Gregor Mendel conducted experiments with pea plants between 1856-1863. Through his experiments, he discovered two fundamental laws of inheritance: the Law of Segregation and the Law of Independent Assortment. The Law of Segregation states that alleles segregate and are passed to gametes independently. The Law of Independent Assortment states that different genes assort independently during gamete formation. Mendel's work laid the foundation for modern genetics although it was not widely recognized until the early 20th century.

1. 1
Mendelian Genetics
Prepared and presented by:
Clarito f. Concepcion

2. 2
A Deck of Genes
 Blending hypothesis holds that the trait
that the children receive from their
parents is simply a mixture of the parental
characters
 Predicts a uniform population of individuals
will result from a freely mating population,
over many generations (not seen!!!)
 Why are some traits observed again after
skipping a generation?

3. -It was generally accepted that the
hereditary traits of the offspring of any
species were merely the diluted blending of
whatever traits were present in the
“parents”
-Similarly, it was also thought that, over
generations, a hybrid would revert to its
original form
-Implication being that a hybrid could not
create new forms
3

4. 4
A Deck of Genes
 “Particulate” Hypothesis of Inheritance
 the GENE idea
 Offsprings receive heritable units (gene)
which retain their separate identities
 An individual’s collection of genes is much
more like a deck of cards than a pail of
paint
 Without getting diluted, genes can be
shuffled and passed along through
generations

5. 5
Mendel stated that physical
traits are inherited as
“particles”
Mendel did not know that the
“particles” were actually
Chromosomes & DNA
Developed decades before
chromosomes were observed
under the microscope and the
significance of their behavior
was understood
Particulate Inheritance

6. 6
Gregor Johann
Mendel
(1822-1884)
- Father of
Modern Genetics
- Responsible for
the Laws
governing
Inheritance of
Traits

7. 1822 - born in Brunn, Austria, now, Brno,(Czech Repub.)
-received agricultural training in school along with basic
education
- as an adolescent, he overcame financial hardship and
illness to excel in high school and, later, at the Olmutz
Philosophical Institute
1843 – entered an Augustinian monastery
- considered becoming a teacher but failed the
necessary exam
1851 – left the monastery to pursue two years of study
in physics and chemistry Univ. of Vienna
- important for his development as a scientist
(under the mentorship of Doppler and Unger)
1857 – began breeding garden peas to study inheritance
7

8. 8
Gregor Johann Mendel
Developed the laws
of inheritance
Presented his findings
in the Natural History
Society of Brunn –
1865
Paper entitled,
“Experiments in Plant
Hybridization” – 1866
German language.

9. 9
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

10. Why pea plant?
1. Pea plant was small and easy to grow
10

11. Why pea plant?
2. Control
mating/fertilization
a. it is a bisexual
flower
b. self fertilizing
11

12. Why pea plant?
3. Produce large number of offspring
12

13. Why pea plant?
4. Short life cycle
13

14. Why pea plant?
5. Available in many contrasting traits
14

15. 15
Site of
Gregor
Mendel’s
experimental
garden in the
Czech
Republic

16. 16
Mendel’s Work place
Fig. 2.5

17. 17
 Mendel's work was not recognized until the
turn of the 20th century
 1900 - Carl Correns, Hugo deVries, and Erich
von Tschermak rediscovered and confirmed
 Called the “Father of Genetics“

18. 18
 Alleles - two forms of a gene (dominant
or 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)
Genetic Terminologies

19. 19
Genetic Terminologies
 Hybridization – mating or crossing of
two true-breeding varieties
 True-breeding – varieties that that
produce only one character over many
generations
 P generation – true-breeding parents or
parental generation

20. 20
 Monohybrid cross - cross involving a
single trait
e.g. flower color
 Dihybrid cross - cross involving two
traits
e.g. flower color & plant height
Genetic Terminologies

21. 21
 Homozygous genotype – When the
two alleles are the same
(dominant or 2 recessive genes)
e.g. TT or tt; also called pure
 Heterozygous genotype – When
the 2 alleles are different- one
dominant & one recessive allele
(e.g. Tt)
Genetic Terminologies

22. 22
 F1 generation – first filial generation
parent (hybrid offspring)
 F2 generation – second filial generation
parent (hybrid offspring)
 Genotype - gene combination for a trait
(e.g. RR, Rr, rr)
Genetic Terminologies

23. 23
 Phenotype - the physical feature
resulting from a genotype (e.g. red,
white); may also refer to physiological
traits that relate directly to appearance
(e.g. non self-pollination)
 Hemizygous – there is only one allele
instead of two
Genetic Terminologies

24. 24

25. 25
Mendel’s Experiments

26. 26
Mendel’s Experiment
 Mendel chose to track only those characters
that occurred in two distinct, alternative
forms
 Also made sure that he started with only
true-breeding varieties
 He cross-pollinated two contrasting, true-
breeding pea varieties (P generation),
producing F1 generation offsprings
 Allowing these F1 hybrids to self-pollinate
(or to cross-pollinate with other F1 hybrids)
produces an F2 generation offsprings

27. 27
Mendel’s Experiment
 Had he stopped with F1 generation, the basic
patterns of inheritance would not have been
discovered
 His quantitative analysis of the F2 plants
from thousands of genetic crosses allowed
him to deduce two fundamental principles of
heredity
 Law of Segregation
 Law of Independent Assortment

28. 28

29. 29
 Mendel reasoned that the heritable factor
for white flowers did not disappear in the
F1 plants, but was somehow hidden, or
masked, when the purple flower was present
 In his jargon, purple flower color is a
dominant trait, and the white flower color is
a recessive trait

30. 30
 According to him, the reappearance of the
white flower plants in the F2 generation was
evidence that the white flowers had not
been diluted or destroyed by coexisting
with the purple flower factor in the F1
hybrids.
 Instead, it had been hidden in the presence
of the purple flower
 He also observed the same pattern of
inheritance in six other characters, each
represented by six different traits

31. 31

32. 32
 First, alternative versions of genes account
for variations in inherited characters
* Alleles – can be related to the DNA and
chromosomes
Mendel’s Model

33. 33
 Second, for each
character, an organism
inherits two copies
(two alleles) of a gene,
one from each parent
* made this
assertion despite not
knowing about the
role, and even
existence of the
chromosomes
Mendel’s Model

34. 34
 Third, if the two
alleles at a locus
differ, then one, the
dominant allele,
determines the
organism’s
appearance; the
other, the recessive
allele, has no
noticeable effect on
the organism’s
appearance
Mendel’s Model

35. 35
 Fourth, is the law of segregation which holds that
the two alleles for a heritable character segregate
(separate from each other) during gamete
formation and end up in different gametes
* thus, an egg or a sperm gets only one of
the two alleles that are present in the
somatic cells of the organism making the
gamete
* in terms of chromosomes, this
segregation corresponds to the distribution
of the two members of a pair of homologous
chromosomes to different gametes in meiosis
Mendel’s Model

36. 36

37. 37

38. 38
Mendel’s First Three Postulates
 1. UNIT FACTORS IN PAIRS
 Genetic characters are controlled by unit
factors existing in pairs in individual organisms.
 In the monohybrid cross involving tall and dwarf
stems, a specific unit factor exists for each
trait
 Each diploid individual receives one factor from
each parent.
 Because the factors occur in pairs, 3 possible
combinations are possible: 2 factors for tall
stems, or 1 of each factor
 Every individual possesses one of 3 combinations,
which determines stem height.

39. 39
 2. DOMINANCE/RECESSIVENESS
 When 2 unlike unit factors responsible for a
single character are present in a single
individual, one 1 unit factor is dominant to the
other, which is recessive.
 The trait not expressed is controlled by the
recessive unit factor.
 The terms dominant and recessive are also used
to designate traits.
 Tall stems are said to be dominant over
recessive dwarf stems.

40. 40
 3. SEGREGATION
 During the formation of gametes, the paired
units separate, or segregate, randomly so that
each gamete receives one or the other with
equal frequency.
 If an individual contains a pair of unit factors
(e.g., both specific for tall), then all the
gametes receive one of that same kind of unit
factor (tall).
 If an individual contains unlike unit factors
(e.g., one for tall and one for dwarf), then
each gamete has a 50% probability of receiving
either the tall or dwarf unit factor.

41. 41
 The postulates provide a suitable explanation
for the results of the monohybrid crosses
 He reasoned that P1 tall plants contained
identical paired unit factors, as did the P1
dwarf plants.
 The gametes of tall plants all receive 1 tall unit
factor (segregation) and so is the case with the
dwarf plants
P1 x
Gametes x
DD dd
D d

42. 42
 Following fertilization, all F1 plants receive one
unit factor from each parent (tall factor from
one and a dwarf factor from the other),
reestablishing the paired relationship.
 But because tall is dominant to dwarf, all F1
plants are tall
P1: x
Gametes x
F1:
DD dd
D d
Dd

43. 43
 When F1 plants form gametes, the postulate of
random segregation demands that each gamete
receives either the tall or the dwarf unit
factor.
 Following fertilization during F1 selfing, four F2
combinations will result with equal frequency:
1. 1 tall/tall
2. 1 tall/dwarf
3. 1 dwarf/tall
4. 1 dwarf/dwarf
Therefore, the F2 is predicted to have ¾ tall and
¼ dwarf, or a ratio of 3:1

44. 44
 4. INDEPENDENT ASSORTMENT
 Two or more genes assort independently, i.e.,
each pair of alleles segregates independently of
each other pair of alleles, during gamete
formation.
 Applies only to genes (allele pairs) located on
different chromosomes (non-homologous), or to
genes that are very far apart on the same
chromosome.

45. Sample Problems
1. Possible colors in cat are black and white, where white is
dominant. List the possible genotypes.
BB/Bb: White bb: black
2. List the possible genoytpes and phenotypes in rose color
flower where red color is dominant over white.
Genotype: WW/Wh: red ww: white
Phenotype: red and white
3. Classify the following as to homozygous and heterozygous.
a. bb: homo- c. Hh: hetero-
b. Rr: hetero- d. GG: homo-
4. Give the genotype for the following using T and t as the
alleles.
a. homozygous recessive: tt c. heterozygous:Tt
b. homozygous dominant:TT
45

46. 46
Types of Crosses

47. 47
 Test cross –
when an organism
of a known
dominant
phenotype but
unknown genotype
is crossed with a
homozygous
recessive
individual;
 to determine
what alleles are
present in the
genotype

48. 48
 Back cross - cross between F1 hybrid with
any of its parents or genetically similar to
its parent;
 to achieve offspring with a genetic identity
which is closer to that of the parent

49. 49
 Reciprocal cross – cross that involves the
reversal of the sex of the parents
 to test the role of parental sex on
inheritance pattern
 Out cross – cross when F1 progeny is
crossed with dominant parents;
 to increase genetic diversity, thus, reducing
the probability of an individual being subject
to disease or genetic abnormalities

50. 50
 Test cross:
P: _ _ x yy
F1: Yellow offsprings
YY x yy
All Yy
(All Yellow)
y y
Y Yy Yy
Y Yy Yy
Yy x yy
½ Yy
½ yy
y y
Y Yy Yy
y yy yy
YY
Yellow
Yy
Using pea plants, a plant with an unknown dominant
genotype yielded yellow offspring. Determine the
genotype of the unknown parent.

51. 51
 Back cross:
P: TT x tt
F1: Tt Tall
Tt x TT
½ TT All Tall
½ Tt
T T
T TT TT
t Tt Tt
Tt x Tt
¼ TT ¾ Tall
½ Tt
¼ tt ¼ short
T t
T TT Tt
t Tt tt

52. 52
A B
AA BB
A A B B
AB
A A B B
B B
A AB AB
A AB AB
B B
A AB AB
A AB AB
Reciprocal
Cross

53. 53
Monohybrid

54. 54
 Monohybrids - Individuals produced from a
cross of true-breeding parents
 meaning, they are heterozygotes for the
trait in question
 Monohybrid Cross - cross between
heterozygotes
 cross between two organisms involving a
single character

55. 55
Monohybrid Cross

56. 56
Monohybrid Cross
 Heterozygous dominant x heterozygous
dominant
 Genotypic ratio = 1 hd:2 het d:1recessive
 Phenotypic raio = 3 dominant phenotype: 1
recessive phenotype

57. 57
Punnett Square
 Each of the possible gametes is assigned a
column or a row:
 The vertical column represents those of the
female parent, and the horizontal represent
those of the male parent
 After assigning the gametes to the rows and
columns, the new generation is predicted by
entering the male and female gametic
information into each box thus, predicting every
possible resulting genotype
 By filling out the the Punnett square, all
possible random fertilization events are listed.

58. 58

59. 59
Punnett Square

60. Sample Problems
1. In dogs, wire hair (S) is dominant to smooth (s). In a cross of a
homozygous wire-haired dog with a smooth-haired dog, what would be
the phenotypes and genotypes of the offspring.
P: SS x ss
G: x
F1: ?
Genotypic ratio: All Ss
Phenotypic ratio: All wire-haired
60
S S
s Ss
Wire-haired
Ss
Wire-haired
s Ss
Wire-haired
Ss
Wire-haired
S s

61. Sample Problems
2. Wood rats are medium-sized rodents with lots of interesting
behaviors. Assuming that the trait of bringing home shiny objects (H) is
dominant to the trait of carrying home dull objects (h). Suppose 2
heterozygous individuals are crossed, what would be the genotypic and
phenotypic ratios?
Genotypic ratio: ¼ DD; 2/4 Dd; ¼ dd (1:2:1)
Phenotypic ratio: ¾ Shiny objects: ¼ Dull objects (3:1)
61
D d
D DD Dd
d Dd dd

62. Sample Problems
3. Saguaro cacti are very tall cylindrical plant with two arms, one on each
side. You have the same species at home where one arm is longer than the
other. Assume that arm length is controlled by a single gene with arms of
the same length (A) being dominant to arms of the different lengths (a),
a. what is the genotype of your cactus?
b. If your cactus fertilizes one that is heterozygous
for arms of the same length, give the genotype of the
offspring.
GR: ½ Vv: ½ vv
PR: ½ same length :
½ different length
62
V v
v Vv vv
v Vv vv

63. Sample Problems
4. Suppose that long tails (L) in a species of blackbird were dominant to
short tails (l). A female short-tailed blackbird mates with a long-tailed
individual which had one true-breeding parent with a long tail and one
true-breeding parent with a short tail.
a. what is the genotype of the male blackbird? Ss
b. what is the genotype of their offspring? Ss : ss (1:1)
c. what is the phenotype of their offspring? 1 long tail :
1 short tail
63
s s
S Ss Ss
s ss ss

64. Sample Problems
5. Tongue-rolling ability is dominant to not being able to roll
the tongue. Suppose that a heterozygous tongue-roller
marries and mates somebody who cannot role his tongue.
What will be the genotypic and phenotypic ratios of the
offspring?
GR- Nn : nn (1:1)
PR- 1 tongue roller : 1
non-tongue roller
64
n n
N Nn Nn
n nn nn

65. 65
Example:
6. In cats, black fur color is dominant. Two
heterozygous cats with black fur mate
together. Show the Punnett square.
a. what is the probability that they will
produce a cat with black fur?
b. what is the probability that the baby cat
will be homozygous?
c. calculate the phenotypic and genotypic
ratios

66. B b
B
b
66
Given: BB – black bb – brown
Bb – black
B b
B BB Bb
b Bb bb

67. 67
Example:
In cats, black fur color is dominant. Two
heterozygous cats with black fur mate
together. Show the Punnett square.
a. what is the probability that they will
produce a cat with black fur?
Black fur = ¾ = 75%

68. 68
Example:
In cats, black fur color is dominant. Two
heterozygous cats with black fur mate
together. Show the Punnett square.
b. what is the probability that the baby cat
will be homozygous?
Probability that baby is homozygous = 2/4
or 50%

69. 69
Example:
In cats, black fur color is dominant. Two
heterozygous cats with black fur mate
together. Show the Punnett square.
c. calculate the phenotypic and genotypic
ratios
Phenotypic ratio: 3 black fur : 1 brown fur
Genotypic ratio: 1:2:1 (1BB:2Bb:1bb)

70. 70
Example:
7. A homozygous wolf with blue eyes
mates with a heterozygous wolf with
brown eyes. Brown being dominant
to blue. Use a Punnet square.
a. what is the probability that they
will produce a wolf with blue eyes?
b. calculate the phenotypic and
genotypic ratios.

71. b b
B
b
71
Given: BB – brown bb – blue
Bb – brown
b b
B Bb Bb
b bb bb

72. 72
a. P = 2/4 or 50%
b. Phenotype ratio:
Brown : Blue
2 : 2
1 : 1
Genotype ratio:
Bb : bb
2 : 2
1 : 1

73. 73
P1 Monohybrid Cross Review
 Homozygous dominant x Homozygous
recessive
 Offspring all Heterozygous
(hybrids)
 Offspring called F1 generation
 Genotypic & Phenotypic ratio is ALL
ALIKE

74. 74
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

75. 75
Practice Your Crosses
Work the P1, F1, and both
F2 Crosses for each of the
other Seven Pea Plant
Traits

76. 76
Dihybrid

77. 77
 As a natural extension of the monohybrid cross,
Mendel also designed experiments in which he
examined two characters simultaneously
(dihybrid cross, or a two-factor cross).
Law of Independent Assortment

78. 78
 Using the result of his dihybrid experiments,
Mendel developed what is now the law of
independent assortment
* two or more genes assort independently, i. e.,
during gamete formation, each pair of alleles
segregates independently of each other pair
of alleles
 This law applies only to genes (allele pairs) located
on different chromosomes (e. i., on non-
homologous chromosomes), or, alternatively, to
genes that are very far apart on the same
chromosome
Law of Independent Assortment

79. 79
 Alleles for different traits are distributed to
sex cells (& offspring) independently of one
another.
* Genes are packaged into gametes in all
possible allelic combinations, as long as each
gamete has one allele for each gene.
 This law can be illustrated using dihybrid
crosses.
Law of Independent Assortment

80. 80
If the hybrids must
transmit their alleles in
the same
combinations in which
the alleles were
inherited from the P
generation, then the
F1 hybrids will
produce only two
classes of gametes:
YR and yr (dependent
assortment)

81. 81
If the two pairs of
alleles segregate
independently of
each other, (genes
are packaged into
gametes in all
possible allelic
combinations, as
long as each gamete
has one allele for
each gene, F1 plant
will produce four
classes of gametes in
equal quantities: YR,
Yr, yR, and yr
(independent
assortment)

82. 82

83. 83

84. 84

85. 85

86. ¼ BB 1/16 AABB yellow, round
¼ AA 2/4 Bb 2/16 AABb yellow, round
¼ bb 1/16AAbb yellow, wrinkled
86
¼ BB 2/16 AaBB yellow, round
2/4Aa 2/4 Bb 4/16 AaBb yellow, round
¼ bb 2/16Aabb yellow, wrinkled
¼ BB 1/16 aaBB green, round
¼ aa 2/4 Bb 2/16 aaBb green, round
¼ bb 1/16 aabb green, wrinkled
9:3:3:1
Forked Line
Method

87. 87

88. 88
Dihybrid Cross
Traits: Seed shape & Seed color
Alleles: W round
w wrinkled
G yellow
g green
WwGg x WwGg
WG Wg wG wg
All possible gamete combinations
FOIL
WG Wg wG wg

89. 89
Dihybrid Cross
WG Wg wG wg
WG
Wg
wG
wg

90. 90
Dihybrid Cross
Round/Yellow: 9
Round/green: 3
wrinkled/Yellow: 3
wrinkled/green: 1
9:3:3:1 phenotypic
ratio
WG Wg wG wg
WG
Wg
wG
wg
WWGG WWGg WwGG WwGg
WWGg WWgg WwGg Wwgg
WwGG WwGg wwGG wwGg
WwGg Wwgg wwGg wwgg

91. 91
Question:
How many gametes will be produced
for the following allele arrangements?
Note: 2n (n = # of heterozygotes)
1. RrYy
2. AaBbCCDd
3. MmNnOoPPQQRrssTtQq

92. 92
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

93. 93

94. 94
Dihybrid Cross
Purple/Round: 9
Round/wrinkled: 3
white/Round: 3
white/wrinkled: 1
9:3:3:1 phenotypic
ratio
PI Pi pI pi
PI
Pi
pI
pi
PPRR PPRi PpRR PpRr
PPRr PPrr PpRr Pprr
PpRR PpRr ppRR ppRr
PpRr Pprr ppRr pprr

95. Exercises
1. Flower color & Pod shape
2. Stem length & Flower position
3. Seed color & Pod color
4. Pod shape & Stem length
5. Flower position & Seed color
6. Seed shape & Pod shape
7. Pod shape & Pod color
8. Flower color & Stem length
9. Flower position & Pod color
10. Seed shape & Flower color
95

96. 96
Summary of Mendel’s laws
LAW
PARENT
CROSS
OFFSPRING
DOMINANCE DD x dd
tall x short
100%
tall
SEGREGATION
Dd x Dd
tall x tall
75% tall
25% short
INDEPENDENT
ASSORTMENT
WwGg x WwGg
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
pods

97. Summary of Mendel’s Hypothesis
1. Genes can have alternate versions called
alleles.
2. Each offspring inherits two alleles, one from
each parent.
3. If the two alleles differ, the dominant allele is
expressed. The recessive allele remains
masked unless the dominant allele is absent.
4. The two alleles for each trait separate during
gamete formation. This now called: Mendel's
Law of Segregation
97

98. 98
◊ The elegance of mendel’s experiments
was partly due to the complete
consistency between his observation and
hypotheses he developed.
◊ However, after Mendel’s work was
rediscovered, it became clear that
simple Medelian model did not
adequately predict experimental
observations in all situations.

99. 99
Dihybrid Cross
TtRr X TtRr
Each parent can produce 4 types of
gametes.
TR, Tr, tR, tr
Cross is a 4 X 4 with 16 possible
offspring.

100. 100
RESULTS
9 Tall, Red flower
3 Tall, white flower
3 short, Red flower
1 short, white flower
Or: 9:3:3:1