Ploidy number question

[StefBall]

Saw this question in the next step test prep MCAT. Can someone explain the answer and ploidy number in general, in layman's terms?

During cyst formation one round of nuclear replication and two additional rounds of DNA replication occur, yielding a cyst with 4 nuclei.

What is the expected ploidy of a cyst that has completed a encystation?

2n

4n

8n

16n

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[bee17]

You can think of ploidy as how many copies of each chromosome are present. The vast majority of human somatic cells have 46 chromosomes total, but these are actually 2 sets of 23. We get one set (of 23) from our mother and one from our father; hence we have 2 copies of each chromosome, so the "ploidy" of humans is 2n, or diploid. Human gametes, containing only 1 set of 23 chromosomes, are haploid, or n. So when egg meets sperm, a simple math problem can represent fertilization: n + n = 2n! Two haploid gametes come together to make a diploid zygote.

According to the question, one round of nuclear replication and two additional rounds of DNA replication have happened. That means that the genome has been copied (or doubled) 3 consecutive times.

1. First round of replication: 2n*2= 4n
2. Second round of replication: 4n*2= 8n
3. Third round of replication: 8n*2= 16n

The answer is 16n (confirmed because I took the same practice exam).

***Important to note: I think this was a poorly written question. The organism in the passage was Giardia, and nowhere in the passage did they specify that the ploidy prior to encystation was 2n. What if the ploidy at that point in the Giardia lifecycle was n? Then the answer would have been 8n. I'm pretty sure the MCAT does not require students to memorize the ploidy of organisms other than humans.

 

[fig]

Late reply.

You can (probably) assume a ploidy of 2n because the nuclei have "similar appearance and DNA content", so each nucleus has at least one full copy of the chromosomes. Thought it is bit tricky given the life stage wasn't specified.

It is accurate though, here's a link to a paper on the Giardia life cycle for anyone else who stumbles into this thread: http://onlinelibrary.wiley.com/doi/10.1046/j.1462-5822.2001.00094.x/abstract

 

[wizzed101]

You can think of ploidy as how many copies of each chromosome are present. The vast majority of human somatic cells have 46 chromosomes total, but these are actually 2 sets of 23. We get one set (of 23) from our mother and one from our father; hence we have 2 copies of each chromosome, so the "ploidy" of humans is 2n, or diploid. Human gametes, containing only 1 set of 23 chromosomes, are haploid, or n. So when egg meets sperm, a simple math problem can represent fertilization: n + n = 2n! Two haploid gametes come together to make a diploid zygote.

According to the question, one round of nuclear replication and two additional rounds of DNA replication have happened. That means that the genome has been copied (or doubled) 3 consecutive times.

1. First round of replication: 2n*2= 4n
2. Second round of replication: 4n*2= 8n
3. Third round of replication: 8n*2= 16n

The answer is 16n (confirmed because I took the same practice exam).

***Important to note: I think this was a poorly written question. The organism in the passage was Giardia, and nowhere in the passage did they specify that the ploidy prior to encystation was 2n. What if the ploidy at that point in the Giardia lifecycle was n? Then the answer would have been 8n. I'm pretty sure the MCAT does not require students to memorize the ploidy of organisms other than humans.

Forgive my ignorance but the problem statement indicated that the there were 4 nuclei in the cyst. So assuming that the original organism was diploid, shouldn't the ploidy number of the cyst be 8n?
I was taught that only the ploidy number is basically the number of homologous chromosomes, no matter what form it takes. For example, in prophase of mitosis, the chromosomes are in sister chromatid forms. Even if the DNA material of each pair of sister chromatids is twice as much as that of a chromosome in G1 or G0, it is still just 1 chromosome. And the only instance during mitosis when a cell is considered 4n is just after anaphase (when the sister chromatids are seperated) AND before cytokinesis is completed.

In that question, I don't see a scenario where the cell can be 16n with 4 nuclei. Even if there were a non-disjunction which resulted in the amount of genetic material being doubled in each of those 4 nuclei, the cell would still tally up to 8n.
Thx

 

[bee17]

Forgive my ignorance but the problem statement indicated that the there were 4 nuclei in the cyst. So assuming that the original organism was diploid, shouldn't the ploidy number of the cyst be 8n?
I was taught that only the ploidy number is basically the number of homologous chromosomes, no matter what form it takes. For example, in prophase of mitosis, the chromosomes are in sister chromatid forms. Even if the DNA material of each pair of sister chromatids is twice as much as that of a chromosome in G1 or G0, it is still just 1 chromosome. And the only instance during mitosis when a cell is considered 4n is just after anaphase (when the sister chromatids are seperated) AND before cytokinesis is completed.

In that question, I don't see a scenario where the cell can be 16n with 4 nuclei. Even if there were a non-disjunction which resulted in the amount of genetic material being doubled in each of those 4 nuclei, the cell would still tally up to 8n.
Thx

A year later, I'm not really remembering this practice exam, but here's a link that talks about ploidy in Giardia: http://www.ncbi.nlm.nih.gov/pubmed/11207620

Basically, each of the four nuclei is 4n, making the whole cyst 16n

 

[wizzed101]

It appears that the authors consider G2 stage of a diploid tetraploid.

But this link http://cyberbridge.mcb.harvard.edu/mitosis_4.html says otherwise...