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Transcript
AP Biology
Chapter 15
The Chromosome Theory of Inheritance


Mendel’s “factors”
are now known to be
genes—segments of
chromosomes.
Where the
chromosomes go
during Meiosis
determines which
traits end up in each
of the gametes.



First described by
Walter S. Sutton in
1902:
Genes have specific
loci (positions) along
chromosomes
It is the
chromosomes that
undergo segregation
and independent
assortment
The Chromosomal Basis of
Mendel’s Laws:
Green-wrinkled
seeds ( yyrr)
Yellow-round
seeds (YYRR)
P Generation
Y
Y
R R
r

y
y
r
Meiosis
Fertilization
Gametes
R Y
y
r
All F1 plants produce
yellow-round seeds (YyRr)
All F1 plants produce
yellow-round seeds (YyRr)
0.5 mm
F1 Generation
R
R
y
r
Y
LAW OF SEGREGATION
The two alleles for each gene
separate during gamete
formation.
y
r
Y
LAW OF INDEPENDENT
ASSORTMENT Alleles of genes
on nonhomologous
chromosomes assort
independently during gamete
formation.
Meiosis
r
R
Y
y
r
R
Metaphase I
Y
y
1
1
r
R
r
R
Y
y
Anaphase I
Y
y
r
R
Metaphase II
R
r
2
2
Gametes
y
Y
Y
R
R
1
4
r
1
YR
3
4
yr
Y
Y
y
r
y
Y
y
Y
r
r
14
Yr
y
y
R
R
14
yR
3
Fertilization recombines the R and r alleles at random.
On a different chromosome, fertilization recombines
theY and y alleles also. YyRr X YyRr produces 9:3:3:1
ratio.
F2 Generation
3
An F1  F1 cross-fertilization
3
9
:3
:3
:1
So, Mendel’s ratios can be explained by examining
the behavior of the chromosomes during meiosis.


The first solid evidence associating a specific
gene with a specific chromosome came from
Thomas Hunt Morgan, an embryologist.
Morgan’s experiments with fruit flies
provided convincing evidence that
chromosomes are the location of Mendel’s
heritable factors.





Thomas Hunt Morgan chose Drosophila
melanogaster, a common insect that feeds on the
fungi growing on fruit. Why?
They are prolific breeders (a single mating can
produce hundreds of offspring)
They can be bred every
two weeks
It has only 4 pairs of
chromosomes
There are many types
of easily identified mutants
which differ from the normal
(wild) type.


When Thomas Hunt Morgan studied fruit
flies, he was looking for naturally occurring
variants. After years of study, he finally found
one male fruit fly with white eyes instead of
the usual red.
The allele for the mutant trait is written as a
lower case letter (ex: white eyes is w). The
wild-type fly (normal phenotype) is shown
with the same letter with a superscript+: w+
Correlating behavior of an allele with
behavior of the chromosome
EXPERIMENT
P
Generation
F1
Generation

All offspring
had red eyes
In one experiment, Morgan mated a white-eyed male with a red-eyed
female.
All of the F1 offspring had red eyes. What does this tell us about the red
eye trait?
Fig. 15-4b
F2 Results:
RESULTS
F2
Generation
When Morgan bred the F1 flies to each other, he observed the
classical 3:1 phenotypic ratio in the F2 offspring.
However, surprisingly, the white-eyed trait showed up only in the
males! Somehow, the fly’s eye color is related to its sex.
CONCLUSION
P
Generation
X
X
w+

w+
X
Y
w
F1
Generation
Eggs
w+
Sperm
w+
w+
w
w+
Eggs
F2
Generation
w
w+
w+
Sperm
w+
w
w
w
w+


In humans and other mammals, there are
two varieties of sex chromosomes: a larger
X chromosome and a smaller Y
chromosome
Only the ends of the Y chromosome have
regions that are homologous with the X
chromosome
X and Y chromosomes →
The SRY gene on the Y
chromosome codes for the
development of the testes.
44 +
XY
In mammals, the sex
of an offspring
depends on whether
the sperm cell
contains an X
chromosome or a Y.
22 +
22 +
or
X
Y
Sperm
44 +
XX
44 +
XX
Parents
22 +
X
+
Egg
or
44 +
XY
Zygotes (offspring)
(a) The X-Y system
Fig. 15-6b
22 +
XX
22 +
X
(b) The X-0 system
In grasshoppers, cockroaches, and some other insects, there is only one
type of sex chromosome, the X. Females are XX, males have only one
sex chromosome (XO). Sex of the offspring is determined by whether
the sperm cell contains an X chromosome or no sex chromosome.
Fig. 15-6c
76 +
ZW
76 +
ZZ
(c) The Z-W system
In birds, some fishes, and some insects, the sex chromosomes present in
the egg (not the sperm) determine the sex of offspring. The sex
chromosomes are designated Z and W. Females are ZW and males are ZZ.
Fig. 15-6d
32
(Diploid)
16
(Haploid)
(d) The haplo-diploid system
There are no sex chromosomes in most species of bees and ants.
Females develop from fertilized eggs and are thus diploid. Males develop
from unfertilized eggs and are haploid; they have no fathers.



The sex chromosomes have genes for
many characters unrelated to sex
A gene located on either sex chromosome
is called a sex-linked gene
In humans, sex-linked usually refers to a
gene on the larger X chromosome
Inheritance Patterns of Sex Chromosomes.



Sex-linked genes follow specific patterns
of inheritance
For a recessive sex-linked trait to be
expressed
◦ A female needs two copies of the allele
◦ A male needs only one copy of the allele
Sex-linked recessive disorders
are much more common in
males than in females
Fig. 15-7
The Transmission of Sex-linked recessive
traits.
XNXN
Sperm Xn

X nY
XNXn
Sperm XN
Y
Eggs XN
XNXn XNY
XN
XNXn XNY
(a)
A color-blind father
will transmit the
mutant allele to all
daughters but not
to sons. Daughters
are carriers.

XNY
XNXn
Sperm Xn
Y

X nY
Y
Eggs XN XNXN XNY
Eggs XN
XNXn XNY
X n XN X n Y
Xn
XnXn XnY
Xn
(b)
If a carrier mates
with a male who
has normal color
vision, there is a
50% chance the son
will be color-blind.
(c)
If a carrier mates
with a color-blind
male, there is a
50% chance their
child will be colorblind.



In mammalian females, one of the two X
chromosomes in each cell is randomly
inactivated during embryonic development
The inactive X condenses into a Barr body.
The Barr body lies along the inside of the
nuclear envelope. Most of the genes in the
Barr body are not expressed.
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings
If a female is heterozygous for a particular
gene located on the X chromosome, she will
be a mosaic for that character
X chromosomes
Early embryo:
Two cell
populations
in adult cat:
Active X
The tortoiseshell gene is
on the X chromosome in
cats.
The tortoiseshell color
requires the presence of
two alleles: one orange
and one black. These are
located on the X
chromosome.
Allele for
orange fur
Allele for
black fur
Cell division and
X chromosome
inactivation
Active X
Inactive X
Black fur
Orange fur
If a female (XX) is
heterozygous, the
orange and black
patches are present
in populations of
cells with that
activated gene.



Genes located near each other on the same
chromosome tend to be inherited together.
These are called linked genes.
These are not to be confused with sex-linked
traits (traits that come from the sex
chromosomes)


Morgan did other experiments with fruit
flies to see how linkage affects inheritance
of two characters
Morgan crossed flies that differed in traits
of body color and wing size
Wild type—normal wings/color
Ebony body
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings
Vestigial (short) wings


In the fruit fly Drosophila melanogaster, flies
reared in the laboratory occasionally exhibit
mutations in their genes.
Two such mutations, affecting body color and
wing structure are linked. Morgan did
experiments studying these two traits.
Grey body and
normal wings are
dominant traits.
Black body and
vestigial wings are
recessive
Fig. 15-UN1
Heterozygous
for grey body,
normal wings
X
b vg
b+ vg+
Parents
in testcross
Most
offspring
Homozygous
for black body,
vestigial wings

b vg
b vg
b+ vg+
b vg
or
b vg
Grey body,
normal wings
b vg
Black body,
vestigical
wings
EXPERIMENT
P Generation (homozygous)
Wild type
(gray body,
normal wings)
b+ b+ vg+ vg+

Double mutant
(black body,
vestigial wings)
b b vg vg
Fig. 15-9-2
EXPERIMENT
P Generation (homozygous)
Wild type
(gray body,
normal wings)

b b vg vg
b+ b+ vg+ vg+
F1 dihybrid
(wild type)
b+ b vg+ vg
Double mutant
(black body,
vestigial wings)
TESTCROSS

Double mutant
b b vg vg
EXPERIMENT
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)
Double mutant
TESTCROSS

b+ b vg+ vg
Testcross
offspring
b b vg vg
b vg
b+ vg
b vg+
Wild type
(gray-normal)
Blackvestigial
Grayvestigial
Blacknormal
b+ b vg+ vg
b b vg vg b+ b vg vg b b vg+ vg
Eggs
b+ vg+
b vg
Sperm
EXPERIMENT
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)
Double mutant
TESTCROSS

b+ b vg+ vg
Testcross
offspring
b b vg vg
b vg
b+ vg
b vg+
Wild type
(gray-normal)
Blackvestigial
Grayvestigial
Blacknormal
b+ b vg+ vg
b b vg vg b+ b vg vg b b vg+ vg
Eggs
b+ vg+
b vg
Sperm
PREDICTED RATIOS
If genes are located on different chromosomes:
1
:
1
:
1
:
1
If genes are located on the same chromosome and
parental alleles are always inherited together:
1
:
1
:
0
:
0
965
:
944
:
206
:
185
RESULTS


Morgan found that body color and wing
size are usually inherited together in
specific combinations (parental
phenotypes)
He noted that these genes do not assort
independently, and reasoned that they
were on the same chromosome
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings
Genetic Recombination


However, nonparental phenotypes were
also produced
Understanding this result involves
exploring genetic recombination, the
production of offspring with combinations
of traits differing from either parent
Original
phenotypes
Recombinant
phenotypes
Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings

http://bcs.whfreeman.com/thelifewire/conte
nt/chp10/1002002.html



When traits appear that are different from
either one of the parents, it is due to
independent assortment when genes are not
on the same chromosome.
Parental types: resemble the parents
Recombinants: contain new combinations of
genes
If genes are located
on different
chromosomes, there
will be a 50%
recombination rate.
Parental
types
Recombinants
Recombinants
Gametes from yellow-round
heterozygous parent (YyRr)
Gametes from greenwrinkled homozygous
recessive parent ( yyrr)
YR
yr
Yr
yR
YyRr
yyrr
Yyrr
yyRr
yr
Parentaltype
offspring
Recombinant
offspring

Recombinations in traits that are located on
the same chromosome (linked genes) are due
to crossing over.


Alfred Sturtevant, one of Morgan’s students,
constructed a genetic map, an ordered list of the
genetic loci along a particular chromosome
Sturtevant predicted that the farther apart two
genes are, the higher the probability that a
crossover will occur between them and therefore
the higher the recombination frequency
Genetic Linkage Maps



A linkage map is a genetic map of a
chromosome based on recombination
frequencies
Distances between genes can be
expressed as map units; one map unit, or
centimorgan, represents a 1%
recombination frequency
Map units indicate relative distance and
order, not precise
locations of genes
A linkage map from Drosophila:
RESULTS
Recombination
frequencies
9%
Chromosome
9.5%
17%
b
cn
vg
Genes that are far apart on a chromosome can have
recombination frequencies close to 50%.
(These behave almost the same as if they
were on difference chromosomes)


Determine which traits are like the original
parents (parental traits). Determine which
traits are new combinations of the genes
(these are the recombinants)
Figure out the total number of recombinant
offspring and divide by the total number of
offspring X 100 = Recombination %
What is the
recombination
frequencies of
the b and vg
genes?






A wild-type fruit fly heterozygous for gray
body color and normal wings, b+b vg+vg, is
mated with a black fly with vestigial wings
bbvgvg. The offspring have the following
phenotypic distribution:
Wild-type (gray, normal wings): 778
Black-vestigial: 785
Black-normal wings: 158
Gray-vestigial: 162
What is the recombination frequency between
these genes for body color and wing size?




Number recombinants = 320
Total offspring = 1883
Recombinant frequency = 320/1883 X 100 =
17%






Determine the sequence of genes along a
chromosome based on the following
recombination frequencies:
A—B , 8 map units
A—C, 28 map units
A—D, 25 map units
B—C, 20 map units
B—D, 33 map units

D
A
B
C

Large-scale chromosomal alterations often
lead to spontaneous abortions
(miscarriages) or cause a variety of
developmental disorders
Children with
Down’s
Syndrome/
Trisomy 21



In nondisjunction, pairs of homologous
chromosomes do not separate normally
during meiosis
As a result, one gamete receives two of the
same type of chromosome, and another
gamete receives no copy
Offspring with
this condition
is called
aneuploidy
Fig. 15-13-1
Meiosis I
Nondisjunction
(a) Nondisjunction of homologous
chromosomes in meiosis I
(b) Nondisjunction of sister
chromatids in meiosis II
Fig. 15-13-2
Meiosis I
Nondisjunction
Meiosis II
Nondisjunction
(a) Nondisjunction of homologous
chromosomes in meiosis I
(b) Nondisjunction of sister
chromatids in meiosis II
Fig. 15-13-3
Meiosis I
Nondisjunction
Meiosis II
Nondisjunction
Gametes
n+1
n+1
n–1
n–1
n+1
n–1
n
Number of chromosomes
(a) Nondisjunction of homologous
chromosomes in meiosis I
(b) Nondisjunction of sister
chromatids in meiosis II
n
Monosomic, Trisomic Zygotes


A monosomic zygote has only one copy of
a particular chromosome
A trisomic zygote has three copies of a
particular chromosome
Turner’s Syndrome Karyotype:
XO (monosomic zygote)
Down’s Syndrome Karyotype:
Trisomy 21
Polyploidy


Polyploidy is a condition in which an
organism has more than two complete
sets of chromosomes
Polyploidy is common in plants, but not
animals. In plants, it may result in hybrids
that are more vigorous.
Polyploidy can result when a
2N zygote fails to divide after
replicating its chromosomes.
This will produce a 4N
embryo. (Note the hybrid
vigor in the middle plant)

Breakage of a chromosome can lead to
four types of changes in chromosome
structure:
◦
◦
◦
◦
Deletion removes a chromosomal segment
Duplication repeats a segment
Inversion reverses a segment within a chromosome
Translocation moves a segment from one
chromosome to another
Cri du chat results from
a deletion of a portion of
chromosome 5.
(a)
(b)
(c)
(d)
A B C D E
F G H
A B C D E
F G H
A B C D E
F G H
A B C D E
F G H
Deletion
Duplication
A B C E
F G H
A B C B C D E
Inversion
A D C B E
R
F G H
M N O C D E
Reciprocal
translocation
M N O P Q
F G H
A B P Q
R
F G H
Translocation
Normal chromosome 9
Normal chromosome 22
Reciprocal
translocation
Translocated chromosome 9
Translocated chromosome 22
(Philadelphia chromosome)
The cancerous cells in nearly all CML (chromic myelogenous leukemia) patients
contain an abnormally short chromosome 22, the so-called Philadelphia
chromosome, and an abnormally long chromosome 9. These altered
chromosomes result from translocation.