SOLUTION: Mendelian genetics worksheet - Studypool - Free Printable

Let's go through the problem step by step and solve it thoroughly, especially focusing on part e (the Punnett squares) and the reasoning behind the correct answer.

---

🔍 Problem Overview:

We are analyzing a family of fish in a tank. The trait being studied is the number of back fins, which follows Mendelian inheritance.

- Dominant phenotype: 2 back fins
- Recessive phenotype: 1 back fin

We are given:
- Two adult fish (parents)
- 8 baby fish (offspring):
- 6 with 2 back fins (dominant phenotype)
- 2 with 1 back fin (recessive phenotype)

We need to determine:
- The genotypes of the parents
- Which Punnett square matches the observed offspring ratio

---

## ✔ Step-by-step Solution:

a. What is the dominant phenotype?

> Answer: 2 back fins
✔ Because it appears more frequently and is expressed even when only one copy of the allele is present.

---

b. What is the recessive phenotype?

> Answer: 1 back fin
✔ This only shows up when an individual has two recessive alleles (ff).

---

c. What are the two possible genotypes of each adult fish?

> Answer: FF and Ff
✔ Since both adults have 2 back fins (dominant phenotype), they could be either:
- Homozygous dominant (FF) or
- Heterozygous (Ff)

But we’ll narrow this down using the offspring.

---

d. How many baby fish have the dominant and recessive phenotypes?

> Answer:
- 6 with dominant phenotype (2 back fins)
- 2 with recessive phenotype (1 back fin)

This gives a ratio of 3:1 (6:2 simplifies to 3:1)

---

e. Complete both Punnett Squares below, one each for all possible genotypes of the upper adult fish.

We’re given four Punnett squares, but we only need to focus on the possible combinations between the two parents.

Let’s analyze each Punnett square:

#### 🟦 Punnett Square 1: FF × FF
```
F F
F FF FF
F FF FF
```
- All offspring: FF → all have 2 back fins
- No recessive offspring → ✘ Not possible (we have 2 with 1 back fin)

#### 🟦 Punnett Square 2: FF × Ff
Wait — actually, the labels above the squares are confusing. Let's clarify:

The four squares represent different parental genotype combinations. But only one is correct based on the offspring.

Let’s look at the last Punnett square (circled in red):

> Parent 1: F f (heterozygous)
> Parent 2: F F (homozygous dominant)

So cross: Ff × FF

Let’s complete that Punnett square:

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

Offspring genotypes:
- FF: 50% → 2 back fins
- Ff: 50% → 2 back fins
→ All offspring have 2 back fins → No 1-back-fin babies → ✘ Not possible

Wait! That doesn't match the data.

But the correct cross must produce some ff offspring (with 1 back fin). So both parents must carry the f allele.

That means both parents must be heterozygous (Ff).

Let’s try:

✔ Correct Cross: Ff × Ff

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

Offspring:
- FF: 1/4 → 2 back fins
- Ff: 2/4 → 2 back fins
- ff: 1/4 → 1 back fin

So:
- 3/4 = 75% have 2 back fins
- 1/4 = 25% have 1 back fin

Now check the actual offspring:
- 6 with 2 back fins
- 2 with 1 back fin
- Total: 8 → 6/8 = 75%, 2/8 = 25% → ✔ Matches perfectly!

So the correct Punnett square should be:

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

But wait — none of the provided squares show this!

Let’s re-express what’s given:

There are four Punnett squares shown:

1. FF × FF → All FF → No recessive → ✘
2. FF × Ff → All F_ → No ff → ✘
3. ff × ff → All ff → All 1 back fin → ✘
4. Ff × Ff → Wait — the last one is labeled as:
- Top: F and f
- Left: F and f
- But written as:
```
F f
F FF Ff
f Ff ff
```

Yes! The last square (circled in red) is actually Ff × Ff, even though it’s labeled incorrectly.

Let’s re-check that one:

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

→ This is Ff × Ff → produces:
- 3/4 with 2 back fins
- 1/4 with 1 back fin → ✔ Matches 6:2 ratio

So the correct Punnett square is the last one (the circled one).

Even though the labeling says "F" and "f" on top and side, and the boxes show FF, Ff, etc., it's consistent with Ff × Ff.

---

✔ Final Answers:

#### e. Complete both Punnett Squares
Only one of them is correct — the last one (Ff × Ff):

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

So circle the last Punnett square.

---

#### f. Circle the Punnett square that is correct for the fish shown here.
> ✔ Circle the last (rightmost) Punnett square (the one showing Ff × Ff)

---

#### g. What are the genotypes of the parents?
> Answer: Ff and Ff
✔ Both parents must be heterozygous to produce 25% recessive offspring.

---

#### h. Why are the other Punnett squares not possible for this family?

> Answer:
The only possible Punnett square is the last one because there is a 75% probability to produce 2-back-fin offspring and a 25% probability to produce 1-back-fin offspring. The other squares cannot be considered because:
- FF × FF → All offspring have 2 back fins → no 1-back-fin babies → ✘
- FF × Ff → All offspring have at least one F → all have 2 back fins → ✘
- ff × ff → All offspring have 1 back fin → no 2-back-fin babies → ✘

Only Ff × Ff produces the 3:1 ratio observed (6:2 = 3:1).

---

✔ Summary:

| Question | Answer |
|--------|--------|
| a. Dominant phenotype | 2 back fins |
| b. Recessive phenotype | 1 back fin |
| c. Possible genotypes of adult fish | FF or Ff |
| d. Offspring counts | 6 dominant, 2 recessive |
| e. Correct Punnett square | Last one (Ff × Ff) |
| f. Circle | Last square |
| g. Parent genotypes | Ff and Ff |
| h. Why others are invalid | They don’t produce both phenotypes in correct ratio |

---

✔ Final Note: The image likely had a labeling issue — the last Punnett square is intended to represent Ff × Ff, even if the alleles aren't clearly labeled as such. Based on the results, it’s the only one that fits the data.

Let me know if you'd like a visual version of the corrected Punnett square!