The answers are:
A. DNA replication in the nucleus of a cell
B. From one helix of DNA in a replication process, we get two: The DNA is a double helix and it consists of two strands of specifically connected amino-acids. When the time for replication comes, a set of enzymes unwind the two strands and leave them as a base for additional two strands attaching to them - the green line is an example of that. The free nucleotides - adenine, guanine, thymine and cytosine are left open and the enzyme called DNA-polymerase helps to produce a new strand on the template of the old parental one (one of the blue ones in the picture)
C. By the location on the smaller picture - replication takes place in the nucleus. And the most important hint are the letters A - adenine, G - guanine, T- thymine, and C-cytosine. A connects with T, and G connects with C.
Answer:
Option A
Explanation:
In this case, there are two varieties of horses: Domestic horses that are raised and bred by humans and Wild horses that lives in the wild. The issue of varieties in coat colour of domestic horses can be explained as what occured by selective breeding, also known as artificial selection which is a technique by which humans develop new offsprings with desirable and suitable characteristics. These breeders select two parents that possess beneficial phenotypic traits to mate, producing offsprings with those desired traits such as strength and also for coat colour as stated.
Answer:
Embedded in the lipid bilayer are large proteins, many of which transport ions and water-soluble molecules across the membrane. Some proteins in the plasma membrane form open pores, called membrane channels, which allow the free diffusion of ions into and out of the cell.
Answer:
p = 0.34
Explanation:
The green allele is recessive, meaning two copies of q (qq) are required to be green. Conversely, animals that are either pp or pq will be blue.
If 44 organisms are green, that means 44 are qq.
For genotype frequencies, the equation is:
homozygous dominant genotype + heterozygous + homozygous recessive = 100%
Which is denoted as
p² + 2pq + q² = 1
We know that q² = 44/100 = 0.44
To work out q, we can do 
= 0.66
For allele frequencies, the total must add up to 100%, so
p + q =1
We know that q= 0.66
So p = 0.34, because 0.66 + 0.34 = 1
They will pass down traits or diseases in the experiment
