Classical Genetics

Why Peas? Read Concept 2 Click Animation and View

Mendel’s Work Traits are passed on by factors (genes) Factors (genes) have more than 1 form called alleles There are at least 2 alleles for each trait

MENDELIAN INHERITANCE Mendelian inheritance patterns Involve genes directly influencing traits Obey Mendel’s laws Law of segregation Law of independent assortment Include Dominant / recessive relationships Gene interactions Phenotype-influencing roles of sex and environment Most genes of eukaryotes follow a Mendelian inheritance pattern 4 4

I. Principles of Genetics Random segregation- dominant and recessive alleles separate randomly during meiosis II

I. Principles of Genetics Independent assortment- parental chromosomes (homologs) separate randomly during meiosis I

II. Types of Crosses Monohybrid Cross- looking at 1 trait in individuals that are mated (AA x Aa). A 4-square Punnett Dihybrid Cross- looking at 2 traits in individuals that are mated (AaBb x AaBB). A 16 Square Punnett. Looking at more than one gene at a time.

II. Breeding Experiments B. The difference between genotype and phenotype 1. Genotype- pattern of alleles (think gene). This is the genetic appearance. AA, Bb, cc 2. Phenotype- physical appearance (phenotype=physical) resulting from the genotype. Ex: A=brown hair, b=no tail 3. Homozygous- 2 copies of the same allele. Either homozygous dominant (AA) or homozygous recessive (aa). 4. Heterozygous- 1 copy of each allele (Aa or Bb).

III. Play with Punnett Squares A. The language of crosses 1. P= parent generation in a cross (what you start with). Example: P=Aa x AA 2. F1= First generation of offspring from the cross 3. F2 = Second generation of offspring, produced by crossing organisms in F1 with themselves.

III. Play with Punnet Squares B. Simple dominance: one allele is completely dominant P=Aa x Aa A=red, a=clear A a   A a F1 Genotype= AA: Aa: aa 1: 2: 1 F1 Phenotype= A (red): a (clear) 3: 1 AA Aa aa

III. Play with Punnett Squares C. Incomplete dominance- a recessive allele is not completely recessive and shows through the dominant allele. Example: breeding red snapdragons and white snapdragons produces pink snapdragons. P= RR x rr R =red, r =white R R r Rr F1 Genotype= Rr Phenotype= pink

III. Play with Punnett Squares D. Codominance- both alleles are equally strong and produce a heterozygous phenotype that is a mixture of both Like in incomplete dominance, heterozygotes have a unique phenotype (appearance). Example: breeding white beans and purple beans produces white beans with purple spots. P=CWCP x CWCP CW=white, CP=purple CW CP CW CP F1 Genotype= AA: Aa: aa 1: 2: 1 Phenotype= A : Aa (spotted): a CWCW CWCP CPCP

Mendel’s Work Alleles are represented by the first letter of the dominant trait Pea plant flower color (trait) Purple or white Purple is dominant over white Alleles are represented as P=purple p=white

Mendel’s Work You try a few… Round seeds are dominant over wrinkled seeds Yellow seeds are dominant over green seeds Tall plants are dominant over short plants 1. R=round r=wrinkled Y=yellow y=green T=tall t= short

Mendel’s Work Describing Traits (Tall or short plants) Genotype The actual genetic make-up of an organism the “genes” Phenotype The physical appearance or form observed the “physical”

Mendel’s Work Possible Genotypes (TT Tt tt) Homozygous The two alleles for the trait are identical TT homozygous dominant (purebred dominant) tt homozygous recessive (purebred recessive) Heterozygous The two alleles for the trait are different Tt heterozygote (hybrid)

Mendel’s Conclusions 1. Law of Dominance Alleles for a trait are either dominant or recessive The dominant form is expressed and the recessive form is hidden The only way to express a recessive trait is if there are two copies of the recessive allele

Mendel’s Conclusions Law of Segregation The two alleles for a trait separate during gametogenesis

II. Types of Crosses C. Test Cross-an individual of unknown genotype (A?) is mated with a homozygous recessive individual (aa) to determine the unknown genotype. -(P= A? x aa)

C. A Test Cross PP Pp pp Pp P ? P ? If all of the offspring have a dominant phenotype, you know the unknown parent genotype was homozygous dominant (left Punnet, AA). If some of the offspring have the recessive phenotype, you know the unknown parent genotype was heterozygous (right Punnett, Aa).

Mendel’s 3rd Law of Independent Assortment Alleles of different genes are assorted independently of one another during the formation of gametes This means that calculating the probability of several traits appearing together is the product of the probability of each trait taken separately The Rule of Multiplication

Studying the inheritance of two characters simultaneously THE DIHYBRID CROSS Studying the inheritance of two characters simultaneously © 2007 Paul Billiet ODWS

Dihybrid Crosses the natural progression for Mendel was to study 2 characteristics at the same time. thus, the study of 2 pairs of contrasting traits at the same time = a dihybrid cross ex.. round yellow seeds X wrinkled green seeds

Mendel’s Fourth Law The Law of Independent Assortment During gamete formation, segregating pairs of unit factors assort independently of each other. the two traits are inherited totally independently of each other. i.e. color is inherited independently of seed shape. Example: Homozygous round green X wrinkled homozygous yellow

Mendel’s peas Character Trait Allele Seed shape Round R Wrinkled r Cotyledon colour Yellow Y Green y © 2007 Paul Billiet ODWS

Combinations Genotype Phenotype RRYY Round Yellow RRYy RrYY RrYy RRyy Round Green Rryy rrYY Wrinkled Yellow rrYy rryy Wrinkled Green © 2007 Paul Billiet ODWS

Is the inheritance of one character affected by the inheritance of another? P Phenotypes Round Yellow x Wrinkled Green (Pure Bred) F1 Phenotypes All Round Yellow (Selfed) F2 Phenotypes Seed colour Yellow Green TOTAL RATIO Round 315 108 423 shape Wrinkled 101 32 133 416 140 556 © 2007 Paul Billiet ODWS

Is the inheritance of one character affected by the inheritance of another? P Phenotypes Round Yellow x Wrinkled Green (Pure Bred) F1 Phenotypes All Round Yellow (Selfed) F2 Phenotypes Seed colour Yellow Green TOTAL RATIO Round 315 108 423 3.18 shape Wrinkled 101 32 133 1 416 140 556 2.97 © 2007 Paul Billiet ODWS

A dihybrid cross can be treated as two separate monohybrid crosses The expected probability of each type of seed can be calculated: Probability of an F2 seed being round = 75% or ¾ Probability of an F2 seed being wrinkled = Probability of an F2 seed being yellow = Probability of an F2 seed being green = © 2007 Paul Billiet ODWS

A dihybrid cross can be treated as two separate monohybrid crosses The expected probability of each type of seed can be calculated: Probability of an F2 seed being round = 75% or ¾ Probability of an F2 seed being wrinkled = 25% or ¼ Probability of an F2 seed being yellow = 75% or ¾ Probability of an F2 seed being green = 25% or ¼ © 2007 Paul Billiet ODWS

Therefore Probability of an F2 seed being round and yellow = ¾ x ¾ = 9/16 = 56.25% Probability of an F2 seed being round and green = Probability of an F2 seed being wrinkled and yellow = Probability of an F2 seed being wrinkled and green = © 2007 Paul Billiet ODWS

Therefore Probability of an F2 seed being round and yellow = ¾ x ¾ = 9/16 = 56.25% Probability of an F2 seed being round and green = ¾ x ¼ = 3/16 = 18.75% Probability of an F2 seed being wrinkled and yellow = ¼ x ¾ = 3/16 = 18.75% Probability of an F2 seed being wrinkled and green = ¼ x ¼ = 1/16 = 6.25% © 2007 Paul Billiet ODWS

Predicting how many seeds we could expect to get in a sample What Mendel observed 556 x 9/16 round yellow 313 315 556 x 3/16 round green 108 556 x 3/16 wrinkled yellow 101 556 x 1/16 wrinkled green 32 © 2007 Paul Billiet ODWS

Predicting how many seeds we could expect to get in a sample What Mendel observed 556 x 9/16 round yellow 313 315 556 x 3/16 round green 104 108 556 x 3/16 wrinkled yellow 101 556 x 1/16 wrinkled green 35 32 © 2007 Paul Billiet ODWS

THE LAW OF INDEPENDENT ASSORTMENT It appears that the inheritance of seed shape has no influence over the inheritance of seed colour The two characters are inherited INDEPENDENTLY The pairs of alleles that control these two characters assort themselves independently © 2007 Paul Billiet ODWS

Mendel & Meiosis The pairs of chromosomes could orientate in different ways at Anaphase 1 © 2007 Paul Billiet ODWS

Dihybrid cross genetic diagram P Phenotypes Round Yellow seed X Wrinkled Green seed (Pure bred) Genotypes RRYY rryy meiosis Gametes RY ry fertilisation F1 RrYy (Selfed) Round Yellow Proportions 100% © 2007 Paul Billiet ODWS

Dihybrid cross genetic diagram F1 Phenotypes RrYy (Selfed) Genotypes Round Yellow Proportions 100% meiosis Gametes Y y R RY Ry r rY ry © 2007 Paul Billiet ODWS

Dihybrid cross genetic diagram Gametes Y y R RY Ry r rY ry fertilisation F2 Genotypes RRYY RRYy RrYY RrYy RRyy Rryy rrYY rrYy rryy © 2007 Paul Billiet ODWS

Dihybrid cross proportions Phenotypes Proportions Round Yellow 9/16 or 56.25% Round Green Wrinkled Yellow Wrinkled Green © 2007 Paul Billiet ODWS

Dihybrid cross proportions Phenotypes Proportions Round Yellow 9/16 or 56.25% Round Green 3/16 or 18.75% Wrinkled Yellow Wrinkled Green 1/16 or 6.25% © 2007 Paul Billiet ODWS

Dihybrid test cross In monohybrid crosses, to know if a dominant trait is homozygous (RR) or heterozygous (Rr) it is necessary to carry out a test cross This is done with a homozygous recessive (rr) individual The same is true for a dihybrid cross where the test cross is made with an individual which is homozygous recessive for both characters (rryy) © 2007 Paul Billiet ODWS

Dihybrid test cross Phenotypes Round Yellow X Wrinkled Green Genotypes RrYy rryy Gametes RY, Ry, rY, ry ry Genotypes RY Ry rY ry RrYy Rryy rrYy rryy Phenotypes Round Yellow Round Green Wrinkled Yellow Wrinkled Green Proportions 1/4 or 25% © 2007 Paul Billiet ODWS

Solving Dihybrid Crosses Follow the steps If you were to cross a Homozygous round green X wrinkled homozygous yellow plant, what would your phenotypic and genotypic ratios be?

Step 1: Make a Key…. G= yellow g = green W = round w = wrinkled

Step 2: Give the genotypes of the parents. . . Homozygous round green X wrinkled homozygous yellow plant P1 = WWgg X wwGG

Step 3: Determine the Gametes . . . WWgg X wwGG WWgg = Wg, Wg, Wg, Wg wwGG = wG, wG, wG, wG HINT: Foil to get the gametes

Step 4: Fill in the Punnett square . . . Wg wG WwGg

Step 5: Answer any questions . . . Genotypic ratio = all WwGg Phenotypic ratio = all round yellow G= yellow g = green W = round w = wrinkled

F1 X F1 = WwGg X WwGg WwGg X WwGg Remember to determine the gametes for your punnett square. WwGg X WwGg

Fill in the Punnett Square WG Wg wG wg WWGG WWGg WwGG WwGg WWgg Wwgg wwGG wwGg wwgg Do you see a pattern?

The PATTERN . . . WG Wg wG wg WWGG WWGg WwGG WwGg WWgg Wwgg wwGG wwGg

What are the ratios? Genotypic Ratio: 1 WWGG 2 WWGg 1 WWgg 2 WwGG

Phenotypic Ratio: Phenotypic Ratio: Genotypic Ratio: 9 Yellow Round G= yellow g = green W = round w = wrinkled Phenotypic Ratio: Genotypic Ratio: 1 WWGG 2 WWGg 1 WWgg 2 WwGG 4 WwGg 2 Wwgg 1 wwGG 2 wwGg 1 wwgg Phenotypic Ratio: 9 Yellow Round Dominant Dominant 3 Yellow Wrinkled Dominant Recessive 3 Green Round Recessive Dominant 1 Green Wrinkled Recessive Recessive ** Explain the patterns

The Short Cut: can only be used if you have (Big little Big little X Big little Big little) BbTt x BbTt Do these fit? BBTt X BbTt BbTt x Bb tt

Try this: C= Inflated Pod c = constricted pod D = Tall d = dwarf The cross: a plant heterozygous for inflated pod and heterozygous tall is crossed with a another heterozygous inflated pod and heterozygous tall plant. What are the genotypic and phenotypic ratios? The parents are: CcDd x CcDd

Example: CcDd x CcDd Does this fit the short cut? Yes C= Inflated Pod c = constricted pod D = Tall d = dwarf Example: Genotypic Ratio: 1 CCDD 2 CCDd 1 CCdd 2 CcDD 4 CcDd 2 Ccdd 1 ccDD 2 ccDd 1 ccdd Genotypic Ratio: 1 2 4 CcDd x CcDd Does this fit the short cut? Yes Phenotypic Ratio: (D) (D) 9 Inflated Tall (D) ® 3 Inflated Dwarf ® (D) 3 Constricted Tall ® ® 1 Constricted Dwarf Phenotypic Ratio: (D) (D) 9 (D) ® 3 ® (D) ® ® 1