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Non-Mendelian Problems. I Sex-linked Traits. These are Traits (genes) that are located on the sex chromosomes. Sex chromosomes are X and Y XX genotype for females XY genotype for males Many sex-linked traits are carried on X chromosome of the sex chromosomes
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I Sex-linked Traits • These are Traits (genes) that are located on the sex chromosomes. • Sex chromosomes are X and Y • XX genotype for females • XY genotype for males • Many sex-linked traits are carried on X chromosome of the sex chromosomes • 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 gene on the X chromosome
fruit fly eye color Xy chromosome – male -the trait will be determined by the gene on the X, none On the Y XX chromosome - female Sex-linked Traits Example: Eye color in fruit flies Sex Chromosomes
Xr Xr XR Y Sex-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
POSSIBLE GENOTYPES IN SEX-LINKED PROBLEMS: XRXR—FEMALE w/ HOMOZYGOUS DOM XRXr—FEMALE w/ HETERZYGOUS XrXr—FEMALE w/ HOMOZYGOUS REC XRY—MALE w/ DOM ALLELE XrY—MALE w/ REC ALLELE
Xr Xr XR Xr XR Xr XR Y Xr Y Xr Y Sex-linked Trait Solution: Genotypic Ratio: 50% XR Xr 50% Xr Y Phenotypic Ratio: 50 % white eyed male 50 % red eyed female
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 Carriers *1/2 filled in box=carrier, filled in box=affected individual
W W R R Incomplete Dominance • F1 hybrids have an appearance somewhat in between the 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:snapdragons (flower) • red (RR) x white (WW) • RW=pink flower • RR = red flower • WW = white flower *Fill in the square to the left
W produces the F1 generation RW RW R R RW RW Genotypic Ratio: 0:4:0—100%RW Phenotypic Ratio: 0:4:0—100%pink Incomplete Dominance 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__ Genotypic ratio: ____ : _____ : _____ Phenotypic ratio: ____ : _____ : _____
P1 = __RR__ x __RW__ R W R RR RW RW R RR 2 2 0 or 50%RR,50%RW Genotypic ratio: ____ : _____ : _____ 2 2 0 or 50%RED,50%ROAN Phenotypic ratio: ____ : _____ : _____
Dihybrid Cross • A breeding experiment that tracks the inheritance of two traits. • Mendel’s “Law of Independent Assortment” • a. Each pair of alleles segregates independently during gamete formation • b. Formula: 2n (n = # of heterozygotes)
Question:How many gametes will be produced for the following allele arrangements? • Remember:2n (n = # of heterozygotes) • 1. RrYy • 2. AaBbCCDd • 3. MmNnOoPPQQRrssTtQq
Answer: 1. RrYy: 2n = 22 = 4 gametes RY Ry rY ry 2. AaBbCCDd: 2n = 23 = 8 gametes ABCD ABCd AbCD AbCd aBCD aBCd abCD abCD 3. MmNnOoPPQQRrssTtQq: 2n = 26 = 64 gametes
Dihybrid Cross • Traits: Seed shape & Seed color • Alleles: R round r wrinkled Y yellow y green RrYy x RrYy RY Ry rY ry RY Ry rY ry All possible gamete combinations by FOIL method
RY Ry rY ry RY Ry rY ry Dihybrid Cross
RY Ry rY ry Round/Yellow: 9 Round/green: 3 wrinkled/Yellow: 3 wrinkled/green: 1 RY RRYY RRYy RrYY RrYy Ry RRYy RRyy RrYy Rryy rY RrYY RrYy rrYY rrYy ry RrYy Rryy rrYy rryy 9:3:3:1 phenotypic ratio Dihybrid Cross
Dihybrid Cross Round/Yellow: 9Round/green: 3wrinkled/Yellow: 3wrinkled/green: 1 9:3:3:1
Multiple Alleles /Codominance • Non-Mendelian Cross where 2 alleles are expressed (multiple alleles) in heterozygous individuals. • Example: blood type Use the genotypes below whenever doing blood type crosses. • 1. type A = AA -pure or AO -hybrid • 2. type B = BB -pure or BO -hybrid • 3. type AB = AB -codominant • 4. type O = OO -pure
IA i IAIB IBi IB Genotypic ratio: 50% IAIB Phenotypic ratio: 50% type AB 50%= IBi 50%= type B IB IAIB IBi Codominance Problem • Example: Cross a male who is homozygous Type B (BB) x a female that is heterozygous Type A (AO)
IA IB i i Another Codominance Problem • Example: Cross a • male Type O (ii) x female type AB (IAIB) *Give the genotypic and phenotypic ratios of the offspring
IA IB IAi IBi i Genotypic Ratio: 50% IAi Phenotypic Ratio: 50% type A 50% IBi 50% type B i IAi IBi Another Codominance Problem • Example:male Type O (ii) x female type AB (IAIB)
Codominance • Question: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)
IA i IAIB IB i ii Codominance • Answer: Parents: genotypes = IAi and IBi phenotypes = A and B