Answers to Conceptual and Experimental Questions (Ch.12-13)

CHAPTER 12 Answers to Conceptual and Experimental Questions Conceptual Questions C1. A. tRNA genes encode tRNA molecules, and rRNA genes encode the rRNAs found in ribosomes. There are also genes for the RNAs found in snRNPs, etc. B. The term template strand is still appropriate because one of the DNA strands is used as a template to make the RNA. The term coding strand is not appropriate because the RNA made from nonstructural genes does not code for a polypeptide sequence. C. Yes. C2. The formation of the open complex, and the release of sigma factor. C3. 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 sequence 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 DNA will begin to unwind to allow transcription to occur. C8. This will not affect transcription. However, it will affect translation by preventing the initiation of polypeptide synthesis. C9. RNA polymerase holoenzyme consists of sigma factor plus the core enzyme, which is a tetramer, a2ßß. The role of sigma factor is to recognize the promoter sequence. The a subunits are necessary for the assembly of the core enzyme and for loose DNA binding. The ß and ß subunits are the portion that catalyze the covalent linkages between adjacent ribonucleotides. C10. Sigma factor can slide along the major groove of the DNA. In this way, it is able to recognize base sequences that are exposed in the groove. When it encounters a promoter sequence, hydrogen bonding between the bases and the sigma factor protein can promote a tight and specific interaction. C11. The mutation would alter the bases so that hydrogen bonding with the sigma factor would not occur, or would not occur in the correct way. From figure 12.4, 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. C12. 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 would be to the left (in the 3 direction) of the template strand. C15. In ρ-dependent termination, the p protein binds to the RNA transcript after the p site has been transcribed. Eventually, RNA polymerase will transcribe a termination stem-loop that will cause it to pause in the transcription process. As it is pausing, the p protein, which functions as a helicase, will catch up to RNA polymerase and knock it off the DNA. In p-independent transcription, there is no p protein. Again, the RNA polymerase transcribes a termination hairpin that causes it to pause. However, when it pauses, an AU-rich region is left base pairing in the open complex. Since 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. C19. A. Ribosomal RNA (5.8S, 18S, and 28S) B. All of the mRNA and certain snRNA genes C. All of the tRNAs and the 5S rRNA C20. Eukaryotic promoters are somewhat variable with regard to the pattern of sequence elements that may be found. In the case of structural 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. C22. It is primarily an accessibility problem. When the DNA is tightly wound around histones, it becomes difficult for large proteins, like RNA polymerase and transcription factors, to recognize the correct base sequence in the DNA and to catalyze the movement of the open complex. It is thought that a partial or complete disruption of the nucleosome structure is necessary for transcription to occur. C29. A gene is colinear when the sequence of bases in the sense 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. C31. In eukaryotes, pre-mRNA can be capped, tailed, and spliced and then exported out of the nucleus. C33. Alternative splicing occurs when exons are spliced out or alternative splice sites are used at intron-exon boundaries. The biological significance is that two or more polypeptide sequences can be derived 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. CHAPTER 13 Answers to Conceptual and Experimental Questions Conceptual Questions C1. The start codon begins at the fifth nucleotide. The amino acid sequence would be Met Gly Asn Lys Pro Gly Gln STOP. C2. When we say the genetic code is degenerate, it means that more than one codon can specify the same amino acid. For example, GGG, GGC, GGA, and GGU all specify glycine. In general, the genetic code is nearly universal, because it is used in the same way by viruses, prokaryotes, fungi, plants, and animals. As discussed in table 13.3, there are a few exceptions, which occur primarily in protozoa and organellar genetic codes. C3. A. True B. False C. False C4. A. This mutant tRNA would recognize glycine codons in the mRNA but would put in tryptophan amino acids where glycine amino acids are supposed to be in the polypeptide chain. B. This mutation tells us that the aminoacyl-tRNA synthetase is primarily recognizing other regions of the tRNA molecule besides the anticodon region. In other words, tryptophanyl-tRNA synthetase (i.e., the aminoacyltRNA synthetase that attaches tryptophan) primarily recognizes other regions of the tRNAtrp sequence (i.e., other than the anticodon region), such as the T- and D-loops. If aminoacyl-tRNA synthetases recognized only the anticodon region, we would expect glycyl-tRNA synthetase to recognize this mutant tRNA and attach glycine. That is not what happens. C7. There are four proline codons, four glycine codons, one methionine codon, and six serine codons. We apply the product rule to solve this problem. 4  4  1  6 = 96 C8. 3–CUU–5 or 3–CUC–5 C9. The codon is 5–CCA–3, which specifies proline. C10. It can recognize 5–GGU–3, 5–GGC–3, and 5–GGA–3. All of these specify glycine. C12. All tRNA molecules have some basic features in common. They all have a cloverleaf structure with three stemloop structures. The second stem-loop contains the anticodon sequence that recognizes the codon sequence in mRNA. At the 3 end, there is an acceptor site, with the sequence CCA, that serves as an attachment site for an amino acid. Most tRNAs also have base modifications that occur within their nucleotide sequences. C13. They are very far apart, at opposite ends of the molecule. C14. The role of aminoacyl-tRNA synthetase enzymes is to specifically recognize tRNA molecules and attach the correct amino acid to them. They are sometimes described as the second genetic code because the specificity of their attachment is a critical step in deciphering the genetic code. For example, if a tRNA has a 3–GGG–5 anticodon, it will recognize a 5–CCC–3 codon, which should specify proline. It is essential that the prolyl-tRNAsynthetase recognizes this tRNA and attaches proline to the 3 end. The other aminoacyl-tRNA synthetases should not recognize this tRNA. C15. In the context of translation, an activated amino acid has had AMP attached to it. This provides necessary energy so that the amino acid can be attached to the correct tRNA. C16. Bases that have been chemically modified can occur at various locations throughout the tRNA molecule. The significance of all of these modifications is not entirely known. However, within the anticodon region, base modification alters base pairing to allow the anticodon to recognize two or more different bases within the codon. C19. Translation requires mRNA, tRNAs, ribosomes, many proteins such as initiation, elongation, and termination factors, and many small molecules. ATP and GTP are small molecules that contain high-energy bonds. The mRNA, tRNAs, and proteins are macromolecules. The ribosomes are a large complex of macromolecules. C23. Initiation: The mRNA, initiator tRNA, and initiation factors associate with the small ribosomal subunit; then the large subunit associates. Elongation: The ribosome moves one codon at a time down the mRNA, adding one amino acid at a time to the growing polypeptide chain. There are three sites on the ribosome, the A, P, and E sites, that are important in this process. The A site is where the tRNA (except for the initiator tRNA) binds to the ribosome and recognizes the codon in the mRNA. The growing polypeptide chain is then transferred to the amino acid attached to this tRNA. The ribosome then translocates so that this tRNA is now moved to the P site. The empty tRNA that was in the P site is moved into the E site. This empty tRNA in the E site is then expelled, and now the next charged tRNA can bind to the A site. Termination: A stop codon is reached and a termination factor binds to the A site. The hydrolysis of GTP initiates a series of events that leads to the disassembly of the ribosomal subunits and the release of the completed polypeptide chain. C24. Most bacterial mRNAs contain a Shine-Dalgarno sequence, which is necessary for the binding of the mRNA to the small ribosomal subunit. This sequence, UAGGAGGU, is complementary to a sequence in the 16S rRNA. Due to this complementarity, these sequences will hydrogen bond to each other during the initiation stage of translation. C28. The A site is the acceptor site. It is the location where a tRNA initially “floats in” and recognizes a codon in the mRNA. The only exception is the initiator tRNA that binds to the P site. The P site is the next location where the tRNA moves. When it first moves to the P site, it carries with it the polypeptide chain. In each round of elongation, the polypeptide chain is transferred from the tRNA in the P site to the amino acid attached to the tRNA in the A site. The third site is the E site. During translocation, the uncharged tRNA in the P site is transferred to the E site. It exits or is released from this site. C29. The amino acid sequence is methionine tyrosine tyrosine glycine alanine. Methionine is at the amino terminus, alanine at the carboxyl terminus. The peptide bonds should be drawn as shown in figure 13.20. C34. A. E site and P site (Note: A tRNA without an amino acid attached is only briefly found in the P site, just before translocation occurs.) B. P site and A site (Note: A tRNA with a polypeptide chain attached is only briefly found in the A site, just before translocation occurs.) C. Usually the A site, except the initiator tRNA can be found in the P site.