Two purebreeding mutant plants were crossed One had large le

Two pure-breeding mutant plants were crossed: One had large leaves (wild-type leaves are small) and the other made purple flowers (wild-type flowers are pink). All F1 individuals had small leaves and purple flowers. Assuming independent assortment, what proportion of the F2 individuals are expected to be phenotypically wild-type ?

Solution

Answer:

SS= small (wild, dominant); ss=Large (mutant, recessive)

PP= purple (mutant, dominant); pp= pink (mutant, recessive)

Large leaves and purple flowers   x small leaves and pink flowers

ssPP x SSpp----------Parens

sP             Sp ---------Gametes

      SsPp (small & purple)----F

SsPp x SsPp

SP

Sp

sP

sp

SP

SSPP(small, purple)

SSPp(small, purple)

SsPP (small, purple)

SsPp (small, purple)

Sp

SSPp(small, purple)

SSpp (small, pink)

SsPp (small, purple)

Sspp (small, pink)

sP

SsPP (small, purple)

SsPp (small, purple)

ssPP (large, purple)

ssPp (large, purple)

sp

SsPp (small, purple)

Sspp (small, pink)

ssPp (large, purple)

sspp (large, pink)

F2 phenotypic ratio:

Small, purple : Small, pink : Large, purple :   Large, pink = 9 : 3 : 3 :

The phenotypically wild-type plants are = small leaves and pink flowers.

The proportion of the F2 individuals are expected to be phenotypically wild-type= 3/16

SP

Sp

sP

sp

SP

SSPP(small, purple)

SSPp(small, purple)

SsPP (small, purple)

SsPp (small, purple)

Sp

SSPp(small, purple)

SSpp (small, pink)

SsPp (small, purple)

Sspp (small, pink)

sP

SsPP (small, purple)

SsPp (small, purple)

ssPP (large, purple)

ssPp (large, purple)

sp

SsPp (small, purple)

Sspp (small, pink)

ssPp (large, purple)

sspp (large, pink)

Two pure-breeding mutant plants were crossed: One had large leaves (wild-type leaves are small) and the other made purple flowers (wild-type flowers are pink).
Two pure-breeding mutant plants were crossed: One had large leaves (wild-type leaves are small) and the other made purple flowers (wild-type flowers are pink).

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