Mendelelian
Genetics
copyright cmassengale
2
Gregor Mendel
(1822-1884)
Responsible
for the Laws
governing
Inheritance of
Traits
copyright cmassengale
3
Gregor Johann Mendel
Austrian monk
Studied the
inheritance of traits in
pea plants
Developed the laws of
inheritance
Mendel's work was
not recognized until
the turn of the 20th
century
copyright cmassengale
4
Gregor Johann Mendel
Between 1856 and
1863, Mendel
cultivated and tested
some 28,000 pea plants
He found that the
plants' offspring
retained traits of the
parents
Called the “Father of
Genetics“
Mendel genetic video
copyright cmassengale
5
Site of Gregor
Mendel’s
experimental
garden in the
Czech Republic
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6
Particulate Inheritance
Mendel stated that
physical traits are
inherited as “particles”
Mendel did not know
that the “particles” were
actually Chromosomes &
DNA
copyright cmassengale
7
What exactly are chromosomes?
How do chromosomes contain genetic
information?
Chromosome structure- DNA
• Chromosomes = long pieces of DNA
Chromosome structure- genes
• Parts of chromosomes are called genes
• Genes help determine your characteristics
– what are some example characteristics?
Human chromosome # 7
• Contains about
1,800 genes
• Contains over 150 million
nucleotides
Link to view genes and DNA sequence
Genetic Terminology
 Trait - any characteristic that can be
passed from parent to offspring
 Heredity - passing of traits from
parent to offspring
 Genetics - study of heredity
copyright cmassengale
11
Types of Genetic Crosses
 Monohybrid cross - cross involving a
single trait
e.g. flower color
 Dihybrid cross - cross involving two
traits
e.g. flower color & plant height
copyright cmassengale
12
Designer “Genes”
 Alleles - two forms of a gene (dominant &
recessive)
 Dominant - stronger of two genes
expressed in the hybrid; represented by a
capital letter (R)
 Recessive - gene that shows up less often
in a cross; represented by a lowercase
letter (r)
copyright cmassengale
13
Chromosomes come in pairs
Cut out from picture and
match each in pairs
How many chromosomes?
• Humans = 23 chromosome pairs, or 46 total. We
think we have about 40,000 genes.
• Pea flower = 7 chromosome pairs (14 total)
• Fruit fly = 4 chromosome pairs (8 total) 13,601
genes identified so far!
• Potato = 24 pairs (total 48)
• Horse = 32 pairs (total 64)
Chromosomes come in pairs
• You get one copy of a chromosome from one
parent and one copy of a chromosome from
the other parent.
• This pair of chromosomes is called a
“homologous pair” because they have the
same genes on them flower color gene
Flower chromosome 3
From
Dad
From
Mom
(from computer simulation)
Chromosomes come in pairs
IMPORTANT: homologous pairs have the same
genes, but they are NOT IDENTICAL
– Alleles can be different
flower color gene
“Blue” allele
of flower color
gene
“Red” allele
of flower color
gene
From
Dad
From
Mom
The combination of alleles determines the
observed characteristics
RED
Chromosome
combination
Allele
combination
RED
allele
allele
red, red
RED
BLUE
allele
allele
red, blue
(or blue, red)
BLUE
allele
BLUE
allele
blue, blue
Observed
characteristic
RED
RED
BLUE
Dominant and recessive alleles
Only one red allele is needed for the flower to
look red.
• Scientists call this a DOMINANT allele
Two blue alleles are needed for the flower to
look blue.
• Scientists call this a RECESSIVE allele
Symbols used for alleles
• The traditional symbols used for alleles are big
and small letters
• The letter chosen usually refers to the
dominant allele
Example: R = Red allele
r = Blue allele
The combination of alleles determines the
observed characteristics
RED
Chromosome
combination
Allele
combination
RED
allele
allele
red, red
RED
BLUE
allele
allele
red, blue
(or blue, red)
BLUE
allele
BLUE
allele
blue, blue
Observed
characteristic
RED
RED
BLUE
The combination of alleles determines the
observed characteristics
Chromosome
combination
Allele
combination
R
R
R
r
RR
Rr
RED
RED
r
r
rr
Observed
characteristic
BLUE
Phenotype – the physical feature resulting from
a genotype (e.g. red, blue)
Genotype – the two alleles that an organism has
for a trait
Example:
Trait
=
Phenotype =
Genotype =
“flower color”
“RED” or “BLUE”
“RR” or “Rr” or “rr”
The combination of alleles determines the
observed characteristics
Chromosome
combination
Allele
combination
R
R
R
r
RR
Rr
RED
RED
r
r
rr
Observed
characteristic
BLUE
The combination of alleles determines the
observed characteristics
Chromosome
combination
GENOTYPE
R
R
R
r
RR
Rr
RED
RED
r
r
rr
PHENOTYPE
BLUE
Homozygous – the two alleles are the same
• RR and rr are homozygous genotypes
Heterozygous – the two alleles are different
• Rr is a heterozygous genotype
Practice: Genotype & Phenotype in
Flowers
Genotype of alleles:
R = red flower
r = yellow flower
All genes occur in pairs, so 2
alleles affect a characteristic
What are the Possible combinations
of the two different alleles?
Genotypes
RR
Rr
rr
Phenotypes
RED
RED
copyright cmassengale
YELLOW
27
Genes and Environment Determine
Characteristics
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28
Genetics Practice Problems – Writing
Alleles
Mendel’s Pea Plant
Experiments
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30
Why peas, Pisum sativum?
Can be grown in a small
area
Produce lots of offspring
Produce pure plants
when allowed to selfpollinate several
generations
Can be artificially crosspollinated
copyright cmassengale
31
Reproduction in Flowering Plants
•Pollen contains sperm
–Produced by the stamen
•Ovary contains eggs
–Found inside the flower
Pollen carries sperm to the
eggs for fertilization
Self-fertilization can
occur in the same flower
Cross-fertilization can
occur between flowers
copyright cmassengale
32
Mendel’s Experimental Methods
•
•Mendel hand-pollinated
flowers using a paintbrush
–He could snip the stamens
to prevent self-pollination
–Covered each flower with
a cloth bag
•He traced traits through the
several generations
copyright cmassengale
33
How Mendel Began
Mendel
produced
pure
strains by
allowing the
plants to
selfpollinate
for several
generations
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34
Eight Pea Plant Traits
•
•
•
•
•
•
•
•
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)
copyright cmassengale
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copyright cmassengale
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copyright cmassengale
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Generation “Gap”
• Parental P1 Generation = the parental generation
in a breeding experiment.
• F1 generation = the first-generation offspring in a
breeding experiment. (1st filial generation)
– From breeding individuals from the P1
generation
• F2 generation = the second-generation offspring
in a breeding experiment.
(2nd filial generation)
– From breeding individuals from the F1
generation
copyright cmassengale
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Following the Generations
Cross 2
Pure
Plants
TT x tt
Results
in all
Hybrids
Tt
Cross 2 Hybrids
get
3 Tall & 1 Short
TT, Tt, tt
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39
Monohybrid
Crosses
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40
Punnett Square
Used to help solve
genetics problems
copyright cmassengale
41
copyright cmassengale
42
Possible combinations of inherited
chromosomes/genes
R = red
r = blue
Dad
Parents
Rr
Mom
Rr
X
RR
Possible
Offspring
Order doesn’t matter,
these two are the same (Rr)
Rr
r R
r r
What are all possible genotypes produced from the parents?
RR,
Rr,
rr
Punnett squares organize all possible
offspring combinations
Dad
Rr
Mom
X
Rr
Your turn:
R
r
R
RR
Rr
r
Rr
rr
What are all
possible genotypes
produced by
crossing Rr with rr?
Answer: 50% Rr & 50% rr
R
r
r
Rr
rr
r
Rr
rr
Punnett squares predict ratios of all genotypes
and phenotypes produced
R
r
R
RR Rr
r
Rr
rr
Genotypic ratio = 1RR: 2Rr:1rr
Phenotypic ratio =
3 Red:1 blue
Predicted ratios determine the probability of
each genotype/phenotype
R
r
R
RR Rr
r
Rr
rr
Genotypic ratio
= 1RR: 2Rr:1rr
Phenotypic ratio
= 3 Red:1 blue
1 out of 4 possibilities is rr = 25% chance of rr
• What is probability of Rr?
• How many out of 4 are red?
• What is the probability of red?
• What is the probability of blue?
Practice Mendel’s Crosses
• Work the P1, F1, and both F2 Crosses for
pure forms of the Pea Plant Trait your
group is assigned on the butcher paper.
• Find the genotypic ratio and phenotypic
ratio for each cross
• Finally, include the percent for each trait
copyright cmassengale
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Back to Mendel: P1 Monohybrid Cross
• Trait: Seed Shape
• Alleles: R – Round
r – Wrinkled
• Cross: Round seeds x Wrinkled seeds
•
RR
x
rr
r
Genotype: Rr
r
Phenotype: Round
R
Rr
Rr
R
Rr
Rr
Genotypic
Ratio: All alike
copyright cmassengale
Phenotypic
Ratio: All alike
49
P1 Monohybrid Cross Review
 Homozygous dominant x Homozygous
recessive
 Offspring all Heterozygous (hybrids)
 Offspring called F1 generation
 Genotypic & Phenotypic ratio is ALL ALIKE
copyright cmassengale
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F1 Monohybrid Cross
• Trait: Seed Shape
• Alleles: R – Round
r – Wrinkled
• Cross: Round seeds x Round seeds
•
Rr
x
Rr
R
r
R
RR
Rr
r
Rr
rr
Genotype: RR, Rr, rr
Phenotype: Round &
wrinkled
G.Ratio: 1:2:1
P.Ratio: 3:1
copyright cmassengale
51
F1 Monohybrid Cross Review
 Heterozygous x heterozygous
 Offspring:
25% Homozygous dominant RR
50% Heterozygous Rr
25% Homozygous Recessive rr
 Offspring called F2 generation
 Genotypic ratio is 1:2:1
 Phenotypic Ratio is 3:1
copyright cmassengale
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What Do the Peas Look Like?
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Mendel’s Experimental Results
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• Did the observed ratio match the theoretical ratio?
The theoretical or expected ratio of
plants producing round or wrinkled seeds
is 3 round :1 wrinkled
Mendel’s observed ratio was 2.96:1
The discrepancy is due to statistical
error
The larger the sample the more nearly
the results approximate to the
theoretical ratio
copyright cmassengale
55
…And Now the Test Cross
• Mendel then crossed a pure & a hybrid
from his F2 generation
• This is known as an F2 or test cross
• There are two possible testcrosses:
Homozygous dominant x Hybrid
Homozygous recessive x Hybrid
copyright cmassengale
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F2 Monohybrid Cross (1st)
• Trait: Seed Shape
• Alleles: R – Round
r – Wrinkled
• Cross: Round seeds x Round seeds
•
RR
x
Rr
R
Genotype: RR, Rr
r
Phenotype: Round
R
RR
Rr
R
RR
Rr
Genotypic
Ratio: 1:1
copyright cmassengale
Phenotypic
Ratio: All alike
57
F2 Monohybrid Cross (2nd)
• Trait: Seed Shape
• Alleles: R – Round
r – Wrinkled
• Cross: Wrinkled seeds x Round seeds
•
rr
x
Rr
R
r
r
Rr
Rr
r
Genotype: Rr, rr
Phenotype: Round &
Wrinkled
rr
G. Ratio: 1:1
rr
P.Ratio: 1:1
copyright cmassengale
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F2 Monohybrid Cross Review
 Homozygous x heterozygous(hybrid)
 Offspring:
50% Homozygous RR or rr
50% Heterozygous Rr
 Phenotypic Ratio is 1:1
 Called Test Cross because the offspring
have SAME genotype as parents
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Mendel’s Laws
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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
copyright cmassengale
61
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)
copyright cmassengale
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Law of Dominance
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63
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.
copyright cmassengale
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Applying the Law of Segregation
copyright cmassengale
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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.
copyright cmassengale
66
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)
copyright cmassengale
67
Dihybrid cross
• Used to predict the combinations of 2 traits
• For example:
– Two double-heterozygous striped, with tail cats
are crossed.
– How many offspring will be striped and have a tail,
how many will be striped with no tail, how many
will be plain with a tail…
Question:
How many gametes will be produced for the
following allele arrangements?
• Remember:
2n (n = # of heterozygotes)
•
1.
RrYy
•
2.
AaBbCCDd
•
3.
MmNnOoPPQQRrssTtQq
copyright cmassengale
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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
copyright cmassengale
70
Dihybrid Cross
• Traits: Seed shape & Seed color
• Alleles: R round
r wrinkled
Y yellow
y green
•
RrYy
x
RrYy
RY Ry rY ry
RY Ry rY ry
All possible gamete combinations
copyright cmassengale
71
Dihybrid Cross
RY
Ry
rY
ry
RY
Ry
rY
ry
copyright cmassengale
72
Dihybrid Cross
RY
Ry
rY
ry
RY
Ry
rY
ry
RRYY
RRYy
RrYY
RrYy
RRYy
RrYY
RrYy
RRyy
RrYy
Rryy
RrYy
rrYY
rrYy
Rryy
rrYy
rryy
copyright cmassengale
Round/Yellow:
Round/green:
9
3
wrinkled/Yellow: 3
wrinkled/green:
1
9:3:3:1 phenotypic
ratio
73
Dihybrid Cross
Round/Yellow: 9
Round/green:
3
wrinkled/Yellow: 3
wrinkled/green: 1
9:3:3:1
copyright cmassengale
74
Practice
•
•
Construct a Punnett Square for each of the
following crosses.
Write the Genotypic and Phenotypic Ratio
below the Punnett Squares.
1. SsTt X SsTt
2. SSTt X SsTt
3. SSTT X SsTt
S = striped
s = plain
T = tail
t = no tail
SsTt X SsTt
ST
S = striped
s = plain
T = tail
t = no tail
SsTt X SsTt
ST
St
S = striped
s = plain
T = tail
t = no tail
SsTt X SsTt
ST
St
sT
S = striped
s = plain
T = tail
t = no tail
SsTt X SsTt
ST
St
sT
st
S = striped
s = plain
T = tail
t = no tail
SsTt X SsTt
ST
ST
St
sT
st
St
sT
st
S = striped
s = plain
T = tail
t = no tail
SsTt X SsTt
ST
St
sT
st
ST
SSTT SSTt SsTT SsTt
St
SSTt SStt
SsTt
sT
SsTT SsTt
ssTT ssTt
st
SsTt
ssTt
Sstt
S = striped
s = plain
T = tail
t = no tail
Sstt
sstt
Genotypes: 1 SSTT: 2 SSTt: 1 SStt: 2 SsTT: 4 SsTt: 2 Sstt: 1 ssTT: 2 ssTt: 1 sstt
Phenotypes: 9 striped, tail : 3 striped, no tail : 3 plain, tail : 1 plain, no tail
SSTt X SsTt
ST
St
sT
st
ST
SSTT SSTt SsTT SsTt
St
SSTt SStt
SsTt
Sstt
ST
Same as
above
Same as
above
Same as
above
Same as
above
St
Same as
above
Same as
above
Same as
above
Same as
above
S = striped
s = plain
T = tail
t = no tail
If they are the same
as above you do
NOT have to
rewrite the
genotype
Genotypes: 1 SSTT: 1SSTt: 1SsTT: 1SsTt: 1SSTt: 1SStt: 1SsTt: 1Sstt
Phenotypes: 6 Striped with Tail: 2 Striped with no tail (3:1 reduced)
SSTT X SsTt
ST
Same
ST
St
sT
st
SSTT SSTt SsTT SsTt
ST
ST
ST
Genotypes: 1 SSTT: 1 SSTt: 1 SsTT: 1 SsTt
Phenotypes: 4 Striped Tail (100%)
S = striped
s = plain
T = tail
t = no tail
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
copyright cmassengale
84
Test Cross
• Possible results:
bc
bC
b___
C
bbCc
bbCc
or
copyright cmassengale
bc
bC
b___
c
bbCc
bbcc
85
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
pods
3/16
pods
3/16
pods
1/16
pods
copyright cmassengale
round seeds & green
round seeds & yellow
wrinkled seeds & green
wrinkled seeds & yellow
86
Incomplete Dominance
and
Codominance
copyright cmassengale
87
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 R
rr = white flower
R
copyright cmassengale
88
Incomplete Dominance
r
r
R
Rr
Rr
R
Rr
Rr
produces the
F1 generation
All Rr = pink
(heterozygous pink)
copyright cmassengale
89
If you cross the F1 generation, what
kind of offspring do you get?
r
r
R
Rr
Rr
R
Rr
Rr
R
r
R
RR
Rr
r
Rr
rr
Incomplete Dominance
copyright cmassengale
92
Codominance
• Two alleles are expressed (multiple alleles) in
heterozygous individuals.
• Example: blood type
•
•
•
•
1.
2.
3.
4.
type A
type B
type AB
type O
=
=
=
=
IAIA or IAi
IBIB or IBi
IAIB
ii
copyright cmassengale
93
Codominance Problem
• Example:
•
homozygous male Type B (IBIB)
x
heterozygous female Type A (IAi)
IA
i
IB
IAIB
IBi
IB
IAIB
IBi
copyright cmassengale
1/2 = IAIB
1/2 = IBi
94
Another Codominance Problem
• Example: male Type O (ii)
x
female type AB (IAIB)
IA
IB
i
IAi
IBi
i
IAi
IBi
copyright cmassengale
1/2 = IAi
1/2 = IBi
95
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) X girl - type AB (IAIB)
copyright cmassengale
96
Codominance
• Answer:
IA
IB
i
i
IAIB
ii
Parents:
genotypes = IAi and IBi
phenotypes = A and B
copyright cmassengale
97
What determines if you are male or
female???
• Females have 2 “X” chromosomes.
• Males have 1 “X” and 1 “y” chromosome.
• “Y” chromosome is very small and carries few
traits
X
X
Female
X
y
Male
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
copyright cmassengale
99
Sex-linked Traits
Example: Eye color in fruit flies
Sex Chromosomes
fruit fly
eye color
XX chromosome - female
Xy chromosome - male
copyright cmassengale
100
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.
• RR = red eyed
Xr
Xr
• Rr = red eyed
• rr = white eyed
• XY = male
XR
• XX = female
Y
copyright cmassengale
101
Sex-linked Trait Solution:
Xr
XR
XR
Xr
Y
Xr Y
Xr
XR
Xr
Xr Y
50% red eyed
female
50% white eyed
male
copyright cmassengale
102
Female Carriers
copyright cmassengale
103
Genetic Practice Problems
copyright cmassengale
104
Breed the P1 generation
• tall (TT) x dwarf (tt) pea plants
t
t
T
T
copyright cmassengale
105
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)
copyright cmassengale
106
Breed the F1 generation
• tall (Tt) vs. tall (Tt) pea plants
T
t
T
t
copyright cmassengale
107
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
copyright cmassengale
108