Mendel`s genetics

Download Report

Transcript Mendel`s genetics

Mendelelian
Genetics
copyright cmassengale
1
*Gregor
Mendel
(1822-1884)
Responsible
for the Laws
governing
Inheritance of
Traits
copyright cmassengale
2
SCI.9-12.B-4 - [Standard] - The student
will demonstrate an understanding of the
molecular basis of heredity.
SCI.9-12.B-4.6 - [Indicator] Predict inherited traits by using the
principles of Mendelian genetics (including
segregation, independent assortment, and
dominance).
SCI.9-12.B-4.7 - [Indicator] Summarize the chromosome theory of
inheritance and relate that theory to
Gregor Mendel's principles of genetics.
copyright cmassengale
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
copyright cmassengale
4
*Gregor Johann Mendel
Called the “Father
of Genetics"
copyright cmassengale
5
Site of
Gregor
Mendel’s
experimental
garden in the
Czech
Republic
copyright cmassengale
6
*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
copyright cmassengale
7
*Terminology
 Genotype - gene combination
for a trait (e.g. RR, Rr, rr)
 Phenotype - the physical
feature resulting from a
genotype (e.g. red, white)
copyright cmassengale
8
*Genotypes
 Homozygous genotype - gene
combination involving 2 dominant
or 2 recessive genes (e.g. RR or
rr); also called pure
Homozygous dominant = TT,
RR, FF, GG
Homozygous recessive = tt, rr,
ff, gg,
copyright cmassengale
9
*Heterozygous genotype gene combination of one
dominant & one recessive
allele
(e.g. Rr); also called
hybrid
Tt, Rr, Ff, Gg
copyright cmassengale
10
SCI.9-12.B-4 - [Standard] - The student
will demonstrate an understanding of the
molecular basis of heredity.
SCI.9-12.B-4.6 - [Indicator] Predict inherited traits by using the
principles of Mendelian genetics (including
segregation, independent assortment, and
dominance).
SCI.9-12.B-4.7 - [Indicator] Summarize the chromosome theory of
inheritance and relate that theory to
Gregor Mendel's principles of genetics.
copyright cmassengale
11
*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
 Trihybrid
 Polyhybrid
copyright cmassengale
12
*Punnett Square
Used to help
solve genetics
problems
copyright cmassengale
13
copyright cmassengale
14
*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)
copyright cmassengale
15
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
copyright cmassengale
16
*Genes and Environment
Determine Characteristics
copyright cmassengale
17
SCI.9-12.B-4 - [Standard] - The student
will demonstrate an understanding of the
molecular basis of heredity.
SCI.9-12.B-4.6 - [Indicator] Predict inherited traits by using the
principles of Mendelian genetics (including
segregation, independent assortment, and
dominance).
SCI.9-12.B-4.7 - [Indicator] Summarize the chromosome theory of
inheritance and relate that theory to
Gregor Mendel's principles of genetics.
copyright cmassengale
18
Mendel’s Pea Plant
Experiments
copyright cmassengale
19
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
copyright cmassengale
20
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
copyright cmassengale
21
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
copyright cmassengale
22
How Mendel Began
Mendel
produced
pure
strains by
allowing the
plants to
selfpollinate
for several
generations
copyright cmassengale
23
SCI.9-12.B-4 - [Standard] - The student
will demonstrate an understanding of the
molecular basis of heredity.
SCI.9-12.B-4.6 - [Indicator] Predict inherited traits by using the
principles of Mendelian genetics (including
segregation, independent assortment, and
dominance).
SCI.9-12.B-4.7 - [Indicator] Summarize the chromosome theory of
inheritance and relate that theory to
Gregor Mendel's principles of genetics.
copyright cmassengale
24
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)
copyright cmassengale
25
copyright cmassengale
26
copyright cmassengale
27
copyright cmassengale
28
Mendel’s Experimental Results
copyright cmassengale
29
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
copyright cmassengale
30
SCI.9-12.B-4 - [Standard] - The student
will demonstrate an understanding of the
molecular basis of heredity.
SCI.9-12.B-4.6 - [Indicator] Predict inherited traits by using the
principles of Mendelian genetics (including
segregation, independent assortment, and
dominance).
SCI.9-12.B-4.7 - [Indicator] Summarize the chromosome theory of
inheritance and relate that theory to
Gregor Mendel's principles of genetics.
copyright cmassengale
31
*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
copyright cmassengale
32
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
copyright cmassengale
33
SCI.9-12.B-4 - [Standard] - The student
will demonstrate an understanding of the
molecular basis of heredity.
SCI.9-12.B-4.6 - [Indicator] Predict inherited traits by using the
principles of Mendelian genetics (including
segregation, independent assortment, and
dominance).
SCI.9-12.B-4.7 - [Indicator] Summarize the chromosome theory of
inheritance and relate that theory to
Gregor Mendel's principles of genetics.
copyright cmassengale
34
Monohybrid
Crosses
copyright cmassengale
35
The original organism
(plant or animal) is called
the P generation or
Parental. The next
generation is called F1
and F2 generation.
P1 Monohybrid Cross
Trait: Seed Shape
Alleles: R – Round
r – Wrinkled
Cross: Round seeds
x Wrinkled seeds
RR
x
rr
r
r
R
Rr
Rr
R
Rr
Rr
Genotype: Rr
Phenotype: Round
Genotypic
Ratio: All alike
Phenotypic
Ratio: All alike
copyright cmassengale
37
P1 Monohybrid Cross Review
 Homozygous dominant x Homozygous
recessive
 Offspring all Heterozygous
(hybrids)
 Offspring called F1 generation
 Genotypic & Phenotypic ratio is ALL
ALIKE
copyright cmassengale
38
F1 Monohybrid Cross
Trait: Seed Shape
Alleles: R – Round
r – Wrinkled
Cross: Round seeds
x Round seeds
Rr
x
Rr
R
r
R
RR
Rr
r
Rr
rr
Genotype: RR, Rr, rr
Phenotype: Round &
wrinkled
G.Ratio: 1:2:1
P.Ratio: 3:1
copyright cmassengale
39
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
copyright cmassengale
40
What Do the Peas Look Like?
copyright cmassengale
41
…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
copyright cmassengale
42
SCI.9-12.B-4 - [Standard] - The student
will demonstrate an understanding of the
molecular basis of heredity.
SCI.9-12.B-4.6 - [Indicator] Predict inherited traits by using the
principles of Mendelian genetics (including
segregation, independent assortment, and
dominance).
SCI.9-12.B-4.7 - [Indicator] Summarize the chromosome theory of
inheritance and relate that theory to
Gregor Mendel's principles of genetics.
copyright cmassengale
43
SCI.9-12.B-4 - [Standard] - The student
will demonstrate an understanding of the
molecular basis of heredity.
SCI.9-12.B-4.6 - [Indicator] Predict inherited traits by using the
principles of Mendelian genetics (including
segregation, independent assortment, and
dominance).
SCI.9-12.B-4.7 - [Indicator] Summarize the chromosome theory of
inheritance and relate that theory to
Gregor Mendel's principles of genetics.
copyright cmassengale
44
*TEST CROSS – done to
determine an unknown
genotype of an organism
that looks dominant. Cross
with a (HR) homozygous
recessive organism
copyright cmassengale
45
F2 Monohybrid Cross
st
(1 )
Trait: Seed Shape
Alleles: R – Round
r – Wrinkled
Cross: Round seeds
x Round seeds
RR
x
Rr
R
r
R
RR
Rr
R
RR
Rr
Genotype: RR, Rr
Phenotype: Round
Genotypic
Ratio: 1:1
Phenotypic
Ratio: All alike
copyright cmassengale
46
F2 Monohybrid Cross (2nd)
Trait: Seed Shape
Alleles: R – Round
r – Wrinkled
Cross: Wrinkled seeds x Round seeds
rr
x
Rr
R
r
r
Rr
Rr
r
rr
rr
Genotype: Rr, rr
Phenotype: Round &
Wrinkled
G. Ratio: 1:1
P.Ratio: 1:1
copyright cmassengale
47
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
copyright cmassengale
48
Practice Your Crosses
Work the P1, F1, and both
F2 Crosses for each of the
other Seven Pea Plant
Traits
copyright cmassengale
49
Mendel’s Laws
copyright cmassengale
50
*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)
copyright cmassengale
51
Law of Dominance
copyright cmassengale
52
*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.
copyright cmassengale
53
Applying the Law of Segregation
copyright cmassengale
54
*Law of Independent
Assortment
Alleles for different traits are
distributed to sex cells (&
offspring) independently of one
another. Not all of mom’s traits
stay together. Mom’s and Dad’s
traits are combined differently,
copyright cmassengale
55
*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)
copyright cmassengale
56
Question:
How many gametes will be produced
for the following allele arrangements?
Remember: 2n (n = # of heterozygotes)
1. RrYy
2. AaBbCCDd
3. MmNnOoPPQQRrssTtQq
copyright cmassengale
57
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
copyright cmassengale
58
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
copyright cmassengale
59
Dihybrid Cross
RY
Ry
rY
ry
RY
Ry
rY
ry
copyright cmassengale
60
Dihybrid Cross
RY
RY RRYY
Ry RRYy
rY RrYY
ry
RrYy
Ry
rY
ry
RRYy
RrYY
RrYy
RRyy
RrYy
Rryy
RrYy
rrYY
rrYy
Rryy
rrYy
rryy
copyright cmassengale
Round/Yellow:
Round/green:
9
3
wrinkled/Yellow: 3
wrinkled/green:
1
9:3:3:1 phenotypic
ratio
61
Dihybrid Cross
Round/Yellow: 9
Round/green:
3
wrinkled/Yellow: 3
wrinkled/green: 1
9:3:3:1
copyright cmassengale
62
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
copyright cmassengale
63
Test Cross
Possible results:
bc
bC
b___
C
bbCc
bbCc
or
bc
copyright cmassengale
bC
b___
c
bbCc
bbcc
64
Summary of Mendel’s laws
LAW
DOMINANCE
SEGREGATION
INDEPENDENT
ASSORTMENT
PARENT
CROSS
OFFSPRING
TT x tt
tall x short
100% Tt
tall
Tt x Tt
tall x tall
75% tall
25% short
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
copyright cmassengale
65
Incomplete Dominance
and
Codominance
copyright cmassengale
66
*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
rr = white flower
R
R
copyright cmassengale
67
Incomplete Dominance
r
r
R Rr
Rr
R Rr
Rr
produces the
F1 generation
All Rr = pink
(heterozygous pink)
copyright cmassengale
68
Incomplete Dominance
copyright cmassengale
69
*Codominance
Two alleles are expressed (multiple
alleles) in heterozygous individuals.
Example: blood type
1.
2.
3.
4.
type
type
type
type
A
B
AB
O
=
=
=
=
IAIA or IAi
IBIB or IBi
IAIB
ii
copyright cmassengale
70
This is codominance. You see red and you
see white. No pink
copyright cmassengale
71
Codominance Problem
Example: homozygous male Type B (IBIB)
x
heterozygous female Type A (IAi)
IA
i
IB
IAIB
IBi
IB
IAIB
IBi
copyright cmassengale
1/2 = IAIB
1/2 = IBi
72
Another Codominance Problem
• Example: male Type O (ii)
x
female type AB (IAIB)
IA
IB
i
IAi
IBi
i
IAi
IBi
copyright cmassengale
1/2 = IAi
1/2 = IBi
73
*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)
AB (IAIB)
X
copyright cmassengale
girl - type
74
Codominance
Answer:
IA
IB
i
i
IAIB
ii
Parents:
genotypes = IAi and IBi
phenotypes = A and B
copyright cmassengale
75
*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
copyright cmassengale
76
*Sex-linked Traits
Example: Eye color in fruit flies
Sex Chromosomes
fruit fly
eye color
XX chromosome - female
Xy chromosome - male
copyright cmassengale
77
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.
Xr
Xr
RR = red eyed
Rr = red eyed
R
X
rr = white eyed
XY = male
Y
XX = female
copyright cmassengale
78
Sex-linked Trait Solution:
Xr
XR
XR
Xr
Y
Xr Y
Xr
XR
Xr
Xr Y
50% red eyed
female
50% white eyed
male
copyright cmassengale
79
*Female Carriers
copyright cmassengale
80
*Examples of human sex-linked traits and
disorders
Hemophilia – missing a protein necessary for
normal blood clotting
Duchenne Muscular dystrophy – doesn’t have the
right protein for muscles (muscle degeneration,
trouble walking and breathing, early death)
Red-Green Colorblindness – cannot tell the
difference between red and green – see as
shades of gray
copyright cmassengale
81
https://www.youtube.com/watch?v=
FS64UXi74lY
copyright cmassengale
82
A woman who is a carrier for Duchenne muscular dystrophy
marries a man who is normal.
a. what are the chances they will have a child that is a
carrier? ________
b. what are the chances they will have a child that has the
disorder? _______
c. what are the chances they will have a child that is totally
normal? ________
d. They have a boy. What are the chances he has the
disorder? ________
e. They have a girl. What are the chances she has the
disorder?_________
copyright cmassengale
83
Color blindness is sometimes acquired. Chronic
illnesses that can lead to color blindness include
Alzheimer's disease, diabetes mellitus, glaucoma,
leukemia, liver disease, chronic alcoholism,
macular degeneration, multiple sclerosis,
Parkinson's disease, sickle cell anemia , and
retinitis pigmentosa. Accidents or strokes that
damage the retina or affect particular areas of
the brain eye can lead to color blindness. Some
medications such as antibiotics , barbiturates,
anti-tuberculosis drugs, high blood pressure
medications, and several medications used to
treat nervous disorders and psychological
problems may cause color blindness. Industrial or
environmental chemicals such as carbon monoxide,
carbon disulfide, fertilizers, styrene, and some
containing lead can cause loss of color vision.
Occasionally, changes can occur in the affected
person's capacity to see colors after age 60.
http://www.encyclopedia.com/topic/color_blindness.aspx
Epistasis – When the presence of a
gene in one location prevents the
expression of a trait somewhere else
Penetrance – The percent of the
population that has the gene and
expresses it. Tay Sachs has 100%
penetrance. Retinoblastoma has 90%
Expressivity – The degree to which a
trait is expressed in an individual.
copyright cmassengale
85
*Types
of mutations
Somatic cells = body cells
Germ cells = gametes
1. Deletion
2. Addition
3. Inversion
4. Translocation
copyright cmassengale
86
*Human Genetic Diseases and Disorders
Sickle Cell Anemia – When the RBC
changes shape because of a
mutation. Causes blood clots and
damages organs. More common in
African-Americans
Cystic Fibrosis – a transport
protein is missing in the lungs and
a thick mucus develops that clogs
lungs and organs common in whites
copyright cmassengale
87
*Tay Sachs – An enzyme is missing
to break dawn fats in the brain.
Brain cells are killed – always
fatal. More common in Jews
PKU – phenylketonuria cannot
break down a specific amino acid,
can lead to brain damage, seizures
and mental retardation.
copyright cmassengale
88
*Marfan syndrome – taller than
average, longer wing span than
height – has weakened heart
valves and aorta. Can lead to
sudden death
copyright cmassengale
89
Pedigree Symbols
Way to visually represent human family
relationships and infer inheritance patterns.
1234
Male
Female
marriage/mating
offspring in order of birth
deceased individuals
individuals showing trait
carriers
90
Pedigree Analysis
A pedigree
Is a family tree that describes the
interrelationships of parents and
children across generations
Can also be used to make predictions
about future offspring
Inheritance patterns of particular traits
Can be traced and described using pedigrees
Ww
ww
Ww ww ww Ww
WW
or
Ww
ww
Ww
ww
Second generation
(parents plus aunts
and uncles)
(a) Dominant trait (widow’s peak)
Ff
FF or Ff
Ff
Ff
Third
generation
(two sisters)
ww
Widow’s peak
Figure 14.14 A, B
Ww
First generation
(grandparents)
No Widow’s peak
Attached earlobe
ff
ff
Ff
Ff
Ff
ff
ff
FF
or
Ff
Free earlobe
(b) Recessive trait (attached earlobe)
Many genetic disorders
Are inherited in a recessive manner
Recessively inherited disorders
Show up only in individuals homozygous for the
allele
Carriers
Are heterozygous individuals who carry the
recessive allele but are phenotypically normal
Genetic Practice
Problems
copyright cmassengale
95
Breed the P1 generation
tall (TT) x dwarf (tt) pea plants
t
t
T
T
copyright cmassengale
96
Solution:
tall (TT) vs. dwarf (tt) pea plants
t
t
T
Tt
Tt
produces the
F1 generation
T
Tt
Tt
All Tt = tall
(heterozygous tall)
copyright cmassengale
97
Breed the F1 generation
tall (Tt) vs. tall (Tt) pea plants
T
t
T
t
copyright cmassengale
98
Solution:
tall (Tt) x tall (Tt) pea plants
T
t
T
TT
Tt
t
Tt
tt
produces the
F2 generation
1/4 (25%) = TT
1/2 (50%) = Tt
1/4 (25%) = tt
1:2:1 genotype
3:1 phenotype
copyright cmassengale
99