ALEL DAN GEN GANDA

MonoHibrid pada Hewan :

Warna Rambut Hitam: (gen A):

However, it is possible to have several

AA (hitam) x aa (albino) different allele possibilities for one gene.

Multiple alleles is when there are more than two

Aa (Hitam) 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 f
This is a controlled by a tri-allelic gene
It can generate 6 genotypes

The ABO antigens on the surface of

• The alleles control the production of
the red blood cells

blood system

Two of the alleles are codominant to one another and both are dominant over the third _A produces antigen A • Allele I _B produces antigen B • Allele I

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) ch 2. c : pigmentasi terang, bulu pigmentasi gelap pada ujung (Chinchilla) h 3. c : pigmentasi bagian ujung-ujung tubuh, bagian lain putih (H= himalaya) ch h

Urutan dominasi alel : C> c > c > 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) _ch c _h causes a chinchilla pattern c causes a Himalayan pattern The alleles are arranged in the following pattern _{ch h}

> c > c  C > c

Full color rabbit

– alleles are dominant to all

Himalayan rabbit – color

_{ch h}

CC, Cc , Cc ,

_{h h h}

or Cc

c c or c c

Chinchilla rabbit

– partial

Albino rabbit – no color

defect in pigmentation

– allele is recessive to all

_ch

Kelinci Gelap: CC, Cc, Cc

; cc

P ; CC x Cch Cch F1 : C Cch x c c F2: Cc

Cch Ch Cch Ch

P; Cch Cch X Ch Ch F1: Cch Ch X Cch Ch F2: Cch Cch

,c Kelinci Albino: cc

; c

c h c h

c Kelinci Himalaya:

, c

h; c

Kelinci lebih terang; Chinchila: c

, C

Cch c

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 or bb

l l 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

_{A A}

I _{A B}

I A

I _A

I AB _{B B} I i A

I _B

I B

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 chromosomes are identical pairs of

– Homologous chromosomes.
– One inherited from mother and one from father sister chromatids joined at the centromere.
– made up of

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
Coat-color Eye-color _{genes genes} _{(homologous pair of} _Tetrad
How crossing over
_{chromosomes in synapsis)} leads to genetic ₁ _{Breakage of homologous chromatids} recombination _{2 Joining of homologous chromatids<Nonsister}
chromatids break _Chiasma in two at the same ₃ _{chromosomes at anaphase I} _{Separation of homologous} spot
The 2 broken
₄ _{Separation of chromatids at} chromatids join anaphase II and completion of meiosis _{Parental type of chromosome} together in a new _{Recombinant chromosome} _{Recombinant chromosome} way _{Parental type of chromosome}
Coat-color Eye-color _{genes genes} _Tetrad
A segment of one
_{(homologous pair of} _{chromosomes in synapsis)} chromatid has ₁ _{Breakage of homologous chromatids} changed places with the equivalent _{2 Joining of homologous chromatids} segment of its _Chiasma nonsister homologue ₃ _{chromosomes at anaphase I} _{Separation of homologous<If there were no}
crossing over meiosis could only produce 2 ₄ _{anaphase II and completion of meiosis} _{Separation of chromatids at} types of gametes _{Recombinant chromosome} _{Parental type of chromosome} _{Parental type of chromosome} Recombinant chromosome
TEORI PELUANG:
The Principles of Probability predict the
The Principles of probability can be used to
outcomes of genetic crosses
Alleles segregate by complete randomness

Similar to a coin flip!
Genetics & Probability
Mendel’s laws:
– segregation

– independent assortment
reflect same laws of probability that apply to tossing coins or rolling dice
50% 100% BB B B Bb B b Determining probability
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?
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
Aa X Aa Kakek (Aa)
F1 Org tua: JTN Org tua: (a) BTN: A mis Peluang muncul aa?
Anak: Aa
Konsep Peluang
Analogi pemindahan satu gen (A/a) dari sepasang Gen (Aa) = pelemparan mata uang yang memiliki dua sisi:
Calculating probability sperm egg 1/2 1/2 offspring = x 1/4 P 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 al e / egg s PP Pp
Pp pp 1/2 1/2 = x 1/4 1/2 1/2 = x 1/4 1/2 +
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 Pp
 pp
Pp x Pp
– probability of
p 1/2 x 1/2 = 1/4

Calculating probability in crosses
Use rule of multiplication to predict crosses

YyRr YyRr

x

yyrr
?% Yy Yy x Rr Rr x

1/16 yy rr Apply the Rule of Multiplication AABbccDdEEFf x AaBbccDdeeFf AabbccDdEeFF 1/2 AA x Aa  Aa Bb x Bb bb 1/4
 Got it? cc x cc cc
1 
Try this ! Dd x Dd  Dd 1/2
1 EE x ee  Ee Rule of addition
Chance that an event can occur 2 or more different ways
– sum of the separate probabilities
Bb x Bb Bb 
– probability of
sperm egg offspring B b Bb 1/4 1/2 x 1/2 = 1/4 1/4 + b B Bb 1/2 1/2 x 1/2 = 1/4
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 masing-masing berdiri = ½ = 50 %. sendiri
P. (X,Y) = P (X) x P (Y) contoh: Peluang dua anak pertama laki-laki P (Kl, LK) = (1/2) x ( ½) = ¼.
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)
Butawarna : gen resesif c
X –linked. P: Cc x C- normal normal F1 : CC:
F, Normal
Cc:
F, 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. ₂ PENGGUNAAN RUMUS BINOMIUM: (a+b)
‘(2G, 2 H)= ? N = 2
2
2
a, b = DUA KEJADIAN YANG TERPISAH
(a 2ab +b + )
n = banyaknya kejadian
2 ab = 2 (1/2) (1/2) = 1/2
1 1 1 1 2 1 1 3 3 1
n=3
1 4 6 4 1
3 ³ ² ² ³ Pelemparan 3 mata uang ( n= 3) ; (a+b) = a + 3 a b + 3 ab + b

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 ?
4
4
3
2
2
3
4 Rumus (a+b) = a + 4 ab +6a b +4ab +b
4
4 Peluang 4 anak normal (a ) = (3/4) = 81/256 Aplikasi lain teori peluang dalam genetika Pada suatu perkawinan: Genotip diketahui, mis : Aa Bb Cc X Aa Bb Cc aa = 1/4 aabbCc 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
Y
X X
X Fertilisation Possible Offsprings PEJANTAN Father Sex cells
Meiosis
INDUK Mother
XX XY Chance of a Female 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,
KASUS KESEIMBANGAN HORMONAL = SEX

Crocodile Sex Determination
Incubating temperature
30 ô C all female
32 ô C all male
31 ô C 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

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
tak ada krom Y
Jantan: XO/ X- ( )
3. UNGGAS/ BURUNG: ZW--------- Betina ZW atau ZO
(Ayam)
Jantan ZZ (burung) atau ZZ
4. LEBAH : haploid/diploid Betina : 2n : 32 buah Jantan : n : 16 buah
Catatan : 1,2,3 dasar kromosom seks 1,3 ada perbedaan (berbalikan)
R I N G K A S A N II
1. JANTAN Heterogametik:
a. Mamalia, Manusia : krom Y == JANTAN betina : XX Jantan : X Y
b. Heminiptera (Kepik, belalang) Betina : XX Jantan : X (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
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 non disjunction , meiosis ,
Abnormal : pembt sel kelamin jantan/betina pd drosophila
X X x XY
Normal:
ND Normal
XX x XY
XX O X Y
X X , Y
XXX XXY XO YO
XX XY sindrom turner : wanita Kelainan kromosom pada manusia : 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 Manusia Drosophila Krom:

X Menentukan sifat wanita Menentukan sifat betina Menentukan kehidupan, YO = lethal Indeks = Jmlh. Kromosome X = X/A Jmlh. pasangan autosom Contoh: Normal BTN 3 AA XX = X/A = 2/2 = 1.0
JTN 3 AA X Y = X/A = ½ = 0.5 Kesimpulan : X/A > 1 = betina super < 1.0
Teori indeks kelamin pada drosophila: krn adanya ND Oleh C.B. BRIDGES: faktor penentu seks jantan pada kromosome, betina pada autosome
– 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
Perbedaan Genetika Individu dan Populasi
INDIVIDU POPULASI 1.banyak tempat/banyak
1.LINGKUNGAN: 1 lingkungan tempat/1 lingkungan
Masa panjang, generasi ke
2.WAKTU: terbatas satu generasi tumpang tindih. generasi
Gen pool satu sampel
3. GENOTIP: genetik khas.
Gen berubah dari generasi Susunan gen tetap ke generasi
Tak ada variasi/ satu ukuran

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 The study of the change of allele Frek Genotip frequencies, genotype frequencies, and
Frek. fenotip 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)
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
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 (frek) A (p) a (q)
A (p) Genotip (frek) AA (p
2 ) Aa
(pq) a (q) Genotip Aa Aa
Only one of the populations below is in genetic equilibrium. Which one? Population sample Genotypes Gene frequencies AA Aa aa A a 100
50
0.6
40
60
0.4 100
0.6
30
20
0.4 100
20
0.6
16
48
36
0.4 100
0.6
80
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)
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)
: 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
2
2 Atau p + 2pq + q = 1 berasal dari ( p + q = 1)
2 Diketahui q = 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
80 Merah RR ???-
50
50 Roan Rr ???-
20 Putih rr ???- 210
90 Total ???-
F(R) ) = 210/300 =
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
Normal allele = A = p = ?
Albino allele = q = a = Albino (no pigment) allele Frequency = q
(0.00005) = 0.007 or 7%
Phenotypes Genotypes Hardy Observed Weinberg frequencies frequencies ₂ Normal AA p 0.99995 Normal Aa 2pq ₂ Albino aa q 0.00005
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% What about multiple alleles?
1 Number
Number of A
Genotype
1
1 A
4
2 X 4
A
1
2 A
41
41
A
2
2 A
84
A
1
3 A
25
25
A
2
3 A
88
A
3
3 A
32
A
274
Total
1
) = ((2 X 4) + 41 + 25)
f(A
÷ (2 X 274)
= (8 +41 + 25)
÷ 548
= 74
÷ 548
= 0.135
SUMMARY
Genetic drift
All can affect the
Mutation
transmission of genes
Mating choice
from generation to generation
Migration • Natural selection
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
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 mempertahankan gen-gen tertentu pd frekuensi tinggi, Prinsip dasar: sementara gen-gen lain dapat dihilangkan (mengekalkan/mempertahankan sifat yang diinginkan)
AA X AA Aa X Aa aa X aa
Aa X Aa
Homosigot AA,AA AA,Aa,Aa,aa aa, aa
2/4 = 50 %
AA Aa
Homosigot
Aa
resesif: ¼ Homosigot : 6/8= 75, %
dan rekombinasi gen: variabilitas meningkat dg perkw. Acak
(pilihan acak dr gen 2 parent, cenderung memprod.
Keturunan lebih bervariasi scr genetik), karena:
4 REPROD. SEXUAL .
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.

ALEL DAN GEN GANDA