Download Chapter 15 - Kenston Local Schools

Chapter 15
Chromosomes, Morgan & Mutations
Figure 15-01
Chromosome Basis of
Inheritance
• Genes have specific positions on
chromosomes
• It is the chromosomes that undergo
segregation and independent
assortment
LE 15-2a
P Generation
Yellow-round
seeds (YYRR)
Green-wrinkled
seeds (yyrr)
Meiosis
Fertilization
Gametes
All F1 plants produce
yellow-round seeds (YyRr)
LE 15-2b
All F1 plants produce
yellow-round seeds (YyRr)
F1 Generation
Meiosis
LAW OF INDEPENDENT ASSORTMENT
LAW OF SEGREGATION
Two equally
probable
arrangements
of chromosomes
at metaphase I
Anaphase I
Metaphase II
Gametes
LE 15-2c
F2 Generation
Fertilization among the F1 plants
Why use Fruit Flies in genetic
experiments?
• Small / vials can hold hundreds
• Short Generation Time
• Many offspring
• Only 8 chromosomes
Morgan originally an Embryologist
• He was the one that came up with the
term “Wild Type” for the trait that was
normally found in nature
• All other traits he called “Mutants”
• Born in Lexington, KY - 1866
• Awarded the Nobel Prize 1933 (Physiology &
Medicine)
• Died in 1945
Figure 15-03
More Morgan
• Sex linked traits
• He was responsible for determining that
genes are located on the “X” chromosome
LE 15-4a
P
Generation
F1
Generation
F2
Generation
LE 15-4b
P
Generation
Ova
(eggs)
Sperm
F1
Generation
Ova
(eggs)
F2
Generation
Sperm
Linked Genes
• Each chromosome has hundreds or even
thousands of genes
• Genes that are on the same chromosome
and are close together are called linked
genes
• Often these are inherited together
Linked Genes – Two Traits
• Morgan did other experiments with fruit
flies to see how linkage affects inheritance
of two or three traits
• Morgan crossed flies that differed in traits
of body color, wing size and/or eye color.
LE 15-UN278-1
Parents
in testcross
Most
offspring
or
LE 15-5
P Generation
(homozygous)
Wild type
(gray body,
normal wings)
Double mutant
(black body,
vestigial wings)
b+ b+ vg+ vg+
b b vg vg
F1 dihybrid
(wild type
(gray body,
normal wings)
Double mutant
(black body,
vestigial wings)
TESTCROSS
b+ b vg+ vg
b b vg vg
Ova
965
944
Wild type
Black(gray-normal) vestigial
206
Grayvestigial
185
Blacknormal
Sperm
Parental-type Recombinant (nonparental-type)
offspring
offspring
Morgan & Crossing Over
• He discovered that genes can be linked,
but sometimes the connection between
genes on the same chromosome appears
to break
• Morgan realized that crossing over of
homologous chromosomes – was
responsible for this
LE 15-6a
Testcross
parents
Gray body,
normal wings
(F1 dihybrid)
Black body,
vestigial wings
(double mutant)
Replication of
chromosomes
Replication of
chromosomes
Meiosis I: Crossing
over between b and vg
loci produces new allele
combinations.
Meiosis I and II:
No new allele
combinations are
produced.
Meiosis II: Separation
of chromatids produces
recombinant gametes
with the new allele
combinations.
Ova
Gametes
Recombinant
chromosomes
Sperm
LE 15-6b
Sperm
Ova
Gametes
Ova
Testcross
offspring
Sperm
965
Wild type
(gray-normal)
944
Blackvestigial
Parental-type offspring
206
Grayvestigial
185
Blacknormal
Recombinant offspring
Recombination
391 recombinants
 100 = 17%
=
frequency
2,300 total offspring
Parentals & Recombinants
• Morgan noticed that sometimes traits in the
offspring were different than 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
recombinants
• A 50% frequency of recombination is
observed for any two genes on different
chromosomes
LE 15-UN278-2
Gametes from yellow-round
heterozygous parent (YyRr)
Gametes from greenwrinkled homozygous
recessive parent (yyrr)
Parental-type
offspring
Recombinant
offspring
Linkage Map
• Alfred Sturtevant: one of Morgan’s
protégés, made a genetic map, a list of
the location of genes “loci” on a
chromosome
• He predicted that the farther apart two
genes are, the higher the chance that a
crossover will occur, so the higher the
recombination frequency
Map units or CentiMorgans = cM
• A linkage map is a genetic map of a
chromosome based on recombination
frequencies
• Distances between genes can be shown as
map units:
– one map unit, or centimorgan = 1%
recombination frequency
• Map units indicate relative distance and
order, not exact locations of genes
LE 15-7
Recombination
frequencies
9%
9.5%
17%
b
Chromosome
cn
vg
Cytogenetic Maps
• Unlike linkage maps that just show
general position –
• Cytogenetic maps show the positions of
genes and the chromosomal features
• “banding patterns” exact locations
LE 15-8
I
II
Y
X
IV
III
Mutant phenotypes
Short
aristae
0
Long aristae
(appendages
on head)
48.5
Gray
body
Vestigial
wings
Cinnabar
eyes
Black
body
57.5
67.0
Red
eyes
Wild-type phenotypes
Brown
eyes
104.5
Normal
wings
Red
eyes
Morgan’s gene representation:
• For example:
– b = black = mutant
– b+ = normal body = WILD
– vg = vestigial = mutant
– vg+ = normal wings = WILD
– w = white = mutant
– w+ = red eyes (normal) = WILD
– ♀ = female
– ♂ = male
Example
• Wild (normal) body & Wild (normal) wings
crossed with a Black body & Vestigial
wings:
– b+ b+ vg+ vg+ X b b vg vg
– AABB x aabb (representing the same thing)
• RESULT:
– b+ b vg+ vg (heterozygote)
– This is sometimes confusing –
– AaBb = same thing – much easier
If on separate chromosomes
& not linked
• Aa Bb X aabb
• Counts will be even:
–
–
–
–
100
100
100
100
AB
Ab
aB
ab
(AaBb)
(Aabb)
(aaBb)
(aabb)
AB
AaBb
ab
ab
ab
ab
Ab
Aabb
aB
aaBb
ab
aabb
If genes located on same
chromosome = LINKED …
• Aa Bb X aabb
• Counts will be NOT even:
– 150
– 50
– 50
– 150
AB
Ab
aB
ab
(AaBb)
(Aabb)
(aaBb)
(aabb)
• To calculate the map distance:
– add the recombinants (50 + 50) / total (400)
100/400 = 25 (x 100) = 25% cM
Three Point Cross
• AABBCC x aabbcc = AaBbCc
• AaBbCc x aabbcc (test cross counts) – to
determine the linkage distance: Example:
– ABC
– ABc
– AbC
– Abc
95
5
700
50
aBC
aBc
abC
abc
50
700
5
95
A few hints …
– ABC
– ABc
– AbC
– Abc
95
5
700
50
aBC
aBc
abC
abc
50
700
5
95
• Parentals: (largest number) AbC & aBc
– Notice they are reciprocals AbC & aBc
• Recombinants: all others
• Double Crossovers: the smallest number &
the “middle” gene (ABc & abC)
– ABC 95
aBC 50
– ABc 5
aBc 700
– AbC 700
abC 5
– Abc 50
abc 95
• Compare Parentals & Double Crossovers:
– ABc
– aBc
5
700
abC
AbC
5
700
• Notice the only gene that is different when
comparing the two is the “A” gene
• That tells you that “A” is in the middle
It is an excellent idea to rewrite the
offspring chart in correct order …
– ABC
– ABc
– AbC
– Abc
95
5
700
50
aBC
aBc
abC
abc
50
700
5
95
• -----------------------------------– BAC
– BAc
– bAC
– bAc
95
5
700
50
BaC
Bac
baC
bac
50
700
5
95
Calculations
– BAC
– BAc
– bAC
– bAc
95
5
700
50
BaC
Bac
baC
bac
50
700
5
95 Total: 1700
• 95 + 95 + 5 + 5 = 200/1700 = 11.8 (bet. B & A)
• 50 + 50 + 5 + 5 = 110/1700 = 6.5 (bet. A & C)
• 95 + 95 + 50 + 50 = 290/1700 =
17.1 (bet. B & C)
–
(all of the above totals are x 100) = 11.8 / 6.5 / 17.1
Order of genes
17.1
B
11.8
A
6.5
C
Sex Determination
• An organism’s sex is determined by the
presence or absence of certain
chromosomes
• In humans and other mammals, there are
two sex chromosomes, X and Y
• Other animals have different chromosome
arrangement to determine male / female
LE 15-UN282
X
Y
LE 15-9a
Parents
Ova
Sperm
Zygotes
(offspring)
The X-Y system
LE 15-9b
The X-0 system
Grasshoppers, Cockroaches, Crickets, Praying
Mantis & some other insects
LE 15-9c
The Z-W system
Birds, some fish and some insects
LE 15-9d
The haplo-diploid system
Bees & Ants
LE 15-9
Parents
Ova
Sperm
Zygotes
(offspring)
The X-Y system
The X-0 system
The Z-W system
The haplo-diploid system
Sex Linked
• The sex chromosomes can have other
genes on them – not related to sex
determination
• A gene located on either sex chromosome
is called a sex-linked gene
• Sex-linked genes follow specific patterns
of inheritance
Sex Linked Diseases
• Some disorders caused by recessive alleles
on the X chromosome in humans:
– Color blindness
– Muscular dystrophy (Duchenne MD)
– Hemophilia
LE 15-10a
Sperm
Ova
LE 15-10b
Sperm
Ova
LE 15-10c
Sperm
Ova
X-Inactivation in Females
• In mammals, one of the two X
chromosomes is randomly inactivated
when “embryo”
• If a female = heterozygous for a gene
located on the X chromosome, she will be
a mosaic for that character
• Very little is understood about how this
works; Alex will tell you more… very soon!
LE 15-11
Two cell populations
in adult cat:
Active X
Early embryo:
Orange
fur
X chromosomes
Cell division
Inactive X
and X
chromosome Inactive X
inactivation
Black
fur
Allele for
orange fur
Allele for
black fur
Active X
Mutations
• In non-disjunction, pairs of homologous
chromosomes do not separate normally
during meiosis
• As a result:
– one gamete receives two chromosomes
– the other gamete receives zero
LE 15-12
Meiosis I
Nondisjunction
Meiosis II
Nondisjunction
Gametes
n+1
n+1
n–1
n–1
n+1
n–1
n
Number of chromosomes
Nondisjunction of homologous
chromosomes in meiosis I
Nondisjunction of sister
chromatids in meiosis I
n
Abnormalities
• Aneuploidy = abnormal number of a
particular chromosome
• Trisomy 3n = three copies of a particular
chromosome
• Monosomic = chromosome missing in
the zygote (only one present)
• Polyploidy = is a condition in which an
organism has more than two complete
sets of chromosomes
• Tetraploidy = 4n = four sets of
chromosomes
Chromosome Breaking Apart
• Breakage = 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
chromosome to another
LE 15-14a
Deletion
A deletion removes a chromosomal
segment.
LE 15-14b
Duplication
A duplication repeats a segment.
LE 15-14c
Inversion
An inversion reverses a segment
within a chromosome.
LE 15-14d
Reciprocal
translocation
A translocation moves a segment
from one chromosome to another,
nonhomologous one.
Genetic Disorders
• Down syndrome is an aneuploid condition
that results from three copies of
chromosome 21 (Trisomy 21)
• It affects about 1 / 700 children born in
the United States
• The frequency of Down syndrome
increases with the age of the mother
(several theories here…)
Down’s Syndrome
Maternal Age
< Risk of
chromosomal
abnormality
Risk of Down’s
Syndrome
15-24
1/500
1/1500
25-29
1/385
1/1100
35
1/178
1/350
40
1/63
1/100
45
1/18
1/25
Figure 15-15
Sex Chromosome Abnormalities
• 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 (survivable)
monosomy in humans
Chromosomal Abnormalities
• One syndrome, cri du chat (“cry of the cat”),
results from a 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
LE 15-16
Normal chromosome 9
Reciprocal
translocation
Translocated chromosome 9
Philadelphia
chromosome
Normal chromosome 22
Translocated chromosome 22
Exceptions to the Rule
• There are two normal exceptions to
Mendelian genetics
• One involves genes located:
– in the nucleus
– involves genes located outside the nucleus
Genes located in Organelles
• Extra-nuclear genes are genes found in
organelles in the cytoplasm
• This depends on the maternal parent
because the zygote’s cytoplasm comes
from the egg
• The first evidence of extra-nuclear genes
came from studies on the inheritance of
yellow or white patches on leaves of an
otherwise green plant
Figure 15-18
Genetic Imprinting
• For a few traits in mammals - the
phenotype depends on which parent the
trait came from
• This is called genetic imprinting
• This involves the silencing of certain genes
that are “stamped” with an imprint during
gamete production
LE 15-17a
Normal lgf2 allele
(expressed)
Paternal
chromosome
Maternal
chromosome
Normal lgf2 allele
(not expressed)
Wild-type mouse
(normal size)
A wild-type mouse is homozygous for the normal lgf2 allele.
LE 15-17b
Normal lgf2 allele
(expressed)
Paternal
Maternal
Mutant lgf2 allele
(not expressed)
Normal size mouse
Mutant lgf2 allele
(expressed)
Paternal
Maternal
Normal lgf2 allele
(not expressed)
Dwarf mouse
When a normal lgf2 allele is inherited from the father, heterozygous
mice grow to normal size. But when a mutant allele is inherited
from the father, heterozygous mice have the dwarf phenotype.
Human Example
• Depends on whether the deletion is inherited
from Mother or Father on chromosome 15q:
• If deletion in Father’s 15q: (Child inherits
both copies of 15q from Mother)
– Prader-Willi syndrome: developmental delay, small
or undescended testes, obesity (never feel satisfied),
short stature, and mild retardation.
• If deletion in Mother’s 15q (Child inherits
both copies of 15q from Father)
– Angelman syndrome: seizures, severe mental
retardation, inappropriate laughter, and a
characteristic face that is small with a large mouth
and prominent chin.
Prader-Willi Syndrome
Angelman’s Syndrome
Figure 15-13