The following Three points a piece 96. Please write the genotypes A and a homozy
ID: 82235 • Letter: T
Question
The following Three points a piece 96. Please write the genotypes A and a homozygous for blood B Then draw out the for a heterozygous induvial of blood type might have please show genotype and Punnett square analysis for the children that these two induvial phenotypes. 97. A red snapdragon is crossed with a white snapdragon white. The flowers seen from this cross are pink. Draw out the genotypes for the parents and the pink flowers. Then draw out the Punnett square analysis for a cross of two pink flowers show genotype and phenotypes.Explanation / Answer
Answer 1
Blood group follows co-dominance
Heterozygous individual with blood type A (IAi)
Homozygous individual with blood type B (IBIB)
Punnet square (IAi X IBIB )
IA
i
IB
IAIB
iIB
IB
IAIB
iIB
Genotypes : IAIB : iIB in 1:1 ratio
Phenotype would be AB type : B type , in ratio 1: 1.
Answer 2
Snapdragon flowers follow incomplete dominance.
Genotypes for Red flowers (RR), White flowers (rr)
Punnet square ( RR X rr )
R
R
r
Rr
Rr
r
Rr
Rr
When two pink flowers are crossed
Punnet square (Rr X Rr)
R
r
R
RR
rR
r
rR
rr
In the F2 generation, the genotypes RR:Rr:rr in the ratio 1:2:1
Similarly, phenotypic ratio would be Red : Pink : White in ratio 1:2:1.
Answer 3 :
The genotypes of both the dogs would be (BbEe) and the
Punnet square (BbEe X BbEe)
BE
Be
bE
be
BE
BBEE
(Black)
BBEe
(Black)
BbEE
(Black)
BbEe
(Black)
Be
BBEe
(Black)
BBee
(Albino)
BbEe
(Black)
Bbee
(Albino)
bE
BbEE
(Black)
BbEe
(Black)
bbEE
(Brown)
bbEe
(Brown)
be
BbEe
(Black)
Bbee
(Albino)
bbEe
(Brown)
Bbee
(Albino)
The genotypes are visible in punnet square.
Phenotypes would be black : no colour (albino) : brown in the ratio 9 : 4 : 3
Solution 4
Homozygous yellow round pea – (YYRR)
Homozygous green wrinkled – (yyrr)
The F1 generation would be YyRr ( yellow round)
When these are crossed , the F2 generation would be
YR
yR
Yr
yr
YR
YYRR
(yellow round)
YyRR
(yellow round)
YYRr
(yellow round)
YyRr
(yellow round)
yR
YyRR
(yellow round)
yyRR
(green round)
YyRr
(yellow round)
yyRr
(green round)
Yr
YYRr
(yellow round)
YyRr
(yellow round)
YYrr
(yellow wrinkled)
Yyrr
(yellow wrinkled)
yr
YyRr
(yellow round)
yyRr
(green round)
Yyrr
(yellow wrinkled)
yyrr
(green wrinkled)
The phenotype would be yellow round : green round : yellow wrinkled : green wrinkled in the ratio 9 : 3 : 3 : 1.
IA
i
IB
IAIB
iIB
IB
IAIB
iIB