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1
Mendelelian Genetics
copyright cmassengale
2
Gregor Mendel
(1822-1884)Responsible for the Laws governing
Inheritance of Traits
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Gregor Johann MendelAustrian monkStudied the inheritance of traits in pea plantsDeveloped the laws of inheritanceMendel's work was not recognized until the turn of the 20th century copyright cmassengale
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Gregor Johann MendelBetween 1856 and 1863, Mendel cultivated and tested some 28,000 pea plantsHe found that the plants' offspring retained traits of the parentsCalled the “Father of Genetics"
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Site of Gregor Mendel’s experimental garden in the Czech Republic
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Mendel stated that physical traits are inherited as “particles”Mendel did not know that the “particles” were actually Chromosomes & DNA
Particulate Inheritance
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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
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Use Your Noodle!Answer the following in your science notebook.
What traits are found in your family?
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You have 2 minutes
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Types of Genetic Crosses
Monohybrid cross - cross involving a single traite.g. flower color
Dihybrid cross - cross involving two traits e.g. flower color & plant height
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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
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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)
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Genotype & Phenotype in Flowers
Genotype of alleles:R = red flowerr = yellow flowerAll genes occur in pairs, so 2 alleles affect a characteristicPossible combinations are:Genotypes RR Rr rrPhenotypesRED RED YELLOW
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Use Your Noodle!Answer the following in
your science notebook.
Determine which allele is dominate and recessive?
RrGGHhff
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You have 2 minutes
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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 copyright cmassengale
Use Your Noodle!Answer the following in
your science notebook.
Determine which allele is Homozygous or Heterozygous?
RrGGHhff
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You have 2 minutes
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Genes and Environment Determine Characteristics
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Mendel’s Pea Plant
Experiments
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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
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Mendel’s Experimental
MethodsMendel hand-pollinated flowers using a paintbrush
He could snip the stamens to prevent self-pollinationCovered each flower with a cloth bag
He traced traits through the several generations
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How Mendel BeganMendel produced pure strains by allowing the plants to self-pollinate for several generations copyright cmassengale
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Eight Pea Plant TraitsSeed 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)
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22copyright cmassengale
23copyright cmassengale
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Mendel’s Experimental Results
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Following the Generations
Cross 2 Pure
PlantsTT x tt
Results in all
HybridsTt
Cross 2 Hybrids
get3 Tall & 1
ShortTT, Tt, tt
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Punnett SquareUsed to help solve genetics problems
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27copyright cmassengale
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Monohybrid Crosses
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Trait: Seed ShapeAlleles: R – Round r – WrinkledCross: Round seeds x Wrinkled
seedsRR x rr
P1 Monohybrid Cross
R
R
rr
Rr
RrRr
Rr
Genotype: RrPhenotype: RoundGenotypicRatio: All alikePhenotypicRatio: All alike
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P1 Monohybrid Cross Review
Homozygous dominant x Homozygous recessive
Offspring all Heterozygous (hybrids)
Offspring called F1 generation Genotypic & Phenotypic ratio
is ALL ALIKE
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Trait: Seed ShapeAlleles: R – Round r – WrinkledCross: Round seeds x Round seeds
Rr x Rr
F1 Monohybrid Cross
R
r
rR
RR
rrRr
Rr
Genotype: RR, Rr, rrPhenotype: Round & wrinkledG.Ratio: 1:2:1P.Ratio: 3:1
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F1 Monohybrid Cross Review
Heterozygous x heterozygous Offspring:
25% Homozygous dominant RR50% Heterozygous Rr25% Homozygous Recessive rr
Offspring called F2 generation Genotypic ratio is 1:2:1 Phenotypic Ratio is 3:1
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What Do the Peas Look Like?
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…And Now the Test CrossMendel 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 HybridHomozygous recessive x Hybrid
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Trait: Seed ShapeAlleles: R – Round r – WrinkledCross: Round seeds x Round seeds
RR x Rr
F2 Monohybrid Cross (1st)
R
R
rR
RR
RrRR
Rr
Genotype: RR, RrPhenotype: RoundGenotypicRatio: 1:1PhenotypicRatio: All alikecopyright cmassengale
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Trait: Seed ShapeAlleles: R – Round r – WrinkledCross: Wrinkled seeds x Round seeds
rr x Rr
F2 Monohybrid Cross (2nd)
r
r
rR
Rr
rrRr
rr
Genotype: Rr, rrPhenotype: Round & WrinkledG. Ratio: 1:1P.Ratio: 1:1
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F2 Monohybrid Cross Review
Homozygous x heterozygous(hybrid)
Offspring:50% Homozygous RR or rr50% Heterozygous Rr
Phenotypic Ratio is 1:1 Called Test Cross because the
offspring have SAME genotype as parents
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Practice Your Crosses
Work the P1, F1, and both F2 Crosses for each of the other
Seven Pea Plant Traits
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Mendel’s Laws
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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 samecopyright cmassengale
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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)
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Law of Dominance
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Law of SegregationDuring 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.
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Applying the Law of Segregation
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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.
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Dihybrid CrossA 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
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Question:How many gametes will be
produced for the following allele arrangements?
Remember: 2n (n = # of heterozygotes)
1. RrYy
2. AaBbCCDd
3. MmNnOoPPQQRrssTtQqcopyright cmassengale
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Answer:1. RrYy: 2n = 22 = 4 gametes
RY Ry rY ry
2. AaBbCCDd: 2n = 23 = 8 gametesABCD ABCd AbCD AbCdaBCD aBCd abCD abCD
3. MmNnOoPPQQRrssTtQq: 2n = 26 = 64 gametes
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Dihybrid CrossTraits: Seed shape & Seed colorAlleles: R round
r wrinkled Y yellow y green
RrYy x RrYy
RY Ry rY ry RY Ry rY ry
All possible gamete combinationscopyright cmassengale
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Dihybrid CrossRY Ry rY ry
RY
Ry
rY
ry
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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: 19:3:3:1 phenotypic ratio
RY Ry rY ry
RY
Ry
rY
ry
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Dihybrid Cross
Round/Yellow: 9Round/green: 3wrinkled/Yellow: 3wrinkled/green: 1
9:3:3:1
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Test CrossA mating between an individual of
unknown genotype and a homozygous recessive individual.
Example: bbC__ x bbcc
BB = brown eyesBb = brown eyesbb = blue eyes
CC = curly hairCc = curly haircc = straight hair
bC b___
bc
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Test CrossPossible results:
bC b___
bc bbCc bbCc
C bC b___
bc bbCc bbccor
c
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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
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Incomplete Dominanceand
Codominance
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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)
RR = red flowerrr = white flower
R
R
r r
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Incomplete Dominance
Rr
Rr
Rr
Rr
R
R
r
All Rr = pink(heterozygous pink)
produces theF1 generation
r
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Incomplete Dominance
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CodominanceTwo alleles are expressed
(multiple alleles) in heterozygous individuals.
Example: blood type
1. type A = IAIA or IAi2. type B = IBIB or IBi3. type AB= IAIB
4. type O = iicopyright cmassengale
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Codominance ProblemExample: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
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Another Codominance Problem
• Example: male Type O (ii) x female type AB (IAIB)
IAi IBi
IAi IBi
1/2 = IAi1/2 = IBi
i
IA IB
i
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CodominanceQuestion:
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)
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CodominanceAnswer:
IAIB
ii
Parents:genotypes = IAi and IBiphenotypes = A and B
IB
IA i
i
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Sex-linked Traits
Traits (genes) located on the sex chromosomes
Sex chromosomes are X and YXX genotype for femalesXY genotype for malesMany sex-linked traits carried
on X chromosome
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Sex-linked Traits
Sex Chromosomes
XX chromosome - female Xy chromosome - male
fruit flyeye color
Example: Eye color in fruit flies
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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 eyedRr = red eyedrr = white eyedXY = maleXX = female
XR
Xr Xr
Y
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Sex-linked Trait Solution:
XR Xr
Xr Y
XR Xr
Xr Y
50% red eyed female
50% white eyed male
XR
Xr Xr
Y
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Female Carriers
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Genetic Practice Problems
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Breed the P1 generation
tall (TT) x dwarf (tt) pea plants
T
T
t t
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Solution:
T
T
t t
Tt
Tt
Tt
Tt All Tt = tall(heterozygous tall)
produces theF1 generation
tall (TT) vs. dwarf (tt) pea plants
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Breed the F1 generation
tall (Tt) vs. tall (Tt) pea plants
T
t
T t
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Solution:
TT
Tt
Tt
tt
T
t
T tproduces theF2 generation1/4 (25%) = TT1/2 (50%) = Tt1/4 (25%) = tt1:2:1 genotype 3:1 phenotype
tall (Tt) x tall (Tt) pea plants
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75copyright cmassengale