Download video slide - Saginaw Valley State University

Chapter 15:
The Chromosomal Basis of
Inheritance
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 15.1 Chromosomes tagged to reveal a specific
gene (yellow)
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 15.2 The chromosomal basis of Mendel’s laws
P Generation
Starting with two true-breeding pea plants,
we follow two genes through the F 1 and F2
generations. The two genes specify seed
color (allele Y for yellow and allele y for
Y
green) and seed shape (allele R for round
and allele r for wrinkled). These two genes are
on different chromosomes. (Peas have seven
chromosome pairs, but only two pairs are
illustrated here.)
Yellow-round
seeds (YYRR)
Green-wrinkled
seeds (yyrr)
Y
R
r
R
y
r
y
Meiosis
Fertilization
y
R Y
Gametes
r
All F1 plants produce
yellow-round seeds (YyRr)
R
R
y
F1 Generation
y
r
r
Y
Y
Meiosis
LAW OF SEGREGATION
r
R
Y
1 The R and r alleles segregate
R
at anaphase I, yielding
two types of daughter
cells for this locus.
Y
y
R
Y
y
LAW OF INDEPENDENT ASSORTMENT
r
Anaphase I
y
Y
r
1 Alleles at both loci segregate
in anaphase I, yielding four
types of daughter cells
R
depending on the chromosome
arrangement at metaphase I.
Compare the arrangement of
the R and r alleles in the cells
y
on the left and right
r
R
Metaphase II
Y
y
R
R
r
1
YR
4
Y
r
r
1
yr
4
F2 Generation
3 Fertilization
recombines the
R and r alleles
at random.
Y
Y
r
1
yr
4
2 Each gamete gets
a long and a short
chromosome in
one of four allele
combinations.
y
y
Y
Y
Y
Gametes
r
r
R
2 Each gamete
gets one long
chromosome
with either the
R or r allele.
Two equally
probable
arrangements
of chromosomes
at metaphase I
y
y
R
R
1
yR
4
Fertilization among the F1 plants
9
:3
:3
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
:1
3 Fertilization results
in the 9:3:3:1
phenotypic ratio in
the F2 generation.
Figure 15.3 Morgan’s first mutant
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 15.4 In a cross between a wild-type female fruit fly and a mutant
white-eyed male, what color eyes will the F1 and F2 offspring have?
EXPERIMENT Morgan mated a wild-type (red-eyed) female
with a mutant white-eyed male. The F1 offspring all had red eyes.
P
Generation
X
F1
Generation
Morgan then bred an F1 red-eyed female to an F1 red-eyed male to
Produce the F2 generation.
RESULTS
The F2 generation showed a typical Mendelian
3:1 ratio of red eyes to white eyes. However, no females displayed the
white-eye trait; they all had red eyes. Half the males had white eyes,
and half had red eyes.
F2
Generation
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
CONCLUSION Since all F offspring had red eyes, the mutant
1
white-eye trait (w) must be recessive to the wild-type red-eye trait (w+).
Since the recessive trait—white eyes—was expressed only in males in
the F2 generation, Morgan hypothesized that the eye-color gene is
located on the X chromosome and that there is no corresponding locus
on the Y chromosome, as diagrammed here.
P
Generation
W+
X
X
X
X
Y
W+
W+
W
W+
W
Ova
(eggs)
F1
Generation
Sperm
W+
W
W+
Ova
(eggs)
F2
Generation
Sperm
W+
W
W+
W+
W+
W
W
W+
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Unnumbered Figure p. 278
b+
vg+
b vg
X
Parents
in testcross
b vg
b vg
Most
offspring
b+
vg+
b
vg
or
b vg
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
b vg
Unnumbered Figure p. 278
Gametes from yellow-round
heterozygous parent (YyRr)
YR
Gametes from greenwrinkled homozygous
recessive parent (yyrr)
yr
Yr
yR
Yyrr
yyRr
yr
YyRr
Parentaltype
offspring
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
yyrr
Recombinant
offspring
Figure 15.5 Are the genes for body color and wing size in fruit flies
located on the same chromosome or different chromosomes?
EXPERIMENT
Morgan first mated true-breeding wild-type flies with black, vestigial-winged flies to produce
heterozygous F1 dihybrids, all of which are wild-type in appearance. He then mated wild-type F1 dihybrid females with
black, vestigial-winged males, producing 2,300 F2 offspring, which he “scored” (classified according to phenotype).
P Generation
(homozygous)
b b vg vg
F1 dihybrid
(wild type)
(gray body,
normal wings)
b+
b
vg+
Double mutant
(black body,
vestigial wings)
x
Wild type
(gray body,
normal wings)
b+ b+ vg+ vg+
Double mutant
(black body,
vestigial wings)
TESTCROSS
x
b b vg vg
vg
RESULTS
b+vg+
b vg
965
944
Wild type
Black(gray-normal) vestigial
b+ vg
b vg+
206
Grayvestigial
185
Blacknormal
b vg
Sperm
b+ b vg+ vg b b vg vg b+ b vg vgb b vg+ vg
Parental-type
offspring
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Recombinant (nonparental-type)
offspring
CONCLUSION
If these two genes were on different chromosomes, the alleles from the F 1 dihybrid
would sort into gametes independently, and we would expect to see equal numbers of the four types of
offspring. If these two genes were on the same chromosome, we would expect each allele combination,
B+ vg+ and b vg, to stay together as gametes formed. In this case, only offspring with parental phenotypes
would be produced. Since most offspring had a parental phenotype, Morgan concluded that the genes for
body color and wing size are located on the same chromosome. However, the production of a small
number of offspring with nonparental phenotypes indicated that some mechanism occasionally breaks the
linkage between genes on the same chromosome.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 15.6 Chromosomal basis for recombination
of linked genes
Testcross
parents
b+ vg+
Gray body,
normal wings
b vg
(F1 dihybrid)
Replication of
chromosomes
b+
Meiosis I: Crossing
over between b and vg
loci produces new allele
combinations.
Black body,
vestigial wings
b vg (double mutant)
Replication of
chromosomes

vg
b vg
b+ vg+
vg
b
b vg
vg
b
b vg
b vg
Meiosis II: Segregation
of chromatids produces
recombinant gametes
with the new allele
combinations.
Gametes
b vg
Meiosis I and II:
Even if crossing over
occurs, no new allele
combinations are
produced.
Recombinant
chromosome
Ova
Sperm
b+vg+
b vg
b+ vg
b
vg+
b vg
b+ vg+
Testcross
offspring
Sperm
b vg
965
Wild type
(gray-normal)
b vg
944
Blackvestigial
b+ vg
b vg+
206
Grayvestigial
185
Blacknormal
Ova
b+ vg+
b vg+
b+ vg+
Recombination
391 recombinants
=
b vg+ frequency
2,300 total offspring
b vg
b vg
b vg
b vg
Parental-type offspring
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Recombinant offspring
 100 = 17%
Figure 15.7 Constructing a Linkage Map
APPLICATION A linkage map shows the relative locations of
genes along a chromosome.
TECHNIQUE A linkage map is based on the assumption that the
probability of a crossover between two genetic loci is proportional to
the distance separating the loci. The recombination frequencies used
to construct a linkage map for a particular chromosome are obtained
from experimental crosses, such as the cross depicted in Figure 15.6.
The distances between genes are expressed as map units (centimorgans),
with one map unit equivalent to a 1% recombination frequency. Genes are
arranged on the chromosome in the order that best fits the data.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
RESULTS
In this example, the observed recombination frequencies
between three Drosophila gene pairs (bcn 9%, cnvg 9.5%, and bvg
17%) best fit a linear order in which cn is positioned about halfway between
the other two genes:
Recombination
frequencies
9.5%
9%
17%
Chromosome b
cn
vg
The bvg recombination frequency is slightly less than the sum of the bcn
and cnvg frequencies because double crossovers are fairly likely to occur
between b and vg in matings tracking these two genes. A second crossover
would “cancel out” the first and thus reduce the observed bvg
recombination frequency.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 15.8 A partial genetic (linkage) map of a
Drosophila chromosome
I
Y
II
X
IV
III
Mutant phenotypes
Short
aristae
Black
body
0
Long aristae
(appendages
on head)
Cinnabar
eyes
Vestigial
wings
48.5 57.5 67.0
Gray
body
Red
eyes
Normal
wings
Wild-type phenotypes
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Brown
eyes
104.5
Red
eyes
Figure 15.9 Some chromosomal systems of sex determination
44 +
XY
22 +
X
Sperm
44 +
XX
44 +
XX
Parents
22 +
Y
Zygotes
(offspring)
Ova
22 +
XY
44 +
XY
(a) The X-Y system
22 +
XX
22 +
X
76 +
ZW
76 +
ZZ
32
(Diploid)
16
(Haploid)
(b) The X-0 system
(c) The Z-W system
(d) The haplo-diploid system
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 15.10 The transmission of sex-linked recessive traits
XAXA
(a) A father with the disorder will transmit the mutant
allele to all daughters but to no sons. When the
mother is a dominant homozygote, the daughters
will have the normal phenotype but will be carriers
of the mutation.
Ova
Xa
(c) If a carrier mates with a male who has the
disorder, there is a 50% chance that each
child born to them will have the disorder,
regardless of sex. Daughters who do not have
the disorder will be carriers, where as males
without the disorder will be completely free of
the recessive allele.
Y
XAXa XAY
XA
XAYa XAY

XA
Ova
XaY
XA
XAXa
(b) If a carrier mates with a male of normal
phenotype, there is a 50% chance that each
daughter will be a carrier like her mother, and
a 50% chance that each son will have the
disorder.

XAY
Y
XA
XAXA XAY
Xa
XaYA XaY
XAXa 
Sperm
Sperm
XaY
Sperm
Xa
Ova
Y
XA
XAXa XAY
Xa
XaYa XaY
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 15.11 X inactivation and the tortoiseshell cat
Two cell populations
in adult cat:
Active X
Early embryo:
X chromosomes
Cell division
Inactive X
and X
chromosome Inactive X
inactivation
Allele for
orange fur Allele for
black fur
Orange
fur
Black
fur
Active X
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 15.12 Meiotic nondisjunction
Meiosis I
Nondisjunction
Meiosis II
Nondisjunction
Gametes
n+1
n+1
n1
n+1
n –1
n–1
Number of chromosomes
(a) Nondisjunction of homologous
chromosomes in meiosis I
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
n
n
(b) Nondisjunction of sister
chromatids in meiosis II
Figure 15.14 Alterations of chromosome structure
(a) A deletion removes a chromosomal
segment.
(b) A duplication repeats a segment.
(c) An inversion reverses a segment within
a chromosome.
(d) A translocation moves a segment from
one chromosome to another,
nonhomologous one. In a reciprocal
translocation, the most common type,
nonhomologous chromosomes exchange
fragments. Nonreciprocal translocations
also occur, in which a chromosome
transfers a fragment without receiving a
fragment in return.
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
Inversion
A B C E
F G H
A B C B C D E
A D C B E
F G H
M N O C D E
Reciprocal
translocation
M N O P Q
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
R
A B P
Q
F G H
R
F G H
Figure 15.15 Down syndrome
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings