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Mendelelian Mendelelian GeneticsGenetics

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Gregor Gregor MendelMendel

(1822-1884)(1822-1884)

Responsible Responsible for the Laws for the Laws governing governing

Inheritance Inheritance of Traitsof Traits

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Gregor Johann Gregor Johann MendelMendel

Austrian monkAustrian monk Studied the Studied the inheritanceinheritance

of traits in of traits in pea plantspea plants Developed the Developed the laws of laws of

inheritanceinheritance Mendel's work was not Mendel's work was not

recognized until the turn recognized until the turn of theof the 20th century 20th century

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Gregor Johann MendelGregor Johann Mendel Between Between 1856 and 1863,1856 and 1863, Mendel Mendel

cultivated and tested some cultivated and tested some 28,000 28,000 pea plantspea plants

He found that the plants' offspring He found that the plants' offspring retained retained traits of the parentstraits of the parents

Called theCalled the “Father of Genetics" “Father of Genetics"

Mendel’s Laws of Inheritance

1. Genes In Pairs: Genetic characters are controlled by genes that exist in pairs of alleles in individual organisms and are passed from parents to their offspring. When two organisms produce offspring, each parent gives the offspring one of the alleles from each pair.

2. Dominance and Recessiveness: When two unlike alleles responsible for a single character are present in a single individual, one allele can mask the expression of another allele. That is, one allele is dominant to the other. The latter is said to be recessive.

3. The Law of Segregation: During the formation of gametes, the paired alleles separate (segregate) randomly so that each gamete receives one allele or the other.

4. The Law of Independent Assortment: During gamete formation, segregating pairs of alleles assort independently of each other. Example: genes on different chromosomes will segregate independently. Linked genes (close together on one chromosome) do not follow this law.

Rule of Dominance

•The trait that is observed in the offspring is the dominant trait (uppercase)

•The trait that disappears in the offspring is the recessive trait (lowercase)

Law of Segregation

•The two alleles for a trait must separate when gametes are formed

•A parent randomly passes only one allele for each trait to each offspring

Law of Independent Assortment•The genes for different

traits are inherited independently of each other.

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Site of Site of Gregor Gregor

Mendel’s Mendel’s experimeexperime

ntal ntal garden in garden in the Czech the Czech RepublicRepublic

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Mendel stated that Mendel stated that physical traits are physical traits are inherited as inherited as “particles”“particles”Mendel did not know Mendel did not know that the “particles” that the “particles” were actually were actually Chromosomes & DNAChromosomes & DNA

Particulate InheritanceParticulate Inheritance

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Genetic TerminologyGenetic Terminology

TraitTrait - any characteristic that - any characteristic that can be passed from parent to can be passed from parent to offspring offspring

HeredityHeredity - passing of traits - passing of traits from parent to offspring from parent to offspring

GeneticsGenetics - study of heredity - study of heredity

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Types of Genetic Types of Genetic CrossesCrosses

Monohybrid cross Monohybrid cross - - cross involving a cross involving a single traitsingle traite.g. flower color e.g. flower color

Dihybrid crossDihybrid cross - - cross involving two cross involving two traits traits e.g. flower color & plant heighte.g. flower color & plant height

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Designer Designer “Genes”“Genes” AllelesAlleles - - two forms of a two forms of a gene gene

(dominant & recessive)(dominant & recessive) DominantDominant - - stronger of two genes stronger of two genes

expressed in the hybrid; represented expressed in the hybrid; represented byby aa capital letter (R)capital letter (R)

RecessiveRecessive - - gene that shows up less gene that shows up less often in a cross; represented by aoften in a cross; represented by a lowercase letter (r)lowercase letter (r)

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More More TerminologyTerminology

GenotypeGenotype - - gene combination for gene combination for a traita trait (e.g. RR, Rr, rr) (e.g. RR, Rr, rr)

PhenotypePhenotype - - the physical feature the physical feature resulting from a genotyperesulting from a genotype (e.g. (e.g. red, white) red, white)

Here are the steps used to solve a Monohybrid Mendelian Genetic Problem:

Step 1: Figure out the genotypes of the parents (Pure Dom, Hybrid, Pure Rec) Use the Capital letter of the DOMINANT trait and then its lower case form to represent the recessive trait.

Step 2: Figure out what kinds of gametes the parents can produce.

Step 3: Set up a Punnett Square for your mating.

Step 4: Fill in the babies inside the table by filling each square by going Down and to the Left.

Step 5: Figure out the genotypic ratio for your predicted babies. How Many of the offspring are--Pure Dom: Hybrid: Pure RecPure Dom: Hybrid: Pure Rec

Step 6: Figure out the phenotypic ratio for your predicted babies. How Many of the offspring are-- Dom : RecDom : Rec

Step 7: Answer the question you've been asked. 19

Solving Genetics Problems I: Monohybrid Crosses

Classical genetics is a science of logic and statistics. While many find the latter intimidating, the mathematical side of most classical genetics puzzles is relatively simple — and there are actually ways to get around most of the math. The logic part is inescapable. All genetics problems are solved using the same basic logic structure. If you learn the sense of the approach, you can solve virtually any genetics problem, provided you are given enough basic information.

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Sample Problem using Steps:

This problem involves two gerbils named Honey and Ritz. The gene in question is a fur color gene which has two alleles — dominant brown (B) and recessive black (b). It's a very good idea to write down the information you are given in a problem so that it will be easy for you to refer to it when necessary. So begin by writing something like this at the top of your work page:

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Each of our parent gerbils is heterozygous for this gene. So here is our mating:

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Step One: Figure out the genotypes of the parents.

Step One: Figure out what kinds of gametes the parents can produce..

Once you've got that settled, you need to address the question of all of the possible kinds of babies they could produce. Before any parent makes babies, of course, that parent makes gametes. So in order to find what kinds of babies they can have, you must first determine what kinds of gametes they can produce. Since Honey is a heterozygote (and paying attention to Rule #1), she can produce two kinds of eggs: B eggs and b eggs.

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Ritz is also a heterozygote, so he can produce two kinds of sperm: B sperm and b sperm. Something like this:

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Step Three: Set up a Punnett Square for your mating

Now you need to determine all the possible ways that his sperm can combine with her eggs. There are several different techniques used for this operation. The most popular among students is the Punnett Square. Punnett Squares are probability tables — a way to do statistics while avoiding as much math as possible.

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Setting up a Punnett Square is easy. You need to create a chart with one column for each of the female's egg types, and one row for each of the male's sperm types. For Honey and Ritz, your table would look like this, then fill in the babies genotype by going down and to the left:

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Step 4: fill in the Punnett Square, down and to the left

BBBB

BbBb bbbb

BbBb

So we have now figured out that, if Honey and Ritz have a lot of babies, we can predict that 1/4 of them should be BB, 1/2 of them (2/4) should be Bb, and 1/4 should be bb.

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Step Five: Figure out the genotypic ratio for your predicted babies.

This conclusion is often expressed as a genotypic ratio:1BB:2Bb:1bb. This means that we are predicting that, for every BB baby, they should have 2 Bb babies (twice as many), and one bb baby.

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Step Six: Figure out the phenotypic ratio for your predicted babies.

To do this, you need to ask yourself one question: do any of these different genotypes produce the same phenotype? In other words, do any of these babies look alike? This is where dominance enters the picture. If B is completely dominant to b, all gerbils with at least one B will look pretty much alike, no matter whether their second allele is B or b. So BB and Bb have the same phenotype, and we can add them together. Thus, our phenotypic ratio is 3 Brown:1 Black. Or, there should be three times as many brown babies as black babies.

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So the answer to our question is, 3/4 of the babies should be brown.

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Step Seven: Answer the question you've been asked.

The mating scheme we've just worked through is called a monohybrid cross. This means that we were paying attention to only one gene (mono=1), and both of our parents were heterozygous for that gene (hybrid=heterozygous).

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Punnett SquarePunnett SquareUsed to help Used to help solve genetics solve genetics problemsproblems

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Genotype & Phenotype in Genotype & Phenotype in FlowersFlowers

Genotype of alleles:Genotype of alleles:RR = red flower= red flowerrr = yellow flower= yellow flower

All genes occur in pairs, so All genes occur in pairs, so 22 allelesalleles affect a characteristic affect a characteristic

Possible combinations are:Possible combinations are:

GenotypesGenotypes RRRR RRrr rrrr

PhenotypesPhenotypesRED RED RED RED YELLOWYELLOW

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GenotypesGenotypes HomozygousHomozygous genotype - gene combination genotype - gene combination

involving 2 dominant or 2 recessive genes involving 2 dominant or 2 recessive genes (e.g. RR or rr);(e.g. RR or rr); also called also called pure pure 

HeterozygousHeterozygous genotype - gene genotype - gene combination of one dominant & one combination of one dominant & one recessive allele    (recessive allele    (e.g. Rr);e.g. Rr); also called also called hybridhybrid

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Genes and Environment Determine Genes and Environment Determine CharacteristicsCharacteristics

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Mendel’s Pea Mendel’s Pea Plant Plant

ExperimentsExperiments

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Why peas,Why peas, Pisum Pisum sativumsativum?? Can be grown in a Can be grown in a small small

areaarea Produce Produce lots of offspring lots of offspring Produce Produce purepure plants when plants when

allowed to allowed to self-pollinateself-pollinate several generations several generations

Can be Can be artificially cross-artificially cross-pollinatedpollinated

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Reproduction in Reproduction in Flowering PlantsFlowering Plants

• Pollen contains spermPollen contains sperm– Produced by the stamenProduced by the stamen

• Ovary contains eggsOvary contains eggs– Found inside the flowerFound inside the flower

Pollen carries sperm to Pollen carries sperm to the eggs for fertilizationthe eggs for fertilization

Self-fertilizationSelf-fertilization can can occur in the same occur in the same flowerflower

Cross-fertilizationCross-fertilization can can occur between occur between flowersflowers

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Mendel’s Experimental Mendel’s Experimental MethodsMethods

• Mendel hand-pollinated Mendel hand-pollinated flowers using a paintbrushflowers using a paintbrush– He could snip the stamens He could snip the stamens

to prevent self-pollinationto prevent self-pollination– Covered each flower with Covered each flower with

a cloth baga cloth bag• He traced traits through the He traced traits through the

several generationsseveral generations

•

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How Mendel BeganHow Mendel BeganMendel Mendel produced produced purepure strains by strains by allowing allowing the plants the plants to to self-self-pollinatepollinate for for several several generatiogenerationsns

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Eight Pea Plant Eight Pea Plant TraitsTraits

• Seed shapeSeed shape --- Round --- Round (R)(R) or Wrinkled or Wrinkled (r)(r)• Seed ColorSeed Color ---- Yellow ---- Yellow (Y)(Y) or  Green or  Green ((yy))• Pod ShapePod Shape --- Smooth --- Smooth (S)(S) or wrinkled or wrinkled ((ss))• Pod ColorPod Color ---  Green ---  Green (G)(G) or Yellow or Yellow (g)(g)• Seed Coat ColorSeed Coat Color ---Gray ---Gray (G)(G) or White or White (g)(g)• Flower positionFlower position---Axial ---Axial (A)(A) or Terminal or Terminal (a)(a)• Plant HeightPlant Height --- Tall --- Tall (T)(T) or Short or Short (t)(t)• Flower color Flower color --- --- Purple Purple (P)(P) or white or white ((pp))

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Mendel’s Experimental Mendel’s Experimental ResultsResults

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• Did the observed ratio match the theoretical ratio?Did the observed ratio match the theoretical ratio?

The theoretical or expected ratio of The theoretical or expected ratio of plants producing round or wrinkled plants producing round or wrinkled seeds is seeds is 3 round :1 wrinkled3 round :1 wrinkled

Mendel’s observed ratio was 2.96:1Mendel’s observed ratio was 2.96:1

The discrepancy is due to The discrepancy is due to statistical statistical errorerror

The The larger the samplelarger the sample the more the more nearly the results approximate to nearly the results approximate to the theoretical ratiothe theoretical ratio

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Generation “Gap”Generation “Gap”• Parental PParental P11 Generation Generation = the parental generation in = the parental generation in

a breeding experimenta breeding experiment..• FF11 generation generation = the first-generation offspring in a = the first-generation offspring in a

breeding experiment. breeding experiment. (1st filial generation)(1st filial generation)– From breeding individuals from the PFrom breeding individuals from the P11 generation generation

• FF22 generation generation = the second-generation offspring in = the second-generation offspring in a breeding experiment. a breeding experiment. (2nd filial generation)(2nd filial generation)– From breeding individuals from the FFrom breeding individuals from the F11 generation generation

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Following the GenerationsFollowing the Generations

Cross 2 Cross 2 Pure Pure

PlantsPlantsTT x ttTT x tt

Results Results in all in all

HybridsHybridsTtTt

Cross 2 Cross 2 HybridsHybrids

getget3 Tall & 1 3 Tall & 1

ShortShortTT, Tt, ttTT, Tt, tt

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Monohybrid Monohybrid CrossesCrosses

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• Trait: Seed ShapeTrait: Seed Shape• Alleles: Alleles: RR – Round – Round rr – Wrinkled – Wrinkled• Cross: Cross: RoundRound seedsseeds xx Wrinkled Wrinkled seeds: Pseeds: P11 = = RRRR xx

rr rr

PP11 Monohybrid Cross Monohybrid Cross

R

R

rr

Rr

RrRr

Rr

Genotype:Genotype: RrRr

PhenotypePhenotype: RoundRound

GenotypicGenotypicRatio:Ratio: All alikeAll alike

PhenotypicPhenotypicRatio:Ratio: All alike All alike

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PP11 Monohybrid Cross Monohybrid Cross

ReviewReview Homozygous dominant x Homozygous dominant x

Homozygous recessiveHomozygous recessive OffspringOffspring allall HeterozygousHeterozygous (hybrids)(hybrids) Offspring calledOffspring called FF11 generation generation

Genotypic & Phenotypic ratio isGenotypic & Phenotypic ratio is ALL ALL ALIKEALIKE

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• Trait: Seed ShapeTrait: Seed Shape• Alleles: Alleles: RR – Round – Round rr – Wrinkled – Wrinkled• Cross: Cross: RoundRound seeds seeds xx Round Round seeds: Pseeds: P1 1 == RrRr

xx Rr Rr

FF11 Monohybrid Cross Monohybrid Cross

R

r

rR

RR

rrRr

RrGenotype:Genotype: RR, Rr, RR, Rr, rrrr

PhenotypePhenotype: Round Round && wrinkled wrinkled

G.Ratio:G.Ratio: 1:2:11:2:1

P.Ratio:P.Ratio: 3:1 3:1

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FF11 Monohybrid Cross Monohybrid Cross

ReviewReview Heterozygous x heterozygousHeterozygous x heterozygous Offspring:Offspring:

25% Homozygous dominant25% Homozygous dominant RRRR50% Heterozygous50% Heterozygous RrRr25% Homozygous Recessive25% Homozygous Recessive rrrr

Offspring calledOffspring called FF22 generation generation Genotypic ratio isGenotypic ratio is 1:2:11:2:1 Phenotypic RatioPhenotypic Ratio is 3:1 is 3:1

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What Do the Peas Look Like?What Do the Peas Look Like?

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……And Now the Test And Now the Test CrossCross

• Mendel then crossed a Mendel then crossed a purepure & a & a hybridhybrid from his from his FF2 2 generationgeneration

• This is known as an This is known as an FF22 or test cross or test cross

• There are There are twotwo possible testcrosses: possible testcrosses:Homozygous dominant x HybridHomozygous dominant x HybridHomozygous recessive x HybridHomozygous recessive x Hybrid

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• Trait: Seed ShapeTrait: Seed Shape

• Alleles: Alleles: RR – Round – Round rr – Wrinkled – Wrinkled

• Cross: Cross: Pure Pure RoundRound seedsseeds xx Hybrid Hybrid Round Round seeds seeds

• P =P =RRRR xx RrRr

FF22 Monohybrid Cross (1 Monohybrid Cross (1stst))

R

R

rR

RR

RrRR

Rr

Genotype:Genotype: RR, RR, RrRr

PhenotypePhenotype: RoundRound

GenotypicGenotypicRatio:Ratio: 1:11:1

PhenotypicPhenotypicRatio:Ratio: All alike All alike

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• Trait: Seed ShapeTrait: Seed Shape

• Alleles: Alleles: RR – Round – Round rr – Wrinkled – Wrinkled

• Cross: Cross: WrinkledWrinkled seedsseeds xx Hybrid Hybrid Round Round seeds seeds = = rrrr xx Rr Rr

FF22 Monohybrid Cross (2nd) Monohybrid Cross (2nd)

r

r

rR

Rr

rrRr

rr

Genotype:Genotype: Rr, rrRr, rr

PhenotypePhenotype: Round & Round & WrinkledWrinkled

G. Ratio:G. Ratio: 1:11:1

P.Ratio:P.Ratio: 1:1 1:1

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FF22 Monohybrid Cross Monohybrid Cross

ReviewReview Homozygous x heterozygous(hybrid)Homozygous x heterozygous(hybrid) Offspring:Offspring:

50% Homozygous 50% Homozygous RR or rrRR or rr50% Heterozygous50% Heterozygous RrRr

Phenotypic RatioPhenotypic Ratio is 1:1 is 1:1 Called Called Test CrossTest Cross because the offspring because the offspring

have have SAMESAME genotype as parents genotype as parents

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Mendel’s LawsMendel’s Laws

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Results of Monohybrid Results of Monohybrid CrossesCrosses

• InheritableInheritable factors or genesfactors or genes are are responsible for all heritable characteristics responsible for all heritable characteristics

• PhenotypePhenotype is based on is based on GenotypeGenotype • Each traitEach trait is based onis based on two genestwo genes, , one from one from

the mother and the other from the fatherthe mother and the other from the father • True-breeding individuals are True-breeding individuals are homozygous homozygous

( both alleles)( both alleles) are the same are the same

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Law of DominanceLaw of Dominance In a cross of parents that are In a cross of parents that are pure for contrasting traitspure for contrasting traits, , only one form of the trait will only one form of the trait will appear in the next generation.appear in the next generation.

All the offspring will be All the offspring will be heterozygous and express heterozygous and express only the only the dominant trait.dominant trait.

RR x rr RR x rr yieldsyields all Rr (round all Rr (round seeds)seeds)

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Law of DominanceLaw of Dominance

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Law of SegregationLaw of Segregation

• During the During the formation of gametesformation of gametes (eggs (eggs or sperm), the or sperm), the two allelestwo alleles responsible responsible for a trait for a trait separateseparate from each other. from each other.

• Alleles for a trait are then Alleles for a trait are then "recombined" at fertilization"recombined" at fertilization, , producing the genotype for the traits producing the genotype for the traits of the offspringof the offspring.

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Applying the Law of Applying the Law of SegregationSegregation

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Law of Independent Law of Independent AssortmentAssortment

• Alleles for Alleles for differentdifferent traits are traits are distributed to sex cells (& offspring) distributed to sex cells (& offspring) independently of one another.independently of one another.

• This law can be illustrated using This law can be illustrated using dihybrid crossesdihybrid crosses..

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Sex-linked TraitsSex-linked Traits• Traits (genes) located on the Traits (genes) located on the sex sex

chromosomeschromosomes• Sex chromosomes are Sex chromosomes are X and YX and Y• XXXX genotype for females genotype for females• XYXY genotype for males genotype for males• Many Many sex-linked traitssex-linked traits carried on carried on XX

chromosomechromosome

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Sex-linked TraitsSex-linked Traits

Sex ChromosomesSex Chromosomes

XX chromosome - female Xy chromosome - male

fruit flyeye color

Example: Example: Eye color in fruit Eye color in fruit fliesflies

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Sex-linked Trait ProblemSex-linked Trait Problem• Example: Eye color in fruit flies• (red-eyed male) x (white-eyed female)

XRY x XrXr

• Remember: the Y chromosome in males does not carry traits.• RR = red eyed• Rr = red eyed• rr = white eyed• XY = male• XX = female

XR

Xr Xr

Y

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Sex-linked Trait Sex-linked Trait Solution:Solution:

XR Xr

Xr Y

XR Xr

Xr Y

50% red eyed female

50% white eyed male

XR

Xr Xr

Y

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Female CarriersFemale Carriers

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Incomplete Incomplete DominanceDominance

andandCodominanceCodominance

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Incomplete DominanceIncomplete Dominance

• F1 hybrids F1 hybrids have an appearance somewhat in in betweenbetween the phenotypes phenotypes of the two parental varieties.

• Example:Example: snapdragons (flower)snapdragons (flower)• red (RR) x white (WW)

• RR = red flowerRR = red flower• WW = white flower

R

R

W W

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Incomplete Incomplete DominanceDominance

RWRW

RWRW

RWRW

RWRW

RR

RR

WW

All RW =All RW = pink pink(heterozygous pink)(heterozygous pink)

produces theproduces theFF11 generation generation

W

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Incomplete Dominance Problem:

• In cattle when a red bull(RR) is mated with white(WW) cow the offspring are roan(RW) a blending of red and white. Mate a red bull with a roan cow. Show the P1, the Punnett Square, and give the genotypic and phenotypic ratios for this cross.

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P1 = __RR__ x __RW__

Phenotypic ratio: ____ : _____ : _____

Genotypic ratio: ____ : _____ : _____

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Incomplete DominanceIncomplete Dominance

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CodominanceCodominance• Two allelesTwo alleles are expressed ( are expressed (multiple allelesmultiple alleles) in ) in

heterozygous individualsheterozygous individuals..• Example:Example: blood type blood type

• 1.1. type Atype A = I= IAAIIAA or I or IAAii• 2.2. type Btype B = I= IBBIIBB or I or IBBii• 3.3. type ABtype AB = I= IAAIIBB

• 4.4. type Otype O = ii= ii

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Codominance Codominance ProblemProblem

• Example:homozygous male Type B (IBIB) x heterozygous female Type A (IAi)

IAIB IBi

IAIB IBi

1/2 = IAIB

1/2 = IBi

IB

IA i

IB

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Another Codominance Another Codominance ProblemProblem

• Example:Example: male Type O (ii) x female type AB (IAIB)

IAi IBi

IAi IBi

1/2 = IAi1/2 = IBi

i

IA IB

i

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CodominanceCodominance• QuestionQuestion::

If a boy has a blood type O and If a boy has a blood type O and his sister has blood type AB, his sister has blood type AB, what are the genotypes and what are the genotypes and phenotypes of their parents?phenotypes of their parents?

• boy - boy - type O (ii) type O (ii) X girl - X girl - type AB (Itype AB (IAAIIBB))

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CodominanceCodominance• Answer:Answer:

IAIB

ii

Parents:Parents:genotypesgenotypes = IAi and IBiphenotypesphenotypes = A and B

IB

IA i

i

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Dihybrid CrossDihybrid Cross• A breeding experiment that tracks the A breeding experiment that tracks the

inheritance of two traitsinheritance of two traits..

• Mendel’s Mendel’s “Law of Independent “Law of Independent Assortment”Assortment”

• a. Each pair of alleles segregates a. Each pair of alleles segregates independentlyindependently during gamete formation during gamete formation

• b. Formula: 2b. Formula: 2nn (n = # of heterozygotes) (n = # of heterozygotes)

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Question:Question:How many gametes will be produced How many gametes will be produced for the following allele arrangements?for the following allele arrangements?

• Remember:Remember: 22nn (n = # of heterozygotes) (n = # of heterozygotes)

• 1.1. RrYyRrYy

• 2.2. AaBbCCDdAaBbCCDd

• 3.3. MmNnOoPPQQRrssTtQqMmNnOoPPQQRrssTtQq

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Answer:Answer:1. RrYy: 21. RrYy: 2nn = 2 = 222 = 4 gametes = 4 gametes

RY Ry rY ryRY Ry rY ry

2. AaBbCCDd: 22. AaBbCCDd: 2nn = 2 = 233 = 8 gametes = 8 gametes

ABCD ABCd AbCD AbCdABCD ABCd AbCD AbCd

aBCD aBCd abCD abCDaBCD aBCd abCD abCD

3. MmNnOoPPQQRrssTtQq: 23. MmNnOoPPQQRrssTtQq: 2nn = 2 = 266 = = 64 gametes64 gametes

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Dihybrid CrossDihybrid Cross

• Traits: Seed shape & Seed colorTraits: Seed shape & Seed color• Alleles:Alleles: R round

r wrinkled Y yellow y green

• RrYy x RrYy

RY Ry rY ryRY Ry rY ry RY Ry rY ryRY Ry rY ry

All possible gamete combinationsAll possible gamete combinations

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Dihybrid CrossDihybrid CrossRYRY RyRy rYrY ryry

RYRY

RyRy

rYrY

ryry

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Dihybrid CrossDihybrid Cross

RRYY

RRYy

RrYY

RrYy

RRYy

RRyy

RrYy

Rryy

RrYY

RrYy

rrYY

rrYy

RrYy

Rryy

rrYy

rryy

Round/Yellow: 9

Round/green: 3

wrinkled/Yellow: 3

wrinkled/green: 1

9:3:3:1 phenotypic ratio

RYRY RyRy rYrY ryry

RYRY

RyRy

rYrY

ryry

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Dihybrid CrossDihybrid Cross

Round/Yellow: 9Round/green: 3wrinkled/Yellow: 3wrinkled/green: 1

9:3:3:1

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Test CrossTest Cross• A mating between an individual of A mating between an individual of unknown genotype unknown genotype

and a and a homozygous recessivehomozygous recessive individual. individual.• Example:Example: bbC__ bbC__ x x bbccbbcc

• BB = brown eyesBB = brown eyes• Bb = brown eyesBb = brown eyes• bb = blue eyesbb = blue eyes

• CC = curly hairCC = curly hair• Cc = curly hairCc = curly hair• cc = straight haircc = straight hair

bCbC b___b___

bcbc

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Test CrossTest Cross

• Possible results:Possible results:bCbC b___b___

bcbc bbCc bbCc

C bCbC b___b___

bcbc bbCc bbccor

c

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Summary of Mendel’s lawsSummary of Mendel’s laws

LAWLAW PARENT PARENT CROSSCROSS OFFSPRINGOFFSPRING

DOMINANCEDOMINANCE TT x ttTT x tt tall x shorttall x short

100% Tt 100% Tt talltall

SEGREGATIONSEGREGATION Tt x TtTt x Tt tall x talltall x tall

75% tall 75% tall 25% short25% short

INDEPENDENT INDEPENDENT ASSORTMENTASSORTMENT

RrGg x RrGgRrGg x RrGg round & round & green x green x round & round & greengreen

9/16 round seeds & green 9/16 round seeds & green pods pods

3/16 round seeds & yellow 3/16 round seeds & yellow pods pods

3/16 wrinkled seeds & 3/16 wrinkled seeds & green pods green pods

1/16 wrinkled seeds & 1/16 wrinkled seeds & yellow podsyellow pods

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Genetic Practice Genetic Practice ProblemsProblems

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Breed the PBreed the P11 generation generation• tall (TT) x dwarf (tt) tall (TT) x dwarf (tt) pea plantspea plants

T

T

t t

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Solution:Solution:

T

T

t t

Tt

Tt

Tt

Tt All Tt = tall(heterozygous tall)

produces theFF11 generation generation

tall (TT) vs. dwarf (tt) pea tall (TT) vs. dwarf (tt) pea plantsplants

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Breed the FBreed the F11

generationgeneration• tall (Tt) vs. tall (Tt) tall (Tt) vs. tall (Tt) pea plantspea plants

T

t

T t

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Solution:Solution:

TT

Tt

Tt

tt

T

t

T tproduces theFF22 generation generation

1/4 (25%) = TT1/2 (50%) = Tt1/4 (25%) = tt1:2:1 genotype1:2:1 genotype 3:1 phenotype3:1 phenotype

tall (Tt) x tall (Tt) pea tall (Tt) x tall (Tt) pea plantsplants

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