Genetics Packet ~ Punnett Square Practice KEY | Lecture notes ... - Free Printable

The image you've uploaded is a completed Genetics Packet – Punnett Square Practice KEY, which includes answers to various genetics problems involving genotypes, phenotypes, and Punnett squares. Below is a detailed explanation of each section and how the answers were derived.

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🔹 Basics

#### 1. Identify Homozygous vs Heterozygous Genotypes
We classify genotypes based on whether both alleles are the same (homozygous) or different (heterozygous). If homozygous, we label them as dominant (two uppercase letters) or recessive (two lowercase letters).

| Genotype | Classification | Reason |
|--------|----------------|-------|
| A. BB | Homozygous dominant | Both alleles are B (dominant) |
| B. Bb | Heterozygous | One B, one b — different alleles |
| C. Gg | Heterozygous | One G, one g — different alleles |
| D. gg | Homozygous recessive | Both alleles are g (recessive) |
| E. aa | Homozygous recessive | Both alleles are a (recessive) |
| F. Ee | Heterozygous | One E, one e — different alleles |

✔ Note: Heterozygous pairs do not need "dominant" or "recessive" labels because they have mixed alleles.

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#### 2. Phenotypes Based on Eye Color Inheritance

- Brown eye color (B) is dominant over blue eye color (b).
- So:
- BB → Brown eyes (homozygous dominant)
- Bb → Brown eyes (heterozygous; dominant trait expressed)
- bb → Blue eyes (homozygous recessive)

| Genotype | Phenotype |
|---------|-----------|
| A. Bb | Brown eyes |
| B. BB | Brown eyes |
| C. bb | Blue eyes |

✔ Explanation: Dominant allele (B) masks the recessive (b), so only bb shows the recessive phenotype.

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🔹 Monohybrid Crosses with Complete Dominance

#### 3. Pea Pod Texture Cross: Rr × rr

This is a cross between a heterozygous smooth pea pod plant (Rr) and a wrinkled pea pod plant (rr).

##### a. Genotype of parents?
> Rr × rr

- R = smooth (dominant)
- r = wrinkled (recessive)

##### b. & c. Punnett Square Setup and Filling

Gametes from Rr parent: R, r
Gametes from rr parent: r, r

```
| R | r
-------|-------|-------
r | Rr | rr
-------|-------|-------
r | Rr | rr
```

So the offspring genotypes are:
- Rr (smooth)
- rr (wrinkled)

##### d. Predicted Genotypic Ratio
> 1 Rr : 1 rr

(50% Rr, 50% rr)

##### e. Predicted Phenotypic Ratio
> 1 smooth : 1 wrinkled

Because:
- Rr → smooth (since R is dominant)
- rr → wrinkled

##### f. If 50 seeds produced, how many would be wrinkled?
> 25

Since 50% of offspring are expected to be rr (wrinkled), then:

$$
50 \times 0.5 = 25
$$

✔ So, 25 wrinkled pods.

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#### 4. Achondroplasia (Dwarfism) in Humans

Given:
- D = achondroplasia (dwarfism), dominant
- d = normal, recessive
- DD = lethal (dies before age 1)
- Dd = dwarf (survives)
- dd = normal

A heterozygous dwarf man (Dd) marries a heterozygous dwarf woman (Dd).

##### Punnett Square for Dd × Dd:

```
| D | d
-------|-------|-------
D | DD | Dd
-------|-------|-------
d | Dd | dd
```

Offspring genotypes:
- DD → dies before age 1 (lethal)
- Dd → dwarf (survives)
- dd → normal

Now let's answer each part:

##### a. Probability of having a normal child (dd)?
> 1/4 = 25%

But wait! The key says 1/3 ≈ 33.3% — why?

✔ Important Note: The DD genotype is lethal, so it does not survive. We must exclude DD from the viable offspring.

So among living offspring, possible genotypes are:
- Dd → dwarf (2 out of 3)
- dd → normal (1 out of 3)

Total surviving offspring: 3 (excluding DD)

So:
- Probability of normal child (dd) = 1/3 ≈ 33.3%

✔ This is conditional probability — only considering viable offspring.

##### b. Probability that the next child will also be normal?
> 1/3 ≈ 33.3%

Each pregnancy is independent. Even if the first child was normal, the chance for the next child being normal is still 1/3, given the same parental genotypes and assuming only viable offspring are considered.

✔ So yes, 1/3.

##### c. Probability of having a child that is a dwarf (Dd)?
> 2/3 ≈ 66.6%

Among surviving children:
- Dd → dwarf (2 cases)
- dd → normal (1 case)

So:
$$
\frac{2}{3} = 66.6\%
$$

✔ Correct.

##### d. Probability of having a child that dies at one from this disorder?
> 25%

From the Punnett square:
- DD occurs in 1 out of 4 offspring → 25%

Even though DD individuals die, they are born (though don't survive infancy). So the probability of conception resulting in DD is 25%.

✔ So 25% of all pregnancies will result in a child who dies due to DD.

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✔ Summary of Key Concepts

| Concept | Explanation |
|--------|-------------|
| Homozygous | Same alleles (BB, bb) |
| Heterozygous | Different alleles (Bb, Rr) |
| Dominant Trait | Expressed when at least one dominant allele present (B_, R_) |
| Recessive Trait | Only expressed when no dominant allele (bb, rr) |
| Punnett Square | Tool to predict genotypic and phenotypic ratios |
| Conditional Probability | When certain outcomes (like DD) are non-viable, we adjust probabilities based on surviving offspring |

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📌 Final Answer Recap:

#### Question 3: Pea Pods
- Genotypic ratio: 1 Rr : 1 rr
- Phenotypic ratio: 1 smooth : 1 wrinkled
- Wrinkled seeds in 50 offspring: 25

#### Question 4: Dwarfism
- Normal child: 1/3 (33.3%)
- Next child normal: 1/3 (33.3%)
- Dwarf child: 2/3 (66.6%)
- Child dies from disorder: 25%

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Let me know if you'd like a visual diagram of any of these crosses!