Prophase
Anaphase
Metaphase
Telophase
Cytokinesis
Answer:
In sample A,
The percentage of thymine or T is 20.5 %. A = T, so the percentage of A will also be 20.5 %.
Thus, A + T = 20.5 + 20.5 = 41 %
Therefore, G + C = 100 - 41 = 59 %
So, G = 59 /2 = 29.5 %
Also. G = C = 29.5 %
In sample B,
The percentage of Thymine or T is 30.7 %. As A = T, so A will also be 30.7 %.
Thus, A + T = 30.7 + 30.7 = 61.4 %
Also, G + C = 100 - 61.4 = 38.6 %
So, G = 38.6 / 2 = 19.3 %
Therefore, C will also be 19.3 %. As GC content is more found in the sample A, that is, 59 %. Hence, the sample A will possess the higher temperature to denature in comparison to B.
Answer:
In the monogastric diet, starch is the primary carbohydrate. In the small intestine, starch is digested by pancreatic amylase in conjunction with other enzymes. The complex polysaccharides are completely digested to monosaccharides. The monosaccharides are readily absorbed into the bloodstream via the small intestine.
Explanation:
Answer:
D
Explanation:
Mitochondria in eukaryotes are believed to have evolved from a symbiotic relationship between protists and prokaryotes. It is proposed that maybe it evolved from protists engulfing prokaryotes as ‘food’. The evidence is due to the fact that mitochondria have their DNA that encodes for about 13 proteins used within the organelle. MtDNA is circular in shape which is very similar to the structure of prokaryotic DNA.
