1) A normal woman who is heterozygous for X-linked Ducheme muscular dystrophy ha
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Question
1) A normal woman who is heterozygous for X-linked Ducheme muscular dystrophy has children with a normal man. What is the probability of a normal son? What is the probability of a normal daughter? 2) A woman with type B blood has a child with type O blood. What are the possible genotypes and blood types of the father? 3) A mutant allele in Drosophila produces a star-eye shape eye when heterozygous. When homozygous, the allele is lethal. What would be the ratio and phenotypes of the offspring if heterozygotes were mated. Show the cress using proper terminology. 4) In chickens, predict the genotypes and phenotypes of the comb shape of offspring from the following matings:Explanation / Answer
1. Xd- allele for muscle dystrophy
X- normal X allele
female XdX .........x............................ XY male
gametes Xd, X ..................................X,Y
F1 progeny XdX .....XdY...... XX......... XY
Hence, probability of a normal son is 1/4
Probability of a normal daughter is 2/4 or 1/2 ( both daughters are phenotypically normal, one is carrier)
2. A woman with Type B blood having child with O blood type must have the genotype IBIO . So she acquires either IO allele or IB allele from her father.
Possible genotype & blood type for her father: IOIO ( blood type O), IBIB (blood type B), IAIB (blood type AB),
IAIO (Blood type A), IBIO (blood type B)- considering the daughter obtains the other allele from her mother.
3. S- star eye allele
s- normal eye allele
SS- dies off; Ss- star-eyed, ss- normal eyed drosophila
parents....... Ss...... x............... Ss
gametes. S,s............................ S,s
gametes
S
s
S
SS (dies off)
Ss (star eyed)
s
Ss (star eyed)
ss (normal eyed)
Phenotypic ratio= Star-eyed: Normal eyed drosophila= 2:1
4. a.
RrPp.................................... x ...............rrpp
gametes RP, Rp, rP,rp..................... rp
F1........ RrPp (walnut) Rrpp (rose) rrPp (pea) rrpp (single)
Phenotypic ratio= walnut: rose: pea : single= 1:1:1:1
Genotypic ratio= RrPp: Rrpp: rrPp: rrpp= 1:1:1:1
b.
Rrpp ....................x........................................... Rrpp
gametes Rp, rp ..........................................Rp, rp
F1 RRpp........... Rrpp................. Rrpp ..........rrpp
Phenotypic ratio= rose: single= 3 :1
Genotypic ratio= RRpp: Rrpp: rrpp= 1:2:1
c.
Rrpp ............................x............................ RrPp
gametes Rp, rp................................. RP, Rp, rP,rp
gametes
RP
Rp
rP
rp
Rp
RRPp walnut
RRpp rose
RrPp walnut
Rrpp rose
rp
RrPp walnut
Rrpp rose
rrPp pea
rrpp single
Phenotypic ratio= Walnut: rose: pea: single= 3:3:1:1
Genotypic ratio= RRPp: RrPp: RRpp: Rrpp: rrPp: rrpp= 1:2:1:2:1:1
d.
Rrpp ..............x...................... rrpp
gametes Rp, rp .........................rp
F1 ......Rrpp.................... rrpp
Phenotypic ratio= Rose: single= 1:1
Genotypic ratio= Rrpp:rrpp= 1:1
gametes
S
s
S
SS (dies off)
Ss (star eyed)
s
Ss (star eyed)
ss (normal eyed)
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