Ribosomal RNA (5.8S, 18S, and 28S) B. All of the mRNA genes, and most of the snRNA genes, long non-coding RNAs, microRNAs, and snoRNAs C. All of the tRNA genes and the 5S rRNA genes, and some of the snRNA genes, long non-coding RNAs, microRNAs, and snoRNAs

Explain the central dogma of genetics at the molecular level.

At the molecular level, a gene is defined as a segment of DNA that is used to make a functional product, either an RNA molecule or a polypeptide. The first step in this process is called transcription, which refers to the process of synthesizing RNA from a DNA template. When a protein-encoding gene is transcribed, the first product is an RNA molecule known as messenger RNA (mRNA). During polypeptide synthesis—a process called translation—the sequence of codons within the mRNA determines the sequence of amino acids in a polypeptide. One or more polypeptides then assemble into a functional protein. DNA replication provides a mechanism for copying that information so it can be transmitted to new daughter cells and from parent to offspring.

What sequence elements are found within the core promoter of protein-encoding genes in eukaryotes? Describe their locations and specific functions.

Core promoters in eukaryotes are somewhat variable with regard to the pattern of sequence elements. In the case of protein-encoding genes that are transcribed by RNA polymerase II, it is common to have a TATA box, which is about 25 bp upstream from a transcriptional start site. The TATA box is important in the identification of the transcriptional start site and the assembly of RNA polymerase and various transcription factors. The transcriptional start site defines where transcription actually begins.

In Chapter 11, we discussed the function of DNA helicase, which is involved in DNA replication. Discuss how the functions of ρ-protein and DNA helicase are similar and how they are different.

DNA helicase and ρ-protein bind to a nucleic acid strand and travel in the 5' to 3' direction. When they encounter a double-stranded region, they break the hydrogen bonds between complementary strands. The ρ-protein differs from DNA helicase in that it moves along an RNA strand, while DNA helicase moves along a DNA strand. The purpose of DNA helicase function is to promote DNA replication, while the purpose of ρ-protein function is to promote transcriptional termination.

What is the consensus sequence of the following six DNA sequences? GGCATTGACT GCCATTGTCA CGCATAGTCA GGAAATGGGA GGCTTTGTCA GGCATAGTCA

GGCATTGTCA

The initiation phase of eukaryotic transcription via RNA polymerase II is considered an assembly and disassembly process. Which types of biochemical interactions—hydrogen bonding, ionic bonding, covalent bonding, and/or hydrophobic interactions—would you expect to drive the assembly and disassembly process? How would temperature and salt concentration affect assembly and disassembly?

Hydrogen bonding is usually the predominant type of interaction when proteins and DNA follow an assembly and disassembly process. In addition, ionic bonding and hydrophobic interactions could occur. Covalent interactions would not occur. High temperature and high salt concentrations tend to break hydrogen bonds. Therefore, high temperature and high salt would inhibit assembly and stimulate disassembly.

In Chapter 9, we considered the dimensions of the double helix (see Figure 9.12). In an α helix of a protein, there are 3.6 amino acids per complete turn. Each amino acid advances the α helix by 0.15 nm; a complete turn of an α helix is 0.54 nm in length. As shown in Figure 12.6, two α 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 α helices occupy the major groove of DNA?

In Figure 12.6, eachαhelix appears to occupy a region that is approximately one-half of a complete turn of the double helix. Because a complete turn of a DNA double helix involves 10 bp, each α helix appears to be bound to about 5 bp in the major groove of the DNA; 5 bp would span a linear distance of approximately 1.7 nm (as described in Chapter 9, Figure 9.12). If we divide 1.7 nm by 0.15 nm/amino acid, we obtain a value of 11.3 amino acids per helix. This would create about 3.15 turns of an α helix; there are three complete turns per helix, for a total of six complete turns for both helices.

What is alternative splicing? What is its biological significance?

In alternative splicing, variation occurs in the pattern of splicing so the resulting mRNAs contain alternative combinations of exons. The biological significance of this variation is that two or more different proteins can be produced from a single gene. This is a more efficient use of the genetic material. In multicellular organisms, alternative splicing is often used in a cell-specific manner.

Describe the processing events that occur during the production of tRNA in E. coli.

In bacteria, the 5 end of the tRNA is cleaved by RNase P. The 3 end is cleaved by a different endonuclease, and then a few nucleotides are digested away by an exonuclease that removes nucleotides until it reaches a CCA sequence.

In eukaryotes, what types of modifications occur to pre-mRNAs?

In eukaryotes, pre-mRNA can be capped, tailed, spliced, edited, and then exported out of the nucleus.

Discuss the differences between ρ-dependent and ρ-independent termination.

In ρ-dependent termination, the ρ protein binds to the rut site in the RNA transcript after the ρ site has been transcribed. Eventually, an RNA sequence forms a stem-loop structure that will cause RNA polymerase to pause in the transcription process. As it is pausing, the ρ protein, which functions as a helicase, will catch up to RNA polymerase and break the hydrogen bonds between the DNA and RNA within the open complex. When this occurs, the completed RNA strand is separated from the DNA along with RNA polymerase. In ρ-independent transcription, there is no ρ protein. The RNA forms a stem-loop that causes RNA polymerase to pause. However, when it pauses, a uracil-rich region is bound to the DNA template strand in the open complex. Because this is holding on by fewer hydrogen bonds, it is rather unstable. Therefore, it tends to dissociate from the open complex and thereby end transcription.

Mutations in bacterial promoters may increase or decrease the rate of gene transcription. Promoter mutations that increase the transcription rate are termed up-promoter mutations, and those that decrease the transcription rate are termed down-promoter mutations. As shown in Figure 12.5, the sequence of the −10 site 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

Mutations that make a sequence more like the consensus sequence are likely to be up promoter mutations, while mutations that cause the promoter to deviate from the consensus sequence are likely to be down promoter mutations. Also, in the -10 region, AT pairs are favored over GC pairs, because the role of this region is to form the open complex. AT pairs are more easily separated, because they form only two hydrogen bonds compared to GC pairs, which form three hydrogen bonds. A. Up promoter B. Down promoter C. Up promoter

A eukaryotic protein-encoding gene contains two introns and three exons: exon 1-intron 1-exon 2-intron 2-exon 3. The 5′ splice site at the boundary between exon 2 and intron 2 has been eliminated by a small deletion in the gene. Describe how the pre-mRNA encoded by this mutant gene will be spliced. Indicate which introns and exons will be found in the mRNA after splicing occurs.

Only the first intron would be spliced out. The mature RNA would be: exon 1-exon 2-intron 2-exon 3.

Discuss the similarities and differences between RNA polymerase (described in this chapter) and DNA polymerase (described in Chapter 11).

RNA and DNA polymerase are similar in the following ways: 1. They both use a template strand. 2. They both synthesize in the 5 to 3 direction. 3. The chemistry of synthesis is very similar in that they use incoming nucleoside triphosphates and make a phosphoester bond between the previous nucleotide and the incoming nucleotide. 4. They are both processive enzymes that slide along a template strand of DNA. RNA and DNA polymerase are different in the following ways; 1. RNA polymerase does not need a primer. 2. RNA polymerase does not have a proofreading function.

What is the subunit composition of bacterial RNA polymerase holoenzyme? What are the functional roles of the different subunits?

RNA polymerase holoenzyme consists of factor plus the core enzyme, which is composed of five subunits, α2ββ'ω. The role of factor is to recognize the promoter sequence. The α subunits are necessary for the assembly of the core enzyme and for loose DNA binding. The β and β' subunits are the portion that catalyzes the covalent linkages between adjacent ribonucleotides. The ω subunit is important for the proper assembly of the core enzyme.

Describe the movement of the open complex along the DNA.

RNA polymerase slides down the DNA and forms an open complex as it goes. The open complex is a DNA bubble of about 17 bp. Within the open complex, the DNA strand running in the 3' to 5' direction is used as a template for RNA synthesis. This occurs as individual nucleotides hydrogen bond to the DNA template strand according to the rules described in question C12. As RNA polymerase slides along, the DNA behind the open complex rewinds back into a double helix.

What is meant by the term self-splicing? What types of introns are self-splicing?

Self-splicing means that an RNA molecule can splice itself without the aid of a protein. Group I and II introns can be self-splicing, although proteins can also enhance the rate of splicing.

Which eukaryotic transcription factor(s) shown in Figure 12.14 play(s) a role that is equivalent to that of σ factor in bacterial cells?

TFIID and TFIIB play equivalent roles to factor, which does two things: it recognizes the promoter (as does TFIID), and it recruits RNA polymerase to the promoter (as does TFIIB).

What is the complementarity rule that governs the synthesis of an RNA molecule during transcription? An RNA transcript has the following sequence: 5′-GGCAUGCAUUACGGCAUCACACUAGGGAUC-3′ What is the sequence of the template and coding strands of the DNA that encodes this RNA? On which side (5′ or 3′) of the template strand is the promoter located?

The complementarity rule for RNA synthesis is DNA-G/RNA-C DNA-C/RNA-G DNA-A/RNA-U DNA-T/RNA-A The template strand is 3'-CCGTACGTAATGCCGTAGTGTGATCCCTAG-5' and the coding strand is 5'-GGCATGCATTACGGCATCACACTAGGGATC-3'. The promoter is located to the left (in the 3' direction) of the template strand.

What is the function of a splicing factor? Explain how splicing factors can regulate the cell-specific splicing of mRNAs.

The function of splicing factors is to influence the selection of splice sites in RNA. In certain cell types, the concentration of particular splicing factors is higher than in others. The high concentration of particular splicing factors, and the regulation of their activities, may promote the selection of particular splice sites and thereby lead to tissue-specific splicing.

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 σ factor? Explain why.

The most highly conserved positions are the first, second, and sixth. In general, when promoter sequences are conserved, they are more likely to be important for binding. That explains why changes are not found at these positions; if a mutation altered a conserved position, the promoter would probably not work very well. By comparison, changes are tolerated occasionally at the fourth position and frequently at the third and fifth positions. The positions that tolerate changes are less important for binding by sigma factor.

Let's suppose a DNA mutation changes the consensus sequence at the −35 site in a way that inhibits σ factor binding. Explain how a mutation could inhibit the binding of σ factor to the DNA. Look at Figure 12.5 and describe two specific base substitutions you think would inhibit the binding of σ factor. Explain why your base substitutions would have this effect.

The mutation would alter the bases so that hydrogen bonding with the factor would not occur or would not occur in the correct way. From Figure 12.5, if we look at the -35 region, we may expect that changing the first two nucleotides to anything besides T might inhibit transcription, changing the third base to an A (perhaps), changing the fourth base to a G or T, or changing the last base to anything besides an A. We would think these substitutions may have an effect, because they would deviate from known functional bacterial promoters.

In bacteria, what event marks the end of the initiation stage of transcription?

The release of factor marks the transition to the elongation stage of transcription.

Describe the structure and function of a spliceosome. Speculate why the spliceosome subunits contain snRNA. In other words, what do you think is/are the functional role(s) of snRNA during splicing?

The spliceosome is composed of several subunits known as snRNPs (containing snRNA and protein). The function of the spliceosome is to cut the RNA in two places, hold the ends together, and then catalyze covalent bond formation between the two ends. During this process, small snRNA could be involved in binding the pre-mRNA and/or it could be catalytic, like the RNA in RNase P.

Describe the allosteric and torpedo models for transcriptional termination of RNA polymerase II. Which model is more similar to ρ-dependent termination in bacteria and which model is more similar to ρ-independent termination?

The two models are presented in Figure 12.15. The allosteric model is more like p-independent termination, whereas the torpedo model is more like -dependent termination. In the torpedo model, a protein knocks RNA polymerase off the DNA, much like the effect of p-protein.

A mutation within a gene sequence changes the start codon to a stop codon. How will this mutation affect the transcription of this gene?

This will not affect transcription. However, it will affect translation by preventing the initiation of polypeptide synthesis.

Describe what happens to the chemical bonding interactions when transcriptional termination occurs. Be specific about the type of chemical bonding.

When transcriptional termination occurs, the hydrogen bonds between the DNA and the part of the newly made RNA transcript that is located in the open complex are broken.

At the molecular level, describe how σ factor recognizes a bacterial promoter. Be specific about the structure of σ factor and the type of chemical bonding.

sigma factor can slide along the major groove of the DNA. The protein contains -helices that fit into the major groove of the DNA. In this way, it is able to recognize base sequences that are exposed in the groove. When factor encounters a promoter sequence, hydrogen bonding between it and the bases can promote a tight and specific interaction.