If a man is homozygous dominant for bend fifth finger and a woman is homozygous recessive what’s the probability that the offspring has a fifth bend finger

There is 100 % probability that the child will have bend finger

Explanation:

According to the given information, the father is homozygous dominant (lets assume BB) for bend finger. The bend fifth finger is dominant trait. And the mother is homozygous recessive (bb) for the same trait.

All the gametes of the father would carry "B" allele while the egg cells from given mother would carry "b" allele. Random fertilization of these gametes will give rise to zygote with Bb genotype that will develop into a child with bend finger.

Hence, all of their children will have bend fifth finger.

hey, i believe your answer would be d. sugar ribose and the base uracil. i believe this is the answer, because rna consists of sugar ribose, adenine, guanine, cytosine, and uracil.

your first option is incorrect, because that's dna. where it consists of sugar deoxyribose, adenine, guanine, cytosine and thymine.

your second option is also incorrect, this is because sugar deoxyribose is in dna, and uracil is in rna. these two do not make sense together.

your third option being incorrect, because sugar ribose is in rna, but not the base of thymine. rna consists of uracil, not thymine.

lastly your fourth option being correct! this answer is correct, because both sugar ribose, and the base uracil are in rna! ~

answer is d. sugar ribose and the base uracil.