GENETICSGENETICS

GeneticsGenetics

• The study of heredityheredity.

• Gregor Mendel (1860’s)Gregor Mendel (1860’s) discovered the fundamental principles principles of genetics genetics by breedingbreeding garden 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 plant

2.2. 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 plant

TT = tall (homozygous dominant)(homozygous dominant)

2.2. dwarf pea plantdwarf pea plant

tt = dwarf (homozygous recessive)(homozygous recessive)

3.3. tall pea planttall pea plant

Tt = 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 generation generation

Breed the FF11 generation generation

• 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 generation generation

1/4 (25%) = TT1/2 (50%) = Tt1/4 (25%) = tt

1:2:1 genotype1:2:1 genotype 3:1 phenotype3:1 phenotype

Monohybrid CrossMonohybrid Cross

• A breeding experiment that tracks the inheritance of a single trait.single trait.

• Mendel’s “principle of segregation”Mendel’s “principle 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 - eye colorMeiosis - eye color

Bb

diploid (2n)

B

b

meiosis I

B

B

b

b

sperm

haploid (n)

meiosis II

Monohybrid CrossMonohybrid Cross

• ExampleExample: Cross between two heterozygotesheterozygotes for brown eyes (Bb)

BB = brown eyes

Bb = brown eyes

bb = blue eyesB

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 genotype 3:1 phenotype

Dihybrid CrossDihybrid Cross

• A breeding experiment that tracks the inheritance of two traits.two traits.

• Mendel’s “principle of independent assortment”Mendel’s “principle of independent assortment”

a. each pair of alleles segregates independently during gamete formation (metaphase I)(metaphase I)

b. formula: 2 2nn (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 gametes

ABCD ABCd AbCD AbCdABCD ABCd AbCD AbCd

aBCD 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 yellowyellow heterozygous pea seeds.

RR = round= round

rr = wrinkled= wrinkled

YY = yellow= yellow

yy = green= greenRY Ry rY ry RY Ry rY ry x RY Ry rY ryRY Ry rY ry possible gametes produced

RrYyRrYy x RrYyRrYy

Dihybrid CrossDihybrid Cross

RYRY 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 eyes

Bb = brown eyes

bb = blue eyes

CC = curly hair

Cc = curly hair

cc = 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 flower

rr = white flower

r

r

R R

Incomplete DominanceIncomplete Dominance

Rr

Rr

Rr

Rr

r

r

R R

All Rr = pink(heterozygous pink)

produces theFF11 generation generation

CodominanceCodominance

• Two allelesTwo alleles are expressed (multiple allelesmultiple alleles) in heterozygous individualsheterozygous individuals.

• Example:Example: blood blood

1. type A = IAIA or IAi

2. type B = IBIB or IBi

3. type AB = IAIB

4. type O = ii

CodominanceCodominance

• Example:Example: homozygous male B (IBIB)

x heterozygous female

A (IAi)

IAIB IAIB

IBi IBi

1/2 = IAIB

1/2 = 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

Sex-linked TraitsSex-linked Traits

• Traits (genes) located on the sex sex chromosomeschromosomes

• Example:Example: fruit fliesfruit flies

(redred-eyed male) X (whitewhite-eyed female)

Sex-linked TraitsSex-linked Traits

Sex ChromosomesSex Chromosomes

XX chromosome - female Xy chromosome - male

fruit flyeye color

Sex-linked TraitsSex-linked 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 eyed

Rr = red eyed

rr = white eyed

Xy = male

XX = female

Xr

XR y

Xr

Sex-linked TraitsSex-linked 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.

• Hardy-Weinberg equilibrium:Hardy-Weinberg equilibrium:

the principle that shuffling of genes that occurs during sexual reproduction, by itself, cannot change the overall genetic makeup of a population.

• Hardy-Wienberg equation:Hardy-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: “Square”“Square” 1 = p + q1 = p + q

1122 = (p + q) = (p + q)22

1 = p1 = p22 + 2pq + q + 2pq + q22

Hardy-Wienberg equationHardy-Wienberg equation

• Five conditions Five conditions are required for Hardy-Wienberg equilibrium.

1.1. large populationlarge population

2.2. isolated populationisolated population

3.3. no net mutationsno net mutations

4.4. random matingrandom mating

5.5. no natural selectionno natural selection

• Need to remember the following:Need to remember the following:

pp2 2 = homozygous dominant= homozygous dominant

2pq = heterozygous2pq = heterozygous

qq22 = 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 - .2 = p1 - .2 = p .8 = p.8 = p

3. for heterozygous use 2pq3. for heterozygous use 2pq

2(.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%

Hardy-Wienberg equationHardy-Wienberg 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%