Understanding the Fundamentals of Genetics

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Sharp Tutor
This presentation gives an introduction to all the fundamentals of genetics. Genetics is a branch of biology concerned with the study of genes, genetic variation, and heredity in organisms. Though heredity had been observed for millennia, Gregor Mendel, Moravian scientist and Augustinian friar working in the 19th century in Brno, was the first to study genetics scientifically.
1. of
Genetics
1
2. Gregor Mendel
(1822-1884)
Responsible
for the Laws
governing
Inheritance of
Traits
2
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
3
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"
4
5. Site of
garden in the
5
6. Particulate Inheritance
 Mendel stated that
physical traits are
inherited as “particles”
 Mendel did not know
that the “particles”
were actually
Chromosomes & DNA
6
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
7
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
8
9. Punnett Square
Used to help
solve genetics
9
10.
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)
11
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)
12
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
13
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
14
15. Genes and Environment
Determine Characteristics
15
16. Mendel’s Pea Plant
Experiments
16
17. Why peas, Pisum sativum?
 Can be grown in a
small area
 Produce lots of
 Produce pure plants
when allowed to self-
pollinate several
 Can be artificially
17
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
18
19. Mendel’s Experimental
Methods
Mendel hand-pollinated
flowers using a
He could snip the
stamens to prevent
self-pollination
He traced traits
through the several
19
20. How Mendel Began
strains by
allowing the
plants to
for several
20
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)
21
22.
23.
24. Mendel’s Experimental Results
24
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
25
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
26
27. Following the Generations
Cross 2 Results Cross 2 Hybrids
Pure in all get
Plants Hybrids 3 Tall & 1 Short
TT x tt Tt TT, Tt, tt
27
28. Crosses
28
29. P1 Monohybrid Cross
Trait: Seed Shape
Alleles: R – Round r – Wrinkled
Cross: Round seeds x Wrinkled seeds
RR x rr
Genotype: Rr
r r
Phenotype:
Phenotype Round
R Rr Rr Genotypic
Ratio: All alike
R Rr Rr Phenotypic
Ratio: All alike
29
30. P1 Monohybrid Cross Review
 Homozygous dominant x
Homozygous recessive
 Offspring all Heterozygous
(hybrids)
Offspring called F1 generation
Genotypic & Phenotypic ratio is
ALL ALIKE
30
31. F1 Monohybrid Cross
Trait: Seed Shape
Alleles: R – Round r – Wrinkled
Cross: Round seeds x Round seeds
Rr x Rr
Genotype: RR, Rr, rr
R r
Phenotype:
Phenotype Round &
RR Rr wrinkled
R
G.Ratio: 1:2:1
r Rr rr P.Ratio: 3:1
31
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
32
33. What Do the Peas Look Like?
33
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
34
35. F2 Monohybrid Cross (1st)
Trait: Seed Shape
Alleles: R – Round r – Wrinkled
Cross: Round seeds x Round seeds
RR x Rr
Genotype: RR, Rr
R r
Phenotype:
Phenotype Round
R RR Rr Genotypic
Ratio: 1:1
R RR Rr Phenotypic
Ratio: All alike
35
36. F2 Monohybrid Cross (2nd)
Trait: Seed Shape
Alleles: R – Round r – Wrinkled
Cross: Wrinkled seeds x Round seeds
rr x Rr
R r Genotype: Rr, rr
Phenotype:
Phenotype Round &
r Rr rr Wrinkled
G. Ratio: 1:1
r Rr rr P.Ratio: 1:1
36
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
37
38. Practice Your Crosses
Work the P1, F1, and both
F2 Crosses for each of the
other Seven Pea Plant
Traits
38
39. Mendel’s Laws
39
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
40
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)
41
42. Law of Dominance
42
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.
43
44. Applying the Law of Segregation
44
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.
45
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)
46
47. Question:
How many gametes will be produced
for the following allele arrangements?
Remember: 2n (n = # of heterozygotes)
1. RrYy
2. AaBbCCDd
3. MmNnOoPPQQRrssTtQq
47
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
48
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
49
50. Dihybrid Cross
RY Ry rY ry
50
51. Dihybrid Cross
RY Ry rY ry
RY RRYY Round/Yellow: 9
RRYy RrYY RrYy
Ry RRYy Round/green: 3
RRyy RrYy Rryy
wrinkled/Yellow: 3
rY RrYY RrYy rrYY rrYy
wrinkled/green: 1
ry RrYy Rryy rrYy rryy 9:3:3:1 phenotypic
ratio
51
52. Dihybrid Cross
Round/Yellow: 9
Round/green: 3
wrinkled/Yellow: 3
wrinkled/green: 1
9:3:3:1
52
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
bC b___
CC = curly hair
Cc = curly hair bc
cc = straight hair
53
54. Test Cross
Possible results:
bC b___
C bC b___
c
bc bbCc bbCc or bc bbCc bbcc
54
55. Summary of Mendel’s laws
PARENT
LAW OFFSPRING
CROSS
DOMINANCE TT x tt 100% Tt
tall x short tall
Tt x Tt 75% tall
tall x tall 25% short
9/16 round seeds & green pods
RrGg x RrGg
INDEPENDENT 3/16 round seeds & yellow
round & green pods
x 3/16 wrinkled seeds & green
pods
round & green 1/16 wrinkled seeds & yellow
pods
55
56. Incomplete Dominance
and
Codominance
56
57. Incomplete Dominance
F1 hybrids have an appearance somewhat
in between the phenotypes of the two
parental varieties.
Example: snapdragons (flower)
red (RR) x white (rr) r r
RR = red flower R
rr = white flower
R
57
58. Incomplete Dominance
r
produces the
R Rr Rr
F1 generation
R Rr Rr All Rr = pink
(heterozygous pink)
58
59. Incomplete Dominance
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
60
61. Codominance Problem
Example: homozygous male Type B (IBIB)
x heterozygous female Type A (IAi)
IA i
IB IAIB IBi
1/2 = IAIB
1/2 = IBi
IB IAIB IBi
61
62. Another Codominance Problem
• Example: male Type O (ii)
x
female type AB (IAIB)
IA IB
i IAi IBi 1/2 = IAi
1/2 = IBi
i IAi IBi
62
63. Codominance
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)
63
64. Codominance
IA i
IB IAIB Parents:
genotypes = IAi and IBi
phenotypes = A and B
i ii
64
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
65
66. Sex-linked Traits
Example: Eye color in fruit flies
Sex Chromosomes
fruit fly
eye color
XX chromosome - female Xy chromosome - male
66
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 Xr Xr
Rr = red eyed
rr = white eyed
Xr
Xy = male
XX = female
Y
67
68. Sex-linked Trait Solution:
Xr Xr
50% red eyed
XR X X X X
R r R r
female
50% white eyed
Y Xr Y Xr Y male
68
69. Female Carriers
69
70. Genetic Practice
Problems
70
71. Breed the P1 generation
tall (TT) x dwarf (tt) pea plants
t t
T
T
71
72. Solution:
tall (TT) vs. dwarf (tt) pea plants
t t
Tt Tt produces the
F1 generation
T Tt Tt All Tt = tall
(heterozygous tall)
72
73. Breed the F1 generation
tall (Tt) vs. tall (Tt) pea plants
T t
T
t
73
74. Solution:
tall (Tt) x tall (Tt) pea plants
T t
produces the
TT Tt F2 generation
1/4 (25%) = TT
Tt tt 1/2 (50%) = Tt
1/4 (25%) = tt
1:2:1 genotype
3:1 phenotype
74
75.