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Transcript of 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
Morgans Experimental Evidence
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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
Morgans Experimental Evidence
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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
(black body,vestigial wings)
Doublemutant
(black body,vestigial wings)
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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