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Chapter 6&7:
Mendel and
Heredity
GREGOR MENDEL
•Austrian Monk
•First to accurately
predict patterns of
heredity - 1866
•Pea Plants
•These patterns
became known as
Genetics
•Known as “Father of
Genetics”
Why Pea Plants
 Have many traits that have only 2
different and distinct forms
 Mating is easy
 Both reproductive parts are
enclosed in the same flower
 Small, Easy to grow, Matures quick,
produces many offspring
Mendel’s Experiment
Mendel look at & compared 7 contrasting
traits of pea plants.
1. Flower Color
2. Seed Color
3. Seed Shape
4. Pod Color
5. Pod Shape
6. Flower Position
7. Plant Height
Pollination types
 Pollen inside the
stamen is
transferred to the
pistil
 Self pollination –
1 plants
 Cross pollination
 2 plants
Experiment Continued


1st experiment – Monohybrid cross

Crosses one pair of contrasting traits
 Green plant crossed with a Yellow plant
 Round peas crossed with shriveled peas
Experiment was conducted in 3 steps
1. Self Pollination (fertilized itself)
 Created true-breeding plants
o Example: All Green, or All Yellow
 P (Parental) generation: fist two individuals that
are crossed in an experiment
2.
Cross-Pollination of two contrasting P
generation plants
• Transfer of pollen from one plant to
another
• F1 generation (first filial)
3.
F1 generation self-pollinated
• Fertilized itself
• F2 generation (second filial)
With every
contrasting trait he
crossed, he
observed an end
result that always
had a 3:1 ratio
RESULTS
Mendel’s Hypotheses
1.
For each trait in humans, an individual has
two genes


2.
There are alternative versions of genes

3.
One from mom
One from dad
Alleles
When two different alleles are together one
will show up, the other may not

Dominant: expressed trait


Written with a capital letter
Recessive: not expressed trait


Only expressed if inherits two recessive alleles
Written with 2 lower case letters
Terms
 Genotype: The set of alleles that an individual
has.
 Homozygous: Two alleles that are the same
 A plant with 2 purple flowers

PP
 A plant with two white flowers

pp
 Heterozygous: Two alleles that are different
 A plant can have a Pp genotype, but still be Purple
 WHY?
 Phenotype: The physical appearance
 All Purple flowers, All white flowers, Some purple and
some white, etc.
Practice Stuff
1.
For each genotype below, state if it is
Heterozygous or Homozygous recessive or
dominant
1.
2.
3.
4.
5.
6.
HH
Hh
Aa
AA
cc
dd
More Practice
 Round pea plants are dominant over wrinkled pea
plants
 What is the genotype of a wrinkled pea plant?
 rr
 What is the phenotype of a plant that is RR
 Round
 What is the phenotype of a plant that is Rr
 Round
& More Practice
 Tall pea plants are dominant over short pea
plants
 What would be the genotype of a short plant?
 tt
 What would be the genotype of a Homozygous
Dominant pea plant?
 TT
 What would be the genotype of a a Heterozygous pea
plant?
 Tt
 What would be phenotype of a tt plant?
 short
Laws of Heredity
 Law of Segregation
 2 alleles for a trait segregate when gametes are formed
 Law of Independent assortment
 Alleles of different genes separate independently of
one another
Predicting
 Punnett square
 Diagram that predicts the expected outcome of a
genetic cross
 Crosses that involve one trait:
 Monohybrid cross
 How to Solve a Punnett Square
 1. Determine the genotypes (letters) of the parents. 2.
Set up the punnett square with one parent on each
side.
3. Fill out the Punnett square middle
4. Analyze the number of offspring of each type
EXAMPLE:
Cross a Homozygous Dominant Round pea plant with a
Heterozygous round pea plant.
Test Cross: RR x Rr
R
r
Genotypes:
•50%: RR
R
R
RR
RR
Rr
•50%: Rr
Rr
Phenotypes:
•100% will be
Round
Dihybrid Cross
 A cross that involves 2 pairs of contrasting traits.
 Example: Predict the results of a cross between two
pea plants that are heterozygous for seed shape
(R=round, and r= wrinkled) and seed color
(Y=yellow, y=green)
 RrYy x RrYy
 1st – determine the possible gametes the two parents
(RrYy) could make
 RY, ry, Ry, rY
 2nd – plug these gametes into the punnett square
Punnett Square
RY
Ry
ry
rY
RRYY
RRYy
RrYy
RrYY
RRYy
RRyy
Rryy
RrYy
RrYy
Rryy
rryy
rrYy
RrYY
RrYy
rrYy
rrYY
RY
Ry
ry
rY
Results
 Genotypes:
 1: RRYY, RRyy, rrYY, & rryy
 2: RRYy, Rryy, RrYY, rrYy
 4: RrYy
 Phenotypes:
 9/16: Round & Yellow
 3/16: Round & Green
 3/16: Wrinkled & Yellow
 1/16: Wrinkled & Green
 9:3:3:1 Ratio always when crossing two
Heterozygous plants!
Problem:
 Being right handed (R) is dominant over being left
handed. Also, Having Freckles (F) is dominant over not
having freckles.
 John and Tonya are getting married, and want to know
the children’s possibilities for displaying these traits.
 Tonya is left handed, and does not have freckles
 John is Homozygous Dominant Right Handed, and has
freckles, but his father did not have freckles.
 What are the chances their children will have
freckles?
 What are the chances their children will be left
handed and have freckles?
Probability
 The likelihood, or chance that an event will occur
 Expressed in
 Fractions (1/4), Ratios (1:4), or
Percents (25%)
P = # of one kind of possible outcome
total # of all possible outcomes
Determining unknown genotypes:
 Test Cross: Homozygous recessive individual is
crossed with a dominant phenotype to find out if it
is Ho or He
 Example: A tall pea plant and a short pea plant
were crossed, find out the genotype of the tall plant
if all the offspring are tall.
Results
 Cross: T___ x tt
t
t
Tt
Tt
50% tall, &
50% short
T
t
t
t
Tt
Tt
Tt
Tt
T
tt
tt
100% Tall
T
The Genotype of the yellow plant is TT, because
this is the only cross that will produce all Tall
offspring
Practice
 If a couple has half freckled children and half not freckled
children. And we know that the father is heterozygous, and
has freckles, what is the genotype for the mother?
 Answer = ff
Pedigrees
 A family history that shows how a trait is inherited
over several generations
 Very helpful in detecting/ determining genetic
disorders
= Male
= Female
Example:
Rules in Pedigree
 Males = squares
 Females = circles
 Horizontal line = marriage line
 Vertical line = children
 Listed oldest to youngest
 Numbered by generation with roman numerals
 Numbered within each generation
Tracing Albinism in a Pedigree
Sex-linked traits
 Located on the X chromosome
 Most are recessive
 Mostly seen in males
 Because males have only one X chromosomes
 Females
 Can be carriers – meaning they have an X
chromosome with the trait, but the other does not
 They will only exhibit the trait if they receive two
recessive alleles
Sex Linked Traits
•Baldness
•Hemophilia
•Colorblindness
Example
 Hemophilia is a sex linked trait, located on the X
chromosome. A female can carry one allele for this
trait and be a carrier, if she carries two she has the
disorder. Males on the other hand, if they carry one
allele with Hemophilia, they have the disorder.
 Cross a Carrier female, with a normal male, and
show the resulting genotypes and phenotypes.
Polygenic Traits
 When several genes influence a trait
 Eye color
 Hair Color
 Height
 Weight
 Skin color
Incomplete Dominance
 A trait that is in between the two
parents alleles
 Example: Red & White Snapdragons
 RR = Red
 rr = white
 Rr = Pink
Codominance
 Occurs when both forms of the traits are displayed
 Examples:
 4’o’clock plants
 Roan Horse coat
Multiple Alleles
 Genes with three or more alleles
 Example:
 Blood Types
A
B
 Alleles: I , I , i

IA & IB are both dominant
o i is recessive
o Neither IA or IB are dominant over each other
o Blood type is controlled by 3 alleles, but can
only express two of the genes
What Alleles Make What Blood Type?
A A
 I I = Type A
A
 I i = Type A
B B
 I I = Type B
B
 I i = Type B
A B
 I I = Type AB
 ii = Type O
More on Blood Types
 Type “O” is known as the “Universal Donor”
 It contains no carbohydrate antigens
 Type AB – “Universal Acceptor”
 Has both A and B antigens, and can accept from O.
Blood Type Problem
 Tracey has blood Type A, and her father, George had
blood type O.
 Pacey, Tracey’s husband has blood type O.
 They are about to have a child, figure out their
offspring’s possible blood types.
RESULTS
IA
i
i
IA i
ii
Phenotypes
50% - IAi = Type A
i
IA i
ii
50% - ii = Type O
Influenced Traits
 Traits Influenced By the Environment:
 Humans
 Weight: nutrition
 Skin Color: sun exposure
 Personality: outside environment
 Animals
 Artic fox – fur color (temperature)
 Types of Turtles
 Sex determined by temperature hatched at
Traits caused by mutations:
 Genetic Disorders





Sickle Cell Anemia
Hemophilia
Cystic Fibrosis
Tay-Sachs Disease
Huntington’s Disease
 Dominant
 Treatments
 Genetic Counseling
 Gene Therapy
Recessive
THE END OF CHAPTER 8!
Applause!