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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. This is
the individual’s GENOTYPE (type of genes)
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 (phenotype) 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
present
A A
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
GENOTYPE
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 assigned 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. Cross a male heterozygous pink
with a female white rose.
Male: Pp
Female: pp
• Ex: In grasshoppers, big hops are dominant
over small hops. Cross a male homozygous
dominant hopper with a female
heterozygous hopper.
Male: BB
Female: Bb
• 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
–This is called simple inheritance
• MOST human traits are complex inheritance
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
PTC (phenylthiocarbamide) Taster (P)
PP or Pp
pp
Thiourea Taster (T)
TT or Tt
tt
Dominant Trait
Recessive Trait
Polydactyl (P)
No Hitchhiker’s Thumb (T)
Tongue rolling (R)
Non-polydactyl (pp)
Hitchhiker’s thumb (tt)
No tongue rolling (rr)
Free-hanging ear lobe (F)
Widow’s Peak (W)
Brown eyes (B)
Left thumb on top (L)
Attached ear lobe (ff)
No widow’s peak (ww)
Blue or green eyes (bb)
Right thumb on top (ll)
Freckles (F)
Dimples (D)
Cleft chin (C)
No freckles (ff)
No dimples (dd)
Smooth chin (cc)
Mid-digit hair (M)
PTC taster (P)
Thiourea taster (T)
No mid-digit hair (mm)
Non-PTC taster (pp)
Non-thiourea taster (tt)
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!
B
• Ex: In tigers, black stripes are dominant
O
over white stripes, and orange fur is
dominant over tan fur. Cross a heterozygote
black striped homozygous orange male with
a white-striped, tan female.
Male: BbOO
Female: bboo
E. 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)
F. Punnett Square
• Easiest way to represent Laws of Segregation and
Independent Assortment is through Punnett Square
– Cross a male homozygous dominant yellow pea with a
female green (homozygous recessive)
– Male genotypes: YY
– Female genotype: 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: 0
– Phenotypic ratio: 4 yellow peas: 0 green
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 Tt: 2 tt
– Phenotypic ratio: 2 two eyes: 2 one eye
• 2. Orange carrots are dominant over purple
carrots. Cross a male purple carrot with a female
heterozygous orange carrot.
O
o Oo
o Oo
o
oo
oo
– Genotypic ratio: 2 Oo: 2 oo
– Phenotypic ratio: 2 orange: 2 purple
Simple monohybrid rules …
• Always will be maximum of 2 phenotypes
– DOMINANT (A) vs. recessive (a)= 2 phenotypes
• AA/Aa, aa
• Heterozygous Aa vs. Heterozygous Aa will always
have same phenotypic ratio
– 3 A trait: 1 a trait
• Homozygous AA vs. Homozygous aa will always
have same phenotypic ratio
– 4 A trait = 4:0
Curi Family Eye Color
• My dad had green eyes
• My mom had brown eyes
– Knowing that I have brown eyes, what is my
GENOTYPE? Let’s do a test cross (two Punnett squares,
one with homozygous dominant, one with
heterozygous).
• 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.
Well what about green eyes AND
attached ears? Dihybrid cross!
Dihybrid Cross Examples
• 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 Example
• 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
rrLl
rl
rrll
rl RrLl
Rrll
rrLl
rrll
rl RrLl
Rrll
rrLl
rrll
rl RrLl
Rrll
rrLl
rrll
rL
• 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
Curi Family Eye Color & Ear Shape
• My father had green eyes and attached ear lobes
(homozygous) while my mother had brown eyes
(heterozygous) and free-hanging ear lobes (homo
or hetero?). What are the possible outcomes for
the children?
– Know that my mother was 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 mom?
–Mom MUST have been heterozygous for freehanging ears!
• Genotypes: bbff & BbFf
• Gametes: bf, bf, bf, bf & BF, Bf, bF, bf
BF
Bf
bF
bf
bf BbFf
Bbff
bbFf
bbff
bf BbFf
Bbff
bbFf
bbff
bf BbFf
Bbff
bbFf
bbff
bf BbFf
Bbff
bbFf
bbff
BRO
SIS
me