2. Mendelian inheritance has its physical
basis in the behavior of chromosomes
• Several researchers proposed in the early 1900s
that genes are located on chromosomes
• The behavior of chromosomes during meiosis
was said to account for Mendel’s laws of
segregation and independent assortment
• The location of a particular gene can be seen by
tagging isolated chromosomes with a
fluorescent dye that highlights the gene
4. • The chromosome theory of inheritance states that:
– Mendelian genes have specific loci (positions) on
chromosomes
– It is the chromosomes that undergo segregation and
independent assortment
• Biologists began to see parallels between the behavior
of Mendel’s proposed hereditary factors and
chromosomes
• Around 1902, Sutton and Boveri and others
independently noted the parallels and the
chromosome theory of inheritance began
to form
5. Figure 15.2
P Generation
F1 Generation
Yellow-round
seeds
(YYRR)
Gametes
Meiosis
Fertilization
Green-wrinkled
seeds (yyrr)
Meiosis
Metaphase
I
Anaphase I
Metaphase
II
All F1 plants produce
yellow-round seeds (YyRr).
LAW OF
SEGREGATION
The two alleles for each
gene separate.
LAW OF INDEPENDENT
ASSORTMENT Alleles of
genes on nonhomologous
chromosomes assort
independently.
Y
Y
Y
Y
R R
R R
R R
R
y
y
y
y y
Y
r
r
r
rr
r r
Y Y
Y Y
YY
y y
y y
y y
RR
R R
r r
r r
rr rr
Y Y Y Y
R R R R
YR yr Yr yR1
4
1
4
1
4
1
4
F2 Generation
Fertilization
recombines the
R and r alleles at random.
An F1 × F1 cross-fertilization Fertilization results
in the 9:3:3:1
phenotypic ratio in
the F2 generation.
9 : 3 : 3 : 1
1
2 2
1
3
3
y
y y y
6. Morgan’s Experimental Evidence: Scientific Inquiry
• The first solid evidence associating a specific
gene with a a specific chromosome came from
Thomas Hunt Morgan, an embryologist
• Morgan’s experiments with fruit flies provided
convincing evidence that chromosomes are the
location of Mendel’s heritable factors
• Morgan noted wild type, or normal, phenotypes
that were common in the fly populations
• Traits alternative to the wild type are called
mutant phenotypes
12. Figure 15.7
(a)
(b) (c)
Sperm
Sperm Sperm
Eggs
Eggs Eggs
XN
XN Xn
Y
XN
Xn
XN
Xn
XN
Y
XN
Y
XN
Y
Xn
Y
XN
Y
Xn
Y
XN
Y Xn
Y
XN
XN
XN
Xn
XN
Xn
XN
Xn
XN
Xn
Xn
Xn
XN
XN
XN
XN
Xn
XN
Xn
Xn
Xn
Y
YY
X-linked genes
produce different
phenotypic ratios in
males and females
13. • Some disorders caused by recessive alleles on
the X chromosome in humans:
– Color blindness
– Duchenne muscular dystrophy
– Hemophilia
14. 1. In cats, a sex-linked gene affects coat color. The O allele produces an
enzyme that converts eumelanin, a black or brown pigment, into
phaeomelanin, an orange pigment. The o allele is recessive to O and
produces a defective enzyme, one that does not convert eumelanin into
phaeomelanin. Which of the following statements is/are accurate?
a. The phenotype of o-Y males is black/brown because the
nonfunctional
allele o does not convert eumelanin into phaeomelanin.
b. The phenotype of OO and Oo males is orange because the
functional allele O converts eumelanin into phaeomelanin.
c. The phenotype of Oo males is mixed orange and black/brown
because the functional allele O converts eumelanin into
phaeomelanin in some cell groups (orange) and because in other
cell groups the nonfunctional allele o does not convert eumelanin
into phaeomelanin.
d. The phenotype of O-Y males is orange because the nonfunctional
allele O does not convert eumelanin into phaeomelanin, while the
phenotype of o-Y males is black/brown because the functional allele
o converts eumelanin into phaeomelanin.
15. 1. In cats, a sex-linked gene affects coat color. The O allele produces an
enzyme that converts eumelanin, a black or brown pigment, into
phaeomelanin, an orange pigment. The o allele is recessive to O and
produces a defective enzyme, one that does not convert eumelanin into
phaeomelanin. Which of the following statements is/are accurate?
a. The phenotype of o-Y males is black/brown because the
nonfunctional
allele o does not convert eumelanin into phaeomelanin.
b. The phenotype of OO and Oo males is orange because the
functional allele O converts eumelanin into phaeomelanin.
c. The phenotype of Oo males is mixed orange and black/brown
because the functional allele O converts eumelanin into
phaeomelanin in some cell groups (orange) and because in other
cell groups the nonfunctional allele o does not convert eumelanin
into phaeomelanin.
d. The phenotype of O-Y males is orange because the nonfunctional
allele O does not convert eumelanin into phaeomelanin, while the
phenotype of o-Y males is black/brown because the functional allele
o converts eumelanin into phaeomelanin.
16. X inactivation in Female Mammals
• Dosage compensation from having an additional
X-chromosome occurs in the female in mammals
by inactivating > 90 % of the genes on one X
• In mammalian females, one of the two X
chromosomes in each cell is randomly
inactivated during embryonic development
(called Barr body)
• If a female is heterozygous for a particular gene
located on the X chromosome, she will be a
mosaic for that character
17. Figure 15.8
X chromosomes
Allele for
orange fur
Allele for
black fur
Cell division and
X chromosome
inactivation
Early embryo:
Two cell
populations
in adult cat:
Active X
Inactive
X
Black fur Orange fur
Active X
18. Abnormal Chromosome Number
• In nondisjunction, pairs of homologous
chromosomes do not separate normally during
meiosis
• As a result, one gamete receives two of the
same type of chromosome, and another gamete
receives no copy
• Aneuploidy results from the fertilization of
gametes in which non-disjunction occurred
• Offspring with this condition have an abnormal
number of a particular chromosome
19. Figure 15.13-3
Meiosis I
Meiosis II
Nondisjunction
Non-
disjunction
Gametes
Number of chromosomes
(a) Nondisjunction of homo-
logous chromosomes in
meiosis I
(b) Nondisjunction of sister
chromatids in meiosis II
n + 1 n + 1 n + 1 n nn − 1 n − 1 n − 1
20. • A trisomic zygote has three copies of a particular
chromosome
• A monosomic zygote has only one copy of a
particular chromosome
• Polyploidy is a condition in which an organism has
more than two complete sets of chromosomes
22. Alterations of Chromosome Structure
• Breakage of a chromosome can lead to four
types of changes in chromosome structure:
– Deletion removes a chromosomal segment
– Duplication repeats a segment
– Inversion reverses a segment within a chromosome
– Translocation moves a segment from one non-
homologous chromosome to another
23. Figure 15.14
(a) Deletion (c) Inversion
(b) Duplication (d) Translocation
A deletion removes a
chromosomal segment.
An inversion reverses a
segment within a chromosome.
A duplication repeats
a segment. A translocation moves a
segment from one chromosome
to a nonhomologous
chromosome.
A B C D E F G H
A B C E F G H
A B C D E F G H
A B C B C D E F G H
A B C D E F G H
A D C B E F G H
A B C D E F G H M N O P Q R
M N O C D E F G H A B P Q R
24. Human Disorders Due to Chromosomal Alterations
• Alterations of chromosome number and structure
are associated with some serious disorders
• Some types of aneuploidy appear to upset the
genetic balance less than others, resulting in
individuals surviving to birth and beyond
• These surviving individuals have a set of symptoms,
or syndrome, characteristic of the type of
aneuploidy
25. Down Syndrome (Trisomy 21)
• Down syndrome is an aneuploid condition
that results from three copies of chromosome
21
• It affects about one out of every 700 children
born in the United States
• The frequency of Down syndrome increases
with the age of the mother, a correlation that
has not been explained
27. Aneuploidy of Sex Chromosomes
• Nondisjunction of sex chromosomes produces a
variety of aneuploid conditions
• Klinefelter syndrome is the result of an extra
chromosome in a male, producing XXY individuals
• Monosomy X, called Turner syndrome, produces X0
females, who are sterile; it is the only known viable
monosomy in humans
28.
29. Disorders Caused by Structurally Altered
Chromosomes
• One syndrome, cri du chat (“cry of the cat”), results
from a specific deletion in chromosome 5
• A child born with this syndrome is mentally
retarded and has a catlike cry; individuals usually
die in infancy or early childhood
• Certain cancers, including chronic myelogenous
leukemia (CML), are caused by translocations of
chromosomes
32. 2. Triploid species are usually sterile (unable to
reproduce), whereas tetraploids are often fertile. Which of
the following is likely a good explanation of these facts?
(Hint: Synapsis.)
a. In mitosis, some chromosomes in triploids have no
partner at synapsis, but chromosomes in tetraploids
do have partners.
b. In meiosis, some chromosomes in triploids have no
partner at synapsis, but chromosomes in tetraploids
do have partners.
c. In mitosis, some chromosomes in tetraploids have no
partner at synapsis, but chromosomes in triploids do
have partners.
d. In meiosis, some chromosomes in tetraploids have no
partner at synapsis, but chromosomes in triploids do
have partners.
33. 2. Triploid species are usually sterile (unable to
reproduce), whereas tetraploids are often fertile. Which of
the following is likely a good explanation of these facts?
(Hint: Synapsis.)
a. In mitosis, some chromosomes in triploids have no
partner at synapsis, but chromosomes in tetraploids
do have partners.
b. In meiosis, some chromosomes in triploids have no
partner at synapsis, but chromosomes in tetraploids
do have partners.
c. In mitosis, some chromosomes in tetraploids have no
partner at synapsis, but chromosomes in triploids do
have partners.
d. In meiosis, some chromosomes in tetraploids have no
partner at synapsis, but chromosomes in triploids do
have partners.
34. Recombination of Unlinked Genes:
Independent Assortment of Chromosomes
• Mendel observed that combinations of traits in some
offspring differ from either parent
• Offspring with a phenotype matching one of the parental
phenotypes are called parental types
• Offspring with nonparental phenotypes (new combinations
of traits) are called recombinant types, or recombinants
• A 50% frequency of recombination is observed for any two
genes on different chromosomes
35. 3. Which of the following is a type of
chromosomal alteration that differ from all
of the others?
a. aneuploidy
b. polyploidy
c. triploidy
d. tetraploidy
e. octaploidy
36. 3. Which of the following is a type of
chromosomal alteration that differ from all
of the others?
a. aneuploidy
b. polyploidy
c. triploidy
d. tetraploidy
e. octaploidy
37. Figure 15.UN02
Gametes from yellow-round
dihybrid parent (YyRr)
Gametes from
testcross
homozygous
recessive
parent (yyrr)
Parental-
type
offspring
Recombinant
offspring
yyRrYyRr Yyrryyrr
YR yr Yr yR
yr
Ratio of 1:1:1:1
For independent
assortment
38. • Each chromosome has hundreds or thousands
of genes
• Genes located on the same chromosome that
tend to be inherited together are called linked
genes
• Morgan did other experiments with fruit flies to
see how linkage affects inheritance of two
characters
• Morgan crossed flies that differed in traits of
body color and wing size
39. Figure 15.9
Experiment
Results
P Generation
(homozygous)
Wild type (gray
body, normal wings)
Double mutant
(black body, vestigial wings)
F1 dihybrid testcross
Wild-type F1 dihybrid
(gray body, normal wings)
Homozygous
recessive (black
body, vestigial wings)
Testcross
offspring Eggs
Sperm
Wild type
(gray-normal)
Black-
vestigial
Gray-
vestigial
Black-
normal
b+
vg+
b vg b+
vg b vg+
b vg
b+
b vg+
vg b b vg vg b+
b vg vg b b vg+
vg
b+
b+
vg+
vg+
b+
b vg+
vg
b b vg vg
b b vg vg
PREDICTED RATIOS
Genes on different
chromosomes:
Genes on the same
chromosome:
1 : 1 : 1 : 1
1 : 1 : 0 : 0
965 : 944 : 206 : 185
40. • From the results, Morgan reasoned that body color
and wing size are usually inherited together in
specific combinations (parental phenotypes)
because the genes are on the same chromosome
• However, nonparental phenotypes were also
produced (non-parental = recombinant)
• Understanding this result involves exploring genetic
recombination, production of offspring with
combinations of traits differing from either parent
41. Figure 15.UN01
Most offspring
F1 dihybrid female
and homozygous
recessive male
in testcross
or
b+
vg+
b+
vg+
b vg
b vg
b vg
b vg
b vg
b vg
42. Recombination of Linked Genes:
Crossing Over
• Morgan discovered that genes can be linked,
but the linkage was incomplete, as evident from
recombinant phenotypes
• Morgan proposed that some process must
sometimes break the physical connection
between genes on the same chromosome
• That mechanism was the crossing over of
homologous chromosomes
43. Figure 15.10 P generation
(homozygous)
Wild type (gray body,
normal wings)
b+
vg+
Double mutant (black body,
vestigial wings)
Wild-type F1
dihybrid (gray body,
normal wings)
F1 dihybrid
testcross
Homozygous recessive
(black body,
vestigial wings)
Replication of
chromosomes
Meiosis I
Meiosis I and II
Meiosis II
Replication of
chromosomes
Recombinant
chromosomes
Eggs
Testcross
offspring
Parental-type
offspring
Recombinant
offspring
Recombination
frequency
391 recombinants
2,300 total offspring
× 100 = 17%=
Sperm
965
Wild type
(gray-normal)
944
Black-
vestigial
206
Gray-
vestigial
185
Black-
normal
b+
vg+
b+
vg+
b+
vg+
b+
vg+
b+
vg+
b+
vg+
b vg
b vg
b vg
b vg
b vg
b vg
b vg
b vg
b vg
b vg
b vg
b vg
b vg
b+
vg
b+
vg
b vg+
b vg+
b vg
b vgb vgb vgb vg
b vgb+
vg+
b+
vg b vg+
44. Linkage Mapping: Using Recombination
Data: Scientific Inquiry
• Alfred Sturtevant, one of Morgan’s students,
constructed a genetic map, an ordered list of
the genetic loci along a particular chromosome
• Sturtevant predicted that the farther apart two
genes are, the higher the probability that a
crossover will occur between them, and
therefore the higher the recombination
frequency
45. • A linkage map is a genetic map of a
chromosome based on recombination
frequencies
• Distances between genes can be expressed as
map units; one map unit, or centimorgan,
represents a 1% recombination frequency
• Map units indicate relative distance and order,
not precise locations of genes
47. • Sturtevant used recombination frequencies to
make linkage maps of fruit fly genes
• Using methods like chromosomal banding,
geneticists can develop cytogenetic maps of
chromosomes
• Cytogenetic maps indicate the positions of
genes with respect to chromosomal features
48. Figure 15.12
Mutant phenotypes
Wild-type phenotypes
Short
aristae
Maroon
eyes
Black
body
Cinnabar
eyes
Vestigial
wings
Down-
curved
wings
Brown
eyes
Long
aristae
(appendages
on head)
Red
eyes
Gray
body
Red
eyes
Normal
wings
Normal
wings
Red
eyes
0 16.5 48.5 57.5 67.0 75.5 104.5
49. Some inheritance patterns are
exceptions to the standard
chromosome theory
• There are two normal exceptions to Mendelian
genetics
• One exception involves genes located in the
nucleus, and the other exception involves genes
located outside the nucleus
50. Genomic Imprinting
• For a few mammalian traits, the phenotype
depends on which parent passed along the
alleles for those traits
• Such variation in phenotype is called genomic
imprinting
• Genomic imprinting involves the silencing of
certain genes that are “stamped” with an
imprint during gamete production
51. Figure 15.17
Normal Igf2 allele
is expressed.
Normal Igf2 allele
is expressed.
Normal Igf2 allele
is not expressed.
Normal Igf2 allele
is not expressed.
Mutant Igf2 allele
is not expressed.
Mutant Igf2 allele
is expressed.
Mutant Igf2 allele
inherited from mother
Mutant Igf2 allele
inherited from father
Normal-sized mouse
(wild type)
Normal-sized mouse (wild type) Dwarf mouse (mutant)
Paternal
chromosome
Maternal
chromosome
(a) Homozygote
(b) Heterozygotes
52. • It appears that imprinting is the result of the
methylation (addition of —CH3) of cysteine
nucleotides
• Genomic imprinting is thought to affect only
a small fraction of mammalian genes
• Most imprinted genes are critical for
embryonic development
53. Inheritance of Organelle Genes
• Extranuclear genes are genes found in organelles in
the cytoplasm
• The inheritance of traits controlled by extranuclear
genes depends on the maternal parent because the
zygote’s cytoplasm comes from the egg
• The first evidence of extranuclear genes came from
studies on the inheritance of yellow or white
patches on leaves of an otherwise green plant
Figure 15.1 Where are Mendel’s hereditary factors located in the cell?
Figure 15.2 The chromosomal basis of Mendel’s laws
Figure 15.3 Morgan’s first mutant
Figure 15.4 Inquiry: In a cross between a wild-type female fruit fly and a mutant white-eyed male, what color eyes will the F1 and F2 offspring have?
Figure 15.7 The transmission of X-linked recessive traits
Answer: A
This focuses on the color of males and the action of the enzyme that converts eumelanin (brown/black pigment) to phaeomelanin (orange pigment). Male genotypes will be either O-Y or o-Y, with phenotypes of either orange or black/brown, respectively. In O-Y males, the eumelanin is converted to phaeomelanin, and in o-Y males, the eumelanin is unchanged. To answer this question, a student must know that males have only one copy of the gene and must understand that a functional allele produces an enzyme that catalyzes the chemical reaction.
Answer: A
This focuses on the color of males and the action of the enzyme that converts eumelanin (brown/black pigment) to phaeomelanin (orange pigment). Male genotypes will be either O-Y or o-Y, with phenotypes of either orange or black/brown, respectively. In O-Y males, the eumelanin is converted to phaeomelanin, and in o-Y males, the eumelanin is unchanged. To answer this question, a student must know that males have only one copy of the gene and must understand that a functional allele produces an enzyme that catalyzes the chemical reaction.
Figure 15.8 X inactivation and the tortoiseshell cat
Figure 15.13-3 Meiotic nondisjunction (step 3)
Figure 15.14 Alterations of chromosome structure
Figure 15.15 Down syndrome
Figure 15.16 Translocation associated with chronic myelogenous leukemia (CML)
Answer: B
The point of this question is to make students think about mitosis and meiosis in relation to polyploids. To answer this question, students should draw chromosomes of a triploid and a tetraploid as they go through mitosis and meiosis. Answers A and C are incorrect because chromosomes do not synapse during mitosis. Answer D is incorrect because tetraploids do have partners at synapsis but triploids do not. Answer B is correct—one-third of the chromosomes do not have a partner.
Answer: B
The point of this question is to make students think about mitosis and meiosis in relation to polyploids. To answer this question, students should draw chromosomes of a triploid and a tetraploid as they go through mitosis and meiosis. Answers A and C are incorrect because chromosomes do not synapse during mitosis. Answer D is incorrect because tetraploids do have partners at synapsis but triploids do not. Answer B is correct—one-third of the chromosomes do not have a partner.
Answer: A
Answer: A
Figure 15.UN02 In-text figure, Punnett square, p. 300
Figure 15.9 Inquiry: How does linkage between two genes affect inheritance of characters?
Figure 15.UN01 In-text figure, testcross, p. 300
Figure 15.10 Chromosomal basis for recombination of linked genes
Figure 15.11 Research method: constructing a linkage map
Figure 15.12 A partial genetic (linkage) map of a Drosophila chromosome
Figure 15.17 Genomic imprinting of the mouse Igf2 gene
Figure 15.18 A painted nettle coleus plant
Figure 15.UN03b Skills exercise: using the chi-square test (part 2)