Sex-Influenced (Autosomes) P Horned x Hornless HH H'H' H H' F 1 Horned x Hornless HH' HH' 1/2 H 1/2 H' 1/2 H 1/2 H' F 2 Genotypes Phenotypes 1/4 HH Horned Horned 2/4 HH' Horned Hornless 1/4 H'H' Hornless Hornless F 2 genotypic ratio - 1:2:1 F 2 phenotypic ratio - - 3:1 - Horned:hornless - 1:3 - Horned:hornless H - horned - dominant in males H' - hornless - dominant in females 155
Sex-influenced dihybrid cross B = bald B = haired F = long index finger length F = short index finger length bald haired BB BB B B bald haired long index fingers short index fingers FF FF F F long index fingers short index fingers B is dominant in B is dominant in F is dominant in F is dominant in Heterozygous bald man, with long index fingers X Bald woman, heterozygous for long index finger length BB FF X BB FF
B B B BB BB B BB BB F F F FF FF F FF FF 1/2 1/2 1/1 bald 1/2 long index fingers, 1/2 short index fingers 1/2 bald, 1/2 haired 1/1 long index fingers 1/2 x 1/1 x 1/2 = 1/4 bald with long index fingers 1/2 x 1/1 x 1/2 = 1/4 bald with short index fingers 1/2 x 1/2 x 1/1 = 1/4 bald with long index fingers 1/2 x 1/2 x 1/1 = 1/4 haired with long index fingers P.R. = 1:1:1:1
Sex-Limited (Autosomes) H - hen feathered h - cock feathered - sex-limited to the male P Hen feathered x hen feathered HH x hh H h F 1 Hen feathered x Hen feathered Hh Hh 1/2 H 1/2 h 1/2 H 1/2 h F 2 Genotypes Phenotypes 1/4 HH Hen feathered Hen feathered 2/4 Hh Hen feathered Hen feathered 1/4 hh Cock feathered Hen feathered F 2 genotypic ratio - 1:2:1 F 2 phenotypic ratio - - 3:1 - hen feathered:cock feathered - 1 - hen feathered 156
Cock- Feathering Pointed Feathers Hen- Feathering Rounded Feathers
Barred Plymouth Rock Cock- Feathered Pointed Feathers hh genotype and male sex Cock feathering is a sex-limited trait! (limited to males) Hen-Feathered Rounded Feathers hh genotype and female sex Or any sex with a H allele in the genotype
Hen-Feathered Rounded Feathers This is a rooster He just has a genotype with a Hen feathered allele, H Possible genotypes: HH, Hh Hen-feathered males are desired in Sebright Chickens Sebright Chickens Hen-Feathered Rounded Feathers Females can be any genotype (HH, Hh, or hh) and will still be Hen-feathered Cock-feathering is a sex limited trait-limited to males
SEX-LINKAGE Morgan (1910): Established the sex-linked inheritance of white eyes in Drosophila melanogaster. Example: P white male x red female, F 1, F 2, and reciprocal. Criteria for identifying sex-linked recessive genes from pedigrees in humans: 1. Occurs more often in males than females. 2. Gene is transmitted from an affected male through his daughters to 1/2 of his grandsons. 3. Never transmitted from father to son. 4. An affected female had to have an affected father and an affected or carrier mother.
Sex-linked Cross (P generation) P homogametic parent x heterogametic parent homozygous x hemizygous AA x ay or aa x AY Genotypes are contrasting F 1 homogametic sex is heterozygous: Aa F 1 heterogametic sex is hemizygous: AY or ay F 2 is produced by crossing + F 1 s. 159
170
Reciprocal crosses - Drosophila Sex-linked 1. P red x white 2. P white x red X W Y x X w X w X w Y x X W X W F 1 white x red F 1 red x red X w Y x X W X w X W Y x X W X w Autosomal P normal x vestigial P vestigial x normal VV x vv vv x VV F 1 normal x normal F 1 normal x normal Vv x Vv Vv x Vv If the genes are on autosomes, reciprocal crosses in the P generation will give similar results (F 1 and F 2 ). If the genes are on x chromosomes (sex-linked), reciprocal crosses in the P generation will not give similar results (F 1 and F 2 ). 160
1. Sex-Linkage in Drosophila melanogaster Male = XY; Female = XX W = red eye color (dominant); w = white eye color (recessive) ph = phenotype; ge = genotype; ga = gametes; circle gametes Male Female P (ph) red x white (ge) X W Y x X w X w (ga) x F 1 (ph) x (ge) x (ga) x F 2 F 2 F 2 Males Females. Phenotype Genotype Phenotype Genotype Genotypic ratio Phenotypic ratio (sex and eye color) (eye color) (sex) 162
1. Sex-Linkage in Drosophila melanogaster Male = XY; Female = XX W = red eye color (dominant); w = white eye color (recessive) ph = phenotype; ge = genotype; ga = gametes; circle gametes Male Female P (ph) red x white (ge) X W Y x X w X w (ga) ½ X W ½ Y x 1/1 X w F 1 (ph) white x red (ge) X w Y x X W X w (ga) ½ X w ½ Y x ½ X W ½ X w F 2 Males Females Phenotype Genotype Phenotype Genotype red X W Y red X W X w white X w Y white X w X w F 2 Genotypic ratio 1:1:1:1 F 2 Phenotypic ratio (sex and eye color) 1:1:1:1 (eye color) 1:1 (sex) 1:1 162
2. Sex-Linkage in Drosophila melanogaster Male = XY; female = XX W = red eye color (dominant); w = white eye color (recessive) ph = phenotype; ge = genotype; ga = gametes; circle gametes Male Female P (ph) white x red (ge) X w Y x X W X W (ga) x F1 (ph) x (ge) x (ga) x F2 Males Females. Phenotype Genotype Phenotype Genotype F2 Genotypic ratio F2 Phenotypic ratio (sex and eye color) (eye color) (sex) 163
2. Sex-Linkage in Drosophila melanogaster Male = XY; female = XX W = red eye color (dominant); w = white eye color (recessive) ph = phenotype; ge = genotype; ga = gametes; circle gametes Male Female P (ph) white x red (ge) X w Y x X W X W (ga) ½ X w ½ Y x 1/1 X W F1 (ph) red x red (ge) X W Y x X W X w (ga) ½ X W ½ Y x ½ X W ½ X w F2 Males Females Phenotype Genotype Phenotype Genotype red X W Y red X W X W white X w Y red X W X w F2 Genotypic ratio 1:1:1:1 F2 Phenotypic ratio (sex and eye color) 2:1:1 (eye color) 3:1 (sex) 1:1 163
F1
Parents Male Female
Male Female Progeny at 4 days
Male Female Progeny at 21 days
Female Progeny at 6 weeks
Male Progeny at 6 weeks
Progeny at 3 months Male Female
Sex-Linkage in Chickens XY = female; XX = male B = barred (dominant); b = nonbarred (recessive) ph = phenotype; ge = genotype; ga = gametes; circle gametes Male Female P (ph) nonbarred x barred (ge) X b X b x X B Y (ga) x F 1 (ph) x (ge) x (ga) x F 2 Males Females Phenotype Genotype Phenotype Genotype F 2 Genotypic ratio F 2 Phenotypic ratio (sex and pattern) (pattern) (sex) 166
Sex-Linkage in Chickens XY = female; XX = male B = barred (dominant); b = nonbarred (recessive) ph = phenotype; ge = genotype; ga = gametes; circle gametes Male Female P (ph) nonbarred x barred (ge) X b X b x X B Y (ga) 1/1 X b x 1/2 X B 1/2 Y F 1 (ph) barred x nonbarred (ge) X B X b x X b Y (ga) 1/2 X B 1/2 X b x 1/2 X b 1/2 Y F 2 Males Females Phenotype Genotype Phenotype Genotype barred X B X b barred X B Y nonbarred X b X b nonbarred X b Y F 2 Genotypic ratio 1:1:1:1 F 2 Phenotypic ratio (sex and pattern) 1:1:1:1 (pattern) 1:1 (sex) 1:1 166
Sex-Linkage in Chickens XY = female; XX = male B = barred (dominant); b = nonbarred (recessive) ph = phenotype; ge = genotype; ga = gametes; circle gametes Male Female P (ph) barred x nonbarred (ge) X B X B x X b Y (ga) x F 1 (ph) x (ge) x (ga) x F 2 Males Females Phenotype Genotype Phenotype Genotype F 2 Genotypic ratio F 2 Phenotypic ratio (sex and pattern) (pattern) (sex) 168
Sex-Linkage in Chickens XY = female; XX = male B = barred (dominant); b = nonbarred (recessive) ph = phenotype; ge = genotype; ga = gametes; circle gametes Male Female P (ph) barred x nonbarred (ge) X B X B x X b Y (ga) 1/1 X B x 1/2 X b 1/2 Y F 1 (ph) barred x barred (ge) X B X b x X B Y (ga) 1/2 X B 1/2 X b x 1/2 X B 1/2 Y F 2 Males Females Phenotype Genotype Phenotype Genotype barred X B X b barred X B Y barred X B X B nonbarred X b Y F 2 Genotypic ratio 1:1:1:1 F 2 Phenotypic ratio (sex and pattern) 2:1:1 (pattern) 3:1 (sex) 1:1 168
1) A cock-feathered rooster is mated with a heterozygous hen-feathered hen. In a clutch of 4 eggs, what is the probability of hatching 3 cock-feathered male and 1 hen-feathered female chicks (no order specified)? 1/2 1/2 + hh x Hh H h 1/2 cock-feather 1/1 hen-feather h Hh hh 1/2 hen-feather h Hh hh P = 4! 3! 1! (1/4) 3 (1/2) 1 = (4)(1/64)(1/2) = 4 = 1 128 32 2) A non-barred rooster is mated with a barred female. In a clutch of 4 eggs, what is the probability of hatching 2 barred male and 2 nonbarred female chicks (no order specified)? P = 4! (1/2) 2 (1/2) 2 2! 2! X b X b X B X B X b X B X b y X b y X b y
3) A bald woman marries a heterozygous bald man. Of 4 children born to the couple, what is the probability of them having 2 bald sons, 1 bald daughter, and 1 haired daughter (born in that specific order)? B=bald (dominant in males); B`=haired (dominant in females). 1/2 1/2 + BB x BB` B B 1/1 bald 1/2 bald B BB BB 1/2 haired B` BB` BB` P = (1/2) 2 (1/4) 1 (1/4) 1 = (1/4) 3 = 1/64
4) A woman heterozygous for long index finger length marries a man also with long index finger length. What is the probability that they will have 1 daughter with long index fingers and 2 sons with long index fingers (no order specified)? long short FF FF F F + long short + x FF F F F FF FF F FF FF FF 1/2 1/2+ 1/2 long 1/1 long 1/2 short P = 3! (1/2) 1 (1/4) 2 1! 2! P = (3)(1/2)(1/16) = 3/32