Dihybrid genetics practice problems with Punnett squares for analyzing trait inheritance in humans, horses, and squash.

Let’s solve each problem step by step.

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Problem 1:

We are told:
- Spotted skin (S) is dominant over non-spotted (s)
- Wooly hair (W) is dominant over non-wooly (w)

Cross:
Man = heterozygous spotted, non-wooly → genotype: Ss ww
Woman = heterozygous wooly-haired, non-spotted → genotype: ss Ww

We need to do a dihybrid cross. First, find the gametes each parent can produce.

Man (Ss ww):
Possible gametes: S w and s w → so Sw and sw

Woman (ss Ww):
Possible gametes: s W and s w → so sW and sw

Now set up the Punnett square (2x2 since each parent has 2 types of gametes):

Gametes from man on top: Sw, sw
Gametes from woman on side: sW, sw

Fill in the squares:

Top row (man’s gametes):
First column (woman’s sW):
→ Sw + sW = Ss Ww → spotted, wooly
Second column (woman’s sw):
→ Sw + sw = Ss ww → spotted, non-wooly

Bottom row (man’s sw):
First column (woman’s sW):
→ sw + sW = ss Ww → non-spotted, wooly
Second column (woman’s sw):
→ sw + sw = ss ww → non-spotted, non-wooly

So offspring genotypes:
- Ss Ww → 1/4
- Ss ww → 1/4
- ss Ww → 1/4
- ss ww → 1/4

Phenotypes:
- Spotted, wooly → Ss Ww → 1/4
- Spotted, non-wooly → Ss ww → 1/4
- Non-spotted, wooly → ss Ww → 1/4
- Non-spotted, non-wooly → ss ww → 1/4

So phenotypic ratio: 1 : 1 : 1 : 1

Genotypic ratio: also 1 : 1 : 1 : 1 (since all four genotypes are unique and equally likely)

✔ Check: Each combination appears once in 4-box grid → correct.

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Problem 2:

Horses:
- Black (B) dominant over chestnut (b)
- Trotting gait (T) dominant over pacing (t)

Cross:
Homozygous black pacer → BB tt
Homozygous chestnut trotter → bb TT

F1 generation: all offspring will be Bb Tt

Because:
Each parent gives one allele per gene.

Black pacer (BB tt) → gametes: B t
Chestnut trotter (bb TT) → gametes: b T

Offspring: Bb Tt → all same genotype.

Phenotype:
B_ → black (since B dominant)
T_ → trotting (since T dominant)

So all F1 horses are black trotters

✔ Check: Homozygous parents → all offspring identical heterozygotes → yes.

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Problem 3:

Summer squash:
- White fruit (W) dominant over yellow (w)
- Disk-shaped (D) dominant over sphere-shaped (d)

Cross:
True-breeding white, disk-shaped → WW DD
True-breeding yellow, sphere-shaped → ww dd

F1 generation: all offspring get W from first parent, w from second → Ww
And D from first, d from second → Dd

So all F1 = Ww Dd

Phenotype:
W_ → white
D_ → disk-shaped

So all F1 plants have white, disk-shaped fruit

✔ Check: True-breeding means homozygous → crossing them gives uniform heterozygous F1 → correct.

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Final Answer:

1. Genotypic ratio: 1 Ss Ww : 1 Ss ww : 1 ss Ww : 1 ss ww
Phenotypic ratio: 1 spotted, wooly : 1 spotted, non-wooly : 1 non-spotted, wooly : 1 non-spotted, non-wooly

2. All F1 horses are black trotters.

3. All F1 squash plants have white, disk-shaped fruit.