Answer Key
Unit 3 Molecular Genetics
Answers to Unit 3 Preparation Questions
Assessing Student Readiness
(Student textbook pages 198–201)
1.
Characteristic
Prokaryotes
Eukaryotes
Relative cell size
small
large
Cell number in typical
organism
single
multiple
Location of genetic material
cytoplasm
nucleus
Membrane-bond genetic
material
no
yes
Number of chromosomes
one
several
2. c
3. Any three of: they are not cells; they have no
cytoplasm, cell membrane, or organelles; they cannot
reproduce outside of host cells; they are dormant
outside of host cells.
4. c
5. d
6. d
7. a. endoplasmic reticulum (ER)
b. Synthesis of proteins and synthesis of lipids and
lipid-containing molecules. For example, in the
liver, the ER helps detoxify the blood of drugs and
alcohol. In the testes and ovaries, the ER produces
testosterone and estrogen.
8. c
9. e
10. d
11. Enzymes help facilitate chemical reactions by acting as
protein catalysts that increase the rate of the reaction.
12. These indentations (active sites) facilitate substrateenzyme binding as the active site changes in shape
to accommodate the substrate in what is called
induced fit.
13. Enzymes are classified according to the type of
reactions they catalyze.
14. nucleotide → nucleic acid → gene → protein
Nucleotides are the building blocks of nucleic acids,
which make up genes, which code for proteins.
15. d
16. a. Hydrogen bonds link nucleotides together between
the two strands that make up a DNA molecule.
These bonds occur between complementary bases.
b. The covalent bonds that link adjacent nucleotides
together within each strand are called
phosphodiester bonds. These bonds occur between
the phosphate group on one nucleotide and a
hydroxyl group on the sugar of the next nucleotide
in the same strand.
17. e
18. An allele is a different form of the same gene.
Homologous chromosomes carry two alleles of the
same gene. Differences in these alleles account for
differences in hair colour.
19. a
20. a
21. e
22. • Deletion—a piece of a chromosome is cut out
• Duplication—a piece of a chromosome appears twice
or more
• Inversion—a piece of a chromosome is flipped
• Translocation—a segment of a chromosome is
attached to another chromosome
23. a. prophase I
b. crossing over
c. During crossing over, the chemical bonds that hold
the DNA together in the chromosome are broken
and reformed. In some cases, the chromosomes do
not reform correctly.
24. The chromosome errors described result in genetic
disorders when they delete, alter, or duplicate genes.
In some case, however, the errors affect regions of
the genome that lack genes or are within non-coding
regions of a gene. These errors are not harmful.
25. d
Biology 12 Answer Key Unit 3 • MHR TR
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26. Sample answer: Genetic engineering can create
transgenic organisms that secrete a human protein for
medical use. First, the human gene that codes for the
protein is injected into an egg from a donor goat. Then
the egg is placed in a host goat where a transgenic goat
develops. Finally, this transgenic goat produces milk
containing the human hormone.
27. c
28. Viruses enter host cells and direct the activity of their
host cell’s DNA. This makes viruses useful tools for
producing a copy of a gene. Researchers can insert the
gene of interest into the virus genome. The virus then
directs the host cell to make multiple copies of the
virus; each new virus that the cell produces will contain
the gene of interest as well.
29. a. transgenic or genetically modified organism
b. It has a different genotype, as it now contains four
new genes
c. It has a different phenotype, as it now displays
higher levels of iron, vitamin A, sulfur, and a
new enzyme.
Chapter 5 The Structure
and Function of DNA
Answers to Learning Check Questions
(Student textbook page 207)
1. Genetic material must contain information that
regulates the production of proteins. It also must be
able to accurately replicate itself to maintain continuity
in future generations. Genetic material must allow
for some mutations so that there is variation within
a species.
2. Griffith used two forms of S. pneumoniae: a pathogenic
S-strain and a non-pathogenic R-strain. After injecting
mice with a mixture of heat-killed S-strain and live
R-strain, the mice died. Griffith concluded that
something from the heat-killed S-strain transferred to
the R-strain to transform it into a pathogenic form.
3. • When they treated heat-killed pathogenic bacteria
with a protein-destroying enzyme, transformation
still occurred.
• When they treated heat-killed pathogenic bacteria
with a DNA-destroying enzyme, transformation did
not occur. These results provided strong evidence for
DNA’s role in transformation.
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MHR TR • Biology 12 Answer Key Unit 3
4. Two different radioactive isotopes were used to trace each
type of molecule. One sample of T2 virus was tagged with
radioactive phosphorus (32P), since phosphorus is present
in DNA and not protein. The other sample of T2 virus
was tagged with radioactive sulfur (35S), since sulfur is
only found in the protein coat of the capsid.
5. The independent variable in the experiment was the
type of radioactive isotope used to tag the virus. The
dependent variable in the experiment was the presence
of radioactivity inside the infected bacterial cells.
Controls include the usage of the same type of virus in
both experiments and the usage of the same protocol
for infecting bacterial cells in both experiments.
6. Bacterial cells that are infected by viruses with
32
P-labelled DNA would not be radioactive. Bacterial
cells infected by viruses with 35S-labelled capsid
proteins would be radioactive.
(Student textbook page 212)
7. Answers should resemble Figure 5.4 on page 208 of
the student textbook, with labels for phosphate group,
sugar group, and nitrogen-containing base.
8. Nucleotides in DNA have a deoxyribose sugar, while
nucleotides in RNA have a ribose sugar with a hydroxyl
group at carbon 2. In addition to the sugar group, each
nucleotide is attached to a phosphate group and a base.
The bases are adenine, cytosine, guanine, and thymine
in the case of DNA, and adenine, cytosine, guanine,
and uracil in the case of RNA.
9. Chargaff ’s rule states that, in the DNA nucleotides,
the amount of adenine will be more or less equal to
the amount of thymine, and the amount of guanine
will be equal to the amount of cytosine. The number
of A-T nucleotides will not necessarily equal the
number of C-G nucleotides. This overturned Levene’s
earlier hypothesis that the nucleotides occurred in
equal amounts and were present in a constant and
repeated sequence.
10. Franklin used X-ray photography to analyze the
structure of DNA. Her observations provided evidence
that DNA has a helical structure with two regularly
repeating patterns. She also concluded that the
nitrogenous bases were located on the inside of the
helical structure, and the sugar-phosphate backbone
was located on the outside, facing toward the watery
nucleus of the cell. Pauling’s methods of assembling
three-dimensional models of compounds led to the
discovery that many proteins had a helical structure.
Watson and Crick also used this information to
propose that DNA had a helix shape.
11. Diagrams should resemble Figure 5.7B on page 213 of
the student textbook. Base pairing and directionality of
strands should be shown.
12. Nucleic acids are soluble in water. Therefore, the
nitrogenous bases, which are somewhat hydrophobic,
must be positioned away from the water found in
the nucleoplasm, and the polar phosphate groups
(which are hydrophilic) must be on the outside of the
molecule, interacting with the water.
(Student textbook page 222)
13. The main objective of DNA replication is to produce two
identical DNA molecules from a parent DNA molecule.
14. DNA replication occurs during the S phase of
interphase, prior to cell division, ensuring that there
is a copy available for each new daughter cell.
15. • Conservative model—Two new daughter strands
form to create a new double helix, and the original
DNA strands re-form into the parent molecule.
• Semi-conservative model—Each new DNA molecule
contains one strand of the original DNA and one
newly synthesized strand.
• Dispersive model—Parental DNA is broken into
fragments. Therefore, the daughter DNA contains a
mix of parental and newly synthesized DNA.
16. Nitrogen is a component of DNA and is incorporated
into newly synthesized daughter strands. Having a
“light” form (14N) and a “heavy” form (15N) allowed
the separation of different DNA strands based on the
amount of isotope present in the newly synthesized
DNA. DNA with more 15N would be denser than
DNA with 14N, and could therefore be separated
by centrifugation.
17. Meselson and Stahl concluded that DNA replication
is semi-conservative. After one round of replication,
DNA appeared as a single band, midway between
the expected positions of 15N-labelled DNA and
14
N-labelled DNA. After the second round of
replication, DNA appeared as two bands, with one
band corresponding to 14N-labelled DNA and the
other band in the position of hybrid DNA (half 14N
and half 15N). In additional rounds of replication, the
same two bands were observed, therefore supporting
the semi-conservative model.
18. Each new cell that is produced must have an exact copy
of parental DNA. The daughter strands of DNA are
part of a DNA molecule that will be in the daughter
cells. This ensures that newly born cells are similar to
parents and maintain their genetic identity.
(Student textbook page 227)
19. Initiation—Helicase enzymes unwind DNA to separate
it into two strands. A replication bubble is formed
when single-strand binding proteins stabilize the
separated strands.
Elongation—New DNA strands are synthesized by
joining free nucleotides together. This is catalyzed by
DNA polymerase, which synthesizes the new strands
that are complementary to the parental strand.
Termination—The two new DNA molecules separate
from one another.
20. Replication takes place in a slightly different way on
each DNA strand because DNA polymerase can only
catalyze elongation in the 5′ to 3′ direction. In order for
both strands of DNA to be synthesized simultaneously,
the method of replication must differ.
21. On the leading strand, DNA synthesis takes place
along the DNA molecule in the same direction as
the movement of the replication fork. On the lagging
strand, DNA synthesis proceeds in the opposite
direction to the movement of the replication fork. The
lagging strand is synthesized in short fragments called
Okazaki fragments.
22. DNA replication requires the use of many enzymes
that have specific roles. The presence of numerous
specialized enzymes may reflect the importance of
having accurate DNA replication, since mutations in
DNA can change the genetic makeup of an organism.
23. Answers may include: DNA polymerases have a
proofreading function during which they excise
incorrect bases and add the correct bases. Mismatch
repair involves a group of enzymes that identify,
remove, and replace incorrect bases.
24. Many tissues and organs require continuous cell
regeneration. Therefore, DNA replication must be
quick and accurate so that new daughter cells receive
exact copies of DNA from the parent cell.
Answers to Caption Questions
Figure 5.2 (Student textbook page 205): If a live strain had
been transferred, the effects would have been due to that
strain, not due to the transfer of a substance form it to the
R-strain, which makes it pathogenic.
Figure 5.3 (Student textbook page 207): The results would
have also shown that protein was not the hereditary
material. However, it would not have directly demonstrated
the role of DNA as the hereditary material since RNA also
contains phosphorus.
Biology 12 Answer Key Unit 3 • MHR TR
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Figure 5.10 (Student textbook page 215): Twisting a
rubber band around itself mimics how DNA supercoiling.
The rubber band becomes compacted due to the coils
that twisting forms. This model is also useful since it
demonstrates the tension that is created by supercoiling.
A rubber band may become linearized, where supercoiling
in bacterial DNA occurs because it is a circular. The rubber
band model also does not reflect the double-stranded
nature of DNA.
Figure 5.16 (Student textbook page 221): If DNA had not
been uniformly labelled with 15N, the banding patterns
would not accurately reflect the presence of parental DNA.
Answers to Section 5.1 Review Questions
(Student textbook page 218)
1. Griffith’s experiments showed the existence of a
transforming principle. That is, something in the heatkilled pathogenic bacteria (S-strain) could transform the
non-pathogenic bacteria (R-strain) into a pathogenic
form. This result led to Avery’s experiments on
Streptococcus pneumoniae to identify the molecules that
caused this transformation. Avery’s research concluded
that DNA was the transforming principle.
2. a. Graphic organizers should include information
about experimental setup (i.e., the use of two
radioactive isotopes to differentially label DNA and
capsid protein), experimental procedure (i.e., the
use of agitation in a blender to dislodge viruses, and
subsequent centrifugation), and results. A summary
of the experiment is found in Figure 5.3 on page 207
of the student textbook.
b. The results showed that DNA is the hereditary material.
3. Miescher isolated nuclein from the nucleus of white
blood cells. He found that this material was present
only in the nuclei of cells. Further experimentation
showed that nuclein was a weakly acidic phosphoruscontaining substance. Nuclein would later be
known as nucleic acid or, more specifically, DNA
(deoxyribonucleic acid).
4. Diagrams should resemble the marginal portion of
Figure 5.4 on page 208 of the student textbook.
5. The nucleotide composition of the human would
be different from the nucleotide composition of the
mouse because the composition of DNA is unique to
each species. However, the percentage of adenine will
remain approximately the same as the percentage of
thymine, and the percentage of cytosine will remain
approximately equal to the percentage of guanine in
each species.
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MHR TR • Biology 12 Answer Key Unit 3
6. C = 26%; G = 26%; T = 24%
7. Diagrams should include labels for sugar-phosphate
molecules (“handrails”), nucleotide base pairing
(“rungs”), and directionality of both strands and
resemble the close up of Figure 5.7 on page 213 of
the student textbook.
8. a. Levene proposed that DNA was composed of
nucleotides, and that each of the four types of
nucleotides contained one of four nitrogen-containing
bases, a sugar molecule, and a phosphate group.
b. Chargaff showed that DNA is composed of repeating
units of nucleotides in fixed proportions (i.e., the
percent composition is of adenine is the same as
thymine, and the percent composition of cytosine is
the same as guanine). Chargaff ’s rule helped Watson
and Crick infer that adenine paired with thymine,
and cytosine paired with guanine.
c. Franklin determined that DNA had a helical
structure, with nitrogenous bases located on the
inside of the structure, and the sugar-phosphate
backbone located on the outside. This information led
to Watson and Crick’s ladder-like double helix model
of DNA, with the sugar-phosphate molecules acting
as “handrails” and the bases making up the “rungs.”
d. Pauling discovered that proteins have a helical
structure. This discovery influenced Watson and
Crick to propose that DNA was shaped like a helix.
9. a. 5′-ATTGAACAT-3′
b. 5′-GATTAACGG-3′
c. 5′-CGGAGCTAA-3′
10. A gene is a functional unit of DNA. It is a specific
sequence that encodes for proteins or RNA molecules.
A genome is an organism’s complete genetic makeup. It
is composed of an organism’s total DNA sequence.
11. Venn diagrams should include:
• Prokaryotes Only—double-stranded, circular DNA
packed in the nucleoid; DNA is compacted via
supercoiling; most are haploid; genomes contain very
little non-essential DNA; contain plasmids
• Prokaryotes and Eukaryotes—chromosomal DNA
is much larger than their cells, and therefore must
be compacted
• Eukaryotes Only—total amount of DNA is much
greater than in prokaryotes, and they therefore have
greater compacting and levels of organization (i.e.,
nucleosomes, chromatin, chromosomes); doublestranded linear DNA is contained in the nucleus;
most are diploid; genomes can vary widely in size and
complexity (i.e., some have large non-coding regions)
12. Diagrams should include the DNA molecule winding
around histones to form nucleosomes, which are
connected to each other by DNA and may resemble
Figure 5.12 on page 216 of the student textbook.
13. a. There is no set relationship between the complexity
of an organism (number of genes in an organism)
and the total size of its genome. An organism may
have an enormous number of base pairs in its
genome and very few genes if the bulk of its genome
consists of non-coding DNA.
b. Comparing the genomes of the two organisms
would show what genes they have in common, and
would indicate their evolutionary relationship—how
closely or distantly related they are.
14. A mutation in a protein-coding region would not
necessarily be more detrimental than a mutation
in a non-coding region since the latter may contain
regulatory sequences (i.e., regions that can influence
the production of proteins and RNA molecules). In
addition, since multiple codons exist for a given amino
acid, a mutation in the protein-coding region of DNA
may not alter the protein sequence.
Answers to Section 5.2 Review Questions
(Student textbook page 229)
1. Daughter cells must have the same genetic information
as parent cells.
2. a. Flowcharts should include a summary of
experimental steps and results shown in Figure 5.16
on page 221 of the student textbook.
b. If DNA replication was conservative, only two
bands would appear following one round of
replication: one band of 14N-only DNA (newly
synthesized DNA), and one band of 15N-only DNA
(old parental DNA). Diagrams should show two
distinct bands, one labelled “light (14N) and the
other labelled heavy (15N).
3. • Initiation—Replication begins at the replication
origin. Helicases bind to the DNA at each replication
origin. The helicases cleave and unravel a section of
the original double helix, creating Y-shaped areas
(replication forks) at the end of the unwound areas,
which form a replication bubble. Single-strand
binding proteins stabilize the separated strands.
These single strands serve as templates for the semiconservative replication of DNA.
• Elongation—New DNA strands are produced when
DNA polymerase inserts into the replication bubble.
A primase synthesizes an RNA primer that serves
as the starting point of new nucleotide attachment
by DNA polymerase. DNA polymerase can only
synthesize the new nucleotide chain in the 5′ to 3′
direction. As a result, one strand (the leading strand)
is replicated continuously in the 5′ to 3′ direction, in
the same direction that the replication fork is moving.
The other strand, known as the lagging strand,
is replicated in short segments, still in the 5′ to 3′
direction, but away from the replication fork. These
fragments, called Okazaki fragments, are joined
together by DNA ligase.
• Termination—When replication is complete, the
two new DNA molecules separate from one another
and the replication machine is dismantled. Each new
molecule of DNA contains one parent strand and one
new strand.
4. Early development is a very rapid process, and
many key molecules are produced during this time.
Therefore, it is expected that more replication origins
would be present in developing embryo cells.
5. A replication bubble is formed as the DNA double
helix unwinds during initiation. The replication forks
are the Y-shaped regions of the replication bubble,
and move along the DNA in opposite directions as
replication proceeds. Diagrams should resemble
the third portion of Figure 5.17 on page 223 of the
student textbook, with labels on the replication bubble,
replication fork, and the double-headed arrow showing
the direction(s) of unwinding.
6. Diagrams should illustrate continuous DNA synthesis
on the leading strand and discontinuous DNA synthesis
on the lagging strand. Diagrams may resemble a
simplified version of Figure 5.19 on page 224 of
the student textbook with labels for leading strand,
lagging strand, Okazaki fragments, RNA primer,
DNA polymerase, DNA ligase, parent DNA, and
directionality of strands.
7. An RNA primer is necessary for DNA synthesis on the
lagging strand. The primer provides a free 3′-hydroxyl
end, which DNA polymerase can extend by adding
new nucleotides.
8. DNA polymerase adds new nucleotides to the 3′ end of
a growing chain during replication. DNA polymerase
also proofreads newly formed base pairs and replaces
any nucleotides that have been incorrectly added.
9. a. No RNA primer would be synthesized. Therefore,
synthesis of the lagging strand cannot be initiated.
b. Okazaki fragments on the lagging strand cannot be
joined together.
c. Unwinding of the DNA double helix during
initiation would not occur.
Biology 12 Answer Key Unit 3 • MHR TR
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10. A—The semi-conservative model states that each new
molecule of DNA would contain one strand of original
parent DNA and one new strand of daughter DNA.
B—The dispersive model states that new molecules of
DNA would be hybrids containing a mixture of old
parent DNA and new daughter DNA strands.
C—The conservative model states that one molecule of
DNA would contain two new daughter DNA strands,
and the other molecule of DNA would contain the
original parent DNA strands.
11. Graphic organizers should include:
• Prokaryotes Only—rate of replication is faster
compared to eukaryotes; five DNA polymerases have
been identified in prokaryotes; circular chromosome
of prokaryotes have a single replication origin
• Prokaryotes and Eukaryotes—require replication
origins; have 5′–3′ elongation; have continuous
synthesis on the leading strand, and discontinuous
synthesis on the lagging strand; require a primer for
Okazaki fragments on the lagging strand; require the
use of DNA polymerase enzymes
• Eukaryotes Only—rate of replication is slower
compared to prokaryotes due to complicated
enzymes complexes and proofreading mechanisms;
13 DNA polymerase enzymes have been identified
in eukaryotes; linear chromosome has multiple
replication origins; presence of telomeres due to
linear nature of eukaryotic chromosome
Answers to Chapter 5 Review Questions
(Student textbook pages 235–9)
1. d
2. b
3. e
4. b
5. c
6. c
7. b
8. d
9. b
10. a
11. d
12. d
13. a
14. a
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MHR TR • Biology 12 Answer Key Unit 3
15. a. While studying DNA in the early 1900s, Phoebus
Levene reported that the nucleotides were present in
equal amounts, and that they appeared in chains in
a constant and repeated sequence of nitrogen bases.
Therefore, most scientists thought that the great
variety of proteins was an important factor, and must
be the hereditary material. Scientists assumed that the
molecular structure of DNA was just too simple to
provide the great variation in inherited traits.
b. Oswald Avery, Colin MacLeod, and Maclyn McCarty
conducted a series of experiments and discovered:
• When they treated heat-killed pathogenic bacteria
with a protein-destroying enzyme, transformation
still occurred.
• When they treated heat-killed pathogenic bacteria
with a DNA-destroying enzyme, transformation
did not occur.
These results provided evidence that genetic
information was carried on DNA.
16. Franklin’s X-ray diffraction images showed that DNA
had a helical structure, with two regularly repeating
patterns. She also concluded that the sugar-phosphate
backbone was located on the outside, and the nitrogencontaining bases protruded inward. Watson and
Crick used these observations to construct the threedimensional model of DNA.
17. a. Each body cell produces two daughter cells from
itself, and one of the two strands could go to each
daughter cell.
b. With bases on the outside, the DNA would not
be uniform width throughout, which all evidence
indicated. Also, the weak hydrogen bonds between
the nitrogen bases could be broken easily. The bonds
between the sugar and phosphate portions of the
nucleotides are much stronger.
c. From Franklin’s X-ray photographs, they reasoned
that DNA was twisted into a spiral, or helix. Since
the spiral consisted of two strands wound around
each other, they called it a double helix.
18. Eukaryotic DNA is compacted in the nucleus through
different levels of organization. DNA associates with
histones to form nucleosomes. It can be further
compacted by the coiling of nucleosomes to produce
30 nm fibres. Additional compacting is achieved
through the formation of loop domains of the 30 nm
fibre on a protein scaffold. This scaffold can condense
further through folding.
19. Purines, such as adenine and guanine, have a doublering structure. Pyrimidines, such as thymine and
cystosine, have a single-ring structure.
20. Hydrogen bonds between the bases hold the two
strands of DNA together.
21. The two strands of a DNA molecule are antiparallel
since each strand has directionality. At each end of the
DNA molecule, the 5′ end of one strand is across from
the 3′ end of the complementary strand.
22. Since nitrogen is a component of DNA, it would
be incorporated into newly synthesized strands of
DNA. Two different isotopes of nitrogen were used to
distinguish between the original parental strand and
the newly synthesized daughter strand.
Furthermore, having a “light” form (14N) and a
“heavy” form (15N) of nitrogen allowed the separation
of different DNA strands based on the amount of
isotope present in the newly synthesized DNA. DNA
with more 15N would be denser than DNA with
14
N, and therefore could be separated by centrifuge
and visualized.
23. Producing exact copies ensures that when a cell
divides, the offspring cells will receive the same genetic
information as the parent cell.
24. An RNA primer is required for discontinuous synthesis
of DNA on the lagging strand. The RNA primer
provides a free 3′ hydroxyl end from which DNA
polymerase can add nucleotides.
25. The replication machine consists of the complex of
proteins and DNA that interact at the replication
fork. These proteins include DNA polymerase, an
enzyme that joins nucleotides together to create a
complementary strand of DNA (elongation); DNA
ligase, an enzyme that joins Okazaki fragments together;
primase, an enzyme that constructs the RNA primer
needed for replication to begin; helicases, a group of
enzymes that cleave and unravel a segment of the double
helix to enable replication; and single-strand binding
proteins, which help stabilize the unwound strands.
division become mutations in the genome, which are
passed on to daughter cells once cell division occurs.
28. Telomeres are present to ensure that important genetic
information is not lost during replication of linear
eukaryotic DNA.
29. Nucleotides can come fully-formed from a variety of
food sources. Nucleotides can also be formed by our
bodies using components that come from our diet
(i.e., sugar, phosphates, nitrogen).
30. Different tissues all develop from the same fertilized
egg cell (zygote). While the tissues have the same
genes, only those genes necessary for a specific tissue’s
functions are active.
31. Sample answer: There are some similarities. However,
DNA “words” are limited to sequences of amino acids.
Each section of code has only “one meaning,” resulting
in one specific protein. This does not compare to the
arrangement of letters in a language, which results in
words that can have a great variety of meanings.
32. a. 5′-GATGTACAG-3′
b. 5′-ATCAGCGAT-3′
c. 5′-AATACGCCG-3′
33. G = 16%, T = 34%
34. Sample B is the viral DNA because the percentages of
adenine and thymine are not the same. Similarly, the
percentages of guanine and cytosine are not the same,
as they are in sample A, which shows complementary
base pairing of these respective bases. Complementary
base pairing does not occur in a single-stranded
DNA virus.
26. Having multiple origins of replication increases the
speed of replication. Instead of starting at one end and
finish at another, having multiple start points increases
the efficiency of replication.
35. a. Sample answer: The small fragments could be
Okazaki fragments that were not joined together
properly. This could be due to a lack of ligase, a
mutant ligase, or a mutation in the ligase gene that
produces a mutant ligase enzyme.
b. Experimental design should include the
experimental setup, proper controls (i.e., one
reaction tube with cells cultured without ligase, and
another with ligase), and expected results.
27. DNA polymerase has a proofreading mechanism,
which identifies and excises an incorrect nucleotide,
and then inserts the correct nucleotide. Another
mechanism of error correction is mismatch repair,
where a group of enzymes can detect deformities in the
newly synthesized strand of DNA caused by mispairing
of nucleotides. This group of enzymes excises the
mispaired nucleotides and inserts the correctly paired
nucleotide. Errors that are not corrected before cell
37. Linker DNA is responsible for joining nucleosomes
together. Micrococcal nuclease preferentially
cuts linker DNA, which would lead to disrupted
formation of 30 nm fibres. Diagrams should include
an illustration of the “beads on a string” appearance
36. A similar base composition does not necessarily
mean that the DNA sequences are similar (i.e., the
order of the nucleotides in the DNA sequence are not
necessarily the same in both organisms).
Biology 12 Answer Key Unit 3 • MHR TR
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of eukaryotic DNA organization with labels for DNA,
histones, nucleosomes, and linker DNA (refer to
Figure 5.12 on page 216 of the student textbook). An
accompanying diagram should illustrate micrococcal
nuclease cutting linker DNA between nucleosomes.
38. Answers and diagrams should include:
• If replication was continuous in a 5′ to 3′ direction,
the two DNA strands would not be antiparallel and
would instead be parallel to each other.
• If continuous replication was able to occur in both
5′ to 3′ and 3′ to 5′ directions, then primase, RNA
primers, Okazaki fragments, and DNA ligase would
not be necessary for synthesis on the lagging strand.
Additionally, telomeres would not be necessary since
there would be complete synthesis of the lagging
strand of eukaryotic DNA.
39. The weak hydrogen bonds in DNA break easily,
making it easier for the two strands in the molecule to
separate during replication. The strong covalent bonds
ensure that the sequence of nucleotides remains fixed
in each strand.
45. Primase would no longer be needed to synthesize a
RNA primer on the lagging strand, since a 3′ hydroxyl
end would already exist. Ligase would not be necessary
since Okazaki fragments would not be synthesized on
the lagging strand.
46. Concept maps should include information on two or
more experiments that addressed a similar hypothesis
(i.e., the transforming principle experiments performed
by Griffith and Avery). The experimental approaches
and results for each experiment should be outlined in the
concept map, in addition to how the different experiments
are related to one another (i.e., did the result from one
experiment help direct subsequent experiments?).
Using a variety of experimental approaches provides
confidence and confirmation of results.
47. Timelines should be a chronological order of
contributions from the scientists presented in the
chapter: Miescher; Levene; Griffith; Avery, MacLeod,
and McCarty; Hershey and Chase; Chargaff; Pauling;
Franklin; Watson and Crick.
48. Sample diagram:
32
40. Radioactive phosphorus ( P) would label newlysynthesized DNA strands. Diagrams should resemble
the semi-conservative model of replication shown
in Figure 5.15 on page 220 of the student textbook,
where the 32P-labelled strands would correspond to the
newly synthesized daughter strands. After two rounds
of replication, both strands of new DNA double helix
should incorporate 32P.
41. 5′-UGACU-3′
42. DNA polymerases are expected to be more active in
continuously dividing skin cells, since DNA replication
would occur more often. Heart muscle cells do not
divide as frequently as skin cells, and therefore DNA
replication (and DNA polymerase activity) occurs
less often.
43. Answers could include targeting bacterial-specific
topoisomerases. These enzymes are essential for
supercoiling and DNA replication, which are both
required for the survival of bacterial cells. Since this
is a bacteria-specific target, the side effects of this
drug on eukaryotic cells could be reduced.
44. A defect in DNA helicase could result in delayed
unwinding of the DNA double helix. Therefore,
the initiation of DNA replication would not occur
efficiently, which could lead to DNA instability and
cell death.
8
MHR TR • Biology 12 Answer Key Unit 3
P
S
P
S
P
S
P
S
P
S
S
3 end
P
5 end
G
G
A
C
T
A
hydrogen
bonding
C
P
5 end
S
C
P
S
T
P
S
G
P
S
A
P
S
T
P
S
3 end
49. Graphic organizers should include:
• Prokaryotes Only—rate of replication is faster
compared to eukaryotes; five DNA polymerases have
been identified in prokaryotes; circular chromosome
of prokaryotes have a single replication origin.
• Prokaryotes and Eukaryotes—require replication
origins; have 5′-3′ elongation; have continuous
synthesis on the leading strand, and discontinuous
synthesis on the lagging strand; require a primer for
Okazaki fragments on the lagging strand; require the
use of DNA polymerase enzymes
• Eukaryotes Only—rate of replication is slower
compared to prokaryotes due to complicated
enzymes complexes and proofreading mechanisms;
13 DNA polymerase enzymes have been identified
in eukaryotes; linear chromosome has multiple
replication origins; presence of telomeres due to
linear nature of eukaryotic chromosome
50. Diagrams should resemble the bottom illustration in
Figure 5.16 on page 221 of the student textbook, with
the first set of parental and new strands labelled first
round, and the next set labelled second round.
51. Flowcharts should include:
• Initiation—Helicase enzymes unwind the DNA
double helix to separate it into two strands. A
replication bubble and replication forks are formed
when single-strand binding proteins stabilize the
separated strands. Topoisomerase enzymes help to
relieve the strain on DNA caused by unwinding.
• Elongation—New DNA strands are synthesized
by joining free nucleotides together. This is
catalyzed by DNA polymerase, which synthesizes
the new strands that are complementary to the
parental strand. Synthesis of the new DNA strands
occurs continuously on the leading strand, and
discontinuously on the lagging strand.
• Termination—The two new DNA molecules, each
composed of one parental strand and one new
daughter strand, separate from one another.
52. Diagrams should resemble Figure 5.17 on page 223
of the student textbook, with labels for replication
bubble, replication forks (each end of the bubble), and
the double-headed arrow showing the direction(s)
of unwinding.
53. DNA can only be synthesized in the 5′ to 3′ direction.
Diagram should therefore illustrate continuous DNA
synthesis on the leading strand and discontinuous
DNA synthesis on the lagging strand. Diagram may
resemble a simplified version of Figure 5.19 on
page 224 of the student textbook with labels for leading
strand, lagging strand, Okazaki fragments, RNA
primer, DNA polymerase, DNA ligase, parent DNA,
and directionality of strands.
54. See Table below.
55. Journal entries should be written in the first person
and summarize the knowledge of heredity related
to his or her research, using scientific terminology.
Challenges could include a lack of research facilities,
lack of financial support, the shortcomings of
the current technology, lack of consensus among
research colleagues, or lack of support for the work
in the broader scientific community. Thoughts
of future significance of the research may reflect
later discoveries.
56. Articles should summarize Watson and Crick’s
findings, including their model of DNA and include
how this discovery would affect the general public (i.e.,
advances in medicine and genetics, ethical issues and
concerns). The article should be written in language
that is aimed at the general public and is appropriate
for the time period.
57. Concept maps should illustrate the different levels
of organization of eukaryotic DNA. Answers may
resemble Figure 5.12 on student textbook page 216,
with the addition of a nucleotide label above the DNA
molecule, nucleosome at the structure illustrating DNA
wrapped around histones, and chromatin on all of the
non-condensed forms of genetic material.
Question 54
Enzyme
Function
Absence of Enzyme
Helicase
Unwinds the double stranded DNA at the
replication fork.
Other enzymes (below) cannot bind to the DNA because
the DNA would remain double-stranded.
Topoisomerase II
Relieves strain on DNA that is generated from
unwinding of the double helix.
Unwinding may not occur efficiently.
Primase
Synthesizes an RNA primer used to generate
Okazaki fragments.
The DNA strands would be open, but synthesis could not
begin because DNA polymerase has to have an existing
chain with a 3′ end to add new nucleotides.
DNA polymerase I, II,
and III
A group of enzymes that:
• Adds new nucleotides to the 3′ end of a
growing chain.
• Proofreads the newly formed base pairs and
cleaves out any nucleotides that do not fit.
• Removes ribonucleotides at the 5′ end (removes
the RNA primer).
Only primer strands would exist on the opened DNA
strands. No new DNA would be synthesized.
DNA ligase
Joins Okazaki fragments together on the
lagging strand.
The leading strand would be normal. However, the
Okazaki fragments making up the lagging strand would
never be joined, and therefore the new DNA would never
be complete and functional.
Biology 12 Answer Key Unit 3 • MHR TR
9
58. This would leave replication errors (i.e., mispairing of
bases) uncorrected. These errors would then become
mutations in the genome, which are then passed onto
daughter cells once cell division occurs.
59. The graphic organizer should effectively represent the
points outlined in the Chapter 5 summary on page 234
of the student textbook.
60. Answers may include:
• Human disease—Many genes that are implicated
in human disease have parallel versions in model
organisms (i.e., yeast, mice, fruit flies), where they
can be studied easily in various experimental settings.
• Gene function—Studying parallel human genes
in model organisms may also provide insight into
gene function.
• Evolutionary biology—Comparative genomics
also allows researchers to study which regions of
a genome have been conserved amongst different
species. These conserved regions are thought to
be essential and important regions of the genome.
Likewise, divergent regions may confer speciesspecific function and contribute to morphological
and functional changes.
61. Answers should include specific examples of how the
chosen scientist’s research would benefit the scientific
research community and the general public. Social,
legal, and ethical implications may be addressed.
The paragraph should be convincing, as this is an
example of writing that would be included in a
grant application.
62. a. When the last RNA primer from the lagging
strand is degraded in linear DNA, the gap that
remains is unable to be filled since there is no
adjacent fragment where nucleotides can be added.
Therefore, the new DNA molecule will be shorter
than the parent DNA molecule. Since telomeres are
at the ends of eukaryotic chromosomes, they are the
sequences which become shorter after each round
of replication.
b. Telomerase is an enzyme that synthesizes telomeres
and replaces sequences that have been lost. In
childhood, telomerase activity in cells is high. As
people age, the activity of telomerase decreases,
which can result in shorter telomeres and therefore
shorter chromosomes. This may lead to loss of
important coding information. Certain lifestyle
factors may influence telomere length based on its
effects on telomerase activity. For example, smoking
may cause accelerated symptoms of aging due to
10
MHR TR • Biology 12 Answer Key Unit 3
decreasing telomerase activity. Exercise, on the other
hand, may delay the symptoms of aging since it
increases telomerase activity.
63. Answers may include:
• If there was a method to increase telomerase activity,
who would have access to “fountain of youth”
technology?
• Should we be interfering with the natural
aging process?
• Why would certain individuals want to delay aging?
• What are the possible benefits and risks of being able
to delay aging?
64. a. Telomerase activity may be higher in cancer cells
since these cells divide rapidly.
b. Telomerase is present and active in normal cells.
Inhibiting telomerase activity as a potential
cancer therapy could therefore cause damage to
normal cells (i.e., shortened or lack of telomeres,
which would lead to more rapid shortening of
chromosomes).
65. a. Students’ opinions should be supported
with examples.
b. Issues may addressed include:
• How would appropriate credit be determined?
• Should contributions be based on the percentage of
work performed for the study or for the analysis?
• Should primary credit go to the scientist who
proposed the hypothesis, or should it go to the
scientist who actually carried out the experiment
(i.e., did the “bench work”)?
66. Answers may be based on cultural, religious, or family
values. Accept any reasoned argument.
67. Answers may include:
• DNA sequencing
• Genetic screening for diseases
• Therapeutic gene targets
• Gene expression (RNA and protein)
• DNA mutations
68. This required a rethinking of many ideas that formed
the basis of other ideas. The lower number of genes
may mean that
• the structures function differently than anticipated
• the number of number genes does not determine an
organism’s complexity
• there are a large number and sizes of introns
• non-coding regions act as regulatory sequences
• genes work in combinations (groups) to perform
diverse function
69. In Linus Pauling’s model, DNA replication would have
to occur without nitrogen base pairing. Accept any
well-reasoned answer.
70. Anti-viral drugs specifically target viral DNA
polymerase to interfere with viral replication. Specificity
is also required to ensure eukaryotic DNA polymerase
activity in not adversely affected.
71. a. Because of the shared features/structure of
their DNA.
b. Graphic organizers (such as a table) should include
the advantages and disadvantages of using the chosen
organism and specific examples of significant research
findings that were obtained using that organism.
c. Ethical issues will vary but may include the harm
done to the animal either by being kept in captivity
or by the experimental process.
72. a. Mispairing of bases. Thymine should be paired
with adenine, and cytosine should be paired
with guanine.
b. Mispairing of bases occurs during replication and
may be due to flexibility in the structure of DNA.
c. Mismatch repair can correct this error. A group of
mismatch repair enzymes recognizes deformities
that are caused by mispairing of bases. These
enzymes then excise the incorrect nucleotide and
insert the correct nucleotide. DNA polymerases
can also correct this error by excising the incorrect
nucleotide in a newly synthesized strand, and adding
the correctly paired nucleotide.
Answers to Chapter 5 Self-Assessment Questions
(Student textbook pages 240–1)
1. b
2. c
3. e
4. c
5. a
6. e
7. c
8. a
9. e
10. b
11. C = 19%, G = 19%, T = 31%
12. a. Two different radioactive isotopes were used to trace
each type of molecule. One sample of T2 virus was
tagged with radioactive phosphorus (32P), since
phosphorus is present in DNA but not in protein.
The other sample of T2 virus was tagged with
radioactive sulfur (35S), since sulfur is only found
in the protein coat of the capsid.
b. In one experiment, Hershey and Chase observed
that most of the radioactively labelled viral DNA
was in bacteria and not in the liquid medium.
In a second experiment, they observed that the
radioactively labelled viral capsid protein was in the
liquid medium and not in the bacteria. These results
demonstrated that viral DNA, not viral protein,
enters the bacterial cell. Therefore, DNA is the
hereditary material.
13. Franklin’s X-ray diffraction images showed that DNA
has a helical structure with two regularly repeating
patterns. She also concluded that the nitrogenous bases
were located on the inside of the helical structure,
and the sugar-phosphate backbone was located on the
outside, facing toward the watery nucleus of the cell.
14. Diagrams should include labels for sugar-phosphate
molecules (“handrails”), nucleotide base pairing
(“rungs”), and directionality of both strands and
resemble the close up part of Figure 5.7 on page 213 of
the student textbook.
15. Prokaryotic DNA is double-stranded, circular,
and packed in the nucleoid. It is compacted
via supercoiling.
Linear eukaryotic DNA is compacted in the nucleus
through different levels of organization. DNA
associates with histones to form nucleosomes. It can
be further compacted by the coiling of nucleosomes
to produce 30 nm fibres. Additional compacting is
achieved through the formation of loop domains of
the 30 nm fibre on a protein scaffold. This scaffold can
condense further through folding.
The differences are due to structure (circular
prokaryotic DNA versus linear eukaryotic DNA) and
size. Since the total amount of DNA in eukaryotes is
much greater than in prokaryotes, they have greater
compacting and levels of organization.
16. Answers should include support for the student’s
interpretation of the statement. While protein-coding
regions include genes which code for proteins, the noncoding regions have regulatory sequences which can
influence and regulate the production of proteins and
RNA molecules.
17. a. Answers could include identity theft or the
identification of an individual’s traits (e.g., disease)
without consent (such as for job or insurance
screening purposes).
Biology 12 Answer Key Unit 3 • MHR TR
11
b. Answers could include forensics, comparative
genomics, screening for diseases, or tracing the
origin or source of an illness.
c. Opinions should be supported with an explanation
that includes evidence of scientific understanding
of the nature of a DNA sequence. Answers may also
consider consent under certain circumstances (i.e.,
genetic screening for disease or identifying lineage).
18. a. Arrow should indicate movement to the left.
b. Okazaki fragment C was made first. Primase
synthesizes an RNA primer, which binds to the
parental strand of DNA. DNA polymerase III adds
new nucleotides to the free 3′-hydroxyl end of the
primer. This newly synthesized fragment is an
Okazaki fragment.
c. Okazaki fragments are necessary for the
discontinuous synthesis of the lagging strand during
DNA replication. Once the Okazaki fragments are
made, DNA polymerase I removes the RNA primer
and the Okazaki fragments are joined together by
DNA ligase to form a complete lagging strand.
19. DNA has weak hydrogen bonds, which exist between
the nucleotides on opposite strands. These weak
hydrogen bonds break easily, making it easier for the
two strands to separate during replication. The strong
covalent bonds that exist on the sugar-phosphate
backbone ensure that the sequence of nucleotides
remains fixed in each strand, since these bonds are
not easily broken.
20. Diagrams should resemble Figure 5.17 on page 223
of the student textbook, with labels identifying the
replication bubble, replication forks at each end of the
bubble, and the double-headed arrow showing the
direction of unwinding.
21. Answers should include the information summarized
in Table 5.2 on page 224 of the student textbook,
which lists the important proteins involved in DNA
replication and their functions (helicase, primase,
single-strand-binding protein, topoisomerase II, DNA
polymerase I, II, and III, and DNA ligase).
22. A person without DNA polymerase would not exist,
because DNA replication could not occur and cells
would not be able to divide and survive.
23. Mutations in Mut genes could lead to an inability to
correct errors in replication, such as mispairing of
bases. Uncorrected errors become mutations in the
genome, which are then passed on to daughter cells
once cell division occurs.
12
MHR TR • Biology 12 Answer Key Unit 3
24. Telomerase activity decreases as we get older, which
results in the shortening of telomeres in somatic cells.
This means that the age of an organism is reflected in
the length of telomeres.
25. Venn diagrams should include:
• Prokaryotes Only—rate of replication is faster; five
DNA polymerases identified; circular chromosome
of prokaryotes have a single replication origin
• Prokayotes and Eukaryotes—require replication
origins; have 5′-3′ elongation; have continuous
synthesis on the leading strand, and discontinuous
synthesis on the lagging strand; require a primer for
Okazaki fragments on the lagging strand; require the
use of DNA polymerase enzymes
• Eukaryotes Only—rate of replication is slower
due to complicated enzymes complexes and
proofreading mechanisms; 13 DNA polymerase
enzymes identified; linear chromosome has multiple
replication origins; presence of telomeres due to
linear nature of eukaryotic chromosome
Chapter 6 Gene Expression
Answers to Learning Check Questions
(Student textbook page 246)
1. The black urine phenotype shown to be caused by a
recessive inheritance factor (gene) that caused that
production of a defective enzyme (protein).
2. Answers could use Figure 6.1 on page 245 of the student
textbook as a guideline. The results of the Beadle and
Tatum experiment showed that a single gene produces
one enzyme (one-gene/one-enzyme hypothesis). This
was later modified to the one-gene/one-polypeptide
hypothesis since not all proteins are enzymes.
3. RNA is found in the nucleus and cytoplasm; the
concentration of RNA in the cytoplasm is correlated
with protein production; RNA is synthesized in the
nucleus and transported to the cytoplasm.
4. Jacob and colleagues saw that bacteria infected with
a virus had a newly synthesized virus-specific RNA
molecule. This RNA molecule associated with bacterial
ribosomes, which are the sites of protein production.
Therefore, the RNA molecule carried the genetic
information for the production of a viral protein.
5. Having DNA continually transport itself could increase
the likelihood of damage to the DNA. Multiple steps
in gene expression provide many opportunities for
regulation. This allows the cell to have increased
control over protein synthesis.
6. 5′-GAUUAACGG-3′
(Student textbook page 254)
7. Answers could include: DNA is the genetic material
that controls protein synthesis, whereas RNA helps
DNA and is involved in protein synthesis. DNA
has the sugar called deoxyribose whereas RNA has
ribose sugar. The bases in DNA are adenine, guanine,
thymine, and cytosine; the bases in RNA include
adenine, guanine, cytosine, and uracil. DNA is a
double-stranded helix, and RNA is single stranded
without a helix.
8. Answers could include information in Table 6.3
on page 252 of the student textbook: mRNA is the
template for translation, while tRNA and rRNA are
involved in the translation of mRNA.
9. Initiation—Transcriptional machinery is assembled
on the sense strand. RNA polymerase binds to the
promoter region of the sense strand.
Elongation—RNA polymerase synthesizes a strand
of mRNA that is complementary to the sense strand
of DNA.
Termination—RNA polymerase detaches from the
DNA strand when it reaches a stop signal. The mRNA
strand is release and the DNA double helix re-forms.
10. To the cytoplasm, for translation
11. During the elongation phase of transcription, a second
RNA polymerase complex can bind to the promoter
region immediately after the previous RNA polymerase
complex starts moving along the DNA. Therefore,
multiple strands of mRNA can be synthesized from one
gene at a time. This is advantageous for the cell since
having more mRNA leads to more protein that can be
produced in a given amount of time.
12. Errors in transcription would be less damaging than
errors in DNA replication. An error in transcription
would results in an error in one protein molecule,
while an unrepaired error in DNA replication would
cause a change in the genetic makeup of an organism.
(Student textbook page 260)
13. Answers may be based on Table 6.4 on page 257 of the
student textbook and should include:
• mRNA—contains genetic information which
determines a protein’s amino acid sequence
• tRNA—a molecule that links mRNA codons with
their corresponding amino acid
• ribosomes—a structure composed of rRNA and
proteins which provides the site for protein synthesis
• translation factors—accessory proteins which are
necessary for each stage of translation
14. A polyribosome is a complex composed of multiple
ribosomes along a strand of mRNA. It can produce
many copies of a protein at the same time.
15. Translation consumes large amounts of energy
since many molecules (i.e., proteins and nuclei acid
components) must be synthesized and assembled.
Energy is also needed to form peptide bonds that exist
between the amino acids in proteins. This energy is
provided by mitochondria, which supply the cell with
most of its energy.
16. Translation is initiated when initiation factors
assemble the translation components. The small
ribosomal subunit attaches to the mRNA near the start
codon (AUG). The initiator tRNA (with anticodon
UAC) binds to the start codon. The large ribosomal
subunit then joins to form the complete ribosome.
Translation is terminated when a stop codon on the
mRNA is reached. This causes the polypeptide and the
translation machinery to separate. The polypeptide is
then cleaved from the last tRNA by a release factor.
17. Answers could be based on Figure 6.14 on page 259 of
the student textbook.
18. The antibiotic would prevent the initiation of
translation of bacterial proteins. Therefore, the
bacterial cells would not survive.
(Student textbook page 265)
19. Mutations that occur in reproductive cells can be
passed on to future generations. Mutations which
occur in somatic cells do not get passed on from one
generation to another.
20. No, since the deletion of nucleotides is divisible
by three.
21. A silent mutation has no effect on the amino acids of
a protein, and therefore would be the least harmful to
an organism. Missense mutations cause alterations in
the amino acids of a protein, and nonsense mutations
results in a shortened protein due to a premature
stop codon.
22. Single-gene mutations involve changes to the
nucleotide sequence of one gene. Chromosomal
mutations involve changes in chromosomes, and
therefore may affect many genes.
23. Spontaneous mutations may arise during DNA
replication if there is incorrect base pairing by DNA
polymerase. DNA transposition can also cause
mutations by the movement of specific DNA sequences
(transposons or “jumping genes”) moving within and
between chromosomes.
Biology 12 Answer Key Unit 3 • MHR TR
13
24. UV radiation can lead to a mutation by covalently
linking thymines to form a thymine dimer. A
specific mechanism for repairing thymine dimers is
photorepair, where an enzyme specifically recognizes
and cleaves thymine dimers. A non-specific
mechanism for repairing UV damage is excision repair,
where a group of excision repair enzymes identify and
correct many different types of DNA damage.
(Student textbook page 269)
25. Constitutive genes code for proteins that needed for
cell survival. Therefore, they are always active and are
expressed at constant levels.
26. The regulation of genes allows for cell specialization.
Gene expression is controlled so that only certain genes
are active at certain levels and at certain times in each
cell type. Therefore, each cell type can have its own set
of proteins.
27. Many genes in bacteria are clustered together in operons
and are therefore under transcriptional control of a
single promoter. This type of transcriptional control
is energy efficient since these genes are transcribed
together into a single polycistronic mRNA, rather than
individual mRNA molecules. Individual proteins are
then synthesized from the polycistronic mRNA.
28. Answer could be based on Figure 6.23 on page 267
of the student textbook. Diagram should include the
regulatory region, promoter region, operator, and the
coding region.
29. The lac operon is inducible since the transcription
of genes is induced when lactose is present. In the
trp operon, the genes are actively transcribed until a
repressor binds to inhibit transcription.
30. A common analogy used is the regulation of
temperature by a thermostat, since the feedback
mechanism is similar (i.e., at low temperatures, the
thermostat turns on to increase the temperature; at
high temperatures, the thermostat turns off to decrease
the temperature).
Answers to Caption Questions
Figure 6.1 (Student textbook page 245): Growth would
occur after each addition of intermediates since glutamate
is the first step of the arginine pathway.
Figure 6.3 (Student textbook page 249): The drugs will also
inhibit translation, since it follows transcription.
Figure 6.8 (Student textbook page 253): The increasing
length of the mRNA strands indicates the direction
of transcription.
14
MHR TR • Biology 12 Answer Key Unit 3
Figure 6.11 (Student textbook page 257): The genes that
code for tRNA produce tRNA molecules, which are not
translated into proteins.
Figure 6.16 (Student textbook page 261): Cysteine
Figure 6.18 (Student textbook page 263): A silent mutation
occurs when a nucleotide substitution has no effect on the
polypeptide sequence (i.e., ACU to ACC)
Figure 6.24 (Student textbook page 268): Both operons
have the same basic structure. The lac operon is inactive
unless lactose is present. The trp operon is normally active,
but becomes inactive if enough tryptophan is made.
Answers to Section 6.1 Review Questions
(Student textbook page 250)
1. While Garrod’s findings suggested a link between genes
and proteins, it was Beadle and Tatum who found
experimental evidence of the relationship between
genes and proteins.
2. a. The arg mutants would be able to grow since all the
required amino acids for growth are provided in
the medium.
b. It would be impossible to determine what enzyme
in the arginine pathway is affected by each mutation
since all of the mutants would be able to grow on
the complete medium (i.e., the medium that is
supplemented with all the required amino acids for
growth). Therefore, these results would not be able
to show that one gene codes for one enzyme.
3. Questions may include:
• Why was the one-gene/one-enzyme hypothesis
updated to the one-gene/one-polypeptide hypothesis?
• Is the one-gene/one-polypeptide hypothesis an
accurate description of gene expression? Explain
your answer.
• If you could update the one-gene/one-polypeptide
hypothesis, what changes would you make? Explain
your answer.
4. Jacob and colleagues showed that messenger RNA
(mRNA) acts as a “genetic messenger.” mRNA is
synthesized from the DNA of genes and carries
information to produce a protein.
5. The triplet hypothesis proposes that the genetic code
consists of a combination of three nucleotides (codon).
6. a. Answers may resemble Figure 6.2 on page 247 of
the student textbook, showing experimental set-up
and results.
b. Crick and Brenner showed that the genetic code
is read continuously in triplets, with no spaces in
the code.
7. Answers should include an explanation of why the
suggested genetic code system could be used. Genetic
codes should be able to code for 20 different amino
acids, with some duplication. For example, a genetic
code of four nucleotides per codon could result in a
maximum of 44 or 256 amino acids. This allows for the
20 different amino acids, with some duplication.
8. • A redundant genetic code means that more than one
codon can code for the same amino acid.
• A continuous genetic code means that the code reads
as a continuous series of three-letter codons without
spaces, punctuation, or overlap.
• A universal genetic code means that almost all living
things use the same genetic code to build proteins.
9. Because the start codon AUG also codes
for methionine.
10. Stop codons, also known as terminator codons, signal
the end of polypeptide growth.
11. Answers should indicate that the genetic code is
redundant—that is, more than one codon can code
for the same amino acid. Only three codons do not
code for any amino acid. The redundancy in the
genetic code is extremely valuable to the organism.
For example, if a mutation occurs to the DNA,
causing an AAT sequence to become AAC, the mRNA
codon transcribed by the original DNA is UUA
while the mutated DNA transcribes a UUG codon.
The mutation, however, will not harm the organism
because these two codons both correspond to leucine.
12. a. The central dogma of genetics.
b. Genetic information flows from DNA to RNA
to protein.
13. Almost all living organisms have the same genetic
code. Everyone has his or her own unique set of
nucleotides (and therefore DNA bases).
14. Amino acid sequence: arg-iso-val-ser-leu
Key: arg = arginine; iso = isoleucine; val = valine;
ser = serine; leu = leucine
15. a. The universality of the genetic code refers to the
discovery that the same genetic code is used by all
living things. This indicates that all living things
share similar ancestry. For example, a codon in a
fruit fly codes for the same amino acid in humans.
b. The universality of the genetic code has significant
implications for molecular biology techniques, such
as cloning. For example, a gene from one organism
can be inserted into a different organism to produce
the same protein.
Answers to Section 6.2 Review Questions
(Student textbook page 256)
1. Each type of RNA has a different function in the cell.
Some RNA are involved in translation (e.g., tRNA,
rRNA), while others act as part of an enzyme (e.g.,
RNA in RNAses). Answers may refer to examples
in Table 6.3 on page 252 of the student textbook,
which lists the different types of RNA molecules and
their functions.
2. Flowcharts should include:
• Initiation—Transcriptional machinery is assembled
on the sense strand. RNA polymerase binds to the
promoter region of the sense strand.
• Elongation—RNA polymerase synthesizes a strand
of mRNA that is complementary to the sense strand
of DNA.
• Termination—RNA polymerase detaches from the
DNA strand when it reaches a stop signal. The mRNA
strand is release and the DNA double helix re-forms.
3. The objective of transcription is to produce an accurate
copy of a small section of genomic DNA in the form
of mRNA. The objective of DNA replication is to
produce two identical DNA molecules from a parent
DNA molecule. While they have defined stages,
different proteins are used in transcription (e.g., RNA
polymerase instead of DNA polymerase).
4. For each gene, only one strand of the DNA molecule
is transcribed. The sense strand is also known as the
coding strand—the strand that is transcribed. The
antisense strand is also known as the non-coding
strand—it is not transcribed.
5. A promoter region is a sequence of nucleotides on
DNA where RNA polymerase complex binds. These
regions are important for transcription because RNA
polymerase must bind to them for transcription to
be initiated.
6. Venn diagrams should include:
• DNA polymerase Only—synthesizes two new
strands of DNA from template DNA; has
proofreading abilities
• DNA and RNA polymerase—work in a 5′-3′
direction; add new nucleotides to the 3′-hydroxyl
group of the previous nucleotide
• RNA polymerase Only—synthesizes new strand of
RNA; transcribes only one strand of template DNA;
catalyzing synthesis of RNA rapidly because it has no
proofreading abilities
Biology 12 Answer Key Unit 3 • MHR TR
15
7. a. 5′-TATGATCAC-3′
b. sense strand
c. 5′-UAUGAUCAC-3′
d. tyrosine-aspartate-histidine
8. Sample answer: Because DNA is double stranded, DNA
polymerase is able to recognize whether hydrogen
bonding is taking place between a base in the newly
synthesized strand of DNA and its complement in
the original strand. The absence of hydrogen bonding
indicates a mismatch between bases, and DNA
polymerase excises the incorrect base and inserts the
correct one. This proofreading ability reduces the
incidence of mutation in the genetic code and the
possible translation of a non-functional protein.
9. Protein synthesis cannot occur at the same time as
transcription since eukaryotic mRNA must undergo
modification before it can be transported from the
nucleus to the cytoplasm.
10. a. 5′ cap, Exon, Exon, Exon, AAA, 3′
b. 5′ cap—a modified G nucleotide covalently
links to the 5′ end; recognized by protein
synthesis machinery
Splicing—introns removed and exons are
joined together
poly-A tail—a series of A nucleotides covalently
linked to the 3′ end; makes mRNA more stable
11. The discrepancy in the number of genes and the
number of proteins in the human proteins may be
explained by alternative splicing, where one gene
can code for more than one protein. Answers should
include a diagram illustrating alternative splicing in
eukaryotic genes.
12. Alternative splicing allows for one gene to code for
more than one protein. For some genes, all exons are
spliced together. In certain cases, only certain exons are
spliced together to form mature mRNA.
13. a. The presence of introns allows a gene to produce
multiple transcripts by alternative splicing. In some
cases introns also contain regulatory sequences to
control gene expression.
b. In prokaryotes, transcription and translation can
occur at the same time since no post-transcriptional
modifications (such as removal of introns)
are needed.
16
MHR TR • Biology 12 Answer Key Unit 3
Answers to Section 6.3 Review Questions
(Student textbook page 266)
1. See Table 6.4 on page 257 of the student textbook.
2. Diagrams should resemble Figure 6.11A on page 257
of the student textbook. Sample caption: Transfer RNA
(tRNA) is a molecule that links codons on mRNA
to the corresponding amino acid during protein
synthesis. Each tRNA molecule contains an anticodon
loop, which has complementary base pairs to a specific
mRNA codon, and an acceptor step, which has the
corresponding amino acid.
3.
Codon (mRNA)
Anticodon (tRNA)
Amino Acid
GCC
3′-CGG-5′
alanine
UUU
3′-AAA-5′
lysine
AAC
3′-UUG-5′
asparagines
4. Ribosomes are the sites of protein synthesis. Each
ribosome is composed of two sub-units: a small
ribosomal sub-unit and a large ribosomal sub-unit. The
small ribosomal sub-unit has a binding site for mRNA,
while the large ribosomal sub-unit contains the three
binding sites for tRNAs.
5. Translation requires large amounts of energy since
many molecules must be synthesized and assembled
during this process. Energy is also needed to form
peptide bonds that exist between the amino acids
in proteins.
6. Answers should resemble Figure 6.14 on page 259 of
the student textbook, showing one amino acid being
added at a time to the growing polypeptide chain.
7. This mutation will not be passed to future offspring.
Mutations in somatic cells, such as skin cells, do not
get passed from one generation to another. Only
mutations in reproductive cells will get passed to
future generations.
8. Single-gene mutations involve changes in the
nucleotide sequence of one gene. Chromosomal
mutations involve changes in chromosomes, and
involve many genes.
9. Transposons would likely cause the most damage
if they inserted into coding regions (exons), since
this would lead to disruption of protein synthesis.
Transposons inserted into non-coding regions
(introns) may also cause damage if they disrupt
regulatory sequences that control gene expression.
10. Physical mutagens such as X-rays and UV radiation
cause mutations by physically changing the structure
of DNA.
Chemical mutagens such as nitrites, gasoline fumes,
and compounds found in cigarette smoke cause
mutations by reacting chemically with DNA.
11. Specific repair mechanisms involve a set of proteins
that are tailored to fix certain types of DNA damage.
Photorepair is a specific repair mechanism that repairs
thymine dimers caused by UV radiation. Photolyase
enzyme recognizes the damage, binds to the dimer, and
then excises the dimer by using visible light.
Non-specific repair mechanisms can fix different types
of DNA damage. For example, excision repair involves
removing a damaged region of DNA and replacing it
with the correct sequence.
12. a. GUU-ACC-UGU-UAU-U; frameshift mutation
b. valine-threonine-cysteine-tyrosine
13. a. 5′-AUGAAUGAGCAGUUGGAA-3′
methionine-asparagine-glutamate-glutamineleucine-glutamate
b. 5′-AUGAAUGAGCAGUAGGAA-3′
methionine-asparagine-glutamate-glutamine
c. nonsense mutation
14. a. DNA damage caused by UV radiation is not
repaired to due defective photorepair mechanisms.
This leads to an accumulation of harmful mutations
that could lead to cancer.
b. Individuals with XP would need to avoid UV
radiation by protecting themselves from sunlight.
Answers to Section 6.4 Review Questions
(Student textbook page 272)
1. Gene regulation allows for cell specificity. Gene
expression is controlled so that only certain genes are
active at certain levels and at certain times in each cell
type. Therefore, each cell type can be specialized and
have its own set of proteins.
2. Constitutive genes include genes that code for proteins
that are required at constant levels for cell survival and
maintenance. Answers may include genes involved in
cell cycle regulation, DNA replication, transcription,
and translation.
3. Certain genes may only be required at certain times,
such as genes involved in early development. If these
genes were always active, they may have unwanted
side effects on cells, since they are not needed in
later development.
4. a. An operon is a cluster of genes that are grouped
together in a region that is under the control of a
single promoter. An operator is a DNA sequence to
which a protein binds to regulate transcription.
b. Activators and repressors alter levels of transcription
by interacting with RNA polymerase, transcription
factors, and enhancers. The activator enhances the
level of transcription, whereas the repressor inhibits
the level.
5. Diagrams should resembled Figure 6.23 on page 267 of
the student textbook.
6. a. Allolactose binds to the repressor so that the
repressor can’t bind to the operator. Therefore,
transcription of genes proceeds to produce the
required enzymes.
b. Lac repressor binds to the operator and prevents
RNA polymerase from binding to the promoter,
inhibiting transcription of genes.
c. Transcription of genes proceeds, but at a slower rate
due to lack of activator protein.
d. Transcription of genes is inhibited.
7. The lac operon is inducible since the transcription
of genes is induced when lactose is present. In
the trp operon, the genes are actively transcribed
and only become inactive if enough tryptophan
is made. This causes the binding of a repressor to
inhibit transcription.
8. Gene regulation is more complex in eukaryotes than
in prokaryotes. In eukaryotes, gene regulation occurs
at five levels: pre-transcriptional, transcription, posttranscription, translational, and post-translational.
In eukaryotes, gene regulation occurs at three levels:
transcriptional, translational, and post-translational.
The most common level of regulation in prokaryotes is
at the transcriptional level, where genes are regulated
through operons.
9. Constitutive genes would more likely be in less
condensed areas since they need to be constantly
transcribed. Being in less condensed areas of
chromatin allows the transcriptional machinery easier
access to DNA.
Biology 12 Answer Key Unit 3 • MHR TR
17
10.
Level of Gene
Regulation
Location in
the Cell
Description
Example
Pre-transcriptional
Nucleus
Regulation occurs at the DNA level
Alteration of chromatin structure and
nucleosome structures
Transcriptional
Nucleus
Regulation is controlled by various proteins
during transcription
Transcription factors enable the initiation
of transcription; activators enhance
transcription initiation
Post-transcriptional
Nucleus
Regulation occurs at mRNA level
Alternative splicing of mRNA; addition of 5′
cap and 3′ poly-A tail
Translational
Cytoplasm
Regulation occurs during protein synthesis
Small RNAs can associate with proteins to
inhibit translation
Post-translational
Cytoplasm
Regulation occurs after the protein has
been synthesized
Protein folding into a three-dimensional
structure; ubiquitin tagging of proteins
for degradation
11. Activators enhance levels of transcription by binding to
transcription factors, RNA polymerase, and enhancers.
Regulation of genes can require many different types
of activators, which allows for greater control of
gene expression.
12. RNA interference could be useful in the development
of new therapeutics since they could potentially control
the expression of genes implicated in diseases. For
example, some diseases are caused by over-production
of certain proteins. Using RNA interference to target
the mRNA transcript could “turn off ” the production
of the protein. This technology has its disadvantages,
however, since it is not always desirable to completely
stop the production of a certain protein.
13. A pre-mRNA strand which lacks a 3′ poly-A tail will
undergo rapid degradation, while lack of a 5′ cap will
result in the mRNA not being transported from the
nucleus to the cytoplasm. Therefore, the mRNA will
not be available for protein synthesis.
14. Mutation in the VHL gene could lead to proteins not
being degraded efficiently. Therefore, a build-up of
certain proteins may lead to unwanted side effects on
cells and lead to disease.
15. C
Answers to Chapter 6 Review Questions
(Student textbook pages 277–81)
1. b
2. d
3. c
4. a
5. b
6. e
18
MHR TR • Biology 12 Answer Key Unit 3
7. c
8. c
9. d
10. a
11. b
12. a
13. c
14. e
15. Beadle and Tatum’s experiment showed that a single
gene produces one enzyme (one-gene/one-enzyme
hypothesis). This was later modified to the one-gene/
one-polypeptide hypothesis since it applies to all
polypeptides, not just to enzymes.
16. The structure of a protein is determined by the
structure of the DNA molecule. The order of the base
pairs in a DNA molecule makes up the genetic code
of an organism. The genetic code determines how the
amino acids are strung together and how the proteins
are made.
17. One-nucleotide and two-nucleotide genetic codes
could not account for the 20 different amino acids.
For example, a one-nucleotide genetic code would
only yield 41, or four possible amino acids and a
two-nucleotide genetic code would only yield 42, or
16 possible amino acids. The triplet hypothesis allows
for 43, or 64 amino acids, which is enough to account
for the 20 different amino acids, with some duplication.
18. • A redundant genetic code means that more than one
codon can code for the same amino acid.
• A continuous genetic code means that the code reads
as a continuous series of three-letter codons without
spaces, punctuation, or overlap.
• A universal genetic code means that almost all living
things use the same genetic code to build proteins.
19. The antisense strand is also known as the non-coding
strand. It is not transcribed. Its counterpart is the sense
strand, the one that is transcribed. That is also known
as the coding strand.
20. a. A codon is a specific sequence of three mRNA
nucleotides, the nitrogen bases of which code for
an amino acid. The mRNA “reads” the genetic
information of the DNA, and transfers that
information to ribosomes in the cytoplasm,
where a corresponding polypeptide is synthesized.
b. An anticodon is a specific sequence of three
tRNA nucleotides, the nitrogen bases of which
complement those on the mRNA. The tRNA carries
a specific amino acid to the ribosome and attaches
the amino acid to the growing polypeptide chain
according to the complementary mRNA codon.
c. The ribosome is the site of protein synthesis, and
moves along the mRNA chain as each codon is read
by a tRNA that carries a specific amino acid to the
polypeptide chain. A polyribosome is composed
of multiple ribosomes that are attached to a
single mRNA.
21. Answers may be based on Table 6.4 on page 257 of the
student textbook and should include:
• mRNA—contains genetic information which
determines a protein’s amino acid sequence
• tRNA—a molecule that links mRNA codons with
their corresponding amino acid
• ribosomes—a structure composed of rRNA and
proteins which provides the site for protein synthesis
• translation factors—accessory proteins which are
necessary for each stage of translation
22. P (peptide)— contains the tRNA with the growing
polypeptide chain
A (amino acid)—contains the tRNA with the
next amino acid to be added to the growing
polypeptide chain
E (exit) site—the uncharged tRNA (tRNA that is no
longer carrying an amino acid) is ejected from this site
23. Sample answer:
• In bacteria, operons regulate expression of a cluster
of genes whose encoded proteins are involved in the
same metabolic pathway.
• In eukaryotic cells, transcription factors regulate
which genes are transcribed and when in a particular
cell type.
• Prokaryotes have three levels of gene regulation:
transcriptional, translational, and post-translational.
Eukaryotes have five levels of gene regulation: pretranscriptional, transcriptional, post-transcriptional,
translational, and post-translational.
24. A physical mutagen physically changes the structure
of DNA. For example, X-rays tear through DNA
molecules, causing random physical changes to DNA
structure. A chemical mutagen causes mutations by
reacting chemically with DNA. Nitrites are an example
found in many cured meats.
25. Specific repair mechanisms involve a set of proteins
that specifically repair certain types of DNA damage.
Photorepair is a specific repair mechanism that fixes
thymine dimers caused by UV radiation. Photolyase
enzyme recognizes the damage, binds to the dimer,
and then excises the dimer by using visible light. Nonspecific repair mechanisms can fix different types of
DNA damage. For example, excision repair involves
removing a damaged region of DNA and replacing it
with the correct sequence.
26. a. In a eukaryotic cell, transcription occurs in the
nucleus and translation occurs in the cytoplasm. In
a bacterial cell, the DNA is in the cytoplasm since
there is no nucleus. Furthermore, in eukaryotes,
newly synthesized mRNA undergoes modification
(removal of introns and addition of 5′ cap and
3′ poly-A tail) to produce a mature mRNA before
translation can begin.
b. The main advantage is that protein synthesis can
occur faster because the mRNA does not have to
leave the nucleus as it does in a eukaryotic cell. A
possible disadvantage is that in bacterial cells, DNA
is not protected by a nucleus and the chance of DNA
mutation may be increased.
27. Because they perform different functions and have
different needs.
28. • Transcription of RNA from DNA (A pairs with U, C
pairs with G)
• mRNA attachment at AUG to the small ribosomal
sub-unit during initiation of translation
• mRNA codon and tRNA anticodon pairing
during translation
• tRNA conformation (intramolecular base pairing)
29. methionine-proline-threonine-threonine
30. If this mRNA came from an eukaryotic source, it is
likely a mature mRNA with its introns excised and
exons joined together. This would explain why the
mRNA has fewer bases compared to its DNA sequence.
However, if the source was a prokaryote, then the
mRNA is either partially degraded or the incorrect
DNA was isolated.
Biology 12 Answer Key Unit 3 • MHR TR
19
31. a. Spliceosomes remove introns from pre-mRNA. If
all the spliceosomes were removed from the yeast
sample, genes would not be correctly expressed since
the introns would remain in the mature mRNA.
b. The introduction of spliceosomes would most likely
not affect gene expression since prokaryotic genes
do not have introns.
32. a. serine
b. threonine
c. valine
d. histidine
33. a. CAA, CAG, CAU, CAC
b. UAC
c. GAA, GAG, GAU, GAC
d. AUA, AUG
34. a. Synthesize an RNA molecule that only contains
G and A nucleotides for translation. This
molecule would then be cultured in a medium to
undergo translation.
b. Supplement the growth medium with
the 20 different amino acids and protein
synthesis machinery.
c. glycine (GGA, GGG)
35. a. 5′-CATATTCGT-3′
b. 5′-CAUAUUCGU-3′
c. histidine-isoleucine-arginine
d. 3′-GUA-5′; 3′-UAA-5′; 3′-GCA-5′
36. a. If the substitution causes coding of the same
amino acid.
b. If the substitution causes coding of a different amino
acid, resulting in a slightly altered protein.
c. If the substitution causes coding of a different amino
acid, and the resulting polypeptide is non-functional.
37. a. The wrong protein would be made since the amino
acid sequence would not correspond with the
mRNA codons.
b. A protein would not be produced since initiation of
translation is inhibited.
c. The protein would not be produced.
d. The protein would be incorrectly translated.
38. • Ribosomes are made up of rRNAs as well as proteins.
• Initiation factors are proteins responsible for
assembling the translational machinery.
• Elongation factors are proteins responsible for
enabling tRNA anticodons to bind to mRNA codons.
• Release factors are proteins responsible for cleaving the
polypeptide from the last tRNA during termination.
20
MHR TR • Biology 12 Answer Key Unit 3
39. It is advantageous for the cell to keep its DNA inside
the nucleus rather than have it move from the nucleus
to the ribosomes in the cytoplasm. This reduces the
chances of a mutation (damage) occurring to the DNA
during this process. As well, if the DNA stays inside
the nucleus, only the gene required to synthesize the
protein has to be exposed. Once again, this reduces the
chances of the DNA being damaged. Other answers
are also possible. For example, the transcription step
allows amplification of protein synthesis since mRNA
is present in much greater copies in the cytoplasm
compared to DNA of a given gene.
40. a. The photorepair mechanism can repair damage to
DNA caused by UV light. A photolyase enzyme
recognizes the thymine dimers caused by UV light
and then uses visible light to repair the dimers.
b. DNA repair is an essential mechanism for cells since
it ensures that potentially harmful mutations do not
accumulate too rapidly.
41. Four: Exon 1, Exon 2, Exon 3, Exon 4; Exon 1, Exon
2, Exon 3; Exon 1, Exon 2, Exon 4; Exon 2, Exon 3,
Exon 4
42. Three bases, or a multiple of three, can be added
or deleted. This results in the code shifting one or
more full frames, which returns the sequence to its
original coding.
43. a. X-ray exposure could disrupt the function of critical
genes resulting in the activation of DNA damage
response in the cell. This will cause production
of new polypeptides that will work to repair
the damage.
b. The newly synthesized polypeptides will persist for
a short period of time. After the cell has repaired the
damage, it is expected that normal protein synthesis
will resume.
44. Transposons which insert themselves in non-coding
regions are not necessarily less harmful if the noncoding regions contain regulatory sequences which are
involved in the control of gene expression.
45. a. Nutrient X acts as an inducer that is most similar to
how lactose acts in the regulation of the lac operon.
When there are high levels of lactose, allolactose
binds to the repressor so that the repressor can’t
bind to the operator. Therefore, transcription of
lactose-metabolizing genes in the operon is high.
When lactose is not present, the lac repressor binds
to the operator and prevents RNA polymerase from
binding to the promoter. Therefore, transcription of
genes is low.
b. Nutrient Y is similar to how tryptophan acts in
the regulation of the trp operon. In the trp operon,
the genes are actively transcribed until a repressor
binds to inhibit transcription. For example, when
there are low levels of tryptophan, it must be
synthesized. Therefore, the repressor does not bind
to the operator, and transcription of tryptophansynthesizing genes is high. When there are high
levels of tryptophan, it binds to a repressor which
leads to lower transcription levels.
46. Genes that are not constitutive (i.e., genes that are
not always expressed and are not needed in constant
amounts to ensure cell survival) are most likely to be
present in highly condensed areas of chromatin. These
genes could be involved in early development, and
therefore are only expressed during certain times.
47. Flowcharts should include:
• Initiation—Transcriptional machinery is assembled
on the sense strand. RNA polymerase binds to the
promoter region of the sense strand.
• Elongation—RNA polymerase synthesizes a strand
of mRNA that is complementary to the sense strand
of DNA.
• Termination—RNA polymerase detaches from the
DNA strand when it reaches a stop signal. The mRNA
strand is release and the DNA double helix re-forms.
48. Answers may resemble Table 6.2 on page 251 of the
student textbook.
Characteristic
DNA
RNA
Base components
Adenine, guanine,
cytosine, thymine
Adenine, guanine,
cytosine, uracil
Sugar component
Deoxyribose
Ribose
Single or
double strand
Double strand
Single strand
49. a. Sample answer: a “protein factory”
• DNA is the blueprint/instruction for
making protein
• RNA is the messenger that delivers instructions to
the protein factory
• Protein is the finished product
b. Questions may include the difference between DNA
and RNA, what protein is (other than a nutrient),
why proteins can’t get information directly from
DNA, what the information looks like, and how long
the process takes.
c. Answers should reflect a cursory level of
understanding presented in language appropriate
to Grade 5.
50. Diagrams of precursor mRNA may resemble 5′, Exon,
Intron, Exon, Intron, Exon, 3′, with labels
for directionality.
Diagrams for mature mRNA may resemble 5′ cap, Exon,
Exon, Exon, 3′ poly-A tail, with labels for directionality.
51. DNA provides the genetic information to produce
proteins, but it is the proteins themselves which carry
out the cell’s functions. Therefore, proteins are often
considered to be the functional units of a cell.
52. Diagrams may resemble Figure 6.13 on page 259 of the
student textbook and should include labels for A site
(amino acid); P site (peptide); E (exit) site; large ribosomal
sub-unit; small ribosomal sub-unit; initiator tRNA with
anti-codon UAC pairing with mRNA start codon AUG.
53. Answers should resemble the first three rows of
Table 6.3 on page 252 of the student textbook and
include a summary for rRNA, mRNA, and tRNA.
54. a. Sample answer: Mutations (changes) to DNA are
caused by natural copying errors, and by chemicals
and radiation. Mutations can be noticeable,
harmless, or cause disease. Only mutations in
reproductive cells are passed on to children.
b.–c. Sample answer:
• Introduction—DNA is the code that tells your
cells how to behave (like eye cells or stomach cells,
for instance). Mutations in DNA simply mean
there is a bad copy of the cell, like a bad scan of
a document in which the colours are off or the
words are garbled. DNA has to be copied to create
new cells to replace old, dead cells.
• Causes of mutations—Mutations occur naturally
when mistakes are made during copying. Some
chemicals such as nitrous acid and benzene can
cause regions of bad code. Radiation such as X-rays
and UV light can change the DNA code too.
• Examples of mutations—DNA mutations are
responsible for a wide range of disorders such
as sickle cell anemia and colour blindness.
Some mutations are harmful, and some have no
noticeable effect—especially those at appear at the
end of a DNA strand. It is possible that a mutation
would be beneficial (such as a colour change that
aids camouflage), but it’s not likely to give you
super powers. Cancer can be caused by mutations
that turn off the signal for a cell to die, causing a
tumour to grow.
• Natural corrections—The process by which DNA
is copied involves steps that check and repair the
code. The cell has its own surveillance and repair
system to ensure that DNA damage is detecting
and fixed.
Biology 12 Answer Key Unit 3 • MHR TR
21
• Conclusion—Your genetic make-up never changes,
but the DNA in some of your cells can change.
You can limit the number of mutations your cells
must deal with by protecting your body from
radiation and some chemicals. Only mutations to
your reproductive cells will be passed on to your
children. There are tests that can tell you whether
you have the gene for a disease, or genes which
are associated with an increased risk of a disease.
Though you may live symptom free despite having
the gene for a disease, such knowledge can give
you the incentive to take preventive measures, and
may factor in your decision to have children.
55. Flowcharts should include:
• pre-transcriptional (in the nucleus)
• transcriptional (in the nucleus)
• post-transcriptional (in the nucleus)
• translational (in the cytoplasm)
• post-translational (in the cytoplasm)
56. Accept any justified arguments for or against the
strength of the thermostat analogy. Graphic organizers
should identify key concepts and relationships relating
to the lac operon, such as the roles of: the CAP in
enhancing transcription of the operon, the repressor
protein in blocking transcription in the absence of
lactose, and allolactose in inducing transcription in
the presence of lactose. Students should also indicate
that the lac operon is like a thermostat in that the
operon is turned “off ” until a signal (the presence
of lactose, rather than a low temperature) turns
it on. The lac operon, however, is not involved in
temperature regulation.
57. Graphic organizers should summarize the points given
on page 276 of the student textbook.
58. a. Answers should include an introduction to the
central dogma and a description of how retroviruses
are able to reverse-transcribe their RNA to DNA
using reverse transcriptase. This DNA then becomes
integrated into the host cell’s genome and undergoes
transcription and translation to produce the
retroviral protein.
b. Diagrams should resemble Figure 6.3 on page 249 of
the student textbook, with the addition of an arrow
showing reverse transcription from RNA to DNA.
59. a. Answers should clearly state the research
question, show evidence of consulting at least two
independent sources, with documentation, and
show evidence of critical thinking that connects the
research with concepts being studied in class. At
22
MHR TR • Biology 12 Answer Key Unit 3
least two or more hurdles should be identified—such
as the difficulty in coming up with tRNA molecules
or ribosomes that will work with new nucleotides.
b. Accept any justified response. Examples include
beneficial impacts, such as the use of new
nucleotides to build new proteins for medical
applications, and drawbacks such as uncertainties
about the safety of this form of genetic engineering.
60. Answers should identify the role of DNA as a carrier
of information as well as the roles of DNA polymerase,
DNA ligase, and other proteins in DNA replication,
and the roles of RNA polymerase and the ribosomal
proteins in gene expression. Inferences should include
that, at one time, RNA molecules might have carried
out similar roles. Answers should be supported by
one or more specific examples such as that mRNA is
a carrier of information and that, through the action
of snRNA molecules, mRNA can be cut and rejoined,
much as DNA ligase rejoins DNA fragments.
61. a. Diagrams should illustrate the mechanism by which
antibiotics on bacterial ribosomes inhibit bacterial
protein synthesis.
b. Antibiotics specifically target bacterial ribosomal
sub-units.
c. This specificity ensures that the antibiotics inhibit
the production of bacterial proteins (which
ultimately lead to death of bacterial cells), and do
not affect eukaryotic protein synthesis.
62. Answers should address how to reduce exposure, as
well as the potential harms of excessive exposure to
these mutagens. Alkylating agents and azides (such as
sodium azide in vehicle air bags) are common chemical
mutagens. Answers may also discuss formaldehyde,
benzene, and nitrous acid. Physical mutagens include
electromagnetic radiation (e.g., gamma rays, X rays,
UV light). Answers may also address safe exposure
levels and government regulations surrounding
each mutagen.
63. Answers should address the importance of
proofreading in preventing mutations. Accept
any reasonable and supported advantages and
disadvantages such as:
a. Researchers could, for example, find the new
form of RNA polymerase useful in experimenting
with methods to reduce mutation. Alternatively,
the new form of RNA polymerase could perhaps
contaminate experiments and interfere with the
action of DNA polymerase.
b. People could, for example, benefit from cancer
prevention techniques that use the new form of
RNA polymerase to prevent mutations. On the
other hand, the new form of RNA could prevent
beneficial mutations from occurring that could
allow populations of certain organisms to evolve.
64. a. Nitrites are a fairly inexpensive preservative used
in many cured meats. By using nitrites, food
manufacturers can produce a product that has
a longer shelf-life, which is less expensive than
producing nitrite-free foods (which have a shorter
shelf-life). Answers may also discuss how foods that
are branded “healthier” or “health conscious” are
often more expensive. Some consumers are willing
to pay this premium.
b. Any supported opinion is acceptable.
65. Answers may discuss:
• The heavy taxation on cigarette sales, which provide
revenue for the government.
• The maximum allowable exposure levels of
formaldehyde versus the mutagens present
in cigarettes.
• The once socially acceptable status of smoking, and
the evolution of attitudes toward it.
• People’s right to make their own health decisions.
• Smoking lobbyists that are more successful than the
hairspray companies at advocating for the sale of
their product.
66. Answers may discuss:
• Proteins are often considered to be the functioning
units of a cell. While genes provide the instructions
to produce proteins, it is the proteins themselves that
are responsible for performing numerous functions
in the cell.
• While the genome of a cell remains relatively static
throughout time, its proteome is constantly changing.
The levels of proteins in a cell depend on factors
such as cellular environment, disease, cell type, and
developmental stage.
• Many diseases can be traced to defects at the
protein level (i.e., protein structure, levels of protein
expression). Therefore, many therapeutic drugs are
either proteins themselves, or have targets at the
protein level.
• Since a cell’s proteome is dynamic, research has been
focused on identifying protein expression profiles
for certain diseases. For example, it has been found
that some diseases are correlated with high levels of
certain proteins. These proteins can therefore be used
as “biomarkers” when screening for disease.
67. Answers may discuss:
• Epigenetic information determines how and when
genes are accessed and transcribed.
• Epigenetic changes are chemical modifications of
DNA that alter gene expression but do not change
the DNA base sequence. These changes can affect
future generations.
• Examples of epigenetic changes include modifications
to chromatin and histone proteins and methylation
of DNA.
• An individual’s epigenome can change based on
environmental factors. Some environmental factors
which have been studied include malnutrition and
epigenetic carcinogens (these substances do not damage
the DNA itself, but can induce epigenetic changes).
• The epigenome provides additional therapeutic
targets which can be used for the diagnosis and
treatment of disease.
Answers to Chapter 6 Self-Assessment Questions
(Student textbook pages 282–3)
1. c
2. e
3. a
4. a
5. a
6. c
7. a
8. d
9. c
10. b
11. Beadle and Tatum proposed the one-gene/one-enzyme
hypothesis, which states that one gene specifies one
enzyme. This hypothesis has since been updated
to the one-gene/one-polypeptide hypothesis, since
not all proteins are enzymes, and many enzymes are
composed of more than one protein.
12. Diagrams may resemble Figure 6.3 on page 249 of the
student textbook.
13. The main objective of transcription is to accurately
produce mRNA from a section of DNA. The main
objective of translation is to synthesize protein from
an mRNA template.
14. Tables should include:
• Ribosomal RNA (rRNA) is found in the ribosomes,
which is where the messenger RNA (mRNA) is
read and the amino acids are assembled to form
a polypeptide.
Biology 12 Answer Key Unit 3 • MHR TR
23
• Messenger RNA (mRNA) transcribes the genes, the
sequence of nitrogen bases in a strand of DNA, and
carries this “message” from the DNA in the nucleus
to the ribosomes in the cytoplasm.
• Transfer RNAs (tRNA) in the cytoplasm bond
to individual amino acids and take them to the
complementary codons of the mRNA at the binding
site on the ribosome, where a growing polypeptide
chain is built.
15. Once the first RNA polymerase complex moves along
the DNA during the elongation phase of transcription,
a second RNA polymerase complex can bind to the
promoter region and synthesize another mRNA
molecule. Therefore, multiple mRNA molecules can by
synthesized at one time.
16. After transcription, pre-mRNA undergoes the
following modifications:
• 5′ cap—A modified G nucleotide is covalently linked
to the 5′ end. This modification is recognized by
protein synthesis machinery.
• Splicing—Introns are removed by spliceosomes and
exons are joined together.
• poly-A tail—A series of A nucleotides is covalently
linked to the 3′ end. This modification makes mRNA
more stable.
17. Many mRNA sequences are possible due to the
redundancy of the genetic code. For example:
5′-GGU-CUU-GUU-AGA-3′
5′-GGC-CUC-GUC-AGG-3′
18. During transcription, DNA serves as a template
for mRNA. After transcription, mRNA undergoes
modifications so that it can leave the nucleus and
provide genetic information to ribosomes for protein
synthesis in the cytoplasm.
19. The discrepancy in the number of genes and mRNA
transcripts can be attributed to alternative splicing
of mRNA.
20. Diagrams should resemble Figure 6.11A on page 257
of the student textbook plus labels for the anticodon
serine (AGA, AGG, AGU, AGC, UCA, or UCG) and
the amino acid (serine).
21. a. frameshift mutation
b. A frameshift mutation causes the reading frame to
be altered. The insertion of the G nucleotide has
caused the reading frame to be shifted, which leads
to a change in the amino acid sequence.
22. a. Reading from left to right—methionine-threoninelysine-glycine-tyrosine. Reading it from right to
left—histitide-tryptophan-lysine-threonine-valine.
24
MHR TR • Biology 12 Answer Key Unit 3
The amino acid sequences in the two interpretations
are different since they would result in the
production of two different proteins.
b. The original amino acid sequence is methioninethreonine-lysine-glycine-tyrosine. The mutated
amino acid sequence would be methionine-lysinelysine-valine.
c. frameshift mutation
23. a. Photorepair is a specific repair mechanism that fixes
thymine dimers caused by UV radiation. Photolyase
enzyme recognizes the damage, binds to the dimer,
and then excises the dimer by using visible light.
b. Not all DNA damage can be repaired, and therefore
harmful mutations caused by UV radiation can
still accumulate.
24. a. high amounts of lactose → allolactose binds to
repressor → repressor no longer binds to operator
→ transcription of genes induced
b. no lactose → lac repressor protein binds to
operator → RNA polymerase inhibited from binding
to promoter → transcription of genes inhibited
25. The regulation of gene expression allows for cell
specificity. Gene expression is controlled so that only
certain genes are active at certain levels and at certain
times in each cell type. For example, there are some
genes which are only expressed during certain periods
of early development. Unwanted side effects on the cell
(and possible developmental abnormalities) may occur
if these genes continue to be actively expressed beyond
their specified time and level.
Chapter 7 Genetic Research
and Biotechnology
Answers to Learning Check Questions
(Student textbook page 291)
1. • Specificity—The cuts made by an endonuclease are
specific and predictable. That is, the same enzyme
will cut a particular strand of DNA the same way
each time, producing an identical set of small DNA
fragments called restriction fragments.
• Staggered cuts—Most restriction endonucleases
produce a staggered cut that leaves a few unpaired
nucleotides on a single strand at each end of the
restriction fragment. These short strands, often
referred to as “sticky ends,” can then form base pairs
with other short strands that have complementary
strands, creating a recombinant DNA molecule.
2. The construction recombinant DNA molecules
require the use of restriction endonucleases. When
a gene of interest and a vector have the same
restriction endonuclease cutting site, “sticky ends”
can be produced when they undergo a restriction
endonuclease reaction. The “sticky ends” in the gene of
interest can form base pairs with the “sticky ends” of
the plasmid to create a recombinant DNA molecule.
3. Gene cloning is a process that produces many identical
copies of a gene. Cloning genes is useful for studying
gene function and for producing large amounts of
mRNA or proteins from a gene for further study.
4. Selectable markers, such as antibiotic resistant genes or
lacZ genes, are used to identify and specifically select
bacterial colonies which have the recombinant DNA.
5. Answers should be based on Figure 7.6 on page 290 of
the student textbook.
6. • Amplification of “ancient DNA” for
evolutionary studies.
• Screening of genetic defects using a DNA sample
from a single cell.
• Amplification of DNA from small crime scene
samples for further analysis.
(Student textbook page 295)
7. DNA fragments are negatively charged. In gel
electrophoresis, the negatively charged DNA fragments
are attracted to and travel toward the positive terminal.
8. The smaller fragments move more easily through the
spaces between the protein molecules of the gel.
9. Gel electrophoresis allows the analysis of DNA
fragments based on size. When cloning a gene, the
size of the plasmid and the size of the gene are known.
Therefore, to check if the gene was successfully cloned
into a plasmid, researchers will cut the recombinant
DNA with the same restriction endonucleases that
were originally used to create the recombinant DNA.
This sample is then run on a gel to analyze the size of
the resulting fragments. If the fragments correspond
to the same sizes of the gene and the plasmid, then
cloning was successful.
10. A technology used to identify individuals by analyzing
the DNA sequence of certain regions of their genome.
11. Short tandem repeats (STR) are short, repeating
sequences of DNA in the genome that vary in length
between individuals. In STR profiling, the STRs of
an individual are amplified by PCR and analyzed
by gel electrophoresis. Since each DNA fragment
is fluorescently labelled, a detector can be used to
measure the amount of fluorescence emitted from each
STR. The resulting printout is a series of peaks that
correspond to STRs of varying molecular mass.
12. Any two of: identifying paternity in court, identifying
the remains of murder or accident victims, tracing the
movement of wildlife, or in plant and animal breeding
programs. Accept any ethical issues identified as they
may be largely linked to family and cultural values.
(Student textbook page 304)
13. The process of specifically altering the genetic material
of an organism such as in the production of transgenic
bacteria, plants, and animals.
14. • Risks— higher costs and exclusivity, therefore
shutting out researchers and other individuals who
may wish to study or use a GMO; ethics surrounding
ownership of organisms
• Benefits—allows researchers to protect their
intellectual property, provides financial incentive for
development of GMOs, and stimulate further research.
15. Expression vectors have sequences which support
transcription and translation of an inserted gene.
Therefore, if a gene that codes for a therapeutic
protein is inserted into an expression vector, it can be
transcribed and translated in bacterial cells in large
quantities. The protein can then be purified from the
bacteria for medicinal use.
16. Bacteria are easy to use and inexpensive to maintain.
Using transgenic bacteria to produce human insulin has
also removed some of the allergic side effects that were
present when insulin was produced in animal sources.
17. Some bacteria can be genetically modified to improve
their ability to break down pollutants, such as pesticides,
herbicides, and oil spills that are found in water.
18. Positive—environmental pollutants can be cleaned up;
can be designed for the purpose
Negative—food supply or habitat of native plants
and animals may be diminished; mutations may
occur in the transgenic bacteria, which would have
unpredictable effects on an ecosystem
(Student textbook page 307)
19. Increased tolerance to pesticides, herbicides, disease,
and pests; increased yield; a reduced spoilage; reduced
pesticide use; cost savings.
20. The biolistic method involves bombarding plant
cells with particles coated with DNA. Since the
bombardment occurs at a high speed, the DNA
can penetrate the plant cells and integrate into the
plant genome.
Biology 12 Answer Key Unit 3 • MHR TR
25
21. Gene of interest inserted in T DNA of Ti plasmid
→ recombinant Ti plasmid is transformed into
A. tumefaciens → plant cells exposed to A. tumefaciens
which transfers and incorporates the T DNA into
the plant cell chromosome → plant cells killed by
kanamycin and A. tumefaciens killed by carbenicillin
→ surviving plant cells transferred to growth media
with hormones that regenerate an entire plant
22. In horizontal gene transfer, transgenic plants may
transfer their introduced traits to other organisms,
such as other plants, animals, bacteria, or fungi. In
vertical gene transfer, the transgenic plant could
transfer its introduced trait into the genomes of the
natural or wild version of the same plant.
23. Health Canada requires 7 to 10 years of health
and safety research for each new GMO intended
for human consumption. This research includes
investigating food safety and how the GMO was
developed. The main concern surrounding GMOs is
the uncertainty surrounding the long-term effects of
GMO consumption.
24. Answers may include: insecticides could affect
unintended organisms; genes may transfer occur,
creating other insecticide secreting organisms; and
targeted insects may move to other crops.
Answers to Caption Questions
Figure 7.1 (Student textbook page 286): AATT
Figure 7.3 (Student textbook page 287): Instead of the two
different molecules being ligated together to produce one
recombinant molecule, each molecule could react with itself
by forming base pairs within the same molecule. Therefore,
this would produce at least two different molecules.
Figure 7.4 (Student textbook page 289): The gene is
cloned in the last step, where the host divides to produce
many cells.
Figure 7.6 (Student textbook page 290):
230 = 1 073 741 824 copies
Answers to Section 7.1 Review Questions
(Student textbook page 300)
1. A restriction endonuclease can cleave the interior
of double stranded DNA. Two features which
make restriction endonucleases helpful to genetic
engineers are:
• Sequence specificity—The cuts made by an
endonuclease are specific and predictable. That is,
the same enzyme will cut a particular strand of DNA
the same way each time, producing an identical set of
small DNA fragments called restriction fragments.
26
MHR TR • Biology 12 Answer Key Unit 3
• Staggered cuts—Most restriction endonucleases
produce a staggered cut that leaves a few unpaired
nucleotides on a single strand at each end of the
restriction fragment. These short strands, often
referred to as “sticky ends,” can then form base pairs
with other short strands that have complementary
strands, creating a recombinant DNA molecule.
2. When working with DNA, it is important to ensure
that contamination of the sample does not occur.
Precautions include wearing gloves, working in a clean
workspace, and careful pipetting.
3. Gene cloning involves making many identical copies of
a gene.
4. Bacteria are inexpensive, grow very quickly in large
amounts, and are easily maintained.
5. Answers should resemble Figure 7.4 on page 289 of the
student textbook.
6. • Gene cloning—involves making several identical
copies of a gene in foreign cells. It is used to produce
enough RNA or protein from the gene of interest for
further study.
• PCR—is an automated method of making many
copies of specific regions of DNA from very small
quantities. It does not rely on the formation of
recombinant DNA molecules or host systems, which
are involved in gene cloning. PCR is used when only
a small section of DNA is needed (e.g., purifying
a fragment of DNA for analysis or for creating a
fragment to be used for cloning), or when a segment
of DNA needs to be amplified from very small DNA
samples (e.g., crime scene investigations).
7. Denaturation of the double-stranded DNA molecule
would not occur if the initial heating step was not
carried out. Therefore, amplification of the DNA would
not occur.
8. DNA is negatively charged, and therefore moves
through the gel toward the positive anode. Small
fragments of DNA will migrate the farthest since their
size lets them can migrate through the gel more easily.
9. a. Restriction enzymes are added to a sample of DNA
from each plant. The enzymes cut the DNA into
fragments. Small amounts of the DNA sample are
placed into gel electrophoresis wells. An electric
charge is attached to the gel, and the DNA segments
migrate in the gel according to their lengths. The
resulting DNA “fingerprints” are analyzed to
determine if segments from the two plants match,
indicating whether the plants could be clones.
b. Diagrams should show identical bands on the DNA
fingerprints of each plant.
10. DNA fingerprinting is a technique that can identify
individuals based on analysis of their DNA at certain
regions. Applications of DNA fingerprinting include
forensic sciences, determining maternity and paternity,
identifying genetic links to individuals or across
species, identifying genes associated with disease, and
creating DNA fingerprint banks.
11. Male 1 is the father of the child. The child’s DNA
fingerprint shows more matches with the DNA
fingerprint of Male 1 compared to Male 2.
12. The objective of Human Genome Project was to
determine the sequence of DNA in the chromosomes
of the entire human genome. The Human Genome
Project is an important step in understanding how
genes determine our genetic characteristics. This
understanding can be applied to medical genetics and
the treatment of disease, as well as to other sciences.
Answers to Section 7.2 Review Questions
(Student textbook page 311)
1. This method is not ideal since most diseases involve
defects in multiple genes. Gene therapy is a possible
therapeutic approach for diseases which are associated
with a defect in a single gene. There is concern about
the possible negative side effects of altering the human
genome with foreign genes.
2. The term transgenic is used to describe an organism
that is produced from the introduction of foreign DNA
into its genome. A genetically modified organism
(GMO) is an organism whose genetic material has
been altered, not always with foreign DNA.
3. Benefits—Patents offer incentive in the form of
potential monetary compensation, and thus promote
innovation in scientific research.
Risks—Patents may discourage research collaboration
and increase competition. There are also ethical issues
surrounding ownership.
4. An expression vector is a plasmid vector that contains
sequences which support transcription and translation
of the introduced gene. Expression vectors are useful
for producing transgenic bacteria which produce
foreign proteins.
5. Diagrams should resemble Figure 7.17 on page 303 of
the student textbook. Advantages include:
• Less expensive to maintain and easier to grow.
• Removes side effects associated with insulin purified
from animal sources.
6. Bioremediation is the use of micro-organisms to
reduce environmental pollutants.
• Benefit—bacteria are inexpensive and easy to
maintain; avoid chemicals
• Risk—transgenic bacteria can also negatively affect
the balance of an ecosystem (i.e., effects on the food
supply or habitat of native plants and animals). There
is also the potential of mutations which may occur in
the transgenic bacteria, which could have undesirable
side effects on an ecosystem.
7. Biolistic method—Plant cells are bombarded at a high
speed with particles coated with DNA; the speeds
aids penetration.
Ti plasmid method—Ti (tumour-inducing) plasmid is
used as a vector for the insertion of a foreign gene into
the plant genome.
8. a. Gene pharming involves the use of animals to
produce therapeutic proteins. For example, a foreign
protein can be made in the mammary cells of a
sheep by using a recombinant DNA vector that has
a promoter which specifically expresses introduced
genes in the mammary cells. The foreign protein
therefore is expressed in the mammary cells and
is secreted in the sheep’s milk, which will undergo
further processing to purify the foreign protein.
b. Sample answer: The therapeutic proteins produced
have all necessary modifications to function.
Additionally, if the proteins are secreted in an
animal’s milk, it is easily collected and purified.
c. Sample answer: Transgenic animals may breed with
non-transgenic animals, resulting in undesired
offspring or introduction of the gene in the wider
population. The transgenic animals may experience
unwanted side effects from producing foreign
human proteins. Therapeutic proteins may enter the
food supply.
d. Answers may include: Is it ethical to use animals as
“production factories” for human proteins? Should
research look for alternatives to using animals for
gene pharming?
e. Answers may suggest methods such as broad
community consultation, consultation with experts,
and educational campaigns to make the issues
and options clear. Answers may suggest that the
variety of religious and culturally based objections
are insurmountable.
9. a. More salmon are produced in a shorter time.
b. More nutrients are required in their short growth
cycle. Those grown in crowded conditions have a
greater risk of contracting and spreading disease to
Biology 12 Answer Key Unit 3 • MHR TR
27
other fish—both the transgenic and the wild stocks.
They may breed with wild salmon and adversely
affect wild salmon stocks or their role in the
ecosystem. They may be susceptible to disease.
c. Sample answer: Commercial fish-farming operations
should have measures in place that prevent diseases
from being transmitted to other fish. Commercially
grown fish should be contained to prevent escape
and breeding with wild fish.
10. a. The keratin protein may not have been modified
properly since it is in a bacteria host. Therefore, due
to improper modifications, the protein may have
quickly degraded.
b. Express the recombinant DNA with the keratin gene
in human or animal cells (i.e., a non-bacteria host)
to see if the protein is properly expressed.
c. If the results of the experimental approach described
in b are correct, then the recombinant DNA with
keratin should be cultured in a human or animal cell
host instead of in bacteria.
11. Flow charts should resemble Figure 7.21 on page 309 of
the student textbook.
12. Answers may include:
• What would be the effects of introducing previously
extinct species into new ecosystems? How could this
positively or negatively affect an ecosystem?
• Would these species survive in presentday environments?
• What benefits would result from reviving
extinct species?
Answers to Chapter 7 Review Questions
(Student textbook pages 319–23)
1. c
2. c
3. e
4. e
5. a
6. a
7. c
8. b
9. a
10. a
11. c
12. a
13. b
28
MHR TR • Biology 12 Answer Key Unit 3
14. c
15. The universality of the genetic code means that
almost all organisms can make proteins using the
same codons. Therefore, a gene that is taken from
one type of organism can be inserted into another
and will produce the same protein. If the genetic code
was not universal, it would be difficult to assess if a
recombinant DNA molecule containing DNA from
two different organisms would be able to express the
proper protein.
16. • Specificity—The cuts made by an endonuclease are
specific and predictable. That is, the same enzyme
will cut a particular strand of DNA the same way
each time, producing an identical set of small DNA
fragments called restriction fragments.
• Staggered cuts—Most restriction endonucleases
produce a staggered cut that leaves a few unpaired
nucleotides on a single strand at each end of the
restriction fragment. These short strands, often
referred to as “sticky ends,” can then form base pairs
with other short strands that have complementary
strands, creating a recombinant DNA molecule.
17. Certain drugs, such as insulin, were previously purified
from natural sources such as animals. This method
was very complex, labour-intensive, and expensive.
Furthermore, individuals who used the insulin purified
from the animal sources often experienced allergic
reactions. Using transgenic bacteria to produce human
insulin has eliminated the allergic side effects. Producing
drugs in transgenic bacteria is also relatively inexpensive
since bacteria is easy to maintain and grows quickly.
18. DNA is negatively charged and will therefore migrate
through the gel toward the positively-charged anode.
19. STRs are repeating short sequences of DNA that
vary in length between individuals. In STR profiling,
STRs of an individual are amplified by PCR and then
separated by gel electrophoresis. These fragments
are fluorescently labelled, which allows a detector to
measure the fluorescence emitted from each STR. The
DNA fingerprint produced from STR profiling is a
unique pattern of peaks which correspond to specific
molecular masses (which correspond to the different
STR lengths). Each individual has a unique series of
these peaks in their STR profile.
20. Dideoxynucleotides lack an –OH group at the 3′ and
2′ carbons on the ribose sugar. A 3′–OH group is
needed for synthesis to occur since it is the site that
reacts with a new nucleotide. Therefore, DNA synthesis
stops when a dideoxynucleotide is incorporated into
the growing strand.
21. It increased the amount of DNA that could be read in
one sequencing reaction, allowed for multiple reactions
to be run at a time, used smaller tubes and smaller gels
to increase efficiency, and provided computerized data.
22. Site-directed mutagenesis is a method that allows
researchers to specifically alter the nucleotide sequence
in a region of DNA. This technique allows researchers
to perform mutational analysis to investigate the
structure and function of genes and proteins.
23. Genetic engineering is the alteration of the genetic
material of an organism—making specific changes
to the organism’s sequence of DNA (i.e., site-directed
mutagenesis). Biotechnology, however, refers to the
application of technologies that involve the use of
an organism, or product of an organism, to benefit
humans; such as the creation and use of transgenic
animals to produce therapeutic human proteins.
24. Creating transgenic animals can be complicated,
time-consuming, and expensive. However, creating
transgenic bacteria is less complicated, since bacteria
grow quickly and are easy and inexpensive to maintain.
25. Answers may include: to produce therapeutic proteins
(gene pharming); to create pest-resistant crops (and
increase yields); to clean up environmental pollutants
(bioremediation).
26. • Biolistic method—Plant cells are bombarded at a
high speed with particles coated with DNA; the
speeds aids penetration of the DNA which has been
selected for pest resistance, which gets integrated into
the corn genome.
• Ti plasmid method—Ti (tumour-inducing)
plasmid is used as a vector for the insertion of the
pest-resistance gene into the plant genome. Pestresistance genes are inserted into the Ti plasmid to
form recombinant DNA. The recombinant DNA is
transformed into a bacterium which then infects the
corn cells, resulting in the integration of the pestresistance genes into the corn genome.
27. • Advantages of transgenic animals—Certain
therapeutic proteins can undergo necessary
modifications for proper functioning in animal cells
compared to bacterial cells.
• Disadvantages of transgenic animals—Complex and
expensive process; safety concerns about products;
concerns about effect of human proteins on the
health of transgenic animals
• Advantages of transgenic bacteria—Less expensive
and easier to grow and maintain bacteria compared
to animals; can produce therapeutic proteins very
quickly due to bacteria growth rate
• Disadvantages of transgenic bacteria—Certain
therapeutic proteins cannot undergo necessary
modifications in bacteria cells
28. a. The donor sheep provided a mammary cell that
was extracted and then cultured in a flask. Another
sheep provided an unfertilized egg. The mammary
cell and the unfertilized egg were then fused
together. The resulting embryo was then transferred
to the third sheep (the surrogate sheep).
b. Dolly was genetically identical to the donor sheep.
Therefore, the donor sheep was Dolly’s clone.
29. • Many genes that are implicated in human disease
have parallel versions in model organisms such as
yeast, mice, and fruit flies, where they can be studied
easily in various experimental settings.
• Studying parallel human genes in model organisms
may also provide insight into gene function.
• Comparing genomes also allows researchers to study
which regions of a genome have been conserved
amongst different species. These conserved regions
are thought to be essential and important regions of
the genome.
30. The genetic code is nearly universal. Nearly all
organisms use the same genetic code to synthesize
proteins. Therefore, a human gene (such as ras) can be
transcribed and translated in mouse cells, since both
humans and mice use the same codons.
31. Two pieces of DNA cut with the same restriction
endonuclease can readily join together to form a
recombinant DNA molecule since complementary
single-stranded sticky ends are produced.
32. a. Diagrams should show the first 4 steps of Figure 7.4
on page 289 of the student textbook, for creating
the desired plasmid. Digest both plasmids with a
restriction endonuclease enzyme to release gene A
and gene B. In the parallel, a new plasmid containing
a selectable marker, such as the ampicillin resistance
gene, is also digested with the same enzyme. When
DNA fragments are mixed together and incubated
in the presence of DNA ligase, in some cases gene
A and gene B will ligate with the linearized new
plasmid leading to the formation of the desired
recombinant plasmid.
b. Three plasmids, two of which will be the modified
parental plasmids lacking gene A and gene B. The
new third plasmid will be the recombinant, since it
contains both genes.
Biology 12 Answer Key Unit 3 • MHR TR
29
33. 215 = 32 768 copies of DNA; 230 = 1 073 741 824
copies of DNA. The amount of DNA doubles with each
replication cycle (i.e., one copy of DNA produces two
copies after one cycle, four copies after two cycles, eight
copies after three cycles, etc.)
34. PCR would be used to increase the amount of ancient
DNA. Only a very small amount of template DNA is
required for amplification by PCR.
35. 1; 2; || ||
2; 4; || || || ||
3; 8; || || || || || || || ||
36. a. Diagrams should show two distinct bars. The top bar
should be labelled 2000 base pairs (vectors) and the
bottom bar should be labelled 500 base pairs (gene)
b. Diagrams should show one distinct bar labelled
2000 base pairs (vectors only).
37. Diagrams should show a bar where each of the
woman’s and man’s showed a bar.
38. • Amplify STRs using primers and PCR
• Separated resulting STR fragments on a gel
(fluorescently labelled)
• Measure fluorescence emitted from each STR
fragment. This is recorded as a series of distinct
peaks, which represent STRs of differing molecular
mass (differing lengths)
39. 5′-GTGAAAAGAG-3′
40. Premature termination of DNA synthesis. Therefore,
only a few (if any) bases would be able to be sequenced.
41. 5′-TCGGATCTTAAC-3′
42. a. If the gene for stoat coat colour could be isolated
with endonuclease, the DNA could be inserted
into a dog egg cell, which could then be artificially
fertilized and implanted into the uterus of a dog.
If the coat colour gene was coded for, the offspring
may show stoat coat colour.
b. Consideration should be given to the biological
characteristics of the transgenic product, compared
with the characteristics of the natural variety. Will
the dog be healthy and able to breed? Will the dog
have adverse behavioural characteristics and how
will these be controlled?
43. a. The protein synthesized in bacteria may not have
undergone the necessary modifications (i.e.,
post-translational modifications) needed for its
proper function.
b. The plant gene could be inserted into a eukaryotic cell
rather than a bacterial cell. The eukaryotic cell would
have the proper cellular environment to perform the
necessary post-translational modifications.
30
MHR TR • Biology 12 Answer Key Unit 3
44. Sample answer:
Sticky ends
5 G
3
Blunt ends
A AT T C
3
C T TA A
G
5 G A T
A T C 3
3
C TA
TAG
5
5
45. Diagrams should resemble Figure 7.4 on page 289 of
the student textbook.
46. Answers should summarize Figure 7.6 on page 290 of
the student textbook.
47. Sample answer:
a. DNA fingerprinting is a chemical process that makes
a map of the genes found in the cells in your body.
Or in a plant or another animal.
b. Who owns your DNA fingerprint? Why would
companies or the government want to collect
everyone’s DNA fingerprint? Should everyone
have their genes checked for potential to develop
diseases? What privacy concerns do you have
about someone being able to access and use your
DNA fingerprint?
c. Procedure, of applications, social and ethical issues,
and data management.
48. Essays should describe the micro-organism and
how it is used in bioremediation, and advantages
and disadvantages of using the micro-organism
for bioremediation.
49. Distribute BLM A-33 Communication Rubric to
show students how the content of the brochure will be
assessed. Brochures should use language and diagrams
that are appropriate for the general public. Creative use
of colour and design should also be assessed in terms
of attractiveness and readability.
50. While answers will be based on students’ beliefs,
the answer should clearly define biotechnology in
relation to food productivity and security. An opinion
agreeing or disagreeing with the statement must be
included with an explanation of the concepts of food
productivity and security using relevant examples. The
author’s conclusion must be clear.
51. Diagrams should resemble Figure 7.17 on page 303 of
the student textbook.
52. Benefits— Produce therapeutic proteins for humans;
therapeutic proteins produced in transgenic animals
have necessary modifications of proteins for proper
functioning; therapeutic proteins produced in a
transgenic animal’s milk are easy to obtain and purify
Risks— Transgenic animals may experience
long-term side-effects due to foreign protein
production; transgenic animals may enter wild
populations, resulting in unknown consequences on
other organisms
53. Sample answer in support of human cloning:
• may contribute to further understanding of the
complexities of gene expression in humans
• may provide a reproductive alternative for
infertile couples
Sample answer against human cloning:
• cloned offspring have a high mortality rate, and high
incidence of disease
• many clones show signs of metabolic disorders, such
as premature aging
• social/ethical issues such as eugenics and
gene discrimination
54. Graphic organizers should include the key concepts
and relationships from the Chapter 7 Summary on
page 318 of the student textbook.
55. Sample answer: Government regulations do require
testing of new products, technology, and procedures
before they are approved for the general public.
For example, Health Canada and the Federal Drug
Administration in the United States regulate the use
of new drugs. While this may be an effective way
to protect the general public from risks associated
with biotechnology, in some instances, personal
risk assessment may be a better option. A person
who is terminally ill may wish to have a “last resort”
treatment, or access to a drug that has not been
approved for use. Thus, in some specific cases,
government regulations may not serve the individual.
56. Objections could focus on concerns about sampling
practices or lab practices relating to the storage and
handling of samples, testing and record keeping. Points
in support should explain the uniqueness of DNA and
the ability to use it to avoid wrongful convictions.
57. a. Answers may take the form of a risk/benefit
analysis or may reflect students’ beliefs. Accept any
supported answer.
b. Answers should indicate an understanding of
the ethical impacts of the genetic testing on an
individual, a society, and a government.
58. a. Answers should reflect the Canadian government’s
most current policies on labelling GMFs. As of
2011, the food labelling standard for genetically
engineered foods (GE foods) was the national
standard developed in 2004. Labelling of GE foods
was voluntary.
b. Opinion should be clear and supported by coherent
and concrete examples for the statement for or against
the policy.
59. a. The significance of the discovery of the herbicide
resistant weed depends on the genetic relationship
and interaction between the weed and the canola.
Will the presence of the weed make it difficult to
grow canola? If this is the case, are there other
herbicides that are safe to use with canola that
are effective against the weed? How did the weed
acquire resistance to the herbicide? Could the
original research that produced the transgenic
canola have predicted the occurrence of the
herbicide resistant weed?
b. Answers may include:
• The farmer worries that the weed will affect the
quality and quantity of the canola crop and worries
about how to control the weed without using
herbicides that will reduce the market value of
the canola.
• An official from the genetic engineering corporation
that created the transgenic canola expresses regrets
that this happened. The company will work on
a solution that may be years away. The farmer is
reimbursed for the cost of the canola seed and
potential crop loss.
• The owner of a nearby organic farm is concerned
about the possible use of herbicides on canola. If
the herbicide should drift to his/her neighbouring
crops, it may result in the loss of his/her organic
crop designation.
• A consumer organization opposed to the
development of genetically modified organisms
wants the public to be informed about the health
risks of consuming GMO foods. The organization
is concerned that the developers of the transgenic
herbicide-resistant canola have not researched the
long-term health effects of this product.
• A genetics researcher wants a grant from the genetic
engineering corporation to study the weed and how it
developed resistance to the herbicide.
60. a. Sample answer: Who owns the genetic information?
Should companies have the right to sell DNA
information to other companies without the
permission of the people who provided the samples?
Should companies that use DNA in medical research
be required to share the results of their work with
the individuals, or communities, whose genetic
information was used? Where is the boundary
between public and private genetic property rights?
Biology 12 Answer Key Unit 3 • MHR TR
31
b. Sample answer: Providing genetic information
should be voluntary. Genetic information can
provide medical benefits to many. Research should
be funded so as to maximize benefit for as many as
possible. It is reasonable for companies to expect a
return on their investment in genetic research. The
outcome of the research should be widely available.
61. Answers should describe and address the two scenarios
(production of human insulin in bacterial cells to treat
diabetes vs mixing animal or plant DNA in agricultural
biotechnology). Answers could take into account the
historical times (early 1920s vs 2000s) and the level of
public knowledge about science and/or trust in science
and medicine in general. Answers may also address
the idea that while the development of insulin quickly
saved lives, the suggested benefits of consuming GM
foods may not be realized and the risks are unknown.
62. a. Either yes or no is acceptable.
b. Answers should list sources or data that support
their position using scientific terms and concepts.
63. Novel foods:
• result from a process not previously used for food
• do not have a history of safe use as a food
• are modified by genetic manipulation or are
biotechnology-derived
The Novel Foods Regulation (under the Food and Drugs
Act) requires that notification be made to the Health
Products and Food Branch (HPFB) by any company
wanting to sell the product prior to marketing or
advertising it. Health Canada conducts a safety
assessment of all biotechnology-derived foods to
demonstrate that a novel food is safe and nutritious
before it is allowed in the Canadian marketplace.
64. a. Answers could include a survey of how other
countries have dealt with the issue. Be aware
that student research into the Canadian decision
may uncover suggestions of controversy within
Health Canada. The final decision was made after
consultation with an independent panel.
b. Answers should state a clear yes or no and be
supported by scientific reasoning.
Using transgenic pigs
as organ donors
65. • an increased amount of DNA that can be read in one
sequencing reaction
• multiple reactions can be run at a time
• all dideoxynucleotides can be put in one tube, since
each is dye labelled
• smaller reaction tubes and smaller gels
increase efficiency
• computerized data collection
66. As of 2011, answers should reflect the fact that cloning
is governed under the Assisted Human Reproduction
Act. According to Health Canada’s web site, “The
AHR Act is one of the most comprehensive pieces
of legislation in the world concerning reproductive
technologies and related research. The legislation has
three objectives: it prohibits human cloning and other
unacceptable activities; it seeks to protect the health
and safety of Canadians who use AHR procedures;
and it requires that AHR related research, which
may help find treatments for infertility and diseases
such as Parkinson’s and Alzheimer’s, takes place in a
controlled environment?
Students using the relevant Health Canada site should
be able to include references to legislation in Australia,
France, Germany, Italy, Japan, Singapore, Sweden,
United Kingdom, the United Nations, and the United
States when they make their comparisons.
67. a. and b. Any one of: DNA fingerprinting/profiling;
the Human Genome Project; gene therapy; genetic
engineering; genetically modified organisms;
bioremediation; or gene pharming.
c. Reports should support opinion using relevant
scientific concepts and terminology.
68. Sample answer:
• Cost—Is it affordable? Should it be covered by OHIP?
• Quality and usefulness of information provided—Is
it medically useful? Are health care professionals
involved? How should an individual interpret data?
• Privacy concerns—How do the companies ensure that
your information is secure, and not shared or sold?
69. Sample answer:
Benefits
Risks
To individual people
• disease treatment
• long term use of anti-rejection drugs and side-effects are unknown
To society
• increased longevity
• better quality of life
• ethical challenges might arise
• possibility of infectious disease transferring from pigs to humans
To the economy
• fewer health care requirements
• individuals remain productive
• bearing the cost of potential adverse effects
To other species
None
None
To the environment
None
None
32
MHR TR • Biology 12 Answer Key Unit 3
Answers to Chapter 7 Self-Assessment Questions
(Student textbook pages 324–5)
1. a
2. e
3. a
4. b
5. d
6. a
7. a
8. c
9. a
10. d
11. a. The cuts made by an endonuclease are specific and
predictable. That is, the same enzyme will cut a
particular strand of DNA the same way each time,
producing an identical set of small DNA fragments
called restriction fragments.
b. Most restriction endonucleases produce a staggered
cut that leaves a few unpaired nucleotides on a
single strand at each end of the restriction fragment.
These short strands, often referred to as “sticky
ends,” can then form base pairs with other short
strands that have complementary strands, creating
a recombinant DNA molecule.
12. DNA ligase is used to covalently join two different
cut fragments of DNA. This creates a stable, doublestranded recombinant DNA molecule.
13. a. Answers should summarize the information in
Figure 7.4 on page 289 of the student textbook. GHI
and the expression vector (with selectable markers)
should be labelled.
b. Selectable markers are used to identify successful
recombinant DNA. For example, if the expression
vector contained the lacZ gene and an ampicillin
resistance gene, the bacteria cells (after transformation)
would be spread onto growth media which contains
ampicillin antibiotic and X-gal (which turns bacterial
colonies blue when bacteria are broken down by the
enzyme coded by the active lacZ gene). Bacterial
colonies that grown on plates containing ampicillin
contain the recombinant DNA or the plasmid only
since both have the ampicillin-resistance gene. From
this population, the bacterial colonies will either be
blue (due to the presence of active lacZ gene, and
therefore no GHI gene) or white (due to inactive
lacZ gene caused by the insertion of the GHI gene).
Therefore, all the white colonies will contain the
recombinant DNA.
14. The two nucleotide primers will not be able to anneal
to the denatured strands of DNA and amplification will
not occur.
15. a. Dideoxynucleotides lack an –OH group at the
3′ and 2′ carbons on the ribose sugar. Therefore,
DNA synthesis stops when a dideoxynucleotide
is incorporated.
b. The 3′–OH group normally reacts with a new
nucleotide during DNA synthesis.
16. Yes, the DNA of an individual is unique and shared
among all of the individual’s cells.
17. Sample answer:
• Privacy concerns—Who will have access to this
information? What types of regulation would be
needed regarding databank access?
• Use—Under what circumstances would national
DNA databank be useful/harmful?
18. Answers should be written with consideration for the
general public. Diagrams may also be used.
19. The Ti plasmid method could be used to produce a
transgenic tomato containing insecticide genes. This
method involves inserting the insecticide gene into
the Ti plasmid to form a recombinant DNA molecule.
The recombinant DNA plasmid is then transformed
into the bacterium A. tumefaciens. The tomato cells
are then exposed to the bacterium. The recombinant
DNA plasmid with the insecticide gene integrates into
the tomato cells. Selection of tomato cells containing
the recombinant DNA plasmid is performed by using
selectable markers.
20. Sample answer: individuals may be denied insurance
coverage and/or employment based on their DNA
profile or DNA fingerprint and any associated potential
for disorders or diseases.
21. Answers may include:
• increased crop yield; increased nutritional value and
food quality due to less spoilage; greater resistance
to pests and diseases; reduction of pesticide usage;
reduction of harvesting costs
• genes may transfer to other plants and animals;
insecticide-resistance plants may release toxins into
ecosystems by affecting non-target organisms; food
safety issues are unknown, as are the long-term
side effects
22. a. Risks—Genes may transfer to other organisms;
disruption to ecosystems (effects on food chains,
toxicity to other organisms); unknown side-effects
of transgenic mosquito bites on humans
Biology 12 Answer Key Unit 3 • MHR TR
33
Benefits—Malaria could be controlled without
individuals having to use expensive drugs or
vaccinations; spread of malaria could be slowed or
halted quickly
b. Answers will be based on students’ beliefs.
23. Graphic organizers should include:
• Transgenic animals—Production is complex and
expensive; concerns about safety of therapeutic
proteins produced; certain therapeutic proteins
can undergo necessary modifications for proper
functioning in animal cells
• Transgenic bacteria—Inexpensive and easy to grow
and maintain; can produce therapeutic proteins
very quickly due to bacteria growth rate; certain
therapeutic proteins cannot undergo necessary
modifications in bacteria cells
24. Answers should resemble Figure 7.22 on page 309 of
the student textbook.
25. Answers may include:
• Cloning as a reproduction alternative for
infertile couples
• Cloning as a way to produce organs for transplants
• Risks of the commodification of human cloning
Answers to Unit 3 Review Questions
(Student textbook pages 329–33)
1. c
2. a
3. e
4. b
5. b
6. d
7. a
8. e
9. c
10. d
11. c
12. c
13. b
14. b
15. c
16. The parent DNA is antiparallel and the enzymes need
to attach at the 3′ end and travel toward the 5′ end.
34
MHR TR • Biology 12 Answer Key Unit 3
17. They both facilitate the sequencing of the new strand
by connecting complimentary bases according to
the template. DNA polymerase produces DNA. RNA
polymerase produces mRNA. DNA polymerase has
proofreading ability and RNA polymerase does not.
18. Proteins can be functional or structural. Proteins are
found in the cell membrane as markers and channels.
They are also found inside the cell and create the
exoskeleton as microtubules and microfibers. Some
proteins are enzymes and others are hormones. The
combination of these proteins will determine what the
cell looks like and what it is able to do.
19. Gene expression refers to the synthesis of protein from
DNA in is two steps: transcription and translation. By
having two discrete stages, more control mechanisms
can be applied to the production of protein. It also
allows for mistakes to be corrected in more than one
stage. Transcription occurs inside the nucleus and can
result in many copies of mRNA travelling to many
different ribosomes for the production of the protein
through translation. If proteins were made directly
from DNA, only one protein could be made at a time.
20. Energy for binding the amino acid to tRNA comes
from ATP conversion to ADP. Proteins are essential for
cell functioning and growth so translation occurs all of
the time and tRNA is constantly picking up new amino
acids for the process.
21. The structure of the codon (3 bases) provides several
different codons for each amino acid. If a mistake is
made during transcription and the variable nucleotide
is the one affected, the same amino acid is coded for.
Transcription only affects the proteins made from
that strand of mRNA. However, if DNA replicates
incorrectly, every cell from that point on will have
the error.
22. • Prokaryotes—rate of replication is faster, have one
origin of replication due to their single chromosome
• Similarities—require origins of replication, have
elongation in the 5′-3′ direction, have continuous
synthesis of a leading strand and discontinuous
synthesis of a lagging strand, require the use
of a primer for the lagging strand, use DNA
polymerase enzymes
• Eukaryotes—13 different DNA polymerases
(compared to five in prokaryotes), have multiple
origins of replication, replication occurs inside
the nucleus
23. • Specific—DNA polymerase II proofreads the newly
synthesized strand to make sure that the correct bases
are added.
• Non-specific—Other enzymes that participate in
the mismatch repair process look for deformities
in the new strand and remove an incorrectly
added nucleotide.
24. AUG, UAC, methionine
25. • Initiation—Transfer RNA picks up amino acids as per
their anticodon. mRNA combines with rRNA and
slips along it until AUG is in the “P” site.
• Elongation—The next anticodon on the mRNA fits
into the “A” site of the ribosome and the two amino
acids are joined by a peptide bond. When methionine
is bonded to the next amino acid, its tRNA is released
and goes to pick up another methionine. The mRNA
is read one codon at a time, with the A site containing
the new amino acid and the P site containing the
tRNA with the polypeptide growing from it.
• Termination—When the stop codon is reached, there
is no tRNA to bind and the polypeptide is released.
26. Multiple copies of mRNA can be made from a gene
and travel to multiple ribosomes. This allows large
amounts of enzymes to be made at once when the cell
requires them.
27. • Gene cloning—Recombinant DNA is inserted into
E. coli cells which are then cultured. The DNA is later
removed and purified. This is used to amplify large
segments of chromosomal DNA.
• PCR—DNA is denatured, cooled, and has DNA
polymerase added. There is no need for a host and
this provides large amounts of DNA very quickly.
This is used to amplify a gene and can be used on
very small traces of DNA.
28. Individualized medicine based upon a person’s genome
sequence and cancer diagnosis and treatment by
tumour profiling.
29. Biolistic method—Plant cells are struck with tiny particles
of gold or platinum that are coated with DNA. The DNA
penetrates the walls of the cells and is incorporated.
Ti plasmid method—The plasmid is placed into
a bacteria which infects the plant. The DNA then
integrates into the host genome.
30. A foreign gene is inserted into the egg which is then
fertilized. The fertilized egg is implanted into the
host and allowed to develop. Applications include
increasing digestive efficiency in pigs which reduces
their environmental impact and increases the size of
salmon to increase the yield at harvest.
31. It would be too difficult to “unzip” them during
replication. ATP would be needed for each step. The
process would be much slower and would consume a
lot more energy.
32. a. produces the antiparallel complementary strand
b. produces the correct mRNA sequence which
is complementary to the DNA sequence
being transcribed
c. attaches tRNA to the correct location on the mRNA
by matching its anticodon to the mRNA codon
33. Although it depends on the type of mutation, in
general, the answer is yes. The exons are the coding
regions which are used in the production of proteins.
Therefore, a mutation could alter the protein sequence.
Introns are non-coding regions and if they are mutated,
will not likely to have an effect on protein production.
34. The DNA is wrapped around proteins called histones.
35. 26% adenine, 23% cytosine, 23% guanine
36. To make more protein at one time.
37. a. The promoter region is the place where an activator
or repressor attaches to control transcription.
Specifically in E. coli, the promotor regions consist
of two specifically positioned sequences that are
necessary for RNA polymerase complex to correctly
bind to the DNA template to promote transcription
in the correct direction. Transcription will not occur
if the mutation does not allow RNA polymerase to
bind to it.
b. Same answer as in a. Transcription of the gene will
not occur if the mutation affects the binding of RNA
polymerase to the promoter region.
38. a. Genes code for all kinds of proteins (which are
polypeptides). The hypothesis was updated when
the genetic code was broken and the process of
protein synthesis was molecularly defined instead
of being defined by examining gene defects and
metabolic disorders.
b. Some proteins are composed of more than one subunit
made by more than one gene. Variation can exist in
genes which code for the same protein. In addition,
there are non-coding regions of genes which are
excised before translation occurs. It would be more
accurate to say that DNA codes for polypeptides.
39. They look for a complementary sequence. They create a
spliceosome. snRNPs are involved in the modification
of mRNA molecules. They recognize the regions where
exons and introns meet. This allows for the removal
of introns.
Biology 12 Answer Key Unit 3 • MHR TR
35
40. a. th edo gwa sma d
b. The entire sequence of amino acids can be altered; a
stop codon can occur that stops translation.
41. Protein synthesis is catalyzed by enzymes. In addition,
the attraction of mRNA for rRNA depends on the
presence of “active sites”—the AUG sequence on
mRNA connects to the P site of the ribosome. Each
time tRNA attaches to mRNA, it occurs due to
the complementary attraction between the codon
and anticodon.
42. a. 5′-CAAUUUCCG-3′
b. GUUAAAGGC
c. glutamine-phenylananine-proline
43. Enzymes (which are made of proteins) assist the
process of synthesizing proteins.
44. An operon has a promoter followed by a coding region.
This allows for simple control of transcription through
positive gene regulation (an activator turns it on) or
negative gene regulation (a repressor turns it off).
45. • Restriction endonucleases—These enzymes specifically
cleave double-stranded DNA in a sequence-specific
manner. They can be used to cut the gene from
purified genomic DNA, and to cut the vector.
• Gene cloning—This process manipulates DNA to
produce many copies of a gene or segment of DNA in
foreign cells. A gene of interest can be inserted into
a vector to form a recombinant DNA molecule. This
recombinant DNA molecule can be taken up into
a bacterial host to make several identical copies for
further study.
• PCR—The polymerase chain reaction is an automated
method for amplifying specific regions of DNA. It can
be used to amplify the gene from genomic DNA.
• Gel electrophoresis—This technique uses an electric
field to separate fragments of DNA, based on size,
as they pass through a gel. It can be used for DNA
fragment analysis and purification, and to check if a
gene has been successfully inserted into a vector.
• Dideoxy sequencing—This method determines the
sequence of DNA using dideoxynucleotides. It can be
used to determine the sequence in a specific fragment
of DNA (such as a gene).
• Site directed mutagenesis—Allows insertion of
mutations in a DNA sequence. It can be used to study
the structure and function of a gene, as well as its
protein products. For example, a specific mutation
can be inserted in a gene to see what affect it has on
the activity of the resulting mutant protein.
36
MHR TR • Biology 12 Answer Key Unit 3
46. a. When lactose is present, the gene will be making
lactase to break it down. The gene would always be
transcribed due to the lack of a repressor.
b. The cell would continue to make lactase even if there is
no substrate since there is no way to turn the gene off.
47. a. Use EcoR 1 because it will release Gene B DNA that
can be cloned into the EcoR 1 site in Plasmid A.
b. I wanted to remove the resistance gene and it has
EcoR 1 sequence to create sticky ends that would
attach to the sticky ends of Gene B. The other
restriction endonuclease Hind III does not disrupt
the resistance gene.
c. The removal of the resistance gene will make testing
for tetracycline resistance possible. If the bacteria is
not resistant, the new gene has been inserted.
48. a. –b. Diagrams should show identical lines in
column A as those in the Evidence column.
Column B should have the same number of lines
but in different locations.
49. a. Isolate the gene for the protein from the bacteria
using restriction endonucleases, then insert the
gene into a Ti plasmid and then insert the plasmid
into A. tumefaciens. Infect plant cells with the
bacteria. Culture the plant cells in kanamycin and
carbenicillin to kill cells that do not contain the new
plasmid. Grow the remaining cells into a plant.
b. Making sure that the fruit is safe to consume and
that the new crops do not harm other organisms/
the environment.
50. a. Isolate the gene and insert it into a plasmid that
has the β-lactoglobulin promoter. Then insert the
resulting recombinant DNA molecule into a cow
oocyte. Fertilize the egg and place the embryo into
a host female cow. The milk of the new cow should
now contain the protein.
b. Diagrams should resemble Figure 7.20 #1 on
page 308 of the student textbook.
c. The production of proteins in bacteria do not always
result in the correct form of the protein. Animal
hosts perform the correct modifications to the
proteins so they are functional. The protein also
does not degrade as rapidly in animal hosts as they
do in bacteria.
51. Diagrams should resemble Figure 5.3 in the student
textbook. Answers should mention the main
procedures and results of the experiments. The terms
bacteriophage, radioactive, DNA, protein, capsid, and
centrifugation should be included.
Results showed that the bacteria were shown to
be radioactive in the pellet when the DNA was
radioactively labelled. When the capsid protein was
radioactively labelled, the bacteriophage “ghosts” were
radioactive. These results were significant because it
provided evidence that DNA was the source of genetic
information for viral reproduction.
54. A—DNA to be copied or 5′ to 3′ DNA strand to
be copied
B—DNA polymerase
C—DNA ligase
D—leading strand
E—Okazaki fragments
F—lagging strand
52. Diagrams should include different symbols to represent
sugar, phosphate, and the nucleotides. Diagrams
should demonstrate an understanding of structure
by showing an alternating of sugar and phosphate for
the sugar-phosphate backbone. The use of nucleotide
symbols should show complementary base pairing (A
and T, G and C). See Figures 5.7 and 6.5 on pages 213
and 251 of the student textbook.
55. Answers might include a rhyme, song, or mnemonic.
Manipulatives such as games and Foldables™ can also
be generated as a study tool.
53. If DNA replication were conservative, the banding
pattern of bands after one round of replication would
have shown two bands: one at 14N and one at 15N.
Subsequent rounds of replication and centrifugation
would show a similar pattern.
Diagrams should resemble Figure 5.16 on page 221
of the student textbook (except with the new banding
pattern). The representation of the strands should
display the original strand of DNA being conserved
throughout rounds of replication, with only 15N
labelling. Subsequently replicated strands should only
have 14N labelling.
Type of Mutation
56. Prokaryotic mRNA does not need modification
before it can be used for protein synthesis. Eukaryotic
precursor mRNA requires several modifications.
One of these modifications is the removal of introns
through the process of splicing. Diagrams should show
that the eukaryotic precursor mRNA has introns and
exons whereas the prokaryotic mRNA does not.
57. Diagrams should illustrate the addition of a 5′ cap
and a 3′ poly-A tail. The removal of introns through
splicing should also be indicated.
58. Diagrams should be similar to Figure 6.25 on
page 269 of the student textbook. Components (DNA,
pre-mRNA, mRNA, ribosome, polypeptide), processes
(transcription, RNA processing, translation), and
location (nucleus, cytosol) should be labelled. Arrows
should be used to show correct sequence of events.
59.
Description
Example
Effect on cell or organism
Point mutation
A change in a DNA sequence resulting
from a substitution, insertion, or
deletion of a single base pair.
Substitution:
AAG CCT AGC to AAG CTT AGC
Variable effects, depending on
the result of the mutation.
Frameshift mutation
An insertion or deletion of nucleotides
in a number not divisible by three,
which affects neighbouring coding
sequences and causes a shift in the
reading frame.
AAG CCT AAC to ATG CCC TAG C Change in amino acid sequence
of the resulting protein.
Silent mutation
AAG CCT AGC to AAG CCC AGC
A change in the DNA sequence that
does not result in a change in amino
acid sequence (e.g., due to redundancy
of the genetic code).
Amino acid sequence is
unaffected and therefore
the protein remains
unaffected. There are no
effects on the organism.
Missense mutation
A change in the DNA sequence that
changes the amino acid sequence.
Substitution of a single base
pair: AAG CCT AGC to
AAG ACT AGC
Results in a change in amino
acid sequence. Effects will vary
due to the extent of amino acid
sequence change.
Nonsense mutation
A change in the DNA sequence that
introduces a premature stop codon
that results in a truncated protein.
Substitution of a single base
pair causing a premature
stop codon: AAG CCT AGC to
AAG CCT ATC
A premature stop codon will
be produced. This will result in
a truncated protein which may
cause loss of function.
Biology 12 Answer Key Unit 3 • MHR TR
37
60. Tables should include:
• mRNA (messenger RNA)—contains genetic
information that determines the amino acid sequence
of a protein, the template for translation
• tRNA (transfer RNA)—contains an anticodon that
base-pairs with a codon on the mRNA and has the
corresponding amino acid attached to it according
to the genetic code, involved in the translation
of mRNA
• rRNA (ribosomal RNA)—involved in the translation
of mRNA, associates with proteins to produce
ribosomes which provide a place for the interaction
of mRNA, tRNAs, and enzymes for protein synthesis
61. Graphic organizers such as charts or concept maps
should include pre-transcriptional, transcriptional,
post-transcriptional, translational, and posttranslational (summarized on pages 269–70 of the
student textbook).
62. Diagrams should illustrate that sticky ends have
complementary overhangs whereas blunt ends do not
produce overhangs and lack specificity. For sticky ends,
see Figure 7.1 on page 286 of the student textbook.
63. Diagrams should show the parts of both operons (see
Figures 6.23 and 6.24 on pages 267 and 268 of the
student textbook) and should identify:
• The CAP-binding site for the binding of the activator
protein CAP is present in the lac operon but not in
the trp operon.
• The lac operon contains three genes whereas the trp
operon contains five genes.
• The two operons use opposite modes of regulation.
In the case of lac operon, transcription is off in
the absence of lactose. Whereas in the case of trp
operon, transcription remains on in the absence of
tryptophan amino acid.
64. a. Therapeutic cloning involves the use of SCNT
(Somatic Cell Nuclear Transfer). A somatic cell is
taken from the patient and the nucleus is removed
and placed into a donor egg whose nucleus was
previous removed. Replication is then stimulated
for the production of stem cells. These stem cells
can then be used for therapeutic treatment of the
patient. Diagrams may include a step by step outline
of the SCNT procedure or possible uses.
b. Therapeutic cloning is not permitted in Canada.
The process of SCNT is banned under the Canadian
Assisted Human Reproduction Act.
c. Benefits—
• Many possibilities for disease treatment
38
MHR TR • Biology 12 Answer Key Unit 3
• Generated stem cells can be used to grow tissue or
organs required
• Host DNA provides genetic information for the cells
Risks—
• There are ethical, legal, and financial concerns
• Difficulties in generating a viable cell after
nuclear transfer
• Research is limited in this area due to bans in
certain countries
65. Flowcharts resemble Figure 7.20 on page 308 of the
student textbook.
66. Students should use ethical, practical, and financial
reasons to support their Point/Counterpoint editorial.
The article should have a clear thesis statement and
should show a clear understanding of cloning practices,
using proper terminology. Articles should also be
supported by points and concrete examples from
several reputable reference sources.
67. a. Inhibiting translation halts translation and thus
prevents bacterial growth, since proteins (the products
of translation) are an essential part of growth.
b. By only binding to bacterial ribosomes, these
antibiotics only inhibit bacterial growth and
therefore do not affect the machinery of the host.
This allows the antibiotic to target only bacteria.
68. I. Patient A will not produce a protein for the parkin
gene if the gene is deleted.
II. lengthen
III. no change
69. a. Since the BLM gene is a member of the DNA
helicase family, it can be hypothesized that
a mutation in the BLM gene could affect the
unwinding action of the helicase during DNA
replication. Defects in the helicase could cause
the high frequency of chromosomal breaks and
rearrangements that are seen in Bloom’s syndrome.
b. • Symptoms could be highly skin related (not
optimal for gene therapy).
• The higher proportion of chromosomal breaks and
rearrangements could cause problems for potential
integration therapy.
• Ethical issues.
70. a. The policy states that the information collected
for the National DNA Data Bank will be used
for law enforcement only. Various safeguards
and procedures are also in place to ensure the
information is not misused. A National DNA Data
Bank Advisory Committee has been established to
work with the RCMP.
b. Any opinion is acceptable if it is includes well
thought-out arguments supported by research.
Answers should show evidence of critical analysis
of the hypothetical situation, possibly supported by
strengths or flaws in the National DNA Data Bank’s
Privacy and Security policy.
71. Answers should show an understanding of patents in
the field of scientific techniques and tools. Answers
should be supported by well thought-out arguments.
For example, if patents exist for techniques and tools,
the process of research will be impeded due to lack
of competition. On the other hand, if patents are not
allowed, how are companies compensated for their
expensive discovery?
72. a. Since certain genes in the Ras family have shown
unregulated overexpression in many cancers, RNAi
therapy could be used to target these genes for
“gene silencing.” RNAi can target specific mRNAs
to stop the expression of certain gene products.
A reduction in gene products will counteract the
overexpression of these genes which are associated
with different cancers.
b. Drawbacks could include a complete silencing of
the Ras family of genes which have functions in
normal cells. It is hard to regulate gene expression to
appropriate levels. Also, multiple genes may need to
be targeted, which could lead to complications.
73. Answers should address the process of “personalized”
medical treatment using appropriate terminology.
Supporting arguments should show evidence of critical
thinking regarding the benefits and drawbacks. Issues
such as effectiveness, feasibility, privacy, security, and
cost may be addressed.
Answers to Unit 3 Self-Assessment Questions
(Student textbook pages 334–5)
1. c
2. b
3. c
4. a
5. a
6. d
11. Hershey and Chase used bacteriophages to find out
whether DNA or protein was the hereditary material.
Bacteriophages are viruses that inject their genetic
material into bacterial cells. Hershey and Chase found
that a virus with radioactively-labelled protein did
not transfer this protein to bacteria, but a virus with
radioactively-labelled DNA did transfer its DNA
to bacteria.
12. The X-ray shows the diffraction pattern of DNA.
From this, Franklin determined that DNA has
a defined helical structure with two repeating
patterns at intervals of 0.34 nm and 3.4 nm. This
evidence supported Watson and Crick’s idea that
the components of DNA fit together in a ladder-like
helix, with sugar-phosphate molecules on the outside
and nitrogenous bases on the inside. In the Watson
and Crick model, the “rungs” of the DNA ladder
have a constant width, as shown by Franklin’s X-ray
diffraction pattern.
13. Beadle and Tatum proposed that a single gene
determines the production of one enzyme, a concept
known as the “one-gene/one-enzyme hypothesis.” This
hypothesis was later updated to account for the fact that
not all proteins are enzymes, and that some enzymes are
composed of two or more polypeptide chains. The onegene/one- polypeptide hypothesis states that one gene
codes for one polypeptide (or protein).
14. Tables should include:
• DNA—deoxyribonucleic acid; polymer of
nucleotides, each with a phosphate group, sugar
group, and base; contains deoxyribose; contains the
nucleotides adenine, thymine, guanine, and cytosine;
double-stranded (double helix); a template for RNA
synthesis (transcription)
• RNA—ribonucleic acid; polymer of nucleotides,
each with a phosphate group, sugar group, and
base (same as DNA); contains ribose; contains the
nucleotides adenine, uracil, guanine, and cytosine;
single-stranded; codes for amino acids; provides
instructions for protein synthesis (translation)
Has various forms with different functions; key types
are messenger RNA (mRNA), transfer RNA (tRNA),
and ribosomal RNA (rRNA)
8. a
15. The structure of a protein depends on its amino acid
sequence. The amino acid sequence is ultimately
encoded in the sequence of nucleotide triplets in DNA.
9. a
16. 18% thymine, 18% guanine, and 18% cytosine
7. b
10. e
Biology 12 Answer Key Unit 3 • MHR TR
39
17.
Enzyme/Protein
Functions
Helicase
Helps unwind the parent DNA
Single-strandbinding protein
Stabilizes single-stranded regions of
DNA during unwinding
Topoisomerase II
Relieves strain on DNA caused by
unwinding
Primase
Synthesizes RNA primer
DNA polymerase III
Adds nucleotides to the 3′ end of
growing DNA strand
DNA polymerase I
Removes RNA primer and fills gaps
between Okazaki fragments; proofreads
newly synthesized DNA
DNA polymerase II
Proofreads newly synthesized DNA
DNA ligase
Joins the ends of Okazaki fragments
18. Graphic organizers should identify key concepts
and relationships illustrated in Figure 5.12 on
page 216 of the student textbook, from the simplest
level of DNA organization (nucleosomes) to its highest
(chromosomes).
19. a. 5′-TACACATGCATC-3′
3′-ATGTGTACGTAG-5′
b. The 3′ end of the segment has the free –OH group.
c. The amino acid sequence of the polypeptide
product of this gene is methionine start-cysteinethreonine-stop.
d. Sample answer: A nucleotide substitution in
the DNA segment 5′-TACACATGCATC-3′ to
5′-TACACATGGATC-3′ would be a silent mutation
because the mutation does not result in an amino
acid change: TGG is transcribed into ACC, which
codes for threonine.
20. Diagrams should resemble Figures 6.12 to 6.15 on
pages 258–60 of the student textbook. Answers should
show an understanding of initiation, elongation,
and termination.
21. The sequence of amino acids in a protein can be traced
back to the sequence of nucleotides in DNA. Because
only complementary bases can pair (A with T or U,
and G with C), the nucleotide sequence in DNA is a
template for mRNA, and the codons in mRNA pair
with specific anticodons in tRNA. Each tRNA carries a
specific amino acid.
40
MHR TR • Biology 12 Answer Key Unit 3
22. Sample answer:
1. Isolate the Bt gene from Bacillus thuringiensis using
restriction enzymes.
2. Insert the Bt gene into the altered T-DNA region of
the Ti plasmid using DNA ligase.
3. Move the recombinant Ti plasmid into
Agrobacterium tumefaciens.
4. Infect plant cells with A. tumefaciens.
5. Select plant cells with selectable markers.
6. Grow plants from the recombinant plant cells.
23. To make the DNA visible it is treated with a chemical,
such as ethidium bromide. → Mixtures of differentsized DNA fragments are added to wells at the end
of a gel. → The gel is placed in a buffer solution and
exposed to an electric current. → The negatively
charged DNA fragments move toward the positively
charged anode. → The electric current is turned off
and the gel viewed under ultraviolet light.
24. a. Sample answer: The transgenic carrot plant can be
grown without the addition of harmful pesticides.
However, prior to approving the transgenic plant for
consumption by people, livestock, or pets, it should
be demonstrated that it is safe to eat. Other risks to
consider are environmental effects, such as negative
impacts on beneficial insects and worms or gene
transfer to wild plants in or near the croplands.
b. Sample answer: Farmers will no longer be exposed
to harmful pesticides when growing this crop
and will not have to spend money on pesticides.
However, the seeds/plant may be expensive and the
farmer may have difficulty controlling the spread of
the pest-resistance gene to other plants.
25. Answers should use relevant formal biology
vocabulary, be free from grammar and spelling errors,
and be substantiated by a reasoned argument. Sample
answer: The potential negative effects of transgenic
plants on ecosystems need to be better understood, so
this area of research will be awarded $20 million. Gene
therapy holds promise for treating devastating genetic
human diseases, so will be awarded $40 million. Cell
cycle regulation holds promise of understanding and
treating cancer and stopping the spread of malignant
tumours, so will be awarded $40 million.