Mendelian Genetics 5/25/2018 Genetika Mendel.

Presentasi berjudul: "Mendelian Genetics 5/25/2018 Genetika Mendel."— Transcript presentasi:

1 Mendelian Genetics 5/25/2018 Genetika Mendel

2 Menemukan hukum pewarisan suatu sifat
Gregor Mendel ( ) Menemukan hukum pewarisan suatu sifat

3 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

4 Mendelian Genetics 5/25/2018 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"

5 Site of Gregor Mendel’s experimental garden in the Czech Republic
Mendelian Genetics 5/25/2018 Site of Gregor Mendel’s experimental garden in the Czech Republic Fig. 5.co

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

7 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

8 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

9 c. FENOTIPE

10

11 Mendel’s Pea Plant Experiments

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

13 Reproduksi pada tanaman
Mendelian Genetics 5/25/2018 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

14 Mendel’s Experimental Methods
Mendelian Genetics 5/25/2018 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

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

17 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)

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20 Mendel’s Experimental Results

21 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

22 Macam Generasi Induk (Parental) P1 = generasi induk dalam suatu percobaan genetika Generasi F1 = Generasi turunan pertama (1st filial generation) dalam suatu percobaan genetika Generasi F2 = Generation turunan kedua generation dalam suatu percobaan genetika. (2nd filial generation)

23 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

24 Monohybrid

25 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

26 P1 Monohybrid Cross r r Rr Rr R R Rr Rr
Sifat: bentuk biji (Seed Shape) Alleles: R – Round r – Wrinkled Cross: Round seeds x Wrinkled seeds RR x rr Genotype: Rr Phenotype: Round Genotypic Ratio: Semua sama Phenotypic Ratio: Semua sama r r Rr Rr R R Rr Rr

27 Homozygous dominant x Homozygous recessive
Keturunan semua Heterozygot (hybrids) Keturunan disebut generasi F1 Rasio Genotip & Phenotip sama semua

28 Persilangan F1 Monohybrid
Trait: Bentuk biji Alleles: R – bundar r – Kisut Cross: Biji Bundar x biji Bundar Rr x Rr Genotip: RR, Rr, rr Phenotip: Bundar & kisut Rasio Gntp: 1:2:1 Rasio Pntp: 3:1 R r RR Rr R r Rr rr

29 Heterozygot x heterozygot Keturunan (Offspring): 25% Dominan Homozygot RR 50% Heterozygot Rr 25% Recessif Homozygous rr Offspring disebut F2 generation Rasio Genotyp adalah 1:2:1 Rasio Phenotyp adalah 3:1

30 Bentuk Biji Yang kelihatan

31 Uji Silang (Test Cross)
Mendel menyilangkan homozigot & hybrida dari generasi F2 Dua kemungkinan test cross: Homozygot dominant x Hybrid Homozygot recessive x Hybrid

32 F2 Monohybrid Cross (1st)
Trait: Seed Shape Alleles: R – Bundar r – Kisut Cross: Biji Bundar x Biji Bundar RR x Rr Genotip: RR, Rr Phenotip: Bundar Rasio Genotip: 1:1 Rasio Phenotip: Semua sama Bundar R r RR Rr R R RR Rr

33 F2 Monohybrid Cross R r Rr rr r r Rr rr Trait: Seed Shape
Alleles: R – Bundar r – Kisut Cross: Biji Kisut x Biji Bundar rr x Rr R r Genotip: Rr, rr Phenotip: Bundar & Kisut Rasio Gntp: 1:1 Rasio pntp: 1:1 Rr rr r r Rr rr

34 F2 Monohybrid Cross Review
Homozygot x heterozygot Keturunan: 50% Homozygot RR or rr 50% Heterozygot Rr Rasio Phenotip adalah 1:1

35 Practice Your Crosses Work the P1, F1, and both F2 Crosses for each of the other Seven Pea Plant Traits

36 Tugas : Buat Persilangan : Antara Homozigot x Homozigot : (tiga macam) Antara Homozigot x Heterozigot : Antara Heterozigot x Heterozigot (1 macam) Buat lengkap dengan skema, perbandingan genotipe dan fenotipe yang dihasilkan baik pada F1 dan F2

37 Mendel’s Laws

38 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

39 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)

40 Law of Dominance

41 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.

42 Applying the Law of Segregation

43 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.

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

45 Remember: 2n (n = # of heterozygotes)
Question: How many gametes will be produced for the following allele arrangements? Remember: 2n (n = # of heterozygotes) 1. RrYy 2. AaBbCCDd 3. MmNnOoPPQQRrssTtQq

46 Answer: 1. RrYy: 2n = 22 = 4 gametes RY Ry rY ry
2. AaBbCCDd: 2n = 23 = 8 gametes ABCD ABCd AbCD AbCd aBCD aBCd abCD abCD 3. MmNnOoPPQQRrssTtQq: 2n = 26 = 64 gametes

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

48 Dihybrid Cross RY Ry rY ry RY Ry rY ry

49 Dihybrid Cross RY Ry rY ry Round/Yellow: 9 Round/green: 3
wrinkled/Yellow: 3 wrinkled/green: 1 9:3:3:1 phenotypic ratio RRYY RRYy RrYY RrYy RRyy Rryy rrYY rrYy rryy

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

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

52 Test Cross Possible results: bC b___ bc bbCc C bC b___ bc bbCc bbcc or

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

54 Incomplete Dominance and Codominance

55 Incomplete Dominance F1 hybrids have an appearance somewhat in between the phenotypes of the two parental varieties. Example: snapdragons (flower) red (RR) x white (rr) RR = red flower rr = white flower r R

56 Incomplete Dominance r R r All Rr = pink (heterozygous pink)
produces the F1 generation Rr

57 Incomplete Dominance

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

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

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

61 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) X girl - type AB (IAIB)

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

63 Sex-linked Traits 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

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

65 Sex-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 Y

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

67 Female Carriers

68 Genetic Practice Problems

69 Breed the P1 generation tall (TT) x dwarf (tt) pea plants t T

70 tall (TT) vs. dwarf (tt) pea plants
Solution: tall (TT) vs. dwarf (tt) pea plants T t All Tt = tall (heterozygous tall) produces the F1 generation Tt

71 Breed the F1 generation tall (Tt) vs. tall (Tt) pea plants T t T t

72 tall (Tt) x tall (Tt) pea plants
Solution: tall (Tt) x tall (Tt) pea plants T t produces the F2 generation 1/4 (25%) = TT 1/2 (50%) = Tt 1/4 (25%) = tt 1:2:1 genotype 3:1 phenotype TT Tt tt