IGA 10e SM Chapter 09

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9

Proteins and Their Synthesis

WORKING WITH THE FIGURES

1. The primary protein structure is shown in Figure 9-3(a). Where in the mRNA(near the 5′ or 3′ end) would a mutation in R 2 be encoded?

Answer: As mRNA is translated 5′ to 3′, polypeptides are assembled fromamino end to carboxyl end (the carboxyl end of the growing polypeptide chainis bound to the amino end of the incoming amino acid) Since R 2 is near theamino end of the protein, a mutation for R 2 would be near the 5′ end of themRNA.

2. In this chapter you were introduced to nonsense suppressor mutations in tRNAgenes. However, suppressor mutations also occur in protein-coding genes.Using the tertiary structure of the β subunit of hemoglobin shown in Figure 9-3(c), explain in structural terms how a mutation could cause the loss of globin

protein function. Now explain how a mutation at a second site in the same protein could suppress this mutation and lead to a normal or near-normal protein.

Answer: Proper folding is dependent on amino acid sequence and necessary for protein function. An amino acid replacement that disrupted folding of the β subunit would cause a loss of function for the protein because the correcttertiary structure could not form. If a subsequent mutation in another region ofβ complemented the first mutation by at least partially reestablishing the normalfolding pattern, adequate tertiary structure could form and the first mutationwould be suppressed. For example, if bonding between points A and B resultedin proper folding, a mutation that changed A would disrupt folding and causeloss of function. A subsequent mutation that changed B so it could bond withmutant A would reestablish the normal folding and normal function.

3. Using the quarternary structure of hemoglobin shown in Figure 9-3(d), explainin structural terms how a mutation in the β subunit protein could be suppressed by a mutation in the α subunit gene.



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Answer: A mutation in the β subunit that prevented bonding of α and β subunitswould prevent formation of the quaternary structure and block protein function.A complementary mutation in the α subunit that established the capacity to bond with the mutant β subunit would allow formation of a normal quaternarystructure and effectively suppress the mutation in β.

4. Transfer RNAs (tRNAs) are examples of RNA molecules that do not encode protein. Based on Figures 9-6 and 9-8, what is the significance of the sequenceof tRNA molecules? What do you predict would be the impact on translation ofa mutation in one of the bases of one of the stems in the tRNA structure? On themutant organism?

Answer: The sequence of the tRNA molecules determines the three dimensionalstructure, producing the characteristic L shape. The conservation of the L shapeamong the different tRNAs implies an important function. If one of the bases inone of the stems were mutant, the formation of the L shape would likely be

impaired, reducing or eliminating the capacity of the tRNA to act as an adaptermolecule. For example, a mutant tRNA might not be able to bind with thesynthetase molecule to become charged, or it might not form a sufficientlystable complex with the ribosome during translation. In the first case, themutant tRNA would not participate in translation at all; in the second, it coulddisrupt translation when it was inserted. The overall effect on the mutantorganism would probably be minimal because there are several copies of tRNAgenes and the normal function would be served by the remaining normal copiesof the gene.

5. Ribosomal RNAs (rRNAs) are another example of a functional RNA molecule.Based on Figure 9-11, what do you think is the significance of the secondarystructure of rRNA?

Answer: The amount of double stranded pairing in the rRNA indicates that alarge part of the molecule will have a double-helical structure. The double-helical regions could potentially interact with ribosomal proteins via their majorgrooves. rRNA could also interact with other RNAs. The size of the rRNA

indicates a complex function. The 16S rRNA contains the ShineDalgarnosequence, which directs the 30S subunit to the start codon, and it could alsostabilize bonding between the 30S subunit and mRNA, and between the 30S

and 50S subunits.

6. The components of prokaryotic and eukaryotic ribosomes are shown in Figure9-10. Based on this figure, do you think that the large prokaryotic ribosomalRNA (23S rRNA) would be able to substitute for the eukaryotic 28S rRNA?Justify your answer.



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Answer: It would be unlikely that the 23S rRNA of prokaryotes would be ableto substitute for the 28S rRNA of eukaryotes. Its different size implies adifferent folding pattern which would affect its affinity for protein componentsof the 60S subunit. Also, the 60S subunit of eukaryotes contains 49 proteins,whereas the 50S of prokaryotes has 31 proteins. Thus, the 23S rRNA would

have to be able to target a related but different set of proteins to functioneffectively. In both prokaryotes and eukaryotes, rRNA and ribosomal proteinshave coevolved as a unit and interact closely during the formation of theribosome and translation. Although specifics about the differences in the rRNAand proteins cannot be determined from the figure, the differences in rRNAsizes and number of protein components between prokaryotes and eukaryotesindicate evolutionary divergence in this intricate interaction. It would not beexpected that a bacterial rRNA would interact functionally with the proteincomponents of a eukaryotic ribosome.

7. In Figure 9-12, is the terminal amino acid emerging from the ribosome encoded

by the 5′ or 3′ end of the mRNA?

Answer: Figure 9-12(b) shows a growing polypeptide chain attached to the Psite. The terminal amino acid shown emerging from the ribosome is at theamino end of the polypeptide and so was the first one added to the polypeptidechain. The amino end of the growing polypeptide chain will be the initialamino acid because the tRNA binds to the carboxyl end, leaving the amino endfree. Since translation begins at the 5′ end of the mRNA and terminates at the 3′ end, this initial amino acid would be encoded at the 5′ end.

8. In Figure 9-12(b), what do you think happens to the tRNA that is released fromthe E site?

Answer: Once the tRNA is released from the E site, it returns to the cytoplasmwhere it will be recharged with another amino acid. In this way, tRNAs arerecycled through the translational machinery.

9. In Figure 9-17, what do you think happens next to the ribosomal subunits afterthey are finished translating that mRNA?

Answer: Once translation is terminated, the ribosomal subunits are releasedfrom the mRNA and the 30S subunit is free to form a new initiation complex.Both subunits will ultimately participate in the translation of another mRNA, but it will not necessarily be the same mRNA.

10. Based on Figure 9-19, can you predict the position of a mutation that wouldaffect the synthesis of one isoform but not the other?



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Answer: Exon 8 (in blue) is present in one isoform only, exon 9 (in green) is present in the other isoform only. A mutation in the 5′ GU in exon 8 would prevent splicing and block synthesis of the blue isoform. This mutation wouldnot affect the green isoform because splicing for the green isoform does not

involve the 5′ GU in exon 8. Likewise, a mutation in the 5′ GU in exon 9 would block synthesis of the green isoform only. Additionally, a nonsense mutation ineither of these exons would block synthesis of that isoform but not the other.

11. Based on Figure 9-24, can you predict the position of a mutation that would produce an active protein that was not directed to the correct location?

Answer: A mutation in the signal sequence could prevent transfer of the proteinto the lumen of the ER, thus preventing it from reaching its proper destination.

BASIC PROBLEMS

Unpacking Problem 12

a. Use the codon dictionary in Figure 9-5 to complete the following table.Assume that reading is from left to right and that the columns representtranscriptional and translational alignments.

C DNA double helix

T G A

C A U mRNA transcribed

G C A Appropriate tRNA anticodon

Tr

p

Tr

p

Amino acids incorporated into

protein

b. Label the 5′ and 3′ ends of DNA and RNA, as well as the amino andcarboxyl ends of the protein.

Answer:3´ CGT ACC ACT GCA 5´ DNA double helix (transcribed strand)5´ GCA TGG TGA CGT 3´ DNA double helix5´ GCA UGG UGA CGU 3´ mRNA transcribed3´ CGU ACC ACU GCA 5´ appropriate tRNA anticodon

NH3 – Ala – Trp – (stop) –COOH amino acids incorporated

13. Consider the following segment of DNA:

5′ GCTTCCCAA 3′ 3′ CGAAGGGTT 5′

Assume that the top strand is the template strand used by RNA polymerase.



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a. Draw the RNA transcribed.b. Label its 5′ and 3′ ends.c. Draw the corresponding amino acid chain.d. Label its amino and carboxyl ends.

Repeat parts a through d , assuming the bottom strand to be the template strand.

Answer:a. and b. 5´ UUG GGA AGC 3´

c. and d. Assuming the reading frame starts at the first base:

NH3 – Leu – Gly – Ser - COOH

For the bottom strand, the mRNA is 5´ GCU UCC CAA 3´ and assumingthe reading frame starts at the first base, the corresponding amino acid chain

is NH3 - Ala - Ser - Gln - COOH.

14. A mutational event inserts an extra nucleotide pair into DNA. Which of thefollowing outcomes do you expect? (1) No protein at all; (2) a protein in whichone amino acid is changed; (3) a protein in which three amino acids arechanged; (4) a protein in which two amino acids are changed; (5) a protein inwhich most amino acids after the site of the insertion are changed.

Answer: (5) With an insertion, the reading frame is disrupted. This will result ina 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 stopsignals in the altered reading frame).

15. Before the true nature of the genetic coding process was fully understood, it was proposed that the message might be read in overlapping triplets. For example,the sequence GCAUC might be read as GCA CAU AUC:

Devise an experimental test of this idea.

Answer: A single nucleotide change should result in three adjacent amino acidchanges in a protein. One and two adjacent amino acid changes would beexpected to be much rarer than the three changes. This is directly the oppositeof what is observed in proteins. Also, given any triplet coding for an aminoacid, the next triplet could only be one of four. For example, if the first is GGG,



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then the next must be GGN (N = any base). This puts severe limits on whichamino acids could be adjacent to each other. You could check amino acidsequences of various proteins to show that this is not the case.

16. In protein-synthesizing systems in vitro, the addition of a specific humanmRNA to the E. coli translational apparatus (ribosomes, tRNA, and so forth)stimulates the synthesis of a protein very much like that specified by themRNA. What does this result show?

Answer: It suggests very little evolutionary change between E. coli and humanswith regard to the translational apparatus. The code is universal, the ribosomesare interchangeable, the tRNAs are interchangeable, and the enzymes involvedare interchangeable. (Initiation of translation in prokaryotes in vivo requiresspecific base-pairing between the 3´ end of the 16s rRNA and a Shine–Dalgarnosequence found in the 5´ untranslated region of the mRNA. A Shine–Dalgarnosequence would not be expected (unless by chance) in a eukaryotic mRNA and

therefore initiation of translation might not occur.)

17. Which anticodon would you predict for a tRNA species carrying isoleucine? Isthere more than one possible answer? If so, state any alternative answers.

Answer: 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´, althoughcomplementary, would also base-pair with 5´ AUG 3´ (methionine) due towobble and therefore would not be an acceptable alternative.

18. a. In how many cases in the genetic code would you fail to know the aminoacid specified by a codon if you knew only the first two nucleotides of thecodon?

b. In how many cases would you fail to know the first two nucleotides of thecodon if you knew which amino acid is specified by it?

Answer:a. By studying the genetic code table provided in the textbook, you will

discover that 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 startingwith UA).

b. If you knew the amino acid, you would not know the first two nucleotides inthe cases of Arg, Ser, and Leu.



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19. Deduce what the six wild-type codons may have been in the mutants that ledBrenner to infer the nature of the amber codon UAG.

Answer: The codon for amber is UAG. Listed below are the amino acids thatwould 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*

In each case, the codon marked by an asterisk would require a single basechange to become UAG.

20. If a polyribonucleotide contains equal amounts of randomly positioned adenineand uracil bases, what proportion of its triplets will encode (a) phenylalanine,(b) isoleucine, (c) leucine, (d) tyrosine?

Answer: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 AUAis (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, ofwhich only UUA can exist. It has a probability of (1/2)(1/2)(1/2) = 1/8.

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.

21. You have synthesized three different messenger RNAs with bases incorporated

in random sequence in the following ratios: (a) 1 U:5 C’s, (b) 1 A:1 C:4 U’s, (c) 1 A:1 C:1 G:1 U. In a protein-synthesizing system in vitro, indicate theidentities and proportions of amino acids that will be incorporated into proteinswhen each of these mRNAs is tested. (Refer to Figure 9-5.)

Answer:a. 1 U : 5 C — The probability of a U is 1/6, and the probability of a C is 5/6.



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Codon Amino acid Probability SumUUU Phe (1/6)(1/6)(1/6) = 0.005 Phe = 0.028UUC Phe (1/6)(1/6)(5/6) = 0.023CCC Pro (5/6)(5/6)(5/6) = 0.578 Pro 0.694CCU Pro (5/6)(5/6)(1/6) = 0.116

UCC Ser (1/6)(5/6)(5/6) = 0.116 Ser = 0.139UCU Ser (1/6)(5/6)(1/6) = 0.023CUC Leu (5/6)(1/6)(5/6) = 0.116 Leu = 0.139CUU Leu (5/6)(1/6)(1/6) = 0.023

1 Phe : 25 Pro : 5 Ser : 5 Leu

b. Using the same method as above, the final answer is 4 stop : 80 Phe : 40 Leu: 24 Ile : 24 Ser : 20 Tyr : 6 Pro : 6 Thr : 5 Asn : 5 His : 1 Lys : 1 Gln.

c. All amino acids are found in the proportions seen in the code table.

22. In the fungus Neurospora, some mutants were obtained that lacked activity for acertain enzyme. The mutations were found, by mapping, to be in either of twounlinked genes. Provide a possible explanation in reference to quaternary protein structure.

Answer: Quaternary structure is due to the interactions of subunits of a protein.In this example, the enzyme activity being studied may be from a proteinconsisting of two different subunits. Both subunits are required for activity. The polypeptides of the subunits are encoded by separate and unlinked genes.

23. A mutant is found that lacks all detectable function for one specific enzyme. Ifyou had a labeled antibody that detects this protein in a Western blot (seeChapter 1), would you expect there to be any protein detectable by the antibodyin the mutant? Explain.

Answer: There are a number of mutational changes that can lead to the absenceof enzymatic function in the product of a gene. Some of these changes wouldresult in the complete absence of protein product and therefore also the absenceof 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 todetectable protein.

24. In a Western blot (see Chapter 1), the enzyme tryptophan synthetase usuallyshows two bands of different mobility on the gel. Some mutants with no



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enzyme activity showed exactly the same bands as the wild type. Other mutantswith no activity showed just the slow band; still others, just the fast band.

a. Explain the different types of mutants at the level of protein structure.b. Why do you think there were no mutants that showed no bands?

Answer: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 withexactly the same bands as wild type) might have mutations that prevent thesubunits from interacting or disrupt the active site of the enzyme.

b. Because the two subunits are encoded by separate genes, the absence of both bands simultaneously would require two independent and raremutagenic events.

25. In the Crick–Brenner experiments described in this chapter, three “insertions”or three “deletions” restored the normal reading frame and the deduction wasthat the code was read in groups of three. Is this deduction really proved by theexperiments? Could a codon have been composed of six bases, for example?

Answer: 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 codonwould have been six bases. Since the mutations were actually adding orsubtracting single bases, the codon is indeed three bases.

26. A mutant has no activity for the enzyme isocitrate lyase. Does this result provethat the mutation is in the gene encoding isocitrate lyase?

Answer: No. The enzyme may require post-translational modification to beactive. Mutations in the enzymes required for these modifications would notmap to the isocitrate lyase gene.

27. A certain nonsense suppressor corrects a nongrowing mutant to a state that is

near, but not exactly, wild type (it has abnormal growth). Suggest a possiblereason why the reversion is not a full correction.

Answer: A nonsense suppressor is a mutation in a tRNA such that its anticodoncan base-pair with a stop codon. In this way, a mutant stop codon (nonsensemutation) 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 inthat position in the original gene. For example, the codon UCG (serine) is



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instead UAG in the nonsense mutant. The suppressor mutation could be intRNA for tryptophan such that its anticodon now recognizes UAG instead ofUGG. During translation in the double mutant, the machinery puts tryptophaninto 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 wildtype.” Another explanation is that translation of the mutant gene is not asefficient and that premature termination still occurs some of the time. Thiswould lead to less product and, again, a state that is “not exactly wild type.”

28. In bacterial genes, as soon as any partial mRNA transcript is produced by theRNA polymerase system, the ribosome jumps on it and starts translating. Drawa diagram of this process, identifying 5′ and 3′ ends of mRNA, the COOH and NH2 ends of the protein, the RNA polymerase, and at least one ribosome. (Whycouldn’t this system work in eukaryotes?)

Answer:

RNA

DNA

5´

RNA polymeraseU

C

C

G G

A

A

A

TT GC C

UA

G G

G C

C

C T A G G C T G C A

ribosome

NH2

polypeptide

In eukaryotes, transcription occurs within the nucleus while translation occursin the cytoplasm. Thus, the two processes cannot occur together.

29. In a haploid, a nonsense suppressor su1 acts on mutation 1 but not on mutation2 or 3 of gene P. An unlinked nonsense suppressor su2 works on P mutation 2 but not on 1 or 3. Explain this pattern of suppression in regard to the nature ofthe mutations and the suppressors.

Answer: Assuming that the three mutations of gene P are all nonsensemutations, 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 ambermutants but not opal or ochre.

30. In vitro translation systems have been developed in which specific RNAmolecules can be added to a test tube containing a bacterial cell extract thatincludes all the components needed for translation (ribosomes, tRNAs, aminoacids). If a radioactively labeled amino acid is included, any protein translated



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from that RNA can be detected and displayed on a gel. If a eukaryotic mRNA isadded to the test tube, would radioactive protein be produced? Explain.

Answer: Initiation of translation in prokaryotes requires specific base-pairing between the 3´ end of the 16s rRNA and a Shine–Dalgarno sequence found in

the 5´ untranslated region of the mRNA. A Shine–Dalgarno sequence would not be expected (unless by chance) in a eukaryotic mRNA and therefore initiationof translation would not occur.

31. In a eukaryotic translation system (containing a cell extract from a eukaryoticcell) comparable with that in Problem 30, would a protein be produced by a bacterial RNA? If not, why not?

Answer: Initiaton of translation in eukaryotes requires initiation factors (eIF4a, b, and G) that associate with the 5´ cap of the mRNA. Because prokaryoticmRNAs are not capped, translation would not initiate.

32. Would a chimeric translation system containing the large ribosomal subunitfrom E. coli and the small ribosomal subunit from yeast (a unicellulareukaryote) be able to function in protein synthesis? Explain why or why not.

Answer: Not likely. Although the steps of translation and the components ofribosomes are similar in both eukaryotes and prokaryotes, the ribosomes are notidentical. The sizes of both subunits are larger in eukaryotes and the manyspecific and intricate interactions that must take place between the small andlarge subunits would not be possible in a chimeric system.

33. Mutations that change a single amino acid in the active site of an enzyme canresult in the synthesis of wild-type amounts of an inactive enzyme. Can youthink of other regions in a protein where a single amino acid change might havethe same result?

Answer: Single amino acid changes can result in changes in protein folding, protein targeting, or post-translational modifications. Any of these changescould give the results indicated.

34. What evidence supports the view that ribosomal RNAs are a more importantcomponent of the ribosome than the ribosomal proteins?

Answer: The first indication of rRNAs importance was the discovery ofribozymes. Recently, structural studies have shown that both the decodingcenter in the 30S subunit and the peptidyl transferase center in the 50S subunit



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are composed entirely of rRNA and that the important contacts in these centersare all tRNA/rRNA contacts.

35. Explain why antibiotics, such as erythromycin and Zithromax, that bind the

large ribosomal subunit do not harm us.

Answer: Antibiotics need to selectively target bacterial structures and functionsthat are essential for life but unique or sufficiently different from the equivalentstructure and functions of their animal hosts. The large bacterial ribosomalsubunit fits these criteria as its function is obviously essential yet its structure issufficiently different from the large eukaryotic ribosomal subunit. While thesteps of protein synthesis are similar overall, eukaryotic ribosomes have largerand more numerous components. These differences make it possible to developdrugs that specifically bind bacterial ribosomes but have little or no affinity foreukaryotic ribosomes.

36. Why do multicellular eukaryotes need to have hundreds of kinase-encodinggenes?

Answer: Recent studies indicate that most proteins function by interacting withother 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 thatcatalyze phosphorylations. Therefore, the complexity of the interactomenecessary for the complexity of multicellularity requires the very large numberof kinase-encoding genes.

37. Our immune system makes many different proteins that protect us from viraland bacterial infection. Biotechnology companies must produce large quantitiesof these immune proteins for human testing and eventual sale to the public. Tothis end, their scientists engineer bacterial or human cell cultures to expressthese immune proteins. Explain why proteins isolated from bacterial culturesare often inactive, whereas the same proteins isolated from human cell culturesare active (functional).

Answer: Bacterial and human cell cultures are both capable of producing the

same polypeptide from the same mRNA, but that does not mean that theresulting protein will be active. Many proteins require posttranslational processing to become functional, and the enzymes and control necessary forsuch processing is not universal. The proteins produced and isolated from ahuman cell culture system will contain the necessary posttranslationalmodifications necessary for human protein function.



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CHALLENGING PROBLEMS

38. A single nucleotide addition and a single nucleotide deletion approximately 15sites apart in the DNA cause a protein change in sequence from

Lys–Ser–Pro–Ser–Leu–Asn–Ala–Ala–LystoLys–Val–His–His–Leu–Met–Ala–Ala–Lys

a. What are the old and new mRNA nucleotide sequences? (Use the codondictionary in Figure 9-5.)

b. Which nucleotide has been added and which has been deleted?

(Problem 38 is from W. D. Stansfield, Theory and Problems of Genetics. McGraw-Hill, 1969.)

Answer:

a. and b. The goal of this type of problem is to align the two sequences. Youare told that there is a single nucleotide addition and single nucleotidedeletion, so look for single base differences that effect this alignment. Theseshould be located where the protein sequence changes (i.e., between Lys-Serand Asn-Ala). Remember also that the genetic code is redundant. (N = any base)

AA UCN CCN UCN CUN AA GCN GCN AA A

G

A

G

AGUC

AGUC

UU AG

Lys Ser Pro Ser Leu Asn Ala Ala Lys

Lys Val His His Leu Met Ala Ala Lys

AA GUN CA CA CUN AUG GCN GCN AA AG

AG

UC

UC

UU AG

U

C

–

+

AA AGU CCA UCA CUU AAU GCN GCN AA AG

AG

Base deleted

Old:

New: AA GUC CAU CAC UUA AUG GCN GCN AA AG

AG

Base added

39. You are studying an E. coli gene that specifies a protein. A part of its sequenceis



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–Ala–Pro–Trp–Ser–Glu–Lys–Cys–His–

You recover a series of mutants for this gene that show no enzymatic activity.By isolating the mutant enzyme products, you find the following sequences:

Mutant 1: –Ala–Pro–Trp–Arg–Glu–Lys–Cys–His–

Mutant 2: –Ala–Pro–

Mutant 3: –Ala–Pro–Gly–Val–Lys–Asn–Cys–His–

Mutant 4: –Ala–Pro–Trp–Phe–Phe–Thr–Cys–His–

What is the molecular basis for each mutation? What is the DNA sequence thatspecifies this part of the protein?

Answer:Mutant 1: A simple substitution of Arg for Ser exists, suggesting a nucleotidechange. Two codons for Arg are AGA and AGG, and one codon for Ser isAGU. The U of the Ser codon could have been replaced by either an A or a G.

Mutant 2: The Trp codon (UGG) changed to a stop codon (UGA or UAG).

Mutant 3: Two frameshift mutations occurred:

5´–GCN CCN (–U)GGA GUG AAA AA(+U or C) UGU(or C) CAU(or C)–3´.

Mutant 4: An inversion occurred after Trp and before Cys. The DNA originalsequence (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–5´ˆ ˆ

Therefore, the complementary RNA sequence was

5´–GCN CCN UGG UUU UUC ACU UGU/C CAU/C–3´ˆ ˆ



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40. Suppressors of frameshift mutations are now known. Propose a mechanism fortheir action.

Answer: 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 wouldsuppress the insertion. Alterations in the ribosome can also induceframeshifting.

41. Consider the gene that specifies the structure of hemoglobin. Arrange thefollowing events in the most likely sequence in which they would take place.

a. Anemia is observed.b. The shape of the oxygen-binding site is altered.c. An incorrect codon is transcribed into hemoglobin mRNA.d. The ovum (female gamete) receives a high radiation dose.

e. An incorrect codon is generated in the DNA of the hemoglobin gene.f. A mother (an X-ray technician) accidentally steps in front of an operating

X-ray generator.g. A child dies.h. The oxygen-transport capacity of the body is severely impaired.i. The tRNA anticodon that lines up is one of a type that brings an unsuitable

amino acid. j. Nucleotide-pair substitution occurs in the DNA of the gene for hemoglobin.

Answer: f, d, j, e, c, i, b, h, a, g

42. An induced cell mutant is isolated from a hamster tissue culture because of its

resistance to -amanitin (a poison derived from a fungus). Electrophoresisshows that the mutant has an altered RNA polymerase; just one electrophoretic band is in a position different from that of the wild-type polymerase. The cellsare presumed to be diploid. What do the results of this experiment tell youabout ways in which to detect recessive mutants in such cells?

Answer: Cells in long-established culture lines usually are not fully diploid. Forreasons 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 recessivealleles.

43. A double-stranded DNA molecule with the sequence shown here produces, invivo, a polypeptide that is five amino acids long.



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TACATGATCATTTCACGGAATTTCTAGCATGTAATGTACTAGTAAAGTGCCTTAAAGATCGTACAT

a. Which strand of DNA is transcribed and in which direction?b. Label the 5′ and the 3′ ends of each strand.

c. If an inversion occurs between the second and the third triplets from the leftand right ends, respectively, and the same strand of DNA is transcribed,how long will the resultant polypeptide be?

d. Assume that the original molecule is intact and that the bottom strand istranscribed from left to right. Give the base sequence, and label the 5 ′ and 3′ ends of the anticodon that inserts the fourth amino acid into the nascent polypeptide. What is this amino acid?

Answer:a. and b. The sequence of double-stranded DNA is as follows:

5´—TAC ATG ATC ATT TCA CGG AAT TTC TAG CAT GTA—3´

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 writtenabove in reverse order from your text.

c. Remember that polarity must be taken into account. The inversion isDNA 5´ TAC ATG CTA GAA ATT CCG TGA AAT GAT CAT GTA 3´RNA 3´ –GAU CUU UAA GGC ACU UUA CUA GUA– 5ámino acids HOOC 7 6 5 4 3 2 1 –NH3

d. DNA 3ÁTG TAC TAG TAA AGT GCC TTA AAG ATC GTA CAT 5´mRNA 5ÚAC 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).

44. One of the techniques used to decipher the genetic code was to synthesize polypeptides in vitro, with the use of synthetic mRNA with various repeating base sequences—for example, (AGA) n, which can be written out asAGAAGAAGAAGAAGA. . . . Sometimes the synthesized polypeptide



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contained just one amino acid (a homopolymer), and sometimes it containedmore than one (a heteropolymer), depending on the repeating sequence used.Furthermore, sometimes different polypeptides were made from the samesynthetic mRNA, suggesting that the initiation of protein synthesis in thesystem in vitro does not always start on the end nucleotide of the messenger.

For example, from (AGA)n, three polypeptides may have been made: aa1 homopolymer (abbreviated aa1 –aa1), aa2 homopolymer (aa2 –aa2), and aa3 homopolymer (aa3 –aa3). These polypeptides probably correspond to thefollowing readings derived by starting at different places in the sequence:

AGA AGA AGA AGA . . .GAA GAA GAA GAA . . .AAG AAG AAG AAG . . .

The following table shows the actual results obtained from the experiment done by Khorana.

SyntheticmRNA

Polypeptide(s) synthesized

(UC) n (Ser–Leu)

(UG) n (Val–Cys)

(AC) n (Thr–His)

(AG) n (Arg–Glu)

(UUC) n (Ser–Ser) and (Leu–Leu) and (Phe–Phe)

(UUG) n (Leu–Leu) and (Val–Val) and (Cys–Cys)

(AAG) n (Arg–Arg) and (Lys–Lys) and (Glu–Glu)

(CAA) n (Thr–Thr) and (Asn–Asn) and (Gln–Gln)

(UAC) n (Thr–Thr) and (Leu–Leu) and (Tyr–Tyr)

(AUC) n (Ile–Ile) and (Ser–Ser) and (His–His)(GUA) n (Ser–Ser) and (Val–Val)

(GAU) n (Asp–Asp) and (Met–Met)

(UAUC) n (Tyr–Leu–Ser–Ile)

(UUAC) n (Leu–Leu–Thr–Tyr)

(GAUA) n None

(GUAA) n None

Note: The order in which the polypeptides or amino acids are listed in the tableis not significant except for (UAUC) n and (UUAC) n.

a. Why do (GUA) n and (GAU) n each encode only two homopolypeptides?b. Why do (GAUA) n and (GUAA) n fail to stimulate synthesis?c. Assign an amino acid to each triplet in the following list. Bear in mind that

there are often several codons for a single amino acid and that the first twoletters in a codon are usually the important ones (but that the third letter isoccasionally significant). Also, remember that some very different-lookingcodons sometimes encode the same amino acid. Try to carry out this taskwithout consulting Figure 9-5.



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AUG GAU UUG AACGUG UUC UUA CAAGUU CUC AUC AGAGUA CUU UAU GAG

UGU CUA UAC GAACAC UCU ACU UAGACA AGU AAG UGA

To solve this problem requires both logic and trial and error. Don’t bedisheartened: Khorana received a Nobel Prize for doing it. Good luck!

(Problem 44 is from J. Kuspira and G. W. Walker, Genetics: Questions and

Problems. McGraw-Hill, 1973.)

Answer:a. (GAU)n codes for Aspn (GAU)n, Metn (AUG)n, and stopn (UGA)n. (GUA)n

codes for Valn (GUA)n, Ser n (AGU)n, and stopn (UAG)n.One reading frame in each contains a stop codon.

b. Each of the three reading frames contains a stop codon.

c. The way to approach this problem is to focus initially on one amino acid at atime. For instance, line 4 indicates that the codon for Arg might be AGA orGAG. Line 7 indicates it might be AAG, AGA, or GAA. Therefore, Arg isat least AGA. That also means that Glu is GAG (line 4). Lys and Glu can beAAG or GAA (line 7). Because no other combinations except the onesalready mentioned result in either Lys or Glu, no further decision can be

made with respect to them. However, taking wobble into consideration, Glumay 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 isSer.

The footnote states that lines 13 and 14 are in the correct order. In line 13, ifUCU 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 anamino acid to each codon.

IGA 10e SM Chapter 09

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