Chapter 5 Self-Quiz
(Page 273)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
False. In eukaryotes transcription occurs in the nucleus and translation occurs in the cytoplasm.
True
True
True
False. When a bacteria’s environment is high in lactose, the LacI protein is bound to a lactose molecule.
False. A missense mutation results in the substitution of one amino acid in the protein chain.
True
False. The role of tRNA is to deliver amino acids to the ribosome to undergo protein synthesis.
True
(e)
(a)
(a)
(c)
(c)
(c)
(a)
Chapter 5 Review
(Page 274–275)
Understanding Concepts
Eukaryotic genomes are divided into two areas. The noncoding regions are known as the introns and the coding
regions are known as the exons. Noncoding regions usually contain a large number of variable number tandem
repeats that consist of many repetitive sequences. If the repetitive region is found in the telomeric or in the
centromeric region of the chromosome, it will likely cause not harm.
(b) The specific DNA regions where an RNA polymerase attaches and initiates RNA synthesis are called promoters. In
prokaryotes, many genes may be under the control of the promoter and an operator. The whole system is known as an
operon.
(c) The transfer of genetic information from DNA to RNA is called transcription. The enzyme that facilitates the process
is called RNA polymerase and can only synthesize messenger RNA in the 5' to 3' direction. Synthesis ceases when
the termination sequence is reached.
2. Messenger RNA (mRNA) represents the product of transcription of a gene. Ribosomes found in the cytoplasm synthesize
proteins using mRNA as the template. Since DNA is transcribed into mRNA, the DNA itself does not have to exit the
nucleus and be subjected to degradation. Transfer RNA (tRNA) delivers amino acids to the ribosome. The amino acids
are used to build protein during the process of translation. Ribosomal RNA (rRNA) binds with ribosomal proteins to form
ribosomes.
3.
Transcription
Translation
Initiated by the enzyme RNA polymerase,
Initiated when ribosome binds to 5' cap (eukaryotes) of
RNA polymerase binds to the promoter
messenger RNA sequence.
region of DNA, causing DNA to unwind.
Ribosomes build polypeptide chains with the aid of transfer
RNA polymerase synthesizes messenger
RNA. The ribosome moves along mRNA, reading the code in
RNA strand in the 5' to 3' direction. RNA
triplets known as codons. When the ribosome encounters the
polymerase adds one complementary
start codon, appropriate tRNA will bring in methionine amino
ribonucleotide at a time.
acid residue. Transfer RNA and messenger RNA must be
complementary for amino acid to be placed on polypeptide
chains. For every triplet of ribonucleotides read, the
corresponding amino acid is added to growing polypeptide
chain.
Terminated when the ribosome encounters a stop codon.
Terminated when RNA polymerase
Enzymes known as release factors cause ribosome to fall of
encounters a termination sequence on the
the mRNA strand.
DNA strand.
1.(a)
108 Unit 2
Molecular Genetics
Copyright © 2003 Nelson
Posttranscriptional modifications are made,
including capping and tailing, and excising
introns if mRNA is eukaryotic.
Transcription takes place in the nucleus.
Polypeptide chain is folded into protein.
Translation takes place in the cytoplasm.
4.
Structural genes can be regulated by prokaryotic systems using the operon system. An operon system comprises a cluster
of genes under the control of a promoter and an operator. The operon control system allows a cell to transcribe and
express genes as they are required. An example of such a control system is the lac operon. To metabolize lactose three
proteins must be present. The genes for these proteins are only transcribed and expressed when lactose levels are high. If
lactose levels are low, there is no need for a cell to waste energy and resources transcribing and translating the genes. The
lac operon is structured as follows. The promoter region is followed by a nucleotide sequence known as the operator.
After the operator nucleotide sequence, the sequences for the three genes required are found. A LacI protein binds to the
operator when lactose levels are low. RNA polymerase can bind to the promoter, but since the LacI protein is bound to the
operator, RNA polymerase cannot get past and transcribe the genes. When lactose levels are high, the LacI protein binds
to lactose, resulting in a conformational change. The LacI–lactose complex cannot bind to the operator, hence RNA
polymerase can now proceed and transcribe the genes required for lactose metabolism. As the levels of lactose decrease,
the concentration of LacI–lactose complex decreases, and the now free LacI protein can bind once again to the operator,
preventing transcription of the lactose metabolizing genes.
5. (a) transcriptional: This type of control regulates which genes are transcribed (DNA to mRNA), or it can control the rate at
which transcription occurs.
(b) posttranscriptional: The mRNA molecules undergo changes in the nucleus before translation occurs. Introns are
removed and exons are spliced.
(c) translational: It controls how often and how rapidly mRNA transcripts will be translated into proteins. This control
affects the length of time it takes for mRNA to be activated and the speed at which cytoplasmic enzymes destroy
mRNA.
(d) posttranslational: Before many proteins become functional, they must pass through the cell membrane. Several control
mechanisms affect the rate at which a protein becomes active and the time that it remains functional, including the
addition of various chemical groups.
6. (a) A substitution would be less harmful than a deletion, because it does not result in a shift of the reading frame. A
substitution can result in a change in one amino acid, while a deletion can result in the translation of many new amino
acids in a growing polypeptide chain.
(b) An addition in an intron region is less harmful than an addition in an exon region. Intron regions are noncoding and
therefore will not be transcribed and translated into protein. An addition in an exon region can result in frameshift
mutations that can lead to different amino acid sequences in a protein.
(c) A substitution in a telomeric region can be less harmful than a substitution in a promoter region. RNA polymerase
recognizes the specific sequences in a promoter region and binds to it. If a substitution takes place, RNA polymerase
may not bind. Telomeric regions are noncoding; therefore, a substitution will not have an effect.
(d) An inversion is more harmful than a substitution. A substitution can result in a change in one amino acid residue in a
protein, whereas an inversion results in many amino acid residue changes in the protein sequence.
7. RNA polymerase transcribes in the 5' to 3' direction, therefore, the 3' to 5' strand will be the template strand.
3'–TTCATGTCGTA–5': template strand
5'–AAGUACAGCAU–3': mRNA strand
lys–tyr–ser: polypeptide chain
8. (a) If splicesomes were inactivated, introns would not be excised from eukaryotic mRNA. The result would be
dysfunctional proteins, as they would contain many extraneous amino acid residues encoded by the introns.
(b) If RNA polymerase were inactivated, no proteins would be synthesized because none of the DNA would be
transcribed.
(c) If poly-A polymerase were inactivated, the mRNA transcript would be subjected to possible degradation on exit into
the cytoplasm. If the transcript were degraded, then the proteins synthesized would be missing the end portion of their
sequences, depending how far along the degradation had proceeded.
(d) If tRNA were inactivated, translation would not take place. Amino acids could not be delivered to the ribosomes that
build polypeptide chains.
(e) If ribosomes were inactivated, proteins would not be synthesized.
9. Coupled transcription-translation is not possible in eukaryotes, because eukaryotes have a nuclear membrane that
separates the ribosomes found in cytoplasm from the mRNA found in the nucleus. Messenger RNA is not allowed to exit
the nuclear membrane until it is completed. Also, eukaryotic mRNA contains introns. Introns must be excised from the
Copyright © 2003 Nelson
Chapter 5 Protein Synthesis 109
mRNA before it can be translated. Because of this posttranscriptional modification, translation cannot take place
concurrently with translation.
10.(a) Student answers will vary. A possible mRNA sequence could be AUG GGU CCU GUU CGU.
(b) More than one sequence may exist because more than one codon encodes for an amino acid. For example, the amino
acid proline is encoded by the codons CCU, CCC, CCA, and CCG.
(c) The variability found in the genetic code can possibly offset mutations. For example, it does not matter what the third
ribonucleotide is in a codon that encodes for proline. If the two first ribonucleotides are intact CC, then proline will be
placed in the polypeptide chain. The variability originates in the third ribonucleotide of a codon.
11. Adenine and thymine share two hydrogen bonds, while guanine and cytosine share three hydrogen bonds. Less energy is
required to break two hydrogen bonds and unwind a DNA molecule as compared with three hydrogen bonds. Since RNA
polymerase must not only transcribe the DNA, but also must cause the molecule to unwind in order to do so, it is
advantageous for it to be high in adenine and thymine residues.
12. Nucleomorphs contain their own genetic material and carry out protein synthesis. Mitochondria divide by binary fission,
contain their own genetic material, which is circular and not membrane bound, produce their own energy, and carry out
their own protein synthesis.
13.(a) If tryptophan levels are high, then the tryptophan molecule is bound to the tryptophan repressor. The repressor–
tryptophan complex binds to the trp operon operator, blocking transcription of the tryptophan synthesizing genes. RNA
polymerase is bound to the promoter but cannot get through.
(b) If the tryptophan levels are low, the tryptophan repressor molecule cannot bind to the trp operon operator. RNA
polymerase is free to transcribe the tryptophan synthesizing genes.
14. Every codon must consist of a triplet of base pairs because there are 20 amino acids. If doublets of base pairs were used to
code for amino acids only 16 possibilities would exist (42 = 16), which would not be enough to code for all the amino
acids. The triplet combination allows for 64 combinations (43 = 64), which are more than enough combinations to code
for all amino acids, as well as the start and stop signals.
15. The structure of mRNA is similar to DNA as follows. Both are polymers of repeating units, both contain the nitrogenous
bases adenine, guanine, and cytosine, and both contain phosphate in their backbone. Messenger RNA does differ from
DNA. Messenger RNA is single stranded, while DNA is double stranded, mRNA contains uracil, while DNA contains
thymine as a complement to adenine, and mRNA contains a ribose sugar, while DNA contains a deoxyribose sugar in its
backbone.
Applying Inquiry Skills
16.(a) From the results of the experiment, it is possible to conclude that enzymes 2, 3, and 4 are the same enzyme or that it
takes more than one enzyme to convert B into C. These conclusions can be made because if there is a defect in what is
believed to be enzyme 2, 3, or 4, then B accumulates.
(b) Assuming that enzyme 2, 3, and 4 are the same enzyme, only two genes are actually present in the coding for this
biochemical pathway.
17. 54 = 625. The minimum number of nucleotides needed to code for 125 amino acids in a 5 nucleotide coding system would
be 4. Each codon would be 4 nucleotides.
18.(a) 5'–GGC CAG AAA CAA GAA–3'
gly–gln–lys–gln–glu
(b) codon GGC for gly, anticodon CCG,CCA, CCC, CCU
codon CAG for gln, anticodon GUC, GUU
codon AAA for lys, anticodon UUU, UUC
codon CAA for gln, anticodon GUU, GUC
codon GAA for glu, anticodon CUU, CUC
Making Connections
19. Transcription involves DNA and RNA. DNA’s backbone consists of a deoxyribose sugar and a phosphate.
Complementary nitrogenous bases are stacked between the two double strands of DNA. Adenine is always bound to
thymine, and cytosine is always bound to guanine. The complementarity of bases extends itself to RNA. Even though
RNA is single stranded and contains a ribose sugar in its backbone, it still consists of the same nitrogenous bases as DNA,
except that uracil replacing thymine. The result of this relationship is that the information stored in DNA can be
transcribed into messenger RNA. The 20 amino acids are in no way chemically related to the four nucleic acids found in
RNA. Despite this, nucleic acids can be used to build a genetic code. A combination of three nucleic acids can be used to
code for one amino acid. Hence, information stored in DNA that is transcribed to mRNA can now be translated into
protein. The simplicity of the system allows for the storage of information. Amino acids cannot themselves be used to
store information. Amino acid structure does not allow for the complementary characteristic of DNA nitrogeneous bases,
and amino acids occur randomly in a protein; therefore, it would be difficult to transfer information from protein to
110 Unit 2
Molecular Genetics
Copyright © 2003 Nelson
protein. Furthermore, 20 amino acids compose proteins, whereas there are only four nucleotides. Devising a transfer of
information from a 20-unit system to a 4-unit system would be highly difficult and complicated; therefore, transferring
information from protein to nucleotide is not feasible.
20. Proteins are the molecules that carry out the biological functions in an organism. In effect, they do all the chemical work
that is required for an organism to survive. If a protein is dysfunctional, it can affect an organism negatively. Given that
DNA encodes protein structure, an error in the DNA sequence results in an inappropriate amino acid(s) placed within the
protein. The protein may malfunction or not function at all because of the amino acid error. To eventually be able to
prevent or “fix” the protein, the DNA sequence that is incorrect and that has led to the erroneous placement of an amino
acid in the protein must be altered. It is therefore not enough to just know the DNA sequence of an organism; researchers
must be able to link DNA sequence to protein to successfully identify the causes of a genetic disorder and eventually be
able to prevent it by altering DNA sequence.
21. If a drug became attached to a specific repressor protein, the result would be disadvantageous. If the attachment of the
drug causes a conformational change in the repressor protein that allows it to bind to its corresponding operator, then the
genes associated with the operon system will not be transcribed even if they are required by the cell. If the drug just binds
to the repressor protein, the repressor protein–drug complex cannot bind to the operator; the genes in the corresponding
operon will be transcribed. Hence, even if the levels of the product of the operon are high, the synthesis of the product
will occur since the repressor protein cannot bind to the operator. Overproduction of the operon product will result.
22. The molecules 5-bromouracil and 2-aminopurine resemble nitrogenous bases found in DNA and therefore can be inserted
into DNA via DNA polymerase III during replication. Their presence in a DNA strand results in a mutation. 5bromouracil resembles thymine, while 2-aminopurine resembles adenine. The molecule 5-bromouracil, when
incorporated into DNA, changes into a form that binds guanine, while 2-aminopurine may pair with adenine or thymine.
Hence, if 5-bromouracil has been added into a DNA strand, it will cause a change from an adenine to a guanine in the
DNA sequence on subsequent replication, while 2-aminopurine may cause an adenine to be added instead of a thymine.
Extensions
23. Student answers will vary. Glutamic acid is very different from valine.
Valine is a nonpolar amino acid, whereas glutamic acid is a negatively charged polar amino acid. Because glutamic acid is
negatively charged, it has the ability to attract other amino acids near it and cause a conformational change in the protein
structure. Other amino acids that could have been substituted instead of glutamic acid are alanine or leucine. They also
would have caused a conformational change, but because they are nonpolar, the effect may not have been as drastic. A
similar amino acid that could have caused a similar side effect to glutamic acid is aspartic acid.
24. Student answers will vary depending on research. Currently, a combination of drugs is prescribed to individuals who are
infected with HIV. The cocktail of drugs is a combination of protease inhibitors and reverse transcriptase inhibitors.
Protease inhibitors and reverse transcriptase inhibitors work in different ways and attack HIV at varying stages of its
lifecycle.
HIV is a retrovirus whose genetic material is found in the form of RNA. Using the DNA building blocks in the cell
(nucleosides), reverse transcriptase converts the viral RNA into a piece of DNA. The enzyme works by using viral RNA
as a template and then assembling a matching DNA molecule of nucleosides from the supply in the cell. The DNA is held
together by chemical bonds between the individual component nucleosides.
AZT is a nucleoside analogue drug. It supplies a cell with nucleosides that differ from the natural nucleosides used to
build DNA. If present, reverse transcriptase uses the nucleoside analogue drug to build the complementary DNA molecule
instead of the natural nucleosides found within the cell. The nucleoside analogues are different from the natural
nucleosides, so the bonds that are needed to hold the DNA molecule together are not made. The result is that the DNA
molecule is left incomplete. Because HIV lacks a correcting mechanism to fix the mistakes made by reverse transcriptase,
the virus cannot further replicate.
Copyright © 2003 Nelson
Chapter 5 Protein Synthesis 111
The second group of drugs, known as protease inhibitors, attack the HIV virus at a different stage of its life cycle.
HIV functions by having its RNA transcribed into DNA and then having the cell’s own machinery translate the DNA into
protein particles. These viral particles are released by the cell and are left to infect additional cells. Protease inhibitors act
against an enzyme that HIV uses to break up large proteins into the smaller proteins from which it builds new viral
particles. New HIV particles produced in the presence of protease inhibitors are immature and noninfectious, thereby
preventing the HIV from infecting more cells.
112 Unit 2
Molecular Genetics
Copyright © 2003 Nelson