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Chapter Nine: Chromosome Variation
COMPREHENSION QUESTIONS
*1. List the different types of chromosome mutations and define each.
Chromosome rearrangements:
Deletion: loss of a portion of a chromosome.
Duplication: addition of an extra copy of a portion of a chromosome.
Inversion: a portion of the chromosome is reversed in orientation.
Translocation: a portion of one chromosome becomes incorporated into a
different (nonhomologous) chromosome.
Aneupoloidy: loss or gain of one or more chromosomes so that the chromosome
number deviates from 2n or the normal euploid complement.
Polyploidy: Gain of entire sets of chromosomes so the chromosome number
changes from 2n to 3n (triploid), 4n (tetraploid), and so on.
*2. Why do extra copies of genes sometimes cause drastic phenotypic effects?
The expression of some genes is balanced with the expression of other genes; the
ratios of their gene products, usually proteins, must be maintained within a narrow
range for proper cell function. Extra copies of one of these genes cause that gene to
be expressed at proportionately higher levels, thereby upsetting the balance of gene
products.
3.
Draw a pair of chromosomes as they would appear during synapsis in prophase I of
meiosis in an individual heterozygous for a chromosome duplication.
In the figure below, adapted from Figure 9.6b, the vertical dashed lines denote the
locations of the genes labeled A, B, C…G. One chromosome has duplicated a
segment containing genes C D and E.
A
B
C
D
E
C
4.
F G
D E
How does a deletion cause pseudodominance?
An individual heterozygous for a deletion has only one copy of the genes that are
missing in the deleted portion of the chromosome. Any alleles, even recessive, of
these genes are therefore expressed phenotypically.
*5. What is the difference between a paracentric and a pericentric inversion?
A paracentric inversion does not include the centromere; a pericentric inversion
includes the centromere.
Chapter Nine: Chromosome Variation 117
6.
How do inversions cause phenotypic effects?
Although inversions do not result in loss or duplication of chromosomal material,
inversions can have phenotypic consequences if the inversion disrupts a gene at one
of its breakpoints or if a gene near a breakpoint is altered in its expression because
of a change in its chromosomal environment, such as relocation to a
heterochromatic region. Such effects on gene expression are called position effects.
*7. Draw a pair of chromosomes as they would appear during synapsis in prophase I of
meiosis in an individual heterozygous for a paracentric inversion.
In the following figure, adapted from Figure 9.12, the inverted sequence with genes
E, F, and G is in the looped region of the chromosomes.
E
A
8.
B
C
D
F
G
H
Explain why recombination is suppressed in individuals heterozygous for
paracentric and pericentric inversions.
A crossover within a paracentric inversion produces a dicentric and an acentric
recombinant chromatid. The acentric fragment is lost, and the dicentric fragment
breaks, resulting in chromatids with large deletions that lead to nonviable gametes
or embryonic lethality. A crossover within a pericentric inversion produces
recombinant chromatids that have duplications or deletions. Again, gametes with
these recombinant chromatids do not lead to viable progeny.
*9. How do translocations produce phenotypic effects?
Like inversions, translocations can produce phenotypic effects if the translocation
breakpoint disrupts a gene or if a gene near the breakpoint is altered in its
expression because of relocation to a different chromosomal environment (a
position effect).
10. Sketch the chromosome pairing and the different segregation patterns that can arise
in an individual heterozygous for a reciprocal translocation.
Chromosome pairing
118 Chapter Nine: Chromosome Variation
Alternate segregation
Adjacent-1 segregation
Adjacent-2 segregation
11. What is a Robertsonian translocation?
The long arms of two acrocentric chromosomes are joined to a common centromere
through translocation, resulting in a large metacentric chromosome and a very
small chromosome with two very short arms. The very small chromosome may be
lost.
12. List four major types of aneuploidy.
Nullisomy: having no copies of a chromosome.
Monosomy: having only one copy of a chromosome.
Trisomy: having three copies of a chromosome.
Tetrasomy: having four copies of a chromosome.
*13. Why are sex-chromosome aneuploids more common in humans than autosomal
aneuploids?
Individuals with more than one X chromosome maintain only one active X
chromosome and inactivate all others in the form of Barr bodies. Thus, the extra
copies of the X chromosome do not cause an imbalance in the dosage of functional
Chapter Nine: Chromosome Variation 119
genes, and people with 1, 2, 3, or 4 X chromosomes all have similar levels of gene
expression for most X-linked genes.
The Y chromosome is small and contains relatively few genes, most of which
appear to be focused on male sex determination and male fertility. There are no
genes on the Y chromosome that are essential for human development or viability
(after all, half of all people appear to get along perfectly well without a Y
chromosome). Thus, having extra copies of the Y chromosome does not affect
human embryonic development.
*14. What is the difference between primary Down syndrome and familial Down
syndrome? How does each arise?
Primary Down syndrome is caused by spontaneous, random nondisjunction of
chromosome 21, leading to trisomy 21. Familial Down syndrome most frequently
arises as a result of a Robertsonian translocation of chromosome 21 with another
chromosome, usually chromosome 14. Translocation carriers do not have Down
syndrome, but their children have an increased incidence of Down syndrome. If the
translocated chromosome segregates with the normal chromosome 21, the gamete
will have two copies of chromosome 21 and result in a child with familial Down
syndrome.
*15. What is uniparental disomy and how does it arise?
Uniparental disomy refers to the inheritance of both copies of a chromosome from
the same parent. This may arise originally from a trisomy condition in which the
early embryo loses one of the three chromosomes, and the two remaining copies are
from the same parent.
16. What is mosaicism and how does it arise?
Mosaicism is a condition in which an individual has patches of cells that have
chromosomal abnormalities and patches of cells with normal chromosome content.
Mosaicism may arise from mitotic nondisjunction during early embryonic divisions.
Mosaicism can also arise from X-inactivation in a heterozygous female.
*17. What is the difference between autopolyploidy and allopolyploidy? How does each
arise?
In autopolyploidy, all sets of chromosomes are from the same species.
Autopolyploids typically arise from mitotic nondisjunction of all the chromosomes
in an early 2n embryo, resulting in an autotetraploid, or from meiotic
nondisjunction that results in a 2n gamete fusing with a 1n gamete to form an
autotriploid. In allopolyploidy, the chromosomes of two different species are
contained in one individual through the hybridization of two related species
followed by mitotic nondisjunction. Fusion of their gametes results in a functionally
diploid hybrid with a haploid set of chromosomes from each parent. If an early
embryonic cell then undergoes mitotic nondisjunction and doubles each
chromosome, then a fertile 4n allotetraploid individual having two copies of each
chromosome from each species may result.
120 Chapter Nine: Chromosome Variation
APPLICATION QUESTIONS AND PROBLEMS
*18. Which types of chromosome mutations:
(a) increase the amount of genetic material on a particular chromosome?
Duplications
(b) increase the amount of genetic material for all chromosomes?
Polyploidy
(c) decrease the amount of genetic material on a particular chromosome?
Deletions
(d) change the position of DNA sequences on a single chromosome without
changing the amount of genetic material?
Inversions
(e) move DNA from one chromosome to a nonhomologous chromosome?
Translocations
*19. A chromosome has the segments labeled below, where 
centromere.
AB
What types of chromosome mutations are required to change this chromosome into
each of the following chromosomes? (In some cases, more than one chromosome
mutation may be required.)
(a) A B A B 
F G: Tandem duplication of AB.
(b) A B 
Displaced duplication of AB.
(c) A B 
Paracentric inversion of DEF.
(d) A 
Deletion of B.
(e) A B 
Deletion of FG.
(f) A B 
Paracentric inversion of CDE.
(g) C 
Pericentric inversion of ABC.
(h) A B 
Duplication and inversion of DEF.
(i) A B 
Duplication of CDEF, inversion of EF.
*20. A chromosome initially has the following segments:
AB
Draw and label the chromosome that would result from each of the following
mutations:
(a) Tandem duplication of DEF: A B  C D E F D E F G.
(b) Displaced duplication of DEF: A B  C D E F G D E F.
(c) Deletion of FG: A B  C D E.
(d) Deletion of CD: A B  E F G.
(e) Paracentric inversion that includes DEFG: A B  C G F E D.
(f) Pericentric inversion of BCDE: A E D C  B F G.
Chapter Nine: Chromosome Variation 121
21. The following diagrams represent two nonhomologous chromosomes:
AB
RS
What type of chromosome mutation would produce the following chromosomes?
(a) A B 
RS
Nonreciprocal translocation of E F G.
(b) A U V B 
RS
Nonreciprocal translocation of U V
(c) A B 
RS
Reciprocal translocation of C D E and T U V.
(d) A B 
RS
Reciprocal translocation of D E F and W.
*22. A species has 2n = 16 chromosomes. How many chromosomes will be found per
cell in each of the following mutants in this species?
(a) Monosomic: 15.
(b) Autotriploid: 24.
(c) Autotetraploid: 32.
(d) Trisomic: 17.
(e) Double monosomic: 14.
(f) Nullisomic: 14.
(g) Autopentaploid: 40.
(h) Tetrasomic: 18.
**23. The Notch mutation is a deletion on the X chromosome of Drosophila
melanogaster. Female flies heterozygous for Notch have an indentation on the
margin of their wings; Notch is lethal in the homozygous and hemizygous
conditions. The Notch deletion covers the region of the X chromosome that contains
the locus for white eyes, an X-linked recessive trait. Give the phenotypes and
proportions of progeny produced in the following crosses:
(a) A red-eyed, Notch female is mated with a white-eyed male.
XNX+ × XwY

¼ XNXw Notch female with white eyes
¼ XNY lethal
¼ X+Xw female with red eyes (wild type)
¼ X+Y wild-type male
Overall, 1/3 of live progeny will be Notch females with white eyes, 1/3 will be
wild-type females, and 1/3 will be wild-type males.
122 Chapter Nine: Chromosome Variation
(b) A white-eyed, Notch female is mated with a red-eyed male.
XNXw × X+Y 
¼ XNX+ Notch females
¼ XNY lethal
¼ X+Xw wild-type females
¼ XwY white-eyed males
The surviving progeny will therefore be 1/3 Notch red-eyed females, 1/3 wildtype females, and 1/3 white-eyed males.
(c) A white-eyed, Notch female is mated with a white-eyed male.
XNXw × XwY 
¼ XNXw Notch, white-eyed females
¼ XNY lethal
¼ XwXw white-eyed females
¼ XwY white-eyed males
Viable progeny are 1/3 Notch white-eyed females, 1/3 white-eyed females, and
1/3 white-eyed males.
24. The green nose fly normally has six chromosomes, two metacentric and four
acrocentric. A geneticist examines the chromosomes of an odd-looking green nose
fly and discovers that it only has five chromosomes; three of them are metacentric
and two are acrocentric. Explain how this change in chromosome number might
have occurred.
A Robertsonian translocation between two of the acrocentric chromosomes would
result in a new metacentric chromosome and a very small chromosome that may
have been lost.
25. Species I is diploid (2n = 8) with chromosomes AABBCCDD; related species II is
diploid (2n = 8) with chromosomes MMNNOOPP. Individuals with the following
sets of chromosomes represent what types of chromosome mutations?
(a) AAABBCCDD: Trisomy A.
(b) MMNNOOOOPP: Tetrasomy O.
(c) AABBCDD: Monosomy C.
(d) AAABBBCCCDDD: Triploidy.
(e) AAABBCCDDD: Ditrisomy A and D.
(f) AABBDD: Nullisomy C.
(g) AABBCCDDMMNNOOPP: Allotetraploidy.
(h) AABBCCDDMNOP: Allotriploidy.
*26. A wild-type chromosome has the following segments:
ABC
.
An individual is heterozygous for the following chromosome mutations. For each
mutation, sketch how the wild-type and mutated chromosomes would pair in
prophase I of meiosis, showing all chromosome strands.
Chapter Nine: Chromosome Variation 123
(a)
(b)
(c)
(d)
27. An individual that is heterozygous for a pericentric inversion has the following two
chromosomes:
ABCD

124 Chapter Nine: Chromosome Variation
(a) Sketch the pairing of these two chromosomes in prophase I of meiosis, showing
all four strands.
F
E
D
A BC
G
H I
(b) Draw the chromatids that would result from a single crossover between the E
and F segments.
ABCD
ABCD


(c) What will happen when the chromosomes separate in anaphase I of
meiosis?
Each cell will have a pair of sister chromatids that consist of one functional,
intact chromatid (containing one copy of all the genes, whether inverted or
normal) joined to a nonfunctional chromatid that contains a large duplication
and deletion.
28. Answer part (b) of problem 27 for a two-strand double crossover between E and F.
ABCD
ABCD


Here all four chromatids will be functional.
*29. An individual heterozygous for a reciprocal translocation possesses the following
chromosomes.
AB
AB
DVWX
RS
RS
Chapter Nine: Chromosome Variation 125
(a) Draw the pairing arrangement of these chromosomes in prophase I of meiosis.
AB
CD
U
V
W
X
G
F
E
T
SR
(b) Diagram the alternate, adjacent-1, and adjacent-2 segregation patterns in
anaphase I of meiosis.
(c) Give the products that result from alternate, adjacent-1, and adjacent-2
segregation.
Alternate: Gametes contain either both normal or both translocation
chromosomes, and are all viable.
AB
+RS
and
AB
+RS
126 Chapter Nine: Chromosome Variation
Adjacent-1: Gametes contain one normal and one translocation
chromosome, resulting in duplication of some genes and deficiency for others.
AB
+RS
and
AB
+RS
WX
Adjacent-2 (rare): Gametes also contain one normal and one translocation
chromosome, with duplication of some genes and deficiency for others.
AB
+AB
and
RS
+RS
30. Red-green color-blindness is an X-linked recessive disorder. A young man with a
47,XXY karyotype (Klinefelter syndrome) is color-blind. His 46,XY brother also is
color-blind. Both parents have normal color vision. Where did the nondisjunction
occur that gave rise to the young man with Klinefelter syndrome?
Because the father has normal color vision, the mother must be the carrier for
color-blindness. The color-blind young man with Klinefelter syndrome must have
inherited two copies of the color-blind X chromosome from his mother. The
nondisjunction event therefore most likely took place in meiosis II of the egg.
31. Some individuals with Turner syndrome are 45,X/46,XY mosaics. Explain how this
mosaicism could arise.
Such mosaicism could arise from mitotic nondisjunction early in embryogenesis, in
which the Y chromosome fails to segregate and is lost. All the mitotic descendents
of the resulting 45,X embryonic cell will also be 45,X. The fetus will then consist of
a mosaic of patches of 45,X cells and patches of normal 46,XY cells.
*32. Bill and Betty have had two children with Down syndrome. Bill’s brother has
Down syndrome and his sister has two children with Down syndrome. On the basis
of these observations, which of the following statements is most likely correct?
Explain your reasoning.
(a) Bill has 47 chromosomes.
(b) Betty has 47 chromosomes.
(c) Bill and Betty’s children each have 47 chromosomes.
(d) Bill’s sister has 45 chromosomes.
(e) Bill has 46 chromosomes.
(f) Betty has 45 chromosomes.
(g) Bill’s brother has 45 chromosomes.
The high incidence of Down syndrome in Bill’s family and among Bill’s
relatives is consistent with familial Down syndrome, caused by a Robertsonian
translocation involving chromosome 21. Bill and his sister, who are unaffected,
are phenotypically normal carriers of the translocation and have 45
chromosomes. Their children and Bill’s brother, who have Down syndrome,
have 46 chromosomes, but one of these chromosomes is the translocation that
has an extra copy of the long arm of chromosome 21. From the information
given, there is no reason to suspect that Bill’s wife Betty has any chromosomal
abnormalities. Therefore, statement (d) is most likely correct.
Chapter Nine: Chromosome Variation 127
*33. Tay-Sachs disease is an autosomal recessive disease that causes blindness, deafness,
brain enlargement, and premature death in children. It is possible to identify carriers
for Tay-Sachs disease by means of a blood test. Mike and Sue have both been tested
for the Tay-Sachs gene; Mike is a heterozygous carrier for Tay-Sachs, but Sue is
homozygous for the normal allele. Mike and Sue’s baby boy is completely normal
at birth, but at age 2 develops Tay-Sachs disease. Assuming that a new mutation has
not occurred, how could Mike and Sue’s baby have inherited Tay-Sachs disease?
Mike and Sue’s baby could have inherited Tay-Sachs disease by uniparental
disomy. A nondisjunction in meiosis II during spermatogenesis could have
produced a sperm carrying two copies of the chromosome bearing the Tay-Sachs
allele. Fertilization of a normal egg would then produce a trisomic zygote. Loss of
the mother’s normal chromosome during the first mitotic division would then
produce an embryo in which both remaining copies of the chromosome bear the
Tay-Sachs allele from Mike.
34.
In mammals, sex-chromosome aneuploids are more common than autosomal
aneuploids but, in fishes, sex-chromosome aneuploids and autosomal aneuploids are
found with equal frequency. Offer an explanation for these differences in mammals
and fishes.
In mammals, the higher frequency of sex chromosome aneuploids compared to
autosomal aneuploids is due to X-chromosome inactivation and the lack of essential
genes on the Y chromosome. If fishes do not have chromosome inactivation, and
both their sex chromosomes have numerous essential genes, then the frequency of
aneuploids should be similar for both sex chromosomes and autosomes.
*35. A young couple is planning to have children. Knowing that there have been a
substantial number of stillbirths, miscarriages, and fertility problems on the
husband’s side of the family, they see a genetic counselor. A chromosome analysis
reveals that, whereas the woman has a normal karyotype, the man possesses only 45
chromosomes and is a carrier for a Robertsonian translocation between
chromosomes 22 and 13.
(a) List all the different types of gametes that might be produced by the man.
If the translocation segregates away from the normal chromosomes 22 and 13,
then the resulting gametes will have either (1) normal chromosome 13 and
normal chromosome 22 or (2) translocated chromosome 13+22.
If the translocation and chromosome 22 segregate away from chromosome
13, then the resulting gametes will have either (3) translocated chromosome
13+22 and normal chromosome 22 or (4) normal chromosome 13.
If the translocation and chromosome 13 segregate away from chromosome
22 (probably rare), then the gametes will have either (5) normal chromosome
13 and translocated chromosome 13+22 or (6) normal chromosome 22.
(b) What types of zygotes will develop when each of the gametes produced by the
man fuses with a normal gamete from the woman?
(1) 13, 13, 22, 22; normal.
(2) 13, 13+22, 22; translocation carrier.
(3) 3, 13+22, 22, 22; trisomy 22.
128 Chapter Nine: Chromosome Variation
(4) 13, 13, 22; monosomy 22.
(5) 13, 13, 13+22, 22; trisomy 13.
(6) 13, 22, 22; monosomy 13.
(c) If trisomies and monosomies involving chromosome 13 and 22 are lethal, what
proportion of the surviving offspring will be carriers of the translocation?
Half, or 50%, from the answer in part (b).
CHALLENGE QUESTION
36. Red-green color-blindness is a human X-linked recessive disorder. Jill has normal
color vision, but her father is color-blind. Jill marries Tom, who also has normal
color vision. Jill and Tom have a daughter who has Turner syndrome and is colorblind.
(a) How did the daughter inherit color-blindness?
The daughter, with Turner syndrome, is 45,XO. A normal egg cell with a colorblind X chromosome was fertilized by a sperm carrying no sex chromosome.
Such a sperm could have been produced by a nondisjunction event during
spermatogenesis.
(b) Did the daughter inherit her X chromosome from Jill or from Tom?
The color-blind daughter must have inherited her X chromosome from Jill. Tom
is not color-blind and therefore could not have a color-blind allele on his single
X chromosome. Jill’s father was color-blind, so Jill must have inherited a colorblind X chromosome from him and passed it on to her daughter.