In cattle, the allele for no horns (p)is dominant to the allele for horns(p). the allele for cloven hooves(c) is dominant to the allele for mule feet(c). cattle that are heterozygous for these traits are crossed. what is the probability that an offspring will have the same phenotype as the parents?

The cattles are heterozygous for both the traits (horns and feet). This means that the genotype of the cattles is PpCc, and the phenotype of these cattles is 'no horns and cloven hooves'. The gametes formed will be : PC , Pc , pc , pC.

The cross between them is shown in the attached punnet square. Now, the number of off springs having the same phenotype as the parents (that is no horns and cloven hooves) are 9.

So, the probability of the off spring having the same phenotype as the parent cattles is (9 ÷ 16) × 100 = 56.25 %.

the cladogram shown here shows similarities and differences among several classes of vertebrates. anatomically, the four have limb structure as their common ancestor. we might separate the mammals from the other three classes because they a) divergent; do not lay eggs b) identical; are warm-blooded c) varied; a four chambered heart d) homologous; have mammary glands

not sure if this is what you'r looking for nnew= nold