Question

In man, assume that spotted skin (S) is dominant over non-spotted skin (s) and that wooly hair (W) is dominant over non-wooly hair (w). Cross a marriage between a heterozygous spotted, non-wooly man with a heterozygous wooly-haired, non-spotted woman. Give genotypic and phenotypic ratios of offspring.​

Answer 1

The phenotypes and genotypes of the progeny can be determined by a dihybrid cross of the parents.

• The heterozygous male will have the genotype 'SSww' and the heterozygous female will have the genotype 'ssWW'.
• When crossed, the F1 offsprings will have a hybrid genotype of 'SsWw'. These offsprings are heterozygous with spotted skin and wooly hair.
• On self-crossing of the F1 hybrids, we find four different combinations of the alleles- SW, Sw, sW and sw. The probability of getting each of these combinations is 1/4.
• Hence, the probability of any dihybrid type is 1 out of the 16 possible genotypes. Using Punnet square, we find

9 SSWW : 3 SSww : 3 ssWW : 1 ssww

This is the phenotypic ratio of the offsprings.

• The ratio of the possible genotypes will be 1:2:1:2:4:2:1:2:1.

Answer 2

Answer:

Phenotypic Ratio = 1:1:1:1

Explanation: As for the man, its heterozygous spotted and wooly hair

                                   So it will be SsWW.

                      In the case of the woman, its heterozygous wooly hair and non-spotted, so it will be ssWw.

  After the punnet square calculations, we will get...

       

                     Phenotypic Ratio = SsWw : ssWw : Ssww : ssww

                                                      4 : 4 : 4 : 4

                                                      1 : 1 : 1 : 1