Practice worksheet for completing Punnett squares to determine genetic probabilities for traits like eye color, hair type, and freckles.

Let’s go step by step through each Punnett square problem. We’ll fill in the parents’ alleles, complete the squares, and calculate probabilities for each genotype or phenotype.

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Problem 1: Brown eyes (B) dominant over blue eyes (b)
Cross: Bb x bb

Parent 1 (Bb) can give either B or b → top of square: B, b
Parent 2 (bb) can only give b → side of square: b, b

Punnett Square:

| | B | b |
|-----|-----|-----|
| b | Bb | bb |
| b | Bb | bb |

Genotypes:
- BB: 0 out of 4 → 0%
- Bb: 2 out of 4 → 50%
- bb: 2 out of 4 → 50%

✔ Final Probabilities:
- BB: 0
- Bb: 1/2 or 50%
- bb: 1/2 or 50%

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Problem 2: Curly hair (C) dominant over straight hair (c)
Cross: CC x Cc

Parent 1 (CC) gives only C → top: C, C
Parent 2 (Cc) gives C or c → side: C, c

Punnett Square:

| | C | C |
|-----|-----|-----|
| C | CC | CC |
| c | Cc | Cc |

Genotypes:
- CC: 2 out of 4 → 50%
- Cc: 2 out of 4 → 50%
- cc: 0 out of 4 → 0%

✔ Final Probabilities:
- CC: 1/2 or 50%
- Cc: 1/2 or 50%
- cc: 0

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Problem 3: Freckles (F) dominant over no freckles (f)

We’re told to identify parent genotypes first based on descriptions.

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Part A: Homozygous recessive freckles X Homozygous dominant freckles

Wait — “homozygous recessive freckles” doesn’t make sense because freckles are dominant. So if someone has *no* freckles, they must be homozygous recessive (ff). But the phrase says “homozygous recessive freckles” — that’s a contradiction. Let’s re-read carefully.

Actually, looking at the worksheet layout:

It says:
> Homozygous recessive freckles
> X
> Homozygous dominant freckles

But since freckles are dominant (F), “homozygous recessive freckles” is impossible — you can’t have freckles and be recessive. This must mean:

→ “Homozygous recessive” = no freckles → ff
→ “Homozygous dominant freckles” = FF

So cross: ff x FF

Parent 1 (ff) → f, f
Parent 2 (FF) → F, F

Punnett Square:

| | F | F |
|-----|-----|-----|
| f | Ff | Ff |
| f | Ff | Ff |

All offspring are Ff → all have freckles (since F is dominant)

Phenotype probability:
- Freckles: 4/4 = 100%
- No freckles: 0%

✔ Final:
- Freckles: 1 or 100%
- No freckles: 0

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Part B: Heterozygous freckles X Homozygous dominant freckles

Heterozygous freckles = Ff
Homozygous dominant freckles = FF

Cross: Ff x FF

Parent 1 (Ff) → F, f
Parent 2 (FF) → F, F

Punnett Square:

| | F | F |
|-----|-----|-----|
| F | FF | FF |
| f | Ff | Ff |

Offspring:
- FF: 2
- Ff: 2
- All have freckles

Phenotype:
- Freckles: 4/4 = 100%
- No freckles: 0%

✔ Final:
- Freckles: 1 or 100%
- No freckles: 0

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Part C: Heterozygous freckles X Heterozygous freckles

Cross: Ff x Ff

Each parent gives F or f

Punnett Square:

| | F | f |
|-----|-----|-----|
| F | FF | Ff |
| f | Ff | ff |

Genotypes:
- FF: 1
- Ff: 2
- ff: 1

Phenotypes:
- Freckles: FF + Ff = 3 out of 4 → 75%
- No freckles: ff = 1 out of 4 → 25%

✔ Final:
- Freckles: 3/4 or 75%
- No freckles: 1/4 or 25%

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Part D: Heterozygous freckles X Homozygous recessive freckles

Again, “homozygous recessive freckles” is misleading — it should mean “homozygous recessive for no freckles” → ff

So cross: Ff x ff

Parent 1 (Ff) → F, f
Parent 2 (ff) → f, f

Punnett Square:

| | F | f |
|-----|-----|-----|
| f | Ff | ff |
| f | Ff | ff |

Offspring:
- Ff: 2 → freckles
- ff: 2 → no freckles

Phenotype:
- Freckles: 2/4 = 50%
- No freckles: 2/4 = 50%

✔ Final:
- Freckles: 1/2 or 50%
- No freckles: 1/2 or 50%

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Final Answer:

Brown eyes (Bb x bb):
- BB: 0
- Bb: 1/2
- bb: 1/2

Curly hair (CC x Cc):
- CC: 1/2
- Cc: 1/2
- cc: 0

Freckles – ff x FF:
- Freckles: 1
- No freckles: 0

Freckles – Ff x FF:
- Freckles: 1
- No freckles: 0

Freckles – Ff x Ff:
- Freckles: 3/4
- No freckles: 1/4

Freckles – Ff x ff:
- Freckles: 1/2
- No freckles: 1/2