NS Diagnostic B/B #6

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NS Diagnostic B/B #6

Post by NS_Tutor_Will » Fri Aug 24, 2018 10:39 am

A student recently wrote to us about this question: "I don't understand how it goes from 2n to 4n to 8n to 16n. I thought it would be an increase of 2 by adding and not multiplying, thus bringing it to 8n total."

In this question, we are asked for the ploidy of a cyst that has completed encystation. The ploidy, at first, is 2n. The passage tells us that the process of cyst formation (encystation) involves "one round of nuclear replication and two additional rounds of DNA replication occur, yielding a cyst with 4 nuclei."

In the first step, the 2n ploidy nucleus replicates, leading to two nuclei of ploidy 2n; this means that the ploidy of the cyst overall will be 4n.

In the second step, the DNA replicates within each of the two nuclei (of 2n each), leading to two nuclei with ploidy 4n; this means the ploidy of the cyst overall will be 8n.

In the third step, the DNA replicates again, leading the two nuclei to be 8n each; this means the ploidy of the cyst overall will now be 16n.

I hope this helps clear up any confusion! Good luck!
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