Answer:
E) Either anaphase I or II
Explanation:
Failure of segregation of homologous chromosomes during anaphase I or failure of segregation of sister chromatids during anaphase II leads to the presence of the abnormal number of chromosomes in resultant gametes. In the given example, the egg mother cell with 48 chromosomes (24 pairs) would enter meiosis I but the failure of one pair of homologous chromosomes to segregate from each other followed by normal meiosis II would result in the formation of two gametes with one extra chromosome and two gametes with one less chromosome.
On the other hand, if the nondisjunction occurs at anaphase II of meiosis II, two normal gametes, one gamete with one extra chromosome and one gamete with one less chromosome will be formed. Therefore, nondisjunction at anaphase I or anaphase II would have resulted in the production of eggs with one extra chromosome.
Answer:
I think this is 2 because there would have been bigger predators in the ocean vs. a lake
<span>In the skeletal muscle cells of vertebrates, as many as 38 molecules of ATP are produced from one molecule of glucose. This is less than might be expected, because electrons from NADH produced during glycolysis must be shuttled through the inner mitochondrial membrane at a cost.
</span>The energy of the electrons can be used to make ATP and in eukaryotes, glycolysis occurs in the <span>cytosol, outside mitochondria. </span>
Because if a hypha is damaged, fungi can seal septa pores to prevent cytoplasmic leakage.
