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2 years ago

Describe what happens by the end of anaphase.

Biology

1 answer:

Vsevolod [243]2 years ago

4 0

Answer:

d. The cell begins to elongate and the two poles have an equivalent collection of chromosomes.

Explanation:

The cell cycle is a fundamental cellular process by which a parent cell divides into two or more daughter cells. In somatic cells, this cycle can be divided into two major phases: interphase, where the cell prepares for its division, and mitosis or 'M phase'. The M phase can in turn be divided into four stages: 1-prophase (also divided into early prophase and prometaphase), 2-metaphase, 3-anaphase, and 4-telophase. During prophase, chromatin condenses, thereby forming visible chromosomes. Subsequently, during metaphase, the sister chromatids (i.e., the two identical halves of a single replicated chromosome) align along the middle of the cell at the metaphase plate by attaching their centromeres to the spindle fibers. Next, during anaphase, sister chromatids are separated and move to opposite poles of the cell, pulled by the mitotic spindle fibers. At the end of anaphase, the microtubules of the mitotic spindle pull the two sister chromatids toward opposite poles, thereby the cell gets begins to lengthen. Finally, during the telophase, daughter chromosomes arrive at opposite poles and uncoil, while daughter nuclei begin to form at the two poles and nuclear envelopes are formed.

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Answer:

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Explanation:

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= 8.5 x $10^{-3}$ m

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∴ Area of the eye, A = $\pi .r^{2}$

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Let $P_{min}$ be the minimum number of photons.

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Thus the minimum number of photons is given by

$N_{min}=P_{min}/E$

where E = $hc/\lambda$

= $\left (6.63\times 10^{-34}\times 3\times 10^{8} \right )/548\times 10^{-9}$

= $3.62\times 10^{-19} J$

Therefore, $N_{min}$ = $\frac{2.8\times 10^{-17}}{3.62\times 10^{-19}}$

= 77.34 photons per second

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