chap 7.doc

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CHAPTER 7 FROM DNA TO PROTEIN: HOW CELLS READ THE GENOME 2004 Garland Science Publishing From DNA to RNA 7-1 RNA in cells differs from DNA in that (a) it contains the base uracil, which pairs with cytosine. (b) it is single-stranded and cannot form base pairs. (c) it is single-stranded and can fold up into a variety of structures. (d) the nucleotides are linked together in a different way. (e) the sugar ribose contains fewer oxygen atoms than does deoxyribose. 7-2 Transcription is similar to DNA replication in that (a) it requires a molecule of DNA helicase to unwind the DNA. (b) it uses the same enzyme as that used to synthesize RNA primers during DNA replication. (c) the newly synthesized RNA remains paired to the template DNA. (d) nucleotide polymerization occurs only in the 5-to-3 direction. (e) an RNA transcript is synthesized discontinuously and the pieces then joined together. 7-3 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. In order for a cell’s genetic material to be utilized, the information is first copied from the DNA into the nucleotide sequence of RNA in a 103

Transcript of chap 7.doc

Page 1: chap 7.doc

CHAPTER 7FROM DNA TO PROTEIN:

HOW CELLS READ THE GENOME 2004 Garland Science Publishing

From DNA to RNA

7-1 RNA in cells differs from DNA in that(a) it contains the base uracil, which pairs with cytosine.(b) it is single-stranded and cannot form base pairs.(c) it is single-stranded and can fold up into a variety of structures.(d) the nucleotides are linked together in a different way.(e) the sugar ribose contains fewer oxygen atoms than does deoxyribose.

7-2 Transcription is similar to DNA replication in that(a) it requires a molecule of DNA helicase to unwind the DNA.(b) it uses the same enzyme as that used to synthesize RNA primers during DNA

replication.(c) the newly synthesized RNA remains paired to the template DNA.(d) nucleotide polymerization occurs only in the 5-to-3 direction.(e) an RNA transcript is synthesized discontinuously and the pieces then joined

together.

7-3 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once.

In order for a cell’s genetic material to be utilized, the information is first copied from the DNA into the nucleotide sequence of RNA in a process called __________________. Various kinds of RNAs are produced, each with different functions. __________________ molecules code for proteins, __________________ molecules act as adaptors for protein synthesis, __________________ molecules are integral components of the ribosome, while __________________ molecules are important for splicing of RNA transcripts.

incorporation rRNA transmembranemRNA snRNA tRNApRNA transcriptionproteins translation

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7-4 Match the following structures with their names:

Figure Q7-4

7-5 Imagine that an RNA polymerase is transcribing a segment of DNA that contains the sequence:

5- AGTCTAGGCACTGA -33- TCAGATCCGTGACT -5

A. If the polymerase is transcribing from this segment of DNA from left to right, which strand (top or bottom) is the template?

B. What will be the sequence of that RNA (be sure to label the 5 and 3 ends of your RNA molecule)?

7-6 The sigma subunit of bacterial RNA polymerase(a) contains the catalytic activity of the polymerase.(b) remains part of the polymerase throughout transcription.(c) recognizes promoter sites in the DNA.(d) recognizes transcription termination sites in the DNA.

7-7 Which of the following might decrease the transcription of only one specific gene in a bacterial cell?(a) A decrease in the amount of sigma factor(b) A decrease in the amount of RNA polymerase(c) A mutation that introduced a stop codon into the DNA preceding the coding

sequence of the gene(d) A mutation that introduced extensive sequence changes into the DNA preceding

the transcription start site of the gene(e) A mutation that moved the transcription termination signal of the gene farther

away from the transcription start site

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7-8 From the list below, pick THREE reasons why the primase that is used to make the RNA primer for DNA replication would not be suitable for gene transcription?(a) Primase initiates RNA synthesis on a single-stranded DNA template.(b) Primase can initiate RNA synthesis without the need for a base-paired primer.(c) Primase synthesizes only RNAs of around 5 to 20 nucleotides in length.(d) The RNA synthesized by primase remains base-paired to the DNA template.(e) Primase uses nucleotide triphosphates.

7-9 Indicate where the following processes take place by adding numbered labeling lines to the schematic diagram of a eucaryotic cell in Figure Q7-9.

Figure Q7-9

1. Transcription2. Translation3. RNA splicing4. Polyadenylation5. RNA capping

7-10 Total nucleic acids are extracted from a culture of yeast cells and are then mixed with resin beads to which the polynucleotide 5-TTTTTTTTTTTTTTTTTTTTTTTTT-3 has been covalently attached. After a short incubation, the beads are then extracted from the mixture. When you analyze the cellular nucleic acids that have stuck to the beads, which of the following will be most abundant?(a) DNA(b) tRNA(c) rRNA(d) mRNA(e) Primary transcript RNA

7-11 Name three modifications that can be made to an RNA molecule in eucaryotic cells before the RNA molecule becomes a mature mRNA.

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7-12 The length of a particular gene in human DNA, measured from the start site for transcription to the end of the protein-coding region, is 10,000 nucleotides, whereas the length of the mRNA produced from this gene is 4000 nucleotides. What is the most likely reason for this discrepancy?

7-13 A fragment of human DNA containing the gene for a protein hormone with its regulatory regions removed is introduced into bacteria; although it is transcribed at a high level into RNA, no protein is made. When this RNA is extracted from the bacteria, mixed with human mRNA encoding the same hormone, and then examined in the electron microscope, you see the following structure (Figure Q7-13). Label each of the statements below as either “consistent” or “inconsistent” with your results and explain your reasoning.

Figure Q7-13

A. The human DNA was inserted in the bacterial DNA next to a bacterial promoter and in its normal orientation.

B. The human DNA was inserted in the bacterial DNA next to a bacterial promoter but in an orientation opposite to normal.

C. The human DNA contained an intron.D. The human DNA acquired a deletion while in the bacterium.

7-14 Why is the old dogma “one gene—one protein” not always true for eucaryotic genes?

7-15 Is this statement TRUE or FALSE? Explain your answer.

“Since introns do not contain protein coding information, they do not have to be removed precisely (meaning, a nucleotide here and there should not matter) from the primary transcript during RNA splicing.”

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7-16 You have discovered a gene (see Figure Q7-16 part A) that is alternatively spliced to produce several forms of mRNA in various cell types, three of which are shown in part B of Figure Q7-16. (Note that splicing is indicated by lines connecting the exons that are included in the mRNA). Your experiments have found that protein translation begins in exon 1. For all forms of the mRNA, the encoded protein sequence is the same in the regions of the mRNA that correspond to exons 1 and 10. Exons 2 and 3 are alternative exons used in different mRNA, as are exons 7 and 8. Which of the following statements about exons 2 and 3 is the most accurate? Explain your answer.

Figure Q7-16

(a) Exons 2 and 3 must have the same number of nucleotides.(b) Exons 2 and 3 must contain an integral number of codons (that is, the number of

nucleotides divided by 3 must be an integer).(c) Exons 2 and 3 must contain a number of nucleotides that when divided by 3,

leaves the same remainder (that is, 0, 1, or 2).

From RNA to Protein

7-17 Which of the following statements about the genetic code are correct?(a) All codons specify more than one amino acid.(b) The genetic code is redundant.(c) All amino acids are specified by more than one codon.(d) The genetic code is different in procaryotes and eucaryotes.(e) All codons specify an amino acid.

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NOTE: The following codon table is to be used for Problems Q7-18 – 7-23, Q7-29, and 7-33.

7-18 The following DNA sequence includes the beginning of a sequence coding for a protein. What would be the result of a mutation that changed the C marked by an asterisk to an A?

5- AGGCTATGAATGGACACTGCGAGCCC.... *

7-19 Which amino acid would you expect a tRNA with the anticodon 5-CUU-3 to carry? (Refer to Codon table provided above Q7-18.)

(a) Lysine(b) Glutamate(c) Glutamine(d) Leucine(e) Phenylalanine

7-20 Which of the following pairs of codons might you expect to be read by the same tRNA as a result of wobble? (Refer to Codon table provided above Q7-18.)

(a) CUU and UUU(b) GAU and GAA(c) CAC and CAU(d) AAU and AGU(e) CCU and GCU

7-21 Below is a segment of RNA from the middle of an mRNA. If you were told that this segment of RNA was part of the coding region of an mRNA for a large protein, give the amino acid sequence for the protein that is encoded by this segment of mRNA. (Refer to Codon table provided above Q7-18.)

5- UAGUCUAGGCACUGA -3

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7-22 (Refer to Codon table provided above Q7-18.) One strand of a section of DNA isolated from the bacterium E. coli reads:

5- GTAGCCTACCCATAGG -3

A. Suppose that an mRNA is transcribed from this DNA using the complementary strand as a template. What will be the sequence of the mRNA in this region (make sure you label the 5 and 3 ends of the mRNA)?

B. How many different peptides could potentially be made from this sequence of RNA, assuming translation initiates upstream of this sequence?

C. What are these peptides? (Give your answer using the one letter amino acid code.)

7-23 A strain of yeast translates mRNA into protein with a high level of inaccuracy. Individual molecules of a particular protein isolated from this yeast have the following variations in the first 11 amino acids compared with the sequence of the same protein isolated from normal yeast cells (Figure Q7-23). What is the most likely cause of this variation in protein sequence? Explain your answer. (Refer to Codon table provided above Q7-18.)

Figure Q7-23

(a) A mutation in the DNA coding for the protein(b) A mutation in the anticodon of the isoleucine tRNA (tRNAIle)(c) A mutation in the isoleucyl-tRNA synthetase that decreases its ability to

distinguish between different amino acids(d) A mutation in the isoleucyl-tRNA synthetase that decreases its ability to

distinguish between different tRNA molecules(e) A mutation in a component of the ribosome that allows binding of incorrect tRNA

molecules to the A-site

7-24 Which of the following statements is TRUE?(a) Ribosomes are large RNA structures composed solely of rRNA.(b) Ribosomes are synthesized entirely in the cytoplasm.(c) rRNA contains the catalytic activity that joins amino acids together.(d) A ribosome consists of two equally sized subunits.(e) A ribosome binds one tRNA at a time.

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7-25 Figure Q7-25A shows the stage in translation when an incoming aminoacyl-tRNA has bound to the A-site on the ribosome. Using the components shown in Figure Q7-25A as a guide, show on Figure Q7-25B and Q7-25C what happens in the next two stages to complete the addition of the new amino acid to the growing polypeptide chain.

Figure Q7-25

7-26 A poison added to an in vitro translation mixture containing mRNA molecules with the sequence 5-AUGAAAAAAAAAAAAUAA-3 has the following effect: the only product made is a Met-Lys dipeptide that remains attached to the ribosome. What is the most likely way in which the poison acts to inhibit protein synthesis?(a) It inhibits binding of the small subunit of the ribosome to mRNA.(b) It inhibits peptidyl transferase activity.(c) It inhibits movement of the small subunit relative to the large subunit.(d) It inhibits release factor.(e) It mimics release factor.

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7-27 In eucaryotes, but not procaryotes, ribosomes find the start site of translation by(a) binding directly to a ribosome-binding site preceding the initiation codon.(b) scanning along the mRNA from the 5 end.(c) recognizing an AUG codon as the start of translation.(d) binding an initiator tRNA.

7-28 Figure Q7-28 shows an mRNA molecule.

Figure Q7-28

A. Match the labels given in the list below with the label lines in Figure Q7-28.(a) ribosome-binding site(b) initiator codon(c) stop codon(d) untranslated 3 region(e) untranslated 5 region(f) protein-coding region

B. Is the mRNA shown procaryotic or eucaryotic? Explain your answer.

7-29 A tRNA for the amino acid lysine is mutated such that the sequence of the anticodon is 5-UAU-3 (instead of 5-UUU-3). Which of the following aberrations in protein synthesis might this tRNA cause? (Refer to Codon table provided above Q7-18.)

(a) Read through of stop codons(b) Substitution of lysine for isoleucine(c) Substitution of lysine for tyrosine(d) Substitution of lysine for phenylalanine(e) Substitution of lysine for the amino-terminal methionine

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7-30 You have discovered a protein that inhibits translation. When you add this inhibitor to a mixture capable of translating human mRNA and centrifuge the mixture to separate polyribosomes and single ribosomes, you obtain the results shown in Figure Q7-30. Which of the following interpretations are consistent with these observations?

Figure Q7-30

(a) The protein binds to the small ribosomal subunit and increases the rate of initiation of translation.

(b) The protein binds to sequences in the 5 region of the mRNA and inhibits the rate of initiation of translation.

(c) The protein binds to the large ribosomal subunit and slows down elongation of the polypeptide chain.

(d) The protein binds to sequences in the 3 region of the mRNA and prevents termination of translation.

7-31 The concentration of a particular protein X in a normal human cell rises gradually from a low point, immediately after cell division, to a high point, just before cell division, and then drops sharply. The level of its mRNA in the cell remains fairly constant throughout this time. Protein X is required for cell growth and survival, but the drop in its level just before cell division is essential for division to proceed. You have isolated a line of human cells that grow in size in culture but cannot divide, and on analyzing these mutants, you find that levels of X mRNA in the mutant cells are normal. Which of the following mutations in the gene for X could explain these results?(a) The introduction of a stop codon that truncates protein X at the fourth amino acid.(b) A change of the first ATG codon to CCA.(c) The deletion of a sequence that encodes sites at which ubiquitin can be attached to

the protein.(d) A change at a splice site that prevents splicing of the RNA.

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7-32 For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once.

Once an mRNA is produced, its message can be decoded on ribosomes. The ribosome is composed of two subunits: the __________________ subunit, which catalyzes the formation of the peptide bonds that link the amino acids together into a polypeptide chain, and the __________________ subunit, which matches the tRNAs to the codons of the mRNA. During the chain elongation process of translating an mRNA into protein, the growing polypeptide chain attached to a tRNA is bound to the __________________ -site of the ribosome. An incoming aminoacyl-tRNA carrying the next amino acid in the chain will bind to the __________________ -site by forming base pairs with the exposed codon in the mRNA. The __________________ enzyme catalyzes the formation of a new peptide bond between the growing polypeptide chain and the newly arriving amino acid. The end of a protein-coding message is signaled by the presence of a stop codon, which binds the __________________ called release factor. Eventually, most proteins will be degraded by a large complex of proteolytic enzymes called the __________________.

A medium proteosomecentral P RNADNA peptidyl transferase smallE polymerase Tlarge protein ubiquitin

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7-33 After treating cells with a mutagen, you isolate two mutants. One carries alanine and the other carries methionine at a site in the protein that normally contains valine. After treating these two mutants again with mutagen, you isolate mutants from each that now carry threonine at the site of the original valine. Assuming that all mutations caused by the mutagen involve single nucleotide changes, deduce the codons that are used for valine, alanine, methionine, and threonine at the affected site. See Figure Q7-33. (Refer to Codon table provided above Q7-18.)

Figure Q7-33

RNA and the Origins of Life

7-34 According to current thinking, the minimum requirement for life to have originated on Earth was the formation of a(a) molecule that could provide a template for the production of a complementary

molecule.(b) double-stranded DNA helix.(c) molecule that could direct protein synthesis.(d) molecule that could catalyze its own replication.

7-35 Which of the following reactions are known to be carried out by a ribozyme?(a) DNA synthesis(b) Transcription(c) RNA splicing(d) Protein hydrolysis(e) Polysaccharide hydrolysis

7-36 You are studying a disease that is caused by a virus, but when you purify the virus particles and analyze them you find they contain no trace of DNA. Which of the following molecules are likely to contain the genetic information of the virus?(a) Protein(b) RNA(c) Lipids(d) Carbohydrates

7-37 Give a reason why DNA makes a better material than RNA for storage of genetic information and explain your answer.

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How We Know: Cracking the Genetic Code

7-38 An extraterrestrial organism (ET) is discovered whose basic cell biology seems pretty much the same as terrestrial organisms except that it uses a different genetic code to translate RNA into protein. You set out to break the code by translation experiments using RNAs of known sequence and cell-free extracts of ET cells to supply the necessary protein-synthesizing machinery. In experiments using the RNAs below, the following results were obtained when the 20 possible amino acids were added either singly or in different combinations of two or three:

RNA 1: 5–GCGCGCGCGCGCGCGCGCGCGCGCGCGC–3RNA 2: 5–GCCGCCGCCGCCGCCGCCGCCGCCGCCGCC–3

Using RNA 1, a polypeptide was produced only if alanine and valine were added to the reaction mixture. Using RNA 2, a polypeptide was produced only if leucine and serine and cysteine were added to the reaction mixture. Assuming that protein synthesis can start anywhere on the template, that the ET genetic code is nonoverlapping and linear, and that each codon is the same length (like the terrestrial triplet code), how many nucleotides does an ET codon contain?(a) 2(b) 3(c) 4(d) 5(e) 6

7-39 NASA has discovered an alien life form. You are called in to help them deduce the genetic code for this alien. Surprisingly, this alien life form shares many similarities with life on Earth: this alien uses DNA as its genetic material, makes RNA from DNA, and reads the information from RNA to make protein using ribosomes and tRNAs. Even more amazing, this alien uses the same 20 amino acids, like the organisms found on Earth, and also codes for each amino acid by a triplet codon. However, the scientists at NASA have found that the genetic code used by the alien life form is different than that used by life on Earth. The experiment that allowed the NASA scientists to draw this conclusion involved creating a cell-free protein synthesis system from alien cells and adding an mRNA made entirely of uracil (poly U). This led to the finding that poly U directs the synthesis of a peptide containing only glycine. NASA scientists have synthesized a poly AU mRNA and find that it codes for a polypeptide of alternating serine and proline residues. From these experiments, can you determine which codons code for serine and proline? Explain why or why not.

Bonus question: Can you propose a mechanism for how the alien’s physiology is altered so that it uses a different genetic code from life on Earth, despite all the similarities?

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Answers

7-1 Choice (c) is the correct answer. Choice (a) is untrue, since although RNA contains uracil, uracil pairs with adenine, not cytosine. Choice (b) is false because RNA can form base pairs with a complementary RNA or DNA sequence. Choice(d) is false. Choice (e) is false because ribose contains one more oxygen atom than deoxyribose.

7-2 Choice (d) is the correct answer. RNA polymerase unwinds only a few base pairs of the double helix at a time and does not need a helicase to do so, which is why choice (a) is incorrect. The enzyme used to make primers during DNA synthesis is indeed an RNA polymerase, but it is a special enzyme, primase, and not the enzyme that is used for transcription, which is why choice (b) is incorrect. Choice (c) is false. Choice (e) is incorrect because an RNA transcript is made by a single polymerase molecule that proceeds from the start site to the termination site without falling off.

7-3 In order for a cell’s genetic material to be utilized, the information is first copied from the DNA into the nucleotide sequence of RNA in a process called transcription. Various kinds of RNAs are produced, each with different functions. mRNA molecules code for proteins, tRNA molecules act as adaptors for protein synthesis, rRNA molecules are integral components of the ribosome, while snRNA molecules are important for splicing of RNA transcripts.

7-4 A—4; B—1; C—2; D—3

7-5 A. The bottom strandB. 5- AGUCUAGGCACUGA -3

7-6 (c)

7-7 Choice (d) is the correct answer. Such changes would probably destroy the function of the promoter, making RNA polymerase unable to bind to it. Decreasing the amount of sigma factor or RNA polymerase (choices (a) or (b)) would affect the transcription of most of the genes in the cell, not just one specific gene. Introducing a stop codon before the coding sequence (choice (c)) would have no effect on transcription of the gene, since the transcription machinery does not recognize translational stops. Moving the termination signal farther away (choice (e)) would merely make the transcript longer.

7-8 Choices (a), (c), and (d) are the correct reasons. Choices (b) and (e) are true for both primase and RNA polymerase.

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7-9 See Figure A7-9.

Figure A7-9

7-10 (d) mRNA is the only type of RNA that is polyadenylated, and this poly(A) tail would be able to base-pair with the strands of poly(T) on the beads and thus stick to them. DNA would not be found in the sample, as the poly(A) tail is not encoded in the DNA and long runs of T are rare in DNA.

7-11 1. A poly(A) tail must be added.2. A 5 cap must be added.3. Introns must be spliced out.

(“Export from nucleus” is also an acceptable answer.)

7-12 The gene contains one or more introns.

7-13 B, C, and D are consistent with the results; A is inconsistent. B must be true for the RNA produced in the bacterium to be complementary to, and thus able to pair with, the mRNA from a human cell. If the human DNA had become inserted in its normal orientation next to the promoter (A), the corresponding portions of RNA would be identical (or at least very similar) in sequence and thus the two RNAs would not be complementary and would not pair. The loop formed in the hybrid tells us that one of the molecules contains sequences that the other is missing. This could come about either because the bacterial RNA was transcribed from human sequences that acquired a deletion (B) or because the human gene contains an intron (C).

7-14 The transcripts from some genes can be spliced in more than one way to give mRNAs containing different sequences, thus encoding different proteins. A single eucaryotic gene, therefore, may encode more than one protein.

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7-15 False. Although it is true that the sequences within the introns are mostly dispensable, the introns must still be removed precisely because an error of one or two nucleotides would shift the reading frame of the resulting mRNA molecule and change the protein it encodes.

7-16 Choice (c) is the only answer that must be true for exons 2 and 3. Although choices (a) and (b) could be true, they don’t have to be. Because the protein sequence is the same in segments of the mRNA corresponding to exons 1 and 10, the choice of either exon 2 or exon 3 would not alter the reading frame. To maintain the normal reading frame, whatever it is, the alternative exons must have a number of nucleotides that when divided by 3 (the number of nucleotides in a codon) give the same remainder.

7-17 Choice (b) is the correct answer. The majority of the amino acids can be specified by more than one codon. Choice (a) is incorrect because each codon specifies only one amino acid. Choice (c) is incorrect because tryptophan and methionine are encoded by only one codon. Choice (d) is incorrect because, with a few minor exceptions, the genetic code is the same in all organisms. Choice (e) is incorrect because some codons specify translational stop signals.

7-18 The change creates a stop codon (TGA, or UGA in the mRNA) very near the beginning of the protein-coding sequence and in the correct reading frame (the beginning of the coding sequence is indicated by the ATG). Thus, translation would terminate after only four amino acids had been joined together, and the complete protein would not be made.

7-19 (a) Lys (lysine). As is conventional for nucleotide sequences, the anticodon is given 5 to 3. The complementary base-pairing occurs between antiparallel nucleic acid sequences, and the codon recognized by this anticodon will therefore be 5-AAG-3.

7-20 Choice (c) is the answer. These two codons differ only in the third position and also encode the same amino acid, which is the definition of wobble. Although the codons GAU and GAA (choice (b)) also differs only in the third position, they are unlikely in normal circumstances to be read by the same tRNA, as they encode different amino acids.

7-21 SLGT is the answer. (Reading frame two is the only reading frame that does not contain a stop codon.)

7-22 A. 5-GUAGCCUACCCAUAGG -3B. Two. (There are three potential reading frames for each RNA. In this case, they

are:GUA GCC UAC CCA UAG ... UAG CCU ACC CAU AGG..... AGC CUA CCC AUA GG?....The center one cannot be used in this case, because UAG is a stop codon.)

C. VAYPSLPIGNote: PTHR will not be a peptide because it is preceded by a stop codon.

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7-23 Choice (c) is the correct answer. A mutation in the isoleucyl-tRNA synthetase that decreases its ability to distinguish between amino acids would allow an assortment of amino acids to be attached to the tRNAIle. These assorted aminoacyl-tRNAs would then base-pair with the isoleucine codon and cause a variety of substitutions at positions normally occupied by isoleucine. Choice (a) is incorrect because a mutation in the gene encoding the protein would cause only a single variant protein to be made. Choice (e) is incorrect because a mutation in the ribosome that allows binding of any amino-acyl-tRNA to the A site would cause substitutions all over the protein, not only at isoleucine residues. Choices (b) and (d) are also incorrect. A mutation in the anticodon loop of tRNAIle (choice (b)) or a mutation in the isoleucine-tRNA synthetase that decreases its ability to distinguish between different tRNA molecules (choice (d)) would cause substitution of isoleucine for some other amino acid (which is the opposite of what is observed).

7-24 Choice (c) is the correct answer. Choice (a) is incorrect because ribosomes contain proteins as well as rRNA. Choice (b) is incorrect because rRNA is synthesized in the nucleus, and ribosomes are partly assembled in the nucleus. Choice (d) is incorrect because a ribosome consists of one small subunit and one large subunit. Choice (e) is incorrect because a ribosome must be able to bind two tRNAs at any one time.

7-25 See Figure A7-25.

Figure A7-25

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7-26 Choice (c) is the correct answer. Either choice (a) or (b) would prevent all peptide bond formation. Choice (d) would have no affect on translation until the stop codon was reached. Choice (e) would be likely to result in a mixture of polypeptides of various lengths; a poison mimicking a release factor could conceivably cause only Met-Lys to be made, but this dipeptide would not remain bound to the ribosome.

7-27 Choice (b) is the correct answer. Choice (a) is true only for procaryotes. Choices (c) and (d) are true for both procaryotes and eucaryotes.

7-28 A. (a)—3; (b)—2; (c)—4; (d)—6;(e)—1; (f)—5B. The mRNA is procaryotic. It contains coding regions for more than one protein,

as shown by the multiple initiation codons, each preceded by a ribosome-binding site. It contains an unmodified 5 end, as shown by the three phosphate groups, and an unmodified 3 end, as shown by the absence of a poly(A) tail.

7-29 (b) The mutant tRNALys will be able to pair with the codon 5-AUA-3, which codes for isoleucine.

7-30 Choice (b) is the correct answer. The results in Figure Q7–30 show a marked decrease in the number of polyribosomes formed relative to normal. Polyribosomes form because the initiation of translation is fairly rapid: ribosomes can bind successively to the free 5 end of an mRNA molecule and start translation before the first ribosome has had a chance to finish translating the message. Therefore, inhibition of the rate of initiation will tend to decrease the number of ribosomes in the polyribosome, and in the extreme case there will be only one ribosome per mRNA. Conversely, increasing the rate of initiation or slowing the rate of elongation would result in an increased number of ribosomes per polyribosome (up to a maximum point), making choices (a) and (c) false. Choice (d) is incorrect, as preventing termination would prevent release of the ribosomes at the end of the coding sequence and would be expected to “freeze” the assembled polyribosomes, so that the ratio of polyribosomes to ribosomes would be much as normal.

7-31 Choice (c) is the correct answer. The drop in level of protein X in the normal cell is most likely due to protein degradation, since levels of mRNA remain constant. The inability of the mutant cell to divide could be due to a mutation that inhibits protein degradation. This would be achieved by removal of sites for attachment of ubiquitin, which targets proteins for destruction. Choices (a), (b), and (d) would probably not produce the result described, as without the production of a functional protein X the mutant cells could not grow in size.

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7-32 Once an mRNA is produced, its message can be decoded on ribosomes. The ribosome is composed of two subunits: the large subunit, which catalyzes the formation of the peptide bonds that link the amino acids together into a polypeptide chain, and the small subunit, which matches the tRNAs to the codons of the mRNA. During the chain elongation process of translating an mRNA into protein, the growing polypeptide chain attached to a tRNA is bound to the P-site of the ribosome. An incoming aminoacyl-tRNA carrying the next amino acid in the chain will bind to the A-site by forming base pairs with the exposed codon in the mRNA. The peptidyl transferase enzyme catalyzes the formation of a new peptide bond between the growing polypeptide chain and the newly arriving amino acid. The end of a protein-coding message is signaled by the presence of a stop codon, which binds the protein called release factor. Eventually, most proteins will be degraded by a large complex of proteolytic enzymes called the proteosome.

7-33 Given that only single nucleotide changes are involved, the only codons consistent with the changes are: GUG for valine, GCG for alanine, AUG for methionine, and ACG for threonine.

7-34 Choice (d) is the correct answer. Choice (a) is incorrect in that although this may have been a step in self-replication, it would not by itself be sufficient. Choices (b) and (c) are incorrect, as these stages in the evolution of the cell must have succeeded the formation of the first self-replicating molecules.

7-35 (c)

7-36 (b)

7-37 Three possible answers are:1. The deoxyribose sugar of DNA makes the molecule much less susceptible to

breakage compared to RNA, due to the lack of the hydroxyl group on carbon 2 of the ribose sugar.

2. DNA is double stranded and therefore the complementary strand provides a template from which damage can be repaired accurately.

3. The use of “T” in DNA instead of “U” (as in RNA) protects against the effect of deamination, a common form of damage. Deamination of T produces an aberrant base (methyl C), whereas deamination of U generates C, a normal base. The presence of an abnormal base eases the cell’s job of recognizing the damaged strand.

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7-38 Choice (d) is the correct answer. An organism having codons with an even number of nucleotides (i.e., 2, 4, or 6) could read 5-GCGCGCGCGC-3 (RNA 1) in either of two ways, namely “GC GC GC GC...” or “CG CG CG CG...” Either one of the two amino acids alone could have supported protein synthesis, so you would not need them in combination (thus eliminating choices (a), (c), and (e)). An organism having three bases per codon could read the sequence 5-GCCGCCGCCGCCGCC-3 (RNA 2) in one of three ways, namely “GCC GCC GCC GCC...,” “CCG CCG CCG CGG...,” or “CGC CGC CGC CGC...,” and so again, any one of the three amino acids could have supported synthesis of a polypeptide, and you would not need to add all three amino acids to produce a polypeptide chain, thus eliminating choice (b). Only a five-nucleotide code gives you two different consecutive codons for RNA 1 and three different consecutive codons for RNA 2.

7-39 No, you cannot definitively determine what the codons that code for serine or valine are because it could be either UAU or AUA.

Bonus: The alien aminoacyl-tRNA synthetases could adapt a different amino acid to each tRNA, thus matching an amino acid with a different codon compared to those codons used by life on Earth.

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