A dihybrid cross SS tt × ss TT is used to make an F1 generation. In the cross S is dominant, but T and t exhibit incomplete dominance. Assuming independent assortment, how many phenotypic classes are expected in F2 when crossing F1 individuals to each other?

Explanation:

Given that:

A dihybrid cross SS tt × ss TT is used to make an F1 generation

where;

S = dominant

T and t exhibit incomplete dominance.

According to Mendelian's Law of Independent Assortment.

The cross between SS tt × ss TT will be as follows:

 If self crossing occurs first, the following traits will be used for the F1 cross

      S          S

t     St          St

t     St           St

         s             s

T       sT           sT

T       sT            sT

                       St                  St                          St                           St

sT                   SstT              SstT                       SstT                      SstT  

sT                   SstT              SstT                       SstT                      SstT

sT                   SstT              SstT                       SstT                      SstT

sT                   SstT              SstT                       SstT                      SstT

Thus; the F1 generation produce offsprings that are all heterozygous SstT

For the F2 generation now;

We have

SsTt × SsTt

           S            s

T         ST          sT

t           St          st

            ST              St                 sT                   st

ST         SSTT         SSTt              SsTT            SstT

St           SSTt         SStt               sSTt              sStt

sT         SsTT          SstT               sSTT              sstT

st          SsTt           Sstt                ssTt               sstt

From the above F2 generation; we have the following offspring.

SSTT   3    = 3/16 = 0.1875 = 18.75 %

SStT   6    = 6/16  = 0.375 = 37.5%

Sstt   3     = 3/16 = 0.1875 = 18.75 %

ssTT 1      = 1/16 = 0.0625   = 6.25%

ssTt   2     = 2/16 =  0.125  = 12.5%

sstt      1      = 1/16 = 0.0625   = 6.25%

after making a punnet square using bv bv bv bv you will find that the ratio will be 9: 7 or a 128.6% probability.

hope this