Mendelian inheritance

1. LESSON # 8
If my father is blond, why am I not?

2. MENDELIAN GENETICS
Inheritance

3. 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.

4. 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

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 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

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
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

9. 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

10. 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

11. 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

12. 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

13. 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

14. P: _____________________ x ______________________
___ ___
Gametes __ __ ___ __
Punnet square:
Gametes
Gametes
Answer:_________________________________________
F1 Genotypes F1 Phenotypes

15. 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.

16. 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.

17. 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

18. 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

19. 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