Chapter 13—Chromosomal Rearrangements and Changes in Chromosome Number
Reshape Eukaryote Genomes
Fill in the Blank
1. Events that reshape genomes by reorganizing the DNA sequences within one or more
chromosomes are known as ____________________.
Ans: rearrangements
Difficulty: 2
2. ____________________, duplications, inversions, translocations, and movements of
transposable elements are all types of rearrangements of chromosomal material.
Ans: Deletions
Difficulty: 2
3. Very large deletions are visible at the relatively low resolution of a
__________________, showing up as the loss of one or more bands from a
chromosome.
Ans: karyotype
Difficulty: 2
4. Changes in gene ____________________, the number of times a given gene is present
in the cell nucleus, can create a genetic imbalance.
Ans: dosage
Difficulty: 3
5. An unpaired bulge of one member of a homologous pair of chromosomes during
prophase of meiosis I is known as a ____________________.
Ans: deletion loop
Difficulty: 3
6. A recessive mutation in the mouse that prevents homozygous animals from walking in a
straight line is known as the ____________________ gene.
Ans: shaker-1
Difficulty: 3
7. When repeats of a region lie adjacent to each other they are called
____________________ duplications.
Ans: tandem
Difficulty: 2
8. Inversions that include the centromere are termed ____________________.
Ans: pericentric
Difficulty: 2
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9. Inversions that exclude the centromere are termed ____________________.
Ans: paracentric
Difficulty: 2
10. A crossover product that lacks a centromere is known as a(n)
____________________fragment.
Ans: acentric
Difficulty: 2
11. The type of large-scale mutation in which parts of two different chromosomes trade
places is a ____________________ translocation.
Ans: reciprocal
Difficulty: 2
12. ____________________ is the enzyme that catalyzes transposition.
Ans: Transposase
Difficulty: 2
13. Organisms with four copies of a particular chromosome (2n+2) are ________________.
Ans: tetrasomic
Difficulty: 3
14. Down syndrome is also known as ____________________.
Ans: trisomy 21
Difficulty: 2
15. If a tetraploid derives all of its chromosome sets from the same species, we call this kind
of polyploid a(n) ____________________.
Ans: autopolyploid
Difficulty: 3
Multiple Choice
16.
A)
B)
C)
D)
E)
Which of the following are considered chromosomal rearrangements?
inversions
duplications
deletions
translocations
all of the above
Ans: E
Difficulty: 1
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17.
A)
B)
C)
D)
E)
Which of the following removes material from the genome?
inversions
duplications
deletions
translocations
none of the above
Ans: C
Difficulty: 1
18.
A)
B)
C)
D)
E)
Which of the following adds material to the genome?
inversions
duplications
deletions
translocations
none of the above
Ans: B
Difficulty: 1
19. The type of chromosomal rearrangement which reorganizes the DNA sequence within
one chromosome is known as a(n):
A) inversion
B) duplication
C) deletion
D) translocation
E) none of the above
Ans: A
Difficulty: 2
20. In general, which of the following usually has a greater chance of lethality than the
others?
A) inversion
B) duplication
C) deletion
D) translocation
E) all have an equal chance
Ans: C
Difficulty: 2
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21. Sometimes a piece of one chromosome attaches to another chromosome. This is known
as a(n):
A) inversion
B) duplication
C) deletion
D) translocation
E) none of the above
Ans: D
Difficulty: 1
22. Sometimes a part of the genome moves from chromosome to chromosome. This is
known generally as a(n):
A) inversion
B) duplication
C) deletion
D) translocation
E) transposable element
Ans: E
Difficulty: 2
23. Rearrangements and changes in chromosome number may affect gene activity or gene
transmission by altering the ________________________ of certain genes in a cell.
A) position
B) order
C) number
D) all of the above
Ans: D
Difficulty: 2
24.
A)
B)
C)
Karyotypes generally remain constant within a species because:
rearrangements occur frequently.
changes in chromosome number occur infrequently.
genetic instabilities produced by genomic changes usually are at a selective
disadvantage.
D) genetic imbalances are often at a selective advantage.
Ans: C
Difficulty: 2
25.
A)
B)
C)
D)
Despite selection against chromosomal variations:
related species almost always have the same karyotype.
related species almost always have a different karyotype.
closely related species diverge by many chromosomal rearrangements.
distantly related species diverge by only a few chromosomal rearrangements.
Ans: B
Difficulty: 2
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26. In higher organisms, using genetic analysis is usually difficult to distinguish small
deletions in one gene from:
A) heterozygotes.
B) small duplications.
C) monosomies.
D) point mutations.
Ans: D
Difficulty: 1
27. For an organism to survive a deletion of more than a few genes, it must carry a
nondeleted homolog of the deleted chromosome. This is known as:
A) a deletion heterozygote.
B) a deletion homozygote.
C) dosage compensation.
D) a triplolethal chromosome.
Ans: A
Difficulty: 2
28. Individuals born heterozygotes for certain deletions have a greatly increased risk of
losing both copies of certain genes and developing cancer. One such disease is:
A) triplolethal.
B) scarlet eyes.
C) retinoblastoma.
D) cataracts.
Ans: C
Difficulty: 1
29. During the pairing of homologs in prophase of meiosis I, the region of a normal,
nondeleted chromosome that has nothing with which to recombine forms a so-called:
A) inversion loop.
B) deletion heterozygote.
C) crossover suppressor.
D) deletion loop.
Ans: D
Difficulty: 2
30. Using Drosophila polytene chromosomes and small deletions, geneticists have been
able to:
A) map the shaker-1 gene in Drosophila.
B) assign genes to regions of one or two polytene chromosome bands.
C) assign genes to regions of 100kb or less of DNA.
D) all of the above
Ans: D
Difficulty: 2
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31. Which of the following molecular techniques could a scientist use to help locate genes
on cloned fragments of DNA with deletion mutants?
A) In situ hybridization
B) Crossover analysis
C) Southern blot analysis
D) all of the above
E) both a and c
Ans: E
Difficulty: 3
32.
A)
B)
C)
D)
Duplications arise by:
chromosomal breakage and faulty repair.
unequal crossing over.
errors in replication.
all of the above
Ans: D
Difficulty: 1
33. During the pairing of homologs in prophase of meiosis I, the region of a chromosome
bearing extra copies of a particular chromosomal region that has nothing with which to
recombine forms a so-called:
A) inversion loop.
B) deletion heterozygote.
C) duplication loop.
D) deletion loop.
Ans: C
Difficulty: 2
34. An inversion may result from:
A) a half-circle rotation of a chromosomal region following two double-strand breaks in a
chromosome's DNA.
B) the action of a transposable element.
C) a crossover between DNA sequences present in two positions on the same chromosome
in inverted orientation.
D) all of the above
E) none of the above
Ans: D
Difficulty: 2
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35.
A)
B)
C)
D)
Inversions may be hard to detect because they:
never visibly change chromosome banding patterns.
increase recombination in heterozygotes.
do not usually cause an abnormal phenotype.
normally are removed immediately in natural populations.
Ans: C
Difficulty: 3
36.
A)
B)
C)
D)
Which of the following does not happen when an intragenic inversion occurs?
One part of the gene is relocated to a distant region of the chromosome.
One part of the gene stays at its original site.
Homozygotes for the inversion do not survive.
The gene's function is not disrupted.
Ans: D
Difficulty: 3
37. When a crossover occurs within the inversion loop of a pericentric inversion each
recombinant chromatid will have:
A) a single centromere.
B) a duplication of one region.
C) a deletion different from the one of duplication.
D) all of the above
Ans: D
Difficulty: 2
38. Robertsonian translocations result from which of the following?
A) Breaks at or near the centromeres of two acrocentric chromosomes followed by the
reciprocal exchange of broken parts.
B) A part of one chromosome becomes attached to a non-homologous chromosome.
C) Unequal crossing over during meiosis.
D) The fusion of two small chromosomes end-to-end such that a double centromere occurs.
Ans: A
Difficulty: 2
39.
A)
B)
C)
D)
Which of the following does not usually show a problem during meiosis?
translocation heterozygotes
translocation homozygotes
paracentric inversion
pericentric inversion
Ans: B
Difficulty: 1
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40. Of the following segregation patterns, which one is most likely to result in a normal
zygote?
A) alternate
B) adjacent-1
C) adjacent-2
D) nondisjunction
Ans: A
Difficulty: 2
41.
A)
B)
C)
D)
The condition of semisterility is most closely associated with:
chromosomal duplications.
pericentric inversions.
translocation heterozygotes.
translocation homozygotes.
Ans: C
Difficulty: 2
42.
A)
B)
C)
D)
Translocations can help:
determine linkage groups.
aid in the diagnosis and treatment of certain cancers.
map important genes.
all of the above
Ans: D
Difficulty: 1
43.
A)
B)
C)
D)
E)
Down Syndrome can result from:
three copies of chromosome 21.
a translocation of a part of chromosome 21.
a reciprocal translocation between any two autosomes.
a and b
a, b, and c
Ans: D
Difficulty: 2
44.
A)
B)
C)
D)
Which of the following do translocations and inversions not have in common?
don't alter the amount of DNA in the genome
ability to alter gene function
use of inversion loops during crossing over
catalysts of speciation
Ans: C
Difficulty: 2
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45. A transposition is considered a cytologically invisible sequence rearrangement. With
which of the following does it share this property?
A) small deletion
B) large duplication
C) inversion
D) translocation
Ans: A
Difficulty: 2
46.
A)
B)
C)
D)
Barbara McClintock is most closely associated with which of the following?
The initial discovery of genetic transposition.
The discovery of transposable elements in corn.
The mutation rate in translocation heterozygotes.
The demonstration of the presence of transposable elements in polytene chromosomes.
Ans: B
Difficulty: 1
47. Transposable elements have many things in common. Which of the following is not a
usual characteristic of them?
A) Typically smaller than 50 bp.
B) May be present in a genome from one to thousands of times.
C) Are found only in a select group of organisms.
D) Need not be sequences that do something for the organism.
Ans: A
Difficulty: 3
48. Retroposons and retro-viruses have structural parallels. Which of the following also
shares structural parallels with them?
A) tRNA
B) DS-DNA
C) rRNA
D) mRNA
Ans: D
Difficulty: 2
49. Which of the following is a possible effect that a transposable element may have on a
gene?
A) Shift the reading frame.
B) Diminish the efficiency of splicing.
C) Provide a transcription stop signal.
D) all of the above
Ans: D
Difficulty: 2
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50.
A)
B)
C)
D)
Which of the following is not an aneuploidy?
monosomy
tetraploid
trisomy
tetrasomy
Ans: B
Difficulty: 1
51. The most common human aneuploidy is trisomy 21, Down syndrome. All of the effects
listed below may be seen in this syndrome except:
A) death always by age 25.
B) mental retardation.
C) skeletal abnormalities.
D) heart defects.
E) increased susceptibility to infection.
Ans: A
Difficulty: 2
52.
A)
B)
C)
D)
E)
Which of the following sex chromosome aneuploidies is not usually seen in live births?
XO
XXY
YO
XXX
None of the above
Ans: C
Difficulty: 2
53. Turner syndrome, XO, is a sex chromosome aneuploidy. Of the effects listed below,
which one is not usually seen in this syndrome?
A) unusually short stature
B) infertility
C) skeletal abnormalities
D) unusually long limbs
Ans: D
Difficulty: 2
54. In Drosophila, a gynandromorph, which is composed of equal parts male and female
tissue, results from:
A) an XX female losing one X chromosome during the first mitotic division after
fertilization.
B) an egg carrying an X chromosome fertilized by a Y-carrying sperm.
C) a normal egg fertilized by both an X-carrying sperm and a Y-carrying sperm.
D) the fusion of a female embryo with a male embryo.
Ans: A
Difficulty: 3
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55.
A)
B)
C)
D)
Which of the following is not an example of a euploid condition?
triploidy
diploidy
Down syndrome
tetraploidy
Ans: C
Difficulty: 1
56.
A)
B)
C)
D)
Triploid organisms usually result from:
the union of haploid and diploid gametes.
unequal disjunction during embryogenesis.
propagation of fused cell lines.
fusion of three gametes simultaneously.
Ans: A
Difficulty: 2
57. During mitosis, if the chromosomes in a diploid tissue fail to separate after replication,
the resulting daughter cells will be:
A) monoploid.
B) tetrasomic.
C) triploid.
D) tetraploid.
Ans: D
Difficulty: 2
58. Hybrids in which the chromosome sets come from two distinct, though related, species
are known as:
A) autopolyploids.
B) allopolyploids.
C) amphiploids.
D) bivalents.
Ans: B
Difficulty: 2
59. The genus Triticale is a new genus of the various allopolyploid hybrids between wheat
and rye. Some of the members of this genus show agricultural promise because:
A) wheat has a high yield.
B) rye adapts well to unfavorable environments.
C) wheat has a high level of protein.
D) rye has a high level of lysine.
E) all of the above
Ans: E
Difficulty: 2
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60.
A)
B)
C)
D)
Which of the following rarely, if ever, results in a positive force for evolution?
polyploidy
allopolyploidy
trisomy
amphidiploidy
Ans: C
Difficulty: 2
Matching
Match the following descriptions with the terms below
a. inversion
b. duplication
c. deletion
d. translocation
e. transposable element
61. A piece of genetic material that moves from place to place in the genome.
Ans: e
Difficulty: 2
62. A change in the genome whereby new material is added to the genome.
Ans: b
Difficulty: 1
63. A change in the genetic material where a DNA sequence changes direction.
Ans: a
Difficulty: 1
64. A decrease of genetic material in the genome.
Ans: c
Difficulty: 1
65. A piece of chromosome attaches to another chromosome.
Ans: d
Difficulty: 2
Match the following descriptions with the terms below
a. retroposon
b. transposon
c. transposable element
d. transposase
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66. Any DNA segment that moves about in the genome.
Ans: c
Difficulty: 3
67. Moves in the genome with the aid of an RNA intermediate.
Ans: a
Difficulty: 2
68. Moves DNA directly.
Ans: b
Difficulty: 3
69. An enzyme that catalyzes a transposition event.
Ans: d
Difficulty: 1
True or False
70. When comparing mouse and human Giemsa-stained karyotypes, we see no conservation
of banding patterns.
Ans: True
Difficulty: 2
71. Karyotypes generally remain constant within a species because rearrangements and
changes in chromosome number occur infrequently.
Ans: False
Difficulty: 3
72. Changes in chromosome number include aneuploidy, monoploidy, polyploidy, and
duplications.
Ans: False
Difficulty: 2
73. Deletion may arise from errors in replication, from faulty meiotic or mitotic
recombination, and from exposure to X-rays.
Ans: True
Difficulty: 2
74. Homozygosity for a deletion is often, but not always, lethal.
Ans: True
Difficulty: 2
75. Recessive mutations can often be covered by deletions in heterozygotes.
Ans: True
Difficulty: 2
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76. Most duplications have no obvious phenotypic consequences and can be detected only
by cytological or molecular means.
Ans: False
Difficulty: 2
77. Duplication of chromosomal segments rarely has an effect on the evolution of genomes.
Ans: True
Difficulty: 2
78. Crossing-over within an inversion loop produces aberrant recombinant chromatids.
Ans: False
Difficulty: 3
79. Reciprocal translocations are usually phenotypically abnormal because they have
neither lost nor gained genetic material.
Ans: True
Difficulty: 1
80. A hallmark of transposons is that their ends are inverted repeats of each other.
Ans: False
Difficulty: 3
81. The mouse genome has high synteny with the human genome since about 170 DNA
blocks are simply rearranged between the two genomes.
Ans: True
Difficulty: 2
82. Euploid cells contain only incomplete sets of chromosomes.
Ans: False
Difficulty: 2
83. Down syndrome is an example of triploidy.
Ans: False
Difficulty: 2
84. Genetic imbalance results from polyploidy.
Ans: False
Difficulty: 1
85. An acentric fragment is an inversion cross-over product lacking a centromere.
Ans: True
Difficulty: 1
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Short Answer
86. Explain how data from the linkage groups of the mouse can be used as a resource for
assessing human linkage groups.
Ans: Because virtually all genes cloned from the mouse genome are conserved in the
human genome and vice versa, it is possible to construct linkage maps for the two
genomes from the same set of markers. Comparisons of the mouse and human
linkage groups allow one to see a picture somewhere between complete
correspondence and unrelatedness. Genes closely linked in the mouse tend to be
closely linked in humans, but genes that are less tightly linked in one species tend
not to be linked at all in the other. This shows that even though mice and humans
diverged about 65 million years ago, the DNA sequences in many regions are very
similar.
Difficulty: 4
87. Explain the differences between chromosomal rearrangements and changes in
chromosome number. Cite at least one example of each.
Ans: Chromosomal rearrangements reorganize the DNA sequences within one or more
chromosomes. Changes in chromosome number involve losses or gains of entire
chromosomes or sets of chromosomes. (Student may cite as an example of
rearrangements: deletion, duplication, inversion, translocation, and transposable
elements. For changes in chromosome number student may cite an aneuploidy
such as a monosomy or trisomy, monoploidy, or polyploidy.)
Difficulty: 4
88. Describe how an inversion heterozygote can reduce the number of recombinant
progeny.
Ans: When inversion heterozygotes have chromosomes pair up during meiosis, an
inversion loop is formed to allow the tightest possible alignment of homologous
regions. This always produces aberrant recombinant chromatids. Two inversion
cases are possible – pericentric and paracentric. In a pericentric crossover within
the inversion loop each recombinant will carry a duplication of one region and a
deletion of another. This abnormal dosage of some genes will result in abnormal
gametes and if they fertilize normal gametes, zygotes may die because of genetic
imbalance. In a paracentric crossover within the inversion loop the recombinant
chromatids will be unbalanced in both gene dosage and centromere number.
(Student may then explain how centromere number can result in genetically
unbalanced gametes such as what acentric and dicentric chromatids would
produce.)
Difficulty: 4
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89. Discuss the several effects that translocations and inversions have in common.
Ans: Both translocations and inversions change genomic position without affecting the
total amount of DNA. If a breakpoint of either one is within a gene, the gene
function may be altered or lost. Both types may produce genetically imbalanced
gametes that may negatively affect a zygote or developing embryo. (Student may
explain at this point the differences between how the imbalanced gametes are
produced.) Because both reduce viable progeny and heterozygotes, they may play
a role in speciation and evolution.
Difficulty: 4
90. Explain the possible effects that a transposable element may have on a gene.
Ans: Insertion of a transposable element near or within a gene can affect gene
expression and alter phenotype. For example, a B type hemophilia occurs after
insertion of Alu into the gene encoding clotting factor IX. Secondly, the effect of
insertion depends on what the element is and where the insertion point is. If
insertion is into a protein-coding exon, the reading frame may shift or a stop
codon may be introduced. Insertion into an intron may lower the efficiency of
splicing, which may result in removal from the transcript that could lower
production of a normal polypeptide. A stop signal could also affect genes
downstream. Upstream insertion into a regulatory gene could affect gene function
in various ways also.
Difficulty: 4
91. Explain the mechanism by which aneuploidy occurs.
Ans: Aneuploidy occurs because of meiotic nondisjunction either in meiosis I or
meiosis II. In meiosis I if homologs do not separate all gametes produced will
contain an error. Two of the gametes will contain both homologs and two will
contain neither. When fertilization of a normal gamete occurs by either of these
abnormal gametes, aneuploidy results. Half of the zygotes will be trisomic and
half will be monosomic. Meiotic nondisjunction during meiosis II will produce
two normal and two abnormal gametes. If fertilization occurs with either of the
abnormal gametes, aneuploid zygotes are produced.
Difficulty: 4
92. Discuss why triploid organisms are almost always sterile.
Ans: (Student may explain how triploids occur.) Triploids are almost always sterile
because meiosis produces mostly unbalanced gametes. During the first meiotic
division in a triploid germ cell, three sets of chromosomes must segregate into two
daughter cells. Most likely one daughter will end up with two chromosomes and
the other will have only one of any one set of homologs. Some cells will have two
of some chromosomes and the normal one of others. Many combinations of
incorrect number of chromosomes will occur with very little chance of the normal
amount. Most gametes will be aberrant and will have a reduced chance of
producing viable offspring.
Difficulty: 4
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93. Discuss how deletions and duplications may contribute to evolution.
Ans: General examples of how chromosomal rearrangements might contribute to
evolution:
Deletions – a small deletion that moves a coding sequence of one gene next to a
promoter or other regulatory element of an adjacent gene may, rarely, allow
expression of a protein at a novel time in development or in a novel tissue. If the
new time or place of expression is advantageous to the organism, it might become
established in the genome. Duplications – a duplication will provide at least two
copies of a gene. If one copy maintains the original function, the other could
conceivably acquire a new function that would probably be related to the original
function. Many examples can be seen in higher plants and animals. (Students may
also write about the evolutionary contributions of the other chromosomal
rearrangements and might even mention the role of changes in chromosome
number.)
Difficulty: 4
94. Why do inversions act as cross-over suppressors?
Ans: Inversions act as cross-over suppressors because only progeny that do not
recombine within an inversion loop will survive.
Difficulty: 4
95. What is a balancer chromosome?
Ans: A balancer chromosome is a special chromosome often created by the use of Xrays for the purpose of genetic manipulation; these chromosomes often carry
multiple, overlapping inversions that enable researchers to follow them through
crosses, and a recessive lethal mutation that prevents the survival of homozygotes.
Difficulty: 4
96. What is the difference between alternate and an adjacent-1 segregation or an adjacent-2
segregation pattern?
Ans: An alternate segregation pattern results in balanced chromosomes while adjacent
1or 2 patterns yield chromosomes that are unbalanced.
Difficulty: 4
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Experimental Design and Interpretation of Data
97. We now know that several organisms have a high degree of synteny at the genomic
level. You wish to test the hypothesis that the laboratory mouse and human share
genomic similarities. What tests would you complete and given that we now know that
the mouse and human genomes are highly syntenic, what results would you expect?
Ans: Karyotype analysis can be used to test the hypothesis of genomic similarities
however, only animals that have high homology will show similar banding
patterns. Therefore, FISH (fluorescence in situ hybridization) would be a more
useful technique to determine synteny. The mouse and human genomes are
similar in that approximately 170 similar fragments an average length of about 18
Mb are simply rearranged (this is not visible in a karyotype).
Difficulty: 4
98. You are mapping traits in your favorite organism but unbeknownst to you, your
laboratory model organism contains a rare deletion. How will your mapping results be
affected?
Ans: The mapping distance will appear smaller than the actual physical distance in the
wild-type organism.
Difficulty: 4
99. You have discovered an altered phenotype and cloned the gene responsible. However,
the gene you cloned appears to have an unusual sequence in it. In order to determine the
chromosomal location of your new gene, you perform FISH, using only the unusual
sequence, on several animals. To your surprise, the FISH results suggest that each
animal contains the gene on a different chromosome. How would you interpret your
results.
Ans: The unusual sequence is a transposon and your “new” phenotype arose via the
disruption of its gene by the transposon.
Difficulty: 4
100. You are a master gardener and your favorite tomato plant is very sensitive to a pesticide
called DEADBUG. You wish to make your special tomato plants resistant to the
pesticide which you spray on other bushes in your garden. Using microbial techniques
give sufficient and complete details of how you would do this (include ploidy status).
Ans: Haploid pollen grains are cold treated and plated on agar plates. The resulting
embryoids are treated with hormone in liquid culture and eventually grown as a
monoploid plant. The plant is treated with a mutagen to induce mutations that can
result in insensitivity to the pesticide. Somatic cells are removed from the treated
plant and plated on agar containing DEADBUG. Only cells resistant to
DEADBUG will grow. Again the embryoid is hormone treated and grown into a
resistant monoploid plant. Treatment with colchicine will allow duplication of
chromosomes without separation resulting in a normal diploid plant.
Difficulty: 4
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