GENETIKA DASAR.

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1 GENETIKA DASAR

2 POKOK BAHASAN 3 Pautan Gen Pindah Silang Pemetaan Kromosom

3 “The Law of Segregation of Allelic Genes” atau Hukum Pemisahan Bebas
HUKUM MENDEL I : “The Law of Segregation of Allelic Genes” atau Hukum Pemisahan Bebas HUKUM MENDEL II : “The law of Independent Assortment of Genes” atau Hukum Pengelompokkan Gen secara Bebas

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6 Pautan Gen (Linkage gene)
Deviation From Independent Assortment Ratios Pautan/Berangkai/Linkage gene : Peristiwa beberapa gen bukan alel terdapat pada satu kromosom yang sama

7 Pautan/Berangkai/Linkage gene :
- RANGKAI/PAUTAN SEMPURNA  Gen-gen yang terangkai letaknya amat berdekatan, maka selama meiosis gen-gen itu tidak mengalami perubahan letak. Sehingga gen-gen itu bersama-sama menuju ke gamet - RANGKAI/PAUTAN TIDAK SEMPURNA  Gen-gen yang terangkai pada satu kromosom letaknya tidak berdekatan satu sama lainnya, sehingga gen-gen itu dapat mengalami perubahan letak yang disebabkan karena ada penukaran segmen dari kromatid-kromatid pada sepasang kromosom homolog

8 Bateson and Punnett crossed a purple, long snapdragon with one that was red and round
The F1 snapdragon was selfed Observed deviation from a 9:3:3:1 ratio

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10 Phenotype (genotype) Number of individuals Observed Approximate number of individuals Expected (from 9:3:3:1) Purple, long (P_L_) 284 215 Purple, round (P_ll) 21 71 Red, long (ppL_) Red, round (ppll) 55 24 Because the parental genes seemed to assort together more then they were expected, Bateson and Punnett said they were coupled

11 Creating a Linkage Hypothesis
Morgan used Drosophila as an experimental organism to prove linkage

12 Morgan crossed Drosophila red eye, normal wing (pr+pr+ vg+vg+) and purple eye vestigal wing (prpr vgvg) The F1 flies were test crossed Observed deviation from the 1:1:1:1 ratio

13 Coupling Cross Chi-Square Test
F1 Gamete Observed Expected (O-E)2/E pr+ vg+ 1339 709.75 557.9 pr+ vg 151 439.9 pr vg+ 154 435.2 pr vg 1195 331.8 Total 2839   2839 X2=1764.8

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15 Proof That Linked Genes Exist
Morgan hypothesized that alleles of two genes close together may not assort indepentently into gametes Parental arrangements may appear together in gametes

16 Morgan performed a second cross to prove the hypothesis
He crossed red eye, vestigal wing (pr+pr+ vgvg) and purple eye, normal wing flies (prpr vg+vg+) The F1 flies were testcrossed

17 Parent : pr+pr+ vgvg x prpr vg+vg+
(red eye, vestigal wing) (purple eye, normal wing) F : pr+pr vgvg+ x testcross

18 Testcross Distribution
F1 Gamete Testcross Distribution Gamete Type pr+ vg+ 157 Recombinant pr+ vg 965 Parental pr vg+ 1067 pr vg 146

19 Repulsion Cross Chi-Square Test
F1 Gamete Observed Expected (O-E)2/E pr+ vg+   157   312.0 pr+ vg   965 249.0 pr vg+ 1067 483.3 pr vg   146 328.3 Total 2335 X2=1372.6

20 Terms Used in Linkage Analysis
Coupling – the F1 configuration where both dominant alleles reside on the same chromosome; also called CIS  gen-gen dominan terangkai pada satu kromosom, sedangkan alel-alel resesifnya terangkai pada kromosom homolognya Penulisannya : AB/ab atau AB ab Repulsion - the F1 configuration where one dominant and one recessive allele reside on the same chromosome; also called TRANS  Gen dominan terangkai dengan alel resesifnya pada satu kromosom, sedangkan alel-alel resesifnya terangkai pada kromosom homolognya Penulisannya : Ab/aB atau Ab aB

21 The Development Of The Coupling Chromosome
Coupling – the F1 configuration where both dominant alleles reside on the same chromosome; also called CIS The Development Of The Coupling Chromosome

22 The Development Of The Repulsion Chromosome
Repulsion - the F1 configuration where one dominant and one recessive allele reside on the same chromosome; also called TRANS The Development Of The Repulsion Chromosome

23 Recombination Occurs Less Frequently Between Closely Linked Genes
Physical crossing over is a normal meiosis event Crossing-over  pertukaran segmen dari kromatid-kromatid bukan non-sister kromatid dari sepasang kromosom homolog The term used to describe crossing over is recombination Recombination can occur between any two genes on a chromosome The farther apart the two genes the more crossing over

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25 Cross over

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29 A B A B meiosis I a b a b A B A B gamet parental A b gamet rekombinasi

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35 Faktor-faktor yang mempengaruhi pindah silang
TEMPERATUR, temperatur kurang atau melebiji temperatur biasa dapat memperbesar kemungkinan pindah silang UMUR, makin tua suatu individu makin kurang mengalami pindah silang ZAT KIMIA tertentu dapat memperbesar kemungkinan terjadinya pindah silang PENYINARAN SINAR X dapat memperbesar kemungkinan pindah silang JARAK ANTAR GEN YANG TERANGKAI, makin jauh letak satu gen dengan gen lainnya, makin besar kemungkinan pindah silang JENIS KELAMIN, umumnya jantan atau betina dapat mengalami pindah silang. Namun pada ulat sutera betina dan Drosophila jantan tidak pernah terjadi pindah silang

36 Determining Linkage Distances
By definition, one map unit is equal to one percent recombinant gametes or phenotypes In honor of Morgan, one map unit is also called one centimorgan (cM) 1 mu = 1% = 1 cM

37 Number of recombinants x 100% Total Number
To determine the distance between two genes, divide the number of recombinant gametes by the total number of gametes Formula : Number of recombinants x 100% Total Number

38 Testcross Distribution
Coupling Data F1 Gamete Testcross Distribution Gamete Type pr+ vg+ 1339 Parental pr+ vg   151 Recombinant pr vg+   154 pr vg 1195 pr vg distance = (( )/2839)*100% = 10.7 m.u = 10.7 cM

39 Testcross Distribution
Repulsion Data F1 Gamete Testcross Distribution Gamete Type pr+ vg+   157 Recombinant pr+ vg   965 Parental pr vg+ 1067 pr vg   146 pr vg distance = (( )/2335)*100% = 13.0 cM

40 Jarak Gen Coupling : pr vg 10,7 cM Repulsion : pr vg 13,0 cM

41 Remember these are estimates; the differences between the two estimates reflect random deviation
Neither estimate is incorrect; repeated experimentation would give a more accurate estimate

42 Deriving Linkage Distance And Gene Order From Three-Point Crosses
Analyzing three genes allows us to determine gene order as well as linkage distance Need to create a F1 and follow deviation from a 1:1:1:1:1:1:1:1 ratio What are the expected gametes when three linked genes are considered?

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44 Analyzing Three-Point Test Cross Data
Genotype Observed ABC    390 abc    374 AbC     27 aBC     30 ABc       5 abC       8 Abc     81     85 Total 1000

45 What are the parental genotypes. What is the gene order
What are the parental genotypes? What is the gene order? What are the linkage distances?

46 What are the parental genotypes?
The genotypes most frequently found are the parental genotypes ABC and abc are the parental genotypes

47 What is the gene order? * Menentukan susunan gen dengan memperhatikan Tipe Parental dan Tipe DCO Pada Tipe Parental di-DCO-kan dan dibandingkan dengan hasil DCO pengamatan The double crossover moves a non-parental allele of the central gene between two parental alleles Gene C is between genes A and B (gene order = A C B)

48 Analyzing Three-Point Test Cross Data
Genotype Observed Type of Gamete ACB    390 Parental acb    374 ACb     27 Single-crossover between genes C and B aCB     30 AcB       5 Double-crossover aCb       8 Acb     81 Single-crossover between genes A and C     85 Total 1000

49 What are the linkage distances?
Linkage distance equals the sum of the appropriate single cross plus all double crosses divided by total number of gametes Formula : Jarak = SCO + DCO x 100% Jumlah Total

50 *Jarak C - B = ((27+30+5+8)/1000)*100 = 7.0 cM
*Jarak A - C = (( )/1000)*100 = 17.9 cM *Jarak C - B = (( )/1000)*100 = 7.0 cM A C B 17,9 cM 7,0 cM

51 Contoh lain : Genotype Observed v cv+ ct+ 580 v+ cv ct 592 v cv ct+ 45
    580    v+ cv ct  592 v cv ct+     45 v+ cv+ ct     40 v cv ct     89 v+ cv+ ct+     94 v cv+ ct       3 v+ cv ct+       5 Total 1448

52 Determine the parental genotypes
The most abundant genotypes are v cv+ ct+ and v+ cv ct.

53 Determine the gene order
Allele ct is paired with v and cv+. This is a different pairing than the parental genotype. Therefore gene ct is in the middle.

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55 v ct+ cv+ v+ ct cv v ct+ cv v+ ct cv+ v ct cv v+ ct+ cv+ v ct cv+
Genotype Observed Type of Gamete v ct+ cv+     580    Parental v+ ct cv  592 v ct+ cv     45 Single-crossover between genes ct and cv v+ ct cv+     40 v ct cv     89 Single-crossover between genes v and ct v+ ct+ cv+     94 v ct cv+       3 Double-crossover v+ ct+ cv       5 Total 1448

56 Determine the linkage distance
v-ct = 100*(( )/1448) = 13.2 cM ct-cv = 100*(( )/1448 = 6.4 cM

57 Determine the linkage distance
v-ct = 100*(( )/1448) = 13.2 cM ct-cv = 100*(( )/1448 = 6.4 cM

58 MENGHITUNG JARAK PETA DAN FREKUENSI REKOMBINASI
Menghitung jarak antar gen dengan menggunakan percobaan Bateson dan Punnett  menggunakan frekuensi rekombinasi dari metode uji silang (testcross)

59 (Purple, long) (red, round)
Parental PL/PL x pl/pl (Purple, long) (red, round) F PL/pl x ppll (purple, long) (red,round) Progeny :

60 Frekuensi rekombinan : rekombinan x 100%
Tipe parental :Purple, long (P_L_) 284 Red, round (ppll) Tipe rekombinan : Purple, round (P_ll) 21 Red, long (ppL_) Frekuensi rekombinan : rekombinan x 100% tipe parental + tipe rekombinan (21+21) x 100% = 11.02% (284+55)+(21+21)

61 Jadi jarak gen untuk warna bunga dan gen untuk pollen adalah : P L 11

62 Rekombinasi dari F1 x F1 Data dari Bateson dan Punnett (F1 x F1 coupling) F PL/pl x PL/pl (Purple, long) (Purple, long) F2 : Purple, long (P_L_) : (a1) Purple, round (P_ll) : (a2) red, long (ppL_) : (a3) red, round (ppll) : (a4)

63 Data dari Bateson dan Punnett (F1 x F1 repulsion)
F1 : Pl/pL x Pl/pL (Purple, long) (Purple, long) F2 : Purple, long (P_L_) : (a1) Purple, round (P_ll) : (a2) red, long (ppL_) : (a3) red, round (ppll) : (a4)

64 Hasil kali tipe tetua * Coupling  Z = a2 x a3 a1 x a4
Rumus menghitung persentase rekombinasi apabila F1 heterosigot Z = Hasil kali tipe rekombinan Hasil kali tipe tetua Sehingga apabila F1 heterosigot dalam keadaan : * Coupling  Z = a2 x a3 a1 x a4 * Repulsion  Z = a1 x a4 a2 x a3

65 Data dari Bateson dan Punnett (F1 x F1 coupling)
F1 : PL/pl x PL/pl (Purple, long) (Purple, long) F2 : Purple, long (P_L_) : 269 (a1) Purple, round (P_ll) : (a2) red, long (ppL_) : (a3) red, round (ppll) : (a4) Coupling  Z = 19 x = 0,0238 269 x 85 Nilai Z = 0,0238**

66 Interference Interference - the reduction in the expected number of crossovers at two adjacent genetic intervals Interferensi – interaksi antar pindah silang Pindah silang pada tempat tertentu mengurangi kemungkinan terjadinya pindah silang pada daerah didekatnya. Pindah silang di daerah I mengurangi terjadinya pindah silang pada daerah II Mengakibatkan frekuensi pindah silang ganda lebih kecil dari yang diharapkan

67 coefficient of coincidence (c.o.c.)
= ratio of observed to expected double crossovers Koefisien koinsidensi adalah ukuran dari kekuatan interferensi dan merupakan nisbah antara frekuensi pindah silang ganda yang diamati dan frekuensi pindah silang ganda yang diharapkan

68 Rumus atau KK = Banyaknya pindah silang ganda (DCO) yang sesungguhnya
Banyaknya pindah silang ganda (DCO) yang diharapkan Interferensi = 1 - KK

69 Apabila interferensi sempurna (1,0) maka tidak ada pindah silang ganda yang dapat diamati
Apabila semua pindah silang ganda yang diharapkan dapat diamati maka interferensinya nol (0)

70 Misalnya diketahui data sebagai berikut:
Genotipe gamet Jumlah Fenotipe Dari F1 heterosigot individu V Gl Va normal v gl va mengkilat, steril sebagian, pucat V gl Va mengkilat v Gl va steril sebagian, pucat V gl va mengkilat, steril sebagian v Gl Va pucat V Gl va steril sebagian v gl Va mengkilat, pucat 726

71 Misalnya : Diketahui peta kromosom : V Gl Va 18,3 13,6 31,9 SCO-1 (V-Gl) : 18,3% SCO-2 (Gl-Va) : 13,6% Pindah silang ganda yang diharapkan = SCO-1 x SCO-2 = 0,183 x 0,136 = 0,025 = 2,5%

72 Pindah silang ganda yang sebenarnya : 7+4 = 1,5% 726 Sehingga Koefisien Koinsidensinya : KK = 1,5% = 0,6 2,5% Jadi Interferensinya : I = 1-KK = 1 – 0,6 = 0,4 = 40%

73 Calculating Interference Values From The Example
v - ct = recombination frequency ct - cv = recombination frequency Expected double crossover frequency equal the product of the two single crossover frequency expected double crossover frequency = x = Total double crossovers = 1448 x = 12 Observed double crossovers = 8 c.o.c = 8/12 I for example = 100 x [1 -(8/12)] = 33% Most often I is between 0 and 1 indicating positive interference Occasionally I is greater than 1 indicating negative interference

74 They performed a testcross to this stock and
Creighton and McClintock's Proof of Chromosomal Exchange During Crossing Over They used corn chromosome 9 markers: c = colorless seed wx = waxy endosperm They created a heterozygote with the following characteristics: repulsion configuration of genetic markers cytological landmarks on both ends of one chromosome They performed a testcross to this stock and analyzed the results.

75 If crossing over involves exchange of chromosomal material each recombinant chromosome would have on of the cytological landmarks. This is the result they obtained.  See the figure below.

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