I have a balloon that can hold 100. liters

Dmitrij [34]

I haven't taken biology in a few years, so I may be wrong, but to me it looks like mitosis, since it only divides once, and if I can see the picture correctly it looks like each cell contains the same amount of chromosomes, but I can't see the picture very well. If it helps here are some differences between meiosis and mitosis:

1. Cell Division

Mitosis: A somatic cell divides once. Cytokinesis (the division of the cytoplasm) occurs at the end of telophase.

Meiosis: A reproductive cell divides twice. Cytokinesis happens at the end of telophase I and telophase II.

2. Daughter Cell Number

Mitosis: Two daughter cells are produced. Each cell is diploid containing the same number of chromosomes.

Meiosis: Four daughter cells are produced. Each cell is haploid containing one-half the number of chromosomes as the original cell.

3. Genetic Composition

Mitosis: The resulting daughter cells in mitosis are genetic clones (they are genetically identical). No recombination or crossing over occur.

Meiosis: The resulting daughter cells contain different combinations of genes. Genetic recombination occurs as a result of the random segregation of homologous chromosomes into different cells and by the process of crossing over (transfer of genes between homologous chromosomes).

4. Length of Prophase

Mitosis: During the first mitotic stage, known as prophase, chromatin condenses into discrete chromosomes, the nuclear envelope breaks down, and spindle fibers form at opposite poles of the cell. A cell spends less time in prophase of mitosis than a cell in prophase I of meiosis.

Meiosis: Prophase I consists of five stages and lasts longer than prophase of mitosis. The five stages of meiotic prophase I are leptotene, zygotene, pachytene, diplotene, and diakinesis. These five stages do not occur in mitosis. Genetic recombination and crossing over take place during prophase I.

5. Tetrad Formation

Mitosis: Tetrad formation does not occur.

Meiosis: In prophase I, pairs of homologous chromosomes line up closely together forming what is called a tetrad. A tetrad consists of four chromatids (two sets of sister chromatids).

6. Chromosome Alignment in Metaphase

Mitosis: Sister chromatids (duplicated chromosome comprised of two identical chromosomes connected at the centromere region) align at the metaphase plate (a plane that is equally distant from the two cell poles).

Meiosis: Tetrads (homologous chromosome pairs) align at the metaphase plate in metaphase I.

7. Chromosome Separation

Mitosis: During anaphase, sister chromatids separate and begin migrating centromere first toward opposite poles of the cell. A separated sister chromatid becomes known as daughter chromosome and is considered a full chromosome.

Meiosis: Homologous chromosomes migrate toward opposite poles of the cell during anaphase I. Sister chromatids do not separate in anaphase I.

I can add the similarities between them if you need. This work is not mine, I got it from thoughtCo.

Good luck :)

3 0

3 years ago

As an FDA physiologist, you need 0.625 L of phosphoric acid acid / dihydrogen phosphate (H3PO4 (aq) / H2PO4 - (aq) ) buffer with

aksik [14]

Answer:

0.4058L of 1.0M H3PO4

0.2192L of 1.5M NaOH

Explanation:

The pKa of the H3PO4 / H2PO4- buffer is 2.12

To solve this question we must use H-H equation for this system:

pH = pKa + log [H2PO4-] / [H3PO4]

2.75 = 2.12 + log [H2PO4-] / [H3PO4]

0.63 = log [H2PO4-] / [H3PO4]

Where [] could be taken as the moles of each reactant

As you have H3PO4 solution, the reaction with NaOH is:

H3PO4 + NaOH → H2PO4- + Na+ + H2O

As you can see, both H3PO4 and H2PO4- comes from the same 1.0M H3PO4 solution

The moles of H3PO4 are:

[H3PO4] = Moles H3PO4 - Moles NaOH

And for H2PO4-:

[H2PO4-] = Moles NaOH added

Replacing in (1):

4.2658 = [Moles NaOH] / [Moles H3PO4 - Moles NaOH]

4.2658 Moles H3PO4 - 4.2658 moles NaOH = Moles NaOH

4.2658 Moles H3PO4 = 5.2658 moles NaOH <em>(1)</em>

In volume:

0.625L = Moles H3PO4 / 1.0M + Moles NaOH / 1.5M

0.625 = Mol H3PO4 + 0.6667 Moles NaOH <em>(2)</em>

Replacing (2) in (1):

4.2658 Moles H3PO4 = 5.2658 (0.625 - Mol H3PO4 / 0.6667)

4.2658 Moles H3PO4 = 5.2658 (0.625 - Mol H3PO4) / 0.6667

4.2658 Moles H3PO4 = 5.2658*(0.9375 - 1.5 mol H3PO4)

4.2658 Moles H3PO4 = 4.9367 -7.8983 mol H3PO4

12.1641 mol H3PO4 = 4.9367

Mol H3PO4 = 0.4058moles * (1L / 1.0moles) =

0.4058L of 1.0M H3PO4

And:

0.625L - 0.4058L =

0.2192L of 1.5M NaOH

5 0

3 years ago

suppose you mix 100.0 g of water at 22.6 oc with 75.0 g of water at 75.4 oc. what will be the final temperature of the mixed wat

Arada [10]

The temperature of mixed water is 45.230C under the conditions stated.

<h3>How can the change in temperature when mixing water be calculated?</h3>

T(final) = (m1 T1 + m2 T2) / (m1 + m2), at which m1 and m2 are indeed the weight training of the water in the initial and 2nd canisters, T1 is the water temperature in the first container, and T2 is the water temperature in the second container, can be used to determine the final the water's temperature mixture.

<h3>Briefing:</h3>

The given parameters;

mass of the cold water, m = 100 g

initial temperature of the water, t₁ = 22.6 ⁰C

initial temperature of the hot water, t₂ = 75.4⁰ C

75 g is the hot water's mass.

specific heat capacity of water is 4.184 J/g⁰C

The mixture's final temperature is estimated as follows;

Based on the principle of conservation of energy;

Heat received by the ice water equals heat lost by the hot water.

mcΔθ₂ = mcΔθ₁

75 x 4.184 x (75.4 - T) = 100 x 4.184 x (T - 22.6)

75 x (75.4 - T) = 100 x (T - 22.6)

(75.4 - T) = 1.333(T - 22.6)

75.4 - T = 1.333T - 30.1258

75.4 + 30.1258 = 1.333T + T

105.5258 = 2.333T

T = 45.2318⁰C

To know more about temperature visit :

brainly.com/question/29386637

#SPJ4

5 0

1 year ago

If an unknown substance CANNOT be broken down into simpler substances, it is

Fynjy0 [20]

If an unknown substance CANNOT be broken down into simpler substances, it is AN ELEMENT made of one kind of atom.

7 0

3 years ago

Read 2 more answers

000 atm of dry nitrogen, placed in a container having a pinhole opening in its side, leaks from the container 3.55 times faster

Ivanshal [37]

We will assume that the question is discussing 1.000 atm of N₂ initially. The question is discussing diffusion rates of two gases and asks us to identify the species. We can use Graham's Law to attempt this problem with the following formula:

Rate₁/Rate₂ = sqrt(M₂/M₁)

We are told that the N₂ is 3.55 times as fast as the unknown species, so rate 1 = 3.55 and rate 2 = 1. We know the molecular weight of N₂ as 28 g/mol. Now we can use the equation above to solve for the molecular weight of the unknown, M₂:

3.55/1 = sqrt(M2/28)
(3.55)² = M₂/28
M₂ = 28 (3.55)₂
M₂ = 353 g/mol

The unknown compound has a molecular mass of roughly 353 g/mol and this is very close to the molecular mass of UF₆ which is 352.02 g/mol. Therefore, it is likely that the unknown gas is UF₆.

7 0

4 years ago