kupdf com lehninger principles of biochemistry test bank ch 27pdf

Chapter 27 Protein Metabolism
Multiple Choice Questions
1. The genetic code
Page: 1038
Difficulty: 2 Ans: C
A certain bacterial mRNA is known to represent only one gene and to contain about 800 nucleotides.
If you assume that the average amino acid residue contributes 110 to the peptide molecular weight,
the largest polypeptide that this mRNA could code for would have a molecular weight of about:
A)
B)
C)
D)
E)

800.
5,000.
30,000.
80,000.
An upper limit cannot be determined from the data given.

2. The genetic code
Page: 1038
Difficulty: 2 Ans: C
Assuming that the average amino acid residue contributes 110 to the peptide molecular weight, what
will be the minimum length of the mRNA encoding a protein of molecular weight 50,000?
A)
B)
C)
D)
E)

133 nucleotides
460 nucleotides
1,400 nucleotides
5,000 nucleotides
A minimum length cannot be determined from the data given.

3. The genetic code
Pages: 1039-1044 Difficulty: 3 Ans: D
Which of the following are features of the wobble hypothesis?
A)
B)
C)
D)
E)

A naturally occurring tRNA exists in yeast that can read both arginine and lysine codons.
A tRNA can recognize only one codon.
Some tRNAs can recognize codons that specify two different amino acids, if both are nonpolar.
The “wobble” occurs only in the first base of the anticodon.
The third base in a codon always forms a normal Watson-Crick base pair.

4. The genetic code
Page: 1039
Difficulty: 2 Ans: C
Which one of the following is true about the genetic code?
A)
B)
C)
D)

All codons recognized by a given tRNA encode different amino acids.
It is absolutely identical in all living things.
Several different codons may encode the same amino acid.
The base in the middle position of the tRNA anticodon sometimes permits “wobble” base pairing
with 2 or 3 different codons.
E) The first position of the tRNA anticodon is always adenosine.


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Chapter 27 Protein Metabolism

5. Protein synthesis
Page: 1045
Difficulty: 1 Ans: D
Which one of the following statements about ribosomes is true?
A)
B)
C)
D)
E)

The large subunit contains rRNA molecules, the small subunit does not.
The RNA in ribosomes plays a structural, not catalytic, role.
There are about 25 of them in an E. coli cell.
There are two major subunits, each with multiple proteins.
They are relatively small, with molecular weights less than 10,000.

6. Protein synthesis
Page: 1049

Difficulty: 2 Ans: A
Which of the following statements about tRNA molecules is false?
A) A, C, G, and U are the only bases present in the molecule.
B) Although composed of a single strand of RNA, each molecule contains several short, doublehelical regions.
C) Any given tRNA will accept only one specific amino acid.
D) The amino acid attachment is always to an A nucleotide at the 3' end of the molecule.
E) There is at least one tRNA for each of the 20 amino acids.
7. Protein synthesis
Page: 1050
Difficulty: 2 Ans: E
Which of the following statements about the tRNA that normally accepts phenylalanine is false?
(mRNA codons for phenylalanine are UUU and UUC.)
A)
B)
C)
D)
E)

It interacts specificially with the Phe synthetase.
It will accept only the amino acid phenylalanine.
Its molecular weight is about 25,000.
Phenylalanine can be specifically attached to an —OH group at the 3' end.
The tRNA must contain the sequence UUU.

8. Protein synthesis
Page: 1051
Difficulty: 2 Ans: E
Which of the following is not true of tRNA molecules?
A)
B)

C)
D)
E)

The 3'-terminal sequence is —CCA.
Their anticodons are complementary to the triplet codon in the mRNA.
They contain more than four different bases.
They contain several short regions of double helix.
With the right enzyme, any given tRNA molecule will accept any of the 20 amino acids.

9. Protein synthesis
Page: 1051
Difficulty: 2 Ans: A
Aminoacyl-tRNA synthetases (amino acid activating enzymes):
A)
B)
C)
D)
E)

“recognize” specific tRNA molecules and specific amino acids.
in conjunction with another enzyme attach the amino acid to the tRNA.
interact directly with free ribosomes.
occur in multiple forms for each amino acid.
require GTP to activate the amino acid.


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315

10. Protein synthesis
Page: 1051
Difficulty: 2 Ans: B
In E. coli, aminoacyl-tRNA synthetases:
A)
B)
C)
D)
E)

activate amino acids in 12 steps.
are amino acid–specific; there is at least one enzyme specific for each amino acid.
fall into two classes, each of which attaches amino acids to different ends of the tRNA.
have no proofreading activities.
require a tRNA, an amino acid, and GTP as substrates.

11. Protein synthesis
Pages: 1051-1054 Difficulty: 2 Ans: D
Which of the following statements about aminoacyl-tRNA synthetases is false?
A)
B)
C)
D)
E)

Some of the enzymes have an editing/proofreading capability.
The enzyme attaches an amino acid to the 3' end of a tRNA.
The enzyme splits ATP to AMP + PPi.
The enzyme will use any tRNA species, but is highly specific for a given amino acid.

There is a different synthetase for every amino acid.

12. Protein synthesis
Page: 1051
Difficulty: 2 Ans: D
The enzyme that attaches an amino acid to a tRNA (aminoacyl-tRNA synthetase):
A)
B)
C)
D)
E)

always recognizes only one specific tRNA.
attaches a specific amino acid to any available tRNA species.
attaches the amino acid at the 5' end of the tRNA.
catalyzes formation of an ester bond.
splits ATP to ADP + Pi.

13. Protein synthesis
Page: 1051
Difficulty: 2 Ans: D
In the “activation” of an amino acid for protein synthesis:
A)
B)
C)
D)
E)

leucine can be attached to tRNAPhe, by the aminoacyl-tRNA synthetase specific for leucine.
methionine is first formylated, then attached to a specific tRNA.

the amino acid is attached to the 5' end of the tRNA through a phosphodiester bond.
there is at least one specific activating enzyme and one specific tRNA for each amino acid.
two separate enzymes are required, one to form the aminoacyl adenylate, the other to attach the
amino acid to the tRNA.

14. Protein synthesis
Page: 1056
Difficulty: 2 Ans: A
Formation of the ribosomal initiation complex for bacterial protein synthesis does not require:
A)
B)
C)
D)
E)

EF-Tu.
formylmethionyl tRNAfMet.
GTP.
initiation factor 2 (IF-2).
mRNA.


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Chapter 27 Protein Metabolism

15. Protein synthesis
Page: 1058
Difficulty: 2 Ans: D
In bacteria the elongation stage of protein synthesis does not involve:

A)
B)
C)
D)
E)

aminoacyl-tRNAs.
EF-Tu.
GTP.
IF-2.
peptidyl transferase.

16. Protein synthesis
Page: 1058
Difficulty: 2 Ans: E
Which one of the following statements about the elongation phase of protein synthesis is true?
A)
B)
C)
D)

At least five high-energy phosphoryl groups are expended for each peptide bond formed.
During elongation, incoming aminoacylated tRNAs are first bound in the P site.
Elongation factor EF-Tu facilitates translocation.
Peptidyl transferase catalyzes the attack of the carboxyl group of the incoming amino acid on an
ester linkage in the nascent polypeptide.
E) Peptidyl transferase is a ribozyme.
17. Protein synthesis
Page: 1059
Difficulty: 2 Ans: C

Which of the following statements about bacterial mRNA is true?
A)
B)
C)
D)
E)

A ribosome usually initiates translation near the end of the mRNA that is synthesized last.
An mRNA is never degraded but is passed on to the daughter cells at cell division.
During polypeptide synthesis, ribosomes move along the mRNA in the direction 5' → 3'.
Ribosomes cannot initiate internally in a polycistronic transcript.
The codon signaling peptide termination is located in the mRNA near its 5' end.

18. Protein synthesis
Page: 1059
Difficulty: 2
Bacterial ribosomes:
A)
B)
C)
D)
E)

Ans: B

bind tightly to specific regions of DNA, forming polysomes.
contain at least one catalytic RNA molecule (ribozyme).
contain three species of RNA and five different proteins.
have specific, different binding sites for each of the 20 tRNAs.
require puromycin for normal function.

19. Protein synthesis
Page: 1062
Difficulty: 1 Ans: B
The large structure consisting of a mRNA molecule being translated by multiple copies of the
macromolecular complexes that carry out protein synthesis is called a:
A)
B)
C)
D)
E)

lysosome.
polysome.
proteosome.
ribosome.
synthosome.


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317

20. Protein synthesis
Page: 1061
Difficulty: 3 Ans: C
It is possible to convert the Cys that is a part of Cys-tRNACys to Ala by a catalytic reduction. If the
resulting Ala-tRNACys were added to a mixture of (1) ribosomes, (2) all the other tRNAs and amino
acids, (3) all of the cofactors and enzymes needed to make protein in vitro, and (4) mRNA for
hemoglobin, where in the newly synthesized hemoglobin would the Ala from Ala-tRNACys be

incorporated?
A)
B)
C)
D)
E)

Nowhere; this is the equivalent of a nonsense mutation
Wherever Ala normally occurs
Wherever Cys normally occurs
Wherever either Ala or Cys normally occurs
Wherever the dipeptide Ala-Cys normally occurs

21. Protein synthesis
Pages: 1066-1067 Difficulty: 2 Ans: C
Which one of the following antibiotics does not function by interfering with the translational process?
A)
B)
C)
D)
E)

Chloramphenicol
Cycloheximide
Penicillin
Puromycin
Streptomycin

22. Protein targeting and degradation
Page: 1069

Difficulty: 2 Ans: D
Which of the following is true about the sorting pathway for proteins destined for incorporation into
lysosomes or the plasma membrane of eukaryotic cells?
A)
B)
C)
D)

Binding of SRP to the signal peptide and the ribosome temporarily accelerates protein synthesis.
The newly synthesized polypeptides include a signal peptide at their carboxyl termini.
The signal peptide is cleaved off inside the mitochondria by signal peptidase.
The signal recognition particle (SRP) binds to the signal peptide soon after it appears outside the
ribosome.
E) The signal sequence is added to the polypeptide in a posttranslational modification reaction.
23. Protein targeting and degradation
Pages: 1069-1070 Difficulty: 2 Ans: A
Glycosylation of proteins inside the endoplasmic reticulum does not involve:
A)
B)
C)
D)
E)

a His residue on the protein.
an Asn residue on the protein.
dolichol phosphate.
glucose.
N-acetylglucosamine.

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Chapter 27 Protein Metabolism

24. Protein targeting and degradation
Page: 1070
Difficulty: 2 Ans: E
Posttranslational glycosylation of proteins is inhibited specifically by:
A)
B)
C)
D)
E)

chloramphenicol.
cycloheximide.
puromycin.
streptomycin.
tunicamycin.

25. Protein targeting and degradation
Page: 1071
Difficulty: 2 Ans: D
The signal sequences that direct proteins to the nucleus are:
A)
B)
C)
D)
E)

always at the amino terminus of the targeted protein.
cleaved after the protein arrives in the nucleus.
glycosyl moieties containing mannose 6-phosphate residues.
not located at the ends of the peptide, but in its interior.
the same as those that direct certain proteins to lysosomes.

26. Protein targeting and degradation
Pages: 1072-1074 Difficulty: 3 Ans: C
The pathway for polypeptides exported from E. coli includes the following steps, which occur in what
order for correct export?
1.
2.
3.
4.
A)
B)
C)
D)
E)

A chaperone, SecA, binds to the polypeptide.
A chaperone, SecB, binds to the polypeptide.
ATP is hydrolyzed by Sec A.
SecA pushes 20 amino acids of the polypeptide into the translocation complex.

1, 2, 3, 4
1, 2, 4, 3
2, 1, 4, 3
2, 3, 1, 4
3, 1, 4, 2

27. Protein targeting and degradation
Page: 1076
Difficulty: 3 Ans: D
Ubiquitin-mediated protein degradation is a complex process, and many of the signals remain
unknown. One known signal involves recognition of amino acids in a processed protein that are
either stabilizing (Ala, Gly, Met, Ser, etc.) or destabilizing (Arg, Asp, Leu, Lys, Phe, etc.), and are
located at:
A)
B)
C)
D)
E)

a helix-turn-helix motif in the protein.
a lysine-containing target sequence in the protein.
a zinc finger structure in the protein.
the amino-terminus of the protein.
the carboxy-terminus of the protein.


Chapter 27 Protein Metabolism

319

Short Answer Questions
28. The genetic code
Pages: 1035-1036 Difficulty: 2
Outline one of the experimental methods providing evidence that the genetic code was a triplet code.
Ans: When one or two nucleotides were added to or deleted from a gene, the resulting mRNA

produced a protein with a different amino acid sequence after the deletion or insertion. When three
nucleotides were added or deleted, the resulting protein had a normal sequence except for the
insertion or deletion of a single amino acid residue.
29. The genetic code
Page: 1036
Difficulty: 3
Describe, succinctly, two ways in which synthetic polynucleotides were used in solving the genetic
code (you need not describe how the synthetic polynucleotides were made).
Ans: (1) When synthetic polymers of only one nucleotide were used as mRNA in vitro, only one of
the 20 amino acids was converted into protein. For example, poly(U) (containing only the codon
UUU) directed the synthesis of polyphenylalanine, showing that UUU encodes Phe. (2)
Trinucleotides of known sequence were used to stimulate aminoacyl-tRNA binding to ribosomes.
Because only that aminoacyl-tRNA whose anticodon matched the trinucleotide “mRNA” was bound,
the coding specificity of each sequence of three bases could be determined by determining which of
the 20 aminoacyl-tRNAs bound. (3) Random polymers of RNA containing known ratios of
nucleotides (e.g., 70% A and 30% T) generate only certain codons in predictable ratios. The
identities and ratios of the amino acids specified by such polymers provided important clues that
helped solve the genetic code. (4) Additional assignments were made possible using synthetic
oligonucleotides containing repeats of specific two, three, or four base pair sequences.
30. The genetic code
Page: 1036
Difficulty: 3
You have isolated a fragment of viral DNA that totally encodes at least two proteins, 120 and 80
amino acids long. The DNA fragment is 400 base pairs long. (a) Why might you consider this
unusual? (b) You sequence the two proteins and find no sequence homology. Propose a model to
account for these findings.
Ans: (a) Two distinct proteins of these sizes should require mRNAs of 360 and 240 base pairs
because each amino acid residue requires 3 base pairs to code for it. (b) No homology means that the
smaller protein cannot be derived from the larger by proteolysis; if it were, there would be 80 amino
acid residues of identical sequence in the two proteins. One possible explanation is that the two genes

coding for these proteins overlap and are read in different reading frames.
31. The genetic code
Page: 1038
Difficulty: 2
The template strand of a segment of double-stranded DNA contains the sequence:
(5')CTT TGA TAA GGA TAG CCC TTC
(a) What is the base sequence of the mRNA that can be transcribed from this strand? (b) What amino
acid sequence could be coded by the mRNA base sequence in (a), using only the first reading frame
starting at the 5' end? (Refer to Fig. 27-7, p. 1038.) (c) Suppose the other (complementary) strand is
used as a template for transcription. What is the amino acid sequence of the resulting peptide, again
starting from the 5' end and using only the first reading frame?


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Chapter 27 Protein Metabolism

Ans:
(a) (5')GAA GGG CUA UCC UUA UCA AAG(3’)
(b) Glu-Gly-Leu-Ser-Leu-Ser-Lys
(c) The codons translate to Leu-Stop-Stop. No peptide would be produced because of the stop
codons.
(See also Fig. 27-6, p. 1038.)
32. The genetic code
Page: 1038

Difficulty: 3

Describe the possible outcomes that could occur because of a single base change in an mRNA
Ans: The most likely result is a single amino acid change in the encoded protein, when a codon is

altered to one of another amino acid. However, some nucleotide changes will be “silent” and not
change the protein, if the altered codon still specifies the original amino acid. Conversion to a
nonsense codon will result in a truncated polypeptide, while alteration of the normal stop codon to a
“sense” codon will result in a lengthened protein. Alternation of the initiaton codon may result in the
total failure to translate, and result in no protein product. Changes in mRNA sequence outside the
protein-coding region may affect translational efficiency, splicing, or mRNA turnover rates.
33. The genetic code
Page: 1038
Difficulty: 3
The following sequence of four amino acids occurred in the structure of a polypeptide found in a
wild-type organism: Leu-Ser-Ile-Arg. Several mutants were isolated, each of which carried a single
base pair change in the region of DNA that coded for this amino acid sequence. Their corresponding
amino acid sequences are:
Mutant
1
2
3
4
5

MET-Ser-Ile-Arg
Leu-TRP-Ile-Arg
Leu-Ser-ARG-Arg
Leu-Ser-Ile-PRO
Leu-Ser-Ile-TRP

What was the nucleotide sequence of the region of mRNA that coded for the amino acid sequence in
the wild-type organism? (Refer to Fig. 27-6, p. 1038.)
Ans: (5')C or (5')U UG UCG AUA CGG
34. The genetic code

Pages: 1039-1044 Difficulty: 2
In protein synthesis, 61 codons specify the 20 amino acids. Base pairing between the codon and the
tRNA anticodon assures that the correct amino acid will be inserted into the nascent polypeptide
chain. Why then does the cell require only 32 different tRNAs to recognize 61 different codons?
Ans: Certain tRNAs have the unusual nucleotide inosinate in the first anticodon position. Because
inosinate can base pair with A, U, or C, a tRNA containing hypoxanthine can recognize three
different codons. In each recognized codon, there is a standard anticodon-codon base pair with the
first two bases of the codon; “wobble” in the third base pair allows one tRNA to read three different
codons. Similarly, tRNAs with U or G in the first anticodon position also exhibit a wobble effect that
permits pairing with two different codons.


Chapter 27 Protein Metabolism

321

35. Protein synthesis
Pages: 1045-1049 Difficulty: 1
Indicate whether the following statements are true (T) or false (F).
___A ribosome is the complex within which protein synthesis occurs.
___Ribosomes contain many separate proteins.
___The three ribosomal RNAs in a bacterial ribosome are distributed in three separate, large
ribosomal subunits.
___There are four binding sites for aminoacyl-tRNAs on a ribosome.
Ans: T; T; F; F
36. Protein synthesis
Pages: 1051-1053 Difficulty: 2
The process of charging tRNAs with their cognate amino acids involves multiple proofreading steps
to increase the overall fidelity. Briefly describe these steps.
Ans: There are two main stages of selection: 1) the synthetase strongly favors activation of the

correct amino acid to become aminoacyl-AMP (incorrect amino acids are very poorly activated) and
2) when the correct uncharged tRNA is bound to the enzyme, only the correct aminoacyl-AMP is
tolerated in the “proofreading” active site (incorrect aminoacyl-AMPs, though they may become
bound, are rapidly hydrolyzed). In addition, for most synthetases, if their tRNA does manage to
become acylated by the wrong amino acid, that product is also rapidly hydrolyzed.
37. Protein synthesis
Pages: 1053-1054 Difficulty: 2
The recognition of an amino acid by its cognate aminoacyl-tRNA synthetase is said to involve a
“second genetic code”. What is meant by this?
Ans: Each of the 20 amino acids has a unique synthetase, but many have multiple cognate tRNAs that
must be charged (aminoacylated). The “second code” refers to features common to all tRNAs that
carry the same amino acid, making them specifically recognizable by the correct synthetase. These
features occur at different places in different tRNA classes, and may be as simple as a single G-U
basepair, or require ten or more specific nucleotides throughout the sequence. (See Fig. 27-16, p.
1053.)
38. Protein synthesis
Page: 1055
Difficulty: 3
In 1961, Howard Dintzis carried out an experiment that defined the direction of polypeptide chain
growth during protein synthesis in cells. The experiment involved the analysis of hemoglobin
molecules that were being synthesized in reticulocytes in the presence of radioactive amino acids.
Describe the analysis and how it demonstrated the direction of chain growth.
Ans: In the experiment, only completed polypeptides were isolated and analyzed for the amount of
redioactivity in different regions. The greatest radioactivity should be located in the region furthest
from the initiation point and the least radioactivity in the region closest to the initiation point. The
opposite is true for the termination point. The analysis showed that polypeptide synthesis was
initiated at the amino terminus and terminated at the carboxyl terminus.


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Chapter 27 Protein Metabolism

39. Protein synthesis
Page: 1056-1058 Difficulty: 3
A given mRNA sequence might be translated in any of three reading frames. Describe how
prokaryotes and eukaryotes determine the correct reading frame.
Ans: In prokaryotes, the Shine-Dalgarno sequence in the mRNA base pairs with a complementary
sequence in the 16S RNA of the ribosome; this positions the correct start codon (AUG) on the 30S
ribosomal subunit. Thus, the initiating AUG is distinguished by its proximity to the Shine-Dalgarno
sequence. In eukaryotes, the initiating AUG codon is the first AUG that the ribosome encounters as it
scans the mRNA from its 5' end. In both cases the initiating AUG also sets the correct reading frame.
40. Protein synthesis
Pages: 1056-1061 Difficulty: 1
Match the factor or enzyme at the right with the stage(s) of protein synthesis at which it acts. If a
factor or enzyme participates in two stages of protein synthesis, indicate both of them.
___
___
___
___

Amino acid activation
Initiation
Elongation
Termination

(a) RF1
(b) EF-Tu
(c) aminoacyl-tRNA
(d) Shine-Dalgarno sequence

Ans: c; d; b; a
41. Protein synthesis
Pages: 1056-1061 Difficulty: 2
Indicate whether each of the following statements is true (T) or false (F).
___Assembly of a complete ribosome onto an mRNA requires ATP hydrolysis.
___Aminoacylation or “charging” of tRNA requires the formation of an aminoacyl-AMP
intermediate.
___Aminoacyl-tRNA binding to the A site of the ribosome requires the accessory factor EF-G
and GTP hydrolysis.
___Translocation of a growing polypeptide from the A to the P site on the ribosome requires EFG and GTP hydrolysis.
___Termination of translation requires release factors, but no NTP hydrolysis.
Ans: F; T; F; T; T
42. Protein synthesis
Pages: 1056-1061 Difficulty: 3
Briefly describe the role of the following components in bacterial protein synthesis.
(a) Initiation factor 2 (IF-2)
(b) 16S RNA
(c) Peptidyl transferase
(d) Release factors
(e) Elongation factor G (EF-G)
(f) N10-formyltetrahydrofolate
(g) ATP
(h) tRNAfMet


Chapter 27 Protein Metabolism

323

Ans: (a) IF-2 is a protein factor that, when bound to GTP, brings the fMet-tRNAfMet to the initiation
complex. (b) 16S RNA is a component of the small (30S) subunit. It contains a sequence
complementary to the Shine-Dalgarno sequence in the mRNA, and helps to line up the mRNA
initiation AUG codon on the ribosome. (c) Peptidyl transferase is a ribozyme in the 50S ribosomal
subunit. It catalyzes formation of each peptide bond as the ribosome moves along the mRNA. (d)
Release factors are proteins that bring about the release of the finished polypeptide when the
ribosome encounters a termination codon in the mRNA. (e) EF-G participates in the translocation of
the ribosome down the mRNA by one codon after each peptide bond is formed. (f) N10-formyltetrahydrofolate is the cofactor that donates a methyl group in the conversion of tRNA-bound Met to
fMet. (g) ATP is the substrate for aminoacyl-tRNA synthetases; it donates an AMP residue in the
formation of aminoacyl adenylate, which donates the aminoacyl group to tRNA. (h) tRNAfMet is the
transfer RNA that initiates protein synthesis by inserting the first amino acid (fMet) in every
prokaryotic protein.
43. Protein synthesis
Pages: 1056-1061 Difficulty: 2
Number the following steps in the proper order with regard to protein synthesis.
___ Aminoacyl-tRNA binds to the A site.
___ Deacylated tRNA is released from ribosome.
___ Peptide bond formation shifts the growing peptide from the P to the A site.
___ The 50S subunit binds to the initiation complex of the 30S subunit and mRNA.
Ans: 2; 4; 3; 1
44. Protein synthesis
Pages: 1056-1062 Difficulty: 2
Indicate whether each of the following statements is true (T) or false (F).
___ Bacterial mRNA is broken down within a few minutes of its formation in E. coli.
___ Bacterial mRNA consists only of the bases that code for amino acids.
___ Polysomes do not necessarily contain mRNA.
___ Bacterial mRNA normally occurs as a double-stranded structure, with one strand containing
codons, the other containing anticodons.
___ Bacterial mRNA can be translated while it is still being synthesized.
Ans: T; F; F; F; T

45. Protein synthesis
Pages: 1057-1058 Difficulty: 2 Ans: B
Regarding translation in eukaryotes versus that in prokaryotes (bacteria), indicate whether each of the
following statements is true (T) or false (F).
___ In eukaryotes the 3' end of the mRNA is associated with the 5' end during initiation whereas
in prokaryotes it is not.
___ In prokaryotes it is initiated at an AUG near a Shine-Dalgarno sequence in the mRNA
whereas in eukaryotes it is initiated at an AUG near the 3' end of the mRNA.
___ In prokaryotes it is initiated with Met whereas in eukaryotes it is initiated with fMet.
___ In prokaryotes translation and transcription are coupled whereas in eukaryotes they are not.
Ans: T, F, F, T


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Chapter 27 Protein Metabolism

46. Protein synthesis
Pages: 1058-1061 Difficulty: 2
Polypeptide chain elongation in E. coli occurs by the cyclical repetition of three steps. What are these
steps and what cellular components are necessary for each of them to occur?
Ans: The three steps are: (1) An aminoacyl-tRNA is brought to the A site by EF-Tu with bound
GTP; (2) peptidyl transferase (a ribozyme) catalyzes peptide-bond formation; (3) the ribosome
translocates three nucleotides down the mRNA, helped by EF-G (translocase). This shifts the
peptidyl-tRNA to the A site and the deacylated tRNA to the E site.
47. Protein synthesis
Page: 1060
Difficulty: 2
A new antibiotic was recently discovered that inhibits prokaryotic protein synthesis. In the presence
of the antibiotic, protein synthesis can be initiated, but only dipeptides that remain bound to the

ribosome are formed. What specific step of protein synthesis is likely to be blocked by this
antibiotic?
Ans: The antibiotic probably blocks translocation.
48. Protein synthesis
Page: 1062
Difficulty: 1 Ans: B
The large structure consisting of a mRNA molecule being translated by multiple copies of the
macromolecular complexes that carry out protein synthesis is called a:
A)
B)
C)
D)
E)

lysosome.
polysome.
proteosome.
ribosome.
synthosome.

49. Protein synthesis
Pages: 1062-1065 Difficulty: 2
Following the synthesis of their polypeptide chain, many proteins require further posttranslational
modifications before they attain their full biological activity or function. List and describe briefly at
least four possible types of modification that can occur.
Ans: 1) Amino-terminal modification (N-acetylation or deacetylation), 2) removal of signal
sequences used for targeting, 3) side-chain modification (phosphorylation, carboxylation,
methylation, etc.), 4) attachment of N-linked (to Asn) or O-linked (to Ser or Thr) oligosaccharide
moieties, 5) isoprenylation of Cys side-chains, 6) incorporation of prosthetic groups (heme, biotin,
etc.), 7) processing of proenzymes or zymogens, and 8) formation of disulfide crosslinks.

50. Protein synthesis
Page: 1065
Difficulty: 3
In no more than three sentences, describe a nonsense suppressor tRNA and how it differs from a
normal tRNA.
Ans: A suppressor tRNA has one or more altered bases in its anticodon, which allows it to base pair
with one of the stop codons. This allows the insertion of an amino acid instead of the chain
termination that would normally have occurred at the point of a nonsense (stop) codon.


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325

51. Protein targeting and degradation
Page: 1068-1069 Difficulty: 2
When first synthesized, proinsulin has an additional leader or signal peptide at its amino terminus.
This complete molecule is called preproinsulin and the signal peptide is cleaved off to give
proinsulin. Briefly, what is the likely function of the signal peptide?
Ans: The leader peptide in proinsulin is a signal sequence directing it to the endoplasmic reticulum,
from which it enters the Golgi complex and is packaged into a secretory vesicle for secretion by
exocytosis.
52. Protein targeting and degradation
Page: 1069
Difficulty: 2
Describe the sequence of events between the transcription of an mRNA for a secreted protein and the
arrival of that protein in the lumen of the endoplasmic reticulum.
Ans: (1) The mRNA forms an initiation complex with a cytoplasmic ribosome and transcription
begins. (2) The amino-terminal portion of the nascent chain, containing a signal sequence, binds to
an SRP (signal recognition particle), interrupting polypeptide elongation. (3) The SRP-ribosomenascent chain complex binds to the ribosome and SRP receptors on the cytosolic face of the ER; the

SRP then dissociates. (4) Chain elongation continues, with the newly synthesized polypeptide
crossing into the ER lumen as it grows. (5) The signal sequence is cleaved.
53. Protein targeting and degradation
Pages: 1071-1073 Difficulty: 2
What are the stages in targeting of nuclear proteins, and why are the targeting sequences not removed
upon arrival of the protein in the nucleus?
Ans: In the cytoplasm, where eukaryotic protein synthesis occurs, proteins carrying nuclear
localization signal (NLS) sequences are bound by a complex of importin α and β, which is then
bound to a nuclear pore. Ran GTPase mediates translocation of this complex into the nucleus, where
importin β dissociates from importin α, and importin α releases the nuclear protein. Importin α and β
are then exported from the nucleus, and available for another cycle of import. The nuclear envelope
of higher eukaryotes breaks down at each cell division, distributing the nuclear contents throughout
the cell. When the nuclear envelope is reestablished, NLS-carrying proteins can be reimported to
fulfill their function.
54. Protein targeting and degradation
Page: 1075
Difficulty: 3
Describe the role of ubiquitin in mediating intracellular protein breakdown.
Ans: Ubiquitin, a protein found in all eukaryotic cells, is covalently joined to proteins targeted for
degradation. The carboxyl-terminal residue of ubiquitin is first activated by the formation of a
thioester with the first enzyme in the pathway in an ATP-dependent reaction. After displacement of
the first enzyme by a second, also yielding a thioester link, ubiquitin is finally joined through its
carboxyl terminus to a lysine ε-amino group in the protein to be degraded. The presence of ubiquitin
targets the protein for degradation by cellular proteases.