Question

In minks, color variation can be attributed to two genes: P and I.  The P locus gives dark in P- and platinum in pp, while the I locus gives dark in I- and imperial platinum in ii. When both the pp and ii mutations are present, the resulting mink has a sapphire colored coat. The traits are inherited in a Mendelian manner.

If true breeding dark and sapphire minks are bred, produce an F₁ which are then allowed to mate among themselves to make an F₂,

Approximately how many of each type would appear in the F₂ from a litter of 80 minks?

Choices:

A. 75 dark, 0 platinum, 0 imperial platinum, 5 sapphire

B. 40 dark, 20 platinum, 20 imperial platinum, 0 sapphire

C. 60 dark, 10 platinum, 5 imperial platinum, 5 sapphire

D. 45 dark, 15 platinum, 15 imperial platinum, 5 sapphire

E. 20 dark, 20 platinum, 20 imperial platinum, 20 sapphire

F. 80 dark, 0 platinum, 0 imperial platinum, 0 sapphire

Answer 1

Answer is: D

Parent: true breeding dark ( PPII) × Saphire(ppii)

Punnet square :

Gametes	pi
PI	PpIi ( dark)

Self crossing of F1:

Parents: PpIi × PpIi

Punnet square:

Gametes	PI	Pi	pI	pi
PI	PPII ( dark)	PPIi ( dark)	PpII ( dark)	PpIi ( dark)
Pi	PPIi( dark)	PPii ( imperial platinum)	PpIi ( dark)	Ppii ( imperial platinum)
pI	PpII ( dark)	PpIi ( dark)	ppII( platinum)	ppIi ( platinum)
pi	PpIi ( dark)	Ppii( imperial platinum)	ppIi ( platinum	ppii ( saphire)

Phenotypic ratio:

Dark = (9/16) ×80=45

Platinum= (3/16) ×80=15

Imperial platinum= (3/16) ×80= 15

Saphire= (1/16) ×80=5