question ①
you are studying a line of cells and need to determine which phase they are in. in one sample, you find that the amount of dna present is twice as much as the normal amount in non - dividing cells. during which stage(s) have you harvested the cells?
select all that apply
□ g₂
□ g₀
□ g₁
□ s

Question

Accepted Answer

# Brief Explanations:
1. Recall the cell cycle phases and DNA content:
   - In $ G_0 $ and $ G_1 $ phases, the DNA content is the normal (diploid) amount as in non - dividing cells.
   - During the S phase, DNA replication occurs, and by the end of S phase, the DNA content doubles.
   - In the $ G_2 $ phase, the cell has already completed DNA replication, so the DNA content is twice the normal (diploid) amount (same as after S phase).
2. Analyze each option:
   - $ G_2 $: After DNA replication (which happens in S phase), the DNA content is doubled. So cells in $ G_2 $ have twice the normal DNA amount.
   - $ G_0 $: Cells in $ G_0 $ are in a non - dividing state, so their DNA content is normal (not doubled).
   - $ G_1 $: Cells in $ G_1 $ have not yet replicated their DNA, so the DNA content is normal (not doubled).
   - S: During the S phase, DNA replication is taking place. By the end of S phase, the DNA content is doubled. So cells harvested during S (especially towards the end) or $ G_2 $ will have twice the normal DNA amount. Wait, actually, during S phase, as replication proceeds, the DNA content increases from the normal to double. But typically, when we say the DNA amount is twice the normal, it's after S phase (in $ G_2 $) or during the later part of S phase. But in standard cell cycle analysis, $ G_2 $ and the end of S phase have double the DNA. However, the main phases with double the normal DNA (compared to $ G_1/G_0 $) are $ G_2 $ (and M phase, but M is not an option here) and the S phase after replication is complete. But more accurately, $ G_2 $ cells have fully replicated DNA (double the normal), and S phase cells are in the process of replicating. But the question says "the amount of DNA present is twice as much as the normal amount in non - dividing cells". So $ G_2 $ cells have double the DNA (since they have completed S phase). Also, in S phase, once replication is complete, the DNA is double. But usually, the answer here is $ G_2 $ and S? Wait, no. Wait, non - dividing cells are in $ G_0 $ or $ G_1 $ (with normal DNA). In S phase, DNA is being replicated, so the DNA content goes from normal to double. So at the start of S, it's normal, at the end, it's double. In $ G_2 $, it's double. So the cells with twice the normal DNA are in $ G_2 $ (and also in S phase after replication is done, but maybe the question considers $ G_2 $ and S? Wait, let's re - check:
   - $ G_0 $: Normal DNA (same as non - dividing), so no.
   - $ G_1 $: Normal DNA (same as non - dividing), so no.
   - S: During S phase, DNA replication occurs. So cells in S phase can have DNA amounts from normal (start of S) to double (end of S). But the question says "the amount of DNA present is twice as much as the normal amount". So cells at the end of S or in $ G_2 $ have double the DNA. But maybe the intended answer is $ G_2 $ and S? Wait, no, let's think again. Non - dividing cells (like $ G_0 $) have a certain amount of DNA (let's say 2n). In $ G_1 $, it's also 2n. In S phase, DNA replicates to 4n (twice 2n). In $ G_2 $, it's 4n. So the phases with twice the normal (2n→4n) are S (after replication, i.e., during S when replication is complete) and $ G_2 $. But maybe the question considers $ G_2 $ and S? Wait, the options are $ G_2 $, $ G_0 $, $ G_1 $, S. So the correct options are $ G_2 $ and S? Wait, no, let's check standard cell cycle DNA content:
     - $ G_0 $: 2n (same as non - dividing)
     - $ G_1 $: 2n
     - S: 2n→4n (during replication)
     - $ G_2 $: 4n
     - M: 4n (until cytokinesis)
   So if the DNA is twice the normal (2n→4n), then the cells are in S (after replication, i.e., when replication is complete, so during S phase, towards the end) or $ G_2 $. But the question is "during which stage(s) have you harvested the cells?". So the stages with DNA amount twice the normal are $ G_2 $ (because in $ G_2 $, DNA is already replicated to 4n) and S phase (when replication is complete, so the DNA is 4n). Wait, but maybe the answer is $ G_2 $ and S? Wait, let's check the options again. The options are $ G_2 $, $ G_0 $, $ G_1 $, S. So the correct options are $ G_2 $ and S? Wait, no, maybe I made a mistake. Wait, non - dividing cells: their DNA content is 2n (for diploid cells). In $ G_1 $, it's 2n. In S phase, DNA is replicated, so the content becomes 4n (twice 2n). In $ G_2 $, it's 4n. So the cells with twice the normal DNA (4n) are in S phase (after replication, i.e., during S when replication is done) and $ G_2 $. So the correct options are $ G_2 $ and S. Wait, but let's confirm with cell cycle knowledge. The S phase is the synthesis phase where DNA is replicated. So at the start of S, DNA is 2n, at the end, it's 4n. $ G_2 $ is the gap after S, so DNA is 4n. So both S (late S) and $ G_2 $ have 4n (twice 2n). So the correct options are $ G_2 $ and S.
# Answer:
$ G_2 $ (the box next to $ G_2 $ should be checked) and S (the box next to S should be checked)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Answer: