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Chapter 14 –
Mendelian
Genetics
Blending vs. Particulate Theory
of Inheritance
Over time,
populations do not
Traitsofover
offspring
Infer
become uniformtime,
were
a “blend”
of
Mendel
His
observations
observed
lead
that
populations
begin
thelook
parental
genes
Often
to
what’s
traits
retain
now
that
their
to
uniform
traits.
separate
seemed
accepted
to
identities
ashave
the
and
look
alike
“disappeared”
Particulate
Theory of
would reappear in
inheritance
subsequent
generations
Vocabulary Review
Genotype
= Genetic
Homozygous
==When
Heterozygous
When
make-up
of of
both
alleles
the
both
alleles
of
the of
Trait:
Characteristic
Allele:
Different
forms
characteristics
genotype
are
the
same
Phenotype
=
expressed
genotype
are different
an
organism
of a gene
represented
by the
2 alleles
trait,
basedresponsible
on
for
different
traits
(Hybrid)
in
organisms
Ex.diploid
Plant Height
genotype
Ex. T = Tall t = short
TT TT
or Tt
Tall
Tt
tt
tt
Short
Single-Factor Crosses
P = Parental Generation
F1 = 1st Filial Generation
X
t
t
T T
Tt Tt
Tt Tt
Memorize these
ratios!
F2 = 2nd Filial Generation
X
T
t
T t
TT Tt
Tt tt
Genotypic
Phenoytypic Ratio
Ratio
of
of monohybrid
monohybrid
cross:
cross
3:1
1:2:1
Two Factor Crosses (2 traits)
Y = yellow, y = green
R = Round, r = wrinkled
X
Dihybrid Cross
X
RrYy
RrYy
Memorize these
ratios!
Test Crosses
If a plant has a dominant phenotype, (for example yellow seeds) and we
are unsure of its genotype (YY, or Yy), you can determine it’s
genotype by crossing it with another with a recessive phenotype
(green seeds) with the genotype yy.
Yy
If F1 = 100% Yellow
Then P must be = YY
X
YY or Yy?
yy
Yy
yy
If F1 = 50% Yellow, 50% green
Then P = Yy (hybrid)
Pedigree
Aa ?
A = tongue roller
a = can not roll tongue
aa
?
?
?
?
AA
aa
Aa
aa
Can you figure out the
rest of the genotypes on
your own?
male
Mating couple
female
Children/Siblings
Shaded = trait being followed
Other Pedigree Symbols
Examples of connected symbols:
• Fraternal twins
• Identical twins
Unique Degrees of Dominance
(exceptions to the rule)
Incomplete Dominance: Dominant trait “blends” when combined
with a recessive allele
Notice how the genotypes are written…
X
CRCR
CRCW
CWCW
Unique Degrees of Dominance
Co-Dominance: When there
are multiple alleles that are dominant and
What’s
are of equal strength
expressed
- then both dominant alleles will be expressed when combined
when
we’re allele for
I’m
the
dominant
- a dominant allele will always mask a recessive allele.
together?
Type A blood!
Example: Blood Type
Don’t
count
me
out
just
I’m the dominant allele
‘cuz
I’m
recesssive.
I’m
for Type
blood !B blood
B or
B i =B type
IiIBAi III=
I
BA =
type O allele!
type
blood
ortype
IA O
i=
AB
type
blood
A blood
Notice how the genotypes are written…
Unique Gene Interactions
Pleiotrophy: The ability of a gene to affect an organism in many
phenoytypic ways
Ex. Sickle Cell Anemia
Blood Clumping
Physical Weakness or Brain Damage
Unique Gene Interactions
Polygenic Inheritance: when multiple genes have an added effect on a single
phenotype (Opposite of Pleiotrophy) – ex. Skin color, height
Notice the range in genotypes…
aabbcc
AABBCC
Unique Gene Interactions
Epistasis: when a gene at one locus alters the expression of a gene at another locus
Alleles for Fur Color:
B = Black Fur
b = brown fur
C = Color
c = albino
BBCC, BBCc, BbCC, BbCc 
bbCC, bbCc 
BBcc, Bbcc, bbcc 
Since cc genotype is albino, the
alleles for fur color (B or b) are not
expressed
Genetics is cool!
But wait…sample
probability
problems to
come!
How do you calculate probability?
Rule of Multiplication
(Alternatives to Punnett Squares)
Sample
CheckProblem
your work!
Brown eyes are dominant
over blue eyes. Parent A
X eyes,
b while
b
has blue
ParentBB isBb
heterozygous
Bb
for brown eyes. What is
b bb bb
the probability that they
will have a child with blue
eyes?
50% probability
for blue eyes
Parent A (blue eyes) = bb
Parent B (brown eyes) = Bb
Process:
Blue-eyed Child has to be bb
Probability of parent A donating
one “b” allele= 1
Probability of parent B donating
the other “b” allele = ½
1 X ½ = ½ (50% probability)
Rule of Multiplication and Addition
Steps for solving:
1. Write out the genotypic
possibilities
2. Use rule of X (multiply
probabilities of each
genotypic combination)
3. Use rule of +
Sample Problem #2:
In a cross between AaBbCc
x Aabbcc, what is the
probability that at least
two of the three recessive
traits is present in the
offspring?
AAbbcc
 AA (1/2) x bb (1/2) x cc (1/2) = 1/8
Aabbcc
 Aa (1/2) x bb (1/2) x cc (1/2) = 1/8
aaBbcc
 Aa (1/4) x Bb (1/2) x cc (1/2) = 1/16
aabbCc
 Aa (1/4) x bb (1/2) x Cc (1/2) = 1/16
aabbcc
 Aa (1/4) x bb (1/2) x cc (1/2) = 1/16
Sum of the
fractions: 6/16 =
3/8