Notice Candium Lab Answer Key & User's Guide Manuals
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. hese indentations (active sites) facilitate substrateenzyme binding as the active site changes in shape
to accommodate the substrate in what is called
induced it.
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.
hese bonds occur between complementary bases.
b. he covalent bonds that link adjacent nucleotides
together within each strand are called
phosphodiester bonds. hese 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 diferent form of the same gene.
Homologous chromosomes carry two alleles of the
same gene. Diferences in these alleles account for
diferences 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 lipped
• 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. he chromosome errors described result in genetic
disorders when they delete, alter, or duplicate genes.
In some case, however, the errors afect regions of
the genome that lack genes or are within non-coding
regions of a gene. hese errors are not harmful.
25. d
13. Enzymes are classiied according to the type of
reactions they catalyze.
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. hen
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. his makes viruses useful tools for
producing a copy of a gene. Researchers can insert the
gene of interest into the virus genome. he 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 modiied organism
b. It has a diferent genotype, as it now contains four
new genes
c. It has a diferent 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. Griith used two forms of S. pneumoniae: a pathogenic
S-strain and a non-pathogenic R-strain. Ater injecting
mice with a mixture of heat-killed S-strain and live
R-strain, the mice died. Griith 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. hese results provided strong evidence for
DNA’s role in transformation.
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MHR TR • Biology 12 Answer Key Unit 3
4. Two diferent 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. he other sample of T2 virus
was tagged with radioactive sulfur (35S), since sulfur is
only found in the protein coat of the capsid.
5. he independent variable in the experiment was the
type of radioactive isotope used to tag the virus. he
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.
he bases are adenine, cytosine, guanine, and thymine
in the case of DNA, and adenine, cytosine, guanine,
and uracil in the case of RNA.
9. Chargaf ’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. he number
of A-T nucleotides will not necessarily equal the
number of C-G nucleotides. his 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. herefore, 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. he 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. herefore, 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 diferent 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. Ater one round of replication,
DNA appeared as a single band, midway between
the expected positions of 15N-labelled DNA and
14
N-labelled DNA. Ater 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. he daughter strands of DNA are
part of a DNA molecule that will be in the daughter
cells. his 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. his is catalyzed by
DNA polymerase, which synthesizes the new strands
that are complementary to the parental strand.
Termination—he two new DNA molecules separate
from one another.
20. Replication takes place in a slightly diferent 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 difer.
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. he
lagging strand is synthesized in short fragments called
Okazaki fragments.
22. DNA replication requires the use of many enzymes
that have speciic roles. he presence of numerous
specialized enzymes may relect 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. herefore, 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 efects 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): he 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.
he rubber band becomes compacted due to the coils
that twisting forms. his 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. he rubber
band model also does not relect 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 relect the presence of parental DNA.
Answers to Section 5.1 Review Questions
(Student textbook page 218)
1. Griith’s experiments showed the existence of a
transforming principle. hat is, something in the heatkilled pathogenic bacteria (S-strain) could transform the
non-pathogenic bacteria (R-strain) into a pathogenic
form. his 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 diferentially 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. he 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 speciically, DNA
(deoxyribonucleic acid).
4. Diagrams should resemble the marginal portion of
Figure 5.4 on page 208 of the student textbook.
5. he nucleotide composition of the human would
be diferent 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. Chargaf showed that DNA is composed of repeating
units of nucleotides in ixed 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). Chargaf ’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. his 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. his discovery inluenced 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 speciic
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. here 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 inluence
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. he 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.
hese 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.
he 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. hese
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.
herefore, 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. he 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
simpliied 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. he 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. herefore,
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—he 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—he dispersive model states that new molecules of
DNA would be hybrids containing a mixture of old
parent DNA and new daughter DNA strands.
C—he 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; ive DNA polymerases have
been identiied 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 identiied
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.
herefore, 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.
hese results provided evidence that genetic
information was carried on DNA.
16. Franklin’s X-ray difraction 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. he 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
diferent levels of organization. DNA associates with
histones to form nucleosomes. It can be further
compacted by the coiling of nucleosomes to produce
30 nm ibres. Additional compacting is achieved
through the formation of loop domains of the 30 nm
ibre on a protein scafold. his scafold 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. he 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 diferent 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 diferent 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 ofspring 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. he RNA primer
provides a free 3′ hydroxyl end from which DNA
polymerase can add nucleotides.
25. he replication machine consists of the complex of
proteins and DNA that interact at the replication
fork. hese 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.
26. Having multiple origins of replication increases the
speed of replication. Instead of starting at one end and
inish at another, having multiple start points increases
the eiciency of replication.
27. DNA polymerase has a proofreading mechanism,
which identiies 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. his group of enzymes excises the
mispaired nucleotides and inserts the correctly paired
nucleotide. Errors that are not corrected before cell
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. Diferent tissues all develop from the same fertilized
egg cell (zygote). While the tissues have the same
genes, only those genes necessary for a speciic tissue’s
functions are active.
31. Sample answer: here are some similarities. However,
DNA “words” are limited to sequences of amino acids.
Each section of code has only “one meaning,” resulting
in one speciic protein. his 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.
35. a. Sample answer: he small fragments could be
Okazaki fragments that were not joined together
properly. his 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.
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).
37. Linker DNA is responsible for joining nucleosomes
together. Micrococcal nuclease preferentially
cuts linker DNA, which would lead to disrupted
formation of 30 nm ibres. Diagrams should include
an illustration of the “beads on a string” appearance
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. he weak hydrogen bonds in DNA break easily,
making it easier for the two strands in the molecule to
separate during replication. he strong covalent bonds
ensure that the sequence of nucleotides remains ixed
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 Griith and Avery). he experimental approaches
and results for each experiment should be outlined in the
concept map, in addition to how the diferent 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
conidence and conirmation of results.
47. Timelines should be a chronological order of
contributions from the scientists presented in the
chapter: Miescher; Levene; Griith; Avery, MacLeod,
and McCarty; Hershey and Chase; Chargaf; 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. Ater 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 oten. Heart muscle cells do not
divide as frequently as skin cells, and therefore DNA
replication (and DNA polymerase activity) occurs
less oten.
43. Answers could include targeting bacterial-speciic
topoisomerases. hese enzymes are essential for
supercoiling and DNA replication, which are both
required for the survival of bacterial cells. Since this
is a bacteria-speciic target, the side efects 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. herefore,
the initiation of DNA replication would not occur
eiciently, 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
C
S
P
5′ end
T
S
P
G
S
P
A
S
P
T
S
P
S
P
3′ end
49. Graphic organizers should include:
• Prokaryotes Only—rate of replication is faster
compared to eukaryotes; ive DNA polymerases have
been identiied 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 identiied
in eukaryotes; linear chromosome has multiple
replication origins; presence of telomeres due to
linear nature of eukaryotic chromosome
resemble a simpliied 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.
50. Diagrams should resemble the bottom illustration in
Figure 5.16 on page 221 of the student textbook, with
the irst set of parental and new strands labelled irst
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. his 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—he 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
54. See Table below.
55. Journal entries should be written in the irst person
and summarize the knowledge of heredity related
to his or her research, using scientiic terminology.
Challenges could include a lack of research facilities,
lack of inancial support, the shortcomings of
the current technology, lack of consensus among
research colleagues, or lack of support for the work
in the broader scientiic community. houghts
of future signiicance of the research may relect
later discoveries.
56. Articles should summarize Watson and Crick’s
indings, including their model of DNA and include
how this discovery would afect the general public (i.e.,
advances in medicine and genetics, ethical issues and
concerns). he 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 diferent 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 eiciently.
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 it.
• 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. his would leave replication errors (i.e., mispairing of
bases) uncorrected. hese errors would then become
mutations in the genome, which are then passed onto
daughter cells once cell division occurs.
59. he graphic organizer should efectively 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 lies), 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 diferent
species. hese conserved regions are thought to
be essential and important regions of the genome.
Likewise, divergent regions may confer speciesspeciic function and contribute to morphological
and functional changes.
61. Answers should include speciic examples of how the
chosen scientist’s research would beneit the scientiic
research community and the general public. Social,
legal, and ethical implications may be addressed.
he 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 illed since there is no
adjacent fragment where nucleotides can be added.
herefore, 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 ater 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. his may lead to loss of
important coding information. Certain lifestyle
factors may inluence telomere length based on its
efects 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 beneits 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
• herapeutic gene targets
• Gene expression (RNA and protein)
• DNA mutations
68. his required a rethinking of many ideas that formed
the basis of other ideas. he lower number of genes
may mean that
• the structures function diferently 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 speciically target viral DNA
polymerase to interfere with viral replication. Speciicity
is also required to ensure eukaryotic DNA polymerase
activity in not adversely afected.
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 speciic examples of signiicant research
indings 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. hymine should be paired
with adenine, and cytosine should be paired
with guanine.
b. Mispairing of bases occurs during replication and
may be due to lexibility 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. hese
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 diferent 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.
he 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. hese results
demonstrated that viral DNA, not viral protein,
enters the bacterial cell. herefore, DNA is the
hereditary material.
13. Franklin’s X-ray difraction 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 diferent levels of organization. DNA
associates with histones to form nucleosomes. It can
be further compacted by the coiling of nucleosomes
to produce 30 nm ibres. Additional compacting is
achieved through the formation of loop domains of
the 30 nm ibre on a protein scafold. his scafold can
condense further through folding.
he diferences 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
inluence and regulate the production of proteins and
RNA molecules.
17. a. Answers could include identity thet or the
identiication 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 scientiic 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 let.
b. Okazaki fragment C was made irst. 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. his 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. hese weak
hydrogen bonds break easily, making it easier for the
two strands to separate during replication. he strong
covalent bonds that exist on the sugar-phosphate
backbone ensure that the sequence of nucleotides
remains ixed 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.
his means that the age of an organism is relected in
the length of telomeres.
25. Venn diagrams should include:
• Prokaryotes Only—rate of replication is faster; ive
DNA polymerases identiied; 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 identiied; 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. he black urine phenotype shown to be caused by a
recessive inheritance factor (gene) that cause
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. hese indentations (active sites) facilitate substrateenzyme binding as the active site changes in shape
to accommodate the substrate in what is called
induced it.
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.
hese bonds occur between complementary bases.
b. he covalent bonds that link adjacent nucleotides
together within each strand are called
phosphodiester bonds. hese 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 diferent form of the same gene.
Homologous chromosomes carry two alleles of the
same gene. Diferences in these alleles account for
diferences 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 lipped
• 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. he chromosome errors described result in genetic
disorders when they delete, alter, or duplicate genes.
In some case, however, the errors afect regions of
the genome that lack genes or are within non-coding
regions of a gene. hese errors are not harmful.
25. d
13. Enzymes are classiied according to the type of
reactions they catalyze.
Biology 12 Answer Key Unit 3 • MHR TR
1
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. hen
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. his makes viruses useful tools for
producing a copy of a gene. Researchers can insert the
gene of interest into the virus genome. he 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 modiied organism
b. It has a diferent genotype, as it now contains four
new genes
c. It has a diferent 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. Griith used two forms of S. pneumoniae: a pathogenic
S-strain and a non-pathogenic R-strain. Ater injecting
mice with a mixture of heat-killed S-strain and live
R-strain, the mice died. Griith 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. hese results provided strong evidence for
DNA’s role in transformation.
2
MHR TR • Biology 12 Answer Key Unit 3
4. Two diferent 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. he other sample of T2 virus
was tagged with radioactive sulfur (35S), since sulfur is
only found in the protein coat of the capsid.
5. he independent variable in the experiment was the
type of radioactive isotope used to tag the virus. he
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.
he bases are adenine, cytosine, guanine, and thymine
in the case of DNA, and adenine, cytosine, guanine,
and uracil in the case of RNA.
9. Chargaf ’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. he number
of A-T nucleotides will not necessarily equal the
number of C-G nucleotides. his 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. herefore, 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. he 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. herefore, 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 diferent 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. Ater one round of replication,
DNA appeared as a single band, midway between
the expected positions of 15N-labelled DNA and
14
N-labelled DNA. Ater 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. he daughter strands of DNA are
part of a DNA molecule that will be in the daughter
cells. his 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. his is catalyzed by
DNA polymerase, which synthesizes the new strands
that are complementary to the parental strand.
Termination—he two new DNA molecules separate
from one another.
20. Replication takes place in a slightly diferent 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 difer.
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. he
lagging strand is synthesized in short fragments called
Okazaki fragments.
22. DNA replication requires the use of many enzymes
that have speciic roles. he presence of numerous
specialized enzymes may relect 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. herefore, 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 efects 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): he 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
3
Figure 5.10 (Student textbook page 215): Twisting a
rubber band around itself mimics how DNA supercoiling.
he rubber band becomes compacted due to the coils
that twisting forms. his 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. he rubber
band model also does not relect 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 relect the presence of parental DNA.
Answers to Section 5.1 Review Questions
(Student textbook page 218)
1. Griith’s experiments showed the existence of a
transforming principle. hat is, something in the heatkilled pathogenic bacteria (S-strain) could transform the
non-pathogenic bacteria (R-strain) into a pathogenic
form. his 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 diferentially 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. he 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 speciically, DNA
(deoxyribonucleic acid).
4. Diagrams should resemble the marginal portion of
Figure 5.4 on page 208 of the student textbook.
5. he nucleotide composition of the human would
be diferent 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.
4
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. Chargaf showed that DNA is composed of repeating
units of nucleotides in ixed 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). Chargaf ’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. his 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. his discovery inluenced 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 speciic
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. here 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 inluence
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. he 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.
hese 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.
he 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. hese
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.
herefore, 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. he 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
simpliied 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. he 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. herefore,
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
5
10. A—he 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—he dispersive model states that new molecules of
DNA would be hybrids containing a mixture of old
parent DNA and new daughter DNA strands.
C—he 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; ive DNA polymerases have
been identiied 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 identiied
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
6
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.
herefore, 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.
hese results provided evidence that genetic
information was carried on DNA.
16. Franklin’s X-ray difraction 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. he 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
diferent levels of organization. DNA associates with
histones to form nucleosomes. It can be further
compacted by the coiling of nucleosomes to produce
30 nm ibres. Additional compacting is achieved
through the formation of loop domains of the 30 nm
ibre on a protein scafold. his scafold 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. he 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 diferent 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 diferent 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 ofspring 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. he RNA primer
provides a free 3′ hydroxyl end from which DNA
polymerase can add nucleotides.
25. he replication machine consists of the complex of
proteins and DNA that interact at the replication
fork. hese 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.
26. Having multiple origins of replication increases the
speed of replication. Instead of starting at one end and
inish at another, having multiple start points increases
the eiciency of replication.
27. DNA polymerase has a proofreading mechanism,
which identiies 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. his group of enzymes excises the
mispaired nucleotides and inserts the correctly paired
nucleotide. Errors that are not corrected before cell
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. Diferent tissues all develop from the same fertilized
egg cell (zygote). While the tissues have the same
genes, only those genes necessary for a speciic tissue’s
functions are active.
31. Sample answer: here are some similarities. However,
DNA “words” are limited to sequences of amino acids.
Each section of code has only “one meaning,” resulting
in one speciic protein. his 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.
35. a. Sample answer: he small fragments could be
Okazaki fragments that were not joined together
properly. his 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.
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).
37. Linker DNA is responsible for joining nucleosomes
together. Micrococcal nuclease preferentially
cuts linker DNA, which would lead to disrupted
formation of 30 nm ibres. Diagrams should include
an illustration of the “beads on a string” appearance
Biology 12 Answer Key Unit 3 • MHR TR
7
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. he weak hydrogen bonds in DNA break easily,
making it easier for the two strands in the molecule to
separate during replication. he strong covalent bonds
ensure that the sequence of nucleotides remains ixed
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 Griith and Avery). he experimental approaches
and results for each experiment should be outlined in the
concept map, in addition to how the diferent 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
conidence and conirmation of results.
47. Timelines should be a chronological order of
contributions from the scientists presented in the
chapter: Miescher; Levene; Griith; Avery, MacLeod,
and McCarty; Hershey and Chase; Chargaf; 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. Ater 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 oten. Heart muscle cells do not
divide as frequently as skin cells, and therefore DNA
replication (and DNA polymerase activity) occurs
less oten.
43. Answers could include targeting bacterial-speciic
topoisomerases. hese enzymes are essential for
supercoiling and DNA replication, which are both
required for the survival of bacterial cells. Since this
is a bacteria-speciic target, the side efects 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. herefore,
the initiation of DNA replication would not occur
eiciently, 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
C
S
P
5′ end
T
S
P
G
S
P
A
S
P
T
S
P
S
P
3′ end
49. Graphic organizers should include:
• Prokaryotes Only—rate of replication is faster
compared to eukaryotes; ive DNA polymerases have
been identiied 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 identiied
in eukaryotes; linear chromosome has multiple
replication origins; presence of telomeres due to
linear nature of eukaryotic chromosome
resemble a simpliied 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.
50. Diagrams should resemble the bottom illustration in
Figure 5.16 on page 221 of the student textbook, with
the irst set of parental and new strands labelled irst
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. his 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—he 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
54. See Table below.
55. Journal entries should be written in the irst person
and summarize the knowledge of heredity related
to his or her research, using scientiic terminology.
Challenges could include a lack of research facilities,
lack of inancial support, the shortcomings of
the current technology, lack of consensus among
research colleagues, or lack of support for the work
in the broader scientiic community. houghts
of future signiicance of the research may relect
later discoveries.
56. Articles should summarize Watson and Crick’s
indings, including their model of DNA and include
how this discovery would afect the general public (i.e.,
advances in medicine and genetics, ethical issues and
concerns). he 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 diferent 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 eiciently.
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 it.
• 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. his would leave replication errors (i.e., mispairing of
bases) uncorrected. hese errors would then become
mutations in the genome, which are then passed onto
daughter cells once cell division occurs.
59. he graphic organizer should efectively 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 lies), 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 diferent
species. hese conserved regions are thought to
be essential and important regions of the genome.
Likewise, divergent regions may confer speciesspeciic function and contribute to morphological
and functional changes.
61. Answers should include speciic examples of how the
chosen scientist’s research would beneit the scientiic
research community and the general public. Social,
legal, and ethical implications may be addressed.
he 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 illed since there is no
adjacent fragment where nucleotides can be added.
herefore, 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 ater 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. his may lead to loss of
important coding information. Certain lifestyle
factors may inluence telomere length based on its
efects 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 beneits 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
• herapeutic gene targets
• Gene expression (RNA and protein)
• DNA mutations
68. his required a rethinking of many ideas that formed
the basis of other ideas. he lower number of genes
may mean that
• the structures function diferently 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 speciically target viral DNA
polymerase to interfere with viral replication. Speciicity
is also required to ensure eukaryotic DNA polymerase
activity in not adversely afected.
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 speciic examples of signiicant research
indings 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. hymine should be paired
with adenine, and cytosine should be paired
with guanine.
b. Mispairing of bases occurs during replication and
may be due to lexibility 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. hese
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 diferent 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.
he 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. hese results
demonstrated that viral DNA, not viral protein,
enters the bacterial cell. herefore, DNA is the
hereditary material.
13. Franklin’s X-ray difraction 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 diferent levels of organization. DNA
associates with histones to form nucleosomes. It can
be further compacted by the coiling of nucleosomes
to produce 30 nm ibres. Additional compacting is
achieved through the formation of loop domains of
the 30 nm ibre on a protein scafold. his scafold can
condense further through folding.
he diferences 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
inluence and regulate the production of proteins and
RNA molecules.
17. a. Answers could include identity thet or the
identiication 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 scientiic 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 let.
b. Okazaki fragment C was made irst. 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. his 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. hese weak
hydrogen bonds break easily, making it easier for the
two strands to separate during replication. he strong
covalent bonds that exist on the sugar-phosphate
backbone ensure that the sequence of nucleotides
remains ixed 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.
his means that the age of an organism is relected in
the length of telomeres.
25. Venn diagrams should include:
• Prokaryotes Only—rate of replication is faster; ive
DNA polymerases identiied; 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 identiied; 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. he black urine phenotype shown to be caused by a
recessive inheritance factor (gene) that cause