mendeliangenetics-110706200307-phpapp02.ppt

1
Mendelelian
Genetics
copyright cmassengale

2
Gregor Mendel
(1822-1884)
Responsible
for the Laws
governing
Inheritance of
Traits

3
Gregor Johann Mendel
Austrian monk
Studied the
inheritance of
traits in pea plants
Developed the laws
of inheritance
Mendel's work was
not recognized until
the turn of the
20th century

4
Gregor Johann Mendel
Between 1856 and
1863, Mendel
cultivated and
tested some 28,000
pea plants
He found that the
plants' offspring
retained traits of
the parents
Called the “Father
of Genetics"

5
Site of
Gregor
Mendel’s
experimental
garden in the
Czech
Republic

6
Mendel stated that
physical traits are
inherited as “particles”
Mendel did not know
that the “particles”
were actually
Chromosomes & DNA
Particulate Inheritance

7
Genetic Terminology
 Trait - any characteristic that
can be passed from parent to
offspring
 Heredity - passing of traits
from parent to offspring
 Genetics - study of heredity

8
Types of Genetic Crosses
 Monohybrid cross - cross
involving a single trait
e.g. flower color
 Dihybrid cross - cross involving
two traits
e.g. flower color & plant height

9
Punnett Square
Used to help
solve genetics
problems

11
Designer “Genes”
 Alleles - two forms of a gene
(dominant & recessive)
 Dominant - stronger of two genes
expressed in the hybrid;
represented by a capital letter (R)
 Recessive - gene that shows up less
often in a cross; represented by a
lowercase letter (r)

12
More Terminology
 Genotype - gene combination
for a trait (e.g. RR, Rr, rr)
 Phenotype - the physical
feature resulting from a
genotype (e.g. red, white)

13
Genotype & Phenotype in Flowers
Genotype of alleles:
R = red flower
r = yellow flower
All genes occur in pairs, so 2
alleles affect a characteristic
Possible combinations are:
Genotypes RR Rr rr
Phenotypes RED RED YELLOW

14
Genotypes
 Homozygous genotype - gene
combination involving 2 dominant
or 2 recessive genes (e.g. RR or
rr); also called pure
 Heterozygous genotype - gene
combination of one dominant &
one recessive allele (e.g. Rr);
also called hybrid

15
Genes and Environment
Determine Characteristics

16
Mendel’s Pea Plant
Experiments

17
Why peas, Pisum sativum?
Can be grown in a
small area
Produce lots of
offspring
Produce pure plants
when allowed to
self-pollinate
several generations
Can be artificially
cross-pollinated

18
Reproduction in Flowering Plants
Pollen contains sperm
Produced by the
stamen
Ovary contains eggs
Found inside the
flower
Pollen carries sperm to the
eggs for fertilization
Self-fertilization can
occur in the same flower
Cross-fertilization can
occur between flowers

19
Mendel’s Experimental
Methods
Mendel hand-pollinated
flowers using a paintbrush
He could snip the
stamens to prevent
self-pollination
Covered each flower
with a cloth bag
He traced traits through
the several generations

20
How Mendel Began
Mendel
produced
pure
strains by
allowing the
plants to
self-
pollinate
for several
generations

21
Eight Pea Plant Traits
Seed shape --- Round (R) or Wrinkled (r)
Seed Color ---- Yellow (Y) or Green (y)
Pod Shape --- Smooth (S) or wrinkled (s)
Pod Color --- Green (G) or Yellow (g)
Seed Coat Color ---Gray (G) or White (g)
Flower position---Axial (A) or Terminal (a)
Plant Height --- Tall (T) or Short (t)
Flower color --- Purple (P) or white (p)

24
Mendel’s Experimental Results

25
Did the observed ratio match
the theoretical ratio?
The theoretical or expected ratio of
plants producing round or wrinkled seeds
is 3 round :1 wrinkled
Mendel’s observed ratio was 2.96:1
The discrepancy is due to statistical
error
The larger the sample the more nearly
the results approximate to the
theoretical ratio

26
Generation “Gap”
Parental P1 Generation = the parental
generation in a breeding experiment.
F1 generation = the first-generation
offspring in a breeding experiment. (1st
filial generation)
From breeding individuals from the P1
generation
F2 generation = the second-generation
offspring in a breeding experiment.
(2nd filial generation)
From breeding individuals from the F1
generation

27
Following the Generations
Cross 2
Pure
Plants
TT x tt
Results
in all
Hybrids
Tt
Cross 2 Hybrids
get
3 Tall & 1 Short
TT, Tt, tt

28
Monohybrid
Crosses

29
Trait: Seed Shape
Alleles: R – Round r – Wrinkled
Cross: Round seeds x Wrinkled seeds
RR x rr
P1 Monohybrid Cross
R
R
r
r
Rr
Rr
Rr
Rr
Genotype: Rr
Phenotype: Round
Genotypic
Ratio: All alike
Phenotypic
Ratio: All alike

30
P1 Monohybrid Cross Review
 Homozygous dominant x Homozygous
recessive
 Offspring all Heterozygous
(hybrids)
 Offspring called F1 generation
 Genotypic & Phenotypic ratio is ALL
ALIKE

31
Trait: Seed Shape
Cross: Round seeds x Round seeds
Rr x Rr
F1 Monohybrid Cross
R
r
r
R
RR
rr
Rr
Rr
Genotype: RR, Rr, rr
Phenotype: Round &
wrinkled
G.Ratio: 1:2:1
P.Ratio: 3:1

32
F1 Monohybrid Cross Review
 Heterozygous x heterozygous
 Offspring:
25% Homozygous dominant RR
50% Heterozygous Rr
25% Homozygous Recessive rr
 Offspring called F2 generation
 Genotypic ratio is 1:2:1
 Phenotypic Ratio is 3:1

33
What Do the Peas Look Like?

34
…And Now the Test Cross
Mendel then crossed a pure & a
hybrid from his F2 generation
This is known as an F2 or test
cross
There are two possible
testcrosses:
Homozygous dominant x Hybrid
Homozygous recessive x Hybrid

35
Trait: Seed Shape
Cross: Round seeds x Round seeds
RR x Rr
F2 Monohybrid Cross (1st)
R
R
r
R
RR
Rr
RR
Rr
Genotype: RR, Rr
Phenotype: Round
Genotypic
Ratio: 1:1
Phenotypic
Ratio: All alike

36
Trait: Seed Shape
Cross: Wrinkled seeds x Round seeds
rr x Rr
F2 Monohybrid Cross (2nd)
r
r
r
R
Rr
rr
Rr
rr
Genotype: Rr, rr
Phenotype: Round &
Wrinkled
G. Ratio: 1:1
P.Ratio: 1:1

37
F2 Monohybrid Cross Review
 Homozygous x heterozygous(hybrid)
 Offspring:
50% Homozygous RR or rr
50% Heterozygous Rr
 Phenotypic Ratio is 1:1
 Called Test Cross because the
offspring have SAME genotype as
parents

38
Practice Your Crosses
Work the P1, F1, and both
F2 Crosses for each of the
other Seven Pea Plant
Traits

39
Mendel’s Laws

40
Results of Monohybrid Crosses
Inheritable factors or genes are
responsible for all heritable
characteristics
Phenotype is based on Genotype
Each trait is based on two genes,
one from the mother and the
other from the father
True-breeding individuals are
homozygous ( both alleles) are the
same

41
Law of Dominance
In a cross of parents that are
pure for contrasting traits, only
one form of the trait will appear in
the next generation.
All the offspring will be
heterozygous and express only the
dominant trait.
RR x rr yields all Rr (round seeds)

42
Law of Dominance

43
Law of Segregation
During the formation of gametes
(eggs or sperm), the two alleles
responsible for a trait separate
from each other.
Alleles for a trait are then
"recombined" at fertilization,
producing the genotype for the
traits of the offspring.

44
Applying the Law of Segregation

45
Law of Independent
Assortment
Alleles for different traits are
distributed to sex cells (&
offspring) independently of one
another.
This law can be illustrated using
dihybrid crosses.

46
Dihybrid Cross
A breeding experiment that tracks
the inheritance of two traits.
Mendel’s “Law of Independent
Assortment”
a. Each pair of alleles segregates
independently during gamete formation
b. Formula: 2n (n = # of heterozygotes)

47
Question:
How many gametes will be produced
for the following allele arrangements?
Remember: 2n (n = # of heterozygotes)
1. RrYy
2. AaBbCCDd
3. MmNnOoPPQQRrssTtQq

48
Answer:
1. RrYy: 2n = 22 = 4 gametes
RY Ry rY ry
2. AaBbCCDd: 2n = 23 = 8 gametes
ABCD ABCd AbCD AbCd
aBCD aBCd abCD abCD
3. MmNnOoPPQQRrssTtQq: 2n = 26 = 64
gametes

49
Dihybrid Cross
Traits: Seed shape & Seed color
Alleles: R round
r wrinkled
Y yellow
y green
RrYy x RrYy
RY Ry rY ry RY Ry rY ry
All possible gamete combinations

50
Dihybrid Cross
RY Ry rY ry
RY
Ry
rY
ry

51
Dihybrid Cross
RRYY
RRYy
RrYY
RrYy
RRYy
RRyy
RrYy
Rryy
RrYY
RrYy
rrYY
rrYy
RrYy
Rryy
rrYy
rryy
Round/Yellow: 9
Round/green: 3
wrinkled/Yellow: 3
wrinkled/green: 1
9:3:3:1 phenotypic
ratio
RY Ry rY ry
RY
Ry
rY
ry

52
Dihybrid Cross
Round/Yellow: 9
Round/green: 3
wrinkled/Yellow: 3
wrinkled/green: 1
9:3:3:1

53
Test Cross
A mating between an individual of unknown
genotype and a homozygous recessive
individual.
Example: bbC__ x bbcc
BB = brown eyes
Bb = brown eyes
bb = blue eyes
CC = curly hair
Cc = curly hair
cc = straight hair
bC b___
bc

54
Test Cross
Possible results:
bC b___
bc bbCc bbCc
C bC b___
bc bbCc bbcc
or
c

55
Summary of Mendel’s laws
LAW
PARENT
CROSS
OFFSPRING
DOMINANCE TT x tt
tall x short
100% Tt
tall
SEGREGATION
Tt x Tt
tall x tall
75% tall
25% short
INDEPENDENT
ASSORTMENT
RrGg x RrGg
round & green
x
round & green
9/16 round seeds & green
pods
3/16 round seeds & yellow
pods
3/16 wrinkled seeds & green
pods
1/16 wrinkled seeds & yellow
pods

56
Incomplete Dominance
and
Codominance

57
F1 hybrids have an appearance somewhat
in between the phenotypes of the two
parental varieties.
Example: snapdragons (flower)
red (RR) x white (rr)
RR = red flower
rr = white flower
R
R
r r

58
Rr
Rr
Rr
Rr
R
R
r
All Rr = pink
(heterozygous pink)
produces the
F1 generation
r

59

60
Codominance
Two alleles are expressed (multiple
alleles) in heterozygous individuals.
Example: blood type
1. type A = IAIA or IAi
2. type B = IBIB or IBi
3. type AB = IAIB
4. type O = ii

61
Codominance Problem
Example:homozygous male Type B (IBIB)
x
heterozygous female Type A (IAi)
IAIB IBi
IAIB IBi
1/2 = IAIB
1/2 = IBi
IB
IA i
IB

62
Another Codominance Problem
• Example: male Type O (ii)
x
female type AB (IAIB)
IAi IBi
IAi IBi
1/2 = IAi
1/2 = IBi
i
IA IB
i

63
Codominance
Question:
If a boy has a blood type O and
his sister has blood type
AB, what are the genotypes
and phenotypes of their
parents?
boy - type O (ii) X girl - type
AB (IAIB)

64
Codominance
Answer:
IAIB
ii
Parents:
genotypes = IAi and IBi
phenotypes = A and B
IB
IA i
i

65
Sex-linked Traits
Traits (genes) located on the sex
chromosomes
Sex chromosomes are X and Y
XX genotype for females
XY genotype for males
Many sex-linked traits carried on
X chromosome

66
Sex-linked Traits
Sex Chromosomes
XX chromosome - female Xy chromosome - male
fruit fly
eye color
Example: Eye color in fruit flies

67
Sex-linked Trait Problem
Example: Eye color in fruit flies
(red-eyed male) x (white-eyed female)
XRY x XrXr
Remember: the Y chromosome in males
does not carry traits.
RR = red eyed
Rr = red eyed
rr = white eyed
XY = male
XX = female
XR
Xr Xr
Y

68
Sex-linked Trait Solution:
XR Xr
Xr Y
XR Xr
Xr Y
50% red eyed
female
50% white eyed
male
XR
Xr Xr
Y

69
Female Carriers

70
Genetic Practice
Problems

71
Breed the P1 generation
tall (TT) x dwarf (tt) pea plants
T
T
t t

72
Solution:
T
T
t t
Tt
Tt
Tt
Tt All Tt = tall
(heterozygous tall)
produces the
F1 generation
tall (TT) vs. dwarf (tt) pea plants

73
Breed the F1 generation
tall (Tt) vs. tall (Tt) pea plants
T
t
T t

74
Solution:
TT
Tt
Tt
tt
T
t
T t
produces the
F2 generation
1/4 (25%) = TT
1/2 (50%) = Tt
1/4 (25%) = tt
1:2:1 genotype
3:1 phenotype
tall (Tt) x tall (Tt) pea plants

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