GENETICS & EVOLUTION: CHROMOSOMAL INHERITANCE & MUTATION Chapter 7.2.

GENETICS & EVOLUTION:CHROMOSOMAL INHERITANCE & MUTATION

Chapter 7.2

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

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


1

2 2

1

Fig. 15-2c

F2 GenerationAn F1 F1 cross-fertilization

9 : 3 : 3 : 1

3 3

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)


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

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 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-UN1

b+ vg+

Parents in testcross

Most offspring

b+ vg+

b vg

b vg

b vg

b vg

b vg

b vg

or

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




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




b b vg vg

b b vg vg


b+ b+ vg+ vg+


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


RESULTS



b b vg vg

b b vg vg


b+ b+ vg+ vg+


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)


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


Gray body, normal wings(F1 dihybrid)

Black body, vestigial wings(double mutant)


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


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


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

Recombinationfrequencies

Chromosome9% 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

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



Meiosis II

Nondisjunction

Fig. 15-13-3

Meiosis I

Nondisjunction



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

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 : 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

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

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


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


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)


Mutant Igf2 alleleis expressed

Mutant Igf2 alleleis not expressed


(b) Heterozygotes

Fig. 15-18Normal Igf2 alleleis expressed

Paternalchromosome

Maternalchromosome


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)


Mutant Igf2 alleleis expressed

Mutant 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

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. 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


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


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.


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


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.


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.


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.


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.


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.


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.


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.



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.

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GENETICS & EVOLUTION: CHROMOSOMAL INHERITANCE & MUTATION Chapter 7.2.

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