1. Mendelelian Genetics copyright cmassengale

2. Gregor Mendel (1822-1884) Responsible for the Laws governing Inheritance of Traits copyright cmassengale 3. Gregor Johann Mendel

Austrian monk

Studied theinheritanceof traits inpea plants

Developed thelaws of inheritance

Mendel's work was not recognized until the turn of the20th century

copyright cmassengale 4. Gregor Johann Mendel

Between1856 and 1863,Mendel cultivated and tested some28,000 pea plants

He found that the plants' offspring retainedtraits of the parents

Called theFather of Genetics"

copyright cmassengale 5. Site of Gregor Mendels experimental garden in the Czech Republic copyright cmassengale 6.

Mendel stated that physical traits are inherited asparticles

Mendel did not know that the particles were actuallyChromosomes & DNA

Particulate Inheritance copyright cmassengale 7. Genetic Terminology

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

Heredity- passing of traits from parent to offspring

Genetics- study of heredity

copyright cmassengale 8. Types of Genetic Crosses

Monohybrid cross-cross involving a single trait e.g. flower color

Dihybrid cross-cross involving two traitse.g. flower color & plant height

copyright cmassengale 9. Punnett Square

Used to help solve genetics problems

copyright cmassengale 10. copyright cmassengale 11. DesignerGenes

Alleles-two forms of agene(dominant & recessive)

Dominant-stronger of two genes expressed in the hybrid; represented by a capital letter (R)

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

copyright cmassengale 12. More Terminology

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

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

copyright cmassengale 13. Genotype & Phenotype in Flowers Genotype of alleles: R = red flower r = yellow flower All genes occur in pairs, so2 allelesaffect a characteristic Possible combinations are: Genotypes RR R r rr Phenotypes REDREDYELLOW copyright cmassengale 14. Genotypes

Homozygous genotype - gene combination involving 2 dominant or 2 recessive genes(e.g. RR or rr);also called pure

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

copyright cmassengale 15. Genes and Environment Determine Characteristics copyright cmassengale 16. Mendels Pea Plant Experiments copyright cmassengale 17. Why peas, Pisum sativum ?

Can be grown in asmall area

Producelots of offspring

Producepureplants when allowed toself-pollinateseveral generations

Can beartificially cross-pollinated

copyright cmassengale 18. Reproduction in Flowering Plants

Pollen contains sperm

• Produced by the stamen

Ovary contains eggs

• Found inside the flower

Pollen carries sperm to the eggs for fertilization

• Self-fertilizationcan occur in the same flower

• Cross-fertilizationcan occur between flowers

copyright cmassengale 19. Mendels Experimental Methods

Mendelhand-pollinatedflowers using apaintbrush

• He couldsnip the stamensto prevent self-pollination

• Covered each flower with a cloth bag

He traced traits through theseveral generations

copyright cmassengale 20. How Mendel Began Mendel producedpurestrains by allowing the plants toself-pollinatefor several generations copyright cmassengale 21. Eight Pea Plant Traits

Seed shape --- Round(R)or Wrinkled(r)

Seed Color ---- Yellow(Y)or Green( y )

Pod Shape --- Smooth(S)or wrinkled( s )

Pod Color --- Green(G)or Yellow(g)

Seed Coat Color ---Gray(G)or White(g)

Flower position ---Axial(A)or Terminal(a)

Plant Height --- Tall(T)or Short(t)

Flower color---Purple(P)or white( p )

copyright cmassengale 22. copyright cmassengale 23. copyright cmassengale 24. Mendels Experimental Results copyright cmassengale 25.

Did the observed ratio match the theoretical ratio?

The theoretical or expected ratio of plants producing round or wrinkled seeds is3 round :1 wrinkled Mendels observed ratio was 2.96:1 The discrepancy is due tostatistical error Thelarger the samplethe more nearly the results approximate to the theoretical ratio copyright cmassengale 26. Generation Gap

Parental P 1Generation = the parental generation in a breeding experiment .

F 1generation = the first-generation offspring in a breeding experiment.(1st filial generation)

• From breeding individuals from the P 1generation

F 2generation = the second-generation offspring in a breeding experiment.(2nd filial generation)

• From breeding individuals from the F 1generation

copyright cmassengale 27. Following the Generations Cross 2 Pure Plants TT x tt Results in all Hybrids Tt Cross 2 Hybrids get 3 Tall & 1 Short TT, Tt, tt copyright cmassengale 28. Monohybrid Crosses copyright cmassengale 29.

Trait: Seed Shape

Alleles:R Round r Wrinkled

Cross:Round seeds x Wrinkledseeds

RR x rr

P 1Monohybrid Cross R R r r Rr Rr Rr Rr Genotype: Rr Phenotype : Round Genotypic Ratio: All alike Phenotypic Ratio:All alike copyright cmassengale 30. P 1Monohybrid Cross Review

Homozygous dominant x Homozygous recessive

Offspring all Heterozygous (hybrids)

Offspring called F 1generation

Genotypic & Phenotypic ratio is ALL ALIKE

copyright cmassengale 31.

Trait: Seed Shape

Alleles:R Round r Wrinkled

Cross:Round seedsx Roundseeds

Rr x Rr

F 1Monohybrid Cross R r r R RR rr Rr Rr Genotype: RR, Rr, rr Phenotype : Round & wrinkled G.Ratio: 1:2:1 P.Ratio: 3:1 copyright cmassengale 32. F 1Monohybrid Cross Review

Heterozygous x heterozygous

Offspring: 25% Homozygous dominant RR 50% Heterozygous Rr 25% Homozygous Recessive rr

Offspring called F 2generation

Genotypic ratio is 1:2:1

Phenotypic Ratiois 3:1

copyright cmassengale 33. What Do the Peas Look Like? copyright cmassengale 34. And Now the Test Cross

Mendel then crossed apure& ahybridfrom hisF 2generation

This is known as anF 2or test cross

There aretwopossible testcrosses: Homozygous dominant x Hybrid Homozygous recessive x Hybrid

copyright cmassengale 35.

Trait: Seed Shape

Alleles:R Round r Wrinkled

Cross:Round seeds x Roundseeds

RR x Rr

F 2Monohybrid Cross (1 st ) R R r R RR Rr RR Rr Genotype: RR, Rr Phenotype : Round Genotypic Ratio: 1:1 Phenotypic Ratio:All alike copyright cmassengale 36.

Trait: Seed Shape

Alleles:R Round r Wrinkled

Cross:Wrinkled seeds x Roundseeds

rr x Rr

F 2Monohybrid Cross (2nd) r r r R Rr rr Rr rr Genotype: Rr, rr Phenotype : Round & Wrinkled G. Ratio: 1:1 P.Ratio:1:1 copyright cmassengale 37. F 2Monohybrid Cross Review

Homozygous x heterozygous(hybrid)

Offspring: 50% HomozygousRR or rr 50% Heterozygous Rr

Phenotypic Ratiois 1:1

CalledTest Crossbecause the offspring haveSAMEgenotype as parents

copyright cmassengale 38. Practice Your Crosses

Work the P 1 , F 1 , and both F 2Crosses for each of the other Seven Pea Plant Traits

copyright cmassengale 39. Mendels Laws copyright cmassengale 40. Results of Monohybrid Crosses

Inheritable factors or genes are responsible for all heritable characteristics

Phenotypeis based onGenotype

Each trait is based on two genes ,one from the mother and the other from the father

True-breeding individuals arehomozygous ( both alleles)are the same

copyright cmassengale 41. Law of Dominance In a cross of parents that arepure for contrasting traits , only one form of the trait will appear in the next generation. All the offspring will be heterozygous and express only thedominant trait. RR x rryieldsall Rr (round seeds) copyright cmassengale 42. Law of Dominance copyright cmassengale 43. Law of Segregation

During theformation of gametes(eggs or sperm), thetwo allelesresponsible for a traitseparatefrom each other.

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

copyright cmassengale 44. Applying the Law of Segregation copyright cmassengale 45. Law of Independent Assortment

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

This law can be illustrated usingdihybrid crosses .

copyright cmassengale 46. Dihybrid Cross

A breeding experiment that tracks theinheritance of two traits .

MendelsLaw of Independent Assortment

a. Each pair of alleles segregatesindependentlyduring gamete formation

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

copyright cmassengale 47. Question: How many gametes will be produced for the following allele arrangements?

Remember: 2 n (n = # of heterozygotes)

1. RrYy

2. AaBbCCDd

3. MmNnOoPPQQRrssTtQq

copyright cmassengale 48. Answer: 1.RrYy:2 n= 2 2= 4 gametes RYRyrYry 2.AaBbCCDd:2 n= 2 3= 8 gametes ABCDABCdAbCDAbCd aBCDaBCdabCDabCD 3.MmNnOoPPQQRrssTtQq:2 n= 2 6= 64 gametes copyright cmassengale 49. Dihybrid Cross

Traits: Seed shape & Seed color

Alleles: R round r wrinkled Y yellow y green

RrYyxRrYy RY Ry rY ry RY Ry rY ry All possible gamete combinations copyright cmassengale 50. Dihybrid Cross copyright cmassengale RY Ry rY ry RY Ry rY ry 51. Dihybrid Cross copyright cmassengale 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 RY Ry rY ry RY Ry rY ry 52. Dihybrid Cross Round/Yellow:9 Round/green:3 wrinkled/Yellow: 3 wrinkled/green:1 9:3:3:1 copyright cmassengale 53. Test Cross

A mating between an individual ofunknown genotypeand ahomozygous recessiveindividual.

Example: bbC__xbbcc

BB = brown eyes

Bb = brown eyes

bb = blue eyes

CC = curly hair

Cc = curly hair

cc = straight hair

copyright cmassengale bC b___ bc 54. Test Cross

Possible results:

copyright cmassengale bC b___ bc bbCc bb C c C bC b___ bc bbCc bb c c or c 55. Summary of Mendels laws copyright cmassengale LAW PARENT CROSS OFFSPRING DOMINANCE TT x tt tall x short 100% Tttall SEGREGATION Tt x Tt tall x tall 75% tall25% short INDEPENDENT ASSORTMENT RrGg x RrGg round & green xround & green 9/16 round seeds & green pods3/16 round seeds & yellow pods3/16 wrinkled seeds & green pods1/16 wrinkled seeds & yellow pods 56. Incomplete Dominance and Codominance copyright cmassengale 57. Incomplete Dominance

F1 hybridshave an appearance somewhatin betweenthephenotypesof the two parental varieties.

Example: snapdragons (flower)

red (RR)xwhite (rr)

RR = red flower

rr = white flower

copyright cmassengale R R r r 58. Incomplete Dominance r copyright cmassengale Rr Rr Rr Rr R R r All Rr =pink (heterozygous pink) produces the F 1generation 59. Incomplete Dominance copyright cmassengale 60. Codominance

Two allelesare expressed ( multiple alleles ) inheterozygous individuals .

Example: blood type

1. type A =I A I Aor I A i

2. type B =I B I Bor I B i

3. type AB =I A I B

4. type O =ii

copyright cmassengale 61. Codominance Problem

Example: homozygous male Type B (I B I B )

xheterozygous female Type A (I A i)

copyright cmassengale I A I B I B i I A I B I B i 1/2 =I A I B 1/2 =I B i I B I A i I B 62. Another Codominance Problem

Example: male Type O (ii) xfemale type AB (I A I B )

copyright cmassengale I A i I B i I A i I B i 1/2 =I A i 1/2 =I B i i I A I B i 63. Codominance

Question : If a boy has a blood type O andhis sister has blood type AB,what are the genotypes andphenotypes of their parents?

boy -type O (ii)Xgirl -type AB (I A I B )

copyright cmassengale 64. Codominance

Answer:

Parents: genotypes =I A iandI B i phenotypes =AandB copyright cmassengale I A I B ii I B I A i i 65. Sex-linked Traits

Traits (genes) located on thesex chromosomes

Sex chromosomes areX and Y

XXgenotype for females

XYgenotype for males

Manysex-linked traitscarried onXchromosome

copyright cmassengale 66. Sex-linked Traits Sex Chromosomes Example:Eye color in fruit flies copyright cmassengale XX chromosome - female Xy chromosome - male fruit fly eye color 67. Sex-linked Trait Problem

Example: Eye color in fruit flies

(red-eyed male) x (white-eyed female) X R YxX r X r

Remember:the Y chromosome in males does not carry traits.

RR = red eyed

Rr= red eyed

rr= white eyed

XY= male

XX = female

copyright cmassengale X R X r X r Y 68. Sex-linked Trait Solution: 50%red eyed female 50%white eyed male copyright cmassengale X RX r X rY X RX r X rY X R X r X r Y 69. Female Carriers copyright cmassengale 70. Genetic Practice Problems copyright cmassengale 71. Breed the P 1generation

tall (TT)xdwarf (tt) pea plants

copyright cmassengale T T t t 72. Solution: tall (TT) vs. dwarf (tt) pea plants copyright cmassengale T T t t Tt Tt Tt Tt All Tt = tall (heterozygous tall) produces the F 1generation 73. Breed the F 1generation

tall (Tt) vs. tall (Tt) pea plants

copyright cmassengale T t T t 74. Solution: tall (Tt)xtall (Tt) pea plants copyright cmassengale TT Tt Tt tt T t T t produces the F 2generation 1/4 (25%) = TT 1/2 (50%) = Tt 1/4 (25%) = tt 1:2:1 genotype 3:1phenotype 75. copyright cmassengale