The parents are AAbbCcdd and AaBBccDD•Predict the outcome (probability) that:

▫Offspring will have at least 2 heterozygous traits

▫Offspring will have at least one recessive trait

▫Offspring will have all heterozygous traits▫Offspring will have at least one

homozygous dominant trait

Chapter 15

The Chromosomal Basis of Inheritance

Chromosome Theory of Inheritance•The chromosome theory of inheritance

states that▫Mendelian genes have specific loci on

chromosomes▫Chromosomes undergo segregation and

independent assortment

•The chromosomal basis of Mendel’s laws

Figure 15.2

Yellow-roundseeds (YYRR)

Green-wrinkledseeds (yyrr)

Meiosis

Fertilization

Gametes

All F1 plants produceyellow-round seeds (YyRr)

P Generation

F1 Generation

Meiosis

Two equallyprobable

arrangementsof chromosomesat metaphase I

LAW OF SEGREGATION LAW OF INDEPENDENT ASSORTMENT

Anaphase I

Metaphase II

Fertilization among the F1 plants

9 : 3 : 3 : 1

14

14

14

14

YR yr yr yR

Gametes

Y

RRY

y

r

r

y

R Y y r

Ry

Y

r

Ry

Y

r

R

Y

r

y

r R

Y y

R

Y

r

y

R

Y

Y

R R

Y

r

y

r

y

R

y

r

Y

r

Y

r

Y

r

Y

R

y

R

y

R

y

r

Y

F2 Generation

Starting with two true-breeding pea plants,we follow two genes through the F1 and F2 generations. The two genes specify seed color (allele Y for yellow and allele y forgreen) and seed shape (allele R for round and allele r for wrinkled). These two genes are on different chromosomes. (Peas have seven chromosome pairs, but only two pairs are illustrated here.)

The R and r alleles segregate at anaphase I, yielding two types of daughter cells for this locus.

1

Each gamete gets one long chromosome with either the R or r allele.

2

Fertilizationrecombines the R and r alleles at random.

3

Alleles at both loci segregatein anaphase I, yielding four types of daughter cells depending on the chromosomearrangement at metaphase I. Compare the arrangement of the R and r alleles in the cellson the left and right

1

Each gamete gets a long and a short chromosome in one of four allele combinations.

2

Fertilization results in the 9:3:3:1 phenotypic ratio in the F2 generation.

3

Alleles of genes on nonhomologous chromosomes assort independently during gamete formation

Two alleles for each gene separate during gamete

formation

Morgan’s Experimental Evidence: Scientific Inquiry•Thomas Hunt Morgan

▫Provided convincing evidence that chromosomes are the location of Mendel’s heritable factors

•Morgan worked with fruit flies▫Wild type, or normal, phenotypes that were

common in the fly populations (w+)•Traits alternative to the wild type

▫Are called mutant phenotypes (w)

Figure 15.3

Huh!Sex matters?!

F2generation

100%red-eye female

50% red-eye male50% white eye male

Discovery of sex linkage

P X

F1generation(hybrids)

100%red eye offspring

true-breeding white-eye male

true-breedingred-eye female

RR rr

What’s up with Morgan’s flies?

x

r r

R

R

Rr

Rr Rr

Rr

100% red eyes

Rr Rr

x

R r

R

r

RR

Rr rr

Rr

3 red : 1 white

Doesn’t workthat way!

•In humans & other mammals, there are 2 sex chromosomes: X & Y▫2 X chromosomes

develop as a female: XX gene redundancy,

like autosomal chromosomes▫an X & Y chromosome

develop as a male: XY no redundancy

Genetics of Sex

X Y

X

X

XX

XY

XY

50% female : 50% maleXX

XRXR XrY

Let’s reconsider Morgan’s flies…

x

Xr Y

XR

100% red eyes

XR

XRXr XRY

XRYXRXr

x

XRXr XRY

XR Y

XR

Xr

XRXr

XRYXRXR

XrY100% red females50% red males; 50% white males

BINGO!

How Linkage Affects Inheritance: Scientific Inquiry

• Linked genes tend to be inherited together because they are located near each other on the same chromosome

• Morgan did other experiments with fruit flies▫To see how linkage affects the inheritance of two

different characters• Body color

▫Wild-type: gray body: b+ ▫Mutant: black body: b

• Wing Size▫Wild type: normal wing: vg+ ▫Mutant: vestigial wing: vg

Double mutant(black body,vestigial wings)

Double mutant(black body,vestigial wings)

Wild type(gray body,

normal wings)

P Generation(homozygous)

b+ b+ vg+ vg+

x

b b vg vg

F1 dihybrid(wild type)(gray body, normal wings)

b+ b vg+ vgb b vg vg

TESTCROSSx

b+vg+ b vg b+ vg b vg+

b vg

b+ b vg+ vg b b vg vg b+ b vg vgb b vg+ vg

965Wild type

(gray-normal)

944Black-

vestigial

206Gray-

vestigial

185Black-normal

Sperm

Parental-typeoffspring

Recombinant (nonparental-type)offspring

RESULTS

EXPERIMENT Morgan first mated true-breedingwild-type flies with black, vestigial-winged flies to produce heterozygous F1 dihybrids, all of which are wild-type in appearance. He then mated wild-type F1 dihybrid females with black, vestigial-winged males, producing 2,300 F2 offspring, which he “scored” (classified according to phenotype).

CONCLUSION If these two genes were on different chromosomes, the alleles from the F1 dihybrid would sort into gametes independently, and we would expect to see equal numbers of the four types of offspring. If these two genes were on the same chromosome, we would expect each allele combination, B+ vg+ and b vg, to stay together as gametes formed. In this case, onlyoffspring with parental phenotypes would be produced. Since most offspring had a parental phenotype, Morganconcluded that the genes for body color and wing sizeare located on the same chromosome. However, the production of a small number of offspring with nonparental phenotypes indicated that some mechanism occasionally breaks the linkage between genes on the same chromosome.

Double mutant(black body,vestigial wings)

Double mutant(black body,vestigial wings)

Wild Type Mutantb+ gray body b black bodyvg+ normal wing vg vestigial wing

•Morgan determined that▫Genes that are close together on the same

chromosome are linked and do not assort independently

▫Unlinked genes are either on separate chromosomes or are far apart on the same chromosome and assort independently

Parentsin testcross

b+ vg+

b vg

b+ vg+

b vg

b vg

b vg

b vg

b vg

Mostoffspring

X

or

Heterozygous gray and normal wings

Homozygous black and vestigial

Heterozygous gray and normal wings

Homozygous black and vestigial wings

Displays gene linkage

•Recombinant offspring▫Are those that show new combinations of the

parental traits•When 50% of all offspring are recombinants

▫Geneticists say that there is a 50% frequency of recombination for any two genes located on different chromosomes.

Gametes from green-wrinkled homozygousrecessive parent (yyrr)

Gametes from yellow-roundheterozygous parent (YyRr)

Parental-type offspring

Recombinantoffspring

YyRr yyrr Yyrr yyRr

YR yr Yr yR

yr

Figure 15.6

Testcrossparents

Gray body,normal wings(F1 dihybrid)

b+ vg+

b vgReplication ofchromosomes

b+ vg

b+vg+

b

vg

vgMeiosis I: Crossingover between b and vgloci produces new allelecombinations.

Meiosis II: Segregationof chromatids producesrecombinant gameteswith the new allelecombinations.

Recombinantchromosome

b+vg+ b   vg b+ vg b vg+

b vg

Sperm

b   vg

b   vgReplication ofchromosomesvg

vg

b

b

bvg

b   vg

Meiosis I and II:Even if crossing overoccurs, no new allelecombinations areproduced.

OvaGametes

Testcrossoffspring

Sperm

b+  vg+ b   vg b+   vg b   vg+

965Wild type

(gray-normal)b+  vg+

b  vg b  vg b  vg b  vg

b  vg+b+  vg+b  vg+

944Black-

vestigial

206Gray-

vestigial

185Black-normal Recombination

frequency =391 recombinants

2,300 total offspring

100 = 17%

Parental-type offspringRecombinant offspring

Ova

b vg

Black body,vestigial wings(double mutant)

b

•Linked genes▫Exhibit recombination frequencies less than

50% due to crossing over

Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings

Linkage Mapping: Using Recombination Data: Scientific Inquiry

• A genetic map

– Is an ordered list of the genetic loci along a particular chromosome

– Can be developed using recombination frequencies

– The further apart two genes are, the higher the probability that a cross-over will occur between them and therefore the higher the recombination frequency

•A linkage map▫Is the actual map of a chromosome based on

recombination frequencies

Recombinationfrequencies

9% 9.5%

17%

b cn vgChromosome

The b–vg recombination frequency is slightly less than the sum of the b–cn and cn–vg frequencies because double crossovers are fairly likely to occur between b and vg in matings tracking these two genes. A second crossoverwould “cancel out” the first and thus reduce the observed b–vg recombination frequency.

In this example, the observed recombination frequencies between three Drosophila gene pairs (b–cn 9%, cn–vg 9.5%, and b–vg 17%) best fit a linear order in which cn is positioned about halfway between the other two genes:

RESULTS

A linkage map shows the relative locations of genes along a chromosome.APPLICATION

TECHNIQUE A linkage map is based on the assumption that the probability of a crossover between twogenetic loci is proportional to the distance separating the loci. The recombination frequencies used to constructa linkage map for a particular chromosome are obtained from experimental crosses, such as the cross depictedin Figure 15.6. The distances between genes are expressed as map units (centimorgans), with one map unitequivalent to a 1% recombination frequency. Genes are arranged on the chromosome in the order that best fits the data.

Figure 15.7

As distance between genes increases, the number of recombinant offspring increases

•Some genes are so far from each other on a chromosome that crossing over between them is inevitable, driving recombination up to 50%▫Same odds as genes on different

chromosomes Genetically unlinked alleles assort

independently as if on different chromosomes Ex: seed color and flower color in pea plants

are on the same chromosome, but so far apart they appear to assort independently

•Many fruit fly genes▫Were mapped initially using recombination

frequencies

Figure 15.8

Mutant phenotypes

Short aristae

Black body

Cinnabareyes

Vestigialwings

Brown eyes

Long aristae(appendageson head)

Gray body

Redeyes

Normalwings

Redeyes

Wild-type phenotypes

IIY

I

X IVIII

0 48.5 57.5 67.0 104.5

Recombination/Mapping Problem

•Use the following recombination frequencies/percentages to determine the order of these genes on the chromosome. Draw out your final map. Genes a, b, c, d

a,d = 8% a,b = 10% b,d= 18% b,c= 16% a,c= 6%

Concept Check 15.2

•Genes A, B, and C are located on the same chromosome. Testcrosses show that the recombination frequency between A and B is 28% and between A and C is 12%. Can you determine the linear order of these genes?

•Sex-linked▫usually means

“X-linked”▫more than

60 diseases traced to genes on X chromosome

Duchenne muscular dystrophyBecker muscular dystrophy

Ichthyosis, X-linkedPlacental steroid sulfatase deficiencyKallmann syndromeChondrodysplasia punctata, X-linked recessive

HypophosphatemiaAicardi syndromeHypomagnesemia, X-linkedOcular albinismRetinoschisis

Adrenal hypoplasiaGlycerol kinase deficiency

Incontinentia pigmentiWiskott-Aldrich syndromeMenkes syndrome

Charcot-Marie-Tooth neuropathyChoroideremiaCleft palate, X-linkedSpastic paraplegia, X-linked, uncomplicatedDeafness with stapes fixation

PRPS-related gout

Lowe syndrome

Lesch-Nyhan syndromeHPRT-related gout

Hunter syndromeHemophilia BHemophilia AG6PD deficiency: favismDrug-sensitive anemiaChronic hemolytic anemiaManic-depressive illness, X-linkedColorblindness, (several forms)Dyskeratosis congenitaTKCR syndromeAdrenoleukodystrophyAdrenomyeloneuropathyEmery-Dreifuss muscular dystrophyDiabetes insipidus, renalMyotubular myopathy, X-linked

Androgen insensitivity

Chronic granulomatous diseaseRetinitis pigmentosa-3

Norrie diseaseRetinitis pigmentosa-2

Sideroblastic anemiaAarskog-Scott syndrome

PGK deficiency hemolytic anemia

Anhidrotic ectodermal dysplasia

AgammaglobulinemiaKennedy disease

Pelizaeus-Merzbacher diseaseAlport syndrome

Fabry disease

Albinism-deafness syndrome

Fragile-X syndrome

Immunodeficiency, X-linked,with hyper IgM

Lymphoproliferative syndrome

Ornithine transcarbamylase deficiency

Human X chromosome

Genes on sex chromosomes•Y chromosome

▫few genes other than SRY sex-determining region master regulator for maleness turns on genes for production of male

hormones many effects = pleiotropy!

•X chromosome▫other genes/traits beyond sex determination

mutations: hemophilia Duchenne muscular dystrophy color-blindness

Map of Human Y chromosome?< 30 genes on Y chromosome

Sex-determining Region Y (SRY)

linked

Channel Flipping (FLP)Catching & Throwing (BLZ-1)

Self confidence (BLZ-2)

Devotion to sports (BUD-E)

Addiction to death &destruction movies (SAW-2)

Scratching (ITCH-E)Spitting (P2E)

Inability to express affection over phone (ME-2) Selective hearing loss (HUH)

Total lack of recall for dates (OOPS)

Air guitar (RIF)

•Sex-linked genes▫Follow specific patterns of inheritance

XAXA XaY

Xa Y

XAXa XAY

XAYXAYa

XA

XA

Ova

Sperm

XAXa XAY

Ova XA

Xa

XAXA XAY

XaYXaYA

XA YSperm

XAXa XaY

Ova

Xa Y

XAXa XAY

XaYXaYa

XA

Xa

A father with the disorder will transmit the mutant allele to all daughters but to no sons. When the mother is a dominant homozygote, the daughters will have the normal phenotype but will be carriers of the mutation.

If a carrier mates with a male of normal phenotype, there is a 50% chance that each daughter will be a carrier like her mother, and a 50% chance that each son will have the disorder.

If a carrier mates with a male who has the disorder, there is a 50% chance that each child born to them will have the disorder, regardless of sex. Daughters who do not have the disorder will be carriers, where as males without the disorder will be completely free of the recessive allele.

(a)

(b)

(c)

Sperm

• Homozygous dominant females appear normal

• Heterozygous females are carriers

• Homozygous recessive females are diseased

• Males are hemizygous• One allele controls gene

expression• Heterozygous males are

diseased

•Some recessive alleles found on the X chromosome in humans cause certain types of disorders▫Color blindness▫Duchenne muscular dystrophy▫Hemophilia

XHYXHXh

XH Ymale / sperm

XH

Xh

fem

ale

/ eg

gs XHXH XHY

XHXh XhY

carrier disease

sex-linked recessive

X-inactivation•Female mammals inherit 2 X chromosomes

▫one X becomes inactivated during embryonic development condenses into compact object = Barr body which X becomes Barr body is random and is

not expressed patchwork trait = “mosaic”

XH

Xh

XHXh

patches of black

patches of orange

2006-2007

Errors of MeiosisChromosomal Abnormalities

Chromosomal abnormalities

•Incorrect number of chromosomes▫nondisjunction

Homologous chromosomes don’t separate properly during meiosis

▫breakage of chromosomes deletion duplication inversion translocation

Nondisjunction • Problems with meiotic spindle cause errors in

daughter cells▫homologous chromosomes do not separate

properly during Meiosis 1▫sister chromatids fail to separate during

Meiosis 2▫too many or too few chromosomes

2n n

n

n-1

n+1

Alteration of chromosome number

all with incorrect number 1/2 with incorrect number

error in Meiosis 1

error in Meiosis 2

trisomy

2n+1

Nondisjunction •Aneuploidy

▫Baby has wrong chromosome number▫trisomy

cells have 3 copies of a chromosome ▫monosomy

cells have only 1 copy of a chromosome n+1 n

monosomy

2n-1

n-1 n

Down syndrome•Trisomy 21

▫3 copies of chromosome 21▫1 in 700 children born in U.S.

•Chromosome 21 is the smallest human chromosome▫but still severe effects

•Frequency of Down syndrome correlates with the age of the mother

Down syndrome & age of mother

Mother’s age

Incidence of Down Syndrome

Under 30 <1 in 1000

30 1 in 900

35 1 in 400

36 1 in 300

37 1 in 230

38 1 in 180

39 1 in 135

40 1 in 105

42 1 in 60

44 1 in 35

46 1 in 20

48 1 in 16

49 1 in 12

Rate of miscarriage due to amniocentesis: 1970s data

0.5%, or 1 in 200 pregnancies

2006 data<0.1%, or 1 in 1600 pregnancies

Genetic testing•Amniocentesis in 2nd trimester

▫sample of embryo cells▫stain & photograph chromosomes

•Analysis of karyotype

•Polyploidy▫Is a condition in which there are more than

two complete sets of chromosomes in an organism

Figure 15.13 A tetraploid mammal. The somatic cells of this burrowing rodent has twice as many chromosomes as those of closely related species.

Sex chromosomes abnormalities

•Human development more tolerant of wrong numbers in sex chromosome

•But produces a variety of distinct syndromes in humans▫XXY = Klinefelter’s syndrome male ▫XXX = Trisomy X female▫XYY = Jacob’s syndrome male▫XO = Turner syndrome female

•XXY male▫one in every 2000 live births▫have male sex organs, but

are sterile▫feminine characteristics

some breast development lack of facial hair

▫tall▫normal intelligence

Klinefelter’s syndrome

Jacob’s syndrome male•XYY Males

▫1 in 1000 live male births

▫extra Y chromosome▫slightly taller than

average▫more active▫normal intelligence, slight learning disabilities▫delayed emotional maturity▫normal sexual development

Trisomy X•XXX

▫1 in every 2000 live births▫produces healthy females

Why? Barr bodies

all but one X chromosome is inactivated

Turner syndrome•Monosomy X or X0

▫1 in every 5000 births▫varied degree of effects ▫webbed neck▫short stature▫sterile

Alterations of Chromosome Structure•Breakage of a chromosome can lead to four

types of changes in chromosome structure▫Deletion▫Duplication▫Inversion▫Translocation

•Occurs during meiosis – generally during crossing over

•An embryo that is homozygous for a large deletion is missing a number of essential genes – such a deletion is generally lethal

•Alterations of chromosome structure

Figure 15.14a–d

A B C D E F G HDeletion

A B C E G HF

A B C D E F G HDuplication

A B C B D EC F G H

A

A

M N O P Q R

B C D E F G H

B C D E F G HInversion

Reciprocaltranslocation

A B P Q R

M N O C D E F G H

A D C B E F HG

(a) A deletion removes a chromosomal segment.

(b) A duplication repeats a segment.- Repeated segment may have been lost in a nearby deletion

(c) An inversion reverses a segment within a chromosome.

(d) A translocation moves a segment fromone chromosome to another,nonhomologous one. In a reciprocal

  translocation, the most common type,nonhomologous chromosomes exchangefragments. Nonreciprocal translocationsalso occur, in which a chromosome transfers a fragment without receiving afragment in return.

Disorders Caused by Structurally Altered Chromosomes

•Cri du chat▫Is a disorder caused by a deletion in a

chromosome 5▫Very severe; infant/childhood death•Certain cancers

▫Are caused by translocations of chromosomes Ex: reciprocal translocation of Chromosome 9 and 22 which is

associated with Chronic Myelogenous Leukemia

Normal chromosome 9 Reciprocaltranslocation

Translocated chromosome 9

Philadelphiachromosome

Normal chromosome 22 Translocated chromosome 22

•Genomic imprinting▫Important for

embryonic development

▫Phenotype depends on whether an allele is supplied by the mother or father

▫Involves the silencing of certain genes that are “stamped” with an imprint during gamete production

(a) A wild-type mouse is homozygous for the normal igf2 allele.

Normal Igf2 allele(expressed)

Normal Igf2 allelewith imprint(not expressed)

Paternalchromosome

Maternalchromosome

Wild-type mouse(normal size)

Normal Igf2 allele

Paternal

Maternal

Mutant lgf2 allele

Mutant lgf2 allele

Paternal

Maternal

Dwarf mouseNormal Igf2 allelewith imprint

Normal size mouse

(b) When a normal Igf2 allele is inherited from the father, heterozygous mice grow to normal size. But when a mutant allele is inherited from the father, heterozygous mice have the dwarf phenotype.

Inheritance of Organelle Genes• Extranuclear genes

▫ Are genes found in organelles in the cytoplasm

• The inheritance of traits controlled by genes present in the chloroplasts or mitochondria▫ Depends solely on the maternal

parent because the zygote’s cytoplasm comes from the egg which contains extranuclear DNA prior to fertilization

▫ Almost all the zygote’s mitochondria are in the cytoplasm of the egg

▫ Mitochondrial genes help make up the protein complexes of the electron transport chain and ATP synthase

•Some diseases affecting the muscular and nervous systems▫Are caused by defects in mitochondrial genes

that prevent cells from making enough ATP