Post on 16-Dec-2015
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
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
P Generation
Gametes
Meiosis
Fertilization
Yellow-roundseeds (YYRR)
Green-wrinkledseeds ( yyrr)
All F1 plants produceyellow-round seeds (YyRr)
y
y
y
rr
r
Y
Y
YR
RR
Fig. 15-2b
0.5 mm
Meiosis
Metaphase I
Anaphase I
Metaphase II
Gametes
LAW OF SEGREGATIONThe two alleles for each gene separate during gamete formation.
LAW OF INDEPENDENTASSORTMENT Alleles of genes on nonhomologous chromosomes assort independently during gamete formation.
14
yr 1 4Yr1
4 YR
3 3
F1 Generation
1 4yR
R
R
R
R
RR
R
R R R R
R
Y
Y
Y Y
Y
YY
Y
YY
YY
yr r
rr
r r
rr
r r r r
y
y
y
y
y
y y
yyyy
All F1 plants produceyellow-round seeds (YyRr)
1
2 2
1
Fig. 15-2P Generation Yellow-round
seeds (YYRR)
Y
F1 Generation
Y
R R
R Y
r
r
r
y
y
y
Meiosis
Fertilization
Gametes
Green-wrinkledseeds ( yyrr)
All F1 plants produceyellow-round seeds (YyRr)
R R
YY
r ry y
Meiosis
R R
Y Y
r r
y y
Metaphase I
Y Y
R Rrr
y y
Anaphase I
r r
y Y
Metaphase IIR
Y
R
y
yyy
RR
YY
rrrr
yYY
R R
yRYryrYR1/41/4
1/41/4
F2 Generation
Gametes
An F1 F1 cross-fertilization
9 : 3 : 3 : 1
LAW OF INDEPENDENTASSORTMENT Alleles of geneson nonhomologouschromosomes assortindependently during gameteformation.
LAW OF SEGREGATIONThe two alleles for each geneseparate during gameteformation.
1
2
33
2
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
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 Parents 44 +
XX
22 + X
22 +X
22 + Yor +
44 + XX or
Sperm Egg
44 + XY
Zygotes (offspring)
(a) The X-Y system
22 + XX
22 + X
(b) The X-0 system
76 + ZW
76 + ZZ
(c) The Z-W system
32(Diploid)
16(Haploid)
(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
(a) (b) (c)
XNXN XnY XNXn XNY XNXn XnY
YXnSpermYXNSpermYXnSperm
XNXnEggsXN
XN XNXn
XNY
XNY
EggsXN
Xn
XNXN
XnXN
XNY
XnY
EggsXN
Xn
XNXn
XnXn
XNY
XnY
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. 15-8X chromosomes
Early embryo:
Allele fororange fur
Allele forblack fur
Cell division andX chromosomeinactivationTwo cell
populationsin adult cat:
Active XActive X
Inactive X
Black fur Orange fur
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
Linked genes tend to be inherited together
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. 15-9-1
EXPERIMENTP Generation (homozygous)
Wild type(gray body,normal wings)
Double mutant(black body,vestigial wings)
b b vg vg b+ b+ vg+ vg+
Fig. 15-9-2
EXPERIMENTP Generation (homozygous)
Wild type(gray body,normal wings)
Double mutant(black body,vestigial wings)
b b vg vg
b b vg vg
Double mutantTESTCROSS
b+ b+ vg+ vg+
F1 dihybrid(wild type)
b+ b vg+ vg
Fig. 15-9-3
EXPERIMENTP Generation (homozygous)
Wild type(gray body,normal wings)
Double mutant(black body,vestigial wings)
b b vg vg
b b vg vg
Double mutantTESTCROSS
b+ b+ vg+ vg+
F1 dihybrid(wild type)
b+ b vg+ vg
Testcrossoffspring Eggs b+ vg+ b vg b+ vg b vg+
Black-normal
Gray-vestigial
Black-vestigial
Wild type(gray-normal)
b vg
Sperm
b+ b vg+ vg b b vg vg b+ b vg vg b b vg+ vg
Fig. 15-9-4
EXPERIMENTP Generation (homozygous)
RESULTS
Wild type(gray body,normal wings)
Double mutant(black body,vestigial wings)
b b vg vg
b b vg vg
Double mutantTESTCROSS
b+ b+ vg+ vg+
F1 dihybrid(wild type)
b+ b vg+ vg
Testcrossoffspring Eggs b+ vg+ b vg b+ vg b vg+
Black-normal
Gray-vestigial
Black-vestigial
Wild type(gray-normal)
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:
If genes are located on the same chromosome andparental alleles are always inherited together:
1
1
1
1
1 1
0 0
965 944 206 185
:
:
:
:
:
:
:
:
:
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
YyRr
Gametes from green-wrinkled homozygousrecessive parent ( yyrr)
Gametes from yellow-roundheterozygous parent (YyRr)
Parental-type
offspring
Recombinantoffspring
yr
yyrr Yyrr yyRr
YR yr Yr yR
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. 15-10Testcrossparents
Replicationof chromo-somes
Gray body, normal wings(F1 dihybrid)
Black body, vestigial wings(double mutant)
Replicationof chromo-somes
b+ vg+
b+ vg+
b+ vg+
b vg
b vg
b vg
b vg
b vg
b vg
b vgb vg
b vg
b+ vg+
b+ vg
b vg+
b vg
Recombinantchromosomes
Meiosis I and II
Meiosis I
Meiosis II
b vg+b+ vgb vgb+ vg+
Eggs
Testcrossoffspring
965Wild type
(gray-normal)
944Black-
vestigial
206Gray-
vestigial
185Black-normal
b+ vg+
b vg b vg
b vg b+ vg
b vg b vg
b vg+
Sperm
b vg
Parental-type offspringRecombinant offspring
Recombinationfrequency =
391 recombinants2,300 total offspring
100 = 17%
Fig. 15-10aTestcrossparents
Replicationof chromo-somes
Gray body, normal wings(F1 dihybrid)
Black body, vestigial wings(double mutant)
Replicationof chromo-somes
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+ vg
b vg+
b vg
Recombinantchromosomes
Meiosis I and II
Meiosis I
Meiosis II
Eggs Sperm
b+ vg+ b vg b+ vg b vgb vg+
Fig. 15-10b
Testcrossoffspring
965Wild type
(gray-normal)
944Black-
vestigial
206Gray-
vestigial
185Black-normal
b+ vg+
b vg b vg
b vg b+ vg
b vg
b vg
b+ vg+
Spermb vg
Parental-type offspringRecombinant offspring
Recombinationfrequency =
391 recombinants2,300 total offspring
100 = 17%
b vg
b+ vg b vg+
b vg+
Eggs
Recombinantchromosomes
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
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
Mutant phenotypes
Shortaristae
Blackbody
Cinnabareyes
Vestigialwings
Browneyes
Redeyes
Normalwings
Redeyes
Graybody
Long aristae(appendageson head)
Wild-type phenotypes
0 48.5 57.5 67.0 104.5
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. 15-13-1
Meiosis I
(a) Nondisjunction of homologous chromosomes in meiosis I
(b) Nondisjunction of sister chromatids in meiosis II
Nondisjunction
Fig. 15-13-2
Meiosis I
Nondisjunction
(a) Nondisjunction of homologous chromosomes in meiosis I
(b) Nondisjunction of sister chromatids in meiosis II
Meiosis II
Nondisjunction
Fig. 15-13-3
Meiosis I
Nondisjunction
(a) Nondisjunction of homologous chromosomes in meiosis I
(b) Nondisjunction of sister chromatids in meiosis II
Meiosis II
Nondisjunction
Gametes
Number of chromosomes
n + 1 n + 1 n + 1n – 1 n – 1 n – 1 n 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
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 : THE CAT ATE THE RAT After deletion:THE ATA TET HER AT After insertion: 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
mRNA
Mutant hemoglobin DNA
mRNA
33
3
3
3
3
55
5
55
5
C CT T TTG GA A AA
A A AGG U
Normal hemoglobin Sickle-cell hemoglobin
Glu Val
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. 17-23Wild-type
3DNA template strand5
5
53
3
Stop
Carboxyl endAmino end
Protein
mRNA
33
3
55
5
A instead of G
U instead of C
Silent (no effect on amino acid sequence)
Stop
T instead of C
33
3
55
5
A instead of G
Stop
Missense
A instead of T
U instead of A
33
3
5
5
5
Stop
Nonsense No frameshift, but one amino acid missing (3 base-pair deletion)
Frameshift causing extensive missense (1 base-pair deletion)
Frameshift causing immediate nonsense (1 base-pair insertion)
5
5
533
3
Stop
missing
missing
3
3
3
5
55
missing
missing
Stop
5
5533
3
Extra U
Extra A
(a) Base-pair substitution (b) Base-pair insertion or deletion
Fig. 17-23a
Wild type
3DNA templatestrand
3
355
5mRNA
Protein
Amino end
Stop
Carboxyl end
A instead of G
33
3
U instead of C
55
5
Stop
Silent (no effect on amino acid sequence)
Fig. 17-23b
Wild type
DNA templatestrand
35
mRNA
Protein
5
Amino end
Stop
Carboxyl end
53
3
T instead of C
A instead of G
33
3
5
5
5
Stop
Missense
Fig. 17-23cWild type
DNA templatestrand
35
mRNA
Protein
5
Amino end
Stop
Carboxyl end
53
3
A instead of T
U instead of A
33
3
5
5
5
Stop
Nonsense
Fig. 17-23d
Wild type
DNA templatestrand
35
mRNA
Protein
5
Amino end
Stop
Carboxyl end
53
3
Extra A
Extra U
33
3
5
5
5
Stop
Frameshift causing immediate nonsense (1 base-pair insertion)
Fig. 17-23e
Wild type
DNA templatestrand
35
mRNA
Protein
5
Amino end
Stop
Carboxyl end
53
3
missing
missing
33
3
5
5
5
Frameshift causing extensive missense (1 base-pair deletion)
Fig. 17-23fWild type
DNA templatestrand
35
mRNA
Protein
5
Amino end
Stop
Carboxyl end
53
3
missing
missing
33
3
5
5
5
No frameshift, but one amino acid missing (3 base-pair deletion)
Stop
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
DeletionA B C D E F G H A B C E F G H(a)
(b)
(c)
(d)
Duplication
Inversion
Reciprocaltranslocation
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 B C D E F G H
A D C B E F G H
M N O C D E F G H
M N O P Q R A B P Q R
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
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
a. Turner syndrome b. Klinefelter syndrome
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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
Reciprocaltranslocation Translocated chromosome 9
Translocated chromosome 22(Philadelphia chromosome)
69
Carcinogenesis
Development of cancer involves a series of mutations
Proto-oncogenes – Stimulate cell cycle
Tumor suppressor genes – inhibit cell cycle
Mutation in oncogene and tumor suppressor gene:
Stimulates cell cycle uncontrollably
Leads to tumor formation
70
Cell Signaling Pathway
Cell signaling pathway that stimulates a mutated tumor suppressor gene
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
receptor
inhibiting growth factor
cytoplasm
plasmamembrane
signaltransducers
transcription factor
nucleus
protein that isunable to inhibitthe cell cycleor promoteapoptosis
mutated tumor suppressor gene
71
Cell Signaling Pathway
Cell signaling pathway that stimulates a proto-oncogene
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
receptor
stimulating growth factor
cytoplasm
plasmamembrane
signaltransducers
transcription factor
nucleus
protein thatoverstimulatesthe cell cycle
oncogene
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
Normal Igf2 alleleis expressed
Paternalchromosome
Maternalchromosome
(a) Homozygote
Wild-type mouse(normal size)
Normal Igf2 alleleis not expressed
Fig. 15-18b
Mutant Igf2 alleleinherited from mother
Mutant Igf2 alleleinherited from father
Normal size mouse(wild type)
Dwarf mouse(mutant)
Normal Igf2 alleleis expressed
Mutant Igf2 alleleis expressed
Mutant Igf2 alleleis not expressed
Normal Igf2 alleleis not expressed
(b) Heterozygotes
Fig. 15-18Normal Igf2 alleleis expressed
Paternalchromosome
Maternalchromosome
Normal Igf2 alleleis not expressed
Mutant Igf2 alleleinherited from mother
(a) Homozygote
Wild-type mouse(normal size)
Mutant Igf2 alleleinherited from father
Normal size mouse(wild type)
Dwarf mouse(mutant)
Normal Igf2 alleleis expressed
Mutant Igf2 alleleis expressed
Mutant Igf2 alleleis not expressed
Normal Igf2 alleleis 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
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. Explain the chromosomal theory of inheritance and its discovery
2. Explain why sex-linked diseases are more common in human males than females
3. Distinguish between sex-linked genes and linked genes
4. Explain how meiosis accounts for recombinant phenotypes
5. Explain how linkage maps are constructed
6. Explain how nondisjunction can lead to aneuploidy
7. Define trisomy, triploidy, and polyploidy8. Define mutation9. Distinguish between different gene
mutations10. Distinguish among deletions, duplications,
inversions, and translocations11. Explain genomic imprinting12. Explain why extranuclear genes are not
inherited in a Mendelian fashion
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.
Morgan’s Experimental EvidenceImagine 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. on the O chromosome of a maleb. on the X chromosome of a malec. on the X chromosome of a femaled. on the Y chromosome of a male
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.
The Chromosomal Basis of SexThink 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. Bee males have half the DNA of bee females whereas human males have nearly the same amount of DNA that human females have.
b. Considered across the genome, harmful (deleterious) recessives will negatively affect bee males more than Drosophila males.
c. Human and Drosophila males have sons but bee males do not.
d. Inheritance in bees is like inheritance of sex-linked characteristics in humans.
e. none of the above
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.
The Chromosomal Basis of SexIn 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 SexImagine 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. in a male because there is no possibility of the presence of a normal, dominant allele
b. in a male because it is haploidc. in a female because there is no possibility of
the presence of a normal, dominant alleled. in a female because all alleles on the W
chromosomes are dominant to those on the Z chromosome
e. none of the above
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.
Inheritance of Sex-Linked GenesIn 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. The phenotype of O-Y males is orange because the functional allele O converts eumelanin into phaeomelanin.
b. The phenotype of o-Y males is black/brown because the non-functional allele o does not convert eumelanin into phaeomelanin.
c. The phenotype of OO and Oo males is orange because the functional allele O converts eumelanin into phaeomelanin.
d. 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.
e. 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 GenesIn 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. The phenotype of O-Y females is orange because the functional allele O converts eumelanin into phaeomelanin.
b. The phenotype of o-Y females is black/brown because the non-functional allele o does not convert eumelanin into phaeomelanin.
c. The phenotype of OO and Oo females is orange because the functional allele O converts eumelanin into phaeomelanin.
d. 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.
e. 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.
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X Inactivation in Female MammalsImagine 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. Both species will have similar sized patches
of orange and black/brown fur.b. Species A will have smaller patches of
orange or black/brown fur than will species B.
c. The females of both species will show the dominant fur color, orange.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.
Mapping the Distance between GenesImagine 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. The recombination rate between locus A and locus C
is either 2% or 68%.b. The recombination rate between locus A and locus C
is probably 2%.c. The recombination rate between locus A and locus C
is either 2% or 50%.d. The recombination rate between locus A and locus C
is either 2% or 39%.e. The recombination rate between locus A and locus C
cannot be predicted.
Copyright © 2008 Pearson Education, Inc., publishing as 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. In mitosis, some chromosomes in triploids have no partner at synapsis, but chromosomes in tetraploids do have partners.
b. In meiosis, some chromosomes in triploids have no partner at synapsis, but chromosomes in tetraploids do have partners.
c. In mitosis, some chromosomes in tetraploids have no partner at synapsis, but chromosomes in triploids do have partners.
d. In meiosis, some chromosomes in tetraploids have no partner at synapsis, but chromosomes in triploids do have partners.
Copyright © 2008 Pearson Education, Inc., publishing as 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. Chromosomal rearrangements are more likely to occur in mammals than in other vertebrates.
b. Translocations and inversions are not deleterious because no genes are lost in the organism.
c. Chromosomal rearrangements are more likely to occur during mitosis than during meiosis.
d. 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.
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. The mitochondrial DNA of the child and “father” did not match.
b. DNA sequencing of chromosome #5 of the child and “father” did not match.
c. The mitochondrial DNA of the child and “mother” did not match.
d. DNA sequencing of chromosome #5 of the child and “mother” did not match.
e. The mitochondrial DNA of the child and “father” matched but the mitochondrial DNA of the child and “mother” did not.