Imagine that dna is composed of equal proportions of six different types of nucleotide, instead of the normal four. the six nucleotides are a, c, g, t, x, and y. in this new type of dna molecule, a pairs with t, c pairs with g, and x pairs with y. what would be the average size of the dna fragments produced when this new dna is cut by a restriction endonuclease with an eight-base recognition sequence?

The average size of the DNA fragments produced when this new DNA is cut by a restriction endonuclease with an eight-base recognition sequence would be 1679616.

Explanation:

Restriction enzymes can be described as the enzymes which are specific to certain DNA sites and cut the DNA from that particular site.

Normally, the average size of a DNA fragment cut by a restriction enzyme is calculated by the formula 4^{no.of base recognition sequence}.

As the number of bases has changed to 6 and the restriction enzyme has eight base recognition site then the formula will be:

6^{no.of base recognition sequence}

i.e

6^{8}  = 1679616

b. heterotroph

explanation:

the oxygenated blood then leaves the lungs through pulmonary veins, which return it to the left heart, completing the pulmonary cycle. this blood then enters the left atrium, which pumps it through the mitral valve into the left ventricle. from the left ventricle, the blood passes through the aortic valve to the aorta.

deoxygenated blood leaves through the right ventricle through the pulmonary artery. from the right atrium, the blood is pumped through the tricuspid valve (or right atrioventricular valve), into the right ventricle.