Chapter 6

Question 1

Which of the following statements is true?

(a) The two genes must be transcribed into RNA using the same strand of DNA.
(b) If gene A is transcribed in a cell, gene B cannot be transcribed.
(c) Gene A and gene B can be transcribed at different rates, producing different amounts of RNA within the same cell.
(d) If gene A is transcribed in a cell, gene B must be transcribed.

Answer 1

c

Question 2

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 sugar ribose contains fewer oxygen atoms than does deoxyribose.

Answer 2

Choice (c) is correct. Choice (a) is untrue because 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 because ribose contains one more oxygen atom than deoxyribose.

Question 3

Transcription is similar to DNA replication in that ___________________.

(a) an RNA transcript is synthesized discontinuously and the pieces are then joined together.
(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.

Answer 3

Choice (d) is correct. Choice (a) 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. 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.

Question 4

Which of the following statements is false?

(a) A new RNA molecule can begin to be synthesized from a gene before the previous RNA molecule’s synthesis is completed.
(b) If two genes are to be expressed in a cell, these two genes can be transcribed with different efficiencies.
(c) RNA polymerase is responsible for both unwinding the DNA helix and catalyzing the formation of the phosphodiester bonds between nucleotides.
(d) Unlike DNA, RNA uses a uracil base and a deoxyribose sugar.

Answer 4

Choice (d) is false. RNA nucleotides contain the sugar ribose.

Question 5

Unlike DNA, which typically forms a helical structure, different molecules of RNA can fold into a variety of three-dimensional shapes. This is largely because ___________________.

(a) RNA contains uracil and uses ribose as the sugar.
(b) RNA bases cannot form hydrogen bonds with each other.
(c) RNA nucleotides use a different chemical linkage between nucleotides compared to DNA.
(d) RNA is single-stranded.

Answer 5

Choice (d) is correct. Choice (a) is true, but is not the main reason why different RNA molecules can form different three-dimensional structures (although ribose does increase potential hydrogen-bonding potentials compared to deoxyribose). Choices (b) and (c) are untrue.

Question 6

Which of the following molecules of RNA would you predict to be the most likely to fold into a specific structure as a result of intramolecular base-pairing?

(a) 5′-CCCUAAAAAAAAAAAAAAAAUUUUUUUUUUUUUUUUAGGG-3′
(b) 5′-UGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUGUG-3′
(c) 5′-AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA-3′
(d) 5′-GGAAAAGGAGAUGGGCAAGGGGAAAAGGAGAUGGGCAAGG-3′

Answer 6

Choice (a) is correct. Choices (b) and (c) do not have any opportunity for intramolecular base-pairing and thus a specific structure is unlikely. Although there is some opportunity for intramolecular base-pairing in choice (d), choice (a) has much more intrastrand complementarity and is a better choice.

Question 7

Which one of the following is the main reason that a typical eukaryotic gene is able to respond to a far greater variety of regulatory signals than a typical prokaryotic gene or operon?

(a) Eukaryotes have three types of RNA polymerase.
(b) Eukaryotic RNA polymerases require general transcription factors.
(c) The transcription of a eukaryotic gene can be influenced by proteins that bind far from the promoter.
(d) Prokaryotic genes are packaged into nucleosomes.

Answer 7

c

Question 8

You have a piece of DNA that includes the following sequence:

5′-ATAGGCATTCGATCCGGATAGCAT-3′
3′-TATCCGTAAGCTAGGCCTATCGTA-5′

Which of the following RNA molecules could be transcribed from this piece of DNA?

(a) 5′-UAUCCGUAAGCUAGGCCUAUGCUA-3′
(b) 5′-AUAGGCAUUCGAUCCGGAUAGCAU-3′
(c) 5′-UACGAUAGGCCUAGCUUACGGAUA-3′
(d) none of the above

Answer 8

7-18 Choice (b) is correct. The molecules listed in choices (a) and (c) have incorrect polarity.

Question 9

The following segment of DNA is from a transcribed region of a chromosome. You know that RNA polymerase moves from left to right along this piece of DNA, that the promoter for this gene is to the left of the DNA shown, and that this entire region of DNA is made into RNA.

5′-GGCATGGCAATATTGTAGTA-3′
3′-CCGTACCGTTATAACATCAT-5′

Given this information, a student claims that the RNA produced from this DNA is:

3′-GGCATGGCAATATTGTAGTA-5′

Give two reasons why this answer is incorrect.

Answer 9

First, the RNA molecule should have uracil instead of thymine bases. Second, the polarity of the molecule is incorrectly labeled. The correct RNA molecule produced, using the bottom strand of the DNA duplex as a template, would be:
5′-GGCAUGGCAAUAUUGUAGUA-3′

Question 10

You have a segment of DNA that contains the following sequence:

5′-GGACTAGACAATAGGGACCTAGAGATTCCGAAA-3′
3′-CCTGATCTGTTATCCCTGGATCTCTAAGGCTTT-5′

You know that the RNA transcribed from this segment contains the following sequence:

5′-GGACUAGACAAUAGGGACCUAGAGAUUCCGAAA–3′

Which of the following choices best describes how transcription occurs?

(a) the top strand is the template strand; RNA polymerase moves along this strand from 5′ to 3′
(b) the top strand is the template strand; RNA polymerase moves along this strand from 3′ to 5′
(c) the bottom strand is the template strand; RNA polymerase moves along this strand from 5′ to 3′
(d) the bottom strand is the template strand; RNA polymerase moves along this strand from 3′ to 5′

Answer 10

(d) The bottom strand can hybridize with the RNA molecule and thus is the template strand. The polymerase moves along the DNA in a 3′-to-5′ direction, because the RNA nucleotides are joined in a 5′-to-3′ polarity.

Question 11

Imagine that an RNA polymerase is transcribing a segment of DNA that contains the following sequence:

5′-AGTCTAGGCACTGA-3′
3′-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)?

Answer 11

A. The bottom strand.

B. 5′-AGUCUAGGCACUGA-3′

Question 12

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.

Answer 12

c

Question 13

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 that precedes the gene’s coding sequence
(d) a mutation that introduced extensive sequence changes into the DNA that precedes the gene’s transcription start site

Answer 13

(d) 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, because the transcription machinery does not recognize translational stops.

Question 14

There are several reasons why the primase used to make the RNA primer for DNA replication is not suitable for gene transcription. Which of the statements below is not one of those reasons?

(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 about 5–20 nucleotides in length.
(d) The RNA synthesized by primase remains base-paired to the DNA template.

Answer 14

Choice (b) is true for both primase and RNA polymerase, so it does not describe why primase cannot be used for gene transcription.

Question 15

You have a bacterial strain with a mutation that removes the transcription termination signal from the Abd operon. Which of the following statements describes the most likely effect of this mutation on Abd transcription?

(a) The Abd RNA will not be produced in the mutant strain.
(b) The Abd RNA from the mutant strain will be longer than normal.
(c) Sigma factor will not dissociate from RNA polymerase when the Abd operon is being transcribed in the mutant strain.
(d) RNA polymerase will move in a backward fashion at the Abd operon in the mutant strain.

Answer 15

(b) Without the termination signal, the polymerase will not halt and release from the DNA template at the normal location when transcribing the Abd operon. Most probably, the polymerase will continue to transcribe RNA until it reaches a sequence in the DNA that can serve as a termination sequence, either from the next downstream operon or in the intervening sequence between the Abd operon and the next operon. Dissociation of sigma factor occurs once an approximately 10-nucleotide length of RNA has been synthesized by RNA polymerase and should not be affected by the lack of a termination signal [choice (c)].

Question 16

Transcription in bacteria differs from transcription in a eukaryotic cell because __________________________.

(a) RNA polymerase (along with its sigma subunit) can initiate transcription on its own.
(b) RNA polymerase (along with its sigma subunit) requires the general transcription factors to assemble at the promoter before polymerase can begin transcription.
(c) the sigma subunit must associate with the appropriate type of RNA polymerase to produce mRNAs.
(d) RNA polymerase must be phosphorylated at its C-terminal tail for transcription to proceed.

Answer 16

Choice (a) is correct. Eukaryotic cells, but not bacteria, require general transcription factors [choice (b)]. There is only a single type of RNA polymerase in bacterial cells [choice (c)]. The general transcription factor TFIIH phosphorylates the C-terminal tail of RNA polymerase in eukaryotic cells but not in bacteria [choice (d)].

Question 17

Which of the following does not occur before a eukaryotic mRNA is exported from the nucleus?

(a) The ribosome binds to the mRNA.
(b) The mRNA is polyadenylated at its 3′ end.
(c) 7-methylguanosine is added in a 5′-to-5′ linkage to the mRNA.
(d) RNA polymerase dissociates.

Answer 17

(a) Ribosomes are in the cytosol and will bind to the mRNA once it has been exported from the nucleus.

Question 18

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 is most abundant?

(a) DNA
(b) tRNA
(c) rRNA
(d) mRNA

Answer 18

(d) mRNA is the only type of RNA that is polyadenylated, and its 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, because the poly-A tail is not encoded in the DNA and long runs of T are rare in DNA.

Question 19

Name three covalent modifications that can be made to an RNA molecule in eukaryotic cells before the RNA molecule becomes a mature mRNA.

Answer 19

1. A poly-A tail is added.
2. A 5′ cap is added.
3. Introns can be spliced out.

Question 20

Which of the following statements about RNA splicing is false?

(a) Conventional introns are not found in bacterial genes.
(b) For a gene to function properly, every exon must be removed from the primary transcript in the same fashion on every mRNA molecule produced from the same gene.
(c) Small RNA molecules in the nucleus perform the splicing reactions necessary for the removal of introns.
(d) Splicing occurs after the 5′ cap has been added to the end of the primary transcript.

Answer 20

(b) The primary transcript of a gene can sometimes be spliced differently so that different exons can be stitched together to produce distinct proteins in a process called alternative splicing.

Question 21

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 difference?

Answer 21

The gene contains one or more introns.

Question 22

Why is the old dogma “one gene—one protein” not always true for eukaryotic genes?

Answer 22

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 eukaryotic gene may therefore encode more than one protein.

Question 23

Genes in eukaryotic cells often have intronic sequences coded for within the DNA. These sequences are ultimately not translated into proteins. Why?

(a) Intronic sequences are removed from RNA molecules by the spliceosome, which works in the nucleus.
(b) Introns are not transcribed by RNA polymerase.
(c) Introns are removed by catalytic RNAs in the cytoplasm.
(d) The ribosome will skip over intron sequences when translating RNA into protein.

Answer 23

a

Question 24

snRNAs ___________________.

(a) are translated into snRNPs.
(b) are important for producing mature mRNA transcripts in bacteria.
(c) are removed by the spliceosome during RNA splicing.
(d) can bind to specific sequences at intron–exon boundaries through complementary base-pairing.

Answer 24

d) snRNAs are part of the snRNPs, which include proteins and RNA molecules. The proteins within the snRNPs are encoded by their own genes and not translated from snRNPs, which is why choice (a) is incorrect. Bacteria do not have introns, which is why choice (b) is incorrect.

Question 25

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.”

Answer 25

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.

Question 26

You have discovered a gene (Figure Q7-36A) that is alternatively spliced to produce several forms of mRNA in various cell types, three of which are shown in Figure Q7-36B. The lines connecting the exons that are included in the mRNA indicate the splicing. From your experiments, you know 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 Q7-36

(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).
(d) Exons 2 and 3 must have different numbers of nucleotides.

Answer 26

Choice (c) is the only answer that must be true for exons 2 and 3. Although choices (a), (b), and (d) could be true, they do not 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.

Question 27

Which of the following statements about the genetic code is 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) All codons specify an amino acid.

Answer 27

b) Most amino acids can be specified by more than one codon. Each codon specifies only one amino acid [choice (a)]. Tryptophan and methionine are encoded by only one codon [choice (c)]. Some codons specify translational stop signals [choice (d)].

Question 28

The piece of RNA below includes the region that codes for the binding site for the initiator tRNA needed in translation.

5′-GUUUCCCGUAUACAUGCGUGCCGGGGGC-3′

Which amino acid will be on the tRNA that is the first to bind to the A site of the ribosome?

(a) methionine
(b) arginine
(c) cysteine
(d) valine

Answer 28

Choice (b) is correct. The initiator methionine is underlined on the RNA molecule below.

5′-GUUUCCCGUAUACAUGCGUGCCGGGGGC-3′

The first tRNA to bind at the A site is the second codon of the protein, because the initiator tRNA is already bound to the P site when translation begins. The codon that follows the binding site for the initiator tRNA is CGU, which codes for arginine.

Question 29

7-43 Below is the sequence from the 3′ end of an mRNA.

5′-CCGUUACCAGGCCUCAUUAUUGGUAACGGAAAAAAAAAAAAAA-3′

If you were told that this sequence contains the stop codon for the protein encoded by this mRNA, what is the anticodon on the tRNA in the P site of the ribosome when release factor binds to the A site?

(a) 5′-CCA-3′
(b) 5′-CCG-3′
(c) 5′-UGG-3′
(d) 5′-UUA-3′

Answer 29

7-43 (a) The stop codon (UAA) is underlined in the mRNA sequence below; this is the only stop codon on this piece of mRNA. The codon (UGG) preceding the stop codon will be binding to a tRNA in the P site of the ribosome when release factor binds to the A site. The anticodon of the tRNA will bind to the codon UGG and will be CCA.

5′-CCGUUACCAGGCCUCAUUAUUGGUAACGGAAAAAAAAAAAAAA-3′

Question 30

7-44 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 that translation initiates upstream of this sequence?
C. What are these peptides? (Give your answer using the one-letter amino acid code.)

Answer 30

7-44 A. 5′-GUAGCCUACCCAUAGG-3′
B. 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. VAYP
SLPIG
Note: PTHR will not be a peptide because it is preceded by a stop codon.

Question 31

7-45 A strain of yeast translates mRNA into protein inaccurately. Individual molecules of a particular protein isolated from this yeast have variations in the first 11 amino acids compared with the sequence of the same protein isolated from normal yeast cells, as listed in Figure Q7-45. What is the most likely cause of this variation in protein sequence?

Figure Q7-45

(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

Answer 31

7-45 (c) 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. A mutation in the gene encoding the protein would cause only a single variant protein to be made [choice (a)]. A mutation in the anticodon loop of tRNAIle [choice (b)] or a mutation in the isoleucyl-tRNA synthetase that decreases its ability to distinguish between different tRNA molecules [choice (d)] would cause the substitution of isoleucine for some other amino acid (which is the opposite of what is observed).

Question 32

Figure Q7-47

7-48 What do you predict would happen if you created a tRNA with an anticodon of 5′-CAA-3′ that is charged with methionine, and added this modified tRNA to a cell-free translation system that has all the normal components required for translating RNAs?

(a) methionine would be incorporated into proteins at some positions where glutamine should be
(b) methionine would be incorporated into proteins at some positions where leucine should be
(c) methionine would be incorporated into proteins at some positions where valine should be
(d) translation would no longer be able to initiate

Answer 32

7-48 (b) The 5′-CAA-3′ anticodon binds to the 5′-UUG-3′ codon, which normally codes for leucine.

Question 33

7-49 In a diploid organism, the DNA encoding one of the tRNAs for the amino acid tyrosine is mutated such that the sequence of the anticodon is now 5′-CTA-3′ instead of 5′-GTA-3′. What kind of aberration in protein synthesis will this tRNA cause? Explain your answer.

Answer 33

7-49 If the DNA sequence specifying the anticodon is changed from 5′-GTA-3′ to 5′-CTA-3′, this tRNA will now pair with the 5′-UAG-3′ codon (instead of 5′ -UAC-3′). The UAG codon normally serves as a stop codon. Thus, this change will result in the amino acid tyrosine being incorrectly incorporated where there is a stop codon, resulting in the addition of amino acids at the end of proteins that normally would come to a stop because of the UAG codon in the mRNA. (Note that the tyrosine codons will NOT cause premature termination of translation, as tyrosine should continue to be incorporated into proteins, as there are additional tyrosine-tRNA genes in the cell that will provide a normal supply of tyrosine-tRNAs.)

Question 34

7-50 The ribosome is important for catalyzing the formation of peptide bonds. Which of the following statements is true?

(a) The number of rRNA molecules that make up a ribosome greatly exceeds the number of protein molecules found in the ribosome.
(b) The large subunit of the ribosome is important for binding to the mRNA.
(c) The catalytic site for peptide bond formation is formed primarily from an rRNA.
(d) Once the large and small subunits of the ribosome assemble, they will not separate from each other until degraded by the proteasome.

Answer 34

7-50 Choice (c) is correct. A ribosome is built from many more proteins than rRNA molecules, although the ribosome is about two-thirds RNA and one-third protein by weight. Thus, (a) is incorrect. The small subunit binds to mRNA, so (b) is incorrect. Choice (d) is incorrect, as the assembly and disassembly of the small subunit with the large subunit occurs every time a protein is produced from an mRNA. When release factor binds to an mRNA, the ribosome will release the mRNA and dissociate into its two subunits, to be recycled during another round of protein synthesis.

Question 35

7-51 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 binds one tRNA at a time.

Answer 35

-51 Choice (c) is correct. Ribosomes contain proteins as well as rRNA [choice (a)]. rRNA is synthesized in the nucleus, and ribosomes are partly assembled in the nucleus [choice (b)]. A ribosome must be able to bind two tRNAs at any one time [choice (d)].

Question 36

7-52 Figure Q7-52A 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 Q7-52A as a guide, show on Figures Q7-52B and Q7-52C what happens in the next two stages to complete the addition of the new amino acid to the growing polypeptide chain.

Figure Q7-52

Answer 36

nope

Question 37

7-53 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 peptidyl transferase activity.
(b) It inhibits movement of the small subunit relative to the large subunit.
(c) It inhibits release factor.
(d) It mimics release factor.

Answer 37

7-53 Choice (b) is correct. Choice (a) would prevent all peptide bond formation. Choice (c) would have no effect on translation until the stop codon was reached. Choice (d) 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.

Question 38

7-54 In eukaryotes, but not in prokaryotes, 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.

Answer 38

7-54 Choice (b) is correct. Choice (a) is true only for prokaryotes. Choices (c) and (d) are true for both prokaryotes and eukaryotes.

Question 39

7-55 Which of the following statements about prokaryotic mRNA molecules is false?

(a) A single prokaryotic mRNA molecule can be translated into several proteins.
(b) Ribosomes must bind to the 5′ cap before initiating translation.
(c) mRNAs are not polyadenylated.

Answer 39

7-55 (b) Bacterial mRNAs do not have 5′ caps. Instead, ribosome-binding sites upstream of the start codon tell the ribosome where to begin searching for the start of translation.

Question 40

7-56 Figure Q7-56 shows an mRNA molecule.

Figure Q7-56

A.	Match the labels given in the list below with the label lines in Figure Q7-56.
(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 prokaryotic or eukaryotic? Explain your answer.

Answer 40

7-56 A. (a)—3; (b)—2; (c)—4; (d)—6;(e)—1; (f)—5
B. The mRNA is prokaryotic. 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.

Question 41

7-57 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 Q7-57. Which of the following interpretations is consistent with these observations?

Figure Q7-57

(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.

Answer 41

7-57 (b) The results in Figure Q7-57 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, because 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.

Question 42

7-58 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

Answer 42

7-58 (c) The decrease in the level of protein X in the normal cell is most probably due to protein degradation, because 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 the removal of sites for attachment of ubiquitin, which targets proteins for destruction. Choices (a), (b), and (d) would probably not produce the results described, because without the production of a functional protein X, the mutant cells could not grow in size.

Question 43

7-60 Which of the following methods is not used by cells to regulate the amount of a protein in the cell?

(a) Genes can be transcribed into mRNA with different efficiencies.
(b) Many ribosomes can bind to a single mRNA molecule.
(c) Proteins can be tagged with ubiquitin, marking them for degradation.
(d) Nuclear pore complexes can regulate the speed at which newly synthesized proteins are exported from the nucleus into the cytoplasm.

Answer 43

7-60 (d) Proteins are synthesized in the cytoplasm and therefore newly synthesized proteins would not be exported from the nucleus into the cytoplasm.

Question 44

7-61 Which of the following statements about the proteasome is false?

(a) Ubiquitin is a small protein that is covalently attached to proteins to mark them for delivery to the proteasome.
(b) Proteases reside in the central cylinder of a proteasome.
(c) Misfolded proteins are delivered to the proteasome, where they are sequestered from the cytoplasm and can attempt to refold.
(d) The protein stoppers that surround the central cylinder of the proteasome use the energy from ATP hydrolysis to move proteins into the proteasome inner chamber.

Answer 44

7-61 (c) Once proteins are sent to the proteasome, proteases degrade them. Chaperone proteins provide a place for misfolded proteins to attempt to refold.