LESSON # 8
If my father is blond, why am I not?
MENDELIAN GENETICS
Inheritance
Gregor Mendel
 Father of modern genetics
 Experimented with pea plants (Pisum sativum) 1856-1863
 Was a monk in the Czech Republic
 His work passed unrecognized until his paper was discovered
in 1900 “Experiments in plant hybridization.” (1865)
 This paper showed that each organism has physical traits that
correspond to invisible elements (genes) within the cell.
 These invisible elements, which we now call genes, exist in
pairs (alleles).
 Mendel showed that only one member of this genetic pair is
passed on to each progeny.
 Mendel's laws form the theoretical basis of our understanding
of the genetics of inheritance.
What he did right
 Picked the pea plant which is easy self or cross pollinate and grows fast.
 Chose to study traits with only two possible phenotypes :
 Pea texture: round or wrinkled seed phenotype
 Pea color: yellow or green seed phenotype
 Flower color: red or white flower phenotype
 Pea plant size: tall or dwarf plant phenotype
 Pea pod color: green or yellow
Genetic studies innovations
 Mendel made two innovations to the science of genetics:
 developed pure lines
 Pure Line - a population that breeds true for a particular trait. All individuals of
the population have the same phenotype for a specific trait they are pure for.
When self-crossing or crossing with individuals of the same pure line population,
the offspring shows the same phenotype.
 Counted his results and kept statistical notes
Law of Dominance or Uniformity of
Hybrids.
 When two pure line individuals of different phenotypes are crossed, all the
resulting offspring show one unique phenotype (the dominant phenotype).
 All the offspring that result from interbreeding pure line individuals of
different phenotypes, have the same phenotype and the same genotype.
All tall
First Filial (F1) generation
P: Parental generation
Representing the cross
 The trait that is maintained in the F1
generation is considered the
Dominant trait.
 The trait the disappears is
considered the Recessive trait.
 The first letter of the dominant
phenotype is used to indicate the
alleles.
 Capital letter for the dominant allele
 Lower case for the recessive allele
P: Parental generation
First Filial (F1) generation
How to begin
Gene studied: Pea stem
Possible alleles:
Possible genotypes and corresponding phenotypes:
Tall (T)
Short (t)
Tall > Short
Genotype TT Tt tt
Phenotype Tall stem Tall stem Short stem
P: Purebred Tall stem x Purebred Short stem
TT tt
Gametes T T t t
p(T) = 1 p(t) = 1
Punnet square:
Gametes
T
T
t t
Tt Tt
Tt Tt
Gametes
F1
Genotype 100% Tt
Phenotype 100% Tall stem
F1 Genotypes F1 Phenotypes
p (Tt) = 1 100 % Tall stem
Solved Exercise 1
A homozygous brown eyed woman and a man with blue eyes want to have
children. What is the probability of the child having blue eyes?
Gene studied: Eye color
Possible alleles:
Possible genotypes and corresponding phenotypes:
Blue (b)
Brown (B)
Brown > Blue
Genotype BB Bb bb
Phenotype Brown eyes Brown eyes Blue eyes
P: Purebred Brown eyes x Purebred Blue eyes
BB bb
Punnet square:
Gametes
B
B
b b
Bb Bb
Bb Bb
Gametes
Answer: The possibility of having a child with blue eyes is 0.
F1 Genotypes F1 Phenotypes
p (Bb) = 1 = 100% 100 % Brown eyes
Gametes B B b b
p(B) = 1 p(b) = 1
Exercise 2
What is the probability of having of getting a pea plant with wrinkled peas if we
cross two hybrids?
Gene studied: Pea texture
Possible alleles:
Possible genotypes and corresponding phenotypes:
Smooth ___
Wrinkled ___
Smooth > Wrinkled
Genotype
Phenotype
Exercise 2
What is the probability of having of getting a pea plant with wrinkled peas if we
cross two hybrids?
Gene studied: Pea texture
Possible alleles:
Possible genotypes and corresponding phenotypes:
Smooth (S)
Wrinkled (s)
Smooth > Wrinkled
Genotype SS Ss ss
Phenotype Smooth peas Smooth peas Wrinkled peas
P: _____________________ x ______________________
___ ___
Gametes __ __ ___ __
Punnet square:
Gametes
Gametes
Answer:_________________________________________
F1 Genotypes F1 Phenotypes
P: Smooth pea plant x Smooth pea plant
Ss Ss
Gametes S s S s
p (S) = p(s) = ½ p (S) = p(s) = ½
Punnet square:
Gametes
Gametes
SS Ss
Ss ss
S
S
s
s F1 Genotypes F1 Phenotypes
p (SS) = ¼ = 25%
¾ Smooth peas or 75 % Smooth peas
p (Ss) = ½ = 50%
p (ss) = ¼ = 25% ¼ Wrinkled peas or 25% Wrinkled peas
Possible answers:
There is a 25% chance of getting a wrinkled pea plant.
The chance of getting wrinkled pea plants is 1 out of 4
For every 3 smooth pea plants we get, we could get 1 wrinkled pea plant.
Developing the
Law of Segregation
 When two hybrids are crossed Mendel observed the
phenotype that disappeared in the F1 generation
reappears in the F2 generation.
 The F2 generation shows 2 different phenotypes in
a 3:1 ratio
 Seeing this he came to the conclusion that two
alleles that the parents carry for a gene segregate
when gametes are formed, and the offspring only
receives one allele from each parent.
Law of Segregation
Mendel’s law of Segregation states:
The two members of a gene pair segregate from each other into
the gametes, so that one-half of the gametes carry one member
of the pair and the other one-half of the gametes carry the other
member of the pair.
 We now know that segregation occurs during meiosis I
Exercise 3
How can I find out if a black hen is purebred or heterozygous?
Gene studied: Color feathers in hens
Possible alleles:
Possible genotypes and corresponding phenotypes:
Black ___
White ___
Black > Black
Genotype
Phenotype
P: Black feathered hen x White feathered rooster
(homozygous recessive)
BB or Bb bb
Gametes ___ ___ or ___ ___ ___ ___
Punnet square:
Gametes
Gametes
Answer: If the black feathered hen is purebred (BB) ________________________________________
Test cross
Gametes
Gametes
or
Answer: If the black feathered hen is hybrid (Bb) _______________________
Therefore, if when we cross a black feathered hen with a white feathered hen, we get a white
feathered hen we know that ___________________
F1 Genotypes F1 Phenotypes
F1 Genotypes F1 Phenotypes
1st Possibility
2nd Possibility
1st Possibility 2nd Possibility
1st Possibility
2nd Possibility

Lesson 8 Mendelian inheritance

  • 1.
    LESSON # 8 Ifmy father is blond, why am I not?
  • 2.
  • 3.
    Gregor Mendel  Fatherof modern genetics  Experimented with pea plants (Pisum sativum) 1856-1863  Was a monk in the Czech Republic  His work passed unrecognized until his paper was discovered in 1900 “Experiments in plant hybridization.” (1865)  This paper showed that each organism has physical traits that correspond to invisible elements (genes) within the cell.  These invisible elements, which we now call genes, exist in pairs (alleles).  Mendel showed that only one member of this genetic pair is passed on to each progeny.  Mendel's laws form the theoretical basis of our understanding of the genetics of inheritance.
  • 4.
    What he didright  Picked the pea plant which is easy self or cross pollinate and grows fast.  Chose to study traits with only two possible phenotypes :  Pea texture: round or wrinkled seed phenotype  Pea color: yellow or green seed phenotype  Flower color: red or white flower phenotype  Pea plant size: tall or dwarf plant phenotype  Pea pod color: green or yellow
  • 5.
    Genetic studies innovations Mendel made two innovations to the science of genetics:  developed pure lines  Pure Line - a population that breeds true for a particular trait. All individuals of the population have the same phenotype for a specific trait they are pure for. When self-crossing or crossing with individuals of the same pure line population, the offspring shows the same phenotype.  Counted his results and kept statistical notes
  • 6.
    Law of Dominanceor Uniformity of Hybrids.  When two pure line individuals of different phenotypes are crossed, all the resulting offspring show one unique phenotype (the dominant phenotype).  All the offspring that result from interbreeding pure line individuals of different phenotypes, have the same phenotype and the same genotype. All tall First Filial (F1) generation P: Parental generation
  • 7.
    Representing the cross The trait that is maintained in the F1 generation is considered the Dominant trait.  The trait the disappears is considered the Recessive trait.  The first letter of the dominant phenotype is used to indicate the alleles.  Capital letter for the dominant allele  Lower case for the recessive allele P: Parental generation First Filial (F1) generation
  • 8.
    How to begin Genestudied: Pea stem Possible alleles: Possible genotypes and corresponding phenotypes: Tall (T) Short (t) Tall > Short Genotype TT Tt tt Phenotype Tall stem Tall stem Short stem
  • 9.
    P: Purebred Tallstem x Purebred Short stem TT tt Gametes T T t t p(T) = 1 p(t) = 1 Punnet square: Gametes T T t t Tt Tt Tt Tt Gametes F1 Genotype 100% Tt Phenotype 100% Tall stem F1 Genotypes F1 Phenotypes p (Tt) = 1 100 % Tall stem
  • 10.
    Solved Exercise 1 Ahomozygous brown eyed woman and a man with blue eyes want to have children. What is the probability of the child having blue eyes? Gene studied: Eye color Possible alleles: Possible genotypes and corresponding phenotypes: Blue (b) Brown (B) Brown > Blue Genotype BB Bb bb Phenotype Brown eyes Brown eyes Blue eyes
  • 11.
    P: Purebred Browneyes x Purebred Blue eyes BB bb Punnet square: Gametes B B b b Bb Bb Bb Bb Gametes Answer: The possibility of having a child with blue eyes is 0. F1 Genotypes F1 Phenotypes p (Bb) = 1 = 100% 100 % Brown eyes Gametes B B b b p(B) = 1 p(b) = 1
  • 12.
    Exercise 2 What isthe probability of having of getting a pea plant with wrinkled peas if we cross two hybrids? Gene studied: Pea texture Possible alleles: Possible genotypes and corresponding phenotypes: Smooth ___ Wrinkled ___ Smooth > Wrinkled Genotype Phenotype
  • 13.
    Exercise 2 What isthe probability of having of getting a pea plant with wrinkled peas if we cross two hybrids? Gene studied: Pea texture Possible alleles: Possible genotypes and corresponding phenotypes: Smooth (S) Wrinkled (s) Smooth > Wrinkled Genotype SS Ss ss Phenotype Smooth peas Smooth peas Wrinkled peas
  • 14.
    P: _____________________ x______________________ ___ ___ Gametes __ __ ___ __ Punnet square: Gametes Gametes Answer:_________________________________________ F1 Genotypes F1 Phenotypes
  • 15.
    P: Smooth peaplant x Smooth pea plant Ss Ss Gametes S s S s p (S) = p(s) = ½ p (S) = p(s) = ½ Punnet square: Gametes Gametes SS Ss Ss ss S S s s F1 Genotypes F1 Phenotypes p (SS) = ¼ = 25% ¾ Smooth peas or 75 % Smooth peas p (Ss) = ½ = 50% p (ss) = ¼ = 25% ¼ Wrinkled peas or 25% Wrinkled peas Possible answers: There is a 25% chance of getting a wrinkled pea plant. The chance of getting wrinkled pea plants is 1 out of 4 For every 3 smooth pea plants we get, we could get 1 wrinkled pea plant.
  • 16.
    Developing the Law ofSegregation  When two hybrids are crossed Mendel observed the phenotype that disappeared in the F1 generation reappears in the F2 generation.  The F2 generation shows 2 different phenotypes in a 3:1 ratio  Seeing this he came to the conclusion that two alleles that the parents carry for a gene segregate when gametes are formed, and the offspring only receives one allele from each parent.
  • 17.
    Law of Segregation Mendel’slaw of Segregation states: The two members of a gene pair segregate from each other into the gametes, so that one-half of the gametes carry one member of the pair and the other one-half of the gametes carry the other member of the pair.  We now know that segregation occurs during meiosis I
  • 18.
    Exercise 3 How canI find out if a black hen is purebred or heterozygous? Gene studied: Color feathers in hens Possible alleles: Possible genotypes and corresponding phenotypes: Black ___ White ___ Black > Black Genotype Phenotype
  • 19.
    P: Black featheredhen x White feathered rooster (homozygous recessive) BB or Bb bb Gametes ___ ___ or ___ ___ ___ ___ Punnet square: Gametes Gametes Answer: If the black feathered hen is purebred (BB) ________________________________________ Test cross Gametes Gametes or Answer: If the black feathered hen is hybrid (Bb) _______________________ Therefore, if when we cross a black feathered hen with a white feathered hen, we get a white feathered hen we know that ___________________ F1 Genotypes F1 Phenotypes F1 Genotypes F1 Phenotypes 1st Possibility 2nd Possibility 1st Possibility 2nd Possibility 1st Possibility 2nd Possibility