Transcript Document

MENDELIAN GENETICS
Laws of Heredity
A. Origins of Genetics
• Passing characteristics from
parent to offspring is called
heredity
• Accurate study of heredity
began with Austrian monk
Gregor Mendel at his
monastery gardens
• Mendel used different
varieties of garden pea
plant
– Could predict patterns of
heredity which form
modern-day genetics
principles
– Garden peas have eight
observable characteristics
with two distinct traits that
Mendel counted and
analyzed with each cross or
breeding
Pea Characteristics & Traits
Characteristic
Trait
Flower color
Purple v. white
Seed color
Yellow v. green
Seed shape
Round v. wrinkled
Seed coat color
Gray v. white
Pod shape
Round v. constricted
Pod color
Green v. yellow
Flower position
Axial v. top
Plant height
Tall v. dwarf
Pictures
• In nature, pea plants
self pollinate since
both reproductive
organs (male stamen
[pollen] & female
pistil) are internal
– Mendel physically
removed stamens &
dusted pistils with
pollen from chosen
plants to observe
results by crosspollination
• Parent plants are P generation
• All offspring are F generation (from Latin filialis for
son/daughter)
– F1 generation = first offspring (children)
– F2 generation = second offspring (grandchildren)
P
F1
F2
P
F1
F1
F2
F2
F2
• Cross-pollinated two
pure-bred plants with one
very different trait (purple
vs. white flowers) in P
generation
– Examined each F1 plant’s
trait & counted them
Monohybrid Cross
P
F1
F1
• Allowed F1 generation to
self-pollinate to produce
F2 generation
– Examined each F2 plant’s
trait & counted them
F2
F2
• Mendel collected tons of data – his results are
reproducible
– Monohybrid cross of white flowers and purple
flowers in P generation produced 100% purple
flowers in F1 generation
– Self-pollination of F1 produced 705 purple flowers &
224 white flowers in F2 generation
P
F1
F2
• Mendel’s ratios still hold true today
– Crossing pure bred traits in monohybrid cross will
ALWAYS express only one trait in F1
– Self-pollinating F1 will ALWAYS result in 3:1 ratio
• 705:224 = 3:1
B. Heredity Theories & Laws
• Mendel knew all ideas about “blending”
characteristics was bogus
– Developed four hypotheses:
• 1. individual has two copies of gene, one
from each parent
From Dad
From Mom
• 2. there exists alternative versions of genes
called alleles represented by letters
A
b
C
A
B
C
From Dad
From Mom
D
e
F
d
e
F
• 3. if two different alleles occur together, one
may be expressed while other is not –
dominant and recessive
– UPPERCASE alleles are dominant alleles
» Trait gets expressed ALWAYS
– lowercase alleles are recessive alleles
» Trait only gets expressed if dominant is not
A A
present
b
C
B trait will be expressed
B
C
From Dad
e trait will be expressed
From Mom
D
e
F
d
e
F
– When both alleles are identical,
individual is considered homozygous
for that trait
• Homozygous dominant = both
dominant (UPPERCASE)
• Homozygous recessive = both
recessive (lowercase)
– When alleles are different, individual is
considered heterozygous for that trait
A A
b B
C C
D d
e e
F F
QUIZ YOURSELF
TT = ?
Homozygous dominant
Tt = ?
Heterozygous
tt = ?
Homozygous recessive
XX = ?
Homozygous dominant
rr = ?
Homozygous recessive
• 4. when gametes (sperm/eggs or spores)
are formed, alleles for each trait separate
independently during meiosis
– Occurs during anaphase
bb
pP
HH
Aa
C. General Rules for Genes
• Each gene is given allele letter
– Letter is always first letter of dominant trait
• Ex: yellow peas are dominant over green peas
Y = yellow, y = green
YY
yy
Yy
• Ex: purple flowers are dominant over white
flowers
P = purple, p = white
PP
Pp
pp
• Ex: In roses, pink petals are dominant over white
petals, and tall stems are dominant over short
stems.
P = pink, p = white
T = tall, t = short
Cross a male heterozygous pink tall with a
female homozygous white short
Male = PpTt
Female = pptt
• Some traits are dominant – only one dominant
allele needed in genome to show phenotype
• Some traits are recessive – both recessive alleles
needed to express phenotype
Dominant Trait
Polydactyl (P)
No Hitchhiker’s Thumb (T)
Tongue rolling (R)
Free-hanging ear lobe (F)
Widow’s Peak (W)
Brown eyes (B)
Left thumb on top (L)
Recessive Trait
Non-polydactyl (p)
Hitchhiker’s thumb (t)
No tongue rolling (r)
Attached ear lobe (f)
No widow’s peak (w)
Blue or green eyes (b)
Right thumb on top (l)
Plus four others!
Polydactyl (PP or Pp)
No Hitchhiker’s Thumb (T)
TT or Tt
tt
Tongue Rolling (R)
RR or Rr
rr
Free-Hanging Earlobes (F)
FF or Ff
ff
Widow’s Peak (W)
WW or Ww
ww
Brown Eyes (B)
BB or Bb
bb
Left Thumb on Top (L)
LL or Ll
ll
Mid-Digit Hair (H)
HH or Hh
hh
Cleft Chin (C)
CC or Cc
cc
Dimples (D)
DD or Dd
dd
Freckles (F)
FF or Ff
ff
D. Laws of Heredity
• During meiosis (forming haploid gametes from
diploid cells), chromatids separate during
anaphase II
– Law of segregation: two alleles for character
separate when gametes are formed
Male Parent
(Tt)
Alleles segregate
(separate) into gametes
Female Parent
(Tt)
T
t
Alleles segregate
(separate) into gametes
Alleles pair up in all combos
Alleles segregate
(separate) into gametes
Alleles pair up in all combos
• Mendel studied whether different characteristics
were inherited together or separately
– Conducted dihybrid crosses where two traits are
studied
– Concluded that traits NOT inherited together &
developed law
• Law of independent assortment: alleles of
different genes separate independently during
gamete formation in meiosis
Law of Independent Assortment
Male Parent
(TtBb)
Traits separate
independently
Female Parent
(TtBb)
Tb
TB
tB
tb
TTBB
TTBb
TtBb
16
total!
E. Punnett Square
• Easiest way to represent Laws of Segregation
and Independent Assortment is through
Punnett Square
– Cross a homozygous dominant yellow pea with a
green (homozygous recessive)
– Genotypes: YY & yy (1 trait, 4 alleles = 4 combos)
Step 1: separate alleles from
genotypes & place on top & down
side
Step 2: determine possible
combinations by crossing alleles
y
y
Y
Yy
Y
Yy
Yy
Yy
y
y
Y
Yy
Y
Yy
Yy
Yy
• Have to analyze findings from crossings
– Genotypic ratio: 4 Yy
– Phenotypic ratio: 4 yellow peas
Curi Family Eye Color
• My dad has green eyes
• My mom has brown eyes
– Knowing that I have brown eyes, what is my
GENOTYPE?
• Brown is dominant (B)
• Green is recessive (b)
–Dad must be bb
b
B Bb
b
Bb
B Bb
Bb
• According to Punnett Square, all my parents’
children should have BROWN eyes
–In reality, my brother has green eyes. What
does this mean?
• Mom’s genotype must be Bb
b
B Bb
b
Bb
b bb
bb
This means that there is a 50%
(2/4 or ½) chance that each
child my parents had could have
green eyes. I lucked out. 
Monohybrid Cross Examples
• 1. Aliens with two eyes are dominant over
aliens with one eye. Cross a heterozygous twoeyed male with a homozygous one-eyed
female.
t
t
T
Tt
Tt
t
tt
tt
– Genotypic ratio: 2:2
– Phenotypic ratio: 2:2
(2 Tt: 2 tt)
(2 two eyes: 2 one eye)
• 2. Orange carrots are dominant over purple
carrots. Cross a male purple carrot with a
heterozygous orange carrot.
O
o Oo
o Oo
– Genotypic ratio: 2:2
– Phenotypic ratio: 2:2
o
oo
oo
(2 Oo: 2 oo)
(2 orange: 2 purple)
• Dihybrid (two traits) cross can be trickier
– Cross heterozygous purple flowers, heterozygous
yellow pea with another of the same.
– Genotypes: PpYy & PpYy (2 traits, 8 alleles = 16
combos!)
Step 1: find possible gametes
PY Py pY py
(two traits each!) for each
PPYy
PpYY
PpYy
PPYY
PY
parent by doing FOIL method
(first, outside, inside, last) &
Py PPYy PPyy PpYy Ppyy
place on top & down side
Step 2: determine possible
pY PpYY PpYy ppYY ppYy
combinations by crossing alleles,
py PpYy Ppyy ppYy ppyy
making sure same alleles are
together
PY Py pY py
PY PPYY PPYy PpYY PpYy
Py PPYy PPyy PpYy Ppyy
pY PpYY PpYy ppYY ppYy
py PpYy Ppyy ppYy ppyy
• Analyze results
– Genotypic ratio: 1:2:2:4:1:2:1:2:1 (1 PPYY: 2 PPYy: 2
PpYY: 4 PpYy: 1 PPyy: 2 Ppyy: 1 ppYY: 2 ppYy: 1 ppyy)
– Phenotypic ratio: 9:3:3:1 (9 purple/yellow: 3 purple
/green: 3 white/yellow: 1 white: green)
Simple dihybrid rules …
• Always will be maximum of 4 phenotypes
– Trait A vs. Trait B, Trait C vs. Trait D = 4
phenotypes
• AC, AD, BC, BD
• Heterozygous AaBb vs. Heterozygous AaBb will
always have same phenotypic ratio
– 9 AB, 3aaB, 3Abb, 1aabb = 9:3:3:1
• Heterozygous AaBb vs. Homozygous aabb will
always have same phenotypic ratio
– 4 AB, 4 aaB, 4 Abb, 4 aabb = 4:4:4:4
Dihybrid Cross Examples
• 1. Red ants are dominant over black ants,
and long antennae are dominant over short
antennae. Cross a black short antennae
male with a heterozygous red long female.
– Genotypes: rrll & RrLl
– Gametes: rl, rl, rl, rl & RL, Rl, rL, rl
RL
rl RrLl
Rl
Rrll
rL
rrLl
rl
rrll
rl RrLl
Rrll
rrLl
rrll
rl RrLl
Rrll
rrLl
rrll
rl RrLl
Rrll
rrLl
rrll
• Genotypic ratio: 4:4:4:4 (4 RrLl: 4 Rrll: 4 rrLl: 4 rrll)
• Phenotypic ratio: 4:4:4:4 (4 red/long: 4 red/short:
4 black/long: 4 black/short
• 2. Green frogs are dominant over brown
frogs, and spots are dominant over no
spots. Cross a heterozygous green spotted
female with the same type of male.
– Genotypes: GgSs & GgSs
– Gametes: GS, Gs, gS, gs & GS, Gs, gS, gs
gS
gs
GS
Gs
GS GGSS GGSs GgSS GgSs
Gs GGSs GGss GgSs Ggss
gS GgSS GgSs ggSS
ggSs
gs GgSs Ggss ggSs ggss
• Genotypic ratio: 1:2:2:4:1:2:1:2:1
• Phenotypic ratio: 9:3:3:1
Curi Family Eye Color & Ear Shape
• My father has green eyes and free-hanging ear
lobes (homo or hetero?) while my mother has
brown eyes (heterozygous) and free-hanging ear
lobes (homo or hetero?). What are the possible
outcomes for the children?
– Know that my mother is heterozygous for brown
eyes since my brother has green eyes
– What about free hanging ear lobes?
• I have free hanging, but my brother and sister are
attached! What does that mean about my parents?
–Both parents MUST be heterozygous for freehanging ears!
• Genotypes: bbFf & BbFf
• Gametes: bF, bf, bF, bf & BF, Bf, bF, bf
BF
Bf
? ?
bf BbFf
? Bbff
bF BbFF
?
? BbFf
bf BbFf Bbff
?
bF
bf
bF BbFF BbFf bbFF bbFf
bbFf
bbff
BRO
SIS
bbFF bbFf
bbFf
bbff
me
F. Probability
• Getting ratios of genotypes & phenotypes is
actually calculating probability
– Probability: likelihood that particular event
(genotype or phenotype) will occur
– Calculated by dividing number of predicted
outcomes by number of total outcomes
– Ex: 3 peas are yellow, 1 is green
• Words: 3 out of 4 are yellow
• Ratios: 3:1 yellow
• Decimals: 0.75 yellow, 0.25 green (add up to 1.0)
• Percentages: 75% yellow, 25% green (add to 100%)
• Fractions: ¾ yellow, ¼ green (add to 4/4)