ALEL DAN GEN GANDA MonoHibrid pada Hewan: Warna Rambut Hitam: (gen A): AA (hitam) x aa (albino) However, it is possible to have several different allele possibilities for one gene.
Aa (Hitam)
Multiple alleles is when there are more than two allele possibilities for a gene.
Gen A: 1 Kali mutasi : -- >alel a Gen Ganda: Bbrp kali mutasi---) bbrp alel: a1,a2,a3, dst
• About 30% of the genes in humans are di-allelic, that is they exist in two forms, (they have two alleles) • About 70% are mono-allelic, they only exist in one form and they show no variation • A very few are poly-allelic having more than two forms
The ABO blood system
• •
This is a controlled by a tri-allelic gene It can generate 6 genotypes
•
The alleles control the production of antigens on the surface of the red blood cells Two of the alleles are codominant to one another and both are dominant over the third
• • • •
© 2007 Paul Billiet ODWS
Allele IA produces antigen A Allele IB produces antigen B Allele i produces no antigen
ALEL
GAN DA
Pengertian: Gen (virgin) kalau bermutasi membentuk Alel ( A -- a) Banyak Gen mengalami mutasi berulang-ulang, menimbulkan banyak macam alel (lebih dari 2, disebut alel Ganda)
ontoh: Gen pigmentasi bulu kelinci (Gen C, pigmentasi hitam), memiliki 3 alel: . c : albino (tak ada pigmentasi) . cch: pigmentasi terang, bulu pigmentasi gelap pada ujung (Chinchilla) . ch: pigmentasi bagian ujung-ujung tubuh, bagian lain putih (H= himalaya) Urutan dominasi alel : C>cch>ch>c
Certain types of rabbits…
…can either be brown, white, have a chinchilla pattern, or have a himalayan pattern C causes fully brown coat cc causes albino (white) cch causes a chinchilla pattern ch causes a Himalayan pattern The alleles are arranged in the following pattern C > cch > ch > c
•
•
Himalayan rabbit – color in certain parts of the body; dominant only to c; chc or chch Albino rabbit – no color – allele is recessive to all other alleles; cc
Full color rabbit – alleles are dominant to all others; CC, Ccch, Cch, or Cc Chinchilla rabbit – partial defect in pigmentation cch allele dominant to all other alleles except C; cchch, cchcch, or cchc
Kelinci Gelap:
Kelinci lebih terang; Chinchila:
CC, Cc, Ccch, Cch
cch cchh; cch, ch; ccchc
Kelinci Himalaya: c h ch; ch,c
Kelinci Albino: cc
P; Cch Cch X Ch Ch
P ; CC x Cch Cch
F1: Cch Ch X Cch Ch
F1 : C Cch
F2: Cch Cch
F2: Cc
Cch Ch Cch Ch Ch Ch
Cch c
xcc
Multiple alleles Each gene locus can have more than 2 alleles. An allele may be dominant to some alleles but recessive to others.
This situation produces more than 2 different phenotypes. Each individual has 2 alleles present in their cells at any one time.
BB or Bb or Bbl
bb or bbl
blbl
In this case both A and B are dominant to O (recessive). A and B are codominant (both expressed) So... there are four human blood types
Genotypes
AA, AO A blood type BB ,BO B blood type AB AB blood type or OO O blood type Phenotypes (Blood types)
IA IA
A
IA IB
AB
IA i
A
IB IB
B
IB i
B
ii
O
Sistem Golongan Darah A-B-O. (K. Landsteiner, 1868 – 1943) Gen Asli I (Isoagglutinogen), : 1. Alelnya : Ia, Ib, I 2. Urutan dominan: Ia = Ib >i Golongan (Fenotip)
Genotip
A
Ia Ia atau Ia i
B
Ib Ib; atau Ib i
Contoh: Gol A x Gol B (Ia Ia; Ia I) x ( Ib Ib; Ib I)
AB
Ia Ib
O
ii
1. Ia Ia x Ib Ib AB 2. Ia Ia x Ib I AB; A 3. Ia I x Ib I AB; B 4. Ia I x Ib I AB; A, B, O
Crossing Over dan Rekombinan • Sometimes in meiosis, homologous chromosomes exchange parts in a process called crossing-over. • New combinations are obtained, called the crossover products.
Structure of Chromosomes – Homologous chromosomes are identical pairs of chromosomes. – One inherited from mother and one from father – made up of sister chromatids joined at the centromere.
Copyright © McGraw-Hill Companies Permission required for reproduction or display
Crossing Over Basics • Occurs at One or More Points Along Adjacent Homologues • Points contact each other • DNA is Exchanged • Menaikkan var.Genetik http://waynesword.palomar.edu/images/cross3.jpg
21 Apr 2002
11
Recombination During Meiosis
Recombinant gametes
Coat-color genes
• How crossing over leads to genetic recombination • Nonsister chromatids break in two at the same spot • The 2 broken chromatids join together in a new way
Eye-color genes Tetrad (homologous pair of chromosomes in synapsis)
1
Breakage of homologous chromatids
2
Joining of homologous chromatids
Chiasma
3
Separation of homologous chromosomes at anaphase I
4
Separation of chromatids at anaphase II and completion of meiosis Parental type of chromosome Recombinant chromosome Recombinant chromosome Parental type of chromosome
Figure 8.18B
Gametes of four genetic types
• A segment of one chromatid has changed places with the equivalent segment of its nonsister homologue • If there were no crossing over meiosis could only produce 2 types of gametes
Coat-color genes
Eye-color genes Tetrad (homologous pair of chromosomes in synapsis)
1
Breakage of homologous chromatids
2
Joining of homologous chromatids
Chiasma
3
Separation of homologous chromosomes at anaphase I
4
Separation of chromatids at anaphase II and completion of meiosis Parental type of chromosome Recombinant chromosome Recombinant chromosome Parental type of chromosome
Figure 8.18B
Gametes of four genetic types
TEORI PELUANG: The Principles of Probability • The Principles of probability can be used to predict the outcomes of genetic crosses • Alleles segregate by complete randomness • Similar to a coin flip!
2013 Kul Genetik Dr. GTC
Genetics & Probability
• Mendel’s laws:
– segregation – independent assortment
reflect same laws of probability that apply to tossing coins or rolling dice
2013 Kul Genetik Dr. GTC
Probability & genetics
• Calculating probability of making a specific gamete is just like calculating the probability in flipping a coin
B 100%
BB B
– probability of tossing heads? – probability making a B gamete?
B 50%
Bb b 2013 Kul Genetik Dr. GTC
Determining probability •
Number of times the event is expected Number of times it could have happened
• Probabilitas pedet lahir jantan dari 10 kelahiran ?. Sex rasio 5:5 The probability is 5:10. • Or you can express it as a fraction: 5/10. Since it's a fraction, why not reduce it? The probability that you will pick an odd number is 1/2. • Probability can also be expressed as a percent...1/2=50% Or as a decimal...1/2=50%=.5 2013 Kul Genetik Dr. GTC
GENETIKA: PERAMALAN KETURUNAN DENGAN HUKUM PELUANG Prinsip dasar: Pemindahan gen dari orang tua kpd keturunannya
Berkumpulnya kembali gen-gen dalam sigot
Aa X Aa
Kakek (Aa) Org tua: JTN (a)
Org tua: BTN: A
Anak: Aa
F1 mis Peluang aa? Konsepmuncul Peluang
Analogi pemindahan satu gen (A/a) dari sepasang Gen (Aa) = pelemparan mata uang yang memiliki dua sisi:
-Gambar
2013 Kul Genetik Dr. GTC
Calculating probability Pp x Pp
sperm
egg
offspring
P
P
PP
P
p
1/2 x 1/2
=
1/4
male / sperm
P
p
female / eggs
1/2 x 1/2
P
PP
Pp
p
Pp
pp
p
=
p
1/2 x 1/2 2013 Kul Genetik Dr. GTC
=
P
1/2 x 1/2
p
Pp 1/4 + 1/4 1/2
pp =
1/4
Rule of multiplication
• Chance that 2 or more independent events will occur together – probability that 2 coins tossed at the same time will land heads up
P
1/2 x 1/2 = 1/4 – probability of Pp x Pp pp
1/2 x 1/2 = 1/4
2013 Kul Genetik Dr. GTC
Pp p
Use rule of multiplication to predict crosses
Calculating probability in crosses YyRr x YyRr
Yy
x
Yy
Rr
yyrr
x
?% 1/16 yy
rr 1/4
2013 Kul Genetik Dr. GTC
x
1/4
Rr
Apply the Rule of Multiplication AABbccDdEEFf
x
AaBbccDdeeFf
AabbccDdEeFF
Got it? Try this!
AA x Aa Aa Bb x Bb bb cc x cc cc Dd x Dd Dd EE x ee Ee GTC Ff2013x KulFfGenetik Dr.FF
1/2 1/4 1 1/2 1 1/4
1/64
Rule of addition
• Chance that an event can occur 2 or more different ways – sum of the separate probabilities – probability of Bb x Bb Bb sperm
egg
offspring
B
b
Bb
1/2 x 1/2 =
b
B
1/4
Bb
1/2 x 1/2 =
1/4
2013 Kul Genetik Dr. GTC
1/4 + 1/4 1/2
DASAR TEORI PELUANG I.
Terjadinga sesuatu yang diinginkan = sesuatu yang diinginkan -------------------------------keseluruhan kejadian
P (X) = X/(X+Y) Contoh : P (gambar) = 1/ 1+1 = ½ = 50 %
P (lahir anak jantan) = lahir jantan/ (lahir JTN + BTN ) ½ masing-masing = 50 %. II. Peluang terjadinya 2 persitiwa /lebih = yang berdiri sendiri
P. (X,Y) = P (X) x P (Y)
contoh: Peluang dua anak pertama laki-laki P (Kl, LK) = (1/2) x ( ½) = ¼. 2013 Kul Genetik Dr. GTC
Aplikasi dalam pewarisan sifat
Butawarna : gen resesif c X –linked.
Contoh: Gen resesif a (Albino)
P: Cc P: Aa normal
F1. AA Aa Aa aa
x
Aa normal
: Normal : Normal ; Normal : albino (1/4)
x C-
normal normal
F1 : CC: Normal
F,
Cc: Normal
F,
C- : 2013 Kul Genetik Dr. GTC Normal Peluang anak laki-laki albino
M,
III. Peluang Terjadinya dua persitiwa /lebih yang saling mempengaruhi
P ( X atau Y) = P (x) + P (Y) Contoh Pelempran dua mata uang bersama Peluang muncul dua gambar atau 2 huruf = ¼ + ¼ = ½.
PENGGUNAAN RUMUS BINOMIUM: (a+b)2
‘(2G, 2 H)= ? N = 2 (a2+2ab+b2)
a, b = DUA KEJADIAN YANG TERPISAH n
= banyaknya kejadian
2 ab = 2 (1/2) (1/2) = 1/2
1 1
1
1 1 1
2 3
4
1 3
6
1 4
n=3 1
Pelemparan 3 mata uang ( n= 3) ; (a+b) 3 = a3 + 3 a2b + 3 ab2 + b3 2 Peluang I G , 2 H = 3 ab2 = 3 ((1/2)(1/2)2013 Kul Genetik Dr. GTC
= 3/8.
Penggunaan Rumus Binomium: Peluang pewarisan sifat Albino
JTN : Aa Aa
x BTN
¾ Normal ¼ Albino Jika suatu perkawinan mempunyai 4 anak ( n = 4) Maka Peluang semua anak normal ? Rumus (a+b)4 = a4+ 4 ab3+6a2b2+4ab3+b4 Peluang 4 anak normal (a4) = (3/4)4 = 81/256
2013 Kul Genetik Dr. GTC
Aplikasi lain teori peluang dalam genetika Pada suatu perkawinan: Genotip diketahui, mis : Aa Bb Cc X Aa Bb Cc
aabbCc
aa = 1/4 bb = ¼ Cc = 1/2
Peluang (aabbCc) = 1/4x1/4x1/2 = 1/32 AaBbCcDdEe X AaBbCcDdEe
AABbccDdEE ? = 1/2x1/2x1/4x1/2x1/4 = 1/256 2013 Kul Genetik Dr. GTC
Contoh Pada dua sifat : GEN: Dominan dan Resesif -mata Merah Dominan thd Putih (M) -Kuliut Albino Resesif (a) Genotip Mm Aa X mm Aa Fenotip Aa X Aa
A a
A a AA Aa Aa aa
F1 ???
= 1/4
Mm x mm M
m
m
m
Bagaimana Peluang Gen Sifat tsb diwariskan pada anak anaknya?
M=½ a=¼=
Mm Mm
2013 Kul Genetik Dr. GTC
1/8
Penentuan Jenis Kelamin (SEKS)
The inheritance of Gender INDUK Mother XX
PEJANTAN Father XY Meiosis
Sex cells
X
Fertilisation
X
X
X
Y
X
XX
XY
X
XX
XY
Y
Possible Offsprings
Chance of a Female 50% Chance of a Male 50% © 2007 Paul Billiet ODWS
Summary: Males and females have different purposes defined by their gametes Development of sexes is dependent on: genes hormones environment Sex is flexible in some species
KASUS KESEIMBANGAN HORMONAL = SEX Mengapa Seks Penting: Kasus Keseimbangan Hormonal,
penentuan jenis kelamin menjadi tidak sederhana Contoh: PIG betina Awal bunting
Lahir : Jantan normal Betina : ??? (alat kelm + Jantan) Testoteron Dewasa
Injeksi hormon betina (Progesteron + Estrogen) Tetap tidak menunjukkan perilaku betina normal
Injeksi hormon jantan (Testoteron) Perilaku jantan jelas, fungsi seks jantan
PENGARUH LINGKUNGAN = SEX
Crocodile Sex Determination Incubating temperature 30oC all female 32oC all male 31oC 50% female, 50% male
http://a.abcnews.com/images/Sports/rt_thailand_ 080514_ssh.jpg
Hasil Analisis Kariotyping: Metode: Disusun besar- kecil Besar,bentuk, homolog Urutan: Besar—kecil Besar dan kesamaan bentuk Letak/bentuk acak Jumlah dapat dihitung
Manfaat : Penentuan Sex
Manfaat: Penentuan normal-abnorma
Penentuan Jenis Kelamin (Krom. SEKS) Dasar: Kariotyping untuk menentukan seks (X-Y Kromosom) Manfaat: Pre-derterminasi seks (deteksi dan manipulasi seks)
RINGKASAN 1. MAMALIA : XY ------- Betina : XX Jantan : XY 2. BELALANG : XO --------- Betina : XX Jantan: XO/ X- (tak ada krom Y) 3. UNGGAS/ BURUNG: ZW--------- Betina ZW atau ZO Jantan ZZ (burung) atau ZZ (Ayam) 4. LEBAH : haploid/diploid Betina : 2n : 32 buah Jantan : n : 16 buah Catatan : 1,2,3 dasar kromosom seks
1,3 ada perbedaan (berbalikan) 4 dasar jumlah kromosom
R I N G K A S A N II 1. JANTAN Heterogametik: a. Mamalia, Manusia : krom Y == JANTAN betina : XX Jantan : XY b. Heminiptera (Kepik, belalang) Betina : XX Jantan : X0 (tak ada krom Y)
2. BETINA Heterogametik : burung, Ikan , Kupu a. Burung : betina kromosom mirip Y spt manusia betina : ZW : bukan penentu seks yg kuat Jantan: ZZ b. Spesies lain (unggas/ayam/itik) : mirip XO Betina : ZO Jantan : ZZ
Tipe XY: Drosophla, manusia, mamalia Sex
Drosophila
Manusia
Jantan
2 XY + 6 A
2 XY + 44 A
Betina
2 XX
2 XX
Contoh : drosophila 6 autosome : bentuk sama 2 seks kromosom: bentuk beda :XX, XY X batang lurus, Y sedikit bengkok di salah satu ujungnya
Munculnya kelainan kromosom
Abnormal: non disjunction, meiosis ,
Normal:
pembt sel kelamin jantan/betina pd drosophila XX x XY
XX
x
X
X,
XX
ND
XY Y
XY
Normal
XX
O
XXX
XXY
B:super
B:Fertil
X
Y
XO
YO
J:Steril
J:Lethal
Kelainan kromosom pada manusia: sindrom turner : wanita sindrom klinefelter: pria sindrom down: autosom/mongolisme XX
X
XY ND
X
XY
XXY
O
XO
Turner (45)
Klinefelter (47) : • testis tak berkembang
-ovary tak berkembang, tak menstruasi
•Mandul dll
- kelj. Mammae tak berkembang baik dll.
Peran Krom:
Manusia
Drosophila
X
Menentukan sifat wanita
Menentukan sifat betina Menentukan kehidupan, YO = lethal
Y
Pemilik gen sifat laki-laki (asal ada Y = laki-laki
Menentukan kesuburan (XO = steril)
Teori indeks kelamin pada drosophila: krn adanya ND Oleh C.B. BRIDGES: faktor penentu seks jantan pada kromosome, betina pada autosome Indeks = Jmlh. Kromosome X = X/A Jmlh. pasangan autosom Contoh: Normal BTN 3 AA XX = X/A = 2/2 = 1.0 JTN 3 AA XY = X/A = ½ = 0.5 Kesimpulan : X/A > 1 = betina super
< 1.0 – 0.5 > : interseks < 0.5 = jantan super
Population Genetics • Predicting inheritance in a population
•mempelajari tingkah laku gen dalam populasi (perubahan frekuensi gen) •Mekanisme pewarisan sifat pada kelompok ternak (populasi), Pada sifat kuantitatif dan kualitatif •how often or frequent genes and/or alleles appear in the population
Populasi: Kelompok ternak t.a. bangsa/spesies yang sama, di daerah tertentu dimana antara anggota terjadi saling kawin satu dgn yang lain
Perlu estimasi frekuensi gen (merugikan) bagi generasi mendatang ( Mis. Ekspresi gen-gen yang mengalami mutasi, dll)
Perbedaan Genetika Individu dan Populasi INDIVIDU
POPULASI
1.LINGKUNGAN: 1
1.banyak tempat/banyak lingkungan
tempat/1 lingkungan
2.WAKTU: terbatas satu generasi
3. GENOTIP: satu sampel genetik khas. Susunan gen tetap Tak ada variasi/ satu ukuran Tidak terjadi evolusi
Masa panjang, generasi ke generasi tumpang tindih.
Gen pool Gen berubah dari generasi ke generasi
Population Genetics • Is simply, the study of Mendelian genetics in populations of animals • Basic foundation is the Hardy-Weinberg law • Usually limited to inheritance of qualitative traits influenced by only a small number of genes • Important to understand why characteristics, desirable or not, can be fixed or continue to exhibit variation in natural populations • Principles applied to the design of selection strategies to increase the frequencies of desirable genes or elimination of deleterious genes
KONSEP-KONSEP DASAR: FREK. GEN Frek Genotip Frek. fenotip
The study of the change of allele frequencies, genotype frequencies, and phenotype frequencies
Konsep Genetik: bahwa setiap indv. mempunyai dua lokus .untuk setiap pasang gen Contoh: Sifat Kualitatif (Warna kulit), dikontrol sepasang Gen R-r Kemungkinan Genotip: RR, Rr, rr (mis sapi Short Horn) (Fenotip: ?)
Pendekatan: : Frek. Gen (R ) = p; alelnya ( r ) = q Frek gen R = p = juml. Gen R/ juml. Gen (R + r) Frek gen r = q = juml. Gen r/Jumlh gen (R + r)
SEBAB SEBAB MODIFIKASI GENETIK Terjadinya modifikasi genetik, perubahan dalam frekuensi gen: -Adaptasi agar dpt survive dlm pop -Lingkungan berubah -Terjadi evolusi Perilaku Gen dalam Populasi: HK. Hardy Weinberg: APAPUN JENIS GENOTIP/FREKUENSI AWAL AKAN TERCAPAI KESEIMBANGAN DARI SATU GENERASI KE GERASI BERIKUTNYA
Syarat Hk. H. Weinberg: 1. Tidak ada kekuatan yang mampu merubah frek.gen (mutasi, dll) 2. Pada pop. Berlaku Hk Mendel 3. Populasi besar 4. Terjadi kawin acak
THE HARDY WEINBERG EQUATION Jadi terjadi keseimbangan, maka frek.gen/alel dll dapat ditentukan dalam populasi Mis : frek A = p, Frek a = q , maka p + q = 1 Jika terjadi perkw. Acak: Jumlah total: p2 (AA)+2pq (Aa) + q2(aa)
Gamet (frek)
A (p)
a (q)
A (p)
Genotip (frek)
AA (p2)
Aa (pq)
a (q)
Genotip (frek)
Aa (pq)
Aa (q2)
Only one of the populations below is in genetic equilibrium. Which one? Population sample
Genotypes
Gene frequencies
AA
Aa
aa
A
a
100
20
80
0
0.6
0.4
100
36
48
16
0.6
0.4
100
50
20
30
0.6
0.4
100
60
0
40
0.6
0.4
© 2008 Paul Billiet ODWS
Contoh Perhitungan Frek . Gen/ (Kodominan):
Fenotip
Merah
Roan
Putih
Genotip
RR
Rr
rr
Jika diketahui dalam populasi sapi short horn: 900 (merah); 450 (Roan) Brp. Frek (RR); Frek (R) ) ? dan 150 (putih) F (RR)) = jml. Indv. RR/ Juml tot indv. = 900/1500 = 0.6 = 60 % F (R ) = jml R/ Total geg = (2x900) + (1x450) + (0 x 150)/ 2 (900+450+150) = 0.75
Contoh : DOMINANSI PENUH: Pada pop 100 ekor sapi FH ditemukan 1 sapi berwarna kemerahan Brp frekuensi FH yang hitam heterosigot? H=p M=q ; maka frek gen HH + HM + MM = 1 Atau p2 + 2pq + q2 = 1 berasal dari ( p + q = 1) Diketahui q2 = 0.01 –maka q = 0.1------p = 0.9 2 pq = 2 (0.1) (0.9) = 0.18
Jadi frekuensi hitam heterosigot adalah: 0.18/ 0.99 = + 0.18 == 18 %.
LATIHAN/ DISKUSI/HOMEWORK: Fenotip
Genotip
j.indv.
j.gen R
Merah
RR
80
???-
Roan
Rr
???-
50
Putih
rr
20
Total F(R) ) = 210/300 = F (r ) = 90 / 300=
???-
J. Gen r
50
???210
90
EXAMPLE ALBINISM IN THE INDO. BUFFALO POPULATION Frequency of the albino phenotype = 1 in 20 000 or 0.00005
A = Normal skin pigmentation allele
Frequency = p
a = Albino (no pigment) allele
Frequency = q
Normal allele = A = p = ? Albino allele = q = (0.00005) = 0.007 or 7%
Phenotypes
Genotypes
Hardy Weinberg frequencies
Normal
AA
p2
Observed frequencies
0.99995 Normal
Aa
2pq
Albino
aa
q2
0.00005
HOW MANY buffalo IN Indonesia/Toraja ARE CARRIERS FOR THE ALBINO ALLELE (Aa)? a allele = 0.007 A allele But p+q Therefore p
=q =p =1 = 1- q = 1 – 0.007 = 0.993 or 99.3% The frequency of heterozygotes (Aa) = 2pq = 2 x 0.993 x 0.007 = 0.014 or 1.4%
© 2008 Paul Billiet ODWS
What about multiple alleles? • • • • • • •
Genotype A1A1 A1A2 A2A2 A1A3 A2A3 A3A3
• Total
Number 4 41 84 25 88 32 274
• f(A1) = ((2 X 4) + 41 + 25) ÷ (2 X 274) • = (8 +41 + 25) ÷ 548 • = 74 ÷ 548 • = 0.135
Number of A1 2X4 41 25
SUMMARY • • • • •
Genetic drift Mutation Mating choice Migration Natural selection
All can affect the transmission of genes from generation to generation
Genetic Equilibrium If none of these factors is operating then the relative proportions of the alleles (the GENE FREQUENCIES) will be constant © 2008 Paul Billiet ODWS
Factors causing genotype frequency changes • • • • •
Selection = variation in fitness; heritable Mutation = change in DNA of genes Migration = movement of genes across populations Recombination = exchange of gene segments Non-random Mating = mating between neighbors rather than by chance • Random Genetic Drift = if populations are small enough, by chance, sampling will result in a different allele frequency from one generation to the next.
FAKTOR-FAKTOR YG MAMPU MERUBAH KESEIMB. FREK GEN 1. MUTASI: Gen mpj sifat “dpt bermutasi”, Gen R ____> r (frekuensi Gen r meningkat dlm pop). Gen-gen terdapat dalam berbagai bentuk sbg alel yang berlainan forward mutation (maju) mengurangi gen tipe liar back mutation (surut) Akibat : menimbulkan polymorfisma : (banyak alel dari gen yg sama) .2. SELEKSI: Kekuatan besar pengaruhnya terhadap frek alel seleksi buatan seleksi alamiah
3. iNBREEDING: Perkawinan Keluarga dan tidak
acak , •
ekspresi gen resesif meningkat
Penurunan variabilitas genetik Peningkatan homosigotik
•
Manfaat : bagi para breeder Hewan yang mempj persamaan ciri dikawinkan (inbreeding) dihasilkan suatu strain/purebreed yang homogen Prinsip dasar: mempertahankan gen-gen tertentu pd frekuensi tinggi, sementara gen-gen lain dapat dihilangkan (mengekalkan/mempertahankan sifat yang diinginkan)
AA X AA Aa X Aa Aa X Aa AA Aa
Aa aa
Homosigot 2/4 = 50 %
AA,AA AA,Aa,Aa,aa
aa X aa aa, aa
Homosigot resesif: ¼
Homosigot : 6/8= 75, %
= 25 %
Homosigot resesif: 3/8 = 37.5 %
4.
REPROD. SEXUAL dan rekombinasi
gen: variabilitas meningkat dg perkw. Acak (pilihan acak dr gen 2 parent, cenderung memprod. Keturunan lebih bervariasi scr genetik), karena: • Adanya pilihan acak sel benih (meiosis) • Fenomena rekombinasi gen dalam kromosom
Adanya berbagai alel dalam pop menentukan variabilitas populasi
5. MIGRASI: perpindahan gen( ke dalam/keluar pop) Mis . Adanya import ternak sapi perah (frekuensi fenotip/genotip sapi perah meningkat dalam pop) Migrasi penduduk (becana alam/perang) merubah frek gen dari populasi yang asli/yang didatangi.
6. ARUS GENETIK: random genetic drift Perubahan scr acak frek.gen dari generasi ke generasi oleh teori PELUANG, A a X Aa mis Aa -- peluang teoritis sama mewaris pada keturunan , tetapi mungkin A>a, sehingga pop kearah frek ttt. Makin kecil populasi maka makin besar dampak arus genetik
