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

(1)

ALEL DAN GEN GANDA

MonoHibrid pada Hewan:

Warna Rambut Hitam: (gen A): AA (hitam) x aa (albino)

Aa (Hitam)

Gen A:

1 Kali mutasi : -- >alel a

Gen Ganda:

Bbrp kali mutasi---) bbrp alel: a1,a2,a3, dst

However, it is possible to have several

different allele possibilities for one gene. Multiple alleles is when there are more than two

▸ Baca selengkapnya: lkpd mutasi gen

(2)

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 IA _{produces antigen A}

• Allele IB _{produces antigen B}

• Allele i produces no antigen

• 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

(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)

Contoh: Gen pigmentasi bulu kelinci (Gen C, pigmentasi hitam), memiliki 3 alel:

1. c : albino (tak ada pigmentasi)

2. cch_{: pigmentasi terang, bulu pigmentasi gelap pada ujung}

(Chinchilla)

3. ch: pigmentasi bagian ujung-ujung tubuh, bagian lain putih (H= himalaya)

(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)

cch _{causes a chinchilla pattern}

ch _{causes a Himalayan pattern}

The alleles are arranged in the following pattern  C > cch _{> c}h _{> c}

• Himalayan rabbit – color in certain parts of the body; dominant only to c; ch_{c or c}h_ch

• Albino rabbit – no color – allele is recessive to all other alleles; cc

Full color rabbit – alleles are dominant to all others; CC, Ccch_{, Cc}h_, or Cc

Chinchilla rabbit – partial defect in pigmentation cch _{allele dominant to all}

other alleles except C;

(5)

Kelinci Gelap: CC, Cc, Ccch_{, C}ch

Kelinci lebih terang; Chinchila: cch _cch_{h; c}ch_{, c}h_{; cc}ch_c Kelinci Himalaya: c h ch; ch,c Kelinci Albino: cc P; Cch Cch X Ch Ch F1: Cch Ch X Cch Ch F2: Cch Cch Cch Ch Cch Ch Ch Ch P ; CC x Cch Cch F1 : C Cch x c c F2: Cc Cch 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 Bbl

blbl

(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

AA, AO A blood type

BB ,BO B blood type

AB AB blood type or

OO O blood type Genotypes Phenotypes (Blood

types) IA _IA _A IA _IB _AB IA_i _A IB _IB _B IB_i _B ii O

(8)

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 AB Ia Ib O 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

(9)

Crossing Over dan

Rekombinan

• Sometimes in meiosis, homologous chromosomes exchange parts in a process called crossing-over.

(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.

(11)

21 Apr 2002 11

Crossing Over Basics

• Occurs at One or More Points Along

Adjacent Homologues

• Points contact each other

• DNA is Exchanged

• Menaikkan var.Genetik

(12)

Recombination During Meiosis

(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 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 Gametes of four genetic types

1 2 3 4 Coat-color genes Eye-color genes

(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

Figure 8.18B

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 Gametes of four genetic types

1 2 3 4 Coat-color genes Eye-color genes

(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!

(16)

Genetics & Probability

• Mendel’s laws:

– segregation

– independent assortment

reflect same laws of probability

that apply to tossing coins or

rolling dice

(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?

50%

100%

BB

B

Bb

B

b

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%=.5

(19)

GENETIKA: PERAMALAN KETURUNAN

DENGAN HUKUM PELUANG

Prinsip dasar: Pemindahan gen dari orang tua kpd keturunannya Berkumpulnya kembali gen-gen dalam sigot

Kakek (Aa) Org tua: JTN (a) Org tua: BTN: A

Anak: Aa _{Konsep Peluang}

Analogi pemindahan satu gen (A/a) dari sepasang Gen (Aa) = pelemparan mata uang yang memiliki dua sisi:

-Gambar -Huruf. Aa X Aa F1 mis Peluang muncul aa? 2013 Kul Genetik Dr. GTC

(20)

Calculating probability

sperm egg

1/2

offspring

=

x

1/4

P

PP

P

p

Pp

1/2

x

1/2

=

1/4

p

pp

p

P

Pp x Pp

P

p

male / sperm

P

p

fem al e / egg s

_PP

_Pp

Pp

pp

1/2

x

1/2

=

1/4

1/2

x

1/2

=

1/4

1/2

+

(21)

• Chance that 2 or more independent events will occur together

– probability that 2 coins tossed at the same time will land heads up – probability of Pp x Pp  pp

Rule of multiplication

1/2

x

1/2

=

1/4

1/2

x

1/2

=

1/4

Pp

P

p

(22)

Calculating probability in crosses

Use rule of multiplication to predict crosses

YyRr

x

YyRr

yyrr

?%

Yy

x

Yy

Rr

x

Rr

1/4

1/16

yy

rr

x

(23)

Apply the Rule of Multiplication

Got it? Try this!

AABbccDdEEFf

x

AaBbccDdeeFf

AabbccDdEeFF

Bb

x

Bb



bb

cc

x

cc



cc

Dd

x

Dd



Dd

EE

x

ee



Ee

Ff

x

Ff



FF

AA

x

Aa



Aa

1/2

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

1/2

x

1/2

=

1/4

B

b

Bb

1/2

x

1/2

=

1/4

b

B

Bb

1/4

+

1/2

(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 ) = ½ = 50 %.

II. Peluang terjadinya 2 persitiwa /lebih yang masing-masing 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.

(26)

Aplikasi dalam pewarisan

sifat

Contoh: Gen resesif a (Albino)

P: Aa x Aa normal normal F1. AA : Normal Aa : Normal Aa ; Normal aa : albino (1/4)

Peluang anak laki-laki albino

???

Butawarna : gen resesif c X –linked. P: Cc x C-normal normal F1 : CC: F, Normal Cc: F, Normal C- : M, Normal c- : M 2013 Kul Genetik Dr. GTC

(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 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1

Pelemparan 3 mata uang ( n= 3) ; (a+b) 3 _{= a}3 _{+ 3 a}2_{b +}_{3 ab}2 _{+ b}3

Peluang I G , 2 H = 3 ab2 _{= 3 ((1/2)(1/2)}2 = 3/8. n=3 ‘(2G, 2 H)= ? N = 2 (a2+2ab+b2) 2 ab = 2 (1/2) (1/2) = 1/2 2013 Kul Genetik Dr. GTC

(28)

Penggunaan Rumus Binomium: Peluang pewarisan sifat 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 JTN : Aa x BTN

Aa

¾ Normal ¼ Albino

(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 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

(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

pada anak anaknya?

M = ½ a = ¼ = 1/8 Mm x mm M m m m Mm Mm mm 2013 Kul Genetik Dr. GTC

(31)

Y X X X Fertilisation Possible Offsprings PEJANTAN Father Sex cells Meiosis INDUK Mother XX XY Chance of a Female 50% Chance of a Male 50% X Y X XX XY X XX XY

The inheritance of Gender

(32)

Summary:

Males and females have different purposes defined by their gametes

Development of sexes is dependent on:

genes

hormones environment

(33)

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

(34)

Crocodile Sex Determination

Incubating temperature

30o_{C all female} 32o_{C all male} 31o_{C 50% female, 50% male} http://a.abcnews.com/images/Sports/rt_thailand_ 080514_ssh.jpg

(35)

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:

(36)

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

Dasar: Kariotyping untuk menentukan seks (X-Y Kromosom)

Manfaat: Pre-derterminasi seks (deteksi dan manipulasi seks)

(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

(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

Munculnya kelainan kromosom

Normal:

XX x XY

X X , Y

XX XY

Abnormal: non disjunction, meiosis ,

pembt sel kelamin jantan/betina pd drosophila X X x XY

ND Normal

XX O X Y

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 -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)

(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

< 1.0 – 0.5 > : interseks < 0.5 = jantan super

(41)

Population Genetics

•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

POPULASI

1.LINGKUNGAN: 1

tempat/1 lingkungan 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

(44)

KONSEP-KONSEP DASAR

:

FREK. GEN Frek Genotip Frek. fenotip

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: ?)

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)

Pendekatan:

:

The study of the change of allele

frequencies, genotype frequencies, and

phenotype frequencies

(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

(46)

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) + q}2_(aa)

Gamet

(frek)

A

(p)

a

(q)

A (p)

Genotip

(frek)

AA

(p

2

)

Aa

(pq)

a (q)

Genotip

(frek)

Aa

(pq)

Aa

(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

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

(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) dan 150 (putih)

Brp. Frek (RR); Frek (R) ) ?

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

(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 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 %.

(50)

LATIHAN/ DISKUSI/HOMEWORK

:

Fenotip

Genotip

j.indv.

j.gen R

J. Gen r

Merah

RR

80

???-Roan

Rr

???-

50 50

Putih

rr

20

???-Total

???-

210 90 F(R) ) = 210/300 = F (r ) = 90 / 300=

(51)

EXAMPLE ALBINISM IN THE INDO. BUFFALO POPULATION

Frequency of the albino phenotype = 1 in 20 000 or 0.00005

Phenotypes Genotypes Hardy Weinberg frequencies Observed frequencies Normal AA p2 0.99995 Normal Aa 2pq Albino aa q2 _0.00005

A = Normal skin pigmentation allele Frequency = p

a = Albino (no pigment) allele Frequency = q

Normal allele = A = p = ?

Albino allele = q =

(52)

HOW MANY buffalo IN Indonesia/Toraja

ARE CARRIERS FOR THE ALBINO

ALLELE (Aa)?

a allele = 0.007

= q

A allele

= p

But

p + q = 1

Therefore

p

= 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%

(53)

• Genotype Number Number of A1 • A1_A1 ₄ _{2 X 4} • A1_A2 ₄₁ ₄₁ • A2_A2 ₈₄ • A1_A3 ₂₅ ₂₅ • A2_A3 ₈₈ • A3_A3 ₃₂ • Total 274 • f(A1_{) = ((2 X 4) + 41 + 25) ÷ (2 X 274)} • = (8 +41 + 25) ÷ 548 • = 74 ÷ 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

(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

(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

(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, %

(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.