Completed biology worksheet using yetis to illustrate Mendelian genetics, including Punnett squares and genotype analysis.

You're looking at a genetics problem involving “yetis” and the inheritance of horns. Let’s break down each part step by step using Mendelian genetics principles.

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Given:

- Two adult yetis (parents) have 4 baby yetis.
- All 4 babies are without horns.
- One parent (left) has horns.
- The other parent (right) is without horns.

We’re to determine dominant/recessive traits, genotypes, and use Punnett squares to verify.

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## a. What is the dominant phenotype?

✔ Answer: Without horns

> Why? Because even though one parent *has* horns (recessive trait), all offspring *lack* horns — meaning the “without horns” trait is masking the “with horns” trait in the offspring. That’s the definition of a dominant phenotype.

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## b. What is the recessive phenotype?

✔ Answer: With horns

> Since “without horns” is dominant, “with horns” must be recessive.

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## c. What is the genotype of the yeti parent on the left?

✔ Answer: hh

> The left parent has horns → which is the recessive phenotype → so it must be homozygous recessive: hh.

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## d. What are the two possible genotypes of the yeti parent on the right?

✔ Answer: HH or Hh

> The right parent is without horns → dominant phenotype → could be either:
> - HH (homozygous dominant)
> - Hh (heterozygous)

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## e. What are the phenotypes of the yeti babies?

✔ Answer: Without horns

> All 4 babies are shown without horns → that’s their phenotype.

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## f. Complete both Punnett Squares below, one each for both possible genotypes of the right adult yeti.

Let’s do this carefully.

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Punnett Square 1: Right parent = HH

Left parent = hh
Right parent = HH

Gametes:
- Left: h, h
- Right: H, H

| | H | H |
|-------|-------|-------|
| h | Hh | Hh |
| h | Hh | Hh |

→ All offspring = Hh → Without horns

✔ Matches the image: all 4 babies without horns.

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Punnett Square 2: Right parent = Hh

Left parent = hh
Right parent = Hh

Gametes:
- Left: h, h
- Right: H, h

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

→ Offspring: 50% Hh (without horns), 50% hh (with horns)

→ So you’d expect 2 without horns, 2 with horns

✘ Does NOT match the image (all 4 babies are without horns).

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## g. Circle the Punnett Square above that is correct for the yetis shown here.

✔ You should circle the first Punnett square (where right parent = HH)

> Because only that one produces all offspring without horns, matching the image.

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## h. What are the genotypes of the parents?

✔ Answer: hh and HH

> Left parent = hh (has horns → recessive)
> Right parent = HH (without horns → homozygous dominant, since if it were Hh, we’d see some horned babies)

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## i. Why can’t the parents have a different genotype? (Assume offspring exactly follow predicted ratios)

✔ Answer:

> If the right parent were Hh (heterozygous), then crossing with hh would produce 50% Hh (without horns) and 50% hh (with horns) → i.e., 2 with horns, 2 without horns out of 4 babies.

> But in the image, ALL 4 babies are without horns → so the right parent must be HH, not Hh.

> Therefore, the only genotype combination that explains the observed offspring is hh × HH.

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## ✔ Final Summary:

- Dominant phenotype: Without horns
- Recessive phenotype: With horns
- Left parent genotype: hh
- Right parent possible genotypes: HH or Hh
- Babies’ phenotype: Without horns
- Correct Punnett square: HH × hh (all Hh → all without horns)
- Actual parental genotypes: hh and HH
- Reason other genotype doesn’t work: Hh × hh would produce 50% horned babies — but none are seen.

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This is a classic example of how observing offspring phenotypes can help deduce parental genotypes — especially when you see all offspring showing the dominant trait from a cross between a recessive and a dominant parent. It tells you the dominant parent must be homozygous.

Let me know if you want a visual diagram of the Punnett squares!