Answer:
Replicated chromosomes at metaphase I = 66
Sister chromatids at metaphase I = 66 x 2 = 132
Sister chromatids at prophase II = 66
Chromosomes in each sperm cells = 33
Explanation:
Metaphase I of meiosis I would have 66 replicated chromosomes in the testicular cells of the bird. Each of the replicated chromosomes would have two sister chromatids. So, a total of 66 replicated chromosomes would have 66 x 2 = 132 sister chromatids.
Due to segregation of homologous chromosomes towards opposite poles in anaphase I, each daughter cell formed by the end of meiosis I would have 33 replicated chromosomes. So, each of the daughter cells would have a total 33 x 2 = 66 sister chromatids at prophase II.
Since meiosis II maintains the chromosome number, each sperm cell formed by the end of meiosis II would have 33 chromosomes.
Answer:
Glycolysis.
Explanation:
Glycolysis is a universal process that provides energy in the form of ATP molecules. It requires two molecules of NAD+, which are reduced to NADH during glycolysis. Thus, regeneration of NAD+ is necessary as if NAD+ is absent, glycolysis cannot be able to continue.
During anaerobic respiration (respiration in the absence of oxygen), fermentation takes place to regenerate NAD+ used in the process of glycolysis.
Answer is B in my opinion
The answer is C) Two. Dipolar means there are two magnetic poles charging with the same magnitude but opposite polarity.