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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


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


•• 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


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


•• 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


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)


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


•• Example:Example: homozygous male B (IBIB)x

heterozygous female A (IAi)

IAIB IAIB

IBi IBi

1/2 = IAIB1/2 = IBi

IA

IB IB

i


•• Example:Example: male O (ii) x female AB (IAIB)

IAi IBi

IAi IBi

1/2 = IAi1/2 = IBi

i

IA IB

i


•• 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)


•• 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%