genetika dasar
DESCRIPTION
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 PresentationTRANSCRIPT
GENETIKA DASAR
POKOK BAHASAN 3
1. Pautan Gen2. Pindah Silang
3. 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” atau Hukum Pengelompokkan
Gen secara Bebas
Pautan Gen (Linkage gene)
Deviation From Independent Assortment Ratios
Pautan/Berangkai/Linkage gene : Peristiwa beberapa gen bukan alel terdapat pada satu kromosom yang sama
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
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
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
Creating a Linkage Hypothesis
Morgan used Drosophila as an experimental organism to prove linkage
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
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
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
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
Parent : pr+pr+ vgvg x prpr vg+vg+
(red eye, vestigal wing) (purple eye, normal wing)
F1 : pr+pr vgvg+ x testcross
F1 Gamete Testcross Distribution Gamete Type
pr+ vg+ 157 Recombinant
pr+ vg 965 Parental
pr vg+ 1067 Parental
pr vg 146 Recombinant
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
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
Coupling – the F1 configuration where both dominant alleles reside on the same chromosome; also called CIS
The Development Of The Coupling 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
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
Cross over
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
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
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
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
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
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
Jarak Gen
Coupling : pr vg 10,7 cM
Repulsion : pr vg
13,0 cM
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
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?
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
What are the parental genotypes?
What is the gene order?
What are the linkage distances?
What are the parental genotypes?
The genotypes most frequently found are the parental genotypes
ABC and abc are the parental genotypes
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)
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
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
*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
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
Determine the parental genotypes
The most abundant genotypes are v cv+ ct+ and v+ cv ct.
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.
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
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
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
MENGHITUNG JARAK PETA DAN FREKUENSI REKOMBINASIMenghitung jarak antar gen dengan
menggunakan percobaan Bateson dan Punnett
menggunakan frekuensi rekombinasi dari metode uji silang (testcross)
Parental PL/PL x pl/pl(Purple, long) (red, round)
F1 PL/pl x ppll (purple, long)
(red,round)
Progeny :
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)
Jadi jarak gen untuk warna bunga dan gen untuk pollen adalah :
P L11.02
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)
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)
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
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*
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
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
Rumus
atauKK = Banyaknya pindah silang ganda (DCO) yang
sesungguhnya Banyaknya pindah silang ganda (DCO) yang diharapkan
Interferensi = 1 - KK
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)
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
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%
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%
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
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.
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.