Genetics Pseudoautosomal Region & Cytokinesis in Plants vs Animals Questionnaire

Chapter 6
Genetic Linkage and Mapping in
Eukaryotes
Johnny El-Rady, Ph.D
University of South Florida
Introduction
• Genes on the same chromosome are said to be linked
•A
chromosome
is a group of genes located on the same
• It corresponds to the
number of chromosomes
• To provide a simplified (hopefully) overview of the major theme
of this chapter, we will discuss the types and relative
proportions of gametes produced by an individual that is
heterozygous for two genes, A and B.
1. Genes A and B are on two different chromosomes
2. Genes A and B are on the same chromosome and very
very close to each other
3. Genes A and B are on the same chromosome and not so
close to each other
4. Genes A and B are on the same chromosome and very
very far from each other
Scenario 1
Genes A and B are
on two different
chromosomes
• This is termed
• The gametes in this
case are produced
in
Independent assortment: Two genes on two different homologous pairs of chromosomes
Scenario 2
Genes A and B are
on the same
chromosome and
very very close to
each other
• This is termed
• The gametes in this
case are produced
in
Linkage: Two genes on a single pair of homologs: no exchange occurs
Scenario 3
Genes A and B are
on the same
chromosome and not
so close to each
other
• This is termed
• The gametes in this
case are produced
in
Linkage: Two genes on a single pair of homologs: exchange occurs between
two nonsister chromatids
Scenario 4
Genes A and B are
on the same
chromosome and
very very far from
each other
• This is termed
• The gametes in this
case are produced
in
Linkage: Two genes on a single pair of homologs: exchange occurs between
two nonsister chromatids
Application Problem
• Consider two completely linked genes, A and B
?
1. What types of gametes can be
produced by an individual with
genotype aaBB?
Application Problem
• Consider two completely linked genes, A and B
?
2. What types of gametes can be
produced by an individual with
genotype AaBb?
Figure 6.2
Alleles and Genetic Symbolism
• Alleles are alternative forms of same gene
• Wild-type allele
• Occurs most frequently in the natural population
• Arbitrarily designated as normal
• Responsible for the normal function and the wild-type
phenotype
• When
gives new alleles
• New alleles often change the activity of the cellular product
of the gene
• => a change in the
Alleles and Genetic Symbolism
• Alleles for simple Mendelian traits are symbolized with
lowercase letters for recessive alleles and uppercase letters for
dominant alleles
NOTE
The name of genes
are usually written
in
• Alleles for another useful system was developed in Drosophila
• The symbol for a gene comes from the first mutant strain found
• The mutant allele is denoted by the initial letter, or a
combination of two or three letters
• The wildtype allele is denoted by the same letter(s) but with a
+ superscript, or simply by the + symbol
• If the mutant trait is recessive, the
form is used
• If the mutant trait is dominant, the
form is used
1. Ebony is a recessive body color mutation
e = ebony
e+ = gray
• e+e+
• e+e
• ee
2. Wrinkled is a dominant wing shape mutation
Wr = Wrinkled
• Wr+ Wr+
• Wr+ Wr
• Wr Wr
Wr + = Normal
Historical Overview
• 1900 = Mendel’s work was independently “rediscovered” by
three botanists
• Hugo de Vries
• Carl Correns
• Erich von Tschermak
• 1902 = William Bateson showed that Mendelian principles apply
to animals
• 1902 – 1903 =
• Independently recognize the parallel behavior between
Mendelian factors and chromosomes during
• They thus proposed the
• 1903 = Sutton pointed out that in any organism the number of
chromosomes is considerably less than the number of
Mendelian factors
• => On each chromosome there are many “factors”
• => These factors would be inherited as a unit, and thus not
• 1905 = Bateson and Reginald Punnett discovered two traits that
did not
• Flower color and pollen shape
• Refer to Figure 6.1
Figure 6.1
!
They suggested that
the transmission of
the traits was
somehow coupled
A much greater proportion
of the two types found in
the parental generation
• 1909 = F.A. Janssens
• Observed chiasmata in
synapsed meiotic homologous
chromosomes
• 1911 = Thomas H. Morgan
discovered the first X-linked gene
in
• 1912 = Morgan studied crosses
involving two X-linked genes
• Morgan conducted many two-factor crosses
• Next, I will discuss two of the earliest such crosses
• Cross A
• y = yellow body
• y+ = gray body
• w = white eyes
• w+ = red eyes
•P
yellow body, white eyes
gray body, red eyes
X
• Cross B
• m = miniature wing
• m+ = wild-type wing
• w = white eyes
• w+ = red eyes
•P
miniature wing, white eyes
wild-type wing, red eyes
X
• Hmmmm!
• Puzzling results that brought up two questions:
?
?
1. What was the source of the
2. Why is the frequency
• Morgan proposed the following answers
1. An exchange of genetic material occurred between the two
mutant genes on the X chromosomes of the
• Morgan used the term
2. The closer the genes are on a chromosome, the less likely
a
will form between them
Based on these experiments, Morgan
hypothesized that genes occurred in a linear
array on a chromosome
And that the linkage could only be modified
by
• 1912 = Alfred H. Sturtevant
• An
student who
spent time in Morgan’s laboratory
• He reasoned that if recombination
frequency (RF) is dependent upon
distance between genes, then RF can
provide information about the physical
location of genes on the chromosomes
• Sturtevant wrote:
“
“In conversation with Morgan … I suddenly realized
that the variations in the length of linkage, already
attributed by Morgan to differences in the spatial
orientation of the genes, offered the possibility of
determining sequences [of different genes] in the
linear dimension of the chromosome. I went home
and spent most of the night (to the neglect of my
undergraduate homework) in producing the first
chromosome map, which included the sex-linked
genes, y, w, v, m, and r, in the order and
approximately the relative spacing that they still
appear on the standard maps.”
”
• Sturtevant concluded from Morgan’s two
aforementioned experiments that
•
is closer to
than it is to
• AND, with one more cross, the relationship between
can be established
• There are two choices:
!
• If first choice
• RF of y – m
RF of w – m
• If second choice
• RF of y – m
RF of w – m
The experiment revealed that
choice was correct!
• Sturtevant considered two other X-linked genes and
constructed the first chromosomal map
• He proposed that the percentage of recombinants be
used as a quantitative measure of the distance
between two gene pairs of a genetic map
• Recombination Frequency =
Genotypic Notations
• Different notations are used to present genotypes:
• yy ww
• Linkage arrangement of the loci is
• yw / yw
• Loci are on
• y/y w/w
• Loci are on
Genotypic Notations
• Two possible configurations exist for a genotype that is
heterozygous for two linked genes
1.
2.
Crossing-Over
• The physical exchange between
• Crossing-over has the following features:
• It occurs in prophase I of meiosis, when the four chromatids
are closely
• It occurs more or less
along the length of a
chromosome pair
• It involves a
exchange of equal and
corresponding segments
Crossing-Over
• Crossing over is NOT detectable if it occurs
region between the two marker genes
the
• Crossing over is NOT detectable if it occurs
between the two marker genes
• In a single tetrad,
are possible
6.3 Genetic Mapping in Plants and Animals
• Genetic mapping is also known as gene mapping or
chromosome mapping
• Its purpose is to determine the linear order of linked genes
along the same chromosome
• Figure 6.8 illustrates a simplified genetic linkage map of
Drosophila melanogaster
Figure 6.8
Figure 6.7 in the 6e
Each gene has its
own unique locus at
a particular site
within a
chromosome
• Genetic maps allow us to estimate the relative distances
between linked genes, based on the likelihood that a crossover
will occur between them
• Experimentally, the percentage of recombinant offspring is
correlated with the distance between the two genes
• If the genes are far apart → more recombinant offspring
• If the genes are close → less recombinant offspring
• Map distance =
• The units of distance are called
• They are also referred to as
• One map unit is equivalent to
• Thus, genetic distance between two loci may be represented in
one of four ways:
1.
of recombination
2.
recombination
3. Map distance in
4. Map distance in
• Genetic mapping experiments are typically accomplished by
carrying out a
• A mating between an individual that is heterozygous for two
or more genes and one that is homozygous recessive for the
same genes
• Figure 6.9 provides an example of a testcross
• This cross concerns two linked genes affecting bristle length
and body color in fruit flies
• s = short bristles
e = ebony body color
• s+ = normal bristles
e+ = gray body color
• One parent displays both recessive traits
• Homozygous recessive (ss ee)
• The other parent is heterozygous for the two genes
• The s and e alleles are linked on one chromosome
• The s+ and e+ alleles are linked on the homologous
chromosome
Figure 6.9
Figure 6.8
in the 6e
Chromosomes are the
product of a crossover
during meiosis in the
heterozygous parent
Recombinant offspring
are fewer in number
than nonrecombinant
offspring
• The data at the bottom of Figure 6.9 can be used to estimate
the distance between the two genes
• Map distance = Number of recombinant offspring X 100
Total number of offspring
=
76 + 75
542 + 537 + 76 + 75
X 100
= 12.3 map units
• Therefore, the s and e genes are 12.3 map units apart from
each other along the same chromosome
Morgan’s Trihybrid Cross
▪ I provided the content of this topic in the form of a
stand-alone “PDF lecture”
▪ Please find it at the beginning of Module 4.2
THREE POINT TEST CROSS
• I lectured on this topic in a stand-alone TUTORIAL
• You can find it in a separate link inside this Module
• Please use slide #s 53-78 of this handout as reference,
while watching the Tutorial
THREE POINT TEST CROSS
• Involves the analysis of three loci, each segregating two alleles
• Three criteria must be met for a successful mapping cross:
1. Genotype of organism producing recombinant gametes
must be
at all loci in question
2. Cross must be constructed so that genotypes can be
deduced from
3. A
number of offspring must be produced
• Consider the following hypothetical example:
•P
aa
• F1
a a + b b + c c+
bb
cc
X
a+a+ b+b+ c+c+
X
aa
bb
cc
• If the three loci are
• The trihybrid produces
• F2 will consist of
types of gametes
phenotypic classes
F2 offspring
Frequency
• If the three loci are
• The trihybrid produces
• F2 will consist of
types of gametes
phenotypic classes
F2 offspring
Frequency
• If the three loci are
• The trihybrid produces
types of gametes
• F2 will consist of
phenotypic classes in various
proportions depending on the
• Well, now that the hypothetical cross is behind us, let’s deal
with some true data …
• A detailed example is provided on pp. 139-141
• For added practice, I will present to you another example
◼
Example = Three X-linked recessive genes in Drosophila
y = yellow
w = white w+ = wild-type
y+ = wild-type
ec = echinus ec+ = wild-type
•P
mutant female
X
wild-type male
◼
y+
w+
ec
207
◼
y
w+
ec
◼
y
w
ec
4685
◼
y+
w
ec
70
◼
y
w
ec+
◼
y+
w
ec+
◼
y+
w+
ec+
4759
◼
y
w+
ec+
80
3
193
3
• Steps to follow:
1.
Arrange data in reciprocal classes
2.
Identify the parental classes as the most frequent phenotypes in the
progeny
3.
Identify the double recombinants as the least frequent phenotypes
in the progeny
4.
Determine the gene order in the following manner
• Compare P to DCO classes
• The gene that is interchanged represents the inside locus
5.
Assign single cross-overs to the region involved
6.
Calculate the frequencies of recombination
7.
Draw the genetic map
•y
w
ec
4685
• y+
w+
ec+
4759
•y
w+
ec+
80
• y+
w
ec
70
•y
w
ec+
193
• y+
w+
ec
207
•y
w+
ec
3
• y+
w
ec+
3
◼
◼
is the inside locus
Recombinant classes are those that
have a
compared to the parental class
◼
RF (
)=
◼
◼
◼
◼
◼
RF (
◼
◼
◼
◼
)=
• Therefore, MAP is
◼
Map distances computed this way are
◼
Distance between
◼
◼
and
=
• This estimate can be verified by directly calculating the average #
of recombinants between the y and ec loci
• RF (y – ec) =
•
•
•
NOTE
DCO are counted twice because
each represents a double
recombinant class
Double Cross-Over
Interference
Are cross-overs in adjacent regions of a
? chromosome independent of one another
or do they affect each other in some way?
• In reality, once one CO event has occurred, the likelihood of a
second CO in the same region can be altered
◼
Let us use the last example
◼ y is 1.56 map units from w
◼ Therefore the probability of a CO
between y and w is
◼
w is 4.06 map units from ec
◼ Therefore the probability of a CO
between w and ec is
◼
Thus the probability that there will be a
CO between y & w and between w & ec
is
◼
The observed number of offspring =
◼
◼
Therefore, the expected number of DCO
recombinants out of
offspring is
◼
SO we see a reduction in DCO from the
expected, or
◼
Quantification of this Interference is
◼
where C is the
coefficient of coincidence
◼
◼
C=
In our example, C =
◼
◼
Therefore, I =
◼
!
Interference is not equivalent in all
parts of a chromosome or among
the different chromosomes of a
given complement
Check out this example!
• Assume a cross that yields the following results:
a
+
b
+
c
+
215
221
a
+
b
+
+
c
223
225
a
+
+
b
c
+
15
17
+
a
b
+
c
+
19
20
• Hmm!
• Neither the Parental nor the DCO classes can be identified
• Therefore, the genes are
• The genes also do not assort independently because the
classes are
• Moreover, the genes are not completely linked because the
cross did not yield
• So what’s up?
• Well, suppose
• Thus, the four largest classes are the
the four lowest are the
and
• Consider genes a and b
a
+
a
+
b
+
+
b
215 + 233 = 448
221 + 225 = 446
15 + 20 = 35
17 + 19 = 36
• This demonstrates that genes a and b are
•
• Consider genes b and c
b
+
b
+
c
+
+
c
215 + 19 = 234
221 + 20 = 241
223 + 17 = 250
225 + 15 = 240
• This demonstrates that genes b and c are
• A similar reasoning can be applied for
2/10/2022
Introduction
Chapter 6
• Genes on the same chromosome are said to be linked
G enetic Linkage and M apping in
Eukaryotes
•A
chromosome
is a group of genes located on the same
• It corresponds to the
Johnny El‐Rady, Ph.D
number of chromosomes
University of South Florida
1
2
Scenario 1
• To provide a simplified (hopefully) overview of the major theme
of this chapter, we will discuss the types and relative
proportions of gametes produced by an individual that is
heterozygous for two genes, A and B.
Genes A and B are
on two different
chromosomes
1. Genes A and B are on two different chromosomes
2. Genes A and B are on the same chromosome and very
very close to each other
3. Genes A and B are on the same chromosome and not so
close to each other
4. Genes A and B are on the same chromosome and very
very far from each other
• This is termed
• The gametes in this
case are produced
in
Independent assortment: Two genes on two different homologous pairs of chromosomes
3
4
Scenario 2
Scenario 3
Genes A and B are
on the same
chromosome and
very very close to
each other
Genes A and B are
on the same
chromosome and not
so close to each
other
• This is termed
• This is termed
• The gametes in this
case are produced
in
• The gametes in this
case are produced
in
Linkage: Two genes on a single pair of homologs: no exchange occurs
5
Linkage: Two genes on a single pair of homologs: exchange occurs between
two nonsister chromatids
6
1
2/10/2022
Application Problem
Scenario 4
Genes A and B are
on the same
chromosome and
very very far from
each other
• Consider two completely linked genes, A and B
?
• This is termed
• The gametes in this
case are produced
in
1. What types of gametes can be
produced by an individual with
genotype aaBB?
Linkage: Two genes on a single pair of homologs: exchange occurs between
two nonsister chromatids
7
8
Application Problem
Figure 6.2
• Consider two completely linked genes, A and B
?
2. What types of gametes can be
produced by an individual with
genotype AaBb?
9
10
Alleles and Genetic Symbolism
Alleles and Genetic Symbolism
• Alleles are alternative forms of same gene
• Alleles for simple Mendelian traits are symbolized with
lowercase letters for recessive alleles and uppercase letters for
dominant alleles
• Wild-type allele
• Occurs most frequently in the natural population
• Arbitrarily designated as normal
• Responsible for the normal function and the wild-type
phenotype
• When
gives new alleles
• New alleles often change the activity of the cellular product
of the gene
• => a change in the
11
NOTE
The name of genes
are usually written
in
12
2
2/10/2022
1. Ebony is a recessive body color mutation
• Alleles for another useful system was developed in Drosophila
e = ebony
• The symbol for a gene comes from the first mutant strain found
e+ = gray
e+e+
•
• e+e
• ee
• The mutant allele is denoted by the initial letter, or a
combination of two or three letters
2. Wrinkled is a dominant wing shape mutation
• The wildtype allele is denoted by the same letter(s) but with a
+ superscript, or simply by the + symbol
• If the mutant trait is recessive, the
form is used
• If the mutant trait is dominant, the
form is used
13
Wr = Wrinkled
Wr + = Normal
Wr+ Wr+
•
• Wr+ Wr
• Wr Wr
14
Historical Overview
• 1902 – 1903 =
• 1900 = Mendel’s work was independently “rediscovered” by
three botanists
• Hugo de Vries
• Carl Correns
• Erich von Tschermak
• 1902 = William Bateson showed that Mendelian principles apply
to animals
15
• Independently recognize the parallel behavior between
Mendelian factors and chromosomes during
• They thus proposed the
16
Figure 6.1
• 1903 = Sutton pointed out that in any organism the number of
chromosomes is considerably less than the number of
Mendelian factors
• => On each chromosome there are many “factors”
• => These factors would be inherited as a unit, and thus not
!
They suggested that
the transmission of
the traits was
somehow coupled
A much greater proportion
of the two types found in
the parental generation
• 1905 = Bateson and Reginald Punnett discovered two traits that
did not
• Flower color and pollen shape
• Refer to Figure 6.1
17
18
3
2/10/2022
• 1909 = F.A. Janssens
• Observed chiasmata in
synapsed meiotic homologous
chromosomes
• Morgan conducted many two-factor crosses
• Next, I will discuss two of the earliest such crosses
• 1911 = Thomas H. Morgan
discovered the first X-linked gene
in
• 1912 = Morgan studied crosses
involving two X-linked genes
19
20
• Cross A
• y = yellow body
• y+ = gray body
• w = white eyes
• w+ = red eyes
•P
21
yellow body, white eyes
gray body, red eyes
X
22
• Cross B
• m = miniature wing
• m+ = wild-type wing
• w = white eyes
• w+ = red eyes
•P
23
miniature wing, white eyes
wild-type wing, red eyes
X
24
4
2/10/2022
25
26
• Morgan proposed the following answers
• Hmmmm!
1. An exchange of genetic material occurred between the two
mutant genes on the X chromosomes of the
• Morgan used the term
• Puzzling results that brought up two questions:
?
?
1. What was the source of the
2. The closer the genes are on a chromosome, the less likely
a
will form between them
2. Why is the frequency
27
28
• 1912 = Alfred H. Sturtevant
• An
student who
spent time in Morgan’s laboratory
Based on these experiments, Morgan
hypothesized that genes occurred in a linear
array on a chromosome
• He reasoned that if recombination
frequency (RF) is dependent upon
distance between genes, then RF can
provide information about the physical
location of genes on the chromosomes
And that the linkage could only be modified
by
29
30
5
2/10/2022
• Sturtevant wrote:
“
• Sturtevant concluded from Morgan’s two
aforementioned experiments that
“In conversation with Morgan … I suddenly realized
that the variations in the length of linkage, already
attributed by Morgan to differences in the spatial
orientation of the genes, offered the possibility of
determining sequences [of different genes] in the
linear dimension of the chromosome. I went home
and spent most of the night (to the neglect of my
undergraduate homework) in producing the first
chromosome map, which included the sex-linked
genes, y, w, v, m, and r, in the order and
approximately the relative spacing that they still
appear on the standard maps.”
•
is closer to
than it is to
• AND, with one more cross, the relationship between
can be established
”
31
32
• There are two choices:
• Sturtevant considered two other X-linked genes and
constructed the first chromosomal map
!
• If first choice
• RF of y – m
RF of w – m
• If second choice
• RF of y – m
RF of w – m
The experiment revealed that
• He proposed that the percentage of recombinants be
used as a quantitative measure of the distance
between two gene pairs of a genetic map
• Recombination Frequency =
choice was correct!
33
34
Genotypic Notations
Genotypic Notations
• Different notations are used to present genotypes:
• Two possible configurations exist for a genotype that is
heterozygous for two linked genes
• yy ww
• Linkage arrangement of the loci is
1.
• yw / yw
• Loci are on
• y/y w/w
• Loci are on
35
2.
36
6
2/10/2022
Crossing-Over
Crossing-Over
• The physical exchange between
• Crossing-over has the following features:
• It occurs in prophase I of meiosis, when the four chromatids
are closely
• It occurs more or less
along the length of a
chromosome pair
• It involves a
exchange of equal and
corresponding segments
37
38
• Crossing over is NOT detectable if it occurs
region between the two marker genes
the
39
• Crossing over is NOT detectable if it occurs
between the two marker genes
40
6.3 Genetic Mapping in Plants and Animals
• In a single tetrad,
are possible
• Genetic mapping is also known as gene mapping or
chromosome mapping
• Its purpose is to determine the linear order of linked genes
along the same chromosome
• Figure 6.8 illustrates a simplified genetic linkage map of
Drosophila melanogaster
41
42
7
2/10/2022
Figure 6.8
Figure 6.7 in the 6e
• Genetic maps allow us to estimate the relative distances
between linked genes, based on the likelihood that a crossover
will occur between them
• Experimentally, the percentage of recombinant offspring is
correlated with the distance between the two genes
• If the genes are far apart  more recombinant offspring
Each gene has its
own unique locus at
a particular site
within a
chromosome
43
• If the genes are close  less recombinant offspring
44
• Thus, genetic distance between two loci may be represented in
one of four ways:
• Map distance =
• The units of distance are called
• They are also referred to as
2.
recombination
4. Map distance in
46
• Figure 6.9 provides an example of a testcross
• This cross concerns two linked genes affecting bristle length
and body color in fruit flies
• s = short bristles
e = ebony body color
• s+ = normal bristles
e+ = gray body color
• Genetic mapping experiments are typically accomplished by
carrying out a
• A mating between an individual that is heterozygous for two
or more genes and one that is homozygous recessive for the
same genes
47
of recombination
3. Map distance in
• One map unit is equivalent to
45
1.
• One parent displays both recessive traits
• Homozygous recessive (ss ee)
• The other parent is heterozygous for the two genes
• The s and e alleles are linked on one chromosome
• The s+ and e+ alleles are linked on the homologous
chromosome
48
8
2/10/2022
Figure 6.9
Figure 6.8
in the 6e
• The data at the bottom of Figure 6.9 can be used to estimate
the distance between the two genes
Chromosomes are the
product of a crossover
during meiosis in the
heterozygous parent
• Map distance = Number of recombinant offspring X 100
Total number of offspring
=
Recombinant offspring
are fewer in number
than nonrecombinant
offspring
76 + 75
542 + 537 + 76 + 75
X 100
= 12.3 map units
• Therefore, the s and e genes are 12.3 map units apart from
each other along the same chromosome
49
50
Morgan’s Trihybrid Cross
THREE POINT TEST CROSS
• I lectured on this topic in a stand-alone TUTORIAL
 I provided the content of this topic in the form of a
stand-alone “PDF lecture”
• You can find it in a separate link inside this Module
 Please find it at the beginning of Module 4.2
• Please use slide #s 53-78 of this handout as reference,
while watching the Tutorial
51
52
THREE POINT TEST CROSS
• Involves the analysis of three loci, each segregating two alleles
• Consider the following hypothetical example:
• Three criteria must be met for a successful mapping cross:
1. Genotype of organism producing recombinant gametes
must be
at all loci in question
•P
aa
bb
cc
• F1
a a+ b b+ c c+
X
a+a+ b+b+ c+c+
X
aa
bb
cc
2. Cross must be constructed so that genotypes can be
deduced from
3. A
53
number of offspring must be produced
54
9
2/10/2022
F2 offspring
Frequency
• If the three loci are
• The trihybrid produces
• F2 will consist of
types of gametes
phenotypic classes
55
56
• If the three loci are
• The trihybrid produces
• F2 will consist of
F2 offspring
types of gametes
phenotypic classes
57
58
• Well, now that the hypothetical cross is behind us, let’s deal
with some true data …
• If the three loci are
• The trihybrid produces
types of gametes
• A detailed example is provided on pp. 139-141
• F2 will consist of
phenotypic classes in various
proportions depending on the
59
Frequency
• For added practice, I will present to you another example
60
10
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
•P
Example = Three X-linked recessive genes in Drosophila
y = yellow
w = white w+ = wild-type
mutant female
X
wild-type male
y+ = wild-type
ec = echinus ec+ = wild-type
61
62

y+
w+
ec
207

y
w+
ec

y
w
ec
4685

y+
w
ec
70

y
w
ec+

y+
w
ec+

y+
w+
ec+
4759

y
w+
ec+
80
• Steps to follow:
3
1.
2.
Identify the parental classes as the most frequent phenotypes in the
progeny
193
3.
Identify the double recombinants as the least frequent phenotypes
in the progeny
3
63
Arrange data in reciprocal classes
64
4.
Determine the gene order in the following manner
• Compare P to DCO classes
5.
Assign single cross-overs to the region involved
6.
Calculate the frequencies of recombination
7.
Draw the genetic map
• The gene that is interchanged represents the inside locus
65
66
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• y
w
ec
4685
• y+
w+
ec+
4759
• y
w+
ec+
80
• y+
w
ec
70
• y
w
ec+
193
• y+
w+
ec
207
• y
w+
ec
3
• y+
w
ec+
3
67


is the inside locus
Recombinant classes are those that
have a
compared to the parental class
68

RF (
)=
• Therefore, MAP is





RF (

Map distances computed this way are

Distance between
)=

and
=





69
70
Double Cross-Over
• This estimate can be verified by directly calculating the average #
of recombinants between the y and ec loci
• RF (y – ec) =
•
•
•
NOTE
DCO are counted twice because
each represents a double
recombinant class
71
72
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Interference

Are cross-overs in adjacent regions of a
? chromosome independent of one another
or do they affect each other in some way?
Let us use the last example
 y is 1.56 map units from w
 Therefore the probability of a CO
between y and w is

• In reality, once one CO event has occurred, the likelihood of a
second CO in the same region can be altered

73
w is 4.06 map units from ec
 Therefore the probability of a CO
between w and ec is
Thus the probability that there will be a
CO between y & w and between w & ec
is
74


The observed number of offspring =


Therefore, the expected number of DCO
recombinants out of
offspring is



Quantification of this Interference is

where C is the
coefficient of coincidence
C=
In our example, C =

SO we see a reduction in DCO from the
expected, or

Therefore, I =

75
76
!
77
Interference is not equivalent in all
parts of a chromosome or among
the different chromosomes of a
given complement
78
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Check out this example!
• Hmm!
• Neither the Parental nor the DCO classes can be identified
• Therefore, the genes are
• Assume a cross that yields the following results:
a
+
b
+
c
+
215
221
a
+
b
+
+
c
223
225
a
+
+
b
c
+
15
17
+
a
b
+
c
+
19
20
• The genes also do not assort independently because the
classes are
• Moreover, the genes are not completely linked because the
cross did not yield
79
80
• Consider genes a and b
• So what’s up?
a
+
a
+
• Well, suppose
• Thus, the four largest classes are the
the four lowest are the
and
b
+
+
b
215 + 233 = 448
221 + 225 = 446
15 + 20 = 35
17 + 19 = 36
• This demonstrates that genes a and b are
•
81
82
• Consider genes b and c
b
+
b
+
c
+
+
c
215 + 19 = 234
221 + 20 = 241
223 + 17 = 250
225 + 15 = 240
• This demonstrates that genes b and c are
• A similar reasoning can be applied for
83
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INTRODUCTION
Chapter 3
 Please refer to the stand-alone PDF lectures I posted at
the beginning of Module 4.1
Chromosome Transmission
During Cell Division and Sexual
Reproduction
Johnny El-Rady, Ph.D
University of South Florida
1
2
Mitosis, Oh My!
Interphase
• The cell cycle consists of:
• A cell that is not actively dividing is said to be in interphase
Interphase
• G1 = Synthesis of proteins and RNA
Mitosis
Cell prepares for DNA replication
• S = Synthesis of DNA
• G2 = Synthesis of proteins and RNA
Cell prepares for mitosis
3
4
Interphase
Figure 3.5
• The total time for interphase varies with the organism, cell type
and environmental conditions
Synthesis
Gap 1
5
• Most variation is seen in the length of
Gap 2
6
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• Late in G1, the cell will follow one of two paths:
It withdraws from the cycle and
enters the
stage
Cells in this stage:
• Then the cell advances to the S phase, where chromosomes are
replicated
It becomes committed to initiate DNA
synthesis and complete the cycle
The cell is to reach a
Either postponed making a decision
to divide
Or made the decision to never divide
again
Figure 3.6
7
8
Mitosis
• Note that at the end of S phase, a cell has
as many
chromatids as there are chromosomes in the G1 phase
• A human cell for example has
• 46 distinct chromosomes in G1 phase
• 46 pairs of sister chromatids in S phase
• Mitosis is usually the
• It is an asexual process, used for ordinary growth, repair and
replacement
• Therefore the term chromosome is relative
• In G1 and late in the M phase, it refers to the equivalent of
one
• In G2 and early in the M phase, it refers to a
9
state of the life cycle
• It occurs in somatic cells and in early germline development
10
Mitosis
• In mitosis, a diploid cell divides once to produce two diploid
daughter cells
• Mitosis is a dynamic process, but for descriptive purposes it is divided into
five stages:
Karyokinesis: Division of the nucleus
Cytokinesis: Division of the cytoplasm
11
1
Prophase
2
Prometaphase
3
Metaphase
4
Anaphase
5
Telophase
12
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Interphase
1 Prophase
• C
progressively shortens
and thickens to form daughter c’somes
Figure 3.8
• Nuclear envelope breaks down;
Nucleolus (or nucleoli) disappears
• Centrosomes (which were replicated
in interphase) migrate to opposite cell
poles
• Spindle apparatus is formed
• Composed of microtubules (MTs)
13
14
Figure 3.7
• Microtubules are formed by rapid polymerization of tubulin proteins
• There are three types of spindle microtubules
1.
microtubules
Important for positioning of spindle apparatus
2.
microtubules
Help to “push” the poles away from each other
3.
microtubules
Attach to the kinetochore, which is bound to the centromere of each individual
chromosome
15
16
2 Prometaphase
3 Metaphase
• Spindle fibers interact with the sister
chromatids
• Chromosomes reach their maximum
contraction
• Kinetochore microtubules grow from
the two poles
• Pairs of sister chromatids align
themselves along a plane called the
metaphase plate
• If they make contact with a
kinetochore, the sister chromatid is
• If not, the microtubule depolymerizes
and retracts to the centrosome
• The two kinetochores on a pair of
sister chromatids
are attached to kinetochore MTs on
opposite poles
17
18
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4 Anaphase
4 Anaphase
• Centromeric region divides
longitudinally and paired sister
chromatids separate
• As anaphase proceeds
• Kinetochore MTs
• Chromosomes move to
opposite poles
• Polar MTs
• Poles themselves move further
away from each other
• Each chromatid, now an individual
chromosome, is linked to only one
pole
• Note: The movement of sister
chromatids to opposite poles is aided
by specific proteins, termed
proteins
• These use the energy generated
from
19
20
Mitosis
5 Telophase
• Daughter chromosomes reach their
respective poles
• They uncoil and become diffuse
once again
• Nuclear membrane reforms to form
two separate nuclei
• Spindle fibers disappear
21
22
Cytokinesis
Figure 3.9
• In animal cells = A cell cleavage furrow develops
• In plants cells = A cell plate develops
• Refer to Figure 3.9
23
24
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Cytokinesis
Animations
• In general cytokinesis follows karyokinesis
• However, in many instances it may be deferred or totally lacking
• These cell are termed
cells
• They are common in fungi
25
26
Meiosis, Oh Me, Oh My!
Significance of Mitosis
Sexual Reproduction
• It maintains a constant chromosome number
• During sexual reproduction, gametes are made that contain half
the amount of genetic material
46
46
• Some simple eukaryotic species are
• They produce gametes that are morphologically similar
46
46
46
46
46
27
28
Meiosis
• Most eukaryotic species are
• Meiosis only occurs in germline cells
• It ensures genetic continuity and variability
These produce gametes that are morphologically different
Sperm cells
Egg cell or ovum
Relatively small and mobile
Usually large and immobile
• Like mitosis, meiosis begins after a cell has progressed through
of the cell cycle
• Unlike mitosis, meiosis involves two successive divisions
Meiosis I: A
division, in which
the chromosome number is halved
Stores a large amount of
nutrients, in animal species
29
Meiosis II: An
division, in which
the chromosome number remains the same
30
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Meiosis I
1 Prophase I
1 Prophase I
• By far the most complex and longest
stage of meiosis
2 Prometaphase I
• Subdivided into five stages:
1.
2.
3.
4.
5.
3 Metaphase I
4 Anaphase I
Leptonema
Zygonema
Pachynema
Diplonema
Diakinesis
5 Telophase I
31
32
1. Leptonema (the leptotene stage)
Figure 3.10
• Chromatin begins to condense, but sister chromatids are not evident yet
A total of 4 chromatids
2. Zygonema (the zygotene stage)
• Chromosomes continue to shorten and thicken
• Homologous chromosomes pair up side-by-side
• In human males, the X and Y chromosomes line up end-to-end
• Each pair of synapsed homologous chromosomes is termed a Bivalent
• Synapsis is mediated by the nucleoprotein “synaptonemal complex”
A recognition
process
33
Bound to
chromosomal DNA
of homologous
chromatids
Provides link between
lateral elements
34
Figure 3.10
3. Pachynema (the pachytene stage)
A total of 4
chromatids
• Chromatin continue to shorten and thicken, so that each is now seen to
consist of two chromatids
• Each bivalent is now a Tetrad of four strands
• Sister vs. Nonsister chromatid
• Crossing-over occurs
4. Diplonema (the diplotene stage)
• Chromosomes continue to shorten and thicken
• Synaptonemal complex breaks down
A recognition
process
35
Bound to
chromosomal
DNA of
homologous
chromatids
• Homologues start to separate
• They remain held together at chiasmata
Provides link between
lateral elements
36
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5. Diakinesis
Figure 3.10
• Chromosomes reach maximum contraction
• Chiasmata move to ends of chromosomes as homologues continue to pull
apart
• Nucleolus(i) disappears; Nuclear envelope degenerates; Spindle fibers
form
37
38
Figure 3.11 – Meiosis I
2
3 Metaphase I
Prometaphase I
• Tetrads are organized along the
metaphase plate
• Pairs of sister chromatids are aligned
in a double row, rather than a single
row (as in mitosis)
• The arrangement is random with
regards to the homologues
• Furthermore:
• A pair of sister chromatids is linked to
one of the poles
• The homologous pair is linked to
opposite pole
Spindle apparatus complete Chromatids
attached via kinetochore microtubules
Figure 3.12
39
40
Interkinesis
A haploid set of chromosomes
is found at each pole
Centromeres do not divide
Homologues separate, and pairs of sister
chromatids move to opposite poles
The details vary between
organisms
Occurs in most species
• A short interphase between the two meiotic divisions
•
DNA synthesis occurs
Dyads
4
41
Anaphase I
5
Telophase I
42
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Meiosis II
Figure 3.11 – Meiosis II
1 Prophase II
1
Prophase II
2
Prometaphase II
3
Metaphase II
4
Anaphase II
5
Telophase II
3 Metaphase II
4 Anaphase II
5 Telophase II
Division is similar
to mitosis
Monads
43
44
Haploid gametes
Diploid cell
PROPHASE
I
Germ-line cell
II
TELOPHASE
I
II
METAPHASE
ANAPHASE
II
I
II
I
45
46
Significance of Meiosis
Significance of Meiosis
2. It generates genetic variability through the various ways in which
maternal and paternal chromosomes are combined into gametes
1. It allows the conservation of the chromosome number in sexuallyreproducing species
• In humans an individual can produce
gametes
Too
much!
different
• So a human can produce
Meiosis reduces genetic content
Just
right!
47
48
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Significance of Meiosis
3. It enhances the potential genetic variation in gametes through the
phenomenon of crossing over between maternal and paternal chromatid
pairs
49
50
Gametogenesis
Spermatogenesis
Takes place in the
Gametogenesis: The production of gametes
of the testes
• In early embryonic development, a group of cells become committed to
form the germ line cells
• These are termed
germ cells
• PGCs proliferate through mitosis, and later undergo meiosis to
produce mature sperm or eggs
51
52
Spermatogenesis
Spermatogenesis
• In the fetus, the PGCs divide by mitosis and differentiate into
spermatogonia
NOTE
• Spermatogonia remain quiescent until puberty
1. Unlike oogenesis, the meiotic divisions
in spermatogenesis are
• At puberty, spermatogonia differentiate into primary spermatocytes
• The primary spermatocytes undergo Meiosis I to produce secondary
spermatocytes
2. Throughout these meiotic divisions,
the “daughter cells” do not completely
separate
• Rather, they remain attached by
narrow
• The secondary spermatocytes undergo Meiosis II to produce spermatids
53
54
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Spermatogenesis
• The last stage of germ cell development is termed
• There is no cell division
(n)
• Spermatids undergo a dramatic change to form the
55
56
Figure 3.13a
Meiosis I yields two
haploid secondary
spermatocytes
57
Meiosis II yields
four haploid
spermatids
Each spermatid
matures into a haploid
sperm cell
58
Oogenesis
• In spermatogenesis, the two meiotic divisions are
• Takes place in the
• Human spermatogonia develop into mature sperm in roughly
• Gamete production in the female differs significantly from that in the male
• Spermatogenesis results in the production of enormous numbers of sperm
• An average adult male produces
sperm per day
• In humans, oogenesis begins in the embryo and continues until
• In humans, the two meiotic divisions of oogenesis are
59
60
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Oogenesis
• At puberty, the primary oocytes complete
Meiosis I to produce
• Secondary oocyte
• Frist polar body
• In the fetus, the PGCs divide by mitosis and differentiate into oogonia
• Oogonia differentiate into primary oocytes
• Cytokinesis in oogenesis is
• The 2o oocyte gets almost all of the
of the “parent cell”
• Primary oocytes undergo Meiosis I
• However, they are arrested in prophase I until
• The First polar body usually degenerates, but
may divide to produce two Second polar
bodies
61
62
Figure 3.13b
• The secondary oocyte enters Meiosis II but arrests in
• It is released into the
• If the secondary oocyte is not
and the secondary oocyte
• If the secondary oocyte is
Asymmetric Divisions
, Meiosis II is not completed,
, Meiosis II is completed
• Another asymmetric division yields the second polar body
• The
2o oocyte becomes the
• The
chromosomes
is diploid, containing a paternal and a maternal set of
63
64
Oogenesis
NOTE
• In humans and other primates, one oocyte
is ovulated each cycle
• => single births are the norm
NOTE
Prior to ovulation,
oocytes form in
65
• In other mammalian species, several
oocytes are ovulated at a time
• =>
66
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3.6 The Chromosome Theory of Inheritance and Sex Chromosomes
• Oogonia begin to develop into primary oocytes by the
gestation
month of
• Primary oocytes are arrested in prophase I by the
gestation
month of
• The 5 month old fetus possesses about
• At birth, the number has dropped to
• At puberty, only about
• The chromosome theory of inheritance describes how the transmission of
chromosomes account for the Mendelian patterns of inheritance
• This theory was independently proposed in 1902-03 by Theodore Boveri
and Walter Sutton
germ cells
remain
• But how many ”eggs” does a human female actually need?
67
68
• The chromosome theory of inheritance is based on five fundamental
principles
4. During the formation of gametes, different types of (nonhomologous)
chromosomes segregate independently
1. Chromosomes contain the genetic material
5. Each parent contributes one set of chromosomes to its offspring
• The sets are functionally equivalent
• Each carries a full complement of genes
2. Chromosomes are replicated and passed along from parent to
offspring
3. The nuclei of most eukaryotic cells contain chromosomes that are
found in homologous pairs
• During meiosis, each homologue segregates into one of the two
daughter nuclei
69
70
Figure 3.15
• The chromosome theory of inheritance allows us to see the relationship
between Mendel’s laws and chromosome transmission
Homologous
chromosomes
segregate from each
other
• Mendel’s law of segregation can be explained by the homologous
pairing and segregation of chromosomes during meiosis
• Refer to Figure 3.15
This leads to the
segregation of the
alleles into
separate gametes
• Mendel’s law of independent assortment can be explained by the
relative behavior of different (nonhomologous chromosomes) during
meiosis
• Refer to Figure 3.16
71
72
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Chromosomes and Sex Determination
Figure 3.16
1. XY system, which is found in mammals, other animals and
some plants
• Females are homogametic (XX)
• Males are heterogametic (XY)
During metaphase I,
the bivalents can
align themeselves in
two different ways
Independent
assortment
of the R/r
and Y/y
alleles
Figure 3.17a
73
74
2. ZW system, which is found in birds, butterflies, moths and
some fishes
• Females are heterogametic (ZW)
• Males are homogametic (ZZ)
Figure 3.17c
75
76
3. XO system, which is found in many species of insects
• Females have two X chromosomes (XX)
• Males have only one (XO)
4. X chromosome-autosome balance system, which is found in
Drosophila
• The main factor in sex determination is the ratio between the
number of X chromosomes and that of the number of sets of
autosomes
• The Y chromosome is essential for male fertility NOT male
development
Figure 3.17b
77
78
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• If the X:A = 1.00 => Fly will be a female
• Drosophila melanogaster has four pairs
of chromosomes
• If the X:A = 0.50 => Fly will be a male
• A pair of sex chromosomes and
three pairs of autosomes
• If the 0.50 < X:A < 1.00 => Fly will be intersex
• Remember though that the three
pairs of autosomes make up two sets
because the organism is diploid
• If the X:A > 1.00 => Fly will be a metafemale
• If the X:A < 0.50 => Fly will be a metamale
79
80
5. The haplo-diploid system in bees
• Males are known as the drones
• They are haploid
• Produced from
eggs
• Females include the workers and queen
• They are diploid
• Produced from
eggs
Figure 3.17d
81
82
Transmission of Genes Located on Human Sex Chromosomes
• Genes that are found on the Y chromosome are called
genes
• Genes that are found on one of the two types of sex
chromosomes but not on both are termed
• The X and Y chromosomes also contain short regions of
homology at one end
• These are termed
• PARS promote the necessary pairing of the two
chromosomes in meiosis I of spermatogenesis
• Genes that are found on the X-chromosome are called
• For these genes, dominance and recessiveness only have
meaning in the female
• The male is said to be
83
• The few genes found in this homologous region follow a
pseudoautosomal pattern of inheritance
• Their inheritance pattern is the same as that of a gene found
on an autosome
84
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Sex Ratio in Humans
Figure 4.12
Primary Sex Ratio
Involved in
antibody
production
The proportion of
males to females
conceived in a
population
Secondary Sex Ratio
The proportion of
males to females
born in a population
In humans, the primary sex ratio heavily favors
while the secondary sex ratio slightly favors
,
Follows a pseudoautosomal pattern
of inheritance
85
86
Dosage Compensation
?
The expected
primary sex ratio is
1:1. Why?
• In mammals, females have two X chromosomes, while males
have one
• Therefore, there could be a “genetic dosage” problem
between males and females for X-linked genes
Males produce equal numbers of
X- and Y- bearing sperms
These are equally viable and motile in
the female reproductive tract
• However, there is a genetic mechanism that allows for dosage
compensation
The egg surface is equally receptive
to these sperms
?
So why is the observed
primary sex ratio
different than the
expected?
87
88
Dosage Compensation
• In 1949, Murray Barr observed a
darkly-staining body in interphase
nerve cells of female cats, but not in
male cats
•
89
or sex chromatin
90
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Figure 5.3
• In 1961, Mary Lyon proposed a hypothesis
• In normal female germline cells, both X chromosomes are active!
• No. of Barr bodies =
1. In female mammalian somatic cells one of the
two X chromosomes is
condensed and inactive
2. Inactivation occurs very early in embryonic life
3. Inactivation occurs independently and
randomly in each cell
4. Once decision is made, the inactivated X
remains inactivated through all subsequent
mitotic events within that cell line
91
92
Figure 5.3
• The mechanism of X inactivation is schematically
illustrated in Figure 5.4
• The result of random X inactivation in females can sometimes
be observed in the external phenotype
• “Calico” pattern of coat coloration in cats
• Example involves a white and black variegated coat
color found in certain mice
• Two X-linked alleles affecting coat color
• B, which confers black color
• b, which confers white color
• A male is
for the X chromosome
• => His coat color is black or white
93
94
• A female can be heterozygous for the white and black alleles
=> Coat color is
The epithelial cells
derived from this
embryonic cell will
produce a patch of
white fur
• In some cell lineages, the X chromosome bearing the black allele is inactive
=> Coat color is
• In some cell lineages, the X chromosome bearing the white allele is inactive
While those from
this will produce a
patch of black fur
=> Coat color is
95
At an early stage of
embryonic development
96
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Other Dosage Compensation Mechanisms
• Caenorhabditis elegans
• Dosage compensation occurs in the
• Animals, other than mammals, also have a genetic dosage
problem
• Genes on the
• However, these accomplish genetic dosage via mechanisms
other than inactivation of a whole chromosome
• Drosophila melanogaster
• Examples
• Dosage compensation occurs in the
• Genes on the
97
98
Mammals Maintain One Active X Chromosome in Their Somatic Cells
• Researchers have found that mammalian cells can count their X
chromosomes and allow only one of them to remain active
• Additional X chromosomes are converted to Barr bodies
99
100
X-Chromosome Inactivation in Mammals Depends on the X-Inactivation
Center
• Reading assignment
• Pages 110-111
101
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Introduction
Chapter 8
Variation in Chromosome
Structure
and Number
Johnny El-Rady, Ph.D
C
: the field that involves the
microscopic examination of chromosomes
• It typically examines the chromosomal composition of a
specific cell or organism
• This allows the detection of individuals with abnormal
chromosome number or structure
• This also provides a way to distinguish between species
University of South Florida
1
2
Anatomy of a Chromosome
Figure 8.1a
2n = 46
2n = 8
2n = 20
Telomere
p
Centromere
– Essential for
chromosome
segregation
q
Light bands
– Less condensed DNA
Dark bands
– Highly condensed DNA
Telomere
– Protects ends and ensures their
proper replication
3
4
Human
chromosome
banding patterns
seen on light
microscopy
A pair of homologous
chromosomes as seen at
metaphase
Alleles
– alternative
forms of a
gene or
marker
Locus
– position of
a gene or
DNA marker
Chromosome 1
Different chromosome banding resolutions can resolve
bands, sub-bands and sub-sub-bands
5
6
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Types of Chromosomes
Figure 8.1
• Chromosomes can be divided into four main types based on
their centromere location
p
p
p
p
q
Metacentric
q
Submetacentric
q
q
Acrocentric
Telocentric
(d) Conventional numbering system of G bands in human chromosomes
Figure 8.1
7
8
• Three main features are used to identify and classify
chromosomes
1. Size
2. Location of the centromere
3. Banding patterns
• These features are all seen in a
• Fluorescence in situ hybridization
9
10
Chromosomal Variation
Chromosomal Variation
• The study of chromosomal variation is important for several reasons
• Chromosomal variation can occur in two basic ways:
1. They can have major effects on the
organism
of an
2. They can have major effects on the phenotype of the
of an organism’s
3. They have been an important force in the
of species
11
12
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• A given abnormality may be present in all body cells, or in some
of them
: the presence of two or more cell lines
in an individual, one or more of which are abnormal
13
14
Changes in Chromosome Number
Polyploid organisms
have three or more
sets of chromosomes
• Chromosome numbers can vary in two ways
Changes in the entire set of
chromosomes
Occurs occasionally in
animals and frequently in
plants
Individual is said
to be trisomic
Changes in individual
chromosomes rather than in
the complete set
Regarded as abnormal
conditions
15
Figure 8.15
Individual is said
to be monosomic
16
Changes in Chromosome Structure
Human
chromosome 1
• Chromosome structure can vary in four ways
•
= Loss of a part of a chromosome
•
= Addition of a part of a chromosome
•
= Part of a chromosome switches 180o
•
Human
chromosome 21
= Movement of a chromosome segment
to another location in the genome
Figure 8.2
17
18
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Aneuploidy
• Includes
• Monosomy – One homologue is missing
• Trisomy = One extra homologue is present
• Tetrasomy = Two extra homologues are present
• Double monosomy = Two different homologues are missing
• Double trisomy = Two different extra homologues are present
• Nullisomy = A pair of chromosomes is missing
In general, changes in chromosome
number have a
severe
effect on survival than changes in
chromosome structure
NOTE
These definitions apply
to a
cell
19
20
Normal Chromosome Complement
2
1
3
Figure 8.16
4
Diploid
(2N)
Aneuploidy
• Aneuploidy commonly causes an
abnormal phenotype
Nullisomic
(2N-2)
Monosomic
(2N-1)
• It leads to an
in
the amount of gene products
Doubly monosomic
(2N-1-1)
Trisomic
(2N+1)
Tetrasomic
(2N+2)
Doubly tetrasomic
(2N+2+2)
21
22
Monosomy
Trisomy
• Tolerated in some plants
• Produces more
than monosomy
• In Drosophila
offspring
• In Drosophila
• Haplo-IV
• Survive but have problems!
• Triplo-IV flies have no visible abnormalities
• Lethal in humans except for the X chromosome
23
24
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Trisomy
• In plants, trisomic individuals are viable but their phenotypes
may be affected
• Jimsom weed Datura stramonium
• Have diploid number of 24
• Each possible thisomy is viable
• But each distinctly alters phenotype of the capsule of the fruit
25
26
Trisomy
Trisomy for Human Autosomes
• In humans, trisomies for chromosomes 13, 18, 21 and X are
compatible with survival
• Trisomy 21 = Down syndrome
• Trisomy 18 = Edwards syndrome
• But they result in major developmental abnormalities
• Trisomy 13 = Patau syndrome
27
28
Down Syndrome
Down Syndrome
Features of Down Syndrome
• First described by John Langdon
Down in 1866
• Chromosomal defect discovered by
Jerome Lejeune in 1959
John Langdon Down
• Short stature
• Prominent epicanthal folds
• Large protruding tongue with distinctive furrowing
• Stubby hands with a crease on the palm
• Hypotonia
• Moderate to severe mental retardation
• Congenital defects
• High susceptibility to leukemia and infections
• Down patients tend to be lively, cheerful and affectionate
Jerome Lejeune
• Karyotype =
• Frequency =
29
live births
30
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31
32
Down Syndrome
Edwards Syndrome
Figure 8.17
• Non-disjunction event is more likely to occur in females
• First described by John Edwards in 1960
• In addition, there is a pronounced maternal age effect
• Karyotype =
• Frequency =
33
live births
34
Edwards Syndrome
Features of Edwards Syndrome
• Growth retardation and intellectual disability
• Characteristic clenched fists
• Small ears; small mouth
• Congenital defects
• Preponderance of females
• Average age < 4 months
• ~ 10% are mosaics
35
36
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Patau Syndrome
Patau Syndrome
Features of Patau Syndrome
• First described by Klaus Patau in 1960
• Growth retardation and intellectual disability
• Characteristic clenched fists
• Deafness
• Polydactyly
• Microophthalmia
• Cleft lip; cleft palate
• Average age < 6 months
• Karyotype =
• Frequency =
live births
37
38
NOTE
• Aneuploidy of sex chromosome
has less severe phenotypic effects
than that of autosomes
WHY?
39
40
Turner Syndrome
• Let’s discuss the various aneuploidies of sex chromosomes
in humans
• First described by Henry Turner
in 1938
• These include:
• Chromosomal abnormality first
described by Ford et al. in 1959
• Turner syndrome
• Klinefelter syndrome
• Triple X syndrome
• Tetra- and Penta-X syndromes
• XYY condition
41
• Karyotype =
• Frequency =
female births
live
42
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Turner Syndrome
Features of Turner Syndrome
•
•
•
•
•
•
•
Short stature
Webbed neck
Underdeveloped ovaries
About normal intelligence
Failure to develop 2o sexual characteristics
40-50% of patients are mosaics
~99% of conceptions are lost prenatally
• Therapy may help
• Anabolic steroids
• Estrogen
43
44
Turner Syndrome
?
Klinefelter Syndrome
• First described by Harry Klinefelter
in 1942
Why does a female with one X
chromosome exhibit a syndrome,
knowing that if two X chromosomes are
present one will be inactive?
Henry Klinefelter
• Chromosomal abnormality first
described by Jacobs and Strong in
1959
• Karyotype =
• Frequency =
male births
45
live
46
Klinefelter Syndrome
• Extreme Klinefelter syndrome karyotypes include:
Features of Klinefelter Syndrome
•
•
•
•
•
•
•
Above average height
Long limbs
Testicular atrophy
Gynecomastia
Feminine-pitched voice
Little body hair
~15% are mosaics
, XXXY
;
, XXYY
;
XXXXY
• Therapy may help
• Testosterone
• Mastectomy
47
48
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47, XXX Syndrome
• First described by Jacobs et al. In 1959
• Frequency =
live female births
• Highly variable expression
• Usually benign consequences
• Under-development of genitalia, mental retardation,
deficiencies in speech development
49
50
Tetra-X and Penta-X Karyotypes
47, XYY Condition
• First described by Sandberg et al. in 1961
• Karyotype =
• Frequency =
• Symptoms tend to be more
47, XXX syndrome
live male births
than those of the
• Features
• Above average height
• Sub-average intelligence
• The relationship between XYY and aggressive, psychopathic or
criminal behavior has aroused great public interest
51
52
Euploidy
• Most species of animals are
• Some euploidy variations are naturally occurring
• Polyploids are organisms in which the number of chromosome
sets exceeds two
53
54
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• Regardless of the number of chromosome sets, the number of
chromosomes in the gametes is half the number of
chromosomes in its somatic cell
• Hexaploid with 60 chromosomes
=>
• For organisms that are regularly polyploid,
• x = the number of chromosomes per set
• n = the number of chromosomes per gamete
=>
?
Consider a hexaploid with 60 chromosomes.
• Solve for x and n?
55
56
• Polyploidy is fairly common in plants
• Polyploidy is rare in animals
• A few vertebrates are polyploids
• In humans, polyploidy is always lethal, usually resulting in
abortion during the first trimester
• Many plants with agricultural and horticultural significance are polyploids
• These tend to be larger or more hardy than their diploid counterpart
Figure 8.18
Figure from previous edition
57
58
Endopolyploidy
?
But wait a minute:
• How can a triploid human be produced?
Endopolyploidy: The phenomenon in which some somatic
cells of an animal have more sets of chromosomes than other
somatic cells
• In humans, some cells are polyploid
59
60
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Natural and Experimental Ways to Produce Variations in Chromosome Number
• In sexually-reproducing species,
•
-numbered sets of chromosomes are usually
maintained reliably generation to generation because they
lead to balanced gametes
•
-numbered sets of chromosomes are usually NOT
maintained reliably because they lead to unbalanced
gametes
• There are three natural mechanisms by which the
chromosome number of a species can vary:
Meiotic Nondisjunction
Mitotic Abnormalities
• Sterility is generally a detrimental trait
• However it can be agriculturally desirable
because it may result in
fruit and flowers
61
Interspecies Crosses
62
Nondisjunction
Meiotic Nondisjunction
• Meiotic nondisjunction can produce haploid cells that have too
many or too few chromosomes
Nondisjunction: The failure of chromosomes to
segregate properly during cell division
Meiotic Nondisjunction
• The result is one cell gets too many and one too few
chromosomes
If such a gamete participates in
fertilization
• Nondisjunction can occur in mitosis or meiosis
• If nondisjunction occurs in mitosis, then the tissue from the
nondisjunction event will have an abnormal number of
chromosomes, but the rest of the body will be normal
63
64
During
fertilization,
these
gametes
produce an
individual
that is
monosomic
for the
missing
chromosome
During
fertilization,
these
gametes
produce an
individual
that is
trisomic for
the missing
chromosome
Figure 8.22
65
The resulting individual will have
an abnormal chromosomal
composition in all of its cells
50 %
Abnormal
gametes
All four gametes are abnormal
50 % Normal
gametes
Figure 8.22
66
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Meiotic Nondisjunction
Mitotic Abnormalities
• Abnormalities in chromosome number often occur after fertilization
(i.e. during mitosis)
• In rare cases, all the chromosomes can undergo nondisjunction
and migrate to one daughter cell
Mitotic Disjunction (Figure 8.23a)
• This is termed complete nondisjunction
• It results in a diploid cell and one without chromosomes
• So what?
• Sister chromatids separate improperly
• Leads to trisomic and monosomic daughter cells
Chromosome Loss (Figure 8.23b)
• One of the sister chromatids does not migrate to a pole
• Leads to normal and monosomic daughter cells
67
68
Mitotic Abnormalities
Figure 8.23
This cell will
be trisomic
This cell will
be monosomic
• Genetic abnormalities that occur after fertilization lead to
• Part of the organism contains cells that are genetically
different from other parts
• The size and location of the mosaic region depends on the
timing and location of the original abnormality
• In the most extreme case, the error could take place during
the first mitotic division
• A bizarre example is shown next
This cell will
be monosomic
This cell will
be normal
Will be degraded if left
outside of the nucleus
when nuclear
envelope reforms
69
70
• Consider a fertilized Drosophila egg that is XX
• One of the X’s is lost during the first mitotic division
• This produces an
cell and an
cell
The XX cell is the
precursor for this side of
the fly, which developed
as a
The X0 cell is the
precursor for this side of
the fly, which developed
as a
Figure from
previous
edition
• This peculiar and rare individual is termed a
71
72
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Interspecies Crosses
Figure 8.24
1. Autopolyploids = contain multiple copies of the same genome
• There are two types of polyploids:
Autopolyploids
ploid
Allopolyploids
ploid
73
74
2. Allopolyploids = contain genomes derived from closely related
species
An allotetraploid:
Contains two
complete sets of
chromosomes
from two different
species
NOTE
• Cultivated bread wheat, Triticum
aestivum, which is an
75
76
Endoreduplication
• Chromosome doubling due to an aborted cell division
Figure 8.26
Figure 8.25
in the 6e
Caused by
complete
nondisjunction
• It may occur:
• Naturally
• Artificially, by treatment with colchicine
• Binds to
77
78
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• Consider two closely-related diploid species
• Each species produces haploid gametes, so the hybrid is
termed an
• However, normal pairing during meiosis I may not be possible
and thus the hybrid is sterile
• The chromosome sets are
• If endoreduplication occurs, the duplicated chromosomes can
now normally pair with each other and the hybrid is no longer
sterile
79
80
Figure 8.25
• Interspecies hybrid between the
road antelope (Hippotragus
equinus) and the sable antelope
(Hippotragus niger)
• The allodiploid is fertile because
the homeologous chromosomes
can properly synapse during
meiosis
81
82
• In 1928, the Russian cytogeneticist Georgi
Karpechenko conducted an interspecies cross between
radish (Raphanus) and cabbage (Brassica)
• Both are diploid and contain 18 chromosomes
• Therefore the interspecies hybrid contains
chromosomes
• However, the radish and cabbage are not closely
related species
• Their chromosomes are distinctly different from
one another and cannot synapse
• Thus, the radish/cabbage hybrid is
• But an allotetraploid would be
• Contains
chromosomes which undergo
proper synapsis
83
No synapsis
between the 9
radish and 9
cabbage
chromosomes
Figure from
previous edition
Proper synapsis
between the 18
radish
chromosomes …
Georgi
Karpechenko
… and the 18
cabbage
chromosomes
(a) Alloploid with a monoploid
set from each species
(b) Alloploid with a diploid
set from each species
84
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Changes in Chromosome Structure
• Result from chromosomal breaks and abnormal rejoining
• The cause may be spontaneous or induced
85
86
Human
chromosome 1
Deletions (deficiencies)
• The loss of a portion of a chromosome
• Deletions may be terminal or interstitial
Human
chromosome 21
Figure 8.2
87
88
Deletions (deficiencies)
• Deletions may produce a mutant phenotype because:
1.
effects
• Break occurs within a gene
2.
effects
• Gene is shifted to a new location
• When deletions are homozygous they are often lethal
• When deletions are heterozygous, they may be lethal
• Deletions may lead to
• Uncovering of the recessive allele
• Deletions upset the
• Deletions “unmask”
89
90
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Figure 8.4
• Cri-du-chat syndrome
• Cri-du-chat syndrome
• Occurs in about 1/50,000 live births
• Individuals are severely impaired mentally and
physically
• Features
• Round face in the newborn → Elongation of
the face in the older child and adult
• Micrognathia
• Multiple organ deformities
• Infants have high-pitched cry, similar to that of
the meowing of a cat
• A human disorder caused by a
deletion
• First described by Lejeune et al.
in 1963
• About half of the 5p arm is
missing
• Karyotype =
91
92
Duplications
Duplications
• The presence of an extra segment of a chromosome
• Categorized by the position and order of the duplicated segment
93
94
Duplications
Duplications
Figure 8.5
• May result from an unequal crossing-over
• As the number of duplicate copies of a segment increases, the
likelihood of unequal crossing over would also increase
• Therefore, once the process has started there is a tendency for
the number of copies to increase over evolutionary time
95
96
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Duplications
Figure 8.6
• The majority of small chromosomal duplications have no
phenotypic effect
• However, they are vital because they provide raw material for
additional genes
Genes derived
from a single
ancestral gene
• This can ultimately lead to the formation of
• Two or more genes that are similar to each other
97
98
Figure 8.7
• A well-studied example is the globin gene family
• It is composed of 14 homologous genes on three different
chromosomes
• All are derived from a single ancestral gene
• Accumulation of different mutations in the members of the
gene family created:
Expressed very early
in embryonic life
Expressed maximally during
Expressed after birth
the second and third trimesters
1. Globin genes that are expressed during different stages
of
Better at
binding and
storing oxygen
in muscle cells
2. Globin proteins that are more specialized in their
99
Duplication
Better at
binding and
transporting
oxygen via red
blood cells
100
Duplications
• Note that plants have a form of globin gene called
• Duplications tend to be
• They generally allow for
• So the ”first” globin gene must have been present in the
common ancestor of both animals and plants
harmful than deletions
• Abnormalities due to
• Upsetting of the genetic balance
• Moreover, invertebrates have myoglobin but NOT hemoglobin
• Therefore, the globin duplication must have occurred after
the divergence between vertebrates and invertebrates
101
102
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Inversions
• A rearrangement that reverses the order of genes in a
chromosome
Centromere lies
outside inverted
region
Centromere lies
within inverted
region
• Two different kinds of relative to the position of the centromere
• Pericentric inversions
• Paracentric inversions
• In
• In
103
inversions, the arm ratio is unchanged
inversions, the arm ratio is often changed
104
• Most inversions do not result in an abnormal phenotype
• In the rare cases that they do, the mutant phenotype is due
to
Inversions
• About
of the human population carries inversions that
are detectable with a light microscope
• Most of these individuals are phenotypically normal
• However, a few can produce offspring with genetic
abnormalities
105
106
Inversions
Figure 8.11
• An inversion heterozygote has one copy of a normal
chromosome and one copy of an inverted chromosome
• They also may have a high probability of producing gametes
that are abnormal in their genetic content
• The abnormality is due to
inverted segment
107
in the
108
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Translocations
Translocations
1. Intrachromosomal
• Chromosomal segments moves from one part of the
chromosome to another
• A chromosomal translocation occurs when a segment of one
chromosome becomes attached to another
• Translocations are of two main types
2. Interchromosomal
a. Non-reciprocal – Chromosomal segment moves from one
chromosome to another
1. Intrachromosomal
2. Interchromosomal
b. Reciprocal = Two non-homologous chromosomes
exchange chromosome segments
109
110
Intrachromosomal translocation
1
2
3
4
5
6
7
Figure 8.12
6
7
1
2
3
4
Telomeres prevent
chromosomal DNA from
sticking to each other
5
• Reciprocal translocations arise from
two different mechanisms
Non-reciprocal interchromosomal translocation
A
B
C
1
2
3
D
4
E
5
6
7
A B
C
1
3
2
D
4
E
6
7
1. Chromosomal breakage and
DNA repair
5

Reciprocal interchromosomal translocation
A
B
C
1
2
3
D
4
E
5
6
7
A B
C
1
3
2
2. Abnormal crossovers
D
4
6
7
5
E
111
112
• Reciprocal translocations, like inversions, are usually without
phenotypic consequences
• Robertsonian translocation is a specific type of reciprocal
translocation
• It involves two non-homologous acrocentric chromosomes,
which fuse at the centromere and lost their short arms
113
114
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Translocations
Translocations: Alter the sizes of chromosomes, as
well as the position of the centromere
• Example: Chronic myelogenous leukemia
• Tumor of certain WBCs
• Typically occurs in adults
• 90% of patients have a reciprocal translocation
• Most translocations do not result in an abnormal phenotype
• In the rare cases that they do, the mutant phenotype is due
to the
effect
115
116
Centromeres and the Genetic Stability of Chromosomes
Figure 24.15
Figure 25.13
in the 6e
• For a chromosome to be reliably transmitted during cell
division it has to be linear and possess only one centromere
• In other words, it has to be a monocentric rod chromosome
• Acentric, dicentric, and ring chromosomes are not reliably
transmitted to daughter cells
117
118
• Chromosomal (or centric) fusion occurs when two nonhomologous chromosomes are joined at their centromeres to
form a large single metacentric chromosome
• Also called Robertsonian translocation
Acentric
Dicentric
Normal chromosome
Two breaks
Reunion of
“sticky” ends
• Chromosomal (or centric) fission occurs when a chromosome
splits at its centromere, thus generating two smaller
chromosomes
Ring chromosome
119
120
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Centromeres and the Genetic Stability of Chromosomes
Often in the process of evolution, the number of
chromosomes arms is conserved whereas the
number of individual chromosomes is not
121
122
• Example 1:
In Drosophila
• Example 2:
123
In higher primates
• Human chromosome #2, a metacentric, is formed by the fusion of the
centromeres of two acrocentric chromosomes from chimpanzees
• D. melanogaster has 4 chromosomes, including two large metacentric ones
• D. virilis has 6 chromosomes, including four large acrocentric ones
124
Centromeres and the Genetic Stability of Chromosomes
• Thus, the number of chromosomal arms rather than the number
of chromosomes, gives a more accurate picture of the affinities
between species
• The number of arms is referred to as the
125
126
21
Chromosome Organization and Molecular
Structure
Introduction
• The genome is the entire genetic complement of a cell of a virus
• In eukaryotes it refers to one complete set of nuclear chromosome
NOTE
• Eukaryotes possess a
mitochondrial genome
• Plants also have a
chloroplast genome
• The genetic material in living organisms is double-stranded DNA
• In eukaryotes it is
• Linear
• In prokaryotes it is
• Circular
• The genetic material in viruses is
• DNA or RNA but not both
• Double-stranded or Single-stranded
• Circular or Linear
• In general, genome size increases roughly with evolutionary complexity
• Viral DNA < Prokaryotic DNA < Unicellular eukaryotic DNA < Multicellular
eukaryotic DNA
The Supercoiling of DNA
• The chromosomal DNA in both prokaryotes and eukaryotes is highly
compacted
• Supercoiling only occurs in DNA molecules with fixes ends
• Prokaryotic and eukaryotic organellar genomes
• Eukaryotic nuclear genomes
The Supercoiling of DNA
• Supercoils can best be explained in the following manner:
• One strand of a double-stranded circular covalently closed DNA molecule is
cut
• One end is rotated a complete turn around its complementary strand, while
the other end is fixed
• → a supercoil is introduced into the molecule
• The enzymes that create or alleviate supercoils are termed
topoisomerases
• There are two main types:
Type I
Type II
• The structure and activity of these enzymes is rather complex
• However, for the purpose of this course, we will use the definition on the following
slide
• Type I = Are usually monomeric proteins
• Induce transient single-strand breaks into the double helix
• Change supercoils one at a time
• Type II = Are usually multimeric proteins
• Induce transient double-strand breaks into the double helix
• Change supercoils two at a time
Figure 10.7
Composed of two A
and two B subunits
The process
introduces two
negative supercoils
into DNA
Remember the
original DNA
contained one
positive supercoil
The B subunits use
ATP to catalyze this
process
by the A subunits
NOTE
Supercoils are eliminated by a
nick in the DNA
Bacterial Chromosomes
Figure 10.2
• The bacterial chromosome is found in a region called the nucleoid
• It is not membrane-bound
Fluorescently labeled nucleoids
are seen as bright, oval-shaped
regions within the cytoplasm. The
DNA is not in a separate
compartment but is in direct
contact with the cytoplasm.
Figure 10.1
A few hundred
nucleotides in length
These play roles in DNA folding, DNA
replication, and gene expression
Bacterial Chromosomes
• The Escherichia coli chromosome consists of about 4.7 X 106 bp
• This is equivalent to a continuous linear duplex of about 1.5 mm
• But the E. coli cell is about 2 m in length (and 1 m in diameter)
• → DNA is ~ 700 X the length of the cell
• Therefore the DNA must be in a highly condensed configuration
Figure 10.3
• The first way to promote compaction is via the formation of loop
domains
The looped
structure compacts
the chromosome
about 10-fold
NOTE
The number of loops varies
according to the size of the
bacterial chromosome and
the species
• E. coli has 50-100 with 40,000 to 80,000 bp of DNA in each loop
• These loops are connected to the central point, which is attached to the cell
membrane
• DNA is condensed with the help of various classes of binding proteins
• These are positively-charged
• Collectively referred to as histone-line
Figure 10.4
• DNA supercoiling is a second important way to compact the
bacterial chromosome
Supercoiling within
loops creates a
more compact DNA
structure
Figure 10.5 provides a schematic illustration of DNA supercoiling
Figure 10.5
Plates preventing
DNA ends from
rotating freely
Less turns
Both overwinding
and underwinding
can induce
supercoiling
These two DNA conformations do not occur in living cells
More turns
These three DNA conformations are topoisomers of each other
Eukaryotic Chromosomes
• Eukaryotic species contain one or more sets of chromosomes
• Chromosomes in eukaryotes are located in the nucleus
• Eukaryotic genomes vary substantially in size
• In multicellular eukaryotes, the amount of DNA does not correlate with
evolutionary complexity, nor is it directly proportional to number of genes
• C-value paradox (C refers to chromosome)
• Differences in genome size among species of
• Arthropods → 250-fold
• Fish → 350-fold
• Angiosperms → 1000-fold
• Genome size of Amphiuma means is ~90 Gb
• Genome size of Fugu rubripes is ~0.4 GB
• Indeed, in higher organisms, most of the DNA does not code for proteins
• In humans, the actual number of genes is 20-25,000, although the genome
can theoretically contain about 1,000,000
• In salamanders, closely-related species may have genomes that differ greatly
in size, yet contain the same number of genes
Figure 10.9
Figure 10.8
• A eukaryotic chromosome contains a long,
linear DNA molecule
• Three types of DNA sequences are needed for
chromosome replication and segregation
• Origins of replication
• Centromeres
• Telomeres
Eukaryotic Chromosomes
• A single chromosome usually has a few hundred to several thousand
genes
• In lower eukaryotes (such as yeast)
• Genes are relatively small
• They contain very few introns
• In higher eukaryotes (such as mammals)
• Genes are long
• They tend to have many introns
The Compaction of Eukaryotic Chromosomes
• Eukaryotic DNA is associated with various protein molecules in a stable
complex termed chromatin
• There are two fundamental classes of chromatin:
Euchromatin
Heterochromatin
The Compaction of Eukaryotic Chromosomes
1. Euchromatin
• Stains lightly by DNA-binding chemicals
• DNA is relatively extended and open
• Contains the vast majority of genes
2. Heterochromatin
•
•
•
•
•
•
Stains deeply by DNA-binding chemicals
DNA is tightly condensed and inaccessible
Replicates later during the S phase than the euchromatin
Contains relatively few genes
Found close to centromeres and telomeres
In some species, could comprise almost an entire chromosome
Figure 10.17
• There are two types of heterochromatin
• Constitutive heterochromatin
• Regions that are always heterochromatic
• Permanently inactive with regard to transcription
• Usually contain highly repetitive sequences
• Facultative heterochromatin
• Regions that can interconvert between euchromatin and heterochromatin
• Example: Barr body formation during development in female
The Compaction of Eukaryotic Chromosomes
• Eukaryotes contain an enormous amount of genetic material
• Humans, for example, have a genome consisting of ~3 X 109 bp
• This corresponds to about 1m of DNA
• The average adult human body contains about 1013 cells
• → it contains 2 X 1013 m of DNA
The Compaction of Eukaryotic Chromosomes
• To put this value in perspective…
• The distance from the earth to the sun is
The Compaction of Eukaryotic Chromosomes
• To put this value in perspective…
• The distance from the earth to the sun is
• 1.5 X 1011 m
• Thus, the DNA of a single human adult can stretch to the sun and back
• More than 50 X!!
The Compaction of Eukaryotic Chromosomes
• A typical eukaryotic chromosome contains from 1-20 cm of DNA
• During metaphase of cell division, this is packaged into a length of only 120 µ
? How is this condensation possible?
NOTE
The nucleus is ~ 6 µ in diameter
Experiment 10A: Nucleosome Structure Revealed
• The model of nucleosome structure was
proposed in 1974 by Roger Kornberg
• Kornberg based his proposal on various
observations about chromatin
• Biochemical experiments
• X-ray diffraction studies
• Electron microscopy images
Experiment 10A: Nucleosome Structure Revealed
• The Markus Noll decided to test Kornberg’s model
• He did this via the following procedure
• Digest DNA with the enzyme DNase I
• Accurately measure the molecular weight of the resulting DNA fragments
• The rationale is that the linker DNA is more accessible than the “core DNA” to
the DNase I
• Thus, the cuts made by DNase I should occur in the linker DNA
The Hypothesis
• The experiment tests the beads-on-a-string model of chromatin structure
• It the model is correct, DNase I should cut in the linker region
• Thereby producing DNA pieces that are about 200 bp long
Testing the Hypothesis
• Refer to Figure 10.12
Figure 10.12
The Data
Interpreting the Data
These longer pieces were all in
multiples of 200 bp
At low concentrations, Dnase I did
not cut at all the linker DNA
And this fragment,
three nucleosomes
This fragment contains
two nucleosomes
All chromosomal DNA digested into
fragments that are ~ 200 bp in
length
The Data
While the previous slides showed a
cartoon of the data, these slides
show actual photographs of the
data
Interpreting the Data
These longer pieces were all in
multiples of 200 bp
At low concentrations, Dnase I did
not cut at all the linker DNA
And this, three
This fragment contains
two nucleosomes
All chromosomal DNA digested
into fragments that are ~ 200 bp
in length
Interpreting the Data
Nucleosomes
• The structure of a nucleosome is essential
invariant in all eukaryotes
• It is composed of histones and DNA
• Histones
• A class of five proteins (H1, H2A, H2B, H3
and H4)
• They share the following features
•
•
•
•
Small
Basic
Ubiquitous
Evolutionary-conseved
Nucleosomes
• The “Beads-on-a-String” Model
• Core particle
• Two molecules each of H2A, H2B, H3 and H4
• 146 bp of DNA wrapped ~ 1.65 times around the octamer
• An H1 monomer
• Linker DNA
• Links beads together
• Length varies
Figure 10.11
Vary in length between 20 to 100 bp,
depending on species and cell type
Diameter of the nucleosome
•
Overall structure of connected nucleosomes resembles “beads on a string”
• This structure shortens the DNA length about seven-fold
Figure 10.11
Play a role in the
organization and compaction
of the chromosome
Figure 10.11
Higher Order Structures
• Nucleosomes associate with each other to form a more compact structure
termed the 30 nm fiber
• Histone H1 plays a role in this compaction
• At moderate salt concentrations, H1 is removed
• The result is the classic beads-on-a-string morphology
• At low salt concentrations, H1 remains bound
• Beads associate together into a more compact morphology
Figure 10.13
Higher Order Structures
• The 30 nm fiber shortens the total length of DNA another seven-fold
• Its structure has proven difficult to determine
• Two models have been proposed
• Solenoid model
• Three-dimensional zigzag model
Figure 10.14
Higher Order Structures
• The 30 nm fiber The two events we have discussed so far have shortened
the DNA about 50-fold
• A third level of compaction involves interaction between the 30 nm fiber
and the nuclear matrix
• The nuclear matrix is composed of two parts
• Nuclear lamina
• Fibers that line the inner nuclear membrane
• Internal matrix proteins
• Connected to nuclear lamina and fills interior of nucleus
• Structural and functional role remains controversial
Higher Order Structures
• The 30 nm fiber The two events we have discussed so far have shortened
the DNA about 50-fold
• A third level of compaction involves interaction between the 30 nm fiber
and the nuclear matrix
• The nuclear matrix is composed of two parts
• Nuclear lamina
• Fibers that line the inner nuclear membrane
• Internal matrix proteins
• Connected to nuclear lamina and fills interior of nucleus
• Structural and functional role remains controversial
Figure 10.15
Figure 10.15
The nuclear matrix is hypothesized to be
an intricate fine network of irregular
protein fibers plus many other proteins
bound to these fibers. This structure is
very dynamic and complex.
Figure 10.15
• The third mechanism of DNA compaction involves the formation of radial
loop domains
Matrix-attachment regions (MARs)
Or
Scaffold-attachment regions (SARs)
MARs are anchored to the nuclear
matrix, thus creating radial loops
Higher Order Structures
• The attachment of radial loops to the nuclear matrix is important in two
ways
1. It plays a role in compaction
2. It serves to organize the chromosomes within the nucleus
• Each chromosome in the nucleus is located in a discrete and nonoverlapping
chromosome territory
• Refer to Figure 10.16
Figure 10.16 – Chromosome Territory
• Each of seven types of chicken chromosomes is labeled a different color.
Each occupies a specific non-overlapping territory during interphase.
Figure 10.18
Figure 10.18
During interphase most chromosomal
regions are euchromatic
Metaphase Chromosomes
• As cells enter M phase, the level of compaction changes dramatically
• By the end of prophase, sister chromatids are entirely heterochromatic
• Two parallel chromatids have an overall diameter of 1,400 nm
• These highly condensed metaphase chromosomes undergo little gene
transcription
Metaphase Chromosomes
• In metaphase chromosomes the radial loops are highly compacted and
stay anchored to a scaffold
• The scaffold is formed from the nuclear matrix
• Histones are needed for the compaction of radial loops
Figure 10.19
• Two multiprotein complexes help to form and organize metaphase
chromosomes
• Condensin = Plays a critical role in chromosome condensation
• Cohesin = Plays a critical role in sister chromatid alignment
• Both contain a category of proteins called SMC proteins
• Acronym = Structural Maintenance of Chromosomes
• SMC proteins use energy from ATP and catalyze changes in chromosome
structure
Figure 10.20 – The Structure of SMC Proteins
Figure 10.21
During interphase,
condensin is in the
cytoplasm
Condesin binds to
chromosomes and
compacts the radial loops
Condesin travels into the
nucleus
The number of loops has not changed. However, the diameter of
each loop is smaller.
Figure 10.22
Cohesins along chromosome
arms are released
Telomere Structure
• Found at the end of linear
chromosomes
Telomere Structure
• Telomeres provide at least three important functions
• Prevent ends from acting in a “sticky” fashion
• Prevent ends from being degraded
• Allow chromosome ends to be properly replicated
• All telomeres so far isolated contain no genes
• Rather, they contain short sequences that are present as direct repeats
• These telomeric DNA sequences are highly conserved throughout evolution
• Mammals, birds & reptiles = 5’-TTAGGG-3’
• Arabidopsis thaliana = 5’-TTTAGGG-3’
• Tetrahymena thermophila = 5’-TTGGGG-3’
• Refer to Figure 11.23 and Table 11.5
Figure 11.23
You have reached the end of the presentation
D
Question 2
In humans, all females have Barr bodies.
O True
O False
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Question 3
The beads-on-a-string-model consists of:
O a histone octamer and linker DNA
a histone hexamer and linker DNA
O DNA loops bound to a scaffold of histones
DNA loops bound to a scaffold of non-histones
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Question 14
The water buffalo Bubalus bubalis has a diploid chromosome number of 50.
Which of the following statements about its MEIOTIC DIVISIONS is TRUE?
There are 200 chromatids in metaphase |
There are 25 tetrads in prophase I
O There are 50 chromosomes in each cell in metaphase II
O Two of the above
O All of the above
I trust that you will observe the honor code!
D
Question 17
In humans, oogenesis ends at menopause.
OTrue
O False
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D
Question 26
Aneuploidy in humans is usually problematic. Why?
O Chromosomal pairing is hampered
O Gene balance is disrupted
Crossing-over is blocked
Chromosomal disintegration is increased
O Nondisjunction is enhanced
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Question 44
The karyotype for Cri-du-Chat syndrome is:
O 46, 5p-
0 g”
46, 5q
O 45, -5p
O 45, 5p
O 45,59
Question 45
Trisomy
is known as Patau syndrome.
07
O 12
O 13
O 18
o 21
Previous
trust that you will observe the honor code!
D
Question 48
In plants, cytokinesis occurs via the formation of a cleavage furrow.
O True
O False
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