Genetika Mendel 1 Gregor Mendel 1822 1884 Menemukan

Genetika Mendel 1

Gregor Mendel (1822 -1884) Menemukan hukum pewarisan suatu sifat 2

§ § Gregor Johann Mendel Seorang pendeta bangsa Austria Meneliti pewarisan sifat pada tanaman arcis (pea) Menghasilkan Hukum Sistem pewarisan Temuan Mendel tak diterima sampai menjelang abad ke 20 3

§ Gregor Johann Mendel Antara 1856 and 1863, Mendel menanam dan menguji 28, 000 tanaman arcis Dia mencatat bahwa selama penelitian sifat induk tetap muncul pada anak Dijuluki “Father of Genetics" § § 4

Site of Gregor Mendel’s experimental garden in the Czech Republic 5

§ § Genetic Terminology Trait - any characteristic that can be passed from parent to offspring Heredity - passing of traits from parent to offspring 6

1). ISTILAH-2 DASAR TERKAIT PEWARISAN SIFAT a. GEN – ALEL Gb. Kromosom mengandung Gen-2 § GEN: FAKTOR GENETIK PENGATUR SIFAT Contoh: gen M mengatur sifat warna bunga, gen K mengatur sifat warna biji § ALEL: BENTUK ALTERNATIF SUATU GEN Contoh: Gen M (pengatur warna bunga) mempunyai 2 alel: § alel M menyebabkan bunga berwarna merah § alel m menyebabkan bunga berwarna putih Contoh: Gen K mengatur sifat warna biji mempunyai 2 alel: § Alel K menyebabkan biji berwarna kuning § Alel k menyebabkan biji berwarna hijau 7

b. GENOTIPE § GENOTIPE (G): susunan genetik (gen-2) organisme § Gen-2 berpasangan (dari tetua jantan dan tetua betina) Contoh: Genotipe organisme berdasar 3 pasang gen (Gambar) mempunyai genotipe: MM KK Bb PASANGAN GEN/ALEL dibedakan atas: § HOMOZIGOT: mempunyai alel sama, contoh: MM, KK § HETEROZIGOT: mempunyai alel tidak sama, contoh: Bb 8

c. FENOTIPE 9

10

Mendel’s Pea Plant Experiments 11

Mengapa arcis, Pisum sativum? § § Dapat ditanam ditempat sempit Menghasilkan banyak keturunan Menghasilkan tanaman homozigot jika menyerbuk sendiri Mudah menyilangkan 12

Reproduksi pada tanaman Pollen contains sperm Produced by the stamen Ovary contains eggs Found inside the flower Pollen membawa sperm ke egg untuk fertilisasi Self-fertilization terjadi pada bunga yg sama Cross-fertilization terjadi antara bunga berbeda 13

Mendel’s Experimental Methods Mendel hand-pollinated flowers using a paintbrush He could snip the stamens to prevent self-pollination He traced traits through the several generations 14

15

Percobaan Mendel membuat tanaman homozigot dengan membiarkannya menyerbuk sendiri beberapa generasi 16

Delapan sifat arcis 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) 17

18

19

Mendel’s Experimental Results 20

Rasio Pengamatan Vs rasio teoritis Rasio teoritis tanaman berbiji bundar (round) atau kisut (wrinkled) adalah 3 round : 1 wrinkled Rasio yang diamati Mendel adalah 2. 96: 1 Perbedaan disebabkan kesalahan statistik (statistical error) Semakin besar sample semakin mendekati rasio teoritis 21

Macam Generasi Induk (Parental) P 1 = generasi induk dalam suatu percobaan genetika Generasi F 1 = Generasi turunan pertama (1 st filial generation) dalam suatu percobaan genetika Generasi F 2 = Generation turunan kedua generation dalam suatu percobaan genetika. (2 nd filial generation) 22

Following the Generations Persilangan 2 induk homozigot TT x tt Hasilnya semua Hybrids Tt Persilangan 2 Hybrid menghasilkan 3 Tinggi & 1 pendek TT, Tt, tt 23

Monohybrid 24

a). Hk. MENDEL I (Hukum Segregasi/Pemisahan Pasangan Gen/Alel): Pada pembentukan gamet*), ALEL DARI PASANGAN-2 GEN AKAN “MEMISAH (BERSEGREGASI) ke dalam Gamet-gamet yg dibentuk” *) Perbiakan generatif melibatkan: § Pembentukan gamet-2 § Penyatuan gamet 25

P 1 Monohybrid Cross Sifat: bentuk biji (Seed Shape) Alleles: R – Round r – Wrinkled Cross: Round seeds x Wrinkled seeds RR x rr r r R Rr Rr Genotype: Rr Phenotype: Phenotype Round Genotypic Ratio: Semua sama Phenotypic Ratio: Semua sama 26

§ § §§ Homozygous dominant x Homozygous recessive Keturunan semua Heterozygot (hybrids) Keturunan disebut generasi F 1 Rasio Genotip & Phenotip sama semua 27

Persilangan F 1 Monohybrid Trait: Bentuk biji Alleles: R – bundar Cross: Biji Bundar Rr R RR Rr rr r – Kisut x biji Bundar x Rr Genotip: RR, Rr, rr Phenotip: Phenotip Bundar & kisut Rasio Gntp: 1: 2: 1 Rasio Pntp: 3: 1 28

§ Heterozygot x heterozygot Keturunan (Offspring): 25% Dominan Homozygot RR 50% Heterozygot Rr 25% Recessif Homozygous rr Offspring disebut F 2 generation Rasio Genotyp adalah 1: 2: 1 Rasio Phenotyp adalah 3: 1 §§ § 29

Bentuk Biji Yang kelihatan 30

Uji Silang (Test Cross) Mendel menyilangkan homozigot & hybrida dari generasi F 2 Dua kemungkinan test cross: Homozygot dominant x Hybrid Homozygot recessive x Hybrid 31

F 2 Monohybrid Cross st (1 ) Trait: Seed Shape Alleles: R – Bundar r – Kisut Cross: Biji Bundar x Biji Bundar RR x Rr R R R r RR Rr Genotip: RR, Rr Phenotip: Phenotip Bundar Rasio Genotip: 1: 1 Rasio Phenotip: Semua sama Bundar 32

F 2 Monohybrid Cross Trait: Seed Shape Alleles: R – Bundar r – Kisut Cross: Biji Kisut x Biji Bundar rr x Rr R r r Rr Rr r rr rr Genotip: Rr, rr Phenotip: Phenotip Bundar & Kisut Rasio Gntp: 1: 1 Rasio pntp: 1: 1 33

F 2 Monohybrid Cross Review §§ § Homozygot x heterozygot Keturunan: 50% Homozygot RR or rr 50% Heterozygot Rr Rasio Phenotip adalah 1: 1 34

Practice Your Crosses Work the P 1, F 1, and both F 2 Crosses for each of the other Seven Pea Plant Traits 35

Tugas : Buat Persilangan : I. Antara Homozigot x Homozigot : (tiga macam) II. Antara Homozigot x Heterozigot : (tiga macam) III. Antara Heterozigot x Heterozigot (1 macam) Buat lengkap dengan skema, perbandingan genotipe dan fenotipe yang dihasilkan baik pada F 1 dan F 2 36

Mendel’s Laws 37

Results of Monohybrid Crosses Inheritable factors or genes are responsible for all heritable characteristics Phenotype is based on Genotype Each trait is based on two genes, one from the mother and the other from the father True-breeding individuals are homozygous ( both alleles) are the same 38

Law of Dominance In a cross of parents that are pure for contrasting traits, only one form of the trait will appear in the next generation. All the offspring will be heterozygous and express only the dominant trait. RR x rr yields all Rr (round seeds) 39

Law of Dominance 40

Law of Segregation During the formation of gametes (eggs or sperm), the two alleles responsible for a trait separate from each other. Alleles for a trait are then "recombined" at fertilization, producing the genotype for the traits of the offspring. 41

Applying the Law of Segregation 42

Law of Independent Assortment Alleles for different traits are distributed to sex cells (& offspring) independently of one another. This law can be illustrated using dihybrid crosses. 43

Dihybrid Cross A breeding experiment that tracks the inheritance of two traits. Mendel’s “Law of Independent Assortment” a. Each pair of independently b. Formula: 2 n alleles segregates during gamete formation (n = # of heterozygotes) 44

Question: How many gametes will be produced for the following allele arrangements? Remember: 2 n (n = # of heterozygotes) 1. Rr. Yy 2. Aa. Bb. CCDd 3. Mm. Nn. Oo. PPQQRrss. Tt. Qq 45

Answer: 1. Rr. Yy: 2 n = 22 = 4 gametes RY Ry r. Y ry 2. Aa. Bb. CCDd: 2 n ABCD ABCd a. BCD a. BCd = 23 = Ab. CD ab. CD 8 gametes Ab. Cd ab. CD 3. Mm. Nn. Oo. PPQQRrss. Tt. Qq: 2 n = 26 = 64 gametes 46

Dihybrid Cross Traits: Seed shape & Seed color Alleles: R round r wrinkled Y yellow y green Rr. Yy RY Ry r. Y ry x Rr. Yy RY Ry r. Y ry All possible gamete combinations 47

Dihybrid Cross RY Ry r. Y ry 48

Dihybrid Cross RY RY RRYY Ry RRYy r. Y Rr. YY ry Rr. Yy Ry r. Y ry RRYy Rr. YY Rr. Yy RRyy Rr. Yy Rryy Rr. Yy rr. YY rr. Yy Rryy rr. Yy rryy Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 9: 3: 3: 1 phenotypic ratio 49

Dihybrid Cross Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 9: 3: 3: 1 50

Test Cross A mating between an individual of unknown genotype and a homozygous recessive individual. Example: bb. C__ x bbcc BB Bb bb = = = brown eyes blue eyes CC = curly hair Cc = curly hair cc = straight hair b. C b___ bc 51

Test Cross Possible results: bc b. C b___ C bb. Cc or bc b. C b___ c bb. Cc bbcc 52

Summary of Mendel’s laws LAW DOMINANCE SEGREGATION INDEPENDENT ASSORTMENT PARENT CROSS OFFSPRING TT x tt tall x short 100% Tt tall x x Tt tall Rr. Gg x Rr. Gg round & green x round & green 75% tall 25% short 9/16 round seeds & green pods 3/16 round seeds & yellow pods 3/16 wrinkled seeds & green pods 1/16 wrinkled seeds & yellow pods 53

Incomplete Dominance and Codominance 54

Incomplete Dominance F 1 hybrids have an appearance somewhat in between the phenotypes of the two parental varieties. Example: snapdragons (flower) red (RR) x white (rr) r r RR = red flower rr = white flower R R 55

Incomplete Dominance r r R Rr Rr produces the F 1 generation All Rr = pink (heterozygous pink) 56

Incomplete Dominance 57

Codominance Two alleles are expressed (multiple alleles) in heterozygous individuals. Example: blood type 1. 2. 3. 4. type A B AB O = = IAIA or IAi IBIB or IBi I AI B ii 58

Codominance Problem Example: homozygous male Type B (IBIB) x heterozygous female Type A (IAi) IA i IB I AI B I Bi 1/2 = IAIB 1/2 = IBi 59

Another Codominance Problem • Example: male Type O (ii) x female type AB (IAIB) IA IB i I Ai I Bi 1/2 = IAi 1/2 = IBi 60

Codominance Question: If a boy has a blood type O and his sister has blood type AB, what are the genotypes and phenotypes of their parents? boy - type O (ii) AB (IAIB) X girl - type 61

Codominance Answer: IA IB i i I AI B ii Parents: genotypes = IAi and IBi phenotypes = A and B 62

Sex-linked Traits (genes) located on the sex chromosomes Sex chromosomes are X and Y XX genotype for females XY genotype for males Many sex-linked traits carried on X chromosome 63

Sex-linked Traits Example: Eye color in fruit flies Sex Chromosomes fruit fly eye color XX chromosome - female Xy chromosome - male 64

Sex-linked Trait Problem Example: Eye color in fruit flies (red-eyed male) x (white-eyed female) X RY x X r Remember: the Y chromosome in males does not carry traits. Xr Xr RR = red eyed Rr = red eyed R X rr = white eyed XY = male Y XX = female 65

Sex-linked Trait Solution: Xr XR XR Xr Y Xr XR Xr Xr Y 50% red eyed female 50% white eyed male 66

Female Carriers 67

Genetic Practice Problems 68

Breed the P 1 generation tall (TT) x dwarf (tt) pea plants t t T T 69

Solution: tall (TT) vs. dwarf (tt) pea plants t t T Tt Tt produces the F 1 generation T Tt Tt All Tt = tall (heterozygous tall) 70

Breed the F 1 generation tall (Tt) vs. tall (Tt) pea plants T t 71

Solution: tall (Tt) x tall (Tt) pea plants T t T TT Tt tt produces the F 2 generation 1/4 (25%) = TT 1/2 (50%) = Tt 1/4 (25%) = tt 1: 2: 1 genotype 3: 1 phenotype 72