Non-Mendelian Problems. I Sex-linked Traits These are Traits (genes) that are located on the sex...

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Non-Mendelian Problems

I Sex-linked TraitsI Sex-linked Traits• These are Traits (genes) that are located on the These are Traits (genes) that are located on the sex chromosomes.sex chromosomes.• Sex chromosomes are Sex chromosomes are X and YX and Y• XXXX genotype for females genotype for females• XYXY genotype for males genotype for males• Many Many sex-linked traitssex-linked traits are carried on are carried on XX chromosome of the sex chromosome of the sex

chromosomeschromosomes• That is why these genetic disorders are found mainly in males, That is why these genetic disorders are found mainly in males,

there is no gene for this trait on the Y chromosome to cancel out a there is no gene for this trait on the Y chromosome to cancel out a bad gene on the X chromosomebad gene on the X chromosome

Sex-linked TraitsSex-linked Traits

Sex ChromosomesSex Chromosomes

XX chromosome - female Xy chromosome – male-the trait will be determinedby the gene on the X, none On the Y

fruit flyeye color

Example: Example: Eye color in fruit Eye color in fruit fliesflies

Sex-linked Trait ProblemSex-linked Trait Problem• Use the same principles used in the Mendelian Monohybrid problems, except that the sex of

the offspring must be included and the Y chromosome will not have an allele• 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 eyed• Rr = red eyed• rr = white eyed• XY = male• XX = female

XR

Xr Xr

Y

POSSIBLE GENOTYPES IN SEX-LINKED POSSIBLE GENOTYPES IN SEX-LINKED PROBLEMS:PROBLEMS:

XXRRXXRR—FEMALE w/ HOMOZYGOUS DOM—FEMALE w/ HOMOZYGOUS DOM

XXRRXXrr—FEMALE w/ HETERZYGOUS —FEMALE w/ HETERZYGOUS

XXrrXXrr—FEMALE w/ HOMOZYGOUS REC—FEMALE w/ HOMOZYGOUS REC

XXRRY—MALE w/ DOM ALLELEY—MALE w/ DOM ALLELE

XXrrY—MALE w/ REC ALLELEY—MALE w/ REC ALLELE

Sex-linked Trait Sex-linked Trait Solution:Solution:

XR Xr

Xr Y

XR Xr

Xr Y

Genotypic Ratio:50% XR Xr

50% Xr Y

Phenotypic Ratio:50 % white eyed male 50 % red eyed female

XR

Xr Xr

Y

Sex-linked Cross ?s from previous problem

1. What % of the males will be red eyed?

2. What % of the offspring will be red eyed?

3. What % of the offspring will be males?

4. What % of the females will be white eyed?

5. What % of the females will be red eyed?

6. What % of the offspring will be white eyed?

Female CarriersFemale Carriers

*1/2 filled in box=carrier, filled in box=affected individual

Incomplete Incomplete DominanceDominance

Incomplete DominanceIncomplete Dominance• F1 hybrids F1 hybrids have an appearance somewhat in betweenin between the

phenotypes phenotypes of the two parental varieties. There is a mixing of the two traits, neither is dominant over the other. Worked like Monohybrid problems except that you will use all capitals letter for each trait, ex. Red=RR, white=WW Pink=RW

• Ex:Ex: snapdragons (flower)snapdragons (flower)• red (RR) x white (WW)• RW=pink flower• RR = red flowerRR = red flower• WW = white flower

R

R

W W

*Fill in the square to the left

Incomplete Incomplete DominanceDominance

RWRW

RWRW

RWRW

RWRW

RR

RR

WW

Genotypic Ratio:Genotypic Ratio:0:4:0—100%RW0:4:0—100%RWPhenotypic Ratio:Phenotypic Ratio:0:4:0—100%pink0:4:0—100%pink

produces theproduces theFF11 generation generation

W

Incomplete Dominance Problem:

• In cattle when a red bull(RR) is mated with white(WW) cow the offspring are roan(RW) a blending of red and white. Mate a red bull with a roan cow. Use the format on the next slide and give the P1, do the Punnett Square, and give the genotypic and phenotypic ratios for F1

generation of this cross.

P1 = __RR__ x __RW__

Phenotypic ratio: ____ : _____ : _____

Genotypic ratio: ____ : _____ : _____

P1 = __RR__ x __RW__

Phenotypic ratio: ____ : _____ : _____

Genotypic ratio: ____ : _____ : _____

R

R

WR

22

RR RW

RR RW

0

0

22

or 50%RR,50%RW

or 50%RED,50%ROAN

Incomplete DominanceIncomplete Dominance

Dihybrid CrossDihybrid Cross• A breeding experiment that tracks the A breeding experiment that tracks the

inheritance of two traitsinheritance of two traits..

• Mendel’s Mendel’s “Law of Independent “Law of Independent Assortment”Assortment”

• a. Each pair of alleles segregates a. Each pair of alleles segregates independentlyindependently during gamete formation during gamete formation

• b. Formula: 2b. Formula: 2nn (n = # of heterozygotes) (n = # of heterozygotes)

Question:Question:How many gametes will be produced How many gametes will be produced for the following allele arrangements?for the following allele arrangements?

• Remember: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 gametes = 4 gametes

RY Ry rY ryRY Ry rY ry

2. AaBbCCDd: 22. AaBbCCDd: 2nn = 2 = 233 = 8 gametes = 8 gametes

ABCD ABCd AbCD AbCdABCD ABCd AbCD AbCd

aBCD aBCd abCD abCDaBCD aBCd abCD abCD

3. MmNnOoPPQQRrssTtQq: 23. MmNnOoPPQQRrssTtQq: 2nn = 2 = 266 = = 64 gametes64 gametes

Dihybrid CrossDihybrid Cross

• Traits: Seed shape & Seed colorTraits: Seed shape & Seed color• Alleles:Alleles: R round

r wrinkled Y yellow y green

• RrYy x RrYy

RY Ry rY ryRY Ry rY ry RY Ry rY ryRY Ry rY ry

All possible gamete combinations All possible gamete combinations by FOIL methodby FOIL method

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

Dihybrid CrossDihybrid Cross

Round/Yellow: 9Round/green: 3wrinkled/Yellow: 3wrinkled/green: 1

9:3:3:1

CODOMINANCECODOMINANCE

Multiple Alleles /CodominanceMultiple Alleles /Codominance

• Non-Mendelian Cross where 2 allelesNon-Mendelian Cross where 2 alleles are expressed are expressed ((multiple allelesmultiple alleles) in ) in heterozygous individualsheterozygous individuals..

• Example:Example: blood type blood typeUse the genotypes below whenever doing blood type Use the genotypes below whenever doing blood type

crosses.crosses.• 1.1. type Atype A = AA -pure or AO -hybrid= AA -pure or AO -hybrid• 2.2. type Btype B = BB -pure or BO -hybrid= BB -pure or BO -hybrid• 3.3. type ABtype AB = AB= AB - -codominantcodominant• 4.4. type Otype O = OO -pure= OO -pure

Codominance Codominance ProblemProblem

• Example:Cross a male who is homozygous Type B (BB) x a female that is heterozygous Type A (AO)

IAIB IBi

IAIB IBi

Genotypic ratio:50% IAIB

50%= IBi

IB

IA i

IB

Phenotypic ratio:50% type AB

50%= type B

Another Codominance ProblemAnother Codominance Problem• Example:Example: Cross a

male Type O (ii) x female type AB (IAIB)

i

IA IB

i

*Give the genotypic and phenotypic ratios of the offspring

Another Codominance Another Codominance ProblemProblem

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

IAi IBi

IAi IBi

Genotypic Ratio:50% IAi50% IBi

i

IA IB

iPhenotypic Ratio:50% type A50% type B

CodominanceCodominance• QuestionQuestion::

If a boy has a blood type O and If a boy has a blood type O and his sister has blood type AB, his sister has blood type AB, What are the genotypes and What are the genotypes and phenotypes of their parents?phenotypes of their parents?

• boy - boy - type O (ii) type O (ii) X girl - X girl - type AB (Itype AB (IAAIIBB))

CodominanceCodominance• Answer:Answer:

IAIB

ii

Parents:Parents:genotypesgenotypes = IAi and IBiphenotypesphenotypes = A and B

IB

IA i

i