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?

Respuesta :

Answer:

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%

The branch of science that deals with genes and inheritance are called genetics. There are two types of an allele in the gene pair, one allele is called recessive and the other is dominant.

The correct answer to the question is mentioned below.

Dihybrid cross

  • A cross in which where two-character are crossed.
  • According to the question, the two-character T and S are crossed together to produced the offspring.

In the cross, the dominant character has a ratio is 3/4 and the recessive character has 1/4.

The ratio of the given character is as follows:-

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%

Hence, the correct answer is mentioned above.

For more information about the genes, refer to the link:-

https://brainly.com/question/264225