Mendelian Patterns of
Inheritance
Chapter 11
BIOL1000
Dr. Mohamad H. Termos
Gregor Mendel
• Austrian monk
 Studied science and math
 Breeding experiments with the garden
pea
 Mathematical data from his
experiments
• Formulated fundamental laws of
heredity in early 1860s
2
One-Trait Inheritance
Cross-breeding experiments
 Used “true-breeding” (homozygous) plants
 Chose pea plants that differed in only one trait (monohybrid cross)
• Parents = P
• First offspring = F1
• Second offspring (result of crossing F1) = F2
Results:
• P= Green pods X Yellow pods (true breeding)
• F1= all green pods
•
Green pod X Green pod
• F2= 3 Green pods : 1 yellow pod
3
Law of Segregation
• Each individual has a pair of
factors (alleles) for each trait
• The factors (alleles) segregate
(separate) during gamete
(sperm & egg) formation
• Each gamete contains only one
factor (allele) from each pair
• Fertilization gives the offspring
two factors for each trait
4
Modern Genetics View and Homologous
Chromosomes
• Each trait in a pea plant is controlled by two alleles
• Dominant allele (capital letter) masks the expression of the recessive allele (lowercase)
• Alleles occur on a homologous pair of chromosomes at a particular gene locus
 Homozygous = identical alleles (TT or tt)
 Heterozygous = different alleles (Tt)
5
Modern Genetics View
• Allele from green pods (G) is dominant to yellow pod allele (g)
• Green pods = ________ or ________
• Yellow pods = _____________
6
Results
• P= Green pods (GG) X Yellow pods (gg)
• F1= all green pods (Gg)
•
Green pod (Gg) X Green pod (Gg)
• F2= 3 Green pods(Gg or GG):1 yellow pod (gg)
7
Mendel’s Monohybrid Crosses:
8
Genotype Versus Phenotype
Genotype
Phenotype
Refers to the two alleles for a
trait
Physical appearance of the
trait
Homozygous
Dominant or
recessive
Heterozygous
9
GG
Homozygous dom.
Green pods
Gg
Heterozygous
Green pods
gg
Homozygous rec.
Yellow pods
Practice question
• In rabbits B= black fur (dominant) and b= white fur (recessive)
• What is the genotype of a rabbit with white fur?
• Black fur?
10
Punnett Square
• Table listing all possible genotypes resulting from a cross
 Sperm genotypes are lined up on one side
 Egg genotypes are lined up on the other side
 Possible zygote genotypes are placed within the squares
11
Monohybrid Cross done by Mendel
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Allele Key
T = tall plant
t = short plant
Phenotypic Ratio
3
1
tall
short
P generation
TT
tt
12
Monohybrid Cross done by Mendel
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Allele Key
T = tall plant
t = short plant
Phenotypic Ratio
3
tall
1
short
P generation
TT
P gametes
T
tt
t
13
Monohybrid Cross done by Mendel
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Allele Key
T = tall plant
t = short plant
Phenotypic Ratio
3
tall
1
short
P generation
TT
P gametes
tt
T
tt
t
F1 generation
Tt
14
Monohybrid Cross done by Mendel
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
T = tall plant
t = short plant
Phenotypic Ratio
3
1
P generation
TT
tall
short
P gametes
tt
t
T
F1 generation
Tt
eggs
F1 gametes
T
t
T
sperm
Allele Key
t
15
Monohybrid Cross done by Mendel
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
P generation
T = tall plant
t = short plant
Phenotypic Ratio
3
1
TT
tt
tall
short
P gametes
t
T
F1 generation
Tt
eggs
F1 gametes
T
t
T
sperm
Allele Key
TT
t
16
Monohybrid Cross done by Mendel
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
F1 generation
T = tall plant
t = short plant
Phenotypic Ratio
3
1
TT
tt
tall
short
t
T
P gametes
F1 generation
Tt
eggs
F1 gametes
T
t
T
sperm
Allele key
TT
Tt
t
17
Monohybrid Cross done by Mendel
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
P generation
T = tall plant
t = short plant
Phenotypic Ratio
3
tall
1
short
TT
P gametes
tt
t
T
F1 generation
Tt
eggs
T
F1 gametes
t
T
TT
sperm
Allele Key
Tt
t
Tt
18
• A punnett square for the true breeding green pods (GG) with yellow
pods (gg)
19
Punnett Square
G
G
G is the allele for green pods
g is the allele for yellow pods
g
Gg
Gg
P= GG X gg (genotypes)
g
Gg
Gg
F1= 4/4 Gg (genotypes)
Phenotypic ratios
4/4 Green pods
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G
g
- F1= Gg X Gg (genotypes)
GG
Gg
- F2= ¼ GG
½ Gg
¼ gg
Gg
gg
G
g
- Phenotypic ratios:
¾ Green pods
¼ yellow pods
21
22
Practice question
• Draw a punnett square for a cross between a homozygous recessive
plant and a heterozygous plant. The dominant allele (F) is for purple
flowers and the recessive allele (f) is for white flowers. What are the
genotypes of the offspring? Phenotypes?
23
Practice question - Answer
• Ff X ff
f
F
f
Ff
ff
Ff
ff
f
24
Practice question - Answer
• Ff X ff
Genotypic ratios?
f
F
f
Ff
ff
50% ff
50% Ff
Phenotypic ratios
f
Ff
25
ff
50% white flowers
50% purple flowers
Dominant Vs Recessive
• Recessive phenotype always = homozygous recessive genotype
• However, dominant phenotype = indeterminate genotype
 May be homozygous dominant, or
 Heterozygous
26
Test Cross
• Test cross determines genotype of individual having dominant
phenotype
• Does a tall plant have genotype TT or Tt?
• Cross with a short plant (tt) and look at their offspring.
27
Practice question - Answer
T
After crossing tall plant with
a short the result is:
t
t
Tt
Tt
-50% short
-50% tall
tt
tt
What is genotype of tall
parent?
t
Tt
28
Practice question - Answer
T
After crossing tall plant with
a short the result is:
t
t
Tt
Tt
-100% tall
Tt
Tt
What is genotype of tall
parent?
T
TT
29
Practice question
• A pea plant that has round seeds (R is dominant)is crossed with a pea
plant that has wrinkled seeds (r is recessive). The cross produces
both round and wrinkled seeds. What must be the genotype of the
pea plant with the round seeds? What are the genotypic and
phenotypic ratios of the offspring?
30
Practice Question - Answer
• Plant with round seeds is either RR or Rr
• Crossed with wrinkled seeds – rr
r
r
R
Rr
Rr
R
Rr
Rr
31
r
r
R
Rr
Rr
r
rr
rr
Practice Question - Answer
• Parent must have Rr genotype
• Offspring are ½ Rr Round seeds
½ rr wrinkled seeds
32
• If 300 offspring were produced from the previous cross, how many of
those will have round seeds?
33
Answer
• Percentage X total offspring
• ½ or 50% X 300= 150 offspring will have round seeds
34
Two-Trait Inheritance
• Dihybrid cross uses true-breeding (homozygous) plants differing in
two traits (TTGG X ttgg)
35
Two-Trait Inheritance
• TTGGxttgg
• Possible Gametes:
• TG and tg
36
tg
TG
TG
TtGg
TtGg
TtGg
TtGg
tg
37
All offspring are tall with
green pods
• TtGg X TtGg
• Possible gametes:
• TG, Tg, tG, tg
• Page 194 in book
38
Two-Trait (Dihybrid) Cross
39
Law of Independent Assortment
• Alleles for one trait segregate independently of other alleles
• All possible combinations of alleles can occur in the gametes
40
Incomplete Dominance
• Heterozygote has phenotype between that of either homozygote
 Homozygous red has red phenotype (R1)
 Homozygous white has white phenotype (R2)
 Heterozygote has pink (intermediate) phenotype (R1 R2)
• Phenotype reveals genotype without test cross
41
Incomplete Dominance
42
Practice question
• If you crossed a pink flowering plant with a red flowering plant what
are the genotypic and phenotypic ratios of the offspring?
43
Answer
• Pink genotype is R1R2
• Red genotype is R1R1
• Genotypic ratio: 2/4 R1R1 2/4 R1R2
• Phenotypic ratio: 2/4 red
2/4 pink
44
R1
R2
R1
R1R1
R1R2
R1
R1R1
R1R2
Multiple Allelic Traits
• Some traits controlled by multiple alleles
• The gene exists in several allelic forms (but each individual only has two)
• ABO blood types
45
ABO Alleles
• The alleles:
 IA
 IB
i
Phenotype
(Blood Type)
A (actually AA or AO)
B (actually BB or BO)
AB
O (actually OO)
46
Genotype
IAIA or IAi
IBIB or IBi
IAIB
ii
Inheritance of Blood Type
47
Practice question
• If a mother is blood type A and a father is blood type B is it possible
for them to have a baby with O blood type?
48
Answer
• Possible genotypes:
father: IA IA or IA i
mother: IB IB or IB i
• If they are IAi X IBi then ¼ offspring will be ii
49
IA
i
IB
IA IB
IB i
i
IA i
ii
Practice question
• A mother has blood type A (both parents are AB) and a father has
parents that both have O blood type. They have a baby with O blood
type- did they bring home the wrong baby, is the father really the
father of the child, or is this their child?
50
Answer
• Mother has A Blood type and both her parents have AB blood type
A
B
A
AA
AB
B
AB
BB
51
Mother must have AA genotype
if her blood type is A
Answer
• Father has both parents with O blood type
i
i
i
ii
ii
i
ii
ii
52
All of their children must be O
blood type so father has O blood
type with genotype ii
Answer
• Mother has genotype AA and father has ii
A baby with O blood type is not
their baby- it is not possible
A
A
i
Ai
Ai
i
Ai
Ai
53
Wrong baby? Or father not
really the father?
Is it possible for the mother to
ever have offspring with O blood
type?
Polygenic Inheritance
• Occurs when a trait is governed by two or more genes having
different alleles
• Result in continuous variation of phenotypes
• height, skin color, eye color
54
Polygenic Inheritance
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P generation
F1 generation
F2 generation
Proportion of Population
20
—
64
15
—
64
6
—
64
1
—
64
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Genotype Examples
• How many pairs of chromosomes in humans?
• Total # chromosomes?
-22 pairs are autosomes
-23rd pair are the sex chromosomes (X and Y)
XY=male
XX=female
• Some traits are located on the X chromosome
56
X-linked
• Color blindness
• The alleles for the red- and green-sensitive pigments are on the X
chromosome.
• Muscular dystrophy
• Wasting away of the muscle due to missing protein
• Hemophilia
• Absence or minimal presence of clotting factor VIII or clotting factor IX
• Affected person’s blood either does not clot or clots very slowly.
57
58
X linked genes
• Eye color in flies - red (R) or white (r)
• Males can be: XRY or XrY
• trait determined by x chromosome (1 allele)
• Females can be XRXR or XRXr or XrXr
• trait determined by 2 alleles
59
• White eyed female X red eyed male
• Mother homozygous recessive
• Father has dominant allele
• Offspring
•
females: 100% heterozygous
(red eyes)
•
males: 100% recessive
(white eyes)
60
Xr
Xr
XR
XRXr
XRXr
Y
X rY
X rY
Genetic disorders
• Many disorders are genetic and can be passed from
parents to offspring
• Can be dominant traits or recessive
• Can also be X linked
61
Autosomal Recessive Disorders
• Methemoglobinemia
• Relatively harmless disorder
• Accumulation of methemoglobin in the blood
causes skin to appear bluish-purple
• Cystic Fibrosis
• Mucus in bronchial tubes and pancreatic ducts is
particularly thick and viscous
62
Methemoglobinemia
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Courtesy Division of Medical Toxicology, University of Virginia
63
Cystic Fibrosis
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
H2O
Cl-
Cl-
ClCl-
H2O
H2O
Cl-
nebulizer
defective
channel
percussion
vest
thick mucus
© Pat Pendarvis
64
Autosomal Dominant Disorders
• Osteogenesis Imperfecta
• Gene mutation  Weakened, brittle bones
• Hereditary Spherocytosis
• Gene mutation  Red blood cells become
spherical, fragile, and burst easily
65
66
Pedigree
• Shows the pattern of inheritance
• Squares are male
• Circles are female
• Shaded shapes have the trait
Autosomal Dominant Pedigree Chart
67
Autosomal Recessive Pedigree Chart
68