ALEL DAN GEN GANDA. Gen A: 1 Kali mutasi : -- >alel a Gen Ganda: Bbrp kali mutasi---) bbrp alel: a1,a2,a3, dst

Transkripsi

1 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

2 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 Allele I A produces antigen A Allele I B produces antigen B Allele i produces no antigen 2007 Paul Billiet ODWS

3 A L E L G A N D A 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) ntoh: Gen pigmentasi bulu kelinci (Gen C, pigmentasi hitam), emiliki 3 alel: c : albino (tak ada pigmentasi) c ch : pigmentasi terang, bulu pigmentasi gelap pada ujung (Chinchilla) c h : pigmentasi bagian ujung-ujung tubuh, bagian lain putih (H= himalaya) rutan dominasi alel : C>c ch >c h >c

4 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) c ch causes a chinchilla pattern c h causes a Himalayan pattern The alleles are arranged in the following pattern C > c ch > c h > c Himalayan rabbit color in certain parts of the body; dominant only to c; c h c or c h c h Albino rabbit no color allele is recessive to all other alleles; cc Full color rabbit alleles are dominant to all others; CC, Cc ch, Cc h, or Cc Chinchilla rabbit partial defect in pigmentation c ch allele dominant to all other alleles except C; c ch c h, c ch c ch, or c ch c

5 Kelinci Gelap: CC, Cc, Cc ch, C ch Kelinci lebih terang; Chinchila: c ch c ch h; c ch, c h ; cc ch c Kelinci Himalaya: c h c h ; c h,c P; Cch Cch X Ch Ch F1: Cch Ch X Cch Ch F2: Cch Cch Cch Ch Cch Ch Ch Ch Kelinci Albino: cc P ; CC x Cch Cch F1 : C Cch F2: Cc Cch c x c c

6 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 Bb l bb or bb l b l b l

7 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 I A I A I A I B I A i I B I B I B i ii AA, AO A blood type BB,BO B blood type AB AB blood type or OO O blood type Phenotypes (Blood types) A AB A B B O

8 Sistem Golongan Darah A-B-O. (K. Landsteiner, ) Gen Asli I (Isoagglutinogen), : 1. Alelnya : Ia, Ib, I 2. Urutan dominan: Ia = Ib >i Golongan (Fenotip) A B AB O Genotip Ia Ia atau Ia i Ib Ib; atau Ib i Ia Ib ii Contoh: Gol A x Gol B (Ia Ia; Ia I) x ( Ib Ib; Ib I) 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

9 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.

10 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

11 Crossing Over Basics Occurs at One or More Points Along Adjacent Homologues Points contact each other DNA is Exchanged Menaikkan var.genetik 21 Apr

12 Recombination During Meiosis Recombinant gametes

13 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 Figure 8.18B Coat-color genes Eye-color genes Tetrad (homologous pair of chromosomes in synapsis) Breakage of homologous chromatids Joining of homologous chromatids Chiasma Separation of homologous chromosomes at anaphase I Separation of chromatids at anaphase II and completion of meiosis Gametes of four genetic types Parental type of chromosome Recombinant chromosome Recombinant chromosome Parental type of chromosome

14 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) Breakage of homologous chromatids Joining of homologous chromatids Chiasma Separation of homologous chromosomes at anaphase I 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

15 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

16 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

17 Probability & genetics Calculating probability of making a specific gamete is just like calculating the probability in flipping a coin probability of tossing heads? probability making a B gamete? BB Bb B B B b 100% 50% 2013 Kul Genetik Dr. GTC

18 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%= Kul Genetik Dr. GTC

19 GENETIKA: PERAMALAN KETURUNAN DENGAN HUKUM PELUANG Prinsip dasar: Pemindahan gen dari orang tua kpd keturunannya Berkumpulnya kembali gen-gen dalam sigot Kakek (Aa) Aa X Aa Org tua: JTN (a) Anak: Aa Org tua: BTN: A mis Peluang Konsep muncul Peluang aa? Analogi pemindahan satu gen (A/a) dari sepasang Gen (Aa) = pelemparan mata uang yang memiliki dua sisi: -Gambar 2013 Kul Genetik Dr. GTC F1

20 female / eggs Calculating probability Pp x Pp sperm egg offspring P p male / sperm P p PP Pp Pp pp 2013 Kul Genetik Dr. GTC P P PP 1/2 x 1/2 = 1/4 P p Pp 1/2 x 1/2 = 1/4 p P + 1/2 x 1/2 = 1/4 1/2 p p pp 1/2 x 1/2 = 1/4

21 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 1/2 x 1/2 = 1/4 probability of Pp x Pp pp 1/2 x 1/2 = 1/4 Pp P p 2013 Kul Genetik Dr. GTC

22 Use rule of multiplication to predict crosses Calculating probability in crosses YyRr x YyRr Yy x Yy yyrr Rr x Rr yy?% 1/16 1/ Kul Genetik Dr. GTC x 1/4 rr

23 Apply the Rule of Multiplication AABbccDdEEFf x AaBbccDdeeFf AabbccDdEeFF Got it? Try this! AA x Aa Aa 1/2 Bb x Bb bb cc x cc cc Dd x Dd Dd EE x ee Ee Ff 2013 xkul Ff Genetik Dr. FFGTC 1/4 1 1/2 1 1/4 1/64

24 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 = 1/4 b B Bb 1/2 x 1/2 = 1/4 1/4 + 1/4 1/ Kul Genetik Dr. GTC

25 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 ) II. Peluang terjadinya 2 persitiwa /lebih = yang ½ masing-masing = 50 %. 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 Kul Genetik Dr. GTC

26 Aplikasi dalam pewarisan sifat Contoh: Gen resesif a (Albino) P: Aa x Aa normal normal Butawarna : gen resesif c X linked. P: Cc x C- normal normal F1. AA : Normal Aa : Normal Aa ; Normal aa : albino (1/4) Peluang anak laki-laki albino F1 : CC: F, Normal Cc: F, Normal C- : M, 2013 Kul Genetik Dr. GTC Normal

27 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 a, b = DUA KEJADIAN YANG TERPISAH n = banyaknya kejadian (2G, 2 H)=? N = 2 (a 2 +2ab+b 2 ) 2 ab = 2 (1/2) (1/2) = 1/2 n=3 Pelemparan 3 mata uang ( n= 3) ; (a+b) 3 = a a 2 b + 3 ab 2 + b 3 Peluang I G, 2 H = 3 ab 2 = 3 ((1/2)(1/2) Kul Genetik Dr. GTC = 3/8.

28 Penggunaan Rumus Binomium: Peluang pewarisan sifat Albino JTN : Aa x BTN Aa ¾ Normal ¼ Albino Jika suatu perkawinan mempunyai 4 anak ( n = 4) Maka Peluang semua anak normal? Rumus (a+b) 4 = a ab 3 +6a 2 b 2 +4ab 3 +b 4 Peluang 4 anak normal (a 4 ) = (3/4) 4 = 81/ Kul Genetik Dr. GTC

29 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 AaBbCcDdEe X AaBbCcDdEe Peluang (aabbcc) = 1/4x1/4x1/2 = 1/32 AABbccDdEE? = 1/2x1/2x1/4x1/2x1/4 = 1/ Kul Genetik Dr. GTC

30 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 F1??? Aa Aa aa = 1/4 Bagaimana Peluang Gen Sifat tsb diwariskan Mm x mm pada anak anaknya? M m Mm Mm m m M = ½ a = ¼ = 1/ Kul Genetik Dr. GTC

31 Penentuan Jenis Kelamin (SEKS) The inheritance of Gender INDUK Mother XX PEJANTAN Father XY Meiosis Sex cells X X X Y Fertilisation X Y X XX XY X XX XY Possible Offsprings 2007 Paul Billiet ODWS Chance of a Female 50% Chance of a Male 50%

32 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

33 KASUS KESEIMBANGAN HORMONAL = SEX Mengapa Seks Penting: Kasus Keseimbangan Hormonal, penentuan jenis kelamin menjadi tidak sederhana Contoh: PIG betina Awal bunting Testoteron Lahir : Jantan normal Betina :??? (alat kelm + Jantan) Dewasa Injeksi hormon betina (Progesteron + Estrogen) Tetap tidak menunjukkan perilaku betina normal Injeksi hormon jantan (Testoteron) Perilaku jantan jelas, fungsi seks jantan

34 PENGARUH LINGKUNGAN = SEX Crocodile Sex Determination Incubating temperature 30 o C all female 32 o C all male 31 o C 50% female, 50% male _ssh.jpg

35 Hasil Analisis Kariotyping: Letak/bentuk acak Jumlah dapat dihitung Metode: Disusun besar- kecil Besar,bentuk, homolog Urutan: Besar kecil Besar dan kesamaan bentuk Manfaat : Penentuan Sex Manfaat: Penentuan normal-abnorma

36 Penentuan Jenis Kelamin (Krom. SEKS) Dasar: Kariotyping untuk menentukan seks (X-Y Kromosom) Manfaat: Pre-derterminasi seks (deteksi dan manipulasi seks) R I N G K A S A N 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

37 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

38 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 Abnormal: non disjunction, meiosis, pembt sel kelamin jantan/betina pd drosophila X X x XY Munculnya kelainan kromosom Normal: XX x XY X X, Y ND Normal XX O X Y XX XY XXX XXY XO YO B:super B:Fertil J:Steril J:Lethal

39 Kelainan kromosom pada manusia: sindrom turner : wanita sindrom klinefelter: pria sindrom down: autosom/mongolisme XX X XY ND X XY O XXY XO Klinefelter (47) : Turner (45) testis tak berkembang Mandul dll Peran Krom: Manusia -ovary tak berkembang, tak menstruasi - kelj. Mammae tak berkembang baik dll. 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)

40 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 < > : interseks < 0.5 = jantan super

41 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)

42 Perbedaan Genetika Individu dan Populasi INDIVIDU 1.LINGKUNGAN: 1 tempat/1 lingkungan POPULASI 1.banyak tempat/banyak lingkungan 2.WAKTU: terbatas satu generasi Masa panjang, generasi ke generasi tumpang tindih. 3. GENOTIP: satu sampel genetik khas. Susunan gen tetap Tak ada variasi/ satu ukuran Tidak terjadi evolusi Gen pool Gen berubah dari generasi ke generasi

43 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

44 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)

45 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

46 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: p 2 (AA)+2pq (Aa) + q 2 (aa) Gamet A a (frek) (p) (q) A (p) Genotip AA Aa (frek) (p 2 ) (pq) a (q) Genotip Aa Aa (frek) (pq) (q 2 )

47 Only one of the populations below is in genetic equilibrium. Which one? Population sample Genotypes Gene frequencies AA Aa aa A a Paul Billiet ODWS

48 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 ( ) = 0.75

49 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 p 2 + 2pq + q 2 = 1 berasal dari ( p + q = 1) Diketahui q 2 = 0.01 maka q = p = pq = 2 (0.1) (0.9) = 0.18 Jadi frekuensi hitam heterosigot adalah: 0.18/ 0.99 = == 18 %.

50 LATIHAN/ DISKUSI/HOMEWORK: Fenotip Genotip j.indv. j.gen R J. Gen r Merah RR 80???- Roan Rr??? Putih rr 20???- Total??? F(R) ) = 210/300 = F (r ) = 90 / 300=

51 EXAMPLE ALBINISM IN THE INDO. BUFFALO POPULATION Frequency of the albino phenotype = 1 in or A = Normal skin pigmentation allele a = Albino (no pigment) allele Frequency = p Frequency = q Normal allele = A = p =? Albino allele = q = ( ) = or 7% Phenotypes Genotypes Hardy Weinberg frequencies Observed frequencies Normal Aa 2pq Albino aa q

52 HOW MANY buffalo IN Indonesia/Toraja ARE CARRIERS FOR THE ALBINO ALLELE (Aa)? a allele = = q A allele = p But p + q = 1 Therefore p = 1- q = = or 99.3% The frequency of heterozygotes (Aa) = 2pq = 2 x x = or 1.4% 2008 Paul Billiet ODWS

53 What about multiple alleles? Genotype Number Number of A 1 A 1 A X 4 A 1 A A 2 A 2 84 A 1 A A 2 A 3 88 A 3 A 3 32 Total 274 f(a 1 ) = ((2 X 4) ) (2 X 274) = ( ) 548 = = 0.135

54 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

55 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.

56 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

57 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 Aa Aa aa Homosigot 2/4 = 50 % Homosigot resesif: ¼ = 25 % AA X AA Aa X Aa aa X aa AA,AA AA,Aa,Aa,aa aa, aa Homosigot : 6/8= 75, % Homosigot resesif: 3/8 = 37.5 %

58 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

59 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