Question

name:

1. now using just the starting parent cell, draw in the chromosomes at the end of meiosis i and meiosis ii and include the alleles.
2. is there another way you could have done #2 to end up with different daughter cells? change how the chromosomes are lined up in the original parent cell and then redo meiosis i and ii so that you end up with different daughter cells.

Explanation:

Step1: Analyze Meiosis I

In meiosis I, homologous chromosomes separate. The parent cell has chromosomes with alleles: \( \text{X}^{H}\text{X}^{H} \), \( \text{X}^{h}\text{X}^{h} \), \( \text{X}^{B}\text{X}^{B} \), \( \text{X}^{B}\text{X}^{B} \), \( \text{X}^{d}\text{X}^{d} \), \( \text{X}^{D}\text{X}^{D} \) (assuming each pair is homologous). During meiosis I, homologous pairs (like \( \text{X}^{H}\text{X}^{H} \) & \( \text{X}^{h}\text{X}^{h} \), \( \text{X}^{d}\text{X}^{d} \) & \( \text{X}^{D}\text{X}^{D} \), and the two \( \text{X}^{B}\text{X}^{B} \) maybe as a pair? Wait, maybe the parent cell is diploid with three pairs? Wait, the parent cell has three types of chromosomes (H/h, B/B, d/D). So in meiosis I, each daughter cell gets one from each homologous pair. Let's group: Pair 1: \( \text{X}^{H}\text{X}^{H} \) and \( \text{X}^{h}\text{X}^{h} \) (homologous), Pair 2: \( \text{X}^{B}\text{X}^{B} \) and \( \text{X}^{B}\text{X}^{B} \) (maybe sister chromatids? No, meiosis I separates homologs. Wait, maybe the parent cell is a diploid with three pairs: (H/h), (B/B) [homozygous], (d/D). So during meiosis I, the cell divides into two cells, each with one homolog from each pair. So for the first pair (H/h), one cell gets \( \text{X}^{H}\text{X}^{H} \), the other \( \text{X}^{h}\text{X}^{h} \). For the B pair (homozygous, so both are \( \text{X}^{B}\text{X}^{B} \), so both daughter cells get \( \text{X}^{B}\text{X}^{B} \). For the d/D pair, one cell gets \( \text{X}^{d}\text{X}^{d} \), the other \( \text{X}^{D}\text{X}^{D} \). So the two cells at end of meiosis I would be: Cell 1: \( \text{X}^{H}\text{X}^{H} \), \( \text{X}^{B}\text{X}^{B} \), \( \text{X}^{d}\text{X}^{d} \); Cell 2: \( \text{X}^{h}\text{X}^{h} \), \( \text{X}^{B}\text{X}^{B} \), \( \text{X}^{D}\text{X}^{D} \).

Step2: Analyze Meiosis II

In meiosis II, sister chromatids separate. Each cell from meiosis I (which is haploid? No, meiosis I produces two haploid cells? Wait, no: meiosis I is reductional, so daughter cells are haploid (n), with one from each homologous pair. Then meiosis II is equational, separating sister chromatids. Wait, the cells after meiosis I have replicated chromosomes (sister chromatids). So for Cell 1: \( \text{X}^{H}\text{X}^{H} \) (sister chromatids), \( \text{X}^{B}\text{X}^{B} \) (sister chromatids), \( \text{X}^{d}\text{X}^{d} \) (sister chromatids). During meiosis II, each of these divides into two cells, each with one sister chromatid. So Cell 1 divides into two cells: \( \text{X}^{H} \), \( \text{X}^{B} \), \( \text{X}^{d} \) (but wait, the original has \( \text{X}^{H}\text{X}^{H} \) – maybe each chromosome is a pair of sister chromatids. Wait, maybe the notation is \( \text{X}^{allele} \) as a chromosome (with two sister chromatids). So parent cell: three chromosomes (each with two sister chromatids): \( \text{X}^{H}\text{X}^{H} \) (chromosome 1, sister chromatids), \( \text{X}^{h}\text{X}^{h} \) (chromosome 2, sister chromatids), \( \text{X}^{B}\text{X}^{B} \) (chromosome 3, sister chromatids), \( \text{X}^{B}\text{X}^{B} \) (chromosome 4, sister chromatids), \( \text{X}^{d}\text{X}^{d} \) (chromosome 5, sister chromatids), \( \text{X}^{D}\text{X}^{D} \) (chromosome 6, sister chromatids). Wait, maybe the parent cell is diploid with three homologous pairs: (1&2: H/h), (3&4: B/B), (5&6: d/D). So meiosis I: separate homologs (1&2, 3&4, 5&6). So two daughter cells: Cell A: 1,3,5; Cell B: 2,4,6. Then meiosis II: separate sister chromatids. So Cell A (1: \( \text{X}^{H}\text{X}^{H} \), 3: \( \text{X}^{B}\text{X}^{B} \), 5: \( \text{X}^{d}\text{X}^{d} \)) divides into two cells: each with \( \text{X}^{H} \), \( \text{X}^{B} \), \( \text{X}^{d} \) (but as single chromosomes, since sister chromatids separate). Similarly, Cell B (2: \( \text{X}^{h}\text{X}^{h} \), 4: \( \text{X}^{B}\text{X}^{B} \), 6: \( \text{X}^{D}\text{X}^{D} \)) divides into two cells: each with \( \text{X}^{h} \), \( \text{X}^{B} \), \( \text{X}^{D} \).

Step3: Draw Chromosomes for Meiosis I

For end of meiosis I, the two cells (let's call them Cell 1 and Cell 2) will have:

Cell 1: \( \text{X}^{H}\text{X}^{H} \), \( \text{X}^{B}\text{X}^{B} \), \( \text{X}^{d}\text{X}^{d} \) (each chromosome has two sister chromatids, so drawn as three chromosomes with two chromatids each, alleles \( H,H \), \( B,B \), \( d,d \))

Cell 2: \( \text{X}^{h}\text{X}^{h} \), \( \text{X}^{B}\text{X}^{B} \), \( \text{X}^{D}\text{X}^{D} \) (alleles \( h,h \), \( B,B \), \( D,D \))

Step4: Draw Chromosomes for Meiosis II

Each of these cells divides into two. So Cell 1 (from meiosis I) divides into two cells:

Cell 1a: \( \text{X}^{H} \), \( \text{X}^{B} \), \( \text{X}^{d} \) (single chromatids, alleles \( H \), \( B \), \( d \))

Cell 1b: \( \text{X}^{H} \), \( \text{X}^{B} \), \( \text{X}^{d} \) (same as 1a, since sister chromatids are identical here as parent was homozygous for B and the H/d pairs? Wait, no, if parent was \( \text{X}^{H}\text{X}^{H} \) and \( \text{X}^{h}\text{X}^{h} \), then sister chromatids are identical. So meiosis II for Cell 1 (with \( \text{X}^{H}\text{X}^{H} \), \( \text{X}^{B}\text{X}^{B} \), \( \text{X}^{d}\text{X}^{d} \)) will produce two cells with \( \text{X}^{H} \), \( \text{X}^{B} \), \( \text{X}^{d} \) (each chromosome is now a single chromatid, so drawn as three single chromosomes with alleles \( H \), \( B \), \( d \))

Similarly, Cell 2 (from meiosis I) divides into two cells:

Cell 2a: \( \text{X}^{h} \), \( \text{X}^{B} \), \( \text{X}^{D} \)

Cell 2b: \( \text{X}^{h} \), \( \text{X}^{B} \), \( \text{X}^{D} \)

Answer (for Drawing):

- **End of Meiosis I Cells**:
- Left Cell: Draw three chromosomes (each with two sister chromatids) with alleles \( \text{X}^{H}\text{X}^{H} \), \( \text{X}^{B}\text{X}^{B} \), \( \text{X}^{d}\text{X}^{d} \)
- Right Cell: Draw three chromosomes (each with two sister chromatids) with alleles \( \text{X}^{h}\text{X}^{h} \), \( \text{X}^{B}\text{X}^{B} \), \( \text{X}^{D}\text{X}^{D} \)
- **End of Meiosis II Cells**:
- Left - Bottom Left: Draw three single chromosomes with alleles \( \text{X}^{H} \), \( \text{X}^{B} \), \( \text{X}^{d} \)
- Left - Bottom Right: Draw three single chromosomes with alleles \( \text{X}^{H} \), \( \text{X}^{B} \), \( \text{X}^{d} \)
- Right - Bottom Left: Draw three single chromosomes with alleles \( \text{X}^{h} \), \( \text{X}^{B} \), \( \text{X}^{D} \)
- Right - Bottom Right: Draw three single chromosomes with alleles \( \text{X}^{h} \), \( \text{X}^{B} \), \( \text{X}^{D} \)

For question 3: Yes, because during meiosis I, the alignment of homologous chromosomes at the metaphase plate is random (independent assortment). So we can change the alignment of the homologous pairs (e.g., pair \( \text{X}^{H}\text{X}^{H} \) could pair with \( \text{X}^{D}\text{X}^{D} \) instead of \( \text{X}^{d}\text{X}^{d} \) during metaphase I, leading to different combinations in daughter cells. So by re - aligning the homologous chromosomes (changing which homologs are on which side of the metaphase plate) in the original parent cell, meiosis I and II will produce different daughter cells due to independent assortment.

Final Answer (for Drawing Guidance and Q3):

- Meiosis I Cells: As drawn with allele combinations \( (H,H; B,B; d,d) \) and \( (h,h; B,B; D,D) \) (each chromosome has two sister chromatids).
- Meiosis II Cells: As drawn with allele combinations \( (H; B; d) \), \( (H; B; d) \), \( (h; B; D) \), \( (h; B; D) \).
- For Q3: Yes, due to independent assortment of homologous chromosomes during meiosis I (changing their alignment at metaphase I leads to different combinations in daughter cells).

Answer:

(for Drawing):

• End of Meiosis I Cells:
• Left Cell: Draw three chromosomes (each with two sister chromatids) with alleles \( \text{X}^{H}\text{X}^{H} \), \( \text{X}^{B}\text{X}^{B} \), \( \text{X}^{d}\text{X}^{d} \)
• Right Cell: Draw three chromosomes (each with two sister chromatids) with alleles \( \text{X}^{h}\text{X}^{h} \), \( \text{X}^{B}\text{X}^{B} \), \( \text{X}^{D}\text{X}^{D} \)
• End of Meiosis II Cells:
• Left - Bottom Left: Draw three single chromosomes with alleles \( \text{X}^{H} \), \( \text{X}^{B} \), \( \text{X}^{d} \)
• Left - Bottom Right: Draw three single chromosomes with alleles \( \text{X}^{H} \), \( \text{X}^{B} \), \( \text{X}^{d} \)
• Right - Bottom Left: Draw three single chromosomes with alleles \( \text{X}^{h} \), \( \text{X}^{B} \), \( \text{X}^{D} \)
• Right - Bottom Right: Draw three single chromosomes with alleles \( \text{X}^{h} \), \( \text{X}^{B} \), \( \text{X}^{D} \)

For question 3: Yes, because during meiosis I, the alignment of homologous chromosomes at the metaphase plate is random (independent assortment). So we can change the alignment of the homologous pairs (e.g., pair \( \text{X}^{H}\text{X}^{H} \) could pair with \( \text{X}^{D}\text{X}^{D} \) instead of \( \text{X}^{d}\text{X}^{d} \) during metaphase I, leading to different combinations in daughter cells. So by re - aligning the homologous chromosomes (changing which homologs are on which side of the metaphase plate) in the original parent cell, meiosis I and II will produce different daughter cells due to independent assortment.

Final Answer (for Drawing Guidance and Q3):

• Meiosis I Cells: As drawn with allele combinations \( (H,H; B,B; d,d) \) and \( (h,h; B,B; D,D) \) (each chromosome has two sister chromatids).
• Meiosis II Cells: As drawn with allele combinations \( (H; B; d) \), \( (H; B; d) \), \( (h; B; D) \), \( (h; B; D) \).
• For Q3: Yes, due to independent assortment of homologous chromosomes during meiosis I (changing their alignment at metaphase I leads to different combinations in daughter cells).