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Let's solve each scenario step by step using X-linked inheritance, where the gene for color vision is located on the X chromosome.

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🔹 Allele Notation Key:

- Colorblind allele = Xᶜ
- Normal vision allele = X^C

> Note: In genetics, we use X^c for colorblind (recessive) and X^C for normal vision (dominant). Since it’s X-linked, males have one X (from mother), and females have two X chromosomes.

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## ✔ Scenario 1: Colorblind Man × Woman with Normal Vision

Given:

- Man: Colorblind → Genotype: X^c Y
- Woman: Non-carrier, normal vision → Genotype: X^C X^C

Step 1: Fill in the Punnett Square

| | X^C | X^C |
|-------|---------|---------|
| X^c | X^c X^C | X^c X^C |
| Y | X^C Y | X^C Y |

> So the offspring are:
- Daughters: All X^c X^C → carriers, but normal vision
- Sons: All X^C Y → normal vision

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Answers:

1. Phenotype: Colorblind
2. Genotype: X^c Y
3. Allele(s): X^c, Y

4. Phenotype: Normal vision
5. Genotype: X^C X^C
6. Allele(s): X^C, X^C

7. Punnett Square (as above)

8. The baby boy has a 0% chance of being colorblind.
> Because all sons get X^C from mother → normal vision.

9. The baby girl has a 0% chance of being colorblind and 100% chance of being a carrier.
> All daughters inherit X^c from father and X^C from mother → X^c X^C → carrier, normal vision.

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## ✔ Scenario 2: Colorblind Woman × Man with Normal Vision

Given:

- Woman: Colorblind → Genotype: X^c X^c
- Man: Normal vision → Genotype: X^C Y

Step 1: Punnett Square

| | X^c | X^c |
|-------|---------|---------|
| X^C | X^C X^c | X^C X^c |
| Y | X^c Y | X^c Y |

> Offspring:
- Daughters: All X^C X^c → carriers, normal vision
- Sons: All X^c Y → colorblind

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Answers:

10. Phenotype: Normal vision
11. Genotype: X^C Y
12. Allele(s): X^C, Y

13. Phenotype: Colorblind
14. Genotype: X^c X^c
15. Allele(s): X^c, X^c

16. Punnett Square (as above)

17. If they have a daughter, what are the chances she will be colorblind?
→ 0%
> She will be X^C X^c → carrier, but since X^C is dominant, she has normal vision.

18. If they have a son, what are the chances he will be colorblind?
→ 100%
> He gets X^c from mother and Y from father → X^c Y → colorblind.

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## ✔ Scenario 3: Colorblind Man × Carrier Woman

Given:

- Man: Colorblind → Genotype: X^c Y
- Woman: Carrier → Genotype: X^C X^c

Step 1: Punnett Square

| | X^C | X^c |
|-------|---------|---------|
| X^c | X^c X^C | X^c X^c |
| Y | X^C Y | X^c Y |

> Offspring:
- Daughters:
- 50% X^c X^C → carrier, normal vision
- 50% X^c X^c → colorblind
- Sons:
- 50% X^C Y → normal vision
- 50% X^c Y → colorblind

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Answers:

19. Phenotype: Colorblind
20. Genotype: X^c Y
21. Allele(s): X^c, Y

22. Phenotype: Carrier with normal vision
23. Genotype: X^C X^c
24. Allele(s): X^C, X^c

25. Punnett Square (as above)

26. If they have a daughter, what are the chances she will be colorblind?
→ 50%
> Because she can get either X^C or X^c from mother, and X^c from father → if she gets X^c from mom → X^c X^c → colorblind.

27. If they have a son, what are the chances he will be colorblind?
→ 50%
> He gets Y from father and X from mother. Mother gives X^c (50%) or X^C (50%). So 50% chance of X^c Y → colorblind.

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✔ Final Summary Table:

| Scenario | Male Parent | Female Parent | Daughter Colorblind? | Son Colorblind? |
|--------|-------------|----------------|------------------------|------------------|
| 1 | X^c Y | X^C X^C | 0% | 0% |
| 2 | X^C Y | X^c X^c | 0% | 100% |
| 3 | X^c Y | X^C X^c | 50% | 50% |

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📌 Key Takeaways:

- Males are more likely to be affected because they only have one X.
- Females need two recessive alleles (X^c X^c) to be colorblind.
- A carrier female (X^C X^c) has normal vision but can pass the allele to children.

Let me know if you'd like this formatted as a printable answer sheet!