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Genetics
Chapter 8
Origins of Genetics
Heredity
The passing of characters from
parents to offspring
Character
Inherited characteristic (ex.
flower color)
Trait
 Single form of a characteristic
(ex. purple flower)
Genes
Control the expression of traits
(i.e. flower color)
Have two parts - alleles
One allele is from Mom other from
Dad.
Alleles
 Dominant - the expressed allele
of the gene
Capital letter
Ex. P
 Recessive - the allele that was not
expressed
Lower case letter
Ex. p
 Homozygous (Homo = same)
Two alleles are the same
Ex. PP or pp
 Heterozygous (Hetero = different)
The alleles are different
Ex. Pp
Modern Terms
 Genotype
The set of alleles for a Gene
 Ex. PP, Pp or pp
 Phenotype
 Physical appearance
 Ex. Purple or White Flowers
The dominant allele determines
phenotype
 Ex. Pp  Purple Flower or PP  Purple
Flower
If both alleles are recessive then the
individual will express the recessive
phenotype.
 Ex. pp  White Flower
Homozygous- genes are the
same HH hh
Heterozygous- genes are
different Hh
Probability
 The likelihood that a specific event will
occur
Expressed in multiples ways:
 Fractions
 Percentages
 Decimals
 Ratios
 Probability =
½
50%
.5
3:1 or 3 to 1
# of one kind of possible outcome
Total # of all possible outcomes
Monohybrid Crosses
Cross 1 character – Seed Color
Yellow Seeds are dominant – Y
Green Seeds are recessive – y
Cross a Homozygous Dominant
Yellow with a Green.
Cross:
 Genotype Ratio:
4Yy
 Phenotype Ratio:
4 Yellow
 Probabilities:
Being Yellow:
4 out of 4
Being Green:
0 out of 4
TT X
Y
tt
Y
y
Y y
Yy
y
Y y
Y y
Monohybrid Crosses
Cross 1 character – Flower Color
Purple Flowers are dominant – P
White Flowers are recessive – p
Cross a Heterozygous with a
homozygous recessive
 Genotype Ratio:
2 Pp; 2 pp
 Phenotype Ratio:
2 Purple
2 White
 Probabilities:
Cross:
Pp X pp
P
p
p
P p
p p
p
P p
p p
Being Purple:
2 out of 4
Being White:
2 out of 4
Incomplete Dominance
Inheritance that is intermediate
between two parents
Ex. Snapdragon with red flowers are
crossed with a snapdragon with white
flowers  a snap dragon with pink
flowers is produced.
Neither the allele for red flowers nor the
allele for white flowers is dominant.
Incomplete Dominance









Genotype for red flowers:
Genotype for white flowers:
Genotype for pink flowers:
Cross a red with a white.
Genotype Ratio:
4 RW
Phenotype Ratio: Cross: RR X
4 Pink Flowers
Probabilities:
 Red Flower:
 0 out of 4
 White Flower
 0 out of 4
 Pink Flower:
 4 out of 4
R
RR
WW,
RW.
WW
R
W
R W
RW
W
R W
R W
Incomplete Dominance









Genotype for red flowers:
Genotype for white flowers:
Genotype for pink flowers:
Cross a pink with a white.
Genotype Ratio:
2 RW; 2WW
Phenotype Ratio: Cross: RW X
2 Pink; 2 White
Probabilities:
 Red Flower:
 0 out of 4
 White Flower
 2 out of 4
 Pink Flower:
 2 out of 4
R
RR
WW,
RW.
WW
W
W
R W
WW
W
R W
WW
Codominance
Case in which both forms of the
character are displayed.
Both traits are displayed
Ex. The ABO blood groups
Neither IA nor IB are dominant is
dominant over the other.
Both are dominant over i.
When both IA and IB are present
they are codominant and the
individual is type AB
Blood Typing
Blood Types
A
IA
IB
i
AB
B
O
IA
IB
i
IAIA
IAIB
IAi
IAIB
IBIB
IBi
IAi
IBi
ii
Blood Typing
Cross a homozygous type A with a type O
 Homozygous type A:
IAIA
 Type O:
ii
 Genotype Ratio:
4 IAi
 Phenotype Ratio:
AIA X ii
I
4 Type A
 Probabilities:
IA
IA
Type A: 4 out of 4
Type O: 0 out of 4
i
i
IAi
IAi
IAi
IAi
Blood Typing
Cross a type AB with a type O
 Homozygous type A:
I AI B
 Type O:
ii
 Genotype Ratio:
2 IAi; 2 IBi
IAIB X ii
 Phenotype Ratio:
2 Type A
2 Type B
 Probabilities:
Type A:
Type B:
2 out of 4
2 out of 4
i
i
IA
IB
IAi
IBi
IAi
IBi
Sex-Linked Traits
Body cell has 46 chromosomes 44 autosomes, 2 sex chromosomes.
Gamete (egg/sperm) has 23
chromosomes –
22 autosomes, 1 sex chromosome
Sex-Linked Gene
A gene located only on the X or Y
Most are carried on the X
chromosome
Males have only one X chromosome
Male who carries a recessive allele
on the X chromosome will exhibit the
sex-linked condition
Hemophilia
(Heme: blood, philia: liking
of)
Condition that impairs the
blood’s ability to clot
A sex-linked trait
Carriers and victims contain
the recessive allele on their
X-chromosomes
Only females can be carriers
 XHXh
Because males only have
one X chromosome
Males CAN NOT be carriers
only victims.
Sex Linked Traits
The normal gene:
XH
The hemophilia gene:
Xh
 Cross a normal male with a carrier
female
Genotype Ratio:
1 XHXh: 1 XHXH: 1 XhY: 1XHY
Phenotype Ratio:
XHY X
XHXh
1 norm. female:
1 carrier female:
h
H
X
X
1 norm. male:
1 hemophilic male
Probabilities
XH
XHXh
XHXH
Normal Female:
1 out of 2
Carrier Female:
1 out of 2
XhY
XHY
Normal Male:
1 out of 2
Y
Hemophilic male:
1 out of 2
Sex Linked Traits
The normal gene:
XH
The hemophilia gene:
Xh
 Cross a hemophilic male with a carrier
female
Genotype Ratio:
1 XHXh: 1 XhXh: 1 XHY: 1XhY
Phenotype Ratio:
XhY X
XHXh
1 carrier female:
1 hemophilic female:
H
h
X
X
1 norm. male:
1 hemophilic male
Probabilities
Xh
XHXh
XhXh
Hemophilic Female:
1 out of 2
Carrier Female:
1 out of 2
XHY
XhY
Normal Male:
1 out of 2
Y
Hemophilic male:
1 out of 2
Genetic Disorders
Sickle Cell Anemia
Recessive disorder among
African Americans
Caused by a mutated allele that
produces a defective hemoglobin
protein.
Hemoglobin in RBC’s bind and
transport oxygen through the body
Genetic Disorders
Cystic Fibrosis
Fatal hereditary recessive
disorder among Caucasians.
Clogging of the airways and lungs
with thick mucus and the blockage
of ducts of the liver and pancreas.
There is no known cure.
Genetic Disorders
Albinism
Recessive
Body is unable to produce an
enzyme necessary to make
melanin.
Melanin is a pigment that gives
color to hair, skin and eyes
Pedigrees
A family history that shows
how a trait is inherited over
several generations
Helps to track down the
carriers (heterozygotes) of
recessive disorders.
Reading Pedigrees
A shaded box –
affected male …………
A shaded circle-
affected female……….
A clear box
-
normal male…………..
A clear circle
-
normal female………...
Pedigree Practice
A man and woman marry. They have five children, 2 girls and
3 boys. The mother is a carrier of hemophilia, an X-linked
disorder. She passes the gene on to two of the boys who died
in childhood and one of the daughters is also a carrier. Both
daughters marry men without hemophilia and have 3 children
(2 boys and a girl). The carrier daughter has one son with
hemophilia. One of the non-carrier daughter’s sons marries a
woman who is a carrier and they have twin daughters. What is
the percent chance that each daughter will also be a carrier?
I
II
III
IV
Legend:
Hemophilic Male: ………
Hemophilic Female: …...
Normal Female: ………
Normal Male: ………
?
?
Pedigree Practice
The non-carrier normal daughter in the 2nd
generation must have a genotype of XHXH her
normal husband must have the genotype XHY;
therefore, it is not possible for any of their
children to have hemophilia or be a carrier of
the trait.
XHXh
XHY XHXh
XhY
XhY
XHY
XHY
XHXh
XhY
XHY
XHXH
Probability of XHXh?
XHXH
XHY
XHY
XHY
XHXH
Probability of XHXh?
Pedigree Practice
However, when their normal son
(genotype XHY) marries a woman who
is a carrier of the trait (genotype
XHXh), then the probability that each of
his twins will be a carrier is 1 out of 2
(50%).
XH
Xh
XH XHXH XHXh
XHXh
XHY XHXh
XhY
XhY
XhY
XHXh
XhY
XHY
XHXH
Probability of XHXh?
XHY
XHY
XhY
Y
XHXH
XHY
XHY
XHY
XHXH
Probability of XHXh?