6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
False. DNA replication is semi-conservative.
True
False. DNA polymerase III builds a complementary DNA strand during DNA replication in the 5' to 3' direction.
False. After one replication of double-stranded DNA, the daughter DNA molecules consist of 50% parental material.
True
(a)
(d)
(d)
(c)
(a)
(c)
Chapter 4 Review
(Page 230–231)
Understanding Concepts
1.
Erwin Chargaff determined the percentage composition of each of the nitrogenous bases in a DNA molecule. He
illustrated that the proportion of adenine is equal to the proportion of thymine and that the proportion of guanine is equal
to the proportion of cytosine. In addition, he determined that the sum of adenine plus guanine is equal to the sum of
thymine plus cytosine. This information led to the incorporation of complementary base pairing in Watson and Crick’s
model of DNA. Complementary base pairing explains the nitrogenous base proportions discovered by Chargaff.
2.
Hydrogen bonds will not form
between adenine and cytosine.
Hydrogen bonds form between an H atom on one molecule and a fluorine, nitrogen, or oxygen atom on another. Cytosine
and adenine do not have the correct molecules available with which to form hydrogen bonds.
3.
Copyright © 2003 Nelson
Chapter 4 DNA: The Molecular Basis of Life
87
4.
The incidence of error during replication is minimized because of the use of a parent strand as a template in building a
new daughter DNA strand (a complement to the original parent DNA). Also since DNA comprises complementary
nucleotides, the new strand that is built contains the same coding information as the parent strand. Another mechanism
that DNA uses to minimize errors during replication is that of exonuclease activity of DNA polymerase III and DNA
polymerase I. Once the new complementary strand is synthesized, these two enzymes can remove any incorrectly paired
nucleotide and insert the appropriate nucleotide in its place.
5.
Enzyme
DNA gyrase
DNA helicase
DNA polymerase I
DNA polymerase III
DNA ligase
Function
The bacterial enzyme that relieves the tension produced by the unwinding of DNA
during replication
The enzyme that unwinds double-helical DNA by disrupting hydrogen bonds
An enzyme that removes RNA primers and replaces them with the appropriate
deoxyribonucleotides during DNA replication
The enzyme responsible for synthesizing complementary strands of DNA during DNA
replication
The enzyme that joins DNA fragments together by catalyzing the formation of a bond
between the 3 hydroxyl group and a 5 phosphate group on the sugar-phosphate
backbones

6.
7.

TTAACGTAT
A deoxyribose sugar differs from a ribose sugar on the 2' carbon. Deoxyribose possesses an H atom on the 2' sugar, while
ribose possesses a hydroxyl group on the 2' sugar.
8.
Rosalind Franklin produced an X-ray diffraction pattern of DNA that suggested it was helical in nature. Her findings also
indicated that the diameter of a DNA molecule was constant at 2 nm. The helical nature of DNA transpired into two
antiparallel DNA strands. The constant diameter supports the complementary base pairing of a purine and pyrimidine
throughout the molecule (if pairing included a purine-purine or pyrimidine–pyrimidine combination, the diameter would
vary from 2 nm), with a deoxyribose sugar and phosphate backbone. The constant width suggested a repeating pattern.
We now know that DNA is a polymer of nucleotides.
9. Both strands of a DNA molecule act as a template during replication. If both new complementary strands were
synthesized in the same direction (toward the replication fork), one strand would built in the 5' to 3' direction (the leading
strand), while the other complementary strand would have to be built in the 3' to 5' direction (because of the antiparallel
nature of DNA strands). However, DNA polymerase III, the enzyme responsible for the synthesis of new complementary
strands, can only build in the 5' to 3' direction. To deal with this problem a mechanism has evolved where one strand, the
lagging strand, is built in small segments away from the replication fork. Initially, small segments of RNA, known as
RNA primers, are laid by primase along the lagging strand template. DNA polymerase III uses these segments to start
synthesizing small sections of DNA away from the replication fork. These small sections of DNA sequence are called
Okazaki fragments. DNA polymerase I will then excise the RNA primers and replace them with the appropriate
nucleotides. DNA ligase joins the gaps in the Okazaki fragments by creating a phosphodiester bond. The result is
discontinuous replication of the 3' to 5' parental strand.
10.(a) If single-stranded binding proteins were malfunctioning during DNA replication, then the two strands of the unwound
DNA molecule would rehybridize. Template strands would not be available for primase to lay down primers and for
DNA polymerase III to build complementary strands. DNA replication would not proceed.
(b) If primase was inactivated, DNA polymerase III would not be able to build a complementary strand. Primase lays down
primers (short sequences of RNA) that DNA polymerase III uses as starting points to start building a complementary
strand. DNA replication would not proceed.
(c) If DNA polymerase I was malfunctioning, the RNA primers that were laid down by primase as starting points for
elongation would remain in the DNA molecule. The result would be a DNA molecule embedded with stretches of RNA
sequence.
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Unit 2 Molecular Genetics
Copyright © 2003 Nelson
11.(a) To form a chemical bond, there must be energy input. Energy is released when a chemical bond is broken. When a
DNA nucleotide is added to an elongating complementary strand, a phosphodiester bond must be formed between the
incoming nucleotide and the complementary strand that has been built. The energy found in bonds between phosphate
groups in a nucleoside triphosphate is used as the source of energy to form the phosphodiester bond. The bond between
the first and second phosphate group in a nucleoside triphosphate is the bond that is broken to produce energy.
(b) The inorganic phosphate is recycled by the cell.
12. A hydrogen bond is very weak on its own. It is the sheer number of hydrogen bonds that are formed within a DNA
molecule that make the molecule very stable.
13. DNA can be classified as a polymer because consists of repeating units known as nucleotides. A nucleotide comprises a
deoxyribose sugar, a nitrogenous base, and a phosphate group.
14. A 3' end is represented by the letter b. The 5' ends are represented by the letters a, c, and d.
15. To ensure replication you would require the enzyme primase. Discontinuous replication would not occur because DNA
polymerase III can only start building a complementary DNA strand if it has a 3' end available as a starting point. The
lagging strand runs in the 3' to 5' direction, toward the replication fork. Primase adds RNA primers along a singlestranded parent DNA strand. DNA polymerase III can then use these primers as starting points and build complementary
strands.
16. The following proportions would be found in the complementary strand: 15% thymine, 30% adenine, 20% cytosine, and
35% guanine. The proportions in the whole molecule would be as follows:
adenine = (15% + 30%)/2 = 22.5%
thymine = (30% + 15%)/2 = 22.5%
cytosine = (35% + 20%)/2 = 27.5%
guanine = (20% + 35%)/2 = 27.5%
17. It is necessary that the A + G percentage be equal to one because the proportion of adenine is equal to the proportion of
C+T
thymine, and the proportion of guanine is equal to the proportion of cytosine. A numerical example illustrates this point.
If the proportion of adenine is 15%, then the proportion of thymine is 15%, the proportion of guanine is 35%, and the
proportion of cytosine is 35%. Plugging into the equation A + G results in 15% + 35% = 1. It is not necessary that the
35% + 15%
C+T
A + T ratio equal one, because the ratio of adenine and thymine does not determine the ratio of guanine and cytosine.
G+C
They do not have any bearing on each other.
18. Student answers will vary. The main point emphasized is that all life, with the exception of some viruses, uses DNA as its
hereditary material. The life can be relatively simple, such as a bacterium, or rather complex, such as a human.
Nevertheless, DNA has been found to act as the hereditary material in all living matter. The main difference between
species is the amount of DNA that they carry.
Applying Inquiry Skills
19.(a) Length in metres of average E .coli chromosome:
3.4 nm
= 1.02 × 106 nm
3 000 000 nucleotides ×
10 nucleotides
6
1
m
(1.02 × 10 nm) ×
= 1.02 × 10í m
109 nm
The average E. coli chromosome is 1.02 × 10í m long.
Length in metres of average E. coli gene:
3.4 nm
= 408 nm
1200 nucleotides ×
10 nucleotides
408 nm × 1 m = 4.08 × 10í m
109 nm
The average E. coli gene is 4.08 × 10í m long.
1 turn
(b) 3 000 000 nucleotides ×
= 300 000 turns
10 nucleotides
There are 300 000 turns in the structure.
Copyright © 2003 Nelson
Chapter 4 DNA: The Molecular Basis of Life
89
(c) (1.02 × 10-3 m) × 10 000 = 10.2 m
10.2 m ×
1000 mm
= 10 200 mm
1m
The strand would be 10 200 mm long.
(d) (4.08 × 10-7 m) × 10 000 = 4.08 × 10-3 m
4.08 × 10-3 m ×
1000 mm
= 4.08 mm
1m
The gene would be 4.08 mm long.
20.(a) According to the experimental results depicted in Figure 2 (page 231 of the Student Text), the optimum wavelength
chosen would be approximately 630 nm. At 630 nm the diphenylamine-purine complex absorbs the most light.
(b) The diphenylamine-purine complex is blue. Matter appears a certain colour depending on which wavelength of light in
the visible spectrum it absorbs and which wavelength of light it reflects. For example, leaves appear green because
chlorophyll absorbs all wavelengths of light with the exception of the wavelength corresponding to the green colour.
This wavelength is reflected, our eyes pick it up, and leaves appear green. If the diphenylamine-purine complex
appears blue, this indicates that it is reflecting the blue wavelength. If a blue wavelength is used to measure its
concentration via absorption, poor results will be obtained since the diphenylamine-purine complex does not absorb the
blue wavelength of light.
(c) Since the purine:pyrimidine concentration in DNA is 1:1, the pyrimidine concentration is also 0.5 mg/mL.
21.(a) It can be hypothesized that a large number of radioactive fragments of DNA would be isolated from the temperaturesensitive strain of E. coli that lacked functioning DNA ligase above 40°C. These radioactive fragments would be
Okazaki fragments.
(b) It was necessary to repeat the experiment with an E. coli strain that contained DNA ligase that worked above 40Û& WR
have a comparison with the E. coli strain that contained DNA ligase that did not work above 40Û& 7KH E. coli strain
that contained DNA ligase that worked above 40Û& DFWHG DV D FRQWURO
(c) The radioactivity would have been found in the Okazaki fragments in one of the complementary strands built.
Radioactivity would have been present in both experiments. More Okazaki fragments would have been found in the
temperature-sensitive E. coli strain, because DNA ligase would not have been able to join, via phosphodiester bonds,
the fragments produced from the lagging strand replication. In the non-temperature-sensitive strain of E. coli, one long
strand would have been produced since DNA ligase was functioning.
(d) Pauling and Hamm could conclude that replication on one of the strands occurred via a different mechanism that joins
short fragments of DNA together via DNA ligase.
Making Connections
22. Student answers will vary. Some possibilities include the following:
DNA ligase could be used to join DNA sequences originating from different species or from different places within a
species. DNA polymerase I could be used as a tool to make changes to a DNA sequence. The effects of such changes can
then be observed.
23. Student answers will vary. Some points that may be made are as follows:
Watson and Crick received the Nobel Prize because they interpreted and derived meaning from experimental data from
various sources. Watson and Crick used information gathered from experimentation performed by Chargaff, Wilkins, and
Franklin. Although Chargaff interpreted his experimental results pertaining to the chemical composition of DNA, he did
not derive any further meaning from them—he simply stated his findings. Therefore, Chargaff should not have been
considered for the Nobel Prize. The same argument can be extended to Franklin and Wilkins. They did not derive any
further meaning from their experimentation. Nevertheless, Wilkins shared the Nobel Prize with Watson and Crick.
Franklin was deceased and therefore did not qualify.
Extension
24. Student answers will vary depending on the research. Adenine and thymine share two hydrogen bonds, whereas guanine
and cytosine share three hydrogen bonds. DNA is denatured using heat. The higher the temperature required to denature
the DNA, the larger the proportion of G–C nucleotide pairs present, as compared with the proportion of A–T nucleotide
pairs. In research a temperature of 92Û& – 95Û& LV XVHG WR HQVXUH WKDW GRXEOH-stranded DNA is fully denatured to singlestranded DNA. A change that accompanies the denaturation of DNA is its absorbance of UV light. This increase in
absorbance is called the hyperchromic shift. At 260 nm, double-stranded DNA has a lower absorbance than singlestranded DNA.
90
Unit 2 Molecular Genetics
Copyright © 2003 Nelson