CH09 - Proteins and Their Synthesis BASIC PROBLEMS 1. NH3 -...

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Unformatted text preview: Proteins and Their Synthesis BASIC PROBLEMS 1. NH3 - Ala - Trp - (stop) - COOH 3’ CGT ACC ACT GCA 5’ 5’ GCA TGG TGA CGT 3’ 5’ GCA UGG UGA CGU 3’ 3’ CGU ACC ACU GCA 5’ DNA double helix (transcribed strand) DNA double helix mRNA transcribed appropriate tRNA anticodon amino acids incorporated 5’ UUG GGA AGC 3’ Assuming the reading frame starts at the first base: a. and b. c. and d. NH3 - Leu — Gly - Ser - COOH For the bottom strand, the mRNA is 5’ GCU UCC CAA 3’ and assuming the reading frame starts at the first base, the corresponding amino acid chain is NH3 - Ala - Ser - Gln — COOH. (5) With an insertion, the reading frame is disrupted. This will result in a drastically altered protein from the insertion to the end of the protein (which may be much shorter or longer than wild type because of the location of stop signals in the altered reading frame). A single nucleotide change should result in three adjacent amino acid changes in a protein. One and two adjacent amino acid changes would be expected to be much rarer than the three changes. This is directly the opposite of what is observed in proteins. Also, given any triplet coding for an amino acid, the next triplet could only be one of four. For example, if the first is GGG, then the next must be GGN (N = any base). This puts severe limits on which amino acids could be adjacent to each other. You could check amino acid sequences of various proteins to show that this is not the case. 182 Chapter Nine 5. It suggests very little evolutionary change between E. coli and humans with regard to the translational apparatus. The code is universal, the ribosomes are interchangeable, the tRNAs are interchangeable, and the enzymes involved are interchangeable. (Initiation of translation in prokaryotes in viva requires specific base-pairing between the 3’ end of the 168 rRNA and a Shine—Delgano sequence found in the 5’ untranslated region of the mRNA. A Shine~Delgano sequence would not be expected (unless by chance) in a eukaryotic mRNA and therefore initiation of translation might not occur.) 6. There are three codons for isoleucine: 5’ AUU 3’, 5’ AUC 3’, and 5’ AUA 3’. Possible anticodons are 3’ UAA 5’ (complementary), 3’ UAG 5’ (complementary), and 3’ UAI 5’ (wobble). 5’ UAU 3’, although complementary, would also base-pair with 5’ AUG 3’ (methionine) due to wobble and therefore would not be an acceptable alternative. 7. a. By studying the genetic code table provided in the textbook, you will discover that there then there are 28 codons that do not specify a particular amino acid with the first two positions (32 if you count Tyr and the stop codons starting with UA). b. If you knew the amino acid, you would not know the first two nucleotides in the cases of Arg, Ser, and Leu. 8. The codon for amber is UAG. Listed below are the amino acids that would have been needed to be inserted to continue the wild-type chain and their codons: glutamine CAA, CAG* lysine AAA, AAG* glutamic acid GAA, GAG* tyrosine UAU*, UAC* tryptophan UGG* serine AGU, AGC, UCU, UCC, UCA, UCG‘k In each case, the codon marked by an asterisk would require a single base change to become UAG. 9. a. The codons for phenylalanine are UUU and UUC. Only the UUU codon can exist with randomly positioned A and U. Therefore, the chance of UUU is (1/2)(1/2)(1/2) = 1/8. b. The codons for isoleucine are AUU, AUC, and AUA. AUC cannot exist. The probability of AUU is (1/2)(1/2)(1/2) = 1/8, and the probability of AUA is (1/2)(1/2)(1/2) = 1/8. The total probability is thus 1/4. c. The codons for leucine are UUA, UUG, CUU, CUC, CUA, and CUG, of which only UUA can exist. It has a probability of ( 1/2)(1/ 2)(1/ 2) = 1/8. 10. 11. 12. 13. Chapter Nine 183 d. The codons for tyrosine are UAU and UAC, of which only UAU can exist. It has a probability of (1/2)(1/2)(1/2) = 1/8, a. 1 U : 5 C —- The probability of a U is 1/6, and the probability of a C is 5/6. Codon Amino acid Probability Sum UUU Phe (1/6)(1/6)(1/6) = 0.005 Phe = 0.028 UUC Phe (1/6)(1/6)(5/6) = 0.023 CCC Pro (5/6)(5/6)(5/6) = 0.578 Pro = 0.694 CCU Pro (5/6)(5/6)(1/6) = 0.116 UCC Ser (1/6)(5/6)(5/6) = 0.116 Ser = 0.139 UCU Ser (1/6)(5/6)(1/6) = 0.023 CUC Leu (5/6)(1/6)(5/6) = 0.116 Leu = 0.139 CUU Leu (5/6)(1/6)(1/6) = 0.023 1Phe:25Pro:SSer:5Leu b. Using the same method as above, the final answer is 4 stop : 80 Phe : 40 Leu:24Ile:24Serz20Tyr:6Pro:6Thr:5Asn:5His:lLys: 'l Gln. c. All amino acids are found in the proportions seen in the code table. Quaternary structure is due to the interactions of subunits of a protein. In this example, the enzyme activity being studied may be from a protein consisting of two different subunits. Both subunits are required for activity. The polypeptides of the subunits are encoded by separate and unlinked genes. There are a number of mutational changes that can lead to the absence of enzymatic function in the product of a gene. Some of these changes would result in the complete absence of protein product and therefore also the absence of a detectable band on a Western blot. Mutations such as deletions of the gene, for example, would result in the lack of detectable protein. Other mutations that destroy function (missense, for example) may not alter the production of the protein and would be detected on a Western blot. Still other mutations (nonsense, frameshift) could alter the size of the protein yet would still lead to detectable protein. a. Tryptophan synthetase is a heterotetramer of two copies each of two different polypeptides, each encoded by a separate gene. Mutations that prevent the synthesis of one subunit would lead to the loss of one of the bands on a Western blot. Mutants that still make both subunits (those with exactly the same bands as wild type) might have mutations that prevent the subunits from interacting or disrupt the active site of the enzyme. 184 Chapter Nine b. Because the two subunits are encoded by separate genes, the absence of both bands simultaneously would require two independent and rare mutagenic events. 14. Yes. It was not known at the time what number of bases the “plus” and “minus” mutations actually were. If each mutation was two bases, then a codon would have been six bases. Since the mutations were actually adding or subtracting single bases, the codon is indeed three bases. 15. No. The enzyme may require post—translational modification to be active. Mutations in the enzymes required for these modifications would not map to the isocitrate lyase gene. 16. A nonsense suppressor is a mutation in a tRNA such that its anticodon can base- pair with a stop codon. In this way, a mutant stop codon (nonsense mutation) can be read through and the polypeptide can be fully synthesized. However, the mutant tRNA may be for an amino acid that was not encoded in that position in the original gene. For example, the codon UCG (serine) is instead UAG in the nonsense mutant. The suppressor mutation could be in tRNA for tryptophan such that its anticodon now recognizes UAG instead of UGG. During translation in the double mutant, the machinery puts tryptophan into the location of the mutant stop codon. This allows translation to continue but does place tryptophan into a position that was serine in the wild-type gene. This may create a protein that is not as active and a cell that is “not exactly wild type.” Another explanation is that translation of the mutant gene is not as efficient and that premature termination still occurs some of the time. This would lead to less product and again, a state that is “not exactly wild type.” 17. DNA RNA polymerase ribosomc In eukaryotes, transcription occurs within the nucleus while translation occurs in the cytoplasm. Thus the two processes cannot occur together. 18. Assuming that the three mutations of gene P are all nonsense mutations, there are three different possible stop codons that might be the cause (amber, ochre, or opal). A suppressor mutation would be specific to one type of nonsense codon. For example, amber suppressors would suppress amber mutants but not opal or ochre. 19. 20. 21. 22. 23. 24. 25. 26. Chapter Nine 185 Initiation of translation in prokaryotes requires specific base-pairing between the 3’ end of the 16s rRNA and a Shine—Delgano sequence found in the 5’ untranslated region of the mRNA. A Shine—Delgano sequence would not be expected (unless by chance) in a eukaryotic mRNA and therefore initiation of translation would not occur. Initiaton of translation in eukaryotes requires initiation factors (elF4a, b, and G) that associate with the 5’ cap of the mRNA. Because prokaryotic mRNAs are not capped, translation would not initiate. Not likely. Although the steps of translation and the components of ribosomes are similar in both eukaryotes and prokaryotes, the ribosomes are not identical. The sizes of both subunits are larger in eukaryotes and the many specific and intricate interactions that must take place between the small and large subunits would not be possible in a chimeric system. Single amino acid changes can result in changes in protein folding, protein targeting, or post-translational modifications. Any of these changes could give the results indicated. The first indication of rRNAs importance was the discovery of ribozymes. Recently, structural studies have shown that both the decoding center in the 308 subunit and the peptidyl transferase center in the SOS subunit are composed entirely of rRNA and that the important contacts in these centers are all tRNA/rRNA contacts. Antibiotics need to selectively target bacterial structures and functions that are essential for life but unique or sufficiently different from the equivalent structure and functions of their animal hosts. The large bacterial ribosomal subunit fits these criteria as its function is obviously essential yet its structure is sufficiently different from the large eukaryotic ribosomal subunit. While the steps of protein synthesis are similar overall, eukaryotic ribosomes have larger and more numerous components. These differences make it possible to develop drugs that specifically bind bacterial ribosomes but have little or no affinity for eukaryotic ribosomes. Recent studies indicate that most proteins function by interacting with other proteins. (The complete set of such interactions is called the interactome.) Most of these essential protein-protein interactions are regulated by phosphorylation/dephosphorylation modifications. Kinases are the enzymes that catalyze phosphorylations. Therefore, the complexity of the interactome necessary for the complexity of multicellularity requires the very large number of kinase—encoding genes. Bacterial and human cell cultures are both capable of producing the same polypeptide from the same mRNA, but that does not mean that the resulting protein will be active. Many proteins require posttranslational processing to 186 Chapter Nine become functional, and the enzymes and control necessary for such processing is not universal. The proteins produced and isolated from a human cell culture system will contain the necessary posttranslational modifications necessary for human protein function. CHALLENGING PROBLEMS 27. a. and b. The goal of this type of problem is to align the two sequences. You are told that there is a single nucleotide addition and single nucleotide deletion, so look for single base differences that effect this alignment. These should be located where the protein sequence changes (i.e., between Lys—Ser and Asn-Ala). Remember also that the genetic code is redundant. (N = any base) Lys Ser Pro Ser Leu Asn Ala Ala Lys _ U U A A ®3C AGC UUG AAG UCN CCN UCN CUN AAU GCN GCN AA .l GUN/tall; .47.. m: Gilli—61> G + Lys Val His His Leu Met Ala Ala Lys 11> Base deleted Old: AA @fiCCA UCA CUU AAU GCN GCN AA GUC CAU CAC UUA AuC)GCN GCN AA / Base added 28. Mutant 1: A simple substitution of Arg for Ser exists, suggesting a nucleotide change. Two codons for Arg are AGA and AGG, and one codon for Ser is AGU. The U of the Ser codon could have been replaced by either an A or a G. New: AA 63> G) 03> (DID Mutant 2: The Trp codon (UGG) changed to a stop codon (UGA or UAG). Mutant 3: Two frameshift mutations occurred: 5’eGCN CCN (—U)GGA GUG AAA AA(+U or C) UGU(or C) CAU(or C)—3'. 29. 30. 31. 32. Chapter Nine 187 Mutant 4: An inversion occurred after Trp and before Cys. The DNA original sequence (with both strands shown for the area of inversion) was 3’—CGN GGN ACC TCA CTT TTT ACA(Or G) GTA(Or G)—5' 5’* —AGT GAA AAA— —3’ Therefore, the complementary RNA sequence was 5’—GCN CCN UGG AGU GAA AAA UGU/C CAU/C—3’ The DNA inverted sequence became 3'—CGN GGN ACC AAA AAG TGA ACA/G GTA/G-S’ Therefore the complementary RNA sequence was 5’—GCN CCN UGG UUU UUC ACU UGU/C CAU/C—3’ If the anticodon on a tRNA molecule also was altered by mutation to be four bases long, with the fourth base on the 5’ side of the anticodon, it would suppress the insertion. Alterations in the ribosome can also induce frameshifting. £¢nghmmag Cells in long—established culture lines usually are not fully diploid. For reasons that are currently unknown, adaptation to culture frequently results in both karyotypic and gene dosage changes. This can result in hemizygosity for some genes, which allows for the expression of previously hidden recessive alleles. a. and b. The sequence of double—stranded DNA is as follows: 5’—TAC ATG ATC ATT TCA CGG AAT TTC TAG CAT GTAw3’ 3'—ATG TAC TAG TAA AGT GCC TTA AAG ATC GTA CAT—5' / First look for stop codons. Next look for the initiating codon, AUG (3 TAC—5’ in DNA). Only the upper strand contains the necessary codons. DNA 3’ TAC GAT CTT TAA GGC ACT 5’ RNA. 5’ AUG CUA GAA AUU CCG UGA 3' protein Met Leu Glu Ile Pro stop The DNA strand is read from right to left as written in your text and is written above in reverse order from your text. c. Remember that polarity must be taken into account. The inversion is 188 Chapter Nine 33. DNA 5' TAC ATG CTA GAA ATT CCG TGA AAT GAT CAT GTA 3’ RNA 3’ —GAU CUU UAA GGC ACU UUA CUA GUAA 5' amino acids HOOC 7 6 5 4 3 2 1 ~NH3 DNA 3’ATG TAC TAG TAA AGT GCC TTA AAG ATC GTA CAT 5’ mRNA 5'UAC AUG AUC AUU UCA CGG AAU UUC UAG 3' 1 2 3 4 5 6 7 stop Codon 4 is 5 ’—UCA—3 ’, which codes for Ser. Anticodon 4 would be 3 ’ —AGU—5’ (or 3’—AGI—5’ given wobble). (GAU)n codes for Aspn (GAU)m Met” (AUG)n, and stop" (UGA)n. (GUA)n codes for Val" (GUA)n, Sern (AGU)n, and stop" (UAG),,. One reading frame in each contains a stop codon. Each of the three reading frames contains a stop codon. The way to approach this problem is to focus initially on one amino acid at a time. For instance, line 4 indicates that the codon for Arg might be AGA or GAG. Line 7 indicates it might be AAG, AGA, or GAA. Therefore, Arg is at least AGA. That also means that Glu is GAG (line 4). Lys and Glu can be AAG or GAA (line 7). Because no other combinations except the ones already mentioned result in either Lys or Glu, no further decision can be made with respect to them. However, taking wobble into consideration, Glu may also be GAA, which leaves Lys as AAG. Next, focus on lines '1 and 5. Ser and Leu can be UCU and CUC. Ser, Leu, and Phe can be UUC, UCU, and CUU. Phe is not UCU, which is seen in both lines. From line 14, CUU is Leu. Therefore, UUC is Phe, and UCU is Ser. The footnote states that line 13 and 14 are in the correct order. In line 13, if UCU is Ser (see above), then Ile is AUC, Tyr is UAU, and Leu is CUA. Continued application of this approach will allow the assignment of an amino acid to each codon. ...
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This note was uploaded on 03/19/2008 for the course BIOL 202 taught by Professor Kieber-hogan during the Spring '08 term at UNC.

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CH09 - Proteins and Their Synthesis BASIC PROBLEMS 1. NH3 -...

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