Download Kuliah 3 Mendelian

Genetika
Mendel
1
Gregor Mendel
(1822-1884)
Menemukan
hukum pewarisan
suatu sifat
2
Gregor Johann Mendel
Seorang pendeta
bangsa Austria
Meneliti pewarisan
sifat pada tanaman
arcis (pea)
Menghasilkan Hukum
Sistem pewarisan
Temuan Mendel tak
diterima sampai
menjelang abad ke 20
3
Gregor Johann Mendel
Antara 1856 and
1863, Mendel
menanam dan menguji
28,000 tanaman arcis
Dia mencatat bahwa
selama penelitian sifat
induk tetap muncul
pada anak
Dijuluki “Father of
Genetics"
4
Site of
Gregor
Mendel’s
experimental
garden in the
Czech
Republic
5
Genetic Terminology
 Trait - any characteristic that
can be passed from parent to
offspring
 Heredity - passing of traits
from parent to offspring
6
1). ISTILAH-2 DASAR TERKAIT PEWARISAN
SIFAT
a. GEN – ALEL
Gb. Kromosom mengandung Gen-2
 GEN: FAKTOR GENETIK PENGATUR SIFAT
Contoh: gen M mengatur sifat warna bunga,
gen K mengatur sifat warna biji
 ALEL: BENTUK ALTERNATIF SUATU GEN
Contoh: Gen M (pengatur warna bunga) mempunyai 2 alel:
 alel M menyebabkan bunga berwarna merah
 alel m menyebabkan bunga berwarna putih
Contoh: Gen K mengatur sifat warna biji mempunyai 2 alel:
 Alel K menyebabkan biji berwarna kuning
 Alel k menyebabkan biji berwarna hijau
7
b. GENOTIPE
 GENOTIPE (G): susunan genetik (gen-2) organisme
 Gen-2 berpasangan (dari tetua jantan dan tetua betina)
Contoh: Genotipe organisme berdasar 3 pasang gen (Gambar) mempunyai
genotipe: MM KK Bb
PASANGAN GEN/ALEL dibedakan atas:
 HOMOZIGOT: mempunyai alel sama, contoh: MM, KK
 HETEROZIGOT: mempunyai alel tidak sama, contoh: Bb
8
c. FENOTIPE
9
M
m
x
M
MM
mm
m
M


m
F1:
Mm
(M dg m alelik)
(M dominan
thd m)
 F1 : keturunan pertama hsl. persil.
 F2 : keturunan kedua hsl persilang.
(hasil selfing antara individu F 1)
 Homosigot: individu dg. 2 alel sama,
MM / mm
 Heterosigot: indv. dg 2 alel berbeda,
Mm
10
Mendel’s Pea Plant
Experiments
11
Mengapa arcis, Pisum sativum?
Dapat ditanam ditempat
sempit
Menghasilkan banyak
keturunan
Menghasilkan tanaman
homozigot jika menyerbuk
sendiri
Mudah menyilangkan
12
Reproduksi pada tanaman
Pollen contains sperm
Produced by the
stamen
Ovary contains eggs
Found inside the
flower
Pollen membawa sperm ke
egg untuk fertilisasi
Self-fertilization terjadi
pada bunga yg sama
Cross-fertilization
terjadi antara bunga
berbeda
13
Mendel’s Experimental
Methods
Mendel hand-pollinated
flowers using a
paintbrush
He could snip the
stamens to prevent
self-pollination
He traced traits
through the several
generations
14
15
Percobaan Mendel
Mendel membuat
tanaman homozigot
dengan
membiarkannya
menyerbuk sendiri
beberapa generasi
16
Delapan sifat arcis
Seed shape --- Round (R) or Wrinkled (r)
Seed Color ---- Yellow (Y) or Green (y)
Pod Shape --- Smooth (S) or wrinkled (s)
Pod Color --- Green (G) or Yellow (g)
Seed Coat Color ---Gray (G) or White (g)
Flower position---Axial (A) or Terminal (a)
Plant Height --- Tall (T) or Short (t)
Flower color --- Purple (P) or white (p)
17
18
19
Mendel’s Experimental Results
20
Rasio Pengamatan Vs rasio teoritis
Rasio teoritis tanaman berbiji bundar
(round) atau kisut (wrinkled) adalah 3
round : 1 wrinkled
Rasio yang diamati Mendel adalah 2.96:1
Perbedaan disebabkan kesalahan statistik
(statistical error)
Semakin besar sample semakin mendekati
rasio teoritis
21
Macam Generasi
Induk (Parental) P1 = generasi induk dalam suatu
percobaan genetika
Generasi F1 = Generasi turunan pertama (1st
filial generation) dalam suatu percobaan
genetika
Generasi F2 = Generation turunan kedua
generation dalam suatu percobaan genetika.
(2nd filial generation)
22
Following the Generations
Persilangan
2 induk
homozigot
TT x tt
Hasilnya
semua
Hybrids
Tt
Persilangan 2 Hybrid
menghasilkan
3 Tinggi & 1 pendek
TT, Tt, tt
23
Monohybrid
24
a). Hk. MENDEL I
(Hukum Segregasi/Pemisahan Pasangan Gen/Alel):
Pada pembentukan gamet*), ALEL DARI PASANGAN-2 GEN AKAN “MEMISAH
(BERSEGREGASI) ke dalam Gamet-gamet yg dibentuk”
*) Perbiakan generatif melibatkan:
 Pembentukan gamet-2
 Penyatuan gamet
25
P1 Monohybrid Cross
Sifat: bentuk biji (Seed Shape)
Alleles: R – Round
r – Wrinkled
Cross: Round seeds
x Wrinkled seeds
RR
x
rr
r
r
R
Rr
Rr
R
Rr
Rr
Genotype: Rr
Phenotype: Round
Genotypic
Ratio: Semua sama
Phenotypic
Ratio: Semua sama
26
 Homozygous dominant x Homozygous
recessive
 Keturunan semua Heterozygot
(hybrids)
 Keturunan disebut generasi F1
 Rasio Genotip & Phenotip sama
semua
27
Persilangan F1 Monohybrid
Trait: Bentuk biji
Alleles: R – bundar
Cross: Biji Bundar
Rr
R
r
R
RR
Rr
r
Rr
rr
r – Kisut
x biji Bundar
x
Rr
Genotip: RR, Rr, rr
Phenotip: Bundar &
kisut
Rasio Gntp: 1:2:1
Rasio Pntp: 3:1
28
 Heterozygot x heterozygot
Keturunan (Offspring):
25% Dominan Homozygot RR
50% Heterozygot Rr
25% Recessif Homozygous rr
 Offspring disebut F2 generation
 Rasio Genotyp adalah 1:2:1
 Rasio Phenotyp adalah 3:1
29
Bentuk Biji Yang kelihatan
30
Uji Silang (Test Cross)
Mendel menyilangkan homozigot &
hybrida dari generasi F2
Dua kemungkinan test cross:
Homozygot dominant x Hybrid
Homozygot recessive x Hybrid
31
F2 Monohybrid Cross
st
(1 )
Trait: Seed Shape
Alleles: R – Bundar r – Kisut
Cross: Biji Bundar x Biji Bundar
RR
x
Rr
R
R
R
r
RR
Rr
RR
Rr
Genotip: RR, Rr
Phenotip: Bundar
Rasio Genotip: 1:1
Rasio Phenotip:
Semua sama Bundar
32
F2 Monohybrid Cross
Trait: Seed Shape
Alleles: R – Bundar r – Kisut
Cross: Biji Kisut x Biji Bundar
rr
x
Rr
R
r
r
Rr
Rr
r
rr
rr
Genotip: Rr, rr
Phenotip: Bundar &
Kisut
Rasio Gntp: 1:1
Rasio pntp: 1:1
33
F2 Monohybrid Cross Review
 Homozygot x heterozygot
 Keturunan:
50% Homozygot RR or rr
50% Heterozygot Rr
 Rasio Phenotip adalah 1:1
34
Practice Your Crosses
Work the P1, F1, and both
F2 Crosses for each of the
other Seven Pea Plant
Traits
35
Tugas :
Buat Persilangan :
I. Antara Homozigot x Homozigot :
(tiga macam)
II. Antara Homozigot x Heterozigot :
(tiga macam)
III. Antara Heterozigot x Heterozigot (1 macam)
Buat lengkap dengan skema, perbandingan genotipe
dan fenotipe yang dihasilkan baik pada F1 dan F2
36
Mendel’s Laws
37
Results of Monohybrid Crosses
Inheritable factors or genes are
responsible for all heritable
characteristics
Phenotype is based on Genotype
Each trait is based on two genes,
one from the mother and the
other from the father
True-breeding individuals are
homozygous ( both alleles) are the
same
38
Law of Dominance
In a cross of parents that are
pure for contrasting traits, only
one form of the trait will appear in
the next generation.
All the offspring will be
heterozygous and express only the
dominant trait.
RR x rr yields all Rr (round seeds)
39
Law of Dominance
40
Law of Segregation
During the formation of gametes
(eggs or sperm), the two alleles
responsible for a trait separate
from each other.
Alleles for a trait are then
"recombined" at fertilization,
producing the genotype for the
traits of the offspring.
41
Applying the Law of Segregation
42
Law of Independent
Assortment
Alleles for different traits are
distributed to sex cells (&
offspring) independently of one
another.
This law can be illustrated using
dihybrid crosses.
43
Dihybrid Cross
A breeding experiment that tracks
the inheritance of two traits.
Mendel’s “Law of Independent
Assortment”
a. Each pair of alleles segregates
independently during gamete formation
b. Formula: 2n (n = # of heterozygotes)
44
Question:
How many gametes will be produced
for the following allele arrangements?
Remember: 2n (n = # of heterozygotes)
1. RrYy
2. AaBbCCDd
3. MmNnOoPPQQRrssTtQq
45
Answer:
1. RrYy: 2n = 22 = 4 gametes
RY
Ry
rY ry
2. AaBbCCDd: 2n = 23 = 8 gametes
ABCD ABCd AbCD AbCd
aBCD aBCd abCD abCD
3. MmNnOoPPQQRrssTtQq: 2n = 26 = 64
gametes
46
Dihybrid Cross
Traits: Seed shape & Seed color
Alleles: R round
r wrinkled
Y yellow
y green
RrYy
RY Ry rY ry
x
RrYy
RY Ry rY ry
All possible gamete combinations
47
Dihybrid Cross
RY
Ry
rY
ry
RY
Ry
rY
ry
48
Dihybrid Cross
RY
RY RRYY
Ry RRYy
rY RrYY
ry
RrYy
Ry
rY
ry
RRYy
RrYY
RrYy
RRyy
RrYy
Rryy
RrYy
rrYY
rrYy
Rryy
rrYy
rryy
Round/Yellow:
Round/green:
9
3
wrinkled/Yellow: 3
wrinkled/green:
1
9:3:3:1 phenotypic
ratio
49
Dihybrid Cross
Round/Yellow: 9
Round/green:
3
wrinkled/Yellow: 3
wrinkled/green: 1
9:3:3:1
50
Test Cross
A mating between an individual of unknown
genotype and a homozygous recessive
individual.
Example: bbC__ x bbcc
BB = brown eyes
Bb = brown eyes
bb = blue eyes
CC = curly hair
Cc = curly hair
cc = straight hair
bC
b___
bc
51
Test Cross
Possible results:
bc
bC
b___
C
bbCc
bbCc
or
bc
bC
b___
c
bbCc
bbcc
52
Summary of Mendel’s laws
LAW
DOMINANCE
SEGREGATION
INDEPENDENT
ASSORTMENT
PARENT
CROSS
OFFSPRING
TT x tt
tall x short
100% Tt
tall
Tt x Tt
tall x tall
75% tall
25% short
RrGg x RrGg
round & green
x
round & green
9/16 round seeds & green
pods
3/16 round seeds & yellow
pods
3/16 wrinkled seeds & green
pods
1/16 wrinkled seeds & yellow
pods
53
Incomplete Dominance
and
Codominance
54
Incomplete Dominance
F1 hybrids have an appearance somewhat
in between the phenotypes of the two
parental varieties.
Example: snapdragons (flower)
red (RR) x white (rr)
r
r
RR = red flower
rr = white flower
R
R
55
Incomplete Dominance
r
r
R Rr
Rr
R Rr
Rr
produces the
F1 generation
All Rr = pink
(heterozygous pink)
56
Incomplete Dominance
57
Codominance
Two alleles are expressed (multiple
alleles) in heterozygous individuals.
Example: blood type
1.
2.
3.
4.
type
type
type
type
A
B
AB
O
=
=
=
=
IAIA or IAi
IBIB or IBi
IAIB
ii
58
Codominance Problem
Example: homozygous male Type B (IBIB)
x
heterozygous female Type A (IAi)
IA
i
IB
IAIB
IBi
IB
IAIB
IBi
1/2 = IAIB
1/2 = IBi
59
Another Codominance Problem
• Example: male Type O (ii)
x
female type AB (IAIB)
IA
IB
i
IAi
IBi
i
IAi
IBi
1/2 = IAi
1/2 = IBi
60
Codominance
Question:
If a boy has a blood type O and
his sister has blood type
AB,
what are the genotypes
and
phenotypes of their
parents?
boy - type O (ii)
AB (IAIB)
X
girl - type
61
Codominance
Answer:
IA
IB
i
i
IAIB
ii
Parents:
genotypes = IAi and IBi
phenotypes = A and B
62
Sex-linked Traits
Traits (genes) located on the sex
chromosomes
Sex chromosomes are X and Y
XX genotype for females
XY genotype for males
Many sex-linked traits carried on
X chromosome
63
Sex-linked Traits
Example: Eye color in fruit flies
Sex Chromosomes
fruit fly
eye color
XX chromosome - female
Xy chromosome - male
64
Sex-linked Trait Problem
Example: Eye color in fruit flies
(red-eyed male) x (white-eyed female)
XRY
x
XrXr
Remember: the Y chromosome in males
does not carry traits.
Xr
Xr
RR = red eyed
Rr = red eyed
R
X
rr = white eyed
XY = male
Y
XX = female
65
Sex-linked Trait Solution:
Xr
XR
XR
Xr
Y
Xr Y
Xr
XR
Xr
Xr Y
50% red eyed
female
50% white eyed
male
66
Female Carriers
67
Genetic Practice
Problems
68
Breed the P1 generation
tall (TT) x dwarf (tt) pea plants
t
t
T
T
69
Solution:
tall (TT) vs. dwarf (tt) pea plants
t
t
T
Tt
Tt
produces the
F1 generation
T
Tt
Tt
All Tt = tall
(heterozygous tall)
70
Breed the F1 generation
tall (Tt) vs. tall (Tt) pea plants
T
t
T
t
71
Solution:
tall (Tt) x tall (Tt) pea plants
T
t
T
TT
Tt
t
Tt
tt
produces the
F2 generation
1/4 (25%) = TT
1/2 (50%) = Tt
1/4 (25%) = tt
1:2:1 genotype
3:1 phenotype
72