2.PROBLEMS ON DIHYBRID CROSS ( dt: 9.4.
25)
1. In goldfish, orange red colour( B) is dominant over blue ( b) and normal
eyes (D) over telescopic eyes (d). If the alleles segregate independently ,
write the most probable genotype for the parents for each of the following
cross by using punnet square method.
SI.No Phenotype of Phenotypes of the offspring
Parents (F2)
0RN 0RT BN BT
1 ORT X ORT
0 320 0 100
2 ORN X ORT 180 190 0 0
3 ORT X BN 200 0 210 0
4 BN X BN 0 0 280 90
ORN – Orange red normal
BN – Blue normal
ORT – Orange Red telescopic
BT – Blue Telescopic
Solution
SI.No Phenotype of Phenotypes of the offspring
Parents (F2)
0RN BN BT
ORT
1 ORT X ORT 0 3 0 1
2 ORN X ORT 1 1 0 0
3 ORT X BN 1 0 1 0
4 BN X BN 0 0 3 1
BBDD – Orange Red Normal
BBdd – Orange Red Telescopic
bbDD – Blue Normal
bbdd – Blue Telescopic
Parental Generation
i) Bbdd x Bbdd
Orange red telescopic (ORT) x Orange red telescopic (ORT)
F2 :
Bd Bd bd bd
Bd BBdd BBdd Bbdd Bbdd
ORT ORT ORT ORT
Bd BBdd BBdd Bbdd BBdd
ORT ORT ORT ORT
bd BBdd Bbdd bbdd bbdd
ORT ORT BT BT
bd Bbdd Bbdd bbdd bbdd
ORT ORT BT BT
2. Parent : Orange Red Telescopic (ORT) x Orange red Normal ( ORN)
Bbdd x BBDd
F2:
BD Bd BD Bd
Bd BBDd BBdd BBDd Bbdd
ORN ORT ORN ORT
Bd BBDd BBdd BbDd BBdd
ORN ORT ORN ORT
bd BbDd Bbdd BbDd Bbdd
ORN ORT ORN ORT
bd BbDd Bbdd BbDd Bbdd
ORN ORT ORT ORT
3. Blue Normal X Orange Red Telescopic
bbDD x Bbdd
F2:
Bd Bd bd bd
bD BbDd BbDd bbDd bbDd
ORN ORN BN BN
bD BbDd BbDd bbDd bbDd
ORN ORN BN BN
bD BbDd BbDd bbDd bbDd
ORN ORN BN BN
bD BbDd BbDd bbDd bbDd
ORN ORN BN BN
4. BN X BN
bbDd x bbDd
Blue Normal x Blue Normal
F2:
bD bd bD bd
bD bbDD bbDd bbDD bbDd
BN BN BN BN
bd bbDd bbdd bbDd bbdd
BN BT BN BT
bD bbDD bbDd bbDD bbDd
BN BN BN BN
bd bbDd bbdd bbDd bbdd
BN BT BN BT
Answers:
The most Probable genotypes of the parents are follows:
i) Bbdd x Bbdd
ii) Bbdd x BBDd
iii) bbDD x Bbdd
iv) bbDd x bbDd
2. Normal spine in guppy is governed by dominant gene ( Sn) and curved spine by
its recessive allele ( Sc). Grey colour results from the action of the dominant
genotype ( GG/Gg) and gold from the recessive genotype ( gg). Determine the
expected genotypic and phenotypic ratios in the progeny for the following crosses.
Given:
SnScGg x SnScGg
ScScGg x SnScGG
SnSngg x SnScgg
ScScgg x ScScgg
Solution :
NSG : Normal Spine Grey
NSGL : Normal spine gold
CSG : Curved spine Grey
CSGL : Curved spine Gold
i) SnScGg x SnScGg
NSG X NSG
F2:
Sn,G Sn,g Sc,G Sc,g
Sn,G SnSnGG SnSnGg SnScGG SnScGg
NSG
NSG NSG NSG
Sn,g SnSnGg SnSngg SnScGg SnScgg
NSG NSG
NSGL NSGL
Sc,G SnScGG SnScGg ScScGG ScScGg
NSG CSG
NSG CSG
Sc,g SnScGg SnScgg ScScGg ScScgg
NSG NSGL CSG CSGL
ii) SnScGg x SnScGG
NSG X CSG
F2:
Sn,G Sn,G Sc,G Sc,G
Sn,G SnSnGG SnSnGG SnScGG SnScGG
NSG NSG CSG CSG
Sn,g SnSnGg SnSnGg SnScGg SnScGg
NSG NSG CSG CSG
Sc,G SnScGG SnScGG ScScGG ScScGG
NSG NSG CSG CSG
Sc,g SnScGg SnScGg ScScGg ScScGg
NSG NSG CSG CSG
iii) SnSngg x SnScgg
NSGL X NSGL
F2:
Sn,g Sn,g Sc,g Sc,g
Sn,g SnSngg SnSngg SnScgg SnScgg
NSGL NSGL NSGL
NSGL
Sn,g SnSngg SnSngg SnScgg SnScgg
NSGL NSGL
NSGL NSGL
Sn,g SnSngg SnSngg SnScgg SnScgg
NSGL NSGL NSGL NSGL
Sn,g SnSngg SnSngg SnScgg SnScgg
NSGL NSGL NSGL NSGL
iv) ScScgg x ScScgg
CSGL X CSGL
F2 :
Sc,g Sc,g Sc,g Sc,g
Sc,g ScScgg ScScgg ScScgg ScScgg
CSGL CSGL CSGL CSGL
Sc,g ScScgg ScScgg ScScgg ScScgg
CSGL CSGL CSGL CSGL
Sc,g ScScgg ScScgg ScScgg ScScgg
CSGL CSGL CSGL CSGL
Sc,g ScScgg ScScgg ScScgg ScScgg
CSGL CSGL CSGL CSGL
Results
a) Phenotypic ratio : Normal spine grey 9 : Normal spine gold 3: Curved spine
grey 3 : curved spine gold 1
Genotypic ratio : SnSnGG ( 1) : SnSnGg (2) : SnScGG (2) : SnScGg ( 4) : SnSngg
(1) : SnSc gg (2) : ScScGG (1) : ScScGg (2) : ScScgg (1)
b) Phenotypic ratio
Normal spine grey : Curved spine grey (1: 1)
Genotypic ratio
SnScGG : SnScGg : SnScGG : ScScGg
1:1:1:1
c) Phenotypic ratio
All normal spine gold
Genotypic ratio
SnSngg : SnScgg (1:1)
d) Phenotypic ratio
All curved spine gold
Genotypic ratio : All ScScgg
3. In the case of minnows Grey – Green ( G) colour is dominant to pink colour.
Normal Fin ( N) is dominant to long fin. If a homozygous Gray – Green with normal
fin male minnow is crossed with a homozygous pink and long fin female . What will
be genotypes and phenotypes of the F1 and F2 offsprings.
Solution
GGN : Gray Green normal
GGL : grey green long fin
PN : Pink Normal fin
PL : Pink Long fin
Genotypes: GG -Gray-Green Dominant
Gg – Pink colour Recessive
NN – Normal fin
nn – Long fin
Parent GGNN X ggnn
GN X gn
F1 : GgNn
F2 Generation:
GgNn x GgNn
Answers:
i) What will be genotypes and phenotypes of the F1 and F2 offsprings.
F1 Phenotype : All grey green Normal fin
F1 Genotype : GgNn
F2 Phenotype : Grey Normal, - 9/16 Grey Long fin – 3/16 , Pink normal – 3/16 Pink -
long fin -1/16.
F2 Genotype : GGNN (1) : GGNn (2) : GgNN (2) : GgNn (4) : GGnn (1) : Ggnn
(2) : ggNN (1) : ggNn (2) : ggnn (1)
4. In the case of Nile Tilapia, two loci A and B are involved in producing normal
body colouration. The recessive allele a and b in combination and only in
ccombination , would result in pearl colouration. Either recessive allele ,
when present alone, would not alter the normal colouration of the fish.
Determine the expected phenotypic ratios in the progeny from the
following mating.
ii aabb x AABB
Solution
i) aabb x AABB
Normal x Pearl
AB AB AB AB
ab AaBb AaBb AaBb AaBb
Pearl Pearl Pearl Pearl
ab AaBb AaBb AaBb AaBb
Pearl Pearl Pearl Pearl
ab AaBb AaBb AaBb AaBb
Pearl Pearl Pearl Pearl
ab AaBb AaBb AaBb AaBb
Pearl Pearl Pearl Pearl
Answers
The expected phenotypic ratios are as follows:
i) All Pearl
Problem 5
In goldfish, the gene ‘ M’ is epistatic with respect to gene S. Genes ‘m
‘and s produce albino.
The following combination of genes are possible
MS and Ms – dark fishes
mS – Light fishes
ms – albino fishes
By using punnet square find out the phenotypic and genotypic ratio for
mating of the gold fish with the following genotypes.
a) Two heterozygous MmSs dark gold fish
b) mmss x MmSs
c) mmss x MMSs
d) MMSs x MmSs
e) MMSs x MmSs
Solution
MS and Ms – dark fishes
mS – light fishes
ms – albino fishes
D) MMSs x MmSs
Dark x Dark
MS Ms mS ms
MS MMSS MMSs MmSs MmSs
dark dark dark dark
Ms MMSs MMss MmSs Mmss
dark dark dark dark
MS MMSS MMSs MmSS MmSs
dark dark dark dark
Ms MMSs MMss MmSs Mmss
dark dark dark dark
e) MMSs x MmSs
Dark x Dark
MS Ms mS ms
Ms MMSs MMss MmSs Mmss
Dark dark dark dark
Ms MMSs MMss MmSs Mmss
dark Dark Dark Dark
Ms MMSs MMss MmSs Mmss
dark dark Dark Dark
Ms MMSs MMss MmSs Mmss
dark Dark Dark dark
Answers:
a) Phenotypic ratios: 12 dark : 3 light : 1 albino
Genotypic ratio : 12 MS or Ms : 3 mS : 1 ms
b) Phenotypic ratio : 2 dark : 1 light : 1 albino
Genotypic ratio : 2 MS or Ms : 1 mS:1 ms
c) Phenotypic ratio : All dark
Genotypic ratio :All MS or Ms
d) Phenotypic ratio : All dark
Genotypic ratio : all MS or Ms
e) Phenotypic ratio : All dark
Genotypic ratio : all MS or Ms
Solution:
F2:
St,R St,r st,R st,r
St,R StSt,RR StSt,Rr Stst,RR Stst,Rr
Normally Normally Normally Normally
pigmented pigmented pigmented pigmented
St,r StSt,Rr StSt,rr Stst,Rr Stst,rr
Normally Gray Normally gray
pigmented pigmented
st,R Stst,RR Stst,Rr stst,RR stst,Rr
Normally Normally gold gold
pigmented pigmented
st,r Stst,Rr Stst,rr stst,Rr stst,rr
Normally gray gold brown
pigmented
Answers
The phenotypic ratio of F2 Progeny resulted from the cross of two
heterozygous ( Stst, Rr) normally pigmented platy fish is 9:3:3:1 (Normally
pigmented : gray: gold : Brown)
