Worksheet on linked genes and inheritance patterns in genetics.

Final Answer:
1. X^NXⁿ and X^NY
2. white eyes: male — XⁿY; red-eyed female (heterozygous): X^NXⁿ; white-eyed female: XⁿXⁿ
3. Offspring genotypes: X^NXⁿ, XⁿXⁿ, X^NY, XⁿY
→ white eyes: 2 (XⁿXⁿ, XⁿY); red-eyed: 2 (X^NXⁿ, X^NY)
4. Parents: XⁿXⁿ (white-eyed female) × X^NY (red-eyed male)
Offspring: X^NXⁿ (red-eyed female), XⁿY (white-eyed male)
→ white eyes: 1 (male); red eyes: 1 (female)
5. Parents: XⁿXⁿ (white-eyed female) × XⁿY (red-eyed male) — *but red-eyed male must be X^NY*, so likely typo; assuming red-eyed male = X^NY, then cross is XⁿXⁿ × X^NY → same as #4
However, if the male is truly red-eyed and genotype given as XⁿY, that’s impossible — so correct cross is XⁿXⁿ × X^NY
Offspring: X^NXⁿ, XⁿY → white eyes: 1 (male); red eyes: 1 (female)
6. Hemophilia is X-linked recessive. Females need two copies (X^hX^h) to have disease. Males with one copy (X^hY) get it.
If 100 males tested, and 20 have hemophilia → 20 are X^hY
So frequency of X^h allele = 20/100 = 0.2
In females, probability of X^hX^h = (0.2)² = 0.04 → 4% of 200 females = 8 females
Males who won’t get it: those with X^HY = 100 − 20 = 80 males

But since the question asks only for final answers in blanks:

For #3: white eyes: 2, red eyes: 2
For #4: white eyes: 1, red eyes: 1
For #5: white eyes: 1, red eyes: 1
For #6: females with disease: 8, males without disease: 80

So filling the blanks as expected:

3. white eyes: 2, red eyes: 2
4. white eyes: 1, red eyes: 1
5. white eyes: 1, red eyes: 1
6. females with disease: 8, males without disease: 80