1. Gregor Mendel

2. *became an ordained priest in 1847, and
* Born in Heinzendorf, Austria on July 22, 1822.
* He died in Brno, Austria January 6, 1884.
* Austrian monk who found actual proof of the
existence of genes...the father of genetics,
*work was relatively unappreciated until
the early 1900's.
*became an ordained priest in 1847, and
in 1851entered into the University of Vienna to
train to be a teacher of Mathematics and Biology.
Failed the elementary teachers exam and returned
to Brno in 1845 ,became a teacher at a technical
high school, where he taughtmathematics,
natural science and general science.

3. * 1868, promoted in the monastery to Abbot.
* During the middle of his life, did work into the
theories of heredity. Using pea pod plants,
studied seven basic characteristics, traced these
and discovered three basic laws which governed
the passage of a trait from one member of a
species to another member of the same species.
First presentation was on his eight years of
experimentation with artificial plant hybridization.
*first two communications were published in 1853 to
1854. Contained information about damage to plants
by insects.
* 1856 to 1863 cultivated, tested almost 28,000 plants

4. *Traits united through fertilization
*Rediscovery of his works brought us close to an
era of speculation on heredity.
*Opened a new pathway of study on heredity to
reveal a new mechanism operating in the sense
of evolution.
work and theories, later became the basis for the
study of modern genetics and are still recognized
and used today.
* His work led to the discovery of particulate
inheritance, dominant and recessive traits,
genotype and phenotype, and the concept of
heterozygousity and homozygousity.

5. Mendel’s Principles ( see other sheets) LAW of SEGREGATION
LAW of INDEPENDENT ASSORTMENT

6. Types of Crosses
1. Monohybrid Cross
(A cross of “one” trait)
A trait is controlled by two alleles, one from each parent. Each can be:
Dominant: TT (homozygous)
Tt (Heterozygous)
Recessive: tt

7. t P1= TT x tt T T Tt Tt Tt Tt Test Cross using a Punnett square:
Homozygous parents: TT for Tall plants x tt (short)
P1= TT x tt
F1 cross (first filial,first generation )
T T t Tt Tt Tt Tt

8. Genotypic ratio( alleles): 4:0
All Heterozygous dominant
All Tt
Phenotypic ration( expression): 4:0
All Tall

9. P2 Tt x Tt (previous F1)
T t T t
F2 cross TT Tt Tt tt

10. Genotypic ratio: 1:2:1
1 TT: 2Tt: 1tt
Phenotypic ration: 3:1
3 Tall : 1 short

11. 2. Dihybrid Cross A cross of two traits (2 sets of alleles)
P TTYY Tall plants, yellow flowers
ttyy Short plants, white flowers
All Heterozygous
Punnett square of ?

12. TY TY TY TY
TtYy
TtYy
TtYy
TtYy ty F1

13. Genotypic ratio: 16:0
All TtYy
Heterozygous dominant
Phenotypic ration:16:0
All tall, yellow

14. F2 Cross : P2 TtYy x TtYy
TtYy TtYy
TY,Ty
tY, ty

15. TY Ty tY ty TY Ty tY ty TTYY TTYy TtYY TtYy TTYy TTyy TtYy Ttyy TtYYF2

16. Genotypic ratio:
Many different
Phenotypic ration:
9:3:3:1
9 Tall, Yellow
3 Tall, white
3 Yellow, short
1 short, white

17. 3. Incomplete Dominance
A heterozygous genotype is a
blending of the dominant and
recessive allele (new phenotype)
The dominant allele isn’t
expressed over the recessive.

18. r Genotype: All Rr Phenotype: All PINK F1 cross
P1 RR red flowers, rr white flowers
R R r Rr Rr Rr Rr
Genotype: All Rr
Phenotype: All PINK

19. R r F2 Cross: P2 Rr (pink) x Rr(pink) Rr rr Rr
Genotype: 1:2: RR: 2 Rr: 1rr
Phenotype: 1:2: Red: 2 Pink: 1white

20. 4. Codominance:
There is more than one
dominate allele (must be at
least three alleles for a trait)
If together, they are both
expressed.
Example: Inheritance of Blood
Types

21. Since Blood type is codominant,
we use a common letter for all.
Use I for dominant A and B,
use i for recessive o.
Phenotypes Genotypes
Type A IAIA, IA io
Type B IBIB, IBio
Type O io io
Type AB IA IB

22. If a woman is heterozygous for type A
Sample problem:
If a woman is heterozygous for type A
blood and her husband is homozygous
for type B, what blood types could
their children be?
Woman: IA io
Man: IBIB
IA io AB Bo IB AB Bo

23. Possible Phenotypes:
½ type AB, ½ type B
Possible Genotypes:
½ IAIB, ½ IBio

24. 5. Sex Linked:
*The trait is carried on the sex chromosome…X- linked
*Must show XX for female
XY for male
*Only “X” carries the gene

25. Female, homozygous dominant XAXA
Female, recessive XaXa
Female, heterozygous, dominant XAXa “Carrier”
Male dominant XAY
Male, recessive XaY

26. If a woman is heterozygous for a
Sample problem:
If a woman is heterozygous for a
Reccessive trait, Red-green colorblindness,
and her husband is not color blind,
What portion of their children could be
Color-blind? What portion of their sons?
Daughters?
Woman: XAXa
Man: XAY
XAXA
XAXa XA
XAY
XaY Y XA Xa

27. Possible Phenotypes:
½ normal female
Possible Genotypes:
¼ normal males
¼ colorblind males

28. red flowers is crossed with a plant recessive for white flowers.
Practice Problem #1
A plant heterozygous dominant for
red flowers is crossed with a plant
recessive for white flowers.
Give the expected phenotype and
genotype for the F1generation.
Show the cross.

29. Practice #2
A strawberry plant that is heterozygous for red, seeded fruits is crossed with a strawberry plant that produces pink, seedless fruits. What is the expected genotype and phenotype of the F1 generation? Show the cross. What proportion of the strawberry plants will produce red, seedless fruits?
(Red is dominant
Seeded is dominant)

30. Practice #3 Coat color in Labs is inherited by
incomplete Dominance. Black is dominant, yellow is Recessive and Chocolate is the heterozygous Phenotype. If a yellow lab is crossed with a chocolate lab, what proportion of the offspring will be chocolate? Show your cross.

31. Homework
*Read chapter 13
*Rest of problems on sheet and
textbook assigned Pg
# 3, 7, 8, 9, 13,14,17-21