Unit 3 Review, pages 414–421
Knowledge
1. d
2. b
3. d
4. a
5. d
6. c
7. a
8. d
9. b
10. c
11. b
12. d
13. d
14. a
15. c
16. a
17. d
18. True
19. True
20. False. Watson and Crick’s model was correct because it solved the consistent width issue
with DNA by pairing purines with pyrimidines.
21. True
22. False. Genetic engineering relies on the ability of eubacteria to incorporate foreign DNA.
23. False. Germ-cell lines have high concentrations of telomerase, which helps them continually
divide.
24. True
25. True
26. True
27. True
28. False. Spontaneous mutations are caused by inaccurate DNA replication.
29. True
30. False. Retroviruses use integrase to incorporate their reverse transcribed double-stranded
DNA into the genome of a host cell.
31. False. When working with restriction enzymes, scientists prefer sticky ends to blunt ends
because the resulting DNA fragments are easy to join together after treatment.
32. True
33. True
34. False. Genetic screening is recommended for parents with genetic disorders in their families,
especially when they are older.
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Unit 3: Molecular Genetics
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35. (a) iii
(b) v
(c) ii
(d) i
(e) vi
(f) iv
36. (a) vi
(b) iv
(c) i
(d) v
(e) iii
(f) ii
37. (a) v
(b) iv
(c) ii
(d) vi
(e) iii
(f) i
38. Although Mendel did not examine the chemistry of genetics he was able to prove that there
was a predictable pattern to the inheritance of traits. The predictability helped other scientists
formulate hypotheses about how this inheritance might occur which led them to investigate
RNA, proteins, and eventually DNA.
39. Pauling and Corey’s model was incorrect because they suggested that the backbone ran down
the centre of the molecule. This would place all the electrically negative phosphates in the centre,
which would be energetically unstable and would make DNA very unstable.
40. Telomeres are the end regions of chromosomes and contain repeated sequences of noncoding DNA. Scientists believe that they prevent damage to coding sequences on the lagging
strand of DNA since there is no primer.
41. Both cancer cells and germ cells have the ability to replicate indefinitely. Both cell types
have elevated concentrations of telomerase that repair telomeres.
42. These four codons are important to protein synthesis as they either initiate (AUG, which
codes for methionine) or terminate (UAA, UAG, UGA) protein synthesis. Without AUG, protein
synthesis would not begin; without the others, the proteins would be too long and not function
properly.
43. Viruses are able to enter a cell and use the cells replication enzymes to replicate. They also
incorporate their own DNA into the host cell's DNA. These features enable them to be used to
transfer DNA from a donor to a host. The process of inserting genetic material into a cell or
bacterium using a viral vector, known as transduction, can potentially repair or even replace
defective genes.
44. Ethidium bromide is a compound that stains the completed electrophoresis gel. It binds inside
the DNA “ladder” structure where it glows when the gel is viewed under a UV light, revealing
the bands of DNA.
Understanding
45. According to Chargaff's rule, determining the base-pair composition of a piece of RNA
should result in percentages of uracil, adenine, guanine, and cytosine that add to 100 % but are
not necessarily in any particular proportion.
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Unit 3: Molecular Genetics
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46. Figure 1 shows supercoiling in a prokaryote primary circular DNA. You can tell this because
the DNA is a loop and has no histones associated with it. This is an important mechanism in
packing the DNA of a prokaryote prior to replication.
47.
48. (a) DNA replication would cease because no RNA primers could be added. DNA polymerase
would have nothing to add to.
(b) The replication fork would not open up and DNA replication would cease or only occur at
the ends. This would cause DNA replication to take a very long time, compared with normal
multiple site replication.
(c) Replication would stop since the DNA would twist up on itself and this tension could not be
relieved by DNA gyrase. This could cause the DNA to break and the strands to separate
uncontrollably.
49. Chromosomes do not exist during interphase because the DNA is needed to maintain and
repair the cell. Chromosomes would make it difficult for the cell to perform transcription and
translation since the packing of the DNA around the histones and solenoids would prevent
replication from occurring properly.
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Unit 3: Molecular Genetics
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50. Student answers will vary. Sample answers:
Six Differences Between DNA in Eukaryotes and Prokaryotes
Eukaryotes
Prokaryotes
linear chromosomes
single circular DNA
no plasmids
smaller plasmids
DNA forms nucleosomes by wrapping
lacks histones and undergoes
with histones, which coil together to
supercoiling to reduce volume
form solenoids
DNA in nucleus
DNA loose in cytosol (no nucleus)
many replication origins
one replication origin
contains exons and introns
no exons and introns
packed using histones
packed using supercoiling
51. Types of RNA Used in Transcription and Translation
Abbreviation Name
Function
Shape
mRNA
messenger transcribes the DNA
single–stranded,
into an RNA strand, in straight line
reverse and
complementary
sequences.
tRNA
transfer
translates the mRNA
using anti-codons and
nucleoside
triphosphates to
assemble amino acids
in the ribosome.
52.
Effects of Errors during Transcription and Translation
Error
Effect
(a) The poly(A) tail is not
The RNA would be subject to attack by RNAadded.
digesting enzymes in the cytosol.
(b) The termination
RNA polymerase might not stop transcribing.
sequence is misread.
As a result, the polypeptide chain might to be
too long and the protein function would be
impaired.
(c) The 5′ cap is not added. Translation would not begin.
(d) The promoter region is
RNA polymerase would not have a site at
not recognized by RNA
which to begin transcription. Transcription
polymerase.
would not begin.
(e) Exons are excised,
mRNA would include only non-coding regions.
rather than introns.
Any transcribed polypeptide would be totally
non-functional.
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Unit 3: Molecular Genetics
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53.
A: polysome, B: chromosome, C: mRNA, D: polypeptide. This micrograph depicts simultaneous
transcription and translation in the cytosol of an E. coli chromosome.
54. Met Phe Pro Pro
55. (a) The A site (aminoacyl site) is where the incoming aminoacyl-tRNA (carrying the next
amino acid to be added to the polypeptide chain) binds to the mRNA. The P site (peptidyl site) is
where the tRNA carrying the growing polypeptide chain is bound.
(b) Start codons initiate the production of a polypeptide chain and code for an amino acid
methionine. Stop codons terminate the sequence and do not code for any amino acids.
56. (a) Post-translational regulation controls when proteins become fully functional, the duration
of their functionality, and when they are degraded.
(b) Post-translational regulation can limit the availability of functional proteins. This prevents
the wasting of energy by the over- or under-production of proteins. This benefits the cell by
having synthesized proteins remain non-functioning until they are needed.
57. For inducible operons, such as the lac operon, the presence of the inducer inactivates the
repressor protein, allowing the gene to be transcribed. This is an advantage when dealing with
potential allergens or foods. The body produces the desired protein when its inducer is present.
The trp operon is classified as a repressor operon because the gene is transcribed only when the
initiator is not present in the cell. This allows the cell to shut down the production of a readily
available protein.
58. original 5′-Met-Ser-Trp-Stop-3′
(a) silent mutation
5′-ATC TCT TGG TAA-3′
5′ Met-Ser-Trp-Stop 3′
(b) missense mutation
5′-ATC TTC TGG TAA-3′
5′ Met-Phe-Trp-Stop 3′
(c) inversion
5′-ATC CTT TGG TAA-3′
5′ Met-Leu-Trp-Stop 3′
(d) deletion
5′-ATC TCT GGT AA-3′
5′ Met-Ser-Gly 3′
(e) nonsense
5′-ATC TAA TGG TAA-3′
5′ Met-Stop-Trp-Stop 3′
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Unit 3: Molecular Genetics
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(f) frameshift (C missing)
5′-ATT CTT GGT AA-3′
5′ Ile-Leu-Gly 3′
59. Answers will vary, but should include the idea that pseudogenes are silenced genes that show
a function in related organisms. They can be used to trace the evolutionary history of a species.
60. (a) A restriction enzyme cuts a plasmid into the same number of fragments as there are cuts
A restriction enzyme cuts a eukaryotic chromosome into the a number of fragments that is equal
to the number of cuts plus 1.
(b) It means that a research scientist must account for this increase in the number of fragments in
the procedure.
61. DNA sequencing plays an important role in genomics because it presents the code to
scientists for investigation. Scientists can then compare how sequences change between species
or even individuals. Scientists can then compare how these changes affect an organism and draw
inferences about the effects of the genes. The first DNA sequencing method was the Sanger
method. Sanger sequencing where ddNTP’s are added to a growing DNA strand. When the
ddNTP is incorporated into DNA it prevents the binding of the next nucleotide, terminating the
elongation of the DNA strand.
62. Answers may vary. Sample answer: I would explain that there are currently no gene therapies
available commercially in Canada. There have been some successful experimental uses of gene
therapy for certain disorders, but there have been enough serious side effects that no therapy has
been officially approved. I would suggest other types of therapy, such as nutritional and lifestyle
counseling, the use of medications, and surgery if necessary. They type of therapy I suggested
would depend on the condition being treated.
63. Nutraceuticals have the potential to deliver medicines in a very inexpensive manner while
providing nutrition. In many developing nations that are struggling with high drug costs and
malnutrition nutraceuticals could have a huge impact on the health of the people.
Analysis and Application
64.
The queen, who is diploid, mates with one of her sons (drone), who is haploid to her, to produce
all the sisters (workers). Therefore, since sisters receive two of the same three chromosomes,
sisters are all at least 50 % related. Males (drones) do not have fathers but instead have
grandfathers, since their father is one generation behind as no males are born as a result of sexual
reproduction. This means that all the males are progeny of the male one generation back, their
grandfather.
65. (a) If you were examining a eukaryote there would be evidence of histones, a nuclear
membrane, and linear chromosomes.
(b) If you were examining a prokaryote you would expect to see circular chromosomes and
supercoiling.
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Unit 3: Molecular Genetics
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66. All four formulas are correct. Formula (a) is correct because no other nucleotides are present.
Formula (b) is correct because the percentage of adenine will always equal the percentage of
thymine and the percentage of guanine will always equal the percentage of cytosine. Therefore
the numerator will always equal the denominator, resulting in a fraction that is equivalent to 1.
Formula (c) is correct for the same reason. Formula (d) is correct because the numerator of each
fraction will be equal to its denominator, so both fractions will be equivalent to 1.
67.
68.
57 min × 60 s • min–1 = 3420 s to replicate
3420 s × 1000 • bp s–1 = 3 420 000 bp replicated in 3420 s (57 min)
3 420 000 bp × 0.34 nm • bp–1 = 1.16 × 106 nm
= 1.16 mm of chromosome
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Unit 3: Molecular Genetics
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69.
The lagging strand ends up with an overhang from the template strand as a result of the last
primer that was added. The leading strand is fully replicated with no overhang.
70.
Turning off Genes in the Production of Compound E
Expected result
Gene turned off
Gene 1
• accumulation of A
• enzyme 1 absent
• no E produced
Gene 2
• accumulation of B
• enzyme 2 absent
• no E produced
Gene 3
• accumulation of C
• enzyme 3 absent
• no E produced
Gene 4
• accumulation of D
• enzyme 4 absent
• no E produced
71. (a) The blanks can represent any codon. There is redundancy in the code.
(b) The sequence explains the wobble hypothesis in which the third codon can be written as any
of the four codons and still code for a single amino acid. The sequence produces the polypeptide
Arg-Pro-Leu-Ser-Gly-Val-Thr
72. (a) Ubiquitin
Eukaryotes 76 bp/6 aa • s–1 = 12.7 s
Prokaryotes 76 bp/20 17 aa • s–1 = 4.5 s
Prokaryotes are 8.2 s faster
Titin
Eukaryotes 34 350 bp/6 aa • s–1 = 5725 s
Prokaryotes 34 350 bp/20 aa • s–1 = 2020.5 s
Prokaryotes are 3704.5 s or 1.03 h faster
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Unit 3: Molecular Genetics
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(b) The formation of polysomes by prokaryotes on the mRNA as they are being translated
drastically increases the speed of transcription and translation. Since translation and transcription
happen almost simultaneously the process is much faster.
73.
Key Differences in Translation
Prokaryotes
Eukaryotes
commence with
commence with
formylmethionine
methionine
ribosomes are
ribosomes are smaller
larger
occurs in nucleus
occurs in cytosol
74. The pathway is enzyme 4-2-1-3.
75. The lacI code has been mutated. Either the operator has been removed or the lacI protein has
been suppressed. Either way this allows for the continual synthesis of β-galactosidase. For the
cell, energy is being wasted on the production of an enzyme that is not required.
76. (a) Having high levels of repeated sequences in this area is advantageous as telomeres lose
genetic information when they are copied. Centromeres are the point where DNA is split during
cell division and having non-coding repeated sequences prevents exons from being deleted.
(b) You would not expect exons in this area as these are regions that are easily damaged and this
could cause major damage to the cell if information is lost.
77. (a) approximately 1000 times faster
(b) 2.5 × 106 bp /2.7 × 10–5 bp • gen–1 = 9.26 ×1010 generations × 20 yr
= 1.85 × 1012 years
(c) 3 × 109 bp /2.7 × 10–8 bp • gen–1 = 1.11x 1017 gen × 20 yr
= 2.22 × 1018 years
(d) Scientists suggest that mitochondria were prokaryotes that entered into a symbiotic
relationship with a eukaryotic cell sometime in the past. Their high mutation rate could help
them adapt very quickly.
78.
79. (a) In nanopore sequencing, a single strand of DNA is pulled through an tiny hole (nanopore)
that measures 2.5 nm in diameter allowing the DNA sequence to be read one base pair at a time.
(b) Nanopore sequencing does not require expensive equipment, so it could allow gene
sequencing for a much wider variety of applications than are available today.
80. (a) DNA microarrays help scientists to identify the location and the function of genes.
(b) The capabilities of DNA microarrays mean that the gene expression of different cells or
organisms can be compared. Microarrays help scientists understand the processes involved in the
development of an organism or to detect genetic mutations by comparing an individual’s DNA
with normal DNA.
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Unit 3: Molecular Genetics
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81. You could use a microarray and use normal DNA as a template for easy comparison with
DNA that might contain genetic mutations. The fluorescence of the labels produces coloured
spots on the microarray that signify which genes were active in the normal cell or in the cancer
cell. It shows how many genes have altered expression in the cancer cells and how that
expression has changed.
Evaluation
82. (a) Answers may vary. Sample answer: They may have predicted that a large number of
radioactive fragments of DNA would be isolated from the temperature sensitive strain of E. coli
that lacked functioning DNA ligase above 40 °C. These radioactive fragments would be Okazaki
fragments.
(b) The strands would have been incubated at 25 °C to establish standard growing conditions
before the temperature treatments were applied.
(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.
(d) Answers may vary. Sample answer: 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 nontemperature-sensitive strain of E. coli, one long strand would have been produced since DNA
ligase was functioning. Pauling and Hamm could have concluded that replication on one of the
strands occurred via a different mechanism that joins short fragments of DNA together via DNA
ligase.
83. Answers may vary. Sample answer: If a vector could be designed to enter a cell and identify
the locus of the mutation, you could then design a template strand that DNA polymerase could
use to verify the coding sequence. DNA polymerase I could then be activated to replace the
nucleotide with the correct one. The DNA ligase would then be activated to anneal the sequence.
The challenges would include developing the vector for infection, the marker, the verification
sequence, and a kill switch for the drug.
84. (a) Answers may vary. Sample answer: Archaea should be closer to prokaryotes since they
both contain circular DNA. It will have a slower DNA replication rate since archaea use histonelike proteins to aid in supercoiling. Thus, the DNA requires extra time to unwind during
replication.
(b) Archaea should be in a distinct domain because this evidence suggests they have a separate
method of passing on their genetic information.
85. The wobble hypothesis serves both of these functions. If a virus or bacteria was able to read
an mRNA and engineer a complementary template, the wobble hypothesis assures that the final
codon can vary. This protects the parental strand DNA as the only three codons that have a single
codon sequence is the start and one stop codon. This means that viruses and bacteria will have
difficulty reverse engineering DNA and ensures the integrity of the DNA by preventing it from
being read. At the same time, DNA will mutate as a result of sexual reproduction and natural
mutation so the wobble hypothesis allows for some mutations and changes without losing the
ability to code for a required amino acid. The ability to vary at the final spot ensures that there
are alternate means of maintaining required protein sequences. This allows for both flexibility
and integrity.
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Unit 3: Molecular Genetics
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86. The gene coding for the protein came from a human, and you replicated it in a prokaryote.
Because the introns were included in the isolated sequence, these were included in the new
protein. Your professor would like you to determine either the locations of the introns and
remove them before they are inserted into E. coli, or to find a eukaryote that you could use as a
host for the sequence.
87. The addition of a second known pathway is used to verify the host organism’s ability to take
up the gene and to verify that the mechanisms of transcription and translation are consistent with
our known models.
88. Based on this information, you can conclude that the rate of mutation in the viruses is related
to the number of immunizations you must receive during your lifetime. The highly error prone
RNA polymerase of the influenza virus can account for the need for multiple vaccinations, while
polio and measles mutate at much lower rates. HIV is proving elusive because of its extremely
high mutation rate. The mutation rate also directly relates to our ability to develop effective
treatments.
89. You would probably want to use a goat. Since spider silk is a large complex protein, an
animal model would be preferred as they are capable of modifying larger proteins more
efficiently. You could isolate the milk production gene and insert the silk gene in place of the
casein protein gene. In this way, you could harvest the milk and refine the protein.
90. Answers will vary. Students may include the following pro and con points.
Pro: more strawberries would be produced at lower cost, and the fruit could be stored at cooler
temperatures once picked and travel farther, increasing the size of markets.
Con: It might also cause allergic reactions in people who consumed the strawberries and were
allergic to fish. It is possible that fish diseases could affect the strawberry or vice versa.
91. (a) Lane A is a series of markers that are used to verify the proper functioning of the gel and
to identify the fragment sizes of the unknown.
(b) You can infer that only sample 3 appears to be from a captive population. Sample 1 shares
two markers with the captive population, and samples 2 and 3 share one marker with the captive
population. Sample 5 shares no markers with the captive population. It appears that only one
animal in this shipment is from a captive population. It is possible that samples 1, 2, and 4 are the
results of a cross with the captive population to make them appear as if they are legal for trade.
The DNA in sample 5 is definitely not from a captive population.
(c) If this was performed using VNTRs you could improve it using SNRPs as they have better
resolution in determining relatedness. This would possibly remove the doubt of samples 1, 2, and
4 and confirm lineage with more accuracy.
(d) Molecular genetics allows customs agents and animal exporters and importers to verify the
source of the animals and the validity of any paperwork that arrives with the animals. This helps
to prevent wild populations from being poached.
(e) DNA microarray technology would drastically improve the efficiency and decrease the cost
of searching for illegally traded animals and animal parts. Since many of the organisms that are
traded illegally do have small captive populations, it would be possible to manufacture cDNA
samples that could isolate the wild and captive populations. Additionally, it would allow border
agents to confirm types and species that are being traded much more quickly, especially when
parts of the organism has been turned into manufactured products.
Copyright © 2012 Nelson Education Ltd.
Unit 3: Molecular Genetics
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Reflect on Your Learning
92. Answers may vary.
93. Answers may vary. Answers should use supporting evidence focussing on the possible
benefits of somatic-line therapy such as the treatment of the disease, as well as the risks such as
problems with the viral vector.
94. Answers may vary. Answers will be based on their interpretations and experiences. As
background Canada is currently reviewing the genetic privacy implications under the Charter of
Rights and Freedoms. Canada has a central DNA database used by law enforcement for people
convicted of a crime. Additionally there is still confusion about what you are required to disclose
about any information that is gleaned from genetic testing to your insurance company. However,
no DNA database exists or is planned to store this information for non-police purposes.
95. Answers may vary. Sample answer: Molecular genetics has drastically altered how we use
medicine and technology, and as a result, how we use law enforcement, and even our food. Many
issues such as genetic engineering, genetic counselling and gene sequencing have each added
their own solutions and issues to society as these technologies challenge societal beliefs and legal
precedents.
96. Answers may vary. Sample answer: I have learned that many of the greatest discoveries have
been made over long periods of time with many people developing parts of theories along the
way. In most cases the initial work establishes a line of thinking by showing possibilities, but in
many cases technology or other knowledge that is required does not exist. The gap in knowledge
and technology must be bridged in order for theories to be tested.
Research
97. Answers may vary. Sample answer:
(a) As telomerase inhibition is increased, cells struggled to reproduce past 50 or 60 cell cycles.
(b) If you can inhibit telomerase activity, you could potentially treat cancer at the cellular level.
(c) My conclusions are similar to Blagoevs’s findings.
98. Answers may vary. In the form of a posting, the following information should be included:
 The Hox gene is located in a homeo box, a group of highly conserved genes. It is responsible
for directing the proper development of segments in segmented animals. It controls their location
and unique identity in an organism.
 The genes are so similar that a fly and a mouse could share the translated hox protein, and it
would function despite the wide range in evolutionary development.
 The Hox gene produces hox proteins that function very similarly in all organisms.
 The Hox gene has helped lay the foundation as it can be traced in all bilateral organisms
starting in Cnidarians. For evolutionary developmental biology the Hox gene has allowed science
to ask questions about the evolutionary relatedness of organisms based on their developmental
biology.
99. Answers may vary. Answers should include the following information:
(a) Myths: rapid sequencing is done on all crime scenes, rapid sequencing takes only a few
hours.
Truths: Rapid sequencing is progressing quickly, rapid sequencing is coming down in price and
makes DNA sequencing cheaper, rapid sequencing provides more information than is needed in
a court case.
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Unit 3: Molecular Genetics
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(b) 1976: Maxim-Gilbert Sequencing. Pros: Purified DNA can be used directly.
Cons: chain termination method uses highly radioactive labels.
1977: Sanger chain termination method. Pros: low level radioactive labels.
Cons: non-specific binding and DNA secondary structures can affect fidelity.
1990: the Lynx Therapeutics’ Massively Parallel Signature Sequencing (MPSS). Pros: extremely
fast. Cons: uses short sequences and requires large amounts of computer power to reassemble the
sequences
2005: Polony sequencing ligation based pairing Pros: very inexpensive and accurate.
Cons: non-uniform amplification can lower the efficiency.
2008: 454 pyrosequencing. Pros: parallel sequencing allows for faster results
Cons: somewhat expensive, cannot handle large repeating sequences
2008: Illumina (Solexa) sequencing. Pros: extended one nucleotide at a time and uses
fluorescence tagging. Cons: Proprietary technology.
2008: SOLiD sequencing. Pros: inexpensive and can handle large repeating sequences.
Cons: proprietary technology.
2010: Ion semiconductor sequencing. Pros: inexpensive and low maintenance cost
Cons: has a hard time handling large repeating sequences
2008: DNA nanoball sequencing. Pros: low reading error rate, Cons: requires multiple PCR
cycles and does not handle large repeated sequences well.
Nanopore DNA sequencing. Pros: does not require PCR, labelling or nucleotides.
Cons: generates only short sequences of DNA.
(c) Even in minor crimes, DNA evidence can play an important role in eliminating or confirming
suspects. However, cases are backlogged and budgets are constrained. There are presently over
1500 cases backlogged in the Canada alone, with an average wait time of over 114 days to
complete a test. The big constraint is budget and space for performing these tests. New
machinery would reduce the backlog but the cost of new technology is a limiting factor.
Currently many law enforcement jurisdictions still use SNRPs which can take time. This method
is still preferred as it is tested and has case law for its use.
100. Answers will vary. Answers should include the following information:
• Henrietta Lacks was an African American woman who died of cervical cancer in 1951; George
Otto Gey was her physician.
• HeLa l cells are a cell line collected from the cancerous lesion on Henrietta Lacks’ cervix. They
were cultured by Dr. Gey who was interested in their ability to not die.
• The interesting part of these cells is that they do not follow the Hayflick limit. Almost all other
cell lines have eventually died out, while HeLa cells are still alive and reproducing. Scientists
believe that the telomerase gene has become turned on permanently allowing these cells to divide
indefinitely without damage to the telomeres.
• The main controversy that surrounds these cells is whether excised tissues from a patient
belong to that patient or are the property of the doctors. Initially, Dr. Gey donated the cell lines
to researchers. As their importance was discovered, these cells developed a major commercial
value. The Lacks family was never informed that these cells were being sold.
• There have been attempts to develop the cells as a paraphyletic species. The scientific
community has valid reasons both for and against developing a classification and species
designation.
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Unit 3: Molecular Genetics
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101. Answers will vary, but students should include the following information:
• Crowdsourcing is the use of large groups of people to solve parts or whole problems outside of
the environment in which they were created. In science, this has helped with the processing of
large volumes of data either through direct investigation by the crowd, like in Fold-it, or through
allowing access to unused computational power in the crowd’s computers.
• Scientist are still needed to develop the problems and interpret the results. Unfortunately, the
public might perceive that they solved a problem that science could not. Some people see this as
a reduction of science. Many others see this as a way to make science important outside of the
lab and a way to get the public involved in a piece of the solution.
• examples: eteRNA: like Fold It but with RNA
Annatathon: Metagenome Annotation uses a distributed grid of undergraduate students’
computers to analyses genomes for coding sequences
PlantGDB: Allows people to design their own genes that might later be patented
102. Answers will vary, but students should include the following information:
• mRNA: messenger RNA that transcribes DNA to RNA
• tRNA: transfer RNA that translates mRNA on ribosomes to proteins
• rRNA: ribosomal RNA that stabilizes the structure of ribosomes
• tmRNA: transfer messenger RNA that identifies proteins that lack stops and terminates the
sequence
• sRNA: short RNA used to stabilize proteins such as telomerase
• miRNA: micro RNA used to decrease production of proteins proteins by cleaving
complimentary mRNA through RNA interference.
• siRNA: small interfering RNA used to regulate gene expression by methylating genes
• CRISPR and AntisenseRNA: used as part of RNAi to form double-stranded RNA that is meant
for degradation
• snRNA: small nuclear RNA that helps bind spliceosomes
• snoRNA: small nucleolar RNA that performs nucleotide modification to mRNA and tRNA
• dsRNA: double-stranded RNA found in some viruses and can trigger RNAi in eukaryotes
• A-DNA: right-handed DNA that is more compact, no function has been found outside of the
lab
• B-DNA: standard DNA involved in biological activity
• c-DNA: complimentary DNA strands made from mRNA and used for genetic research
• z-DNA: left-handed DNA that winds in a zig-zag pattern that appears to relieve torsion during
supercoiling; it is short lived
• DNA and RNA are found in every kingdom including viruses but in various roles.
• DNA and RNA are similar except that thymine is found in DNA and uracil in RNA and that
most RNA is single stranded. This allows RNA to form 3D structures outside of helixes,
allowing it to perform in the role of a protein, enzyme, or other structural component.
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Unit 3: Molecular Genetics
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103. Answers will vary, but students should include the following information:
 Types
• Non-processed or duplicated: areas that have become non-functional but still contain all their
introns and exons.
• Transposed or processed: LINES and SINES which contain genes but have their introns spliced
out.
• Unitary or disabled: mutated genes that prevent the transcription of a gene in a population.
• In some bioinformatics research, scientists have been able to insert pseudogenes into introns
and have had the genes transcribed. Since much of the research in pseudogenes relies on
computers and algorithms, many scientists think that more analysis is needed to verify
pseudogene function. Other researchers have found that pseudogenes play an important role in
siRNA and RNAi.
• Scientists are not sure about the function of pseudogenes. Some believe that they are the last
stop for non-coding genes before they are removed. Other scientists believe pseudogenes offer
some genetic immunity from viruses by presenting them with non coding DNA.
• Scientists are learning how pseudogenes work, their evolutionary importance and how they
affect gene expression and regulation.
104. Answers will vary, but students should include the following information:
(a) In Canada the mortality rate from measles is approximately 1/1000. The rate of mortality or
serious complicatino from the measles vaccine is 1/1 000 000. 1/1 000 000 is 1000 times greater
than 1/1000, so the risk of dying from measles is 1000 times greater than the risk of dying or
having serious complications as a result of being vaccinated against it.
(b) Bad science, like the publication of tampered data, in the case of the measles vaccine has
increased the risk of getting the disease.
(c) Vaccination rates dropping even below 90 % increase the risk of the disease returning and
mutating outside of the genetic knowledge of the vaccine. In Britain, where vaccination rates
went from over 90 % to below 88 % for initial vaccination and now below 75 % for the booster,
the rate of disease went from less than 10 cases in the early 1990s to greater the 300 currently.
105. (a) Knockout mice helped scientist discover that the FoxP2 gene played an important role in
vocalizations and communication. They believe that the gene plays a role in the development of
neuronal tissue and in the Purkinje layer of the brain. Birds and other vertebrates that have been
investigated suggest that the protein coded by this gene helps nerves remain plastic.
(b) The FoxP2 gene, discovered in 1998, is conserved across the vertebrate kingdom in most
organisms (there is debate about bats), suggesting a common ancestor. In primates it is highly
conserved, and scientists believe a small AA substitution in the protein allowed for the
development of speech in humans. People who have alterations to this gene appear to have no
cognitive disability but cannot communicate effectively using oral language.
(c) Examples include Vacanti mice for regrowing organs, sudden death mosquitos, glass frogs,
ruby puppies, land mine detecting plants, enviro pigs and cows, and various bacteria.
(d) Answers will vary depending on the GMO selected.
Copyright © 2012 Nelson Education Ltd.
Unit 3: Molecular Genetics
U3-19
106. Answers will vary, but should include the following information:
Proteomics
• The large scale study of proteins, their structure and function
Technology
• Methylation, computational modeling software, High throughput sequencing, SAGE, Micro
Arrays, AP/MS
Goal
• To develop a database of proteins to allow for comparative studies of form and function
Genomics
• Technology to catalogue and decipher the DNA sequence of an organism
Technology
• High throughput sequencing, microarray, BLAST
Goal
• To inexpensively determine the genome of all organisms on the planet
Functional Genomics
• Studies the gene-gene and protein-protein interactions. This focuses on genome-wide questions
rather than individual genes.
Technology
• It relies on High throughput sequencing, SAGE, Microarrays, and AP/MS.
Goal
• To discover a key to determining gene function by examining conserved and amalgamated
genes
Structural Genomics
• Aims to describe the 3-D structure of every protein in a genome
Technology
• High throughput sequencing, NMR, X-Ray crystallography, ab initio modelling, sequence
based modelling, and threading.
Goal
• To identify protein folding factors and establish a proteome database
Copyright © 2012 Nelson Education Ltd.
Unit 3: Molecular Genetics
U3-20