1 mendelelian genetics copyright cmassengale. 2 gregor mendel ( 1822-1884) responsible for the laws...
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Gregor Mendel(1822-1884)
Responsible for the Laws governing
Inheritance of Traits
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Gregor Johann Mendel
Austrian monkStudied the inheritance of traits in pea plantsDeveloped the laws of inheritanceMendel's work was not recognized until the turn of the 20th century
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Gregor Johann Mendel
Between 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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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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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)
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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 rrPhenotypes RED RED YELLOW
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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
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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)
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Trait: Seed ShapeAlleles: R – Round r – WrinkledCross: Round seeds x Wrinkled seeds RR 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 recessiveOffspring all Heterozygous (hybrids)Offspring called F1 generationGenotypic & 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 heterozygousOffspring:25% Homozygous dominant RR50% Heterozygous Rr25% Homozygous Recessive rrOffspring called F2 generationGenotypic ratio is 1:2:1Phenotypic Ratio is 3:1
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…And Now the Test Cross
Mendel then crossed a pure & a hybrid from his F2 generationThis is known as an F2 or test crossThere 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 alike
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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 RrPhenotypic Ratio is 1:1Called 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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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
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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 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.
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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 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 formationb. Formula: 2n (n = # of heterozygotes)
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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. MmNnOoPPQQRrssTtQq
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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
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Dihybrid Cross
Traits: 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 combinations
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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: 1
9:3:3:1 phenotypic ratio
RY Ry rY ry
RY
Ry
rY
ry
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Test CrossA 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
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Test Cross
Possible results:
bC b___
bc bbCc bbCc
C bC b___
bc bbCc bbccor
c
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Summary of Mendel’s lawsLAW PARENT
CROSSOFFSPRING
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 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 flower rr = 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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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
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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
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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
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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)
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Codominance
Answer:
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 chromosomesSex 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 XrXrRemember: 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 female50% white eyed male
XR
Xr Xr
Y
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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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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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