TLS Online TPP Program

#Question id: 4772


A black Labrador, Three phenotype of black, brown and yellow is control by non allelic gene interaction. Following some genotype

 The following conclusions were made

A. The 9:3:4 ratio obtained upon selfing of AaBb here indicates recessive epistasis

B. This is a recessive epistasis where homozygous bb is epistatic to A and a.

C. When Gene A and B incompletely link, It will be produced all phenotype black, brown and yellow upon selfing of AaBb

D. When Gene A and B incompletely link, test cross of AaBb will be produced all phenotype black, brown and yellow in ratio 1:1:2 respectively

Which of the following conclusions are correct?

#Unit 8. Inheritance Biology
  1. A and B only             

  2. A, B and D

  3. B and C only        

  4. A, B and C