Chapter 6 Review, pages 304–309
Knowledge
1. c
2. d
3. b
4. c
5. b
6. d
7. b
8. a
9. b
10. c
11. c
12. d
13. a
14. False. Bacteria do not possess membrane-bound organelles to store their DNA.
15. False. Purines pair with pyrimidines.
16. True
17. True
18. False. During DNA replication, Okazaki fragments form only on the lagging strand of DNA.
19. False. After the addition of a primer to a DNA strand, DNA polymerase I builds in the Æ
direction on both strands.
20. False. DNA replication is described as semiconservative.
21. False. Telomeres are repeating sequences of DNA at the ends of a chromosome that help to
protect the coding regions of the DNA.
22. False. Okazaki fragments on the lagging strand of DNA cause the copied DNA to be shorter
than the parental strand.
23. True
24. (a) ii
(b) iii
(c) i
(d) iv
25. (a) ii
(b) iv
(c) iii
(d) i
26. (a) i
(b) ii
(c) iv
(d) iii
27.
A: nitrogenous bases; their sequence is the code for all genetic information.
B: base pairs; A bonds to T with two hydrogen bonds and G bonds to C with three hydrogen
bonds.
C: backbone; it is made up of a deoxyribose sugar and a phosphate, linked by a phosphodiester
bond. The bases attach to the backbone forming a nucleotide.
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Chapter 6: DNA Hereditary Molecules of Life
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D: double helix, the structure of DNA
28. (a) Griffith found that the virulence factor from heat-treated pathogenic bacteria could be
passed to non-pathogenic bacteria so that they become virulent. He called this process
transformation.
(b) The significance of his findings was the proof that genetic material from nearby sources
could be incorporated into the genome of certain bacteria.
29. The newly formed complementary strand will have the following portions of nucleotides:
33 % adenine; 14 % thymine; 32 % cytosine; 21 % guanine.
30. Technology contributed to Watson and Crick’s discovery of the double helix structure of
DNA through their use of crystallography photos produced by Franklin and Wilkins to help
conceptualize a double helix with inward-facing bases.
31. DNA polymerases assemble the nucleotides that build the new DNA strand and proofread the
new strand for errors.
32. DNA in eukaryotes wraps around histone proteins. DNA is negatively charged so it is
attracted to the positively-charged histones, which allows the DNA to wrap tightly around the
proteins.
33. Plasmids are important in genetic engineering because they are small circular sections of
prokaryotic DNA that float in the cytoplasm, and that can move in and out of a cell and even be
incorporated into other cells’ DNA. These characteristics allow genetic engineers to insert
segments of DNA into host bacteria.
34. The cell’s ability to replicate past 50 cell divisions is limited by the length of the telomeres
on the cell’s chromosomes. When the telomeres become too short, they cease to protect the
coding regions of the chromosome. When coding regions are removed the cell may no longer be
able to divide successfully.
Understanding
35. Proteins cannot be directly synthesized from DNA because DNA cannot move outside the
nucleus. DNA must be transcribed onto RNA because mRNA can move outside the nucleus,
allowing protein synthesis on the ribosomes, which are outside the nucleus.
36. Avery, McLeod, and McCarty isolated DNA as the molecule that had caused transformation
in Griffith’s experiment by treating the bacteria with enzymes that destroyed one of the three
options for the genetic molecule—DNA, RNA, or proteins. This allowed them to isolate the
molecule that caused transformation.
37. (a) DNA is classified as a polymer because it is a chain made up by linking of singular
monomers.
(b) The monomers that make up the DNA polymer chain are nucleotides.
38. Major experiments that led to the discovery of DNA as the molecule of heredity:
1865: Gregor Mendel identified “factors”.
1868: Meischer isolated acidic nuclein from pus.
1920s: Levene showed DNA contained deoxyribose, phosphate groups, and nitrogenous bases.
1928: Griffith described transformation in bacteria.
1944: Avery, McLeod, and McCarty identified DNA as the most probable molecule of heredity.
1949: The four nitrogenous bases in DNA were identified.
1950: Chargaff discovered that bases always occurred in definite ratios, A:T and C:G.
1952: Hershey and Chase confirmed DNA as the molecule of heredity.
1952: Franklin and Wilkins used X-ray crystallography to show the shape of DNA.
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Chapter 6: DNA Hereditary Molecules of Life
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1952: Watson and Crick determined the structure of DNA.
39. The factors that give DNA symmetry include the binding of adenine with thymine using two
hydrogen bonds and binding of guanine with cytosine using three hydrogen bonds. This
complementary pairing between the purines and pyrimidines maintains an equal distance
between the two backbones of DNA and ensures that one strand is the exact complement of the
other.
40. (a) The 3' end of the molecule is indicated by “A”, “B”, and “D”.
(b) The 5' end of the molecule is indicated by “C”.
(c) The answers for part (a) and part (b) were developed by examining which strand is the
leading or lagging component during DNA replication by noting the Okazaki fragments. The
lagging strand is the lower blue strand. DNA replication always occurs in the 5' to 3' direction.
41. The bonding between the base pairs consists of hydrogen bonds while the sugars of the
nucleotides are bound to the phosphate groups to form the backbone with phosphodiester bonds.
42. The rule of base pairing allows a complementary strand to be made from just one other
strand, not two strands contained in a double helix. Since each base can be paired to only one
other base, one of the original strands serves as a template for the copy.
43. DNA ligase is the enzyme that fills in gaps left by the removal of the RNA primers. If a cell
was damaged and DNA ligase could no longer be produced, replication would be affected
because the gaps between the Okazaki fragments of the lagging strand would remain.
44. A common error that occurs during DNA replication is base-pair mismatch. This occurs
when, for example, a thymine is mismatched with a guanine. Mismatches are usually found and
corrected by enzymes in the cell.
45. A cellular solenoid is made up of six nucleosomes that are wrapped around histones. The
nucleosomes resemble the wire in the physical solenoid and the histone resembles the metal core.
46. The pink structures are telomeres. They are found on the ends of chromosomes and are noncoding regions of DNA. They ensure that important coding regions are not lost during
replication.
47. DNA replication will be affected when the telomeres are destroyed. Since the ends of
chromosomes are not copied during replication, some essential coding regions at the ends of the
DNA will not be copied. The new strands will be incomplete. Over several cell divisions, much
of the original DNA molecule may be missing.
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Chapter 6: DNA Hereditary Molecules of Life
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Analysis and Application
48. Given:
average length of nucleotide, LN = 0.34 × 10–9 m/bp
number of base pairs in human diploid cell, NH = 6 × 109 bp/cell
approximate number of cells in a mature human, NC = 5.0 × 1013 cell/human
approximate distance to Sun, LS = 1.5 × 1011 m
Required:
number of time the DNA from one human could reach the sun and return, NR
Solution:
Step 1. Find out the total length of all the nucleotide base pairs in the human body combined.
Multiply the average length of nucleotide by the number of base pairs in human diploid cell.
Multiply this by the approximate number of cells in a mature human to get the total length of the
base pairs in the human body, LTotal bp
LTotal bp = LN × NH × NC
= (0.34 × 1010 m/bp) × (6 × 109 bp/cell) × (5.0 × 1013 cell/human)
LTotal bp = 1.02 × 1014 m/human
Step 2. Divide the total length of the base pairs in the human body by the twice the distance
between the earth and the sun.
NR
= LTotal bp ÷ LS
= (1.02 × 1014 m) ÷ (2 × 1.5 × 1011 m)
NR
= 340/human
Statement:
The DNA from one human could reach the Sun and return 340 times.
49. For this molecule, the percentage of guanine is also 24 %, while the percentage of thymine is
26 % and the percentage of adenine is 26 %.
50. (a) Given: average length of double helical turn, Lt = 3.4 × 10–9 m/turn
number of base pairs in turn, Nt = 10 bp/turn
approximate number of base pairs in a DNA molecule, Nbp = 170 × 106 bp
Required:
length of a DNA molecule containing 170 million base pairs, LM
Solution:
Step 1. Find the number of turns in the molecule, NT. Divide the total number of base pairs by the
number of base pairs per turn.
NT
= Nbp ÷ Nt
= (170 × 106 bp) ÷ (10 bp/turns)
NT
= 170 × 105 turns
Step 2. Find out the total length of the molecule. Multiply the average length of turn by the
number of turns in the molecule.
LM
= NT × L N
= (170 × 105 turns) × (3.4 × 10–9 m/turn)
LM
= 5.8 × 10–2 m
Statement: A DNA molecule that contained 170 million base pairs would be 5.8 × 10–2 m in
length.
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Chapter 6: DNA Hereditary Molecules of Life
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(b) Given:
number of base pairs in turn, Nt = 10 bp/turn
approximate number of base pairs in a DNA molecule, Nbp = 170 × 106 bp
number turns in the DNA molecule containing 170 million base pairs, NT
Required:
Solution:
Find the number of turns in the molecule, NT. Divide the total number of base pairs by the
number of base pairs per turn.
NT
= Nbp ÷ Nt
= (170 × 106 bp) ÷ (10 bp/turns)
NT
= 170 × 105 turns
Statement: A DNA molecule that contained 170 million base pairs would have 170 × 105 turns.
51. (a) Given: length of DNA molecule, LM = 75 × 103 m
length of a 10 base pair turn, Lturn = 3.4 × 109 m/turn
number of base pairs per turn, Nbp = 10 bp/turn
Required:
number nucleotide pairs in molecule, NM
Solution:
Find the number of nucleotide pairs in the molecule, NM. Divide the total length of the molecule
by the number of length of a turn and multiply by the number of nucleotide pairs in a turn.
NM
= LM ÷ Lturn × Nbp
= (75 × 10-3 m) ÷ (3.4 × 10–9 m /turn) × (10 bp/ turn)
NM
= 2.2 × 108 bp
Statement: A DNA molecule that is 75 mm in length contains 2.2 × 108 nucleotide pairs.
(b) Given:
length of DNA molecule, LM
= 75 × 103 m
length of a 10 base pair turn, Lturn
= 3.4 × 109 m/turn
Required:
number nitrogenous bases in the molecule, NB
Solution:
Step 1. Find the number of nucleotide pairs in the molecule, NM. Divide the total length of the
molecule by the length of a turn and multiply by the number of nucleotide pairs in a turn.
NM
= LM ÷ Lturn × Nbp
= (75 × 10-3 m) ÷ (3.4 × 10–9 m /turn) ×(10 bp/ turn)
NM
= 2.2 × 108 bp
Step 2. Multiply the number of nucleotide pairs in the molecule by the number of nitrogenous
bases in a pair, Np
NB
= NM × Np
= (2.2 × 108 bp) × (2 nitrogenous bases/bp)
NB
= 4.4 × 108 nitrogenous bases
Statement: The DNA molecule contains 4.4 × 108 nitrogenous bases.
52. Since purines always pair with pyrimidines, for every A there should be a T and for every G
there should be a C. Therefore, if there are 10 purines (A + G), there should be 10 pyrimidines
(C + T) and 10/10 = 1. The number of A-T pairs compared to G-C pairs is irrelevant and can be
any ratio.
53. Griffith’s discovery rules out the possibility that the live R-strain cells were just using the Sstrain capsules to become infections because the transformed cells were able to reproduce and
pass on the trait to the next generation of cells, Griffith knew that it was not a one-time-only
situation.
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Chapter 6: DNA Hereditary Molecules of Life
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54. (a)
(b) can be determined by the locations of the phosphate
55. (a) Their proposed model did not match the observation by Chargaff that adenine and
thymine always appeared in the same ratio and cytosine and guanine always appeared in the
same ratio. Watson and Crick’s observations suggest that adenine (a pyrimidine) pairs with
thymine (a purine), since they appear it the same ratio. Similarly, their observations suggest that
guanine (a pyrimidine) pairs with cytosine (a purine), since they appear in the same ratio. As a
result, Watson and Crick changed their model to reflect this.
(b) Franklin and Wilkins’ X-ray crystallography results showed the strands of DNA were evenly
spaced. If Watson and Crick’s original model was right the pyrimidine-pyrimidine pairs would
produce bulges in the structure.
56. The molecular structure of DNA illustrated how scientists use communication to build upon
the work of other scientists’ discoveries. Watson and Crick were able to determine the structure
of the DNA molecule because they read the published discoveries of Pauling, Corey, Chargaff,
and others and they discussed Franklin and Wilkins’s research results with them. This helped
both teams interpret research results and draw conclusions from them.
57. (a) Answers may vary. Sample answer: The centrifuge was essential for allowing physical
separation of various cellular components and aiding in the isolation of DNA. Meselson and
Stahl used a centrifuge to prove DNA replicated in a semi-conservative fashion. Centrifuges are
commonly used in molecular biology labs.
(b) Answers may vary. Sample answer: X-ray crystallography was essential to the discovery of
the double helical structure of DNA. Analyzing the patterns in X-ray crystallography provided
Franklin and Wilkins, and later Watson and Crick, with evidence that the DNA molecule had a
helical shape and that its backbone was on the outside.
(c) Answers may vary. Sample answer: Radiation detection was essential for experiments of
Hershey and Chase, Meselson and Stahl, and Franklin and Wilkins. Meselson and Stahl labeled
both strands of parent DNA with radioisotopes and showed that DNA replication was
semiconservative, because the isotopes ended up in both copied strands. Hershey and Chase used
radioisotopes to label determine that it was DNA in a virus, and not the protein coat, that
transformed bacteria. Being able to detect radio-labeled molecules and to use high energy
electromagnetic radiation was critical to our understanding of DNA.
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58. Watson and Crick determined the structure of the DNA molecule by observing that the bases
adenine and thymine were always found in the same ratio and the bases cytosine and guanine
were always found in the same ratio. This led them to the conclusion that adenine paired with
thymine and guanine paired with cytosine.
59. (a) Given: rate of replication in E. coli, RE = 1000 bp/s
number of base pairs in E. coli, NE = 4.6 × 106 bp
Required:
time to replicate the E. coli genome, tE
Solution:
Divide the total number of base pairs in the E. coli genome by its rate of replication.
tbE
= NE ÷ RE
= (4.6 × 106 bp) ÷ (1000 bp/s)
tE
= 4.6 × 103 s
Statement: It will take 4.6 × 103 s or 78.3 min to replicate E. coli’s genome.
(b) Given:
rate of replication in human, RH = 50 bp/s
number of base pairs in human genome, NH = 150 × 106 bp
Required:
time to replicate the genome, tH
Solution:
Divide the total number of base pairs in the E. coli genome by its rate of replication.
tH = NH ÷ RH
= (150 × 106 bp) ÷ (50 bp/s)
tH = 3 × 106 s
Statement: It will take 3 × 106 s or 50 000 min, roughly 34.7 days to replicate the human
genome.
(c) Prokaryotes usually have only one replication origin and one replication bubble, while
eukaryotic DNA replicates with many replication origins and bubbles simultaneously. Thus,
despite being slower per base pair, the process is occurring in many locations, increasing the
speed at which the entire genome can be replicated.
60.
61. (a) You would still need RNA primers, RNA primase, and DNA ligase for DNA to be
replicated.
(b) No, there would be no replication forks since this is only single-stranded DNA. There might
still be Okazaki fragments depending on the length of the strands and the specificity of the RNA
primer. If the RNA primer fits only the opening base pairs of the sequence, then no fragments
will exist and ligase will not be needed. If the primer can attach several times on the strands then
there will still be Okazaki fragments and ligase will be needed.
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62. (a) Answers may vary. Sample answer: UV rays can cause chemical changes to the bases,
which might undergo reactions that break bonds and form new ones incorrectly. Also, exposure
to UV rays may increase the incidence of base-pair mismatch, which is a common error during
replication.
(b) The cell combats damage from UV rays by producing enzymes, such as DNA polymerases,
that seek out replication errors and correct them. Other DNA repair mechanisms are also used for
other types of damage.
(c) Skin cells will be most at risk for damage caused by UV rays because they are exposed to the
Sun more than any other part of the body.
63. Answers may vary. Sample answer:
1) Base pairing relies on recognition. The base on the template strand must be recognized
accurately by the DNA polymerases in order to be matched with its correct base.
2) RNA primers must recognize starting points of replication in order to begin replication.
3) DNA polymerases and other error-control mechanisms must recognize errors in the DNA
sequence when proofreading new DNA strands.
64. At a replication bubble, there are two replication forks that replicate in opposite directions.
At one fork the template builds the leading strand but at the opposite fork, it builds the lagging
strand.
65. Steps in DNA replication:
Step 1: Helicase unwinds the DNA molecule, producing the replication fork.
Step 2: RNA primase produces and places RNA primers on the template strand to initiate
replication.
Step 3: DNA polymerase III adds nucleotides to the RNA primer to form the new DNA strand
along the template.
Step 4: DNA polymerase I replaces the RNA nucleotides of the RNA primers with DNA
nucleotides between the Okazaki fragments on the lagging strand.
Step 5: DNA ligase forms phosphodiester bonds between the phosphates and the sugars on the
new strand’s backbone.
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Chapter 6: DNA Hereditary Molecules of Life
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66. Most likely the geneticist forgot to include DNA polymerase I and DNA ligase in the
mixture. Her results appear to be many lagging strands with Okazaki fragments that were not
filled in. DNA polymerase I removes the RNA primers and replaces them with DNA in the
lagging strand and DNA ligase forms the phosphodiester bond that fills the gaps left at the ends
of those DNA fragments.
67. Answers may vary. Sample answer: The graph shows that the incidence of the disease and
the age of the patient are related. Many diseases are associated with aging because as cells reach
their Hayflick limit, they stop dividing and eventually die. I would hypothesize that this
particular disease is caused by cellular senescence, which is when the cells begin to decline
because they can no longer divide correctly. It may be the result of the cell’s telomeres becoming
too short for proper DNA replication.
Evaluation
68. Answers may vary. Sample answer: It would be reasonable to propose a triple helix with
three inward pointing bases at each step. Bases R, S, and T are always grouped together, as are
bases X, Y, and Z.
69. Answers may vary. Sample answer:
1) Prokaryotic DNA plasmids could be used as vectors to insert DNA into another organism.
2) DNA polymerase I could replace RNA primers with DNA when constructing plasmid vectors.
3) DNA ligase could seal the backbone in DNA segments, which could be used to form the
completed recombinant DNA.
70. Topoisomerase enzymes are involved with prokaryotic supercoiling by keeping the DNA
strands from gaining too much tension as they are unwound and by keeping them from becoming
tangled and binding to themselves and other strands. It is important that DNA maintain its
correct structure even when it is twisted and bent over on itself. This is particularly important in
prokaryotic cells because they lack nuclear membranes, which means that transcription and
translation can occur simultaneously.
71. Answers may vary. Sample answer: Eukaryotes have more than one replication site because
in general they have larger chromosomes than prokaryotes. The large size means that without
multiple replication sites the time required for replication in eukaryotes would be extremely long.
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Chapter 6: DNA Hereditary Molecules of Life 6-10
72. Answers may vary. Sample answer: The carapace is coded for in the animals’ DNA, which
is passed down to offspring. A scientist might try to determine the sequence of base pairs in the
portions of DNA responsible for this feature by comparing the DNA sequences of tortoises with
similar carapace morphologies.
73. Benefits of the nucleotide sequence of the telomere being repeated include telomerase’s
ability to build the telomere section without a corresponding template; it must already have an
RNA pattern of the base sequence. The base sequence never changes, so telomerase can
repeatedly add these nucleotides without requiring instruction from a template strand.
Reflect on Your Learning
74. Answers may vary. Sample answer: DNA and its structures and processes are like a poem.
The nucleotides are the letters that form words. The structure of the poem is precise, like the
double helix of DNA. The original manuscript that stays with the poet is the DNA, but copies of
the poem made from the original go out into the world, like RNA travelling outside the nucleus.
The spoken words of the poem are its performance, just like the proteins that DNA codes for are
its performance.
75. Answers may vary. Students’ answer should include the approximately 80 years between
when Mendel suggested the role of “factors” in heredity and the discoveries of Watson and
Crick. Between that time there were a large number of people involved including Chargaff,
Franklin, Wilkins, Meischer, Griffiths, Avery, McLeod, McCarty, Hershey, and Chase.
76. Answers may vary. Sample answer: The discovery of DNA relied on the efforts of many
scientists, some of whom never would see their impact on the understanding of DNA. Even in
the few years prior to Watson and Crick’s publication many scientists did not realize the
significance of their work (puzzle pieces) or were not able to develop their ideas enough to
elucidate DNA’s structure (the puzzle). In many ways data is just a set of disunited facts until the
vision to see how it all fits together comes together. Watson and Crick were able to fit together
all the others’ results and for that they won the Nobel Prize.
77. Answers may vary. Sample answer: The use of radionucleotides in scientific study was a
mechanism of interest. This has helped me think about the role that radionucleotides could play
in other areas of biology. For example, they could use radionucleotides to investigate how DNA
is damaged through dietary choices.
78. Answers may vary. Sample answer: I would not increase my telomerase levels if it were
possible because the risks outweigh the benefits in my opinion. Increasing telomerase levels
could result in cancer because cells will continue to divide without stopping, which could cause
the formation of a tumour. Some people may choose to increase their telomerase because
shortened telomeres lead to aging. If many people chose that option, the human population
would explode and planet would be incapable of sustaining such a population. It is not a
responsible option.
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Chapter 6: DNA Hereditary Molecules of Life 6-11
Research
79. (a) The various types of DNA polymerases include the following:
Family A: responsible for repairing Okazaki fragments in the lysing and ligating base pairs in the
lagging strand;
Family B: responsible for extending primers and fixing terminal mismatches;
Family C: primary bacterial polymerase;
Family D: found in Archaea and are responsible for replication;
Family X: responsible for joining end breaks in DNA;
Family Y: able to replicate DNA through damaged DNA; and
Family RT: reverse transcriptase found in viruses.
(b) DNA slippage occurs during replication where the DNA polymerase slips from the DNA
template strand and causes the new DNA to contain a large repeated section of DNA.
(c) Freidreich’s ataxia is a genetic disorder that causes sections of DNA to be repeated as much
as 1000 times. The disorder affects the heart and muscles and gets progressively worse through
the generations as the repeated sequences are elongated by further cellular division.
80. Answers will vary. Multimedia overviews should include: After their discovery, Watson and
Crick did more research with DNA and how it controls protein synthesis. Their work has helped
to form the basis of proteomics. The establishment of the Cold Springs Harbor Institute has
helped educate many graduate students and helped discover many aspects of molecular genetics.
81. Answers may vary. Multimedia productions should include: DNA repair mechanisms are
specific to the type of repair needed. In the case of UV damage, photolyase is activated to undo
the damage caused in a process called photoreactivation. Methylation of guanine is counteracted
by a protein called methyl guanine methyl transferase (MGMT). Base pair excision removes a
specific base and is completed by DNA glycosylase, which excises the identified error. DNA
ligase repairs the gap with the proper base. Mismatch repair looks specifically for mismatched
nucleotides and removes them with DNA polymerase. Diseases associated with DNA repair
include: Werners syndrome: premature aging; Bloom’s syndrome: hypersensitivity to UV light;
ataxia telangiectasia: hypersensitivity to ionizing radiation; trichothiodystrophy: brittle hair, nails
and sensitive skin; cockayne syndrome: sensitivity to UV light; and breast cancer caused by
BRCA1 and BRCA2.
82. Answers may vary. Students’ answer may mention that scientists have shown telomerase
activity is increased in cancer cells, turning them into immortal cells capable of perpetual
replication like stem cells. Telomerase activity is high in stem cells and increased in cells that
undergo cellular division at relatively high rates but is in low concentrations in many other cells.
Current research into cancer drug therapy looks at turning telomerase activity off causing cells to
return to normal and shorten their telomeres as they replicate, and theoretically making the
cancer cells mortal once again.
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