Question

1. A pea plant with wrinkled seeds is crossed with a plant hybrid for seed shape. Use P.S to demonstrated the number of plants with different phenotypes for an F1 generation with 120 plants

2. A heterozygous woman with type B blood has eight children with a man heterozygous for type A blood. Use a Punnett square to show the possible genotypes of their children and write all the possible blood types (phenotypes) of these children. What is the probability that all the children have the same blood type?

Answer 1

Answer 1:

As per the question, the two two parent plants are; a plant with wrinkled seeds and a plant with hybrid seeds.

Round seeds of a pea plant are dominant over the wrinkled (recessive) seeds.

Thus, the genotype of parent plants would be rr for wrinkled seeds and Rr for hybrid seeds.  

The four alleles would be produced by these parents. They are r, r, R, and r, respectively.

The crosses among these alleles are shown below:

The four progeny are produced in F-1 generation with the following two genotypes Rr nad rr. The phenotype of these offspring would be round and wrinkled, respectively.

The probability of getting both of these type of plants is half (50%). Thus out of 120 plants, 60 will be of round seeds and another 60 will be of wrinkled seeds.

Answer 2:

As per the question, the genotype of woman and man along with the crosses among their alleles are shown below:

The probability of occurrence of all four progeny is 25% (each). Thus, the probability of all the children with the same blood type would be 25%.