Post on 07-Apr-2018
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Mendels Principles of
HeredityChapter 2
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The Units of Inheritance
Gene a region of DNA thatencodes a specific protein or a
particular type of RNA First described by Mendel in
1866
Mendel hypothesized that observabletraits are determined by independent unitsof inheritance (genes) not visible by thenaked eye
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Four General Themes ofMendel's Work
1. Variation is widespread in nature and providesfor continuously evolving diversity
2. Observable variation is essential for followinggenes from one generation to the next
3. Variation is inherited by genetic laws (notchance), which can explain why like begets like
and unlike4. Mendel's laws apply to all sexually reproducing
organisms (from protozoans to peas tohumans!)
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The Historical Puzzle ofInheritance
Artificial selection was the first appliedgenetic technique
Purposeful control of mating by choice ofparents for the next generation
Domestication of plants and animals wasa key transition in human civilization
Domestication of dogs from wolves
Domestication of rice, wheat, barley, andlentils from weed like plants
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Two Theories of Inheritance
at the Time of Mendel's Studies
1. Inherited features of offspring are
contributed mainly by only one parent
(e.g., a "homunculus" inside the sperm)
2. Parental traits become mixed and changed in
the offspring (i.e., "blended inheritance")
Neither theory could explain why some traitswould appear, disappear, and then reappear
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Three Basic QuestionsMendel Wanted to Answer
Regarding Selective Breeding
What is inherited? How is it inherited?
What is the role of chance inheredity?
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Keys to Mendels Experiments1. The garden pea was an ideal organism
Vigorous growth Self-fertilization Easy to cross-fertilize Produced large number of offspring each
generation
2. Mendel analyzed traits with discretealternative forms (antagonistic pairs)
Purple vs. white flowers Yellow vs. green peas Round vs. wrinkled seeds
3. Mendel established pure breeding lines to
conduct his experiments
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Keys to Mendels Experiments
(cont.)4. Mendel carefully controlled his matings
He performed reciprocal crosses (reversingmale and female traits)
Demonstrated that parents contribute EQUALLY to
inheritance5. Mendel worked with large numbers of
offspring, counted all offspring, subjectedhis findings to numerical analysis, and
compared his results with his predictions6. Mendel was a practical experimentalist
He focused on one trait (variable) at a time He controlled his experiments carefully
Considered sunlight exposure and watering
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Types of Plant CrossesSelf-fertilization Cross-fertilization
Doesnt matter which traits are
associated with the egg and whichare associated with the sperm
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Pea PlantCharacters
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Important Definitions
Pure-breeding Lines Organismsthat produce offspring with specificparental traits that remain constantfrom generation to generationSelf-fertilization of pure-breeding lines
ALWAYS results in offspring exactlylike the parent
Hybrids Offspring of geneticallydissimilar parents
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Other Important Definitions
Parental (P) Generation individualswhose progeny (in subsequentgenerations) will be studied for specifictraits
First Filial (F1) Generation progenyresulting from the controlled cross ofindividuals from the parental generation
Second Filial (F2) Generation progeny
resulting from the controlled self-cross orintercross of individuals from the F1generation
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Example of a Monohybrid Cross
Green traitdidnt
disappear it
was simplyhidden
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Mendels Hypothesis Based
on Monohybrid Crosses
Traits have two forms that can each breedtrue (e.g., seed shape could be eitherround or wrinkled)
Progeny inherit one unit from the maternalparent and the other unit from the paternalparent
Units of inheritance are now known as "genes Therefore, each organism has two genes for each trait
Alternative forms of a single gene are "alleles
For the seed shape gene, there is a round allele
(designated R) and a wrinkled allele (designated r)
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Important Information Learnedfrom Monohybrid Crosses
Mendel called the version of the trait thatappeared in all of the F1 hybridsdominant
The version of the trait that remainedhidden in the F1 hybrids (the otherversion in the antagonistic pair) was
called recessive This version of the trait re-appears in the
F2 generationWhy did the F2 generation progeny always have
dominant:recessive ratio of 3:1?
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R
r
Conversion
Noconversion
Gene Biochemical Change of Unbranched Starch Molecules Pea Shape
Activeenzyme
Unbranched starch Branched starch Round pea
X
Inactiveenzyme
Dominantallele
Recessiveallele
How Can Two Different
Versions of a Gene Give Rise toTwo Different Appearances?
Unbranched starch Wrinkled peaUnbranched starch
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Mendels Law of SegregationPart 1
If a plant has two copies of every gene, how does it pass only one copy to
its progeny?
The two alleles for each trait separate (segregate) duringgamete formation (i.e., meiosis).
Eachgametehas only
ONE
allele foreachtrait!
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Mendels Law of SegregationPart 2
How does an offspring end up with two copies of the same gene, one from
each parent?
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Mendels Law of Segregation
The two alleles for each trait
separate (segregate) during gameteformation, and then unite at random,one from each parent, at fertilization.
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The Punnett SquareA Monohybrid Cross
Gene symbols areitalicized
Dominant alleles areCAPITALIZED
Recessive alleles arelowercase
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Still Other ImportantDefinitions
Phenotype the observable expressionof an organisms genes
Genotype pair of alleles present in anindividual
Homozygous two alleles of trait are the
same (YY or yy) Heterozygous two alleles of trait are
different (Yy)
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Genotypes vs Phenotpyes
YyYy
1:2:1YY:Yy:yy
3:1yellow: green
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Laws of Probability
Product Rule the probability of two ormore independent events occurringTOGETHER is the PRODUCT of the
probabilities of each individual eventProbability of Event #1 AND Event #2 = Probability of Event #1 X Probability of Event #2
Sum Rule the probability EITHER of two
mutually exclusive events occurring is theSUM of the probabilities of each individualevent
Probability of Event #1 OR Event #2 = Probability of Event #1 + Probability of Event #2
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Example of the Product Ruleof Probability
What is the probability of the cross
Rr X rr
producing an offspring with the genotypeRr?
The only way an Rr offspring can be produced
is if1. Parent 1 donates an R gamete
and
2. Parent 2 donates an r gamete
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Example of the Product Ruleof Probability (cont.)
Since Parent 1 is a heterozygote, he/she canmake two different types of gametes (an Rgamete or an r gamete).
Therefore, the probability of him/her making an Rgamete is
Since Parent 2 is a homozygote, he/she can
make only one type of gamete (an r gamete) Therefore, the probability of him/her making an r
gamete is 1
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Example of the Product Ruleof Probability (cont.)
Since the only way an Rr offspring can beproduced is to have Parent 1 donate an R
gamete and Parent 2 donate an r gamete, the
probability of BOTH things occurring togetheris the product of their probabilities
Probability ofRr
=Probability of
R gamete from
Parent 1
XProbability ofr gamete from
Parent 2
= X 1
=
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Example of the Sum Rule ofProbability
What is the probability of the cross
RR X Rr
producing a ROUND offspring?
The offspring can be round by having the
genotype1. RR
or
2. Rr
E l f h S R l f
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Example of the Sum Rule ofProbability (cont.)
The only way an RR offspring can be producedis if
1.Parent 1 donates an R gamete
and2.Parent 2 donates an R gamete
The only way an Rr offspring can be produced
is if
1.Parent 1 donates an R gamete
and
2.Parent 2 donates an r gamete
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Example of the Sum Rule ofProbability (cont.)
Lets consider the RR offspring first
Since Parent 1 is a homozygote, he/she canmake only one type of gametes (an R gamete).
Therefore, the probability of him/her making an Rgamete is 1
Since Parent 2 is a heterozygote, he/she canmake two different types of gametes (an Rgamete or an r gamete)
Therefore, the probability of him/her making an R
gamete is
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Example of the Sum Rule ofProbability (cont.)
Since the only way an RR offspring can beproduced is to have Parent 1 donate an R
gamete and Parent 2 donate an R gamete, the
probability of BOTH things occurring togetheris the product of their probabilities
Probability ofRR
=Probability of
R gamete from
Parent 1
XProbability of
R gamete from
Parent 2
= 1 X
=
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Example of the Sum Rule ofProbability (cont.)
Now, lets consider the Rr offspring
Since Parent 1 is a homozygote, he/she canmake only one type of gametes (an R gamete).
Therefore, the probability of him/her making an Rgamete is 1
Since Parent 2 is a heterozygote, he/she can
make two different types of gametes (an Rgamete or an r gamete)
Therefore, the probability of him/her making an r
gamete is
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Example of the Sum Rule ofProbability (cont.)
Since the only way an Rr offspring can beproduced is to have Parent 1 donate an R
gamete and Parent 2 donate an r gamete, the
probability of BOTH things occurring togetheris the product of their probabilities
Probability ofRr
=Probability of
R gamete from
Parent 1
XProbability ofr gamete from
Parent 2
= 1 X
=
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Example of the Sum Rule ofProbability (cont.)
Since a round offspring can have either thegenotype RR OR Rr, the probability of
producing a round offspring is the sum of theprobability of producing an RR and an Rr
offspring.
Probability ofR
=Probability of
an RR offpsring +Probability of
an Rr offspring
= +
= 1
In other words, a RR X Rr cross will ALWAYS produce round offspring!
G Ph
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Genotype/PhenotypeRelationship
If you know the genotype of anorganism and the dominancerelationship of the alleles, you can
accurately predict the phenotype of theorganism.
However, if an organism has a
dominant phenotype, you cannot becertain of that organisms genotype
Poses a problem for selective breeders!
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Problem for Selective Breeders:How Can You Determine the Genotype of
a Dominantly Phenotyped Organism?
The genotype of a dominantlyphenotyped organism could be either
homozygous (or true-breeding) orheterozygous (or NOT true-breeding)
How could a breeder be certain that
his/her dominantly phenotypedorganism was a true-breeder?
ANSWER: Perform a testcross
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Testcross
A mating in which an individual showing adominant phenotype is crossed with anindividual expressing the recessivephenotype
The organism whos genotype is in question
MUST have a dominant phenotype
Must be a dominant homozygote or a
heterozygote The other organism MUST have a recessive
phenotype
Must be a recessive homozygote
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Example Testcross
You have a yellow plant.
The plants genotype could be either YY or Yy
Cross that plant with a green plant (yy)
Examination of the offspring will tell thegenotype of the yellow plant
Testcross Reveals Unkown
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Testcross Reveals UnkownGenotype
Yellow plant is either YYor Yy
If the unknown parent is YY If the unknown parent is Yy
Therefore, if ANY progeny have a recessive phenotype (e.g., green),
then the parent MUST have been Yy!
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After describing the Law of
Segregation as it applies tothe inheritance of single
traits, Mendel next turned toexamining the simultaneousinheritance of two (or more)
apparently unrelated traitsusing DIHYBRID CROSSES
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Dihybrid Crosses
Dihybrid - an individual that is heterozygous attwo distinct genes
Mendel designed experiments to determine if
two genes segregate independently of oneanother in dihybrids
First constructed true breeding lines for bothtraits, crossed them to produce dihybridoffspring, and examined the F2 progeny forparental or recombinant types (newcombinations not present in the parents)
Dih brid Crosses Sho Both
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Dihybrid Crosses Show BothParental And Recombinant Types
Parental typesyellow round
green wrinkled
Recombinant typesyellow wrinkled
green round
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Dihybrid Cross Produces APredictable Ratio Of Phenotypes
Th L f I d d t
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The Law of IndependentAssortment
The detection of recombinant progenyin a dihybrid cross provides evidencethat the alleles for the different genes
can shuffle Creates new combinations of alleles
This is the basis of Mendels Second
Law - The Law of IndependentAssortment
The Law Of Independent
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The Law Of IndependentAssortment
During gameteformation differentpairs of allelessegregateindependently ofeach other Yis just as likely to assort
with Ras it is with r
yis just as likely to assortwith Ras it is with r
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Testcrosses with Dihybrids
The organism to be tested MUST bydominantly phenotyped for BOTH traitsand is crossed with an organism that is
homozygous recessive for BOTH traits
For example:
Yellow Round X green wrinkled
YYRRYYRrYyRRYyRr
yyrr
T i h Dih b id
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Cross A
P
1F
YY RR yy rr
y r
Y RYy Rr
Cross C
P
1F
Yy RR yy rr
Y r
Y RYy Rr
y Ryy Rr
Yy rrYy RrP
1Fy r
Yy Rr
Yy rr
Y R
yy Rr
yy rry r
y R
Y r
Cross DYy rr
Yy Rr
y r
yy rrYY Rr
1F
Y R
Y r
P
Cross B
Testcrosses with Dihybrids
If the recessivephenotypeshows up in the
progeny of thetest cross, thenthe parent inquestion MUSThave been aheterozygote forthat trait!
Laws of Probability for
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Laws of Probability forMultiple Genes
P RRYYTTSS X rryyttss
F1 RrYyTtSs X RrYyTtSs
F2 What is the ratio of different genotypes
and phenotypes?
gametes RYTS ryts
gametesRYTS
RyTS
rYTS
RYTs
RyTs
rYTs
RYtS
RytS
rYts
Ryts
RYts
rYTS
ryTs rYtS rYts ryts
RYTS
RyTS
rYTS
RYTs
RyTs
rYTs
RYtS
RytS
rYts
Ryts
RYts
rYTS
ryTs rYtS rYts ryts
A ( h h d )
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Answer 1 (the hard way) Punnet Square method:
24 = 16 possible gamete combinations for each F1
progeny
Therefore, you will have to perform a 16 16 PunnetSquare
Gives 256 genotypes!
Answer 2 (the easier way) Look at each locus independently
Calculate the probability of the genotype of
interest for each trait (locus)
Multiply the probabilities of the independent traits(loci)
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F1 RrYyTtSs RrYyTtSs
What is the probability of obtaining the genotypeRrYyTtss?
P RRYYTTSS rryyttss
Rr
Rr
1RR:2Rr:1rr
2/4 Rr
Yy X Yy
1YY:2Yy:1yy
2/4 Yy
Tt
Tt
1TT:2Tt:1tt
2/4 Tt
Ss
Ss
1SS:2Ss:1ss
1/4 ss
Calculate the probability of obtaining individual with Rr AND YyAND Tt AND ss
(Remember: each of these events MUST occur together)
= 2/4
2/4
2/4
1/4 = 8/256 = 1/32
P RRYYTTSS rryyttss
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F1 RrYyTtSs RrYyTtSs
P RRYYTTSS rryyttss
What is the probability of obtaining a completelyhomozygous genotype?
(The genotype could be RRYYTTSS or rryyttss)
Rr Rr
1RR:2Rr:1rr
1/4 RR1/4 rr
Yy Yy
1YY:2Yy:1yy
1/4 YY1/4 yy
Tt Tt
1TT:2Tt:1tt
1/4 TT1/4 tt
Ss Ss
1SS:2Ss:1ss
1/4 SS1/4 ss
Calculate the probability of obtaining individual with RR AND YY ANDTT AND SS OR an individual with rr AND yy AND tt AND ss
(Remember: EITHER of these events could occur)
(1/4
1/4
1/4
1/4) + (1/4
1/4
1/4
1/4) = 2/256 = 1/128
Mendelian Inheritance In
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Mendelian Inheritance InHumans
Most traits in humans are due to theinteraction of multiple genes and do notshow a simple Mendelian pattern of
inheritance A few traits represent single-genes
Examples: sickle-cell anemia, cystic fibrosis,Tay-Sachs disease, and Huntingtons disease
Because we can not do breedingexperiments on humans, we use modelorganisms
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Pedigrees Are Used to StudyInheritance
Pedigrees are an orderly diagramof a familys relevant genetic
features extending throughmultiple generations
Pedigrees help us infer if a trait is
from a single gene and if the traitis dominant or recessive
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Pedigree Analysis
A Vertical Pattern of Inheritance
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A Vertical Pattern of InheritanceIndicates a Rare Dominant Trait
Huntingtons disease: A rare dominant trait Assign the genotypes by working backward through
the pedigree
R g i i g D i t T it i
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Recognizing a Dominant Trait ina Pedigree
Affected children ALWAYS have atleast one affected parent
See at least one affected person in each
generation Shows a vertical pattern of inheritance
Two affected parents can have
unaffected children (i.e., both parentsare heterozygotes)
A Horizontal Pattern of Inheritance
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A Horizontal Pattern of InheritanceIndicates a Rare Recessive Trait
Cystic fibrosis: a recessive condition
Assign the genotypes for each pedigree
Recognizing a Recessive Trait in
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Recognizing a Recessive Trait ina Pedigree
Affected individuals can be the children oftwo unaffected carriers (i.e., heterozygotes) Especially if the carriers are related
All of the children of two affected parents
should be affected Rare recessive traits show a horizontal
pattern of inheritance
Trait first appears in several members of onegeneration and was not seen in earliergenerations
Common recessive traits MAY show a