ALEL DAN GEN GANDA - Universitas · PDF fileALEL DAN GEN GANDA MonoHibrid pada Hewan: Warna...

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

allele possibilities for a gene.

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

© 2007 Paul Billiet ODWS

• 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

http://www.saburchill.com/IBbiology/bio_hp.html

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)

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:

CC, Cc, Ccch, Cch

Kelinci lebih terang; Chinchila:

cch cchh; cch, ch; ccchc

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

Multiple allelesEach 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

bb or bbl

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 typesAA, 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

IAi A

IB IB B

IBi 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

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

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

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

http://waynesword.palomar.edu/images/cross3.jpg

Recombination During Meiosis

Recombinant gametes

• 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 ofchromosomes in synapsis)

Breakage of homologous chromatids

Joining of homologous chromatids

Chiasma

Separation of homologouschromosomes at anaphase I

Separation of chromatids atanaphase II and completion of meiosis

Parental type of chromosome

Recombinant chromosome



Gametes of four genetic types

1

2

3

4

Coat-colorgenes

Eye-colorgenes

• 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 ofchromosomes in synapsis)

Breakage of homologous chromatids

Joining of homologous chromatids

Chiasma

Separation of homologouschromosomes at anaphase I

Separation of chromatids atanaphase II and completion of meiosis





Gametes of four genetic types

1

2

3

4

Coat-colorgenes

Eye-colorgenes

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


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

B

Bb

B

b


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


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?


Calculating probabilitysperm egg

1/2 1/2

offspring

=x 1/4P P PP

P p Pp

1/2 1/2 =x 1/4

p p pp

p P

Pp x Pp

P p

male / sperm

P

p

fem

ale

/ e

gg

s PP Pp

Pp pp

1/2 1/2 =x 1/4

1/2 1/2 =x 1/4

1/2

+


• 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


Calculating probability in crosses

Use rule of multiplication to predict crosses

YyRr YyRrx

yyrr

?%

Yy Yyx Rr Rrx

1/4 1/4

1/16

yy rr

x2013 Kul Genetik Dr. GTC

Apply the Rule of Multiplication

Got it?Try this!

AABbccDdEEFf AaBbccDdeeFfx

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 1/642013 Kul Genetik Dr. GTC

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 1/2 =x 1/4

B b Bb

1/2 1/2 =x 1/4

b B Bb

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 = 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) = ¼.2013 Kul Genetik Dr. GTC

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


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 = a3 + 3 a2b + 3 ab2 + b3

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


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


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


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


YXX X

Fertilisation Possible

Offsprings

PEJANTAN

Father

Sex cells

Meiosis

INDUK

MotherXX XY

Chance of a Female 50%

Chance of a Male 50%

X Y

X XX XY

X XX XY


The inheritance of Gender

Penentuan Jenis Kelamin (SEKS)


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

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

KASUS KESEIMBANGAN HORMONAL = 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

PENGARUH LINGKUNGAN = SEX

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-abnormal

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)

Penentuan Jenis Kelamin (Krom. SEKS)

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

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

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)

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

•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)

• Predicting inheritance in a population

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

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

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

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

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)

THE HARDY WEINBERG EQUATION

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



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

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 J. Gen r

Merah RR 80 ???-

Roan Rr ???- 50 50

Putih rr 20 ???-

Total ???- 210 90

F(R) ) = 210/300 =

F (r ) = 90 / 300=

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

0.99995Normal 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 =

(0.00005) = 0.007 or 7%

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%



• Genotype Number Number of A1

• A1A1 4 2 X 4

• A1A2 41 41

• A2A2 84

• A1A3 25 25

• A2A3 88

• A3A3 32

• Total 274

• f(A1) = ((2 X 4) + 41 + 25) ÷ (2 X 274)

• = (8 +41 + 25) ÷ 548

• = 74 ÷ 548

• = 0.135

What about multiple alleles?

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



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

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 %

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

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