Answer:
Let "L" be the dominant allele for little finger shape and "l" be the recessive allele for the same.
It is given that a person is homozygous dominant for blood group B and little finger shape. Hence, the genotype would be <em>LL</em>
The other person is heterozygous for blood group A and homozygous recessive for little finger shape. Hence, the genotype would be
The type of gametes formed by the first parent would be <em> </em>only.
The type of gametes formed by second parent would be and <em>il.</em>
The cross is shown in the attached figure.
According to the cross, we get two phenotype:
- offspring with blood group AB and dominant little finger shape, and
- offspring with blood group group B and dominant little finger shape.
- They are present in 4 : 4 = 1 : 1
According to the cross, we get two genotype:
- They are also present in 4 : 4 = 1 : 1.