Punnett Square practice - Free Printable

Let’s solve each part step by step.

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Problem 1a:

We’re told:
- Dominant allele W = long whiskers
- Recessive allele w = short whiskers
- One parent is homozygous dominant → that means WW
- The other parent is heterozygous → that means Ww

We need to find the probability of offspring with short whiskers (which only happens if genotype is ww).

Set up Punnett square:

Parent 1 (WW) gametes: W, W
Parent 2 (Ww) gametes: W, w

Punnett Square:

| | W | W |
|-------|-------|-------|
| W | WW | WW |
| w | Ww | Ww |

Offspring genotypes:
- WW → 2 out of 4 → 50%
- Ww → 2 out of 4 → 50%
- ww → 0 out of 4 → 0%

Phenotypes:
- Long whiskers: WW or Ww → all 4 → 100%
- Short whiskers: only ww → 0%

✔ So:
- % long whiskers = 100%
- % short whiskers = 0%

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Problem 1b:

One parent is heterozygous long-whiskered → Ww
Other parent is short-whiskered → must be ww (since short is recessive)

Gametes:
- Parent 1 (Ww): W, w
- Parent 2 (ww): w, w

Punnett Square:

| | w | w |
|-------|-------|-------|
| W | Ww | Ww |
| w | ww | ww |

Offspring:
- Ww → 2 out of 4 → 50% → long whiskers
- ww → 2 out of 4 → 50% → short whiskers

✔ So:
- % long whiskers = 50%
- % short whiskers = 50%

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Problem 2:

Purple people eaters:
- H = one horn (dominant)
- h = no horns (recessive)
- Cross two hybrid parents → hybrid means heterozygous → Hh x Hh

Gametes from each parent: H, h

Punnett Square:

| | H | h |
|-------|-------|-------|
| H | HH | Hh |
| h | Hh | hh |

Genotypes:
- HH → 1 out of 4
- Hh → 2 out of 4
- hh → 1 out of 4

So possible genotypes of offspring: HH, Hh, hh

Phenotypes:
- One horn: HH or Hh → 3 out of 4 → 75%
- No horns: hh → 1 out of 4 → 25%

So possible phenotypes: one horn, no horns

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Final Answer:

1a.
% long whiskers: 100%
% short whiskers: 0%

1b.
% long whiskers: 50%
% short whiskers: 50%

2.
Possible genotypes of offspring: HH, Hh, hh
Possible phenotypes of offspring: one horn, no horns