Download - Non-Mendelian Problems
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, there 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 bad is no gene for this trait on the Y chromosome to cancel out a bad gene on the X chromosomegene 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 gametesABCD ABCd AbCD AbCdABCD ABCd AbCD AbCdaBCD 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: 19:3:3:1 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
i Phenotypic 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 his sister has blood type AB, sister has blood type AB, What are What are the genotypes and the genotypes and phenotypes of phenotypes of their parents?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