Pedigree Practice Worksheet Collection For Teaching & Learning - Free Printable

Let's solve this step-by-step based on the pedigree practice worksheet about albinism, a recessive genetic disorder.

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🔬 Background: Albinism

- Albinism is caused by a recessive allele (a).
- The gene has two alleles:
- A = normal pigment (dominant)
- a = no pigment (recessive, causes albinism)
- Genotypes:
- AA → Normal (not albino), homozygous dominant
- Aa → Normal (carrier), heterozygous
- aa → Albino (affected), homozygous recessive

Now let’s go through each part of the worksheet.

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## ✔ Problem 1: Describe the phenotypes

Given genotypes:

- AA = Normal
- Aa = Normal (carrier, but not albino)
- aa = Albino

> Explanation: Since albinism is recessive, only individuals with aa will show the trait. Aa and AA individuals have normal pigmentation.

---

Pedigree 1: First Family Tree

```
O —— □
\
\
O □ O □
\ / \
□ O
\
O (shaded)
```

- Shaded squares/circles = affected (albino)
- Unshaded = normal

#### a) How many children does the couple have?

- The first couple (top circle and square) have 3 children: one daughter (unshaded), one son (shaded), and another daughter (unshaded).
- So, 3 children.

✔ Answer: 3

#### b) What is the sex of the oldest child?

- In pedigrees, children are listed from left to right in order of birth.
- The leftmost child is the oldest.
- The first child is a circle → female.

✔ Answer: Female

#### c) How many grandchildren does the couple have?

- Only the daughter (on the right) has a child.
- She married a male (square), and they had one child, who is shaded → albino.
- So, the original couple has 1 grandchild.

✔ Answer: 1

---

## ✔ Problem 2: Fill out the pedigree with genotypes (AA, Aa, aa)

We now analyze the second pedigree.

```
O —— □
\
\
O □ O □
\ / \
□ O
\
O
```

Wait — actually, here's the correct layout (from image):

```
O —— □
\
\
O □ O □
\ / \
□ O
\
O
```

But looking at the actual diagram in the image:

It shows:

- Original couple: mother (O), father (□)
- They have 4 children:
1. Daughter (O) → married to a man (□), had one son (□)
2. Son (■) → shaded → albino
3. Daughter (■) → shaded → albino
4. Son (□) → married to an albino woman (■), they have two children: one shaded (■), one unshaded (O)

So, let's label the genotypes.

---

Step 1: Assign genotypes

#### Start with affected individuals (shaded): must be aa

- Son 2 (■) → aa
- Daughter 3 (■) → aa
- Granddaughter (■) → aa
- Child of last couple (■) → aa

Now work backward.

#### Parents (original couple):
- Both parents are normal (unshaded), but they have albino children → so both must be carriers (Aa)

✔ So:
- Mother (O) → Aa
- Father (□) → Aa

#### Their children:

1. First daughter (O) → normal, but we don’t know if carrier.
- But her son is normal (□), and she married a normal man (□). We can't determine her genotype yet.
- But since she has no albino children, and her husband is normal, possible genotypes: AA or Aa.
- But we can’t tell for sure without more data. However, since her parents are Aa × Aa, possible genotypes for her: AA (1/4), Aa (1/2), aa (1/4). But she is not albino → not aa → so either AA or Aa.
- But we cannot determine unless we see offspring.

Let’s assume she is Aa? Wait — not necessary. For now, just note she is not albino, so AA or Aa.

But since her son is normal, and his father is normal, we can’t determine much.

But we don’t need to assign her genotype unless asked.

2. Second child (son, ■) → albino → aa

3. Third child (daughter, ■) → albino → aa

4. Fourth child (son, □) → normal → could be AA or Aa

But he married an albino woman (aa), and they had:
- One albino child (■) → aa
- One normal child (O) → must be Aa

So the father (fourth child) must have contributed an a allele to the albino child, and an A to the normal child.

Therefore, he must be Aa (carrier).

So:
- Fourth son → Aa

Now back to third child (albino daughter) → aa

She married a normal man (□). We don’t know his genotype, but their child is albino (■) → aa.

So the mother gave a, and the father must have given a → so the father must be Aa (carrier).

But he is unshaded → normal → so yes, Aa

So:
- Husband of albino daughter → Aa

Their child → aa (albino)

Now, the first daughter (unshaded) married a normal man (□). They have one son (□) → normal.

We cannot determine if she is AA or Aa.

But since her parents were both Aa, and she is normal, she has a 2/3 chance of being Aa (since among normal offspring, 1/3 are AA, 2/3 are Aa).

But we don’t need to assign unless required.

---

Now fill in blanks:

Let’s write genotypes:

- Original mother (O) → Aa
- Original father (□) → Aa
- Child 1 (daughter, O) → ? (could be AA or Aa) → we'll leave as A_ unless specified
- Child 2 (son, ■) → aa
- Child 3 (daughter, ■) → aa
- Child 4 (son, □) → Aa
- Husband of child 3 (□) → Aa (because they have albino child)
- Child of child 3 and husband → aa
- Husband of child 1 (□) → unknown → probably AA or Aa
- Child of child 1 (□) → normal → could be AA or Aa

But the question asks us to fill out the blanks.

So let’s write the genotypes where possible.

#### Final genotypes:

| Individual | Genotype |
|----------|---------|
| Original mother (O) | Aa |
| Original father (□) | Aa |
| Child 1 (daughter, O) | AA or Aa (unknown) |
| Child 2 (son, ■) | aa |
| Child 3 (daughter, ■) | aa |
| Child 4 (son, □) | Aa |
| Husband of child 3 (□) | Aa |
| Grandchild (child of child 3 & husband) | aa |
| Husband of child 1 (□) | ? (assume normal, could be AA or Aa) |
| Grandchild (child of child 1) | ? (normal, could be AA or Aa) |

But likely, the blank spaces are meant to be filled in for the individuals shown.

Looking at the diagram:

We should label:

- Original parents: Aa, Aa
- Child 1: ? → not determined
- Child 2: aa
- Child 3: aa
- Child 4: Aa
- Husband of child 3: Aa
- Their child: aa
- Husband of child 1: ? → assume AA or Aa, but not specified
- Their child: ? → normal → AA or Aa

But perhaps the worksheet wants us to fill only what we know.

So:

- Child 2: aa
- Child 3: aa
- Child 4: Aa
- Husband of child 3: Aa
- Their child: aa

And original parents: Aa, Aa

Now answer the questions.

---

Question 2: Fill out blanks

#### a) How many children did the original couple have?

- 4 children → 2 sons, 2 daughters

✔ Answer: 4

#### b) How many grandchildren?

- Child 1 → has 1 child → 1 grandchild
- Child 2 → no spouse shown → no kids
- Child 3 → has 1 child → 1 grandchild
- Child 4 → has 2 children → 2 grandchildren

Total grandchildren: 1 + 0 + 1 + 2 = 4

✔ Answer: 4

#### c) How many of their grandchildren are albino?

- Child 1’s child → normal → not albino
- Child 3’s child → albino → yes
- Child 4’s children: one albino, one normal → one albino

So total albino grandchildren: 2

✔ Answer: 2

#### d) Create an extension to show the albino granddaughter marrying and having a son who is NOT albino.

The albino granddaughter is the child of child 3 → aa

She marries someone who is NOT albino → so AA or Aa

They have a son who is NOT albino → so AA or Aa

But since the mother is aa, she gives a allele.

So the son must have gotten A from father → so father must be Aa or AA

To have a non-albino son: son must be Aa (if father is Aa) or AA (if father is AA)

But since mother is aa, the son must be Aa (he gets a from mom, A from dad)

So the father must be AA or Aa, but to have a non-albino child, it’s possible.

Let’s draw the extension:

```
■ (albino granddaughter) —— □ (normal man)
\
\
□ (son, normal)
```

Label:
- Albino granddaughter → aa
- Her husband → AA or Aa (but to have normal child, he must contribute A)
- Their son → Aa (normal, carrier)

So the son is not albino, so unshaded.

✔ This satisfies the condition.

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✔ Final Answers

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📝 Problem 1

- AA = Normal
- Aa = Normal
- aa = Albino

#### a) How many children? → 3
#### b) Sex of oldest child? → Female
#### c) How many grandchildren? → 1

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📝 Problem 2

#### a) How many children did original couple have? → 4
#### b) How many grandchildren? → 4
#### c) How many grandchildren are albino? → 2
#### d) Extension:

Draw:

```
■ (aa) —— □ (AA or Aa)
\
\
□ (Aa, normal)
```

Label:
- Albino granddaughter: aa
- Her husband: AA or Aa (e.g., Aa)
- Son: Aa, unshaded (normal)

This shows a non-albino son.

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✔ All done!

Let me know if you'd like a visual version of the completed pedigree!