Chapter 2 Mendel s Principles of Heredity

8/3/2019 Chapter 2 Mendel s Principles of Heredity

1/57

Mendels Principles of

HeredityChapter 2


2/57

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


3/57

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!)


4/57

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


5/57

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


6/57

Three Basic QuestionsMendel Wanted to Answer

Regarding Selective Breeding

What is inherited? How is it inherited?

What is the role of chance inheredity?


7/57

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


8/57

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


9/57

Types of Plant CrossesSelf-fertilization Cross-fertilization

Doesnt matter which traits are

associated with the egg and whichare associated with the sperm


10/57

Pea PlantCharacters


11/57

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


12/57

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


13/57

Example of a Monohybrid Cross

Green traitdidnt

disappear it

was simplyhidden


14/57

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)


15/57

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?


16/57

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


17/57

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!


18/57

Mendels Law of SegregationPart 2

How does an offspring end up with two copies of the same gene, one from

each parent?


19/57

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.


20/57

The Punnett SquareA Monohybrid Cross

Gene symbols areitalicized

Dominant alleles areCAPITALIZED

Recessive alleles arelowercase


21/57

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)


22/57

Genotypes vs Phenotpyes

YyYy

1:2:1YY:Yy:yy

3:1yellow: green


23/57

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


24/57

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


25/57

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


26/57

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

=


27/57

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


28/57

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


29/57


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


30/57


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 ofRR

=Probability of

R gamete from

Parent 1

XProbability of

R gamete from

Parent 2

= 1 X

=


31/57


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


32/57


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 ofRr

=Probability of

R gamete from

Parent 1

XProbability ofr gamete from

Parent 2

= 1 X

=


33/57


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


34/57

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!


35/57

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


36/57

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


37/57

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


38/57

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!


39/57

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


40/57

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


41/57

Dihybrid Crosses Show BothParental And Recombinant Types

Parental typesyellow round

green wrinkled

Recombinant typesyellow wrinkled

green round


42/57

Dihybrid Cross Produces APredictable Ratio Of Phenotypes

Th L f I d d t


43/57

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


44/57

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


45/57

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


46/57

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


47/57

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 )


48/57

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)


49/57

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


50/57

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


51/57

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


52/57

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


53/57

Pedigree Analysis

A Vertical Pattern of Inheritance


54/57

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


55/57

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


56/57

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


57/57

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

Chapter 2 Mendel s Principles of Heredity

Documents

Transcript of Chapter 2 Mendel s Principles of Heredity