1406 chromosomal basis 1

8/9/2019 1406 Chromosomal Basis 1

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The Chromosomal Basis of Inheritance


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Chromosomal Behavior

Mendelian inheritance has its physical basisin the behavior of chromosomes

The behavior of chromosomes during

meiosis was said to account for Mendelslaws of segregation and independentassortment

Several researchers proposed in the early1900s that genes are located onchromosomes


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Chromosome Theory Of Inheritance Mendelian genes have specific loci on chromosomes

Chromosomes undergo segregation and independent assortment

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 OFINDEPENDENTASSORTMENT

Anaphase I

Metaphase II

FertilizationamongtheF1 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 F2generations. The two genes specify seedcolor (allele Yfor yellow and allele yforgreen) and seed shape (allele Rfor roundand allele r for wrinkled). These two genes areon different chromosomes. (Peas have sevenchromosome pairs, but only two pairs areillustrated here.)

The R and ralleles segregateat anaphase I, yieldingtwo types of daughtercells for this locus.

1

Each gametegets one longchromosomewith either theRorrallele.

2

Fertilizationrecombines theRand rallelesat random.

3

Alleles at both loci segregatein anaphase I, yielding fourtypes of daughter cellsdepending 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 getsa long and a shortchromosome inone of four allelecombinations.

2

Fertilization resultsin the 9:3:3:1phenotypic ratio inthe F2 generation.

3


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Morgans Experimental Evidence

Thomas Hunt Morgan provided convincingevidence that chromosomes are the location ofMendels heritable factors

Morgan worked with fruit flies

Because they breed at a high rate

A new generation can be bred every two weeks

They have only four pairs of chromosomes


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Morgan first observed and noted wild type(normal), phenotypes that were common in the flypopulations

Traits alternative to the wild type are called mutantphenotypes



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Correlating the Behavior of a Genes Alleleswith a Chromosome Pair

In one experiment Morgan mated male flies withwhite eyes (mutant) with female flies with red eyes(wild type)

The F1 generation all had red eyes The F2 generation showed the 3:1 red:white eye

ratio, but only males had white eyes

Morgan determined that the white-eye mutant

allele must be located on the X chromosome


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Sex Linkage

Sex in humans (and other organisms) isdetermined by genes located on a pair ofchromosomes called the sex chromosomes

All other chromosomes are calledautosomes.

Even though the sex chromosomes pairduring synapsis, they are not homologous.

The larger chromosome is called the X andthe smaller is the Y.

In humans, XX is female and XY is male.


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Parents

Parental

Gametes

F1 Offspring:

Y

XX XY

XX XY

XY XX

X

X

Y

X

X X X


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Sex-linked Traits

Other traits, besides sex

, are controlled by genes on thesex chromosomes.

Traits controlled by the X are X-linked.

Traits controlled by the Y are Y-linked.

Since most sex-linked traits are controlled by the X, its

assumed X-linkage X-linked traits are an exception to Mendels laws because

females have 2 alleles for each X-linked trait, but maleshave only 1.

In humans, hemophilia is caused by a recessive allele onthe X chromosome. Two normal parents have a son withhemophilia. What is the probability their next child will alsohave hemophilia?


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ParentsParental

Gametes

F1 Offspring:

Y

XHXH XHY

XHXh XhY

XHYX

H

X

h

XH

Xh

Y

XH

XH XH Xh


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The F2 generation showed a typical Mendelian3:1 ratio of red eyes to white eyes. However, no females displayed the

white-eye trait; they all had red eyes. Half the males had white eyes,and half had red eyes.

Morgan then bred an F1 red-eyed female to an F1 red-eyed male toproduce the F2 generation.

RESULTS

P

Generation

F1Generation

X

F2Generation

Morgan mated a wild-type (red-eyed) femalewith a mutant white-eyed male. The F1 offspring all had red eyes.

EXPERIMENT



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CONCLUSION Since all F1 offspring had red eyes, the mutantwhite-eye trait (w) must be recessive to the wild-type red-eye trait (w+).

Since the recessive traitwhite eyeswas expressed only in males inthe F2 generation, Morgan hypothesized that the eye-color gene islocated on the X chromosome and that there is no corresponding locuson the Y chromosome, as diagrammed here.

P

Generation

F1Generation

F2Generation

Ova(eggs)

Ova(eggs)

Sperm

Sperm

XX

XXY

WW+

W+

W

W+W+ W+

W+

W+

W+

W+

W+

W

W+

W W

W


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Morgans discovery

Transmission of the X chromosome in fruitflies correlates with inheritance of the eye-color trait

First solid evidence indicating that a specificgene is associated with a specificchromosome


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Linked genes

Linked genes tend to be inherited togetherbecause they are located near each other onthe same chromosome

Each chromosome has hundreds orthousands of genes

Morgan did other experiments with fruit fliesto see how linkage affects the inheritance oftwo different characters


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Morgans Dihybred Experiment

At one gene locus controlling body color, he founda dominant allele for a gray body (b+) and arecessive allele for a black body (b).

At another gene locus controlling wing length, he

found a dominant allele for normal wings (vg+) anda recessive allele for vestigial (short) wings (vg).

Morgan crossed a female heterozygous at bothgene loci with a male homozygous for bothrecessive alleles.


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DrosophilaRed vs Purple Eyes, Normal vs Short Wings

Genotype of parents: b+b vg+vg X bb vg vg

Genotype testcross gametes:

(b+vg+) (b+vg) (bvg+) (bvg) X (b vg) Based on Mendels hypothesis, what is the

expected outcome?

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

b vg


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Morgans Results Offspring:

Genotype b+b vg+vg b+b vgvg bb vg+ vg bb vgvg

Phenotype gray,normal

gray,short

black,normal

black,short

Expected 25% 25% 25% 25%


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Morgans Results

Why dont the observed results agree with the predictedresults?

Is it chance or is Mendels hypothesis wrong? A statistical test gives a p< 0.01 - There is a very small

probability the difference was caused by chance alone.

Offspring:

Genotype b+b vg+vg b+b vgvg bb vg+ vg bb vgvg

Phenotype gray, normal gray, short black, normal black, short

Expected 25% 25% 25% 25%

Observed47% 5% 5% 43%


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Morgans Results

The female gametes are mostly b+vg+ or bvg. Why?

b+ and vg+ are linked together on one chromosome, whileb and vg are linked together on the homologouschromosome.

Genotype b+b vg+vg b+b vgvg b+b vgvg bb vgvgPhenotype gray, normal gray, short black, normal black, short

Expected 25% 25% 25% 25%

Observed 47% 5% 5% 43%

Offspring:

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

b vg b+b vg+vg b+b vgvg b+b vgvg bb vgvg


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Morgans Dihybrid Cross

Double mutant(black body,vestigial wings)

Doublemutant

(black body,vestigial wings)

Wild type(gray body,

normal wings)

P Generation

(homozygous)

b+ b+ vg+ vg+

x

bbvgvg

F1 dihybrid

(wildtype)

(gray body,normal wings)

b+ bvg+ vg

bbvgvg

TESTCROSS

x

b+vg+ bvg b+ vg bvg+

bvg

b+ bvg+ vg bbvgvg b+ bvgvgbbvg+ vg

965Wild type

(gray-normal)

944Black-

vestigial

206Gray-

vestigial

185Black-normal

Sperm

Parental-typeoffspring

Recombinant (nonparental-type)offspring

RESULTS

EXPERIMENTMorgan first mated true-breeding

wild-type flies with black, vestigial-winged flies to produceheterozygous F1 dihybrids, all of which are wild-type inappearance. He then mated wild-type F1 dihybrid females withblack, vestigial-winged males, producing 2,300 F2 offspring, whichhe scored (classified according tophenotype).

CONCLUSION

If these two genes were ondifferent chromosomes, the alleles from the F1 dihybridwould sort into gametes independently, and we wouldexpect 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 bvg,to stay together as gametes formed. In this case, only

offspring 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, theproduction of a small number of offspring withnonparental phenotypes indicated that some mechanismoccasionally breaks the linkage between genes on thesame chromosome.

Doublemutant


Doublemutant



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Morgans Conclusion

Genes that are close together on the same chromosomeare linked and do not assort independently

Unlinked genes are either on separate chromosomes of arefar apart on the same chromosome and assortindependently

Parentsin testcross

b+ vg+

bvg

b+ vg+

bvg

b vg

bvg

b vg

b vg

Mostoffspring

X

or


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Unlinked genes

Located on different chromosomes assortindependently:

B b

Y y

By byBY

bY

25% 25% 25% 25%


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Linked genes

Located on the same chromosome often doNOT assort independently:

B

n

B b

N n

b

n

B

N

b

N

47% 5% 5% 43%

Recombinant types

Parental type Parental type


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Genetic Recombination and Linkage

Recombinant offspring are those that show newcombinations of the parental traits

Morgan discovered that genes can be linked butdue to the appearance of recombinant phenotypes,

the linkage appeared incomplete Morgan proposed that some process must

occasionally break the physical connectionbetween genes on the same chromosome

Crossing over of homologous chromosomes wasthe mechanism


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Recombinant Types

Produced when a crossover occurs

between the 2 genes being studied:

A a

Bb

A a

B b

A a

B b

a A

Bb

Recombinant types Non recombinant types


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Genetic Maps

Because crossovers occur along the lengthof a chromosome at random, the fartherapart 2 genes are, the larger the chance acrossover will occur between them, and thehigher the frequency of recombinant types.

Therefore, the frequency of recombinanttypes can be used to construct genetic

maps:


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Testcross

parentsGray body,normal wings(F1 dihybrid)

b+ vg+

b vg

Replication ofchromosomes

b+vg

b+ vg+

b

vg

vgMeiosis I: Crossingover between b and vg

loci produces new allelecombinations.

Meiosis II: Segregationof chromatids producesrecombinant gameteswith the new allelecombinations.

v

Recombinantchromosome

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

b vg

Sperm

b vg

b vg

Replication ofchromosomesvg

vg

b

b

bvg

b vg

Meiosis IandII:

Even if crossing overoccurs, no new allelecombinations areproduced.

OvaGametes

Testcross

offspringSperm

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 offspringv 100 = 17%

Parental-type offspring Recombinant offspring

Ova

b vg

Black body,vestigial wings(double mutant)

b

Linked genes

Exhibit recombination frequencies less than 50%


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Linkage Mapping: Using Recombination Data

A genetic map Is an ordered list of the genetic loci along

a particular chromosome

Can be developed using recombinationfrequencies

A linkage map

Is the actual map of a chromosome basedon recombination frequencies


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Map Units The farther apart genes are on a chromosome the

more likely they are to be separated duringcrossing over

2 genes on the same chromosome can be locatedso far apart that the frequency of recombinanttypes reaches 50%

Same as for genes located on differentchromosomes.

These genes will assort independently, eventhough they are on the same chromosome.


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One map unit is defined as the distance between 2genes that will produce 1% recombinant types.

What is the distance in map units between the 2gene loci Morgan studied:

b+b vg+vg b+b vgvg bb vg+vg bb vgvg

47% 5% 5% 43%

b+b vgvg + bb vg+vg = 10% = 10 map units

Genetic Maps Units


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Genetic Mapping

Many fruit fly genes were mapped initiallyusing recombination frequencies

Figure15.8

Mutantphenotypes

Shortaristae

Blackbody

Cinnabareyes

Vestigialwings

Browneyes

Long aristae(appendageson head)

Graybody

Redeyes

Normalwings

Redeyes

Wild-typephenotypes

IIY

I

X IVIII

0 48.5 57.5 67.0 104.5


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Abnormal Chromosome Number or Structure

Alterations of chromosome number orstructure cause some genetic disorders

Large-scale chromosomal alterations often

lead to spontaneous abortions(miscarriages) or cause a variety ofdevelopmental disorders


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Abnormal Chromosome Number

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

MeiosisI

Nondisjunction

MeiosisII

Nondisjunction

Gametes

n +1

Numberofchromosomes

Nondisjunctionofhomologous

chromosomesinmeiosisI

n +1 n 1 n 1 n +1 n 1 n n

Nondisjunctionofsister

chromatidsinmeiosisI


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Aneuploidy results from the fertilization of gametesin which nondisjunction occurred

Offspring with this condition have an abnormalnumber of a particular chromosome

A trisomic zygote has three copies of a particularchromosome

A monosomic zygote has only one copy of a particularchromosome

Polyploidy is a condition in which an organism has

more than two complete sets of chromosomes

Aneuploidy


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Aneuploidy

Some types of aneuploidy appear to upset thegenetic balance less than others, resulting inindividuals surviving to birth and beyond

These surviving individuals have a set of

symptoms, or syndrome, characteristic of the typeof aneuploidy


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Down Syndrome

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


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AneuploidyofSexChromosomes

Nondisjunction of sex chromosomes produces avariety of aneuploid conditions

Klinefelter syndrome is the result of an extrachromosome in a male, producing XXY individuals

These individuals have male sex organs, but have abnormallysmall testes and are sterile.

Although the extra X is inactivated, some breast enlargement andother female characteristics are common.

Monosomy X, called Turner syndrome, producesX0 females, who are sterile; it is the only knownviable monosomy in humans


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Alterations of Chromosome Structure

Breakage of a chromosome can lead to fourtypes 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


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Alterations of chromosome structure

Figure15.14ad

A B C D E F G H Deletion 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 H Inversion

Reciprocaltranslocation

A B P Q R

M N O C D E F G H

A D C B E F H G

(a)A deletion removes a chromosomalsegment.

(b)A duplication repeats a segment.

(c)An inversion reverses a segment withina chromosome.

(d)A translocation moves a segment fromone chromosome to another,

nonhomologous one. In a reciprocaltranslocation, the most common type,nonhomologous chromosomes exchangefragments. Nonreciprocal translocationsalso occur, in which a chromosometransfers a fragment without receiving afragment in return.


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Alterations of Chromosome Structure

One syndrome, criduchat(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 earlychildhood

Certain cancers, including chronic myelogenous leukemia(CML), are caused by translocations of chromosomes

Normal chromosome 9Reciprocal

translocation

Translocated chromosome 9

Philadelphiachromosome

Normal chromosome 22 Translocated chromosome 22

1406 chromosomal basis 1

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