Question

1. How many different phenotypes are found among the progeny of the cross listed below?

a.)

Aa

Bb

dd

Ee

x

AA

bb

Dd

ee

b.)

aa

bb

Dd

ee

x

AA

Bb

Dd

Ee

c.)

aa

Bb

dd

Ee

Answer 1

Answer a)

Aa

Bb

dd

Ee

x

AA

bb

Dd

ee

As phenotype for A would be dominant only whether it is AA or Aa , for B it would be dominant (Bb) or recessive(bb), for D it would be dominant (Dd) or recessive (dd), for E it would be dominant (Ee) or recessive (ee)

So, Total possible phenotypes = 1x2x2x2 = 8 i.e.

1. Dominant A, Dominant B, Dominant D, Dominant E
2. Dominant A, Dominant B, Dominant D, Recessive E
3. Dominant A, Dominant B, Recessive D, Dominant E
4. Dominant A, Recessive B, Dominant D, Dominant E
5. Dominant A, Recessive B, Recessive D, Dominant E
6. Dominant A, Recessive B, DominantD, Recessive E
7. Dominant A, Dominant B, Recessive D, Recessive E
8. Dominant A, Recessive B, Recessive D, Recessive E

	ABdE	aBdE	AbdE	abdE	ABde	aBde	Abde	abde
AbDe	AABbDdEe Type 1	AaBbDdEe Type 1	AAbbDdEe Type 4	AabbDdEe Type 4	AABbDdee Type 2	AaBbDdee Type 2	AabbDdee Type 6	AabbDdee Type 6
Abde	AABbddEe Type 3	AaBbddEe Type 3	AAbbddEe Type 5	AabbddEe Type 5	AABbddee Type 7	AaBbddee Type 7	AAbbddee Type 8	Aabbddee Type 8

Answer b)

aa

bb

Dd

ee

x

AA

Bb

Dd

Ee

As phenotype for A would be dominant (Aa) , for B it would be dominant (Bb) or recessive(bb), for D it would be dominant (DD, Dd) or recessive (dd), for E it would be dominant (Ee) or recessive (ee)

So, Total possible phenotypes = 1x2x2x2 = 8  

1. Dominant A, Dominant B, Dominant D, Dominant E
2. Dominant A, Dominant B, Dominant D, Recessive E
3. Dominant A, Dominant B, Recessive D, Dominant E
4. Dominant A, Recessive B, Dominant D, Dominant E
5. Dominant A, Recessive B, Recessive D, Dominant E
6. Dominant A, Recessive B, DominantD, Recessive E
7. Dominant A, Dominant B, Recessive D, Recessive E
8. Dominant A, Recessive B, Recessive D, Recessive E

	ABDE	ABDe	ABde	ABdE	AbDe	AbdE	Abde	AbdE
abDe	AaBbDDEe Type 1	AaBbDDee Type 2	AaBbDdee Type 2	AaBbDdEe Type 1	AabbDDee Type 6	AabbDdEe Type 4	AabbDdee Type 6	AabbDdEe Type 4
abde	AaBbDdEe Type 1	AaBbDdee Type 2	AaBbddee Type 7	AaBbddEe Type 3	AabbDdee Type 6	AabbddEe Type 5	Aabbddee Type 8	AabbddEe Type 5

Answer c) Only one genotype is given, So, if self crossing is assumed, aaBbddEe x aaBbddEe

Gametes are aBdE, aBde, abdE, abde

As phenotype for A would be recessive only (aa) , for B it would be dominant (Bb) or recessive(bb), for D it would be only recessive (dd), for E it would be dominant (Ee) or recessive (ee)

So, Total possible phenotypes = 1x2x1x2 = 4 i.e.

1. Recessive A, Dominant B, Recessive D, Dominant E i.e. aaBbddEe
2. Recessive A, Dominant B, Recessive D, Recessive E i.e. aaBbddee
3. Recessive A, Recessive B, Recessive D, Dominant E i.e. aabbddEe
4. Recessive A, Recessive B, Recessive D, Recessive E i.e. aabbddee