Transcript MF011_fhs_lnt_002b_May11 - MF011 General Biology 2 (May

GENETICS & EVOLUTION:
CHROMOSOMAL INHERITANCE &
MUTATION
Chapter 2.2
Overview




Chromosomal Inheritance
Sex-linked Genes
Gene linkage and analysis
Mutations
 Gene
Mutations
 Chromosomal Abberations
Overview: Locating Genes Along
Chromosomes



Mendel’s “hereditary factors” were genes, though
this wasn’t known at the time
Today we can show that genes are located on
chromosomes
The location of a particular gene can be seen by
tagging isolated chromosomes with a fluorescent dye
that highlights the gene
Fig. 15-1
Mendelian inheritance has its physical
basis in the behavior of chromosomes



Mitosis and meiosis were first described in the late
1800s
The chromosome theory of inheritance states:

Mendelian genes have specific loci (positions) on chromosomes

Chromosomes undergo segregation and independent
assortment
The behavior of chromosomes during meiosis was
said to account for Mendel’s laws of segregation
and independent assortment
Fig. 15-2a
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)
Fig. 15-2b
F1 Generation
All F1 plants produce
yellow-round seeds (YyRr)
0.5 mm
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
r
R
Y
y
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
YR
r
1
3
4
yr
Y
Y
y
r
y
Y
y
Y
r
r
14
Yr
y
y
R
R
14
yR
3
Fig. 15-2c
F2 Generation
An F1  F1 cross-fertilization
3
3
9
:3
:3
:1
Fig. 15-2
P Generation
Yellow-round
seeds (YYRR)
Y
Y
R
r

R
y
Green-wrinkled
seeds ( yyrr)
y
r
Meiosis
Fertilization
y
R Y
Gametes
r
All F1 plants produce
yellow-round seeds (YyRr)
F1 Generation
R
R
y
r
Y
Y
LAW OF SEGREGATION
The two alleles for each gene
separate during gamete
formation.
y
r
LAW OF INDEPENDENT
ASSORTMENT Alleles of genes
on nonhomologous
chromosomes assort
independently during gamete
formation.
Meiosis
R
r
Y
y
r
R
Y
y
Metaphase I
1
1
R
r
Y
y
r
R
Y
y
Anaphase I
R
r
Y
y
Metaphase II
r
R
Y
y
2
2
Y
Y
Gametes
R
1/
4 YR
F2 Generation
R
y
r
Y
Y
y
r
r
1/ yr
4
r
1/
4 Yr
y
R
y
R
1/ yR
4
An F1  F1 cross-fertilization
3
3
9
:3
:3
:1
The Chromosomal Basis of Sex



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
The SRY gene on the Y chromosome codes for the
development of testes
Fig. 15-5
X
Y



Females are XX, and males are XY
Each ovum contains an X chromosome, while a sperm
may contain either an X or a Y chromosome
Other animals have different methods of sex
determination
Fig. 15-6
44 +
XY
44 +
XX
Parents
22 +
22 +
or
X
Y
Sperm
+
44 +
XX
or
22 +
X
Egg
44 +
XY
Zygotes (offspring)
(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
Inheritance of Sex-Linked Genes




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 chromosomeSex-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
XNXN
Sperm Xn

XnY
Sperm XN
Y
Eggs XN
XNXn XNY
XN
XNXn XNY
(a)
XNXn

XNY
XNXn
Sperm Xn
Y

XnY
Y
Eggs XN XNXN XNY
Eggs XN
XNXn XNY
XnXN XnY
Xn
XnXn XnY
Xn
(b)
(c)

Some disorders caused by recessive alleles on the X
chromosome in humans:
 Color
blindness
 Duchenne muscular dystrophy
 Hemophilia
X Inactivation in Female Mammals



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
If a female is heterozygous for a particular gene
located on the X chromosome, she will be a mosaic
for that character
Fig. 15-8
X chromosomes
Early embryo:
Two cell
populations
in adult cat:
Active X
Allele for
orange fur
Allele for
black fur
Cell division and
X chromosome
inactivation
Active X
Inactive X
Black fur
Orange fur
Linked genes tend to be inherited together




Each chromosome has hundreds or thousands of
genes
Genes located on the same chromosome that tend to
be inherited together are called linked genes
Morgan did other experiments with fruit flies to see
how linkage affects inheritance of two characters
Morgan crossed flies that differed in traits of body
color and wing size
How Linkage Affects Inheritance




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
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
Fig. 15-UN1
b vg
b+ vg+
Parents
in testcross
Most
offspring

b vg
b vg
b+ vg+
b vg
or
b vg
b vg
Fig. 15-9-1
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
Fig. 15-9-3
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
Eggs
b+ vg+
Wild type
(gray-normal)
b vg
b+ vg
b vg+
Blackvestigial
Grayvestigial
Blacknormal
b vg
Sperm
b+ b vg+ vg
b b vg vg b+ b vg vg b b vg+ vg
Fig. 15-9-4
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
Eggs
b+ vg+
Wild type
(gray-normal)
b vg
b+ vg
b vg+
Blackvestigial
Grayvestigial
Blacknormal
b vg
Sperm
b b vg vg b+ b vg vg b b vg+ vg
b+ b vg+ vg
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
Genetic Recombination and Linkage

The genetic findings of Mendel and Morgan relate
to the chromosomal basis of recombination
Recombination of Unlinked Genes:
Independent Assortment of Chromosomes




Mendel observed that combinations of traits in some
offspring differ from either parent
Offspring with a phenotype matching one of the
parental phenotypes are called parental types
Offspring with nonparental phenotypes (new
combinations of traits) are called recombinant
types, or recombinants
A 50% frequency of recombination is observed for
any two genes on different chromosomes
Fig. 15-UN2
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
Recombination of Linked Genes: Crossing
Over



Morgan discovered that genes can be linked, but the
linkage was incomplete, as evident from recombinant
phenotypes
Morgan proposed that some process must sometimes
break the physical connection between genes on the
same chromosome
That mechanism was the crossing over of
homologous chromosomes
Fig. 15-10
Testcross
parents
Gray body, normal wings
(F1 dihybrid)
Black body, vestigial wings
(double mutant)
b+ vg+
b vg
b vg
b vg
Replication
of chromosomes
Replication
of chromosomes
Meiosis I
b+ vg+
b vg
b+ vg+
b vg
b vg
b vg
b vg
b vg
b+ vg+
Meiosis I and II
b+ vg
b vg+
b vg
Meiosis II
Recombinant
chromosomes
Eggs
Testcross
offspring
b vg
b+ vg+
b+ vg
b vg+
965
944
206
185
Wild type
(gray-normal)
Blackvestigial
Grayvestigial
Blacknormal
b+ vg+
b vg
b+ vg
b vg+
b vg
b vg
b vg
b vg
Parental-type offspring
Recombination
frequency
=
Recombinant offspring
391 recombinants
2,300 total offspring
 100 = 17%
b vg
Sperm
Fig. 15-10a
Testcross
parents
Black body, vestigial wings
(double mutant)
Gray body, normal wings
(F1 dihybrid)
Replication
of chromosomes
Meiosis I
b+ vg+
b vg
b vg
b vg
b+ vg+
b vg
b+ vg+
b vg
b vg
b vg
b vg
b vg
b+ vg+
b+
Meiosis I and II
vg
b vg+
b vg
Meiosis II
Recombinant
chromosomes
b+ vg+
b vg
Eggs
b+ vg
b vg+
b vg
Sperm
Replication
of chromosomes
Fig. 15-10b
Recombinant
chromosomes
Eggs
Testcross
offspring
b+ vg+
965
Wild type
(gray-normal)
b vg
944
Blackvestigial
b+ vg
206
Grayvestigial
b vg+
185
Blacknormal
b+ vg+
b vg
b+ vg
b vg+
b vg
b vg
b vg
b vg
Parental-type offspring
Recombination
frequency
=
Recombinant offspring
391 recombinants
2,300 total offspring
 100 = 17%
b vg
Sperm
Mapping the Distance Between Genes
Using Recombination Data


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”



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
Fig. 15-11
RESULTS
Recombination
frequencies
9%
Chromosome
9.5%
17%
b
cn
vg





Genes that are far apart on the same chromosome
can have a recombination frequency near 50%
Such genes are physically linked, but genetically
unlinked, and behave as if found on different
chromosomes
Sturtevant used recombination frequencies to make
linkage maps of fruit fly genes
Using methods like chromosomal banding, geneticists
can develop cytogenetic maps of chromosomes
Cytogenetic maps indicate the positions of genes
with respect to chromosomal features
Fig. 15-12
Short
aristae
0
Long aristae
(appendages
on head)
Mutant phenotypes
Black
body
48.5
Gray
body
Cinnabar
eyes
57.5
Red
eyes
Vestigial
wings
67.0
Normal
wings
Wild-type phenotypes
Brown
eyes
104.5
Red
eyes
Abnormal Chromosome Number



Large-scale chromosomal alterations often lead to
spontaneous abortions (miscarriages) or cause a variety
of developmental disorders
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
Fig. 13-8a
Prophase I
Metaphase I
Centrosome
(with centriole pair)
Sister
chromatids
Chiasmata
Spindle
Sister chromatids
remain attached
Centromere
(with kinetochore)
Metaphase
plate
Homologous
chromosomes
separate
Homologous
chromosomes
Fragments
of nuclear
envelope
Telophase I and
Cytokinesis
Anaphase I
Microtubule
attached to
kinetochore
Cleavage
furrow
Fig. 13-8d
Prophase II
Metaphase II
Anaphase II
Telophase II and
Cytokinesis
Sister chromatids
separate
Haploid daughter cells
forming
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




Aneuploidy results from the fertilization of gametes
in which nondisjunction occurred
Offspring with this condition have an abnormal
number of a particular chromosome
A monosomic zygote has only one copy of a
particular chromosome
A trisomic zygote has three copies of a particular
chromosome

Polyploidy is a condition in which an organism has
more than two complete sets of chromosomes
 Triploidy
(3n) is three sets of chromosomes
 Tetraploidy (4n) is four sets of chromosomes


Polyploidy is common in plants, but not animals
Polyploids are more normal in appearance than
aneuploids
Fig. 15-14
Mutagens



Spontaneous mutations can occur during DNA
replication, recombination, or repair
Mutagens are physical or chemical agents that can
cause mutations
Mutations are changes in the genetic material of a
cell or virus
 Gene
mutation
 Chromosomal abberation
Causes of Mutations

Spontaneous mutation
DNA can undergo a chemical change
 Movement of transposons from one chromosomal location to
another
 Replication Errors



1 in 1,000,000,000 replications
DNA polymerase



Proofreads new strands
Generally corrects errors
Induced mutation:
Mutagens such as radiation, organic chemicals
 Many mutagens are also carcinogens (cancer causing)
 Environmental Mutagens



Ultraviolet Radiation
Tobacco Smoke
Effect of Mutations on Protein Activity

Point Mutations
Involve change in a single DNA nucleotide
 Changes one codon to a different codon
 Affects on protein vary:





Nonfunctional
Reduced functionality
Unaffected
Frameshift Mutations
One or two nucleotides are either inserted or deleted from
DNA
 Protein always rendered nonfunctional




Normal :
After deletion:
After insertion:
THE CAT ATE THE RAT
THE ATA TET HER AT
THE CCA TAT ETH ERA T
Point mutations can affect protein
structure and function


Point mutations are chemical changes in just one
base pair of a gene
The change of a single nucleotide in a DNA template
strand can lead to the production of an abnormal
protein
Fig. 17-22
Wild-type hemoglobin DNA
C T T
3
5
G A A
Mutant hemoglobin DNA
C A T
5 3
G T A
3 5
mRNA
5
5
3
mRNA
G A A
Normal hemoglobin
Glu
3
5
G U A
Sickle-cell hemoglobin
Val
3
Types of Point Mutations

Point mutations within a gene can be divided into
two general categories
 Base-pair
substitutions
 Base-pair insertions or deletions
Fig. 17-23
Wild-type
DNA template strand
3
5
5
3
mRNA 5
3
Protein
Stop
Amino end
Carboxyl end
A instead of G
3
5
Extra A
5
3
3
5
3
5
U instead of C
5
5
3
Extra U
3
Stop
Stop
Silent (no effect on amino acid sequence)
Frameshift causing immediate nonsense (1 base-pair insertion)
T instead of C
missing
3
5
5
3
3
5
3
5
5
3
A instead of G
missing
5
3
Stop
Missense
Frameshift causing extensive missense (1 base-pair deletion)
missing
A instead of T
5
3
3
5
U instead of A
5
5
3
3
5
missing
3
5
Stop
Stop
Nonsense
(a) Base-pair substitution
3
No frameshift, but one amino acid missing (3 base-pair deletion)
(b) Base-pair insertion or deletion
Fig. 17-23a
Wild type
DNA template 3
strand 5
5
3
mRNA 5
3
Protein
Stop
Amino end
Carboxyl end
A instead of G
5
3
3
5
U instead of C
5
3
Stop
Silent (no effect on amino acid sequence)
Fig. 17-23b
Wild type
DNA template 3
strand 5
5
3
mRNA 5
3
Protein
Stop
Amino end
Carboxyl end
T instead of C
5
3
3
5
A instead of G
3
5
Stop
Missense
Fig. 17-23c
Wild type
DNA template 3
strand 5
5
3
mRNA 5
3
Protein
Stop
Amino end
Carboxyl end
A instead of T
3
5
5
3
U instead of A
5
3
Stop
Nonsense
Fig. 17-23d
Wild type
DNA template 3
strand 5
5
3
mRNA 5
3
Protein
Stop
Amino end
Carboxyl end
Extra A
5
3
3
5
Extra U
5
3
Stop
Frameshift causing immediate nonsense (1 base-pair insertion)
Fig. 17-23e
Wild type
DNA template 3
strand 5
5
3
mRNA 5
3
Protein
Stop
Amino end
Carboxyl end
missing
5
3
3
5
missing
5
Frameshift causing extensive missense (1 base-pair deletion)
3
Fig. 17-23f
Wild type
DNA template 3
strand 5
5
3
mRNA 5
3
Protein
Stop
Amino end
Carboxyl end
missing
5
3
3
5
missing
5
3
Stop
No frameshift, but one amino acid missing (3 base-pair deletion)
Substitutions




A base-pair substitution replaces one nucleotide
and its partner with another pair of nucleotides
Silent mutations have no effect on the amino acid
produced by a codon because of redundancy in the
genetic code
Missense mutations still code for an amino acid, but
not necessarily the right amino acid
Nonsense mutations change an amino acid codon
into a stop codon, nearly always leading to a
nonfunctional protein
Insertions and Deletions



Insertions and deletions are additions or losses of
nucleotide pairs in a gene
These mutations have a disastrous effect on the
resulting protein more often than substitutions do
Insertion or deletion of nucleotides may alter the
reading frame, producing a frameshift mutation
Alterations of Chromosome Structure

Breakage of a chromosome can lead to four types
of changes in chromosome structure:
 Deletion
removes a chromosomal segment
 Duplication repeats a segment
 Inversion reverses a segment within a chromosome
 Translocation moves a segment from one chromosome to
another
Fig. 15-15
(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
Inversion
A B C E
F G H
A B C B C D E
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
Human Disorders Due to Chromosomal
Alterations



Alterations of chromosome number and structure are
associated with some serious disorders
Some types of aneuploidy appear to upset the
genetic balance less than others, resulting in
individuals surviving to birth and beyond
These surviving individuals have a set of symptoms,
or syndrome, characteristic of the type of
aneuploidy
Down Syndrome (Trisomy 21)



Down syndrome is an aneuploid condition that
results from three copies of chromosome 21
It affects about one out of every 700 children born
in the United States
The frequency of Down syndrome increases with the
age of the mother, a correlation that has not been
explained
Fig. 15-16
Aneuploidy of Sex Chromosomes



Nondisjunction of sex chromosomes produces a
variety of aneuploid conditions
Klinefelter syndrome is the result of an extra
chromosome in a male, producing XXY individuals
Monosomy X, called Turner syndrome, produces X0
females, who are sterile; it is the only known viable
monosomy in humans
Changes in Sex Chromosome
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
a. Turner syndrome
b. Klinefelter syndrome
a: Courtesy UNC Medical Illustration and Photography; b: Courtesy Stefan D. Schwarz, http://klinefeltersyndrome.org
Disorders Caused by Structurally Altered
Chromosomes



The syndrome cri du chat (“cry of the cat”), results
from a specific deletion in chromosome 5
A child born with this syndrome is mentally retarded
and has a catlike cry; individuals usually die in
infancy or early childhood
Certain cancers, including chronic myelogenous
leukemia (CML), are caused by translocations of
chromosomes
Fig. 15-17
Normal chromosome 9
Normal chromosome 22
Reciprocal
translocation
Translocated chromosome 9
Translocated chromosome 22
(Philadelphia chromosome)
Some inheritance patterns are exceptions
to the standard chromosome theory


There are two normal exceptions to Mendelian
genetics
One exception involves genes located in the nucleus,
and the other exception involves genes located
outside the nucleus
Genomic Imprinting



For a few mammalian traits, the phenotype depends
on which parent passed along the alleles for those
traits
Such variation in phenotype is called genomic
imprinting
Genomic imprinting involves the silencing of certain
genes that are “stamped” with an imprint during
gamete production
Fig. 15-18a
Paternal
chromosome
Normal Igf2 allele
is expressed
Maternal
chromosome
Normal Igf2 allele
is not expressed
(a) Homozygote
Wild-type mouse
(normal size)
Fig. 15-18b
Mutant Igf2 allele
inherited from mother
Mutant Igf2 allele
inherited from father
Normal size mouse
(wild type)
Dwarf mouse
(mutant)
Normal Igf2 allele
is expressed
Mutant Igf2 allele
is expressed
Mutant Igf2 allele
is not expressed
Normal Igf2 allele
is not expressed
(b) Heterozygotes
Fig. 15-18
Paternal
chromosome
Normal Igf2 allele
is expressed
Maternal
chromosome
Normal Igf2 allele
is not expressed
Wild-type mouse
(normal size)
(a) Homozygote
Mutant Igf2 allele
inherited from mother
Mutant Igf2 allele
inherited from father
Normal size mouse
(wild type)
Dwarf mouse
(mutant)
Normal Igf2 allele
is expressed
Mutant Igf2 allele
is expressed
Mutant Igf2 allele
is not expressed
Normal Igf2 allele
is not expressed
(b) Heterozygotes



It appears that imprinting is the result of the
methylation (addition of –CH3) of DNA
Genomic imprinting is thought to affect only a small
fraction of mammalian genes
Most imprinted genes are critical for embryonic
development
Fig. 15-UN3
Inheritance of Organelle Genes




Extranuclear genes (or cytoplasmic genes) are genes
found in organelles in the cytoplasm
Mitochondria, chloroplasts, and other plant plastids
carry small circular DNA molecules
Extranuclear genes are inherited maternally because
the zygote’s cytoplasm comes from the egg
The first evidence of extranuclear genes came from
studies on the inheritance of yellow or white patches
on leaves of an otherwise green plant

Some defects in mitochondrial genes prevent cells
from making enough ATP and result in diseases that
affect the muscular and nervous systems
 For
example, mitochondrial myopathy and Leber’s
hereditary optic neuropathy
You should now be able to:
1.
2.
3.
4.
5.
Explain the chromosomal theory of inheritance and
its discovery
Explain why sex-linked diseases are more common
in human males than females
Distinguish between sex-linked genes and linked
genes
Explain how meiosis accounts for recombinant
phenotypes
Explain how linkage maps are constructed
6.
7.
8.
9.
10.
11.
12.
Explain how nondisjunction can lead to aneuploidy
Define trisomy, triploidy, and polyploidy
Define mutation
Distinguish between different gene mutations
Distinguish among deletions, duplications, inversions,
and translocations
Explain genomic imprinting
Explain why extranuclear genes are not inherited in
a Mendelian fashion
Morgan’s Experimental Evidence
Imagine that Morgan had used a grasshopper
(2N = 24 and an XX, XO sex determination
system). Predict where the first mutant would
have been discovered.
a.
b.
c.
d.
on the O chromosome of a male
on the X chromosome of a male
on the X chromosome of a female
on the Y chromosome of a male
Copyright © 2008 Pearson
Education, Inc., publishing as
Pearson Benjamin Cummings.
The Chromosomal Basis of Sex
Think about bees and ants, groups in which males are haploid.
Which of the following are accurate statements about bee and ant
males when they are compared to species in which males are XY
and diploid for the autosomes?
a.
b.
c.
d.
e.
Bee males have half the DNA of bee females whereas human
males have nearly the same amount of DNA that human females
have.
Considered across the genome, harmful (deleterious) recessives will
negatively affect bee males more than Drosophila males.
Human and Drosophila males have sons but bee males do not.
Inheritance in bees is like inheritance of sex-linked characteristics in
humans.
none of the above
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Education, Inc., publishing as
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The Chromosomal Basis of Sex
In some Drosophila species there are genes on the Y chromosome that
do not occur on the X chromosome. Imagine that a mutation of one
gene on the Y chromosome reduces the size by half of individuals with
the mutation. Which of the following statements is accurate with
regard to this situation?





This mutation occurs in all offspring of a male with the mutation.
This mutation occurs in all male but no female offspring of a male
with the mutation.
This mutation occurs in all offspring of a female with the mutation.
This mutation occurs in all male but no female offspring of a female
with the mutation.
This mutation occurs in all offspring of both males and females with
the mutation.
Copyright © 2008 Pearson
Education, Inc., publishing as
Pearson Benjamin Cummings.
The Chromosomal Basis of Sex
Imagine that a deleterious recessive allele occurs on the W
chromosome of a chicken (2N = 78). Where would it be most likely
to appear first in a genetics experiment?
a.
b.
c.
d.
e.
in a male because there is no possibility of the presence
of a normal, dominant allele
in a male because it is haploid
in a female because there is no possibility of the
presence of a normal, dominant allele
in a female because all alleles on the W chromosomes
are dominant to those on the Z chromosome
none of the above
Copyright © 2008 Pearson
Education, Inc., publishing as
Pearson Benjamin Cummings.
Inheritance of Sex-Linked Genes
In cats, a sex-linked gene affects coat color. The O allele produces an enzyme
that converts eumelanin, a black or brown pigment, into phaeomelanin, an
orange pigment. The o allele is recessive to O and produces a defective
enzyme, one that does not convert eumelanin into phaeomelanin. Which of the
following statements is/are accurate?
a.
b.
c.
d.
e.
The phenotype of O-Y males is orange because the functional allele O converts
eumelanin into phaeomelanin.
The phenotype of o-Y males is black/brown because the non-functional allele o
does not convert eumelanin into phaeomelanin.
The phenotype of OO and Oo males is orange because the functional allele O
converts eumelanin into phaeomelanin.
The phenotype of Oo males is mixed orange and black/brown because the
functional allele O converts eumelanin into phaeomelanin in some cell groups
(orange) and because in other cell groups the non-functional allele o does not
convert eumelanin into phaeomelanin.
The phenotype of O-Y males is orange because the non-functional allele O does
not convert eumelanin into phaeomelanin while the phenotype of o-Y males is
black/brown because the functional allele o converts eumelanin into phaeomelanin.
Copyright © 2008 Pearson
Education, Inc., publishing as
Pearson Benjamin Cummings.
Inheritance of Sex-Linked Genes
In cats, a sex-linked gene affects coat color. The O allele produces an enzyme
that converts eumelanin, a black or brown pigment, into phaeomelanin, an
orange pigment. The o allele is recessive to O and produces a defective
enzyme, one that does not convert eumelanin into phaeomelanin. Which of the
following statements is/are accurate?
a.
b.
c.
d.
e.
The phenotype of O-Y females is orange because the functional allele O converts
eumelanin into phaeomelanin.
The phenotype of o-Y females is black/brown because the non-functional allele o does
not convert eumelanin into phaeomelanin.
The phenotype of OO and Oo females is orange because the functional allele O
converts eumelanin into phaeomelanin.
The phenotype of Oo females is mixed orange and black/brown because the functional
allele O converts eumelanin into phaeomelanin in some cell groups (orange) and
because in other cell groups the non-functional allele o does not convert eumelanin into
phaeomelanin.
The phenotype of O-Y females is orange because the non-functional allele O does not
convert eumelanin into phaeomelanin while the phenotype of o-Y males is black/brown
because the functional allele o converts eumelanin into phaeomelanin.
Copyright © 2008 Pearson
Education, Inc., publishing as
Pearson Benjamin Cummings.
X Inactivation in Female Mammals
Imagine two species of mammals that differ in the timing of Barr body formation during
development. Both species have genes that determine coat color, O for the dominant
orange fur and o for the recessive black/brown fur, on the X chromosome. In species A,
the Barr body forms during week 1 of a 6-month pregnancy whereas in species B, the
Barr body forms during week 3 of a 5-month pregnancy. What would you predict
about the coloration of heterozygous females (Oo) in the two species?
a.
b.
c.
Both species will have similar sized patches of orange
and black/brown fur.
Species A will have smaller patches of orange or
black/brown fur than will species B.
The females of both species will show the dominant fur
color, orange.
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Education, Inc., publishing as
Pearson Benjamin Cummings.
Mapping the Distance between Genes
Imagine a species with three loci thought to be on the same chromosome. The
recombination rate between locus A and locus B is 35% and the recombination
rate between locus B and locus C is 33%. Predict the recombination rate
between A and C.
a.
b.
c.
d.
e.
The recombination rate between locus A and locus C is either 2% or
68%.
The recombination rate between locus A and locus C is probably
2%.
The recombination rate between locus A and locus C is either 2% or
50%.
The recombination rate between locus A and locus C is either 2% or
39%.
The recombination rate between locus A and locus C cannot be
predicted.
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Pearson Benjamin Cummings.
Triploid species are usually sterile (unable to
reproduce) whereas tetraploids are often fertile.
Which of the following are likely good explanations
of these facts?
a.
b.
c.
d.
In mitosis, some chromosomes in triploids have no partner at
synapsis, but chromosomes in tetraploids do have partners.
In meiosis, some chromosomes in triploids have no partner at
synapsis, but chromosomes in tetraploids do have partners.
In mitosis, some chromosomes in tetraploids have no partner at
synapsis, but chromosomes in triploids do have partners.
In meiosis, some chromosomes in tetraploids have no partner at
synapsis, but chromosomes in triploids do have partners.
Copyright © 2008 Pearson
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Pearson Benjamin Cummings.
Chromosomal rearrangements can occur after
chromosomes break. Which of the following
statements are most accurate with respect to
alterations in chromosome structure?
a.
b.
c.
d.
Chromosomal rearrangements are more likely to occur in mammals
than in other vertebrates.
Translocations and inversions are not deleterious because no genes
are lost in the organism.
Chromosomal rearrangements are more likely to occur during
mitosis than during meiosis.
An individual that is homozygous for a deletion of a certain gene is
likely to be more damaged than is one that is homozygous for a
duplication of that same gene because loss of a function can be
lethal.
Copyright © 2008 Pearson
Education, Inc., publishing as
Pearson Benjamin Cummings.
Imagine that you could create medical policy for a country. In this country it is known that
the frequency of Down syndrome babies increases with increasing age of the mother
and that the severity of characteristics varies enormously and unpredictably among
affected individuals. Furthermore, financial resources are severely limited, both for
testing of pregnant women and for supplemental training of Down syndrome children.
The graph on the next slide shows the incidence of Down syndrome as a function of
maternal age. Which of the following policies would you implement?
a.
No testing of pregnant women should be conducted and all the health care money
should be used for training of Down syndrome children.
b.
The health care system should provide testing only for women over 30.
c.
The health care system should provide testing only for women over 40.
d.
The health care system should require termination of all Down syndrome fetuses.
e.
The health care system should provide training for the 30% most seriously affected
children only.
Copyright © 2008 Pearson
Education, Inc., publishing as
Pearson Benjamin Cummings.
Copyright © 2008 Pearson
Education, Inc., publishing as
Pearson Benjamin Cummings.
The lawyer for a defendant in a paternity suit asked for DNA
testing of a baby girl. Which of the following set of results
would demonstrate that the purported father was not actually
the genetic father of the child?
a.
b.
c.
d.
e.
The mitochondrial DNA of the child and “father” did not match.
DNA sequencing of chromosome #5 of the child and “father” did
not match.
The mitochondrial DNA of the child and “mother” did not match.
DNA sequencing of chromosome #5 of the child and “mother” did
not match.
The mitochondrial DNA of the child and “father” matched but the
mitochondrial DNA of the child and “mother” did not.
Copyright © 2008 Pearson
Education, Inc., publishing as
Pearson Benjamin Cummings.