Mitosis and Meiosis Worksheet | PDF - Free Printable

Let’s solve this step by step.

We are given a parent cell with 4 chromosomes (2 pairs of homologous chromosomes). Each chromosome is duplicated (has two sister chromatids), so we see:

- One pair: Chromosome with alleles A and A (sister chromatids) — let’s call this Chromosome 1
- Its homolog: Chromosome with alleles a and a — Chromosome 2
- Another pair: Chromosome with B and B — Chromosome 3
- Its homolog: Chromosome with b and b — Chromosome 4

So, diploid number = 4 chromosomes (2n=4), each with 2 chromatids → total 8 chromatids in the starting cell.

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PART I: MITOSIS

Mitosis produces 2 identical daughter cells. The chromosome number stays the same (diploid → diploid).

Steps:

1. Prophase: Chromosomes condense. Nuclear envelope breaks down. Spindle forms.
→ Draw all 4 chromosomes (each as X-shaped, since they’re duplicated) floating in the cell. Label them: AA, aa, BB, bb.

2. Metaphase: Chromosomes line up at the equator (middle) of the cell.
→ Draw all 4 X-shaped chromosomes lined up single file along the center line.

3. Anaphase: Sister chromatids separate and move to opposite poles.
→ Draw 4 single-chromatid chromosomes going to left pole, and 4 going to right pole. So each side gets: A, a, B, b.

4. Telophase: Two nuclei form around the separated chromosomes. Cell starts to pinch.
→ Draw two groups of 4 single-chromatid chromosomes (A, a, B, b on each side), with nuclear membranes forming.

5. Daughter Cells: Two identical cells, each with 4 chromosomes (same as parent).
→ Draw two circles. In each, draw 4 single-line chromosomes: one labeled A, one a, one B, one b.

✔ Mitosis result: 2 diploid daughter cells, genetically identical to parent.

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PART II: MEIOSIS

Meiosis reduces chromosome number by half. Produces 4 haploid gametes. Two divisions: Meiosis I (separates homologs) and Meiosis II (separates sister chromatids).

Parent cell same as above: 4 chromosomes (AA, aa, BB, bb) — all duplicated.

Meiosis I:

1. Prophase I: Homologous chromosomes pair up (synapsis), may cross over.
→ Draw 2 tetrads (pairs of homologous chromosomes):
- Tetrad 1: AA paired with aa
- Tetrad 2: BB paired with bb
You can show them tangled or side-by-side.

2. Metaphase I: Tetrads line up at equator. Orientation is random (independent assortment).
→ Draw both tetrads lined up at middle. For example:
Left side of tetrad 1: AA; Right side: aa
Left side of tetrad 2: BB; Right side: bb
(Note: It could be mixed — e.g., AA with bb on one side — but for simplicity, we’ll keep parental combination unless told otherwise.)

3. Anaphase I: Homologous chromosomes separate (sister chromatids stay together!).
→ Draw AA and BB moving to left pole; aa and bb moving to right pole.
(Each “chromosome” still has 2 chromatids!)

4. Telophase I: Two cells form, each with 2 chromosomes (but each chromosome has 2 chromatids).
→ Draw two cells.
Cell 1: AA and BB (each X-shaped)
Cell 2: aa and bb (each X-shaped)

Meiosis II (no DNA replication between I and II):

Now each of the two cells from Telophase I divides again.

Start with Cell 1 (AA, BB):

5. Prophase II: Chromosomes re-condense.
→ In each of the two cells, draw the 2 X-shaped chromosomes.

6. Metaphase II: Chromosomes line up singly at equator.
→ In each cell, draw AA and BB lined up at center.

7. Anaphase II: Sister chromatids separate.
→ In each cell, split AA into two A’s going to opposite poles; same for BB → two B’s.
So each pole gets: A and B.

8. Telophase II: Four nuclei form.
→ Draw four cells. Each has 2 single-chromatid chromosomes: A and B (from first division) OR a and b (from second division).

Wait — actually, depending on how homologs aligned in Metaphase I, you might get different combinations. But since no crossing over was shown, and we assumed parental alignment, we get:

→ Two daughter cells with AB
→ Two daughter cells with ab

But note: In reality, independent assortment means it could also be Ab and aB — but since the problem doesn’t specify, and the original cell shows AA/aa and BB/bb as separate pairs, we assume standard segregation.

Actually — looking back at the original diagram: the parent cell has chromosomes labeled A, a, B, b — meaning two homologous pairs. In meiosis I, homologs separate. So if in Metaphase I, the orientation was:

Left: A and B → then Right: a and b

Then after Meiosis I:

Cell 1: A and B (each with 2 chromatids)
Cell 2: a and b (each with 2 chromatids)

Then Meiosis II separates sisters:

From Cell 1: → two cells with A and B (single chromatids)
From Cell 2: → two cells with a and b (single chromatids)

So final daughter cells: two with AB, two with ab.

But wait — in the original drawing, the chromosomes are drawn with specific positions. Let me check:

In the very first circle (parent cell for mitosis), we have:

- Top: chromosome with a,a (so homozygous recessive for gene A)
- Bottom left: A,A (homozygous dominant for gene A)
- Bottom right: B,B (homozygous dominant for gene B)
- Middle right: b,b (homozygous recessive for gene B)

Actually — looking closely, the labels are on the chromatids. So:

One chromosome has two chromatids both labeled 'a' → so genotype aa
Another has two 'A' → AA
Another has two 'B' → BB
Another has two 'b' → bb

And they are arranged such that AA and aa are homologs? Probably yes — same size/shape implied. Similarly BB and bb.

So in meiosis I, homologs separate: AA goes with BB or with bb? Independent assortment.

But since the problem says “same parent cell”, and doesn’t specify crossing over or orientation, we should show one possible outcome — typically, we assume the homologs segregate without mixing unless specified.

However, in many textbooks, when drawing meiosis, they often show the homologous pairs aligning randomly. To be accurate, let’s choose one common way:

Assume in Metaphase I:

Tetrad 1 (A/a): AA on left, aa on right
Tetrad 2 (B/b): BB on left, bb on right

Then Anaphase I: left gets AA and BB; right gets aa and bb

Then Meiosis II: each splits sisters → so four gametes: two with AB, two with ab.

Alternatively, if in Metaphase I, the B/b tetrad flipped:

Tetrad 1: AA left, aa right
Tetrad 2: bb left, BB right

Then left gets AA and bb → gametes Ab
Right gets aa and BB → gametes aB

Both are valid. Since the problem doesn’t specify, either is acceptable. But to match typical worksheet expectations, perhaps they want the parental combinations.

Looking at the original cell drawing: the chromosomes are positioned with A and B on bottom, a and b on top/right — but that’s probably just layout.

I think safest is to show separation of homologs, and in Meiosis II, separation of sisters, resulting in haploid cells with one allele per gene.

Final daughter cells for meiosis: 4 cells, each with 2 chromosomes (haploid, n=2), each chromosome single chromatid.

Possible genotypes: AB, AB, ab, ab — or Ab, Ab, aB, aB — depending on alignment.

Since the problem likely expects a standard answer, and no crossing over indicated, I’ll go with:

After Meiosis I:
Cell 1: chromosomes AA and BB
Cell 2: chromosomes aa and bb

After Meiosis II:
From Cell 1: two cells with A and B
From Cell 2: two cells with a and b

So four daughter cells: two with AB, two with ab.

Now, to draw:

For Meiosis:

- Prophase I: two tetrads (AA+aa together, BB+bb together)
- Metaphase I: tetrads at equator — say AA and BB on left side of metaphase plate, aa and bb on right
- Anaphase I: AA and BB move to left pole; aa and bb to right pole (each still X-shaped)
- Telophase I: two cells — left cell has AA and BB; right cell has aa and bb
- Then for each of those cells, do Meiosis II:
- Prophase II: in each cell, two X-shaped chromosomes
- Metaphase II: line up at equator
- Anaphase II: sisters separate — so in left cell: A and A go apart, B and B go apart → each new cell gets one A and one B
- Same for right cell: each gets one a and one b
- Telophase II: four cells — two with A,B; two with a,b
- Daughter Cells: four small circles, each with two single-line chromosomes: two labeled A and B, two labeled a and b.

Wait — in the daughter cells, since it's haploid, each cell has only one copy of each chromosome type. So for gene A locus: either A or a; for gene B locus: either B or b.

Yes.

So final answer drawings:

MITOSIS:

- Prophase: 4 X-shaped chromosomes (AA, aa, BB, bb) in cell
- Metaphase: 4 X-shaped in a row at center
- Anaphase: 4 single chromatids to each pole (A,a,B,B on each side? No — wait: sisters separate, so from AA → two A’s; from aa → two a’s; etc. So each pole gets: A, a, B, b — one of each)
- Telophase: two groups, each with A,a,B,b (single lines)
- Daughter Cells: two cells, each with 4 single chromosomes: A, a, B, b

MEIOSIS:

- Prophase I: two tetrads: (AA+aa) and (BB+bb)
- Metaphase I: tetrads at equator — e.g., AA and BB on left, aa and bb on right
- Anaphase I: AA and BB to left pole; aa and bb to right pole (each still X-shaped)
- Telophase I: two cells — left: AA, BB; right: aa, bb
- Prophase II: in each cell, two X-shaped chromosomes
- Metaphase II: in each cell, chromosomes line up at equator
- Anaphase II: sisters separate — in left cell: A and A separate, B and B separate → each daughter gets one A and one B; similarly right cell: each gets one a and one b
- Telophase II: four cells — two with A,B; two with a,b
- Daughter Cells: four cells, each with two single chromosomes: two cells have A and B; two cells have a and b

This is correct.

Now, to write the final answer as requested — since the user wants the solution filled in, but we can't draw, we describe what to draw in each stage.

But the instruction says: "Solve the problem accurately." and "Final Answer" should be provided.

Since it's a drawing task, the final answer is the description of what to draw in each blank circle.

But perhaps for the purpose of this format, we summarize the key points.

Actually, re-reading the user request: "Solve the problem accurately." and the image has blanks to fill with drawings.

Since we can't provide images, we must describe what to draw in each stage.

But the Final Answer section should contain the essential outcome.

Perhaps:

Final Answer:
For Mitosis:
- Prophase: 4 duplicated chromosomes (AA, aa, BB, bb)
- Metaphase: 4 duplicated chromosomes aligned at equator
- Anaphase: 8 single chromatids (4 to each pole: A, a, B, b per pole)
- Telophase: 2 nuclei, each with 4 single chromosomes (A, a, B, b)
- Daughter Cells: 2 identical cells, each with 4 chromosomes (A, a, B, b)

For Meiosis:
- Prophase I: 2 tetrads (AA+aa and BB+bb)
- Metaphase I: tetrads aligned at equator (e.g., AA and BB on one side, aa and bb on other)
- Anaphase I: homologous chromosomes separate (AA and BB to one pole, aa and bb to other)
- Telophase I: 2 cells, each with 2 duplicated chromosomes (one has AA,BB; other has aa,bb)
- Prophase II: in each cell, 2 duplicated chromosomes
- Metaphase II: chromosomes align at equator in each cell
- Anaphase II: sister chromatids separate (each pole gets one chromatid from each chromosome)
- Telophase II: 4 cells, each with 2 single chromosomes
- Daughter Cells: 4 haploid cells — two with A and B, two with a and b

This is the accurate solution.

To box it as final answer:

Final Answer:
Mitosis results in two diploid daughter cells, each with chromosomes A, a, B, b (identical to parent).
Meiosis results in four haploid daughter cells: two with chromosomes A and B, two with chromosomes a and b.