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GENETICSGENETICS
GeneticsGenetics
• The study of heredityheredity.
•• Gregor Mendel (1860Gregor Mendel (1860’’s)s) discovered the fundamental principles principles of geneticsgenetics by breedingbreedinggarden peasgarden peas.
GeneticsGenetics•• AllelesAlleles
1. Alternative forms of genes.genes.
2. Units that determine heritable traits.
3. Dominant alleles Dominant alleles (TTTT - tall pea plantstall pea plants)a. homozygous dominanta. homozygous dominant
4. Recessive alleles Recessive alleles (tt tt - dwarf pea plantsdwarf pea plants)a. homozygous recessivea. homozygous recessive
5. HeterozygousHeterozygous (TtTt - tall pea plantstall pea plants)
PhenotypePhenotype
•• Outward appearanceOutward appearance•• Physical characteristicsPhysical characteristics
•• Examples:Examples:1.1. tall pea planttall pea plant2.2. dwarf pea plantdwarf pea plant
GenotypeGenotype•• Arrangement of genes that produces the Arrangement of genes that produces the
phenotypephenotype•• Example:Example:
1.1. tall pea planttall pea plantTT = tall (homozygous dominant)(homozygous dominant)
2.2. dwarf pea plantdwarf pea planttt = dwarf (homozygous recessive)(homozygous recessive)
3.3. tall pea planttall pea plantTt = tall (heterozygous)(heterozygous)
Punnett squarePunnett square
• A Punnett square Punnett square is used to show the possible combinationscombinations of gametesgametes.
Breed the P generationP generation
•• tall (TT) vs. dwarf (tt) pea plantstall (TT) vs. dwarf (tt) pea plants
t
t
T T
tall (TT) vs. dwarf (tt) pea plantstall (TT) vs. dwarf (tt) pea plants
t
t
T T
Tt
Tt
Tt
Tt All Tt = tall(heterozygous tall)
produces theFF11 generationgeneration
Breed the FF11 generationgeneration
•• tall (Tt) vs. tall (Tt) pea plantstall (Tt) vs. tall (Tt) pea plants
T
t
T t
tall (Tt) vs. tall (Tt) pea plantstall (Tt) vs. tall (Tt) pea plants
TT
Tt
Tt
tt
T
t
T t
produces theFF22 generationgeneration
1/4 (25%) = TT1/2 (50%) = Tt1/4 (25%) = tt
1:2:1 genotype1:2:1 genotype3:1 phenotype3:1 phenotype
Monohybrid CrossMonohybrid Cross
• A breeding experiment that tracks the inheritance of a single trait.single trait.
•• MendelMendel’’s s ““principle of segregationprinciple of segregation””a. pairs of genes separate during gamete gamete
formation (meiosis).(meiosis).
b. the fusion of gametesgametes at fertilization pairs genes once again.
Homologous ChromosomesHomologous Chromosomes
eye color locusb = blue eyes
eye color locusB = brown eyes
Paternal Maternal
This person would have brown eyes (Bb)
Meiosis Meiosis -- eye coloreye color
Bb
diploid (2n)
B
b
meiosis I
B
B
b
b
sperm
haploid (n)
meiosis II
Monohybrid CrossMonohybrid Cross
•• ExampleExample: Cross between two heterozygotesheterozygotesfor brown eyes (Bb)
BB = brown eyesBb = brown eyesbb = blue eyes
B
b
B b
Bb x Bb
malegametes
female gametes
Monohybrid CrossMonohybrid Cross
BB
Bb
Bb
bb
B
b
B b
Bb x Bb
1/4 = BB - brown eyed1/2 = Bb - brown eyed1/4 = bb - blue eyed
1:2:1 genotype3:1 phenotype
Dihybrid CrossDihybrid Cross
• A breeding experiment that tracks the inheritance of two traits.two traits.
•• MendelMendel’’s s ““principle of independent assortmentprinciple of independent assortment””
a. each pair of alleles segregates independently during gamete formation (metaphase I)(metaphase I)
b. formula: 22nn (n = # of heterozygotes)(n = # of heterozygotes)
Independent AssortmentIndependent Assortment•• Question: Question: How many gametes will be produced
for the following allele arrangements?
• Remember: 22nn (n = # of heterozygotes)(n = # of heterozygotes)
1.1. RrYyRrYy
2.2. AaBbCCDdAaBbCCDd
3.3. MmNnOoPPQQRrssTtQqMmNnOoPPQQRrssTtQq
Answer:Answer:1. RrYy: 21. RrYy: 2nn = 2= 222 = = 4 gametes4 gametes
RY Ry rY ryRY Ry rY ry
2. AaBbCCDd: 22. AaBbCCDd: 2nn = 2= 233 = = 8 gametes8 gametesABCD ABCd AbCD AbCdABCD ABCd AbCD AbCdaBCD aBCd abCD abCD aBCD aBCd abCD abCD
3. MmNnOoPPQQRrssTtQq: 23. MmNnOoPPQQRrssTtQq: 2nn = 2= 266 = = 64 gametes64 gametes
Dihybrid CrossDihybrid Cross
•• Example:Example: cross between roundround and yellowyellowheterozygous pea seeds.
RR = round= roundrr = wrinkled= wrinkledYY = yellow= yellowyy = green= green
RY Ry rY ry RY Ry rY ry x RY Ry rY ryRY Ry rY rypossible gametes produced
RrYyRrYy x RrYyRrYy
Dihybrid CrossDihybrid CrossRYRY RyRy rYrY ryry
RYRY
RyRy
rYrY
ryry
Dihybrid CrossDihybrid 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
RYRY RyRy rYrY ryry
RYRY
RyRy
rYrY
ryry
Test CrossTest Cross• A mating between an individual of unknown genotypeunknown genotype
and a homozygous recessivehomozygous recessive individual.•• Example:Example: bbC__ bbC__ x bbccbbcc
BB = brown eyesBb = brown eyesbb = blue eyes
CC = curly hairCc = curly haircc = straight hair
bCbC b___b___
bcbc
Test CrossTest Cross
•• Possible results:Possible results:
bCbC b___b___
bcbc bbCc bbCc
C bCbC b___b___
bcbc bbCc bbccor
c
Incomplete DominanceIncomplete Dominance•• F1 hybrids F1 hybrids have an appearance somewhat in in
betweenbetween the phenotypes phenotypes of the two parental varieties.
•• Example:Example: snapdragons (flower)snapdragons (flower)• red (RR) x white (rr)
RR = red flowerRR = red flowerrr = white flower
r
r
R R
Incomplete DominanceIncomplete Dominance
Rr
Rr
Rr
Rr
r
r
R R
All Rr = pink(heterozygous pink)
produces theFF11 generationgeneration
CodominanceCodominance
•• Two allelesTwo alleles are expressed (multiple allelesmultiple alleles) in heterozygous individualsheterozygous individuals.
•• Example:Example: bloodblood
1. type A = IAIA or IAi2. type B = IBIB or IBi3. type AB = IAIB4. type O = ii
CodominanceCodominance
•• Example:Example: homozygous male B (IBIB)x
heterozygous female A (IAi)
IAIB IAIB
IBi IBi
1/2 = IAIB1/2 = IBi
IA
IB IB
i
CodominanceCodominance
•• Example:Example: male O (ii) x female AB (IAIB)
IAi IBi
IAi IBi
1/2 = IAi1/2 = IBi
i
IA IB
i
CodominanceCodominance
•• QuestionQuestion: 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)
CodominanceCodominance
•• Answer:Answer:
IAIB
ii
Parents:Parents:genotypesgenotypes = IAi and IBiphenotypesphenotypes = A and B
IB
IA i
i
SexSex--linked Traitslinked Traits
• Traits (genes) located on the sex sex chromosomeschromosomes
•• Example:Example: fruit fliesfruit flies
(redred-eyed male) X (whitewhite-eyed female)
SexSex--linked Traitslinked TraitsSex ChromosomesSex Chromosomes
XX chromosome - female Xy chromosome - male
fruit flyeye color
SexSex--linked Traitslinked Traits•• Example:Example: fruit fliesfruit flies
(red-eyed male) X (white-eyed female)•• Remember:Remember: the Y chromosomeY chromosome in males
does not carry traits.
RR = red eyedRr = red eyedrr = white eyed
Xy = maleXX = female
Xr
XR y
Xr
SexSex--linked Traitslinked Traits
XR Xr
XR Xr
Xr y
Xr y
1/2 red eyed and female1/2 white eyed and male
Xr
XR y
Xr
Population GeneticsPopulation Genetics• The study of genetic changesgenetic changes in populationspopulations.
• The science of microevolutionary changesmicroevolutionary changes in populationspopulations.
•• HardyHardy--Weinberg equilibrium:Weinberg equilibrium:the principle that shuffling of genes that occurs during sexual reproduction, by itself, cannot change the overall genetic makeup of a population.
•• HardyHardy--Wienberg equation:Wienberg equation: 1 = p1 = p22 + 2pq + q+ 2pq + q22
Question:Question:
•• How do we get this equation?How do we get this equation?
Answer:Answer: ““SquareSquare”” 1 = p + q1 = p + q
↓↓1122 = (p + q)= (p + q)22
↓↓1 = p1 = p22 + 2pq + q+ 2pq + q22
HardyHardy--Wienberg equationWienberg equation
•• Five conditions Five conditions are required for Hardy-Wienberg equilibrium.1.1. large populationlarge population2.2. isolated populationisolated population3.3. no net mutationsno net mutations4.4. random matingrandom mating5.5. no natural selectionno natural selection
•• Need to remember the following:Need to remember the following:
pp2 2 = homozygous dominant= homozygous dominant2pq = heterozygous2pq = heterozygousqq22 = homozygous recessive= homozygous recessive
ImportantImportant
• Iguanas with webbed feet (recessive trait) (recessive trait) make up 4% of the population. What in the population is heterozygousheterozygous and homozygoushomozygous dominantdominant.
Question:Question:
Answer:Answer:
1. q1. q2 2 = 4% or .04= 4% or .04 qq22 = .04= .04 q = .2q = .2
2. then use 1 = p + q2. then use 1 = p + q1 = p + .21 = p + .2 1 1 -- .2 = p.2 = p .8 = p.8 = p
3. for heterozygous use 2pq3. for heterozygous use 2pq2(.8)(.2) = .32 or 32%2(.8)(.2) = .32 or 32%
4. For homozygous dominant use p4. For homozygous dominant use p22
.8.822 = .64 or 64%= .64 or 64%
HardyHardy--Wienberg equationWienberg equation
1 = p1 = p22 + 2pq + q+ 2pq + q22
•• 64% = p64% = p2 2 = homozygous dominant= homozygous dominant•• 32% = 2pq32% = 2pq = heterozygous= heterozygous•• 04% 04% = q= q22 = homozygous recessive= homozygous recessive•• 100%100%
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