Problem: What percentage of the female's daughters will be albino and colorblind if she is a carrier for both albinism and color blindness and marries a man who is a carrier for albinism but colorblind?

Solution:
To solve this problem, we need to use the principles of Mendelian genetics and Punnett squares.

Step 1: Determine the genotypes of the parents.
The female is a carrier for both albinism and color blindness, so her genotype is AaXBb. The male is a carrier for albinism but colorblind, so his genotype is AAXBb.

Step 2: Create a Punnett square.
To determine the probability of their daughter being albino and colorblind, we need to create a Punnett square.

| | A | a |
|---|---|---|
| A | AA | Aa |
| A | AA | Aa |
| B | AB | Ab |
| b | Ab | bb |

Step 3: Determine the possible genotypes of their daughters.
From the Punnett square, we can see that there are four possible genotypes for their daughters:

- AAxBb: not albino, not colorblind
- AaxBb: not albino, not colorblind
- AAxbb: albino, colorblind
- Aaxbb: albino, colorblind

Step 4: Calculate the probability of each genotype.
We can calculate the probability of each genotype by counting the number of squares with that genotype and dividing by the total number of squares.

- AAxBb: 2/8 or 25%
- AaxBb: 4/8 or 50%
- AAxbb: 1/8 or 12.5%
- Aaxbb: 1/8 or 12.5%

Step 5: Answer the question.
The question asks for the percentage of their daughters who will be albino and colorblind, which is the sum of the probabilities of the two genotypes that have those traits:

- AAxbb: 1/8 or 12.5%
- Aaxbb: 1/8 or 12.5%
- Total: 25%

Conclusion:
Therefore, the percentage of the female's daughters who will be albino and colorblind is 25%.