meiosis question

[joonkimdds]

1) After meiosis 1 (_) daugheter cells have been produced, and each cell is (_).

answer is two; haploid

why isn't it two; diploid?

2n => 2n(diploid) after meiosis 1 but then it divides again without being replicated this time so 2n =>1n(haploid).

The question asked after meiosis1, not meiosis2 so isn't it diploid?

2) crossing over involves the exchange of chromosomal segments between
a) sister chromatids of a chromosome
b) chromatids of nonhomologues.
c) nonsister chromatids of homologues
d) sex chromosomes and autosomes
e) sex chromosomes only.

answer is C but can't e be the answer too because cross over happens in meiosis only and meiosis uses sex chromosomes.
by the way, what's autosomes?

---

Members don't see this ad.

 

[Streetwolf]

xsome = chromosome

1) Meiosis I pulls the xsome pairs apart so that each daughter cell (2 of them) contains half the # of xsomes = haploid. In the second division, these xsomes are divided again into the separate strands.

2) Meiosis occurs in sex CELLS not sex xsomes. So both autosomes (non sex xsomes) and sex xsomes can undergo crossing over.

There are 23 pairs of xsomes in a human, 22 are autosomes and the 23rd pair is the sex xsome (XY or XX).

 

[joonkimdds]

xsome = chromosome

1) Meiosis I pulls the xsome pairs apart so that each daughter cell (2 of them) contains half the # of xsomes = haploid. In the second division, these xsomes are divided again into the separate strands.

if it becomes haploid, and the second division divide these, then shouldn't it be half of the half? thus. if we started with 2n it becomes => 1n and then the second division makes this into => 0.5n?

 

[Streetwolf]

Let n be the number of xsome pairs. What happens is that in Meiosis I you stay with 2n distinct xsomes but everything is duplicated. Now what normally happens in MITOSIS is that the sister chromatids (the duplicate strands) are pulled apart from the original strands in all 2n xsomes, resulting in 2 identical cells each with 2n xsomes. In MEIOSIS I, the homologous xsomes separate. This leads to the haploid, or n, state in the beginning of Meiosis II. From there, the sister chromatids separate in each cell.

Here's the human example. We have 23 pairs of xsomes. So there are 46 distinct xsomes. In Meiosis I they all duplicate so we have 92 xsomes now. Call them a1, a2, ... , a46 and b1, b2, ... , b46. So the b's are our duplicates and the a's are our originals. Let's also assume that a1 and a2 are a xsome pair, a3 and a4 are a xsome pair, ... , and a45 and a46 are a xsome pair (same goes for the b's).

What happens in MITOSIS is that a1 and b1, a2 and b2, ... , a46 and b46 are pulled apart in Anaphase and so we end up with 2 cells, one with a1, ... , a46 and the other with b1, ... , b46. What happens in MEIOSIS I is that a1 and b1 will be attached to each other and separate from a2 and b2 which are also attached to each other. So after meiosis I we have one cell that contains a1/b1, a3/b3, a5/b5, ... , a45/b45 and the second cell that contains a2/b2, a4/b4, a6/b6, ... , a46/b46. So since the b's are just duplicates of the a's, you can see that each cell contains 23 DISTINCT xsomes, the haploid number. In the subsequent division in Meiosis II, the paired up a's and b's will separate from each other, as in mitosis, to form the 4 haploid cells.

If you still don't get that you could try googling the following:

Meiosis
Sister Chromatids
Homologues

 

[thejamespark]

Let n be the number of xsome pairs. What happens is that in Meiosis I you stay with 2n distinct xsomes but everything is duplicated. Now what normally happens in MITOSIS is that the sister chromatids (the duplicate strands) are pulled apart from the original strands in all 2n xsomes, resulting in 2 identical cells each with 2n xsomes. In MEIOSIS I, the homologous xsomes separate. This leads to the haploid, or n, state in the beginning of Meiosis II. From there, the sister chromatids separate in each cell.

Here's the human example. We have 23 pairs of xsomes. So there are 46 distinct xsomes. In Meiosis I they all duplicate so we have 92 xsomes now. Call them a1, a2, ... , a46 and b1, b2, ... , b46. So the b's are our duplicates and the a's are our originals. Let's also assume that a1 and a2 are a xsome pair, a3 and a4 are a xsome pair, ... , and a45 and a46 are a xsome pair (same goes for the b's).

What happens in MITOSIS is that a1 and b1, a2 and b2, ... , a46 and b46 are pulled apart in Anaphase and so we end up with 2 cells, one with a1, ... , a46 and the other with b1, ... , b46. What happens in MEIOSIS I is that a1 and b1 will be attached to each other and separate from a2 and b2 which are also attached to each other. So after meiosis I we have one cell that contains a1/b1, a3/b3, a5/b5, ... , a45/b45 and the second cell that contains a2/b2, a4/b4, a6/b6, ... , a46/b46. So since the b's are just duplicates of the a's, you can see that each cell contains 23 DISTINCT xsomes, the haploid number. In the subsequent division in Meiosis II, the paired up a's and b's will separate from each other, as in mitosis, to form the 4 haploid cells.

If you still don't get that you could try googling the following:

Meiosis
Sister Chromatids
Homologues

you sir are very good at explaining things. i was starting to get confused again with this stuff because of a question i ran into.
Is it possible if you can take a quick look at my thread regarding 2 questions i did?


So I was doing a problem and wanted to see if my logic is correct

1) If a cell has 46 chromosomes at the beginning of mitosis, then at anaphase there would be a total of...
a) 23 chromatids
b) 23 chromosomes
c) 46 chromosomes
d) 46 chromatids
e) 92 chromosomes
The answer is E.
In beginning of Anaphase, the two sister chromatids are split apart and are now considered EACH chromosome so that is why its 92 chromosomes yes?

The book answer explains: "Metaphase ends when each chromosome separates into a pair of chromatids. During anaphase, chromatids from each pair move to opposite poles. Each chromatid is now considered a complete chromosome, since it consists of a complete DNA molecule. To count the number of chromosomes at any point during the cell cycle, count the centromeres."


2) If a cell has 46 chromosomes at the beginning of meiosis, then at anaphase I there would be a total of...
a) 23 chromatids
b) 23 chromosomes
c) 46 chromosomes
d) 46 chromatids
e) 92 chromosomes
Answer: C

Could you explain #2's answer??

From my knowledge, the homologue pairs are now splitting apart with half the number of chromosomes (n =23) as you explained with the a1/b1, a3/b3.... and so forth.
Then since we have two new rising cells with haploid number(now EACH considered a chromosome), the TOTAL = 46 chromosomes. is that correct?

Just want to get my logic clear.
Any help would be appreciated.

 

[jefff]

you sir are very good at explaining things. i was starting to get confused again with this stuff because of a question i ran into.
Is it possible if you can take a quick look at my thread regarding 2 questions i did?


So I was doing a problem and wanted to see if my logic is correct

1) If a cell has 46 chromosomes at the beginning of mitosis, then at anaphase there would be a total of...
a) 23 chromatids
b) 23 chromosomes
c) 46 chromosomes
d) 46 chromatids
e) 92 chromosomes
The answer is E.
In beginning of Anaphase, the two sister chromatids are split apart and are now considered EACH chromosome so that is why its 92 chromosomes yes?

The book answer explains: "Metaphase ends when each chromosome separates into a pair of chromatids. During anaphase, chromatids from each pair move to opposite poles. Each chromatid is now considered a complete chromosome, since it consists of a complete DNA molecule. To count the number of chromosomes at any point during the cell cycle, count the centromeres."


2) If a cell has 46 chromosomes at the beginning of meiosis, then at anaphase I there would be a total of...
a) 23 chromatids
b) 23 chromosomes
c) 46 chromosomes
d) 46 chromatids
e) 92 chromosomes
Answer: C

Could you explain #2's answer??

From my knowledge, the homologue pairs are now splitting apart with half the number of chromosomes (n =23) as you explained with the a1/b1, a3/b3.... and so forth.
Then since we have two new rising cells with haploid number(now EACH considered a chromosome), the TOTAL = 46 chromosomes. is that correct?

Just want to get my logic clear.
Any help would be appreciated.

when a pair sister chromatids 'X' --separate--> '| |' you have an increase in number of chromosomes

when tetrads/homologous pairs separate there is still the same number of chromosomes and chromatids until cytokensis

when tetrads aka homologous pairs separate and after cytokensis occurs you have a decrease in the number of chromosomes and chromatids


in meiosis I you have tetrads/homologous pair which consists of 2 sets of sister chromatids (XX) *4 chromatids total*

since #2 asks @ anaphase there are still 46 chromosomes

(XX) - meiosis 46 chromosomes (92 chromatids) ---1 cell

(X X) - meiosis I @ anaphase 46 chromosomes (92 chromatids) ---1 cell *(pre-cytokensis)

(X) (X) - meiosis I @ end of telophase/cytokensis 23 chromosomes (46 chromatids) ---2 cells ... therefore these #s are per cell

let me know if this helps, good luck 👍

 

[DrRoyal Pains]

Diploid cells have nothing to do with Meiosis. The main purpose is to produce 4 genetically distinct daughter cells. Thus, after Meiosis I you have two haploid daughter cells, which each undergo an additional round of Meiosis, giving rise to 4 haploid daughter cells. When these gametes fuse with other sex cells, a zygote forms (a diploid cell).