Autosomes. The somatic chromosomes are referred to as autosomes.
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.
<span>"they form a cyst: a hard covering is formed around the protozoan and metabolic rate is slowed"</span>
Answer and Explanation: In enzyme kinetics, one constant describing enzyme activity is <em>Maximal Velocity</em> (Vmax). It indicates how fast an enzyme can catalyze the reaction. It is dependent on substrate concentration.
As the muscle is an organ which needs a great amount of energy, the enzyme glycogen phosphorilase is very active on the organ, compared to the liver, where glicose is stored. So, the Vmax of glycogen phosphorylase expressed in muscle is faster than when expressed in the liver, means the enzyme in muscle has a bigger concentration of substrate and therefore will reach Vmax faster, i.e. will be significantly larger.
