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:
4. The correct pathway for the flow of electron during photosynthesis is mentioned in option D. 
> <em>NADP </em>> <em>Calvin cycle.</em>
- Electron is first provided to the photo-system II by the phtolysis of water.
- Then it is passed to photo-system I with the help of plastoquinone (PQ) and cytochrome
. - Finally, electron from the photo-system II is used to reduce NADP to NADPH.
- NADPH and ATP are used in light independent phase or Calvin cycle to synthesize carbohydrate from carbon dioxide.
5. The correct answer is B.) photosynthesis.
- Photosynthesis is the process by which green plants synthesize glucose or carbohydrate from carbon dioxide and water in presence of sunlight and chlorophyll.
- In light independent phase of photosynthesis, it requires ATP and NADPH in order to synthesize food.
- Lastly, oxidation or breaking down of glucose releases energy and photosynthesis is a anabolic process instead of catabolic one.
I believe it would be C) Cellulose. As this is one of the essential components found in the cell wall of plant cells.
Layer c is the oldest and then it’s b and a
The answer to this is true
