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GENETIKA DASAR

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GENETIKA DASAR. POKOK BAHASAN 3 Pautan Gen Pindah Silang Pemetaan Kromosom. HUKUM MENDEL I : “ The Law of Segregation of Allelic Genes” atau Hukum Pemisahan Bebas HUKUM MENDEL II : “ The law of Independent Assortment of Genes” a tau Hukum Pengelompokkan Gen secara Bebas. - PowerPoint PPT Presentation

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Page 1: GENETIKA  DASAR

GENETIKA DASAR

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POKOK BAHASAN 3

1. Pautan Gen2. Pindah Silang

3. Pemetaan Kromosom

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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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Pautan Gen (Linkage gene)

Deviation From Independent Assortment Ratios

Pautan/Berangkai/Linkage gene : Peristiwa beberapa gen bukan alel terdapat pada satu kromosom yang sama

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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

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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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Because the parental genes seemed to assort together more then they were expected, Bateson and Punnett said they were coupled

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_) 21 71

Red, round (ppll) 55 24

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Creating a Linkage Hypothesis

Morgan used Drosophila as an experimental organism to prove linkage

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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

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Coupling Cross Chi-Square Test

F1 Gamete Observed Expected (O-E)2/E

pr+ vg+ 1339 709.75 557.9

pr+ vg 151 709.75 439.9

pr vg+ 154 709.75 435.2

pr vg 1195 709.75 331.8

Total 2839   2839 X2=1764.8

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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

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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

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Parent : pr+pr+ vgvg x prpr vg+vg+

(red eye, vestigal wing) (purple eye, normal wing)

F1 : pr+pr vgvg+ x testcross

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F1 Gamete Testcross Distribution Gamete Type

pr+ vg+ 157 Recombinant

pr+ vg 965 Parental

pr vg+ 1067 Parental

pr vg 146 Recombinant

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Repulsion Cross Chi-Square Test F1 Gamete Observed Expected (O-E)2/E

pr+ vg+   157   583.75 312.0

pr+ vg   965   583.75 249.0

pr vg+ 1067   583.75 483.3

pr vg   146   583.75 328.3

Total 2335 2335 X2=1372.6

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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 homolognyaPenulisannya : 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 homolognyaPenulisannya : Ab/aB atau Ab

aB

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Coupling – the F1 configuration where both dominant alleles reside on the same chromosome; also called CIS

The Development Of The Coupling Chromosome

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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

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Recombination Occurs Less Frequently Between Closely Linked Genes Physical crossing over is a normal meiosis

eventCrossing-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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Cross over

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A B A B meiosis I

a b a b

A B A B gamet parental

A b gamet rekombinasi

a B gamet rekombinasi

a b a b gamet parental

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Faktor-faktor yang mempengaruhi pindah silang

1. TEMPERATUR, temperatur kurang atau melebiji temperatur biasa dapat memperbesar kemungkinan pindah silang

2. UMUR, makin tua suatu individu makin kurang mengalami pindah silang

3. ZAT KIMIA tertentu dapat memperbesar kemungkinan terjadinya pindah silang

4. PENYINARAN SINAR X dapat memperbesar kemungkinan pindah silang

5. JARAK ANTAR GEN YANG TERANGKAI, makin jauh letak satu gen dengan gen lainnya, makin besar kemungkinan pindah silang

6. JENIS KELAMIN, umumnya jantan atau betina dapat mengalami pindah silang. Namun pada ulat sutera betina dan Drosophila jantan tidak pernah terjadi pindah silang

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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

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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

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Coupling Data F1 Gamete Testcross Distribution Gamete Type

pr+ vg+ 1339 Parental

pr+ vg   151 Recombinant

pr vg+   154 Recombinant

pr vg 1195 Parental

pr vg distance = ((151 +154)/2839)*100% = 10.7 m.u

= 10.7 cM

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Repulsion Data F1 Gamete Testcross Distribution Gamete Type

pr+ vg+   157 Recombinant

pr+ vg   965 Parental

pr vg+ 1067 Parental

pr vg   146 Recombinant pr vg distance = ((157 + 146)/2335)*100% = 13.0

cM

Page 40: GENETIKA  DASAR

Jarak Gen

Coupling : pr vg 10,7 cM

Repulsion : pr vg

13,0 cM

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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

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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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Analyzing Three-Point Test Cross DataGenotype Observed

ABC    390

abc    374

AbC     27

aBC     30

ABc       5

abC       8

Abc     81

aBC     85

Total 1000

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What are the parental genotypes?

What is the gene order?

What are the linkage distances?

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What are the parental genotypes?

The genotypes most frequently found are the parental genotypes

ABC and abc are the parental genotypes

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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)

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Analyzing Three-Point Test Cross Data Genotype Observed Type of Gamete

ACB    390 Parental

acb    374 Parental

ACb     27 Single-crossover between genes C and B

aCB     30 Single-crossover between genes C and B

AcB       5 Double-crossover

aCb       8 Double-crossover

Acb     81 Single-crossover between genes A and C

aCB     85 Single-crossover between genes A and C

Total 1000

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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

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*Jarak A - C = ((81+85+5+8)/1000)*100 = 17.9 cM

*Jarak C - B = ((27+30+5+8)/1000)*100 = 7.0 cM

A C B17,9 cM 7,0 cM

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Contoh lain :Genotype Observed

v cv+ ct+     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

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Determine the parental genotypes

The most abundant genotypes are v cv+ ct+ and v+ cv ct.

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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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Genotype Observed Type of Gamete

v ct+ cv+     580    Parental

v+ ct cv  592 Parental

v ct+ cv     45 Single-crossover between genes ct and cv

v+ ct cv+     40 Single-crossover between genes ct and cv

v ct cv     89 Single-crossover between genes v and ct

v+ ct+ cv+     94 Single-crossover between genes v and ct

v ct cv+       3 Double-crossover

v+ ct+ cv       5 Double-crossover

Total 1448

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Determine the linkage distance

v-ct = 100*((89+94+3+5)/1448) = 13.2 cM

ct-cv = 100*((45+40+3+5)/1448 = 6.4 cM

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Determine the linkage distance v-ct = 100*((89+94+3+5)/1448) = 13.2

cM ct-cv = 100*((45+40+3+5)/1448 = 6.4

cM

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MENGHITUNG JARAK PETA DAN FREKUENSI REKOMBINASIMenghitung jarak antar gen dengan

menggunakan percobaan Bateson dan Punnett

menggunakan frekuensi rekombinasi dari metode uji silang (testcross)

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Parental PL/PL x pl/pl(Purple, long) (red, round)

F1 PL/pl x ppll (purple, long)

(red,round)

Progeny :

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Tipe parental :Purple, long (P_L_) 284 Red, round (ppll) 55

Tipe rekombinan : Purple, round (P_ll) 21 Red, long (ppL_) 21

Frekuensi rekombinan :

rekombinan x 100%tipe parental + tipe rekombinan

(21+21) x 100% = 11.02%(284+55)+(21+21)

Page 61: GENETIKA  DASAR

Jadi jarak gen untuk warna bunga dan gen untuk pollen adalah :

P L11.02

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Rekombinasi dari F1 x F1Data dari Bateson dan Punnett (F1 x F1 coupling)

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)

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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)

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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

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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) : 19(a2) red, long (ppL_) : 27(a3) red, round (ppll) : 85(a4)

Coupling Z = 19 x 27 = 0,0238269 x 85

Nilai Z = 0,0238*

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Interference Interference - the reduction in the expected number of crossovers at two adjacent genetic intervals

Interferensi – interaksi antar pindah silangPindah 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

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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

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Rumus

atauKK = Banyaknya pindah silang ganda (DCO) yang

sesungguhnya Banyaknya pindah silang ganda (DCO) yang diharapkan

Interferensi = 1 - KK

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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)

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Misalnya diketahui data sebagai berikut:

Genotipe gamet Jumlah FenotipeDari F1 heterosigot individu

V Gl Va 235 normal

v gl va 270 mengkilat, steril sebagian, pucat

V gl Va 7 mengkilat

v Gl va 4 steril sebagian, pucat

V gl va 62 mengkilat, steril sebagian

v Gl Va 60 pucat

V Gl va 40 steril sebagian

v gl Va 48 mengkilat, pucat

726

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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%

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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%

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Calculating Interference Values From The Example

v - ct = 0.132 recombination frequency ct - cv = 0.064 recombination frequency

Expected double crossover frequency equal the product of the

two single crossover frequency expected double crossover frequency = 0.132 x 0.064 = 0.0084 Total double crossovers = 1448 x 0.0084 = 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

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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 andanalyzed the results.

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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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