Mendelelian
Genetics
1
Gregor Mendel
(1822-1884)
Responsible
for the Laws
governing
Inheritance of
Traits
2
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
3
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"
4
Mendel’s Laws
of Inheritance
1. Genes In Pairs: Genetic characters are
controlled by genes that exist in pairs of
alleles in individual organisms and are
passed from parents to their offspring.
When two organisms produce offspring,
each parent gives the offspring one of the
alleles from each pair.
2. Dominance and Recessiveness: When
two unlike alleles responsible for a single
character are present in a single
individual, one allele can mask the
expression of another allele. That is, one
allele is dominant to the other. The latter
is said to be recessive.
3. The Law of Segregation: During the
formation of gametes, the paired alleles
separate (segregate) randomly so that
each gamete receives one allele or the
other.
4. The Law of Independent Assortment:
During gamete formation, segregating
pairs of alleles assort independently of
each other. Example: genes on different
chromosomes will segregate
independently. Linked genes (close
together on one chromosome) do not
follow this law.
Rule of Dominance
• The trait that is observed in
the offspring is the dominant
trait (uppercase)
• The trait that disappears in
the offspring is the
recessive trait (lowercase)
Law of Segregation
• The two alleles for a trait
must separate when gametes
are formed
• A parent randomly passes
only one allele for each trait
to each offspring
Law of Independent
Assortment
• The genes for different
traits are inherited
independently of each other.
Site of
Gregor
Mendel’s
experime
ntal
garden in
the Czech
Republic
13
Particulate Inheritance
Mendel stated that
physical traits are
inherited as “particles”
Mendel did not know
that the “particles”
were actually
Chromosomes & DNA
14
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
15
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
16
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)
17
More
Terminology
 Genotype - gene combination for
a trait (e.g. RR, Rr, rr)
 Phenotype - the physical feature
resulting from a genotype (e.g.
red, white)
18
Here are the steps used to solve a
Monohybrid Mendelian Genetic Problem:
Step 1:
Figure out the genotypes of the parents (Pure Dom, Hybrid, Pure
Rec) Use the Capital letter of the DOMINANT trait and then its
lower case form to represent the recessive trait.
Step 2:
Figure out what kinds of gametes the parents can produce.
Step 3:
Set up a Punnett Square for your mating.
Step 4:
Fill in the babies inside the table by filling each square by going
Down and to the Left.
Step 5:
Figure out the genotypic ratio for your predicted babies.
How Many of the offspring are--Pure Dom: Hybrid: Pure Rec
Step 6:
Figure out the phenotypic ratio for your predicted babies.
How Many of the offspring are-- Dom : Rec
Step 7:
Answer the question you've been asked.
19
Solving Genetics Problems I:
Monohybrid Crosses
Classical genetics is a science of logic and
statistics. While many find the latter intimidating,
the mathematical side of most classical genetics
puzzles is relatively simple — and there are
actually ways to get around most of the math.
The logic part is inescapable. All genetics
problems are solved using the same basic logic
structure. If you learn the sense of the approach,
you can solve virtually any genetics problem,
provided you are given enough basic information.
20
Sample Problem using Steps:
This problem involves two gerbils named
Honey and Ritz. The gene in question is a
fur color gene which has two alleles —
dominant brown (B) and recessive black (b).
It's a very good idea to write down the
information you are given in a problem so
that it will be easy for you to refer to it when
necessary. So begin by writing something
like this at the top of your work page:
21
22
Step One: Figure out the genotypes of
the parents.
Each of our parent gerbils is heterozygous for this
gene. So here is our mating:
23
Step One: Figure out what kinds of
gametes the parents can produce..
Once you've got that settled, you need to
address the question of all of the possible kinds
of babies they could produce. Before any parent
makes babies, of course, that parent makes
gametes. So in order to find what kinds of
babies they can have, you must first determine
what kinds of gametes they can produce. Since
Honey is a heterozygote (and paying attention to
Rule #1), she can produce two kinds of eggs: B
eggs and b eggs.
24
Ritz is also a heterozygote, so he can
produce two kinds of sperm: B sperm and b
sperm. Something like this:
25
Step Three: Set up a Punnett Square for
your mating
Now you need to determine all the possible
ways that his sperm can combine with her
eggs. There are several different techniques
used for this operation. The most popular
among students is the Punnett Square.
Punnett Squares are probability tables — a
way to do statistics while avoiding as much
math as possible.
26
Step 4: fill in the Punnett Square,
down and to the left
Setting up a Punnett Square is easy. You need to
create a chart with one column for each of the
female's egg types, and one row for each of the
male's sperm types. For Honey and Ritz, your
table would look like this, then fill in the babies
genotype by going down and to the left:
BB
Bb
Bb
bb
27
Step Five: Figure out the genotypic ratio
for your predicted babies.
So we have now figured out that, if
Honey and Ritz have a lot of
babies, we can predict that 1/4 of
them should be BB, 1/2 of them
(2/4) should be Bb, and 1/4 should
be bb.
28
This conclusion is often expressed
as a genotypic ratio:1BB:2Bb:1bb.
This means that we are predicting
that, for every BB baby, they
should have 2 Bb babies (twice as
many), and one bb baby.
29
Step Six: Figure out the phenotypic ratio
for your predicted babies.
To do this, you need to ask yourself one question:
do any of these different genotypes produce the
same phenotype? In other words, do any of these
babies look alike? This is where dominance enters
the picture. If B is completely dominant to b, all
gerbils with at least one B will look pretty much
alike, no matter whether their second allele is B or
b. So BB and Bb have the same phenotype, and
we can add them together. Thus, our phenotypic
ratio is 3 Brown:1 Black. Or, there should be
three times as many brown babies as black
30
babies.
So the answer to our
question is, 3/4 of the babies
should be brown.
31
Step Seven: Answer the
question you've been asked.
The mating scheme we've just worked
through is called a monohybrid cross. This
means that we were paying attention to only
one gene (mono=1), and both of our parents
were heterozygous for that gene
(hybrid=heterozygous).
32
Punnett Square
Used to
help
solve
genetics
problems
33
34
Genotype & Phenotype in
Flowers
Genotype of alleles:
R = red flower
r = yellow flower
All genes occur in pairs, so 2
alleles affect a characteristic
Possible combinations are:
Genotypes
RR
Rr
rr
Phenotypes
RED
RED
YELLOW
35
Genotypes
 Homozygous genotype - gene
combination involving 2 dominant or
2 recessive genes (e.g. RR or rr);
also called pure
 Heterozygous genotype - gene
combination of one dominant & one
recessive allele (e.g. Rr); also
called hybrid
36
Genes and Environment Determine
Characteristics
37
Mendel’s Pea
Plant
Experiments
38
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
cross-pollinated
39
Reproduction in
Flowering Plants
• Pollen contains
sperm
– Produced by
the stamen
• Ovary contains
Pollen carries sperm to the
eggs
eggs for fertilization
– Found inside
Self-fertilization can
the flower
occur in the same flower
Cross-fertilization can
occur between flowers
40
Mendel’s Experimental
Methods
•
• Mendel hand-pollinated
flowers using a paintbrush
– He could snip the
stamens to prevent selfpollination
– Covered each flower with
a cloth bag
• He traced traits through the
several generations
41
How Mendel Began
Mendel
produced
pure
strains by
allowing the
plants to
selfpollinate
for several
generations
42
•
•
•
•
•
•
•
•
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)
43
44
45
Mendel’s Experimental Results
46
• 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
47
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 secondgeneration offspring in a breeding
experiment.
(2nd filial generation)
– From breeding individuals from the
48
F1 generation
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
49
Monohybrid
Crosses
50
P1 Monohybrid Cross
• Trait: Seed Shape
• Alleles: R – Round
r – Wrinkled
• Cross: Round seeds x Wrinkled seeds:
P = RR x
rr
1
r
r
R
Rr
Rr
R
Rr
Rr
Genotype: Rr
Phenotype: Round
Genotypic
Ratio: All alike
Phenotypic
Ratio: All alike
51
P1 Monohybrid Cross
Review
 Homozygous dominant x
Homozygous recessive
 Offspring all Heterozygous
(hybrids)
 Offspring called F1 generation
 Genotypic & Phenotypic ratio is
ALL ALIKE
52
F1 Monohybrid Cross
• Trait: Seed Shape
• Alleles: R – Round
r – Wrinkled
• Cross: Round seeds x Round
seeds: P = Rr
x
Rr
1
R
r
R
RR
Rr
r
Rr
rr
Genotype: RR, Rr, rr
Phenotype: Round &
wrinkled
G.Ratio: 1:2:1
P.Ratio: 3:1
53
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
54
What Do the Peas Look Like?
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
56
F2 Monohybrid Cross (1st)
• Trait: Seed Shape
• Alleles: R – Round
r – Wrinkled
• Cross: Pure Round seeds x
Hybrid Round seeds
• P =RR x Rr
R
r
R
RR
Rr
R
RR
Rr
Genotype: RR, Rr
Phenotype: Round
Genotypic
Ratio: 1:1
Phenotypic
Ratio: All alike
57
F2 Monohybrid Cross (2nd)
• Trait: Seed Shape
• Alleles: R – Round
r – Wrinkled
• Cross: Wrinkled seeds x Hybrid
Round seeds = rr
x
Rr
R
r
r
Rr
Rr
r
rr
rr
Genotype: Rr, rr
Phenotype: Round &
Wrinkled
G. Ratio: 1:1
P.Ratio: 1:1
58
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
59
Mendel’s Laws
60
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
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)
62
Law of Dominance
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.
64
Applying the Law of
Segregation
65
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.
66
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
67
Sex-linked Traits
Example: Eye color in fruit flies
Sex Chromosomes
fruit fly
eye color
XX chromosome - female
Xy chromosome - male
68
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
• RR = red eyed
• Rr = red eyed
XR
• rr = white eyed
• XY = male
Y
• XX = female
Xr
69
Sex-linked Trait
Solution:
Xr
XR
XR
Xr
Y
Xr Y
Xr
XR
Xr
Xr Y
50% red eyed
female
50% white eyed
male
70
Female Carriers
71
Incomplete
Dominance
and
Codominance
72
Incomplete Dominance
• F1 hybrids have an appearance somewhat in
between the phenotypes of the two parental
varieties.
• Example: snapdragons (flower)
• red (RR) x white (WW)
W
• RR = red flower
• WW = white flower
W
R
R
73
Incomplete
Dominance
W
W
R RW
RW
R RW
RW
produces the
F1 generation
All RW = pink
(heterozygous pink)
74
Incomplete Dominance Problem:
• In cattle when a red bull(RR) is mated with
white(WW) cow the offspring are
roan(RW) a blending of red and white.
Mate a red bull with a roan cow. Show the
P , the Punnett Square, and give the
genotypic and phenotypic ratios for this
cross.
1
75
P1 = __RR__ x __RW__
Genotypic ratio: ____ : _____ : _____
Phenotypic ratio: ____ : _____ : _____
76
Incomplete Dominance
77
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
78
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
79
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
80
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)
81
Codominance
• Answer:
IA
IB
i
i
IAIB
ii
Parents:
genotypes = IAi and IBi
phenotypes = A and B
82
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)
83
Question:
How many gametes will be produced
for the following allele arrangements?
• Remember: 2n (n = # of heterozygotes)
•
1.
RrYy
•
2.
AaBbCCDd
•
3.
MmNnOoPPQQRrssTtQq
84
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
85
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
86
Dihybrid Cross
RY
Ry
rY
ry
RY
Ry
rY
ry
87
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
88
Dihybrid Cross
Round/Yellow: 9
Round/green:
3
wrinkled/Yellow: 3
wrinkled/green: 1
9:3:3:1
89
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
•
•
•
bc
CC = curly hair
Cc = curly hair
cc = straight hair
bC
b___
90
Test Cross
• Possible results:
bc
bC
b___
C
bbCc
bbCc
or
bc
bC
b___
c
bbCc
bbcc
91
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
92
pods
Genetic Practice
Problems
93
Breed the P1 generation
• tall (TT) x dwarf (tt)
pea plants
t
t
T
T
94
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)
95
Breed the F1
generation
• tall (Tt) vs. tall (Tt)
pea plants
T
t
T
t
96
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
97
98