How many meiotic divisions are essential in formation of 100 seeds in cyperaceae family? (Given ans = 200)

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Please find below the solution to the asked query

The answer is 200 meiotic division. For producing 100 seeds, 100 female gametes (eggs) and 100 male gametes (pollen grains) are needed. In cyperaceae family one functional microspore (pollen) is produced during single meiotic division and similarly one functional megaspore (egg) is produced during single meitotic division. Hence, 200 meiotic divisions (100 each to produce 100 megaspore and 100 microspore) are needed to produce 100 seeds. 

This can be explained in detail as:

For the production of 100 microspores (pollen grains) and 100 megaspores (eggs) in a member of family Cyperaceae, 200 meiotic divisions occur 100 in each anther and carpel. It is because, out of the 4 microspores and 4 megaspores produced in anther and carpel respectively from one meiotic division each, only one survives and three degenerates. This is an abnormal behavior shown by the member of Cyperaceae. It is against of n+n/4 division rule.

n+n/4 division rule is the rule for determining the number of meiotic divisions required to produce seeds in normal cases of angiospermous plants. n represents number of seeds. For example - for 100 seeds production, total number of meiotic divisions required are $100 + \frac{100}{4} = 125$. But in case of Cyperaceae, abnormal behaviour is shown while pollen development. Of the four nuclei formed after meiosis during microsporogenesis, only one is functional. So, this results in exceptional case of n+n/4 division rule and the rule  becomes 2n for Cyperaceae.

Hope this information will clear your doubts about the Cyperaceae.
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