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.

Option B, Transcription that originates at the core promoter and +1 site

Explanation:

In basal transcription  process a special class of transcription factors which  are nothing but the protein that binds to specific sites  on DNA (which are also known as promoter sites) with an objective of activating transcription of genetic information from DNA to RNA , specifically messenger RNA. The transcription initiation requires association at promoter site(or regions of DNA that are found in the upstream of the coding region of  a gene)  along with a  σ sub unit

Hence, option B is correct